The proposal is that the maths from Einstein+co for Special Relativity is full of maths mistakes and when those maths mistakes are corrected it turns back into the maths of Galilean relativity.

i.e. they would be the same theory but Establishments' maths for SR is full of errors. And the paradoxes of relativity - Twin paradox etc are just result of maths being wrong.

Time dilation arises from math mistakes

Re: latest article at: http://www.wbabin.net/science/anderton18.pdf

No, the math employed by SR is correct, it is just that you are an old imbecile that can't understand it.

Roger,

First, that Wikipedia article is the worst article I've seen on the subject. Yes,they reversed the mirror disgination as you say, but they also reverse or at least describe in a very confused manner. Usually the first tdiagram is said to be what one moving with the mirror sees, that is stataionary with respect to the mirror. The second diagram is usually described as what a stationary observer sees as the mirror moves past him. I think they actually get it right but it is very confusing.

In any case, you describe the relativstic analysis as a mathematical mistake when the mathematics are perfectly correct for the assumptions. Einstein's second postulate is that all observers will measure light at the same velocity, c. You assumed something else. The math isn't wrong, your assumption is. Experimentally, nobody has ever found the speed of light in vacuo to be anything but the same value.

Nice try though.

cinci

Here is a correct explanation of the MMX:

http://en.wikibooks.org/wiki/Special_Relativity/Aether#Mathematical_analysis_of_the_Michelson_Morley_Experiment

Not all wiki is bad

Light travels for duration x.
An observer(s), any observer(s), can not measure the duration to be greater than or lessor than x irrespective of the motion of the observer(s), or his (their) distance from the source of the light pulse and any mathematical equation that derives differently is wrong!

bob s

Re: Special Relativity is Galilean Relativity December 14 2008, 10:40 AM

Cinci,

Cinci>First, that Wikipedia article is the worst article I've seen on the subject.

They are all bad in some way or other.

Cinci>Yes,they reversed the mirror disgination as you say, but they also reverse or at least describe in a very confused manner. Usually the first tdiagram is said to be what one moving with the mirror sees, that is stataionary with respect to the mirror. The second diagram is usually described as what a stationary observer sees as the mirror moves past him. I think they actually get it right but it is very confusing.

Still wrong; if you get the correct answer by doing things wrong - that's still wrong as far as should be concerned.

>>In any case, you describe the relativstic analysis as a mathematical mistake when the mathematics are perfectly correct for the assumptions. Einstein's second postulate is that all observers will measure light at the same velocity, c. You assumed something else. The math isn't wrong, your assumption is. Experimentally, nobody has ever found the speed of light in vacuo to be anything but the same value.


No, I worked from the assumption of speed of light as constant and did the maths correctly; others such as the article cited in wikipedia did it wrong.

>>Nice try though.

Thanks. And the maths when I did it was correct.

Roger

Roger, SR and Galilean relativity give DIFFERENT predictions for MMX.
Since the MMX measurements agree with SR and disagree with Galilean relativity, you are barking up the wrong tree.

Ted as per normal you are just talking nonsense.

There are lots of mistakes made in special relativity, I was dealing with the time dilation bit. There is little point going into the other errors at this stage.

Cinci: First, that Wikipedia article is the worst article I've seen on the subject.

Roger: They are all bad in some way or other.

cinci: No, some are quite good. I suggest you look for one of them. Or look at Albert's 1905 paper.
**************************

Cinci: Yes,they reversed the mirror disgination as you say, but they also reverse or at least describe it in a very confused manner. Usually the first diagram is said to be what one moving with the mirror sees, that is stataionary with respect to the mirror. The second diagram is usually described as what a stationary observer sees as the mirror moves past him. I think they actually get it right but it is very confusing.

Roger: Still wrong; if you get the correct answer by doing things wrong - that's still wrong as far as should be concerned.

cinci: Using confusing terminology is not the same as making a mistake.
**************************

cinci: In any case, you describe the relativstic analysis as a mathematical mistake when the mathematics are perfectly correct for the assumptions. Einstein's second postulate is that all observers will measure light at the same velocity, c. You assumed something else. The math isn't wrong, your assumption is. Experimentally, nobody has ever found the speed of light in vacuo to be anything but the same value.

Rober: No, I worked from the assumption of speed of light as constant and did the maths correctly; others such as the article cited in wikipedia did it wrong.


cinci: Roger, if we're going to have a serious discussion, you can't deny what's in your paper. It says: "It is just that the velocity to cover the distance L is less than c, let us call it c." You are assuming different values of c for different observers.
*******************

Roger: Thanks. And the maths when I did it was correct.

cinci: Your math is fine, assuming that c has different values for different inertial observers is not. No experiment has ever found this to be correct and many find that it is not correct. In fact virtually every time anyone measures the speed of light you are proven incorrect.
*************************************************************

Re: Special Relativity is Galilean Relativity December 14 2008, 5:44 PM

Cinci: First, that Wikipedia article is the worst article I've seen on the subject.
Roger: They are all bad in some way or other.
cinci: No, some are quite good. I suggest you look for one of them. Or look at Albert's 1905 paper.

ol' Al is full of errors. ref Ohanian: http://discovermagazine.com/2008/sep/01-einstein.s-23-biggest-mistakes


**************************

Cinci: Yes,they reversed the mirror disgination as you say, but they also reverse or at least describe it in a very confused manner. Usually the first diagram is said to be what one moving with the mirror sees, that is stataionary with respect to the mirror. The second diagram is usually described as what a stationary observer sees as the mirror moves past him. I think they actually get it right but it is very confusing.
Roger: Still wrong; if you get the correct answer by doing things wrong - that's still wrong as far as should be concerned.
cinci: Using confusing terminology is not the same as making a mistake.

I don't know how you can judge if you are confused by the wikipedia article.


**************************

cinci: In any case, you describe the relativstic analysis as a mathematical mistake when the mathematics are perfectly correct for the assumptions. Einstein's second postulate is that all observers will measure light at the same velocity, c. You assumed something else. The math isn't wrong, your assumption is. Experimentally, nobody has ever found the speed of light in vacuo to be anything but the same value.
Rober: No, I worked from the assumption of speed of light as constant and did the maths correctly; others such as the article cited in wikipedia did it wrong.
cinci: Roger, if we're going to have a serious discussion, you can't deny what's in your paper. It says: "It is just that the velocity to cover the distance L is less than c, let us call it c." You are assuming different values of c for different observers.

What I said was:--" It is just that the velocity to cover the distance L is less than c, let us call it c."
That's not the same thing; it is not assuming different values for c for different observers. Both observers in O and O' frame measure c.



*******************

Roger: Thanks. And the maths when I did it was correct.
cinci: Your math is fine, assuming that c has different values for different inertial observers is not. No experiment has ever found this to be correct and many find that it is not correct. In fact virtually every time anyone measures the speed of light you are proven incorrect.

My maths is fine for c being the same as measured by different observers.


*************************************************************

Roger Anderton: There are lots of mistakes made in special relativity, I was dealing with the time dilation bit. There is little point going into the other errors at this stage.

Ted: The only errors are in your head due to your inability to understand some simple math/physics. I just showed you that you have no clue in understanding MMX.
How old are you? 70+?

Cinci: First, that Wikipedia article is the worst article I've seen on the subject.
Roger: They are all bad in some way or other.
cinci: No, some are quite good. I suggest you look for one of them. Or look at Albert's 1905 paper.

ol' Al is full of errors. ref Ohanian: http://discovermagazine.com/2008/sep/01-einstein.s-23-biggest-mistakes

cinci: Anybody could make a list like the one in that article. There is no data in the article...they are trying to sell you a book.

Waste your money on the book and present some data here.
**************************

Cinci: Yes,they reversed the mirror disgination as you say, but they also reverse or at least describe it in a very confused manner. Usually the first diagram is said to be what one moving with the mirror sees, that is stataionary with respect to the mirror. The second diagram is usually described as what a stationary observer sees as the mirror moves past him. I think they actually get it right but it is very confusing.

Roger: Still wrong; if you get the correct answer by doing things wrong - that's still wrong as far as should be concerned.

cinci: Using confusing terminology is not the same as making a mistake.

Roger: I don't know how you can judge if you are confused by the wikipedia article.

cinci: I didn't say I was confused by it, I said it was confusing....I figured it out. Why didn't you?
**************************

cinci: In any case, you describe the relativstic analysis as a mathematical mistake when the mathematics are perfectly correct for the assumptions. Einstein's second postulate is that all observers will measure light at the same velocity, c. You assumed something else. The math isn't wrong, your assumption is. Experimentally, nobody has ever found the speed of light in vacuo to be anything but the same value.

Rober: No, I worked from the assumption of speed of light as constant and did the maths correctly; others such as the article cited in wikipedia did it wrong.

cinci: Roger, if we're going to have a serious discussion, you can't deny what's in your paper. It says: "It is just that the velocity to cover the distance L is less than c, let us call it c'." You are assuming different values of c for different observers.

Roger: What I said was:--" It is just that the velocity to cover the distance L is less than c, let us call it c'."

That's not the same thing; it is not assuming different values for c for different observers. Both observers in O and O' frame measure c.

cinci: You are very confused. Who is measuring the value c'. (I inadvertently left the "'" out in my previous message.
*******************

Roger: Thanks. And the maths when I did it was correct.

cinci: Your math is fine, assuming that c has different values for different inertial observers is not. No experiment has ever found this to be correct and many find that it is not correct. In fact virtually every time anyone measures the speed of light you are proven incorrect.

Roger: My maths is fine for c being the same as measured by different observers.

cinci: Except that you indicate somebody is measuring c' or that c' actually exists as a velocity of light different than c. If youassume both observers measure the speed of light at c, you get Einstein's answer.
****************************

Re: Special Relativity is Galilean Relativity December 14 2008, 7:56 PM


Ted: The only errors are in your head

Roger: You got that wrong, its your head; you are the nut.Go away.

Re: Special Relativity is Galilean Relativity December 14 2008, 11:25 PM



cinci: Anybody could make a list like the one in that article. There is no data in the article...they are trying to sell you a book.
Waste your money on the book and present some data here.


Roger: Data is in the book; you buy the book, if you are interested, and don't waste my time.


**************************



Roger: I don't know how you can judge if you are confused by the wikipedia article.
cinci: I didn't say I was confused by it, I said it was confusing....I figured it out. Why didn't you?

Roger: I did figure it out. I think you were confused by it.

**************************



cinci: You are very confused. Who is measuring the value c'. (I inadvertently left the "'" out in my previous message.

Roger: For some reason or other its important to distinguish observers, so the "'" was very important, so its you confused.


*******************


cinci: Except that you indicate somebody is measuring c' or that c' actually exists as a velocity of light different than c. If youassume both observers measure the speed of light at c, you get Einstein's answer.

Roger: both observers measure c, one of them also measures c'. And I point out the errors in the time dilation article. Einstein made lots of mistakes, as pointed out earlier. So, if its Einstein's mistake or whether it is mistakes by someone trying to correct Einstein - well pass.




****************************

Roger: both observers measure c, one of them also measures c'. And I point out the errors in the time dilation article. Einstein made lots of mistakes, as pointed out earlier

Ted: No , he didn't. But you do, old fart.

Re: Special Relativity is Galilean Relativity December 15 2008, 12:00 PM



Ted: No

You are in denial. Go away.

You are an old deluded fart, stay away.

Re: Special Relativity is Galilean Relativity December 14 2008, 11:25 PM



cinci: Anybody could make a list like the one in that article. There is no data in the article...they are trying to sell you a book.
Waste your money on the book and present some data here.


Roger: Data is in the book; you buy the book, if you are interested, and don't waste my time.


cinci: So you're wusing a book as a source that you don'tthinkis worth buying. That makes your reference to the book meaningless.
**************************

cinci: You are very confused. Who is measuring the value c'. (I inadvertently left the "'" out in my previous message.

Roger: For some reason or other its important to distinguish observers, so the "'" was very important, so its you confused.

cinc: No, it was a typogrphical error.
*******************


cinci: Except that you indicate somebody is measuring c' or that c' actually exists as a velocity of light different than c. If you assume both observers measure the speed of light at c, you get Einstein's answer.

Roger: both observers measure c, one of them also measures c'. And I point out the errors in the time dilation article. Einstein made lots of mistakes, as pointed out earlier. So, if its Einstein's mistake or whether it is mistakes by someone trying to correct Einstein - well pass.

cinci: Both measure c but one mesures c'? You need to back up and try again; you're stuck. And you haven't found any errors in relativity.
****************************

Re: Ted December 15 2008, 4:00 PM

Ted: You are an old deluded fart, stay away.

The smell is actually coming from you, that's why I am asking you to go.

Roger: Data is in the book; you buy the book, if you are interested, and don't waste my time.
cinci: So you're wusing a book as a source that you don'tthinkis worth buying. That makes your reference to the book meaningless.

No, you thinking is faulty. I am aware of such errors. Since you seem unaware of such errors, I was referring you to a source so that you might enlighten yourself. But I infer you are just a waster, and not really interested.


**************************

cinci: You are very confused. Who is measuring the value c'. (I inadvertently left the "'" out in my previous message.
Roger: For some reason or other its important to distinguish observers, so the "'" was very important, so its you confused.
cinc: No, it was a typogrphical error.

ok

*******************


cinci: Except that you indicate somebody is measuring c' or that c' actually exists as a velocity of light different than c. If you assume both observers measure the speed of light at c, you get Einstein's answer.
Roger: both observers measure c, one of them also measures c'. And I point out the errors in the time dilation article. Einstein made lots of mistakes, as pointed out earlier. So, if its Einstein's mistake or whether it is mistakes by someone trying to correct Einstein - well pass.
cinci: Both measure c but one mesures c'? You need to back up and try again; you're stuck. And you haven't found any errors in relativity.

Your thinking is faulty again. Both observers measure speed of light as c, while c' is something else.


****************************

Roger, Iggy Ligetty is on a high, smelling other peoples' farts. I have never heard of this complex before, It's historical, and histerical.

Roger: Your thinking is faulty again. Both observers measure speed of light as c, while c' is something else.


cinci: Something else? And "something else" proves Einstein's math is faulty?

Roger, I have to admit that you are "something else".
**********************

Re: Special Relativity is Galilean Relativity December 15 2008, 11:33 PM

Roger: Your thinking is faulty again. Both observers measure speed of light as c, while c' is something else.
cinci: Something else? And "something else" proves Einstein's math is faulty?

Well it shows someone made a maths mistake, Yes. I'm not attacking the theory per se, just pointing out some maths mistakes made in the context of that theory. Einstein made a lot of mistakes in his maths as pointed out by the Ohanian link I gave you. Ideally those should have been corrected in relativity texts, so whoever made the mistake I am pointing out, need not be Einstein.
**********************

Roger: Your thinking is faulty again. Both observers measure speed of light as c, while c' is something else.
cinci: Something else? And "something else" proves Einstein's math is faulty?

Well it shows someone made a maths mistake, Yes. I'm not attacking the theory per se, just pointing out some maths mistakes made in the context of that theory. Einstein made a lot of mistakes in his maths as pointed out by the Ohanian link I gave you. Ideally those should have been corrected in relativity texts, so whoever made the mistake I am pointing out, need not be Einstein.


cinci: There isn't any math mistake in the article.
****************************

Re: Special Relativity is Galilean Relativity December 16 2008, 11:32 AM

Roger: Your thinking is faulty again. Both observers measure speed of light as c, while c' is something else.
cinci: Something else? And "something else" proves Einstein's math is faulty?
Roger:Well it shows someone made a maths mistake, Yes. I'm not attacking the theory per se, just pointing out some maths mistakes made in the context of that theory. Einstein made a lot of mistakes in his maths as pointed out by the Ohanian link I gave you. Ideally those should have been corrected in relativity texts, so whoever made the mistake I am pointing out, need not be Einstein.
cinci: There isn't any math mistake in the article.

You must have had your eyes closed.

cinci: There isn't any math mistake in the article.

Roger: You must have had your eyes closed.

cinci: And you must be kidding.
**********************

Re: Special Relativity is Galilean Relativity December 16 2008, 5:58 PM


Roger: You must have had your eyes closed.

cinci: And you must be kidding.


You are the one with your eyes closed.

The issue raised is about c and c', and if you can't debate as to whether they are equal or not, then you are just wasting time.

Roger: You must have had your eyes closed.

cinci: And you must be kidding.


You are the one with your eyes closed.

The issue raised is about c and c', and if you can't debate as to whether they are equal or not, then you are just wasting time.


cinci: Here's your debate: The gist of your article is that Eisntein made a mistake in his calculations. His calculations say that both the observer moving with tthe clock and the one who is watching it pass by will measure the speed of light as c....that's what his second postulate says and that's how the math is done.

Now you may want to question the second postualte, but the math is correct.

Here is what you do: "That is done so blatantly that we can be seduced by it--- and we need to add the BUT-
-- the time interval for covering the distance L must be the same as the time interval
for covering the distance D in the O frame. It is just that the velocity to cover the
distance L is less than c, let us call it c."

So you're not repeating Einstein's calculations or showing they are wrong, you're analyzing it according to some postulate that says t and t' are the same. Since the object of the analysis is to compare t and t', it doesn't make any sense to declare they are equal. You don't bother to tell us who is measuring c' but it looks like it has to be the observer travelling with the clock. And if you force t = t', well yes, you could calculate that c' is different than c.

Here is what yousay: "So O and O are disagreeing on light speed measurement. O claims you measure it along D not L, because along L thats a reduced velocity, but O claims you measure
it along L."

Since you didn't specify your theory beyond the idea that t = t' you could have reached the conclusion that L is different for each observer. How do you know that isn't the answer?

There are experiments with muons and with actual clocks that say t will not equal t' and many that show all observers regardless of their velocity through space measure light speed at c. On what basis do you assume something else?
**********************

Re: Special Relativity is Galilean Relativity December 17 2008, 4:51 AM

Roger:The issue raised is about c and c', and if you can't debate as to whether they are equal or not, then you are just wasting time.


cinci: Here's your debate: The gist of your article is that Eisntein made a mistake in his calculations. His calculations say that both the observer moving with tthe clock and the one who is watching it pass by will measure the speed of light as c....that's what his second postulate says and that's how the math is done.

Roger: No, a mistake is made with the maths by Einstein+co (relativists); working from the two postulates, a consequence is derived which is wrong.

cinci:Now you may want to question the second postualte, but the math is correct.

No

cinci quoting Roger:Here is what you do: "That is done so blatantly that we can be seduced by it--- and we need to add the BUT-
-- the time interval for covering the distance L must be the same as the time interval
for covering the distance D in the O frame. It is just that the velocity to cover the
distance L is less than c, let us call it c."


No, you misquote me, what I said was:

That is done so blatantly that we can be seduced by it--- and we need to add the BUT-
-- the time interval for covering the distance L must be the same as the time interval
for covering the distance D in the O frame. It is just that the velocity to cover the
distance L is less than c, let us call it c.

You keep missing out the dashes. There is a dash with O and with second use of c.


cinci:So you're not repeating Einstein's calculations or showing they are wrong, you're analyzing it according to some postulate that says t and t' are the same.

I disagree, I derive t = t' but since you miss out the dashes, maybe you misinterpret the article due to that.

Roger

correction:



What this computer system does is seem to miss out the dashes when posting them.

I'm attempting to put the dashes in---


--------------


cinci December 17 2008, 6:34 AM

Re: Special Relativity is Galilean Relativity December 17 2008, 4:51 AM

Roger:The issue raised is about c and c', and if you can't debate as to whether they are equal or not, then you are just wasting time.


cinci: Here's your debate: The gist of your article is that Eisntein made a mistake in his calculations. His calculations say that both the observer moving with tthe clock and the one who is watching it pass by will measure the speed of light as c....that's what his second postulate says and that's how the math is done.

Roger: No, a mistake is made with the maths by Einstein+co (relativists); working from the two postulates, a consequence is derived which is wrong.

cinci:Now you may want to question the second postualte, but the math is correct.

No

cinci quoting Roger:Here is what you do: "That is done so blatantly that we can be seduced by it--- and we need to add the BUT-
-- the time interval for covering the distance L must be the same as the time interval
for covering the distance D in the O frame. It is just that the velocity to cover the
distance L is less than c, let us call it c."


No, you misquote me, what I said was:

That is done so blatantly that we can be seduced by it--- and we need to add the BUT-
-- the time interval for covering the distance L must be the same as the time interval
for covering the distance D in the O [**dash should be here] frame. It is just that the velocity to cover the
distance L is less than c, let us call it c [** dash should be here].

You keep missing out the dashes. There is a dash with O and with second use of c.


cinci:So you're not repeating Einstein's calculations or showing they are wrong, you're analyzing it according to some postulate that says t and t' are the same.

I disagree, I derive t = t' but since you miss out the dashes, maybe you misinterpret the article due to that.

Roger



cinci December 17 2008, 6:34 AM

Re: Special Relativity is Galilean Relativity December 17 2008, 4:51 AM

Roger:The issue raised is about c and c', and if you can't debate as to whether they are equal or not, then you are just wasting time.


cinci: Here's your debate: The gist of your article is that Eisntein made a mistake in his calculations. His calculations say that both the observer moving with tthe clock and the one who is watching it pass by will measure the speed of light as c....that's what his second postulate says and that's how the math is done.

Roger: No, a mistake is made with the maths by Einstein+co (relativists); working from the two postulates, a consequence is derived which is wrong.

cinci:Now you may want to question the second postualte, but the math is correct.

No

cinci quoting Roger:Here is what you do: "That is done so blatantly that we can be seduced by it--- and we need to add the BUT-
-- the time interval for covering the distance L must be the same as the time interval
for covering the distance D in the O frame. It is just that the velocity to cover the
distance L is less than c, let us call it c."


No, you misquote me, what I said was:

That is done so blatantly that we can be seduced by it--- and we need to add the BUT-
-- the time interval for covering the distance L must be the same as the time interval
for covering the distance D in the O frame. It is just that the velocity to cover the
distance L is less than c, let us call it c.

You keep missing out the dashes. There is a dash with O and with second use of c.


cinci:So you're not repeating Einstein's calculations or showing they are wrong, you're analyzing it according to some postulate that says t and t' are the same.

I disagree, I derive t = t' but since you miss out the dashes, maybe you misinterpret the article due to that.

Roger

-------------------

It seems a waste of time talking on this computer setup, if it arbitrarily edits out the dashes when sending.

Best stop debating cinci; gets too difficult.

regards

Roger

Roger:The issue raised is about c and c', and if you can't debate as to whether they are equal or not, then you are just wasting time.


cinci: Here's your debate: The gist of your article is that Eisntein made a mistake in his calculations. His calculations say that both the observer moving with tthe clock and the one who is watching it pass by will measure the speed of light as c....that's what his second postulate says and that's how the math is done.

Roger: No, a mistake is made with the maths by Einstein+co (relativists); working from the two postulates, a consequence is derived which is wrong.



cinci: The consequence is time dialtion and it is experimentally verified.
******************************

cinci:Now you may want to question the second postulate, but the math is correct.

Roger: No


cinci: Yes, it is. It's simple geometry and algebra and I've done it myself. If you think there is an error, then point out exactly what it is. Debating includes explaining your assertions...you haven't.
*********************

cinci quoting Roger:Here is what you do: "That is done so blatantly that we can be seduced by it--- and we need to add the BUT-
-- the time interval for covering the distance L must be the same as the time interval
for covering the distance D in the O frame. It is just that the velocity to cover the
distance L is less than c, let us call it c'."


Roger: No, you misquote me, what I said was:

That is done so blatantly that we can be seduced by it--- and we need to add the BUT-
-- the time interval for covering the distance L must be the same as the time interval
for covering the distance D in the O [**dash should be here] frame. It is just that the velocity to cover the distance L is less than c, let us call it c [** dash should be here].

You keep missing out the dashes. There is a dash with O and with second use of c.

cinci: I cut and pasted; I dont' know why it doesn't get it all. In any case, the dahes aren't the problem. The problem is that you say: "It is just that the velocity to cover the
distance L is less than c, let us call it c'." This c' speed of light arises because you require t = t' and there is no reason to do that.
*************


cinci: So you're not repeating Einstein's calculations or showing they are wrong, you're analyzing it according to some postulate that says t and t' are the same.

Roger: I disagree, I derive t = t' but since you miss out the dashes, maybe you misinterpret the article due to that.


cinci: Well no, you didn't derive it at all. You "derivation" is all tied up with assuming the c' is different that c and the difference makes t = t'. If that's your theory, that's your theory. Now all youhave to do is prove that light has two different velocities. Good luck. But the fact that you have a theory different than SR doesn't mean that Eisntein made a math mistake. I doubt that you can even get AAF to go along with this line of reasoning.
************************


Roger: Best stop debating cinci; gets too difficult.



cinci: So far it's not a debate. I keep telling you what you've done wrong and you keep repeating that Einstein made a math error that yoiu don't seem to be able to describe.
************************************

Re: Special Relativity is Galilean Relativity December 17 2008, 2:28 PM

Roger:The issue raised is about c and c', and if you can't debate as to whether they are equal or not, then you are just wasting time.


cinci: Here's your debate: The gist of your article is that Eisntein made a mistake in his calculations. His calculations say that both the observer moving with tthe clock and the one who is watching it pass by will measure the speed of light as c....that's what his second postulate says and that's how the math is done.
Roger: No, a mistake is made with the maths by Einstein+co (relativists); working from the two postulates, a consequence is derived which is wrong.
cinci: The consequence is time dialtion and it is experimentally verified.

Mistakes can be made by experimentalists when they do their maths.




******************************

cinci:Now you may want to question the second postulate, but the math is correct.
Roger: No
cinci: Yes, it is. It's simple geometry and algebra and I've done it myself. If you think there is an error, then point out exactly what it is. Debating includes explaining your assertions...you haven't.

The article explained it.
*********************



cinci: I cut and pasted; I dont' know why it doesn't get it all.

Yes some sort of mess happens, when I tried it.

cinci: In any case, the dahes aren't the problem.

They are to me; if I can't have my dashes I'm not happy talking about it.

cinci:The problem is that you say: "It is just that the velocity to cover the distance L is less than c, let us call it c'."

That's in the O' frame as explained in the article. With L being something different in the O frame.

cinci:This c' speed of light arises because you require t = t' and there is no reason to do that.

No I don't require that, I end up with the result that t=t'

(hopefully I am not losing dashes)

*************


cinci: So you're not repeating Einstein's calculations or showing they are wrong, you're analyzing it according to some postulate that says t and t' are the same.

Roger: I disagree, I derive t = t' but since you miss out the dashes, maybe you misinterpret the article due to that.


cinci: Well no, you didn't derive it at all. You "derivation" is all tied up with assuming the c' is different that c and the difference makes t = t'.

I did derive it; For delta t = delta t' Then c does not equal c'.

Cinci:If that's your theory, that's your theory. Now all youhave to do is prove that light has two different velocities. Good luck.

Thankyou

cinci>But the fact that you have a theory different than SR doesn't mean that Eisntein made a math mistake.

It's actually Einstein+co, and they didn't cover the possibility I mention so they are in error.

************************


cinci: So far it's not a debate. I keep telling you what you've done wrong and you keep repeating that Einstein made a math error that yoiu don't seem to be able to describe.

Roger: I did describe it - it is about c and c'. If we consider that possibility things are different, "they" didn't consider so "they" are in error.

(if dashes don't come out properly, then it will just be more of a mess.)

To measure c of a pulse of light emitted in another frame, the travel time of the light pulse must be measured over a distance in the opposite frame. To reflect it from mirrors in that frame is to convert it to the local speed of light.

curt>To measure c of a pulse of light emitted in another frame, the travel time of the light pulse must be measured over a distance in the opposite frame. To reflect it from mirrors in that frame is to convert it to the local speed of light.

Where's the reference for that, please?

Answer to Roger: It is just logic. To rule out the possibility that the mirrors convert the speed to the local speed, a portion of the pulse from the other frame must "stop" the time on two locally synchronized clocks, directly, a known distance apart. The difference in time between the clocks, a known distance apart, gives the true speed of the incoming light pulse.

There are no local effects in the usual scenario that SR considers - inertial frames, no gravity etc., so light speed is constant.

Re: Rodger, curt December 19 2008, 8:13 PM

Rodger,
"There are no local effects in the usual scenario that SR considers - inertial frames, no gravity etc., so light speed is constant."

Rodger, light can only propagate from a gravitational body and gravity does effect light speed so I think that questioning the constancy of the speed of light is still an issue for debate. A ponderable body in motion at the speed of light can leave a trail but not light the way.

bob s

Special Relativity (SR) is usually modelled as considering the scenario where there is no gravity; when gravity is considered then you are supposed to use General Relativity (GR). So SR then is considering light speed constant, with no influences altering that.

Re: Rodger, Bob December 19 2008, 10:20 PM

Rodger,
"Special Relativity (SR) is usually modelled as considering the scenario where there is no gravity;..."

That is true Rodger, but! is it possible to consider (ponder) a scenario in which light exists absent a gravitational source? Yes! But! the scenario is just fanciful (SiFi).

Rodger,
"...when gravity is considered then you are supposed to use General Relativity (GR)."

General Relativity is not valid unless Newtonian Mechanics fail and then, iff Special Relativity is valid, which it is not, as previously stated.

bob s

roger: Roger: I did describe it - it is about c and c'. If we consider that possibility things are different, "they" didn't consider so "they" are in error.


cinci: Einstein makes the assumption that there isn't an c'; that is, everybody in any inertial frame will mesaure c at the same value. So not considreing c' is not an error in approach and even if it were it wouldn't be a math error.

The reason for the accuption is that every bmeasurement we make of the sol comes out c. The Bureau of Standards fixed it a particular value decades ago. That speed is a constant on nature. There simply isn't any c'.
********************************

Bob>>That is true Rodger, but! is it possible to consider (ponder) a scenario in which light exists absent a gravitational source? Yes!

Yes

Bob>>But! the scenario is just fanciful (SiFi).

If you want to mislabel it as "science fiction" when it's really "science"; so what.


Bob>>General Relativity is not valid unless Newtonian Mechanics fail and then, iff Special Relativity is valid, which it is not, as previously stated.

The point of my articles are-- there are maths mistakes made by many in the context of SR, correct those maths mistakes and it's back to Newtonian physics. Similarly with GR, correct those and its back to Newtonian physics.

cinci: Einstein makes the assumption that there isn't an c'; that is, everybody in any inertial frame will mesaure c at the same value.

No he does not. Yes he assumes c is constant, but c' is not c.


....|...../
....|..../
....|.../
....|../
....|./
....|/________


y direction vertically up
x direction horizontal
suppose the hyp of this triangle is the direction light goes at with velocity c
let v be the velocity in the x direction
let c' be the velocity in y direction
then by Pythagoras
c^2 = v^2 + c'^2

v and c' do not equal c.











********************************

cinci: Einstein makes the assumption that there isn't an c'; that is, everybody in any inertial frame will mesaure c at the same value.

Roger: No he does not. Yes he assumes c is constant, but c' is not c.


....|...../
....|..../
....|.../
....|../
....|./
....|/________


y direction vertically up
x direction horizontal
suppose the hyp of this triangle is the direction light goes at with velocity c
let v be the velocity in the x direction
let c' be the velocity in y direction
then by Pythagoras
c^2 = v^2 + c'^2

v and c' do not equal c.

cinci: We're talking about a single ray of light here. The observer moving with thte clock sees it go up and back along a vertical line at c.

For the stationary observer, the light has to encounter the same two point on the mirrors as the moving observer. That means he sees the ray travel along the hypotenuse as you have shown. But the stationary observer doesn't see light travelling along the vertical line; that would mean he is seeing two rays where only one exists.

So the moving observer sees the light travel 2L while the stationary observer sees it travel 2D. Both observers will say the light travels at c; therefore they will measure different amounts of time between the same two events; that is, the event when the ray leaves the bottom mirror and the event when it returns. Only clock rate differences will explain this.
****************************

cinci: We're talking about a single ray of light here. The observer moving with thte clock sees it go up and back along a vertical line at c.

So, you moved on, and accepted c does not equal (?)

cinci:The observer moving with thte clock sees it go up and back along a vertical line at c.

Yes. Observer in O frame observes delta t = 2L/c

cinci:For the stationary observer, the light has to encounter the same two point on the mirrors as the moving observer.

The observer in the O and O' yes.

cinci:That means he sees the ray travel along the hypotenuse as you have shown.

The observer in O' frame sees light travel along the hyp yes.

cinci:But the stationary observer doesn't see light travelling along the vertical line; that would mean he is seeing two rays where only one exists.

Who is being referred to as stationary observer (?) both observers in their frames O and O' believe themselves stationary.

The observer in O' frame sees light travel along the hyp of the triangle, and does not see light travel along the vertical of that triangle. He is using time interval 1/2 delta t'. The velocity along the horizontal is v, and along the vertical is c'.

Both c and c' do not equal the hyp of the triangle which is c; unless v =0. And by Pythagoras-

c^2 = v^2 + c'^2

if multiply through by time interval (1/2 delta t')^2, we get the distances-

c^2 (1/2 delta t')^2 = v^2 (1/2 delta t')+ c'^2 (1/2 delta t')^2

the vertical distance is c'^2 (1/2 delta t')^2

that equated to distance light travelled by light as seen by observer in O frame

c^2 (1/2 delta t)^2

i.e. c'^2 (1/2 delta t')^2 = c^2 (1/2 delta t)^2

and from that relativistic factor is obtained using c'^2 = c^2 - v^2

--- that's standard maths in SR,

And as noted that is in error because the vertical of the triangle c'^2 (1/2 delta t')^2 , no one observes light travelling that distance.


the error is not to equate distance light seen travelled by O' observer with distance light seen travelled by O observer.

instead of doing it that way.

it takes a distance that observer in O' frame does not see light travel along and equates it with distance light seen travelled along by observer in O frame.

Do the maths correctly and there is no time dilation.

Roger: Both c and c' do not equal the hyp of the triangle which is c; unless v =0. And by Pythagoras-

c^2 = v^2 + c'^2

Ted: This is the stupidity that you keep repeating. The stationary observer calculates that light travels along the hypotenuse according to the equation:

(ct)^2=(vt)^2+L^2

From this, any sensible scientist (not Roger Anderton, who is an imbecile) concludes that:

t=L/sqrt(c^2-v^2)

cinci December 20 2008, 12:20 PM


cinci: We're talking about a single ray of light here. The observer moving with thte clock sees it go up and back along a vertical line at c.

So, you moved on, and accepted c does not equal (?)

cinci:The observer moving with thte clock sees it go up and back along a vertical line at c.

Yes. Observer in O frame observes delta t = 2L/c

cinci:For the stationary observer, the light has to encounter the same two point on the mirrors as the moving observer.

The observer in the O and O' yes.

cinci:That means he sees the ray travel along the hypotenuse as you have shown.

The observer in O' frame sees light travel along the hyp yes.

cinci:But the stationary observer doesn't see light travelling along the vertical line; that would mean he is seeing two rays where only one exists.

Who is being referred to as stationary observer (?) both observers in their frames O and O' believe themselves stationary.

The observer in O' frame sees light travel along the hyp of the triangle, and does not see light travel along the vertical of that triangle. He is using time interval 1/2 delta t'. The velocity along the horizontal is v, and along the vertical is c'.

(correction) c' does not equal the hyp of the triangle which is c; unless v =0. And by Pythagoras-

c^2 = v^2 + c'^2

if multiply through by time interval (1/2 delta t')^2, we get the distances-

c^2 (1/2 delta t')^2 = v^2 (1/2 delta t')+ c'^2 (1/2 delta t')^2

the vertical distance is c'^2 (1/2 delta t')^2

that equated to distance light travelled by light as seen by observer in O frame

c^2 (1/2 delta t)^2

i.e. c'^2 (1/2 delta t')^2 = c^2 (1/2 delta t)^2

and from that relativistic factor is obtained using c'^2 = c^2 - v^2

--- that's standard maths in SR,

And as noted that is in error because the vertical of the triangle c'^2 (1/2 delta t')^2 , no one observes light travelling that distance.


the error is not to equate distance light seen travelled by O' observer with distance light seen travelled by O observer.

instead of doing it that way.

it takes a distance that observer in O' frame does not see light travel along and equates it with distance light seen travelled along by observer in O frame.

Do the maths correctly and there is no time dilation.

Roger> And by Pythagoras-

c^2 = v^2 + c'^2

Ted: This is the stupidity that you keep repeating.

No you are being stupid.

Ted>>The stationary observer calculates that light travels along the hypotenuse according to the equation:

(ct)^2=(vt)^2+L^2

That is an error as I have pointed out and which you ignore.

Observer in O frame observes L

Observer in O' frame observes c'(1/2 delta t') along the vertical.

So forms by Pythagoras-

(c 1/2 delta t')^2 = (v 1/2 delta t')^2 + (c'(1/2 delta t')^2

cinci: We're talking about a single ray of light here. The observer moving with thte clock sees it go up and back along a vertical line at c.

Roger: So, you moved on, and accepted c does not equal (?)

cinci: c = 300,000 kn/sec. I have not idea what you mean by "So, you moved on". I am repeating the analysis just like relativsts always do it.
*************************************

cinci:The observer moving with thte clock sees it go up and back along a vertical line at c.

Roger: Yes. Observer in O frame observes delta t = 2L/c

cinci: Yes.
***********************

cinci:For the stationary observer, the light has to encounter the same two point on the mirrors as the moving observer.

Roger: The observer in the O and O' yes.

cinci: Yes, the light ray reaching to points on the mirrors must be agreed to by each observer.
***************************

cinci: That means he sees the ray travel along the hypotenuse as you have shown.

Roger: The observer in O' frame sees light travel along the hyp yes.

cinci: The example youused on the internet has confusing terminology so i'm not sure which is primed. Normally, people don't prime the statioanry frame and do prime the moving frame. What I said was, the stationary observer sees the light travel the hypotneuse.
********************************

cinci:But the stationary observer doesn't see light travelling along the vertical line; that would mean he is seeing two rays where only one exists.

Roger: Who is being referred to as stationary observer (?) both observers in their frames O and O' believe themselves stationary.

cinci: Yes, they do. Usually, people set this up with an observer perhpas standing by a railrod track and call him stationary. The light clock would be going by on a railroad car and would be moving relative to the statioanry observer.
*******************************

Roger: The observer in O' frame sees light travel along the hyp of the triangle, and does not see light travel along the vertical of that triangle. He is using time interval 1/2 delta t'. The velocity along the horizontal is v, and along the vertical is c'.

cinci: Again, you have primed the stationary frame which will confuse you with almost every analysis of this situation I have ever seen. So I will stipulate here that your O' observer is the one standing by the train tracks watching the light clock move by. You say "The velocity ......along the vertical is c'. There is no velocity along the vertical as far as this observer is concerned. There is one ray of light and it travels from the center of the bottom mirror to the center of the top mirror. Since the mirrors are moving together, the line between their centers is not a vertical line; it is the hypotenuse of the triangle whose sides are vt/2 and L. And time to traverse it is t' = [((vt/2)^2 + L^2)^.5]/c
**********************************

Roger; Both c and c' do not equal the hyp of the triangle which is c; unless v =0. And by Pythagoras-

c^2 = v^2 + c'^2

if multiply through by time interval (1/2 delta t')^2, we get the distances-

c^2 (1/2 delta t')^2 = v^2 (1/2 delta t')+ c'^2 (1/2 delta t')^2

the vertical distance is c'^2 (1/2 delta t')^2

that equated to distance light travelled by light as seen by observer in O frame

c^2 (1/2 delta t)^2

i.e. c'^2 (1/2 delta t')^2 = c^2 (1/2 delta t)^2

and from that relativistic factor is obtained using c'^2 = c^2 - v^2

--- that's standard maths in SR,


cinci: No, that is not standard math in SR. There is no velocity c'.

The SR math is done this way

The hypotenuse along which the stationary observer O' sees the light travel is

D = (L^2 + (vt')^2)^.5

The time to traverse this distance at c is

ct' = D

ct' = (L^2 + (vt')^2)^.5

(ct')^2 = L^2 + (vt')^2

L^2 = t'^2(c^2 - v^2)

L = t'(c^2-v^2)^.5

The time for the O observer to see the ray traverse the veritcal distance is

t = L/c

Substituting yields

t = t'((c^2 -v^2)^.5)/c

t = t'(1 - (v/c)^2)^.5

Which says taht while O', the statioanry observer, sees t' pass on his clock, the moving observer O see t'(1 - (v/c)^2)^.5 pass on his clock. Since (1 - (v/c)^2)^.5 is always less than 1, the moving observers clock is running slower than the stationary clock.
*********************************

Roger: And as noted that is in error because the vertical of the triangle c'^2 (1/2 delta t')^2 , no one observes light travelling that distance.

cinci: The error is that there is no c'.
***********************

Roger: the error is not to equate distance light seen travelled by O' observer with distance light seen travelled by O observer.

cinci: The distance seen by O' is not the same distance as seen by O. Equating them is an obvious error.
**************************

Roger: instead of doing it that way.

it takes a distance that observer in O' frame does not see light travel along and equates it with distance light seen travelled along by observer in O frame.

cinci: I have no idea what this sentence means. Why would you use teh distance O' does not see?
**************************

Roger: Do the maths correctly and there is no time dilation.

cinci: I did the maths correctly and there is time dilation. Experiments confrim the phenomenon. If you have a theory that says there isn't time dilation then the theory is in coflict with reality....why would you support that?
******************************

Re: Special Relativity is Galilean Relativity December 20 2008, 1:38 PM

cinci: We're talking about a single ray of light here. The observer moving with thte clock sees it go up and back along a vertical line at c.

Roger: So, you moved on, and accepted c does not equal (?)

cinci: c = 300,000 kn/sec. I have not idea what you mean by "So, you moved on". I am repeating the analysis just like relativsts always do it.

If you still believe c' = c then there is no point to continuing the analysis. And c is only approximately that value you give.


*************************************

cinci:The observer moving with thte clock sees it go up and back along a vertical line at c.

Roger: Yes. Observer in O frame observes delta t = 2L/c

cinci: Yes.

ok
***********************

cinci:For the stationary observer, the light has to encounter the same two point on the mirrors as the moving observer.

Roger: The observer in the O and O' yes.

cinci: Yes, the light ray reaching to points on the mirrors must be agreed to by each observer.

ok

***************************

cinci: That means he sees the ray travel along the hypotenuse as you have shown.

Roger: The observer in O' frame sees light travel along the hyp yes.

cinci: The example youused on the internet has confusing terminology so i'm not sure which is primed. Normally, people don't prime the statioanry frame and do prime the moving frame. What I said was, the stationary observer sees the light travel the hypotneuse.

yes this forum edits out dashes if we are not careful.

The observer in O frame observes

-------
...|
...|... delta t = 2L/c
...|
___|___

light bouncing up and down.


The observer in O' frame is looking at this light clock in O frame and sees the triangle.

cinci>>What I said was, the stationary observer sees the light travel the hypotneuse.

But in SR, both observers believe themselves in stationary frame.


********************************

cinci:But the stationary observer doesn't see light travelling along the vertical line; that would mean he is seeing two rays where only one exists.

Roger: Who is being referred to as stationary observer (?) both observers in their frames O and O' believe themselves stationary.

cinci: Yes, they do. Usually, people set this up with an observer perhpas standing by a railrod track and call him stationary. The light clock would be going by on a railroad car and would be moving relative to the statioanry observer.

But that would be bad terminology, because in SR both observers believe themselves stationary. In the train, the observer if following SR believes himself stationary. An apocryphal story of Einstein on a train is him asking the question "what time does the next station arrive on this train?" -- treating the train as stationary and the stations are coming to the train. Motion is relative in SR.



*******************************

Roger: The observer in O' frame sees light travel along the hyp of the triangle, and does not see light travel along the vertical of that triangle. He is using time interval 1/2 delta t'. The velocity along the horizontal is v, and along the vertical is c'.

cinci: Again, you have primed the stationary frame which will confuse you with almost every analysis of this situation I have ever seen.

I think those analysis might be confused; as already pointed out they seem to use bad terminology if they refer to one frame as stationary and the other as not stationary; that is not conistent with SR.

cinci:So I will stipulate here that your O' observer is the one standing by the train tracks watching the light clock move by.

I think that is not helpful bringing trains into it; unless you accept the pecularity that from the train frame it is stationary with the other things that go with it.

cinci:You say "The velocity ......along the vertical is c'. There is no velocity along the vertical as far as this observer is concerned.

That is incorrect, I presume you are referring to the case where we have velocity c and v, from those two things we can form a velocity vector diagram - standard maths.


cinci: There is one ray of light and it travels from the center of the bottom mirror to the center of the top mirror.

Yes


cinci:Since the mirrors are moving together, the line between their centers is not a vertical line; it is the hypotenuse of the triangle whose sides are vt/2 and L.

No, if you are referring to the triangle with v and c, the sides of the velocity triangle are v, c , c' which we can multiply each by 1/2 delta t' to form distances.




**********************************

Roger; Both c and c' do not equal the hyp of the triangle which is c; unless v =0. And by Pythagoras-

c^2 = v^2 + c'^2

if multiply through by time interval (1/2 delta t')^2, we get the distances-

c^2 (1/2 delta t')^2 = v^2 (1/2 delta t')+ c'^2 (1/2 delta t')^2

the vertical distance is c'^2 (1/2 delta t')^2

that equated to distance light travelled by light as seen by observer in O frame

c^2 (1/2 delta t)^2

i.e. c'^2 (1/2 delta t')^2 = c^2 (1/2 delta t)^2

and from that relativistic factor is obtained using c'^2 = c^2 - v^2

--- that's standard maths in SR,


cinci: No, that is not standard math in SR. There is no velocity c'.

Some texts miss out using c' and use other letters.

cinci:The SR math is done this way

The hypotenuse along which the stationary observer O' sees the light travel is

D = (L^2 + (vt')^2)^.5

And I have pointed out it is not L.

it is-

c^2 (1/2 delta t')^2 = v^2 (1/2 delta t')+ c'^2 (1/2 delta t')^2

Roger: Observer in O' frame observes c'(1/2 delta t') along the vertical.

Ted: No, he doesn't but there is no way to get this idiocy out of your brain

Ted: No, he doesn't but there is no way to get this idiocy out of your brain.

The idiocy is in your brain.

Considering light bouncing between two mirrors A and B.

In SR standard maths we have the following-

........B.____
...........|
...........|...... h1
...........|
........A__|__

A person in O frame observing its own light clock, light travels distance h1

(taking time interval as up and return as delta t, one way trip is then time interval 1/2 delta t)

thus h1= 1/2 c delta t

A person in O' frame observing O

triangle being-

...................B.
................../.|
.............h2../..|.. b2
................/...|
.............A./____|
.................a2

Standard maths of SR makes the mistake of saying b2= h1
When really it is h2 = h1.

Because should be - Light seen by O' frame covers distance h2 and that equates to O frame seeing light cover distance h1.


h2 = 1/2 c delta t
b2= 1/2 c' delta t
a2= 1/2 v delta t

c = speed of light
c' = projection of speed of light onto the vertical, and is less than c
v = velocity of motion of O according to O'

Considering light bouncing between two mirrors A and B.

In SR standard maths we have the following-

........B.____
...........|
...........|...... h1
...........|
........A__|__

A person in O frame observing its own light clock, light travels distance h1

(taking time interval as up and return as delta t, one way trip is then time interval 1/2 delta t)

thus h1= 1/2 c delta t

A person in O' frame observing O

triangle being-

...................B.
................../.|
.............h2../..|.. b2
................/...|
.............A./____|
.................a2

Standard maths of SR makes the mistake of saying b2= h1
When really it is h2 = h1.

Because should be - Light seen by O' frame covers distance h2 and that equates to O frame seeing light cover distance h1.


h2 = 1/2 c delta t'
b2= 1/2 c' delta t'
a2= 1/2 v delta t'

c = speed of light
c' = projection of speed of light onto the vertical, and is less than c
v = velocity of motion of O according to O'

Neither observer in either frame can directly see the light bouncing between the mirrors. The only light that can be seen is the light from one of the mirrors ending up in the observer's eyes. The whole scenario is imagined. A transverse pulse of light, when recorded in a moving frame traveling in exactly the opposite direction, will appear as a line exactly as long as the duration of the pulse divided by the relative speed between the frames.

At the speed of light, neither frame will see the other coming, but the pulse, transformed into a line, will be identical as recorded in either frame.

curt>>Neither observer in either frame can directly see the light bouncing between the mirrors.

That's a diversion from real issue of the original posting.

Whether light can directly be seen or whether light excites matter which is then seen; path of light can be seen using filters etc.

cinci: We're talking about a single ray of light here. The observer moving with thte clock sees it go up and back along a vertical line at c.

Roger: So, you moved on, and accepted c does not equal (?)

cinci: c = 300,000 kn/sec. I have not idea what you mean by "So, you moved on". I am repeating the analysis just like relativsts always do it.

Roger: If you still believe c' = c then there is no point to continuing the analysis. And c is only approximately that value you give.

cinci: Roger, there is nothing travelling at c'. You're mixing up what one observer sees with what the other on sees. The observer with the clock sees light travelling along the lenght L. The other observer does not and this is where you define the c'.

If you want the exact number for c, look it up.
*************************************

cinci:The observer moving with the clock sees it go up and back along a vertical line at c.

Roger: Yes. Observer in O frame observes delta t = 2L/c

cinci: Yes.

ok


cinci:For the stationary observer, the light has to encounter the same two point on the mirrors as the moving observer.

Roger: The observer in the O and O' yes.

cinci: Yes, the light ray reaching to points on the mirrors must be agreed to by each observer.

ok


cinci: That means he sees the ray travel along the hypotenuse as you have shown.

Roger: The observer in O' frame sees light travel along the hyp yes.

cinci: The example you used on the internet has confusing terminology so I'm not sure which is primed. Normally, people don't prime the stationary frame and do prime the moving frame. What I said was, the stationary observer sees the light travel the hypotneuse.

yes this forum edits out dashes if we are not careful.

The observer in O frame observes

-------
...|
...|... delta t = 2L/c
...|
___|___

light bouncing up and down.


The observer in O' frame is looking at this light clock in O frame and sees the triangle.

cinci>>What I said was, the stationary observer sees the light travel the hypotneuse.

But in SR, both observers believe themselves in stationary frame.

cinci: Yes the each may consider themselves stationary, but I picked one situation here to analyze and the one I picked has the clock moving. Throwing in this comment is just to cause unnecessary confusion.
********************************

cinci: But the stationary observer doesn't see light travelling along the vertical line; that would mean he is seeing two rays where only one exists.

Roger: Who is being referred to as stationary observer (?) both observers in their frames O and O' believe themselves stationary.

cinci: Once again, I explained that I chose to consider the clock is moving so that makes the observer with the clock the movign observer and the other is the stationary observer. Is your problem that you can't keep these choices in mind?

The problem can be worked the other way, but youcan't make both assumptions at once. Perhaps this confusion is why youthink there is a c'.
***********************

cinci: Yes, they do. Usually, people set this up with an observer perhaps standing by a railroad track and call him stationary. The light clock would be going by on a railroad car and would be moving relative to the statioanry observer.

Roger: But that would be bad terminology, because in SR both observers believe themselves stationary.

cinci: It doesn't make any difference what the observers think, they aren't doing this analysis, YOU are. YOU are analyzing the problem and YOU get to pick which one YOU consider stationary. So the terminology is perfectly valid.
**************************************

Roger: In the train, the observer if following SR believes himself stationary. An apocryphal story of Einstein on a train is him asking the question "what time does the next station arrive on this train?" -- treating the train as stationary and the stations are coming to the train. Motion is relative in SR.

cinci: The theory doesn't say that everybody considers himself stationary; this is another of those apochryphal oversimplifications. What the theory says is that you can't perform an experiment that proves you are moving. Both observers can know that they have relative motion. But they are free to assume that they are moving or stationary. The issues work out the same either way. But anybody analyzing the situation including the observers can make either assumpiton and the fact they might have made the alternate assumption doesn't mean they can't do the analysis. You're either very confused about this or throwing it in as a red herring.
*******************************

Roger: The observer in O' frame sees light travel along the hyp of the triangle, and does not see light travel along the vertical of that triangle. He is using time interval 1/2 delta t'. The velocity along the horizontal is v, and along the vertical is c'.

cinci: Here is the prrof of your confusion. First you say this observer sees the light travel along the hypotenuse (at c I assume, you don't say) abd then you say "The velocity....along the vertical is c'" The velocity of what....you just said the light was along the hypotenuse. what's going along the vertical?
******************************

cinci: Again, you have primed the stationary frame which will confuse you with almost every analysis of this situation I have ever seen.

I think those analysis might be confused; as already pointed out they seem to use bad terminology if they refer to one frame as stationary and the other as not stationary; that is not conistent with SR.

cinci:So I will stipulate here that your O' observer is the one standing by the train tracks watching the light clock move by.

Roger: I think that is not helpful bringing trains into it; unless you accept the pecularity that from the train frame it is stationary with the other things that go with it.

cinci: I used trains to try to clairy your confusion about statioanry and moving observers......apparently it didn't work.
****************************

cinci:You say "The velocity ......along the vertical is c'. There is no velocity along the vertical as far as this observer is concerned.

Roger: That is incorrect, I presume you are referring to the case where we have velocity c and v, from those two things we can form a velocity vector diagram - standard maths.

cinci: Yes, and 2 + 2 = 4 but it doesn't happen to have anything to do with this situation. For the observer who sees light moving at c along the hypotenuse, nothing is moving along the vertical length L. So you can perform that calsulation but it doesn't pertain to anything physical in this model. This observer see the length L there, nothing more.
***************************

cinci: There is one ray of light and it travels from the center of the bottom mirror to the center of the top mirror.

Roger: Yes

cinci: Since the mirrors are moving together, the line between their centers is not a vertical line; it is the hypotenuse of the triangle whose sides are vt/2 and L.

Roger: No, if you are referring to the triangle with v and c, the sides of the velocity triangle are v, c , c' which we can multiply each by 1/2 delta t' to form distances.

cinci: No, the sides are ct, vt, and L. There is nothing moving along L for this oberver and therefore no velocity c'
**********************************

Roger; Both c and c' do not equal the hyp of the triangle which is c; unless v =0. And by Pythagoras-

c^2 = v^2 + c'^2

if multiply through by time interval (1/2 delta t')^2, we get the distances-

c^2 (1/2 delta t')^2 = v^2 (1/2 delta t')+ c'^2 (1/2 delta t')^2

the vertical distance is c'^2 (1/2 delta t')^2

that equated to distance light travelled by light as seen by observer in O frame

c^2 (1/2 delta t)^2

i.e. c'^2 (1/2 delta t')^2 = c^2 (1/2 delta t)^2

and from that relativistic factor is obtained using c'^2 = c^2 - v^2

--- that's standard maths in SR,


cinci: No, that is not standard math in SR. There is no velocity c'.

Roger: Some texts miss out using c' and use other letters.

cinci: The texts don't use c' because that would indicate a velocity and nothing is moving along that leg. They use L or some other letter implying a distance.
***************************

cinci:The SR math is done this way

The hypotenuse along which the stationary observer O' sees the light travel is

D = (L^2 + (vt')^2)^.5

Roger: And I have pointed out it is not L.

it is-

c^2 (1/2 delta t')^2 = v^2 (1/2 delta t')+ c'^2 (1/2 delta t')^2

cinci: Well go back and look at the diagram in http://www.wbabin.net/science/anderton18.pdf.

L is the distance between the mirrors. c'^2(1/2delta t')^2 = L^2.

You analysis simply goes wrong here. If you can't tell me what is travelling at c', then you should acknowledge that it's wrong. Somehow I doubt that you will.
***************************

cinci: Roger, there is nothing travelling at c'. You're mixing up what one observer sees with what the other on sees. The observer with the clock sees light travelling along the lenght L. The other observer does not and this is where you define the c'.

Yes, I suppose you could look on it as nothing travelling at c', it is a projection of c along the vertical part of the triangle as seen by the person in the O' frame. It is not me mixing this c' up with c it is the usual maths presented in SR texts doing that; they mix up the length that c' acts on with the length that c acts on, and that is what I am saying they do wrong.


*************************************


cinci: Yes the each may consider themselves stationary, but I picked one situation here to analyze and the one I picked has the clock moving. Throwing in this comment is just to cause unnecessary confusion.

No it wasn't. It was about the relativity of observers, and it was important that you should be aware of it. Your way of talking as if there was only one stationary observer was implying you did not know relativity.


********************************


cinci: Yes, they do. Usually, people set this up with an observer perhaps standing by a railroad track and call him stationary. The light clock would be going by on a railroad car and would be moving relative to the statioanry observer.

Roger: But that would be bad terminology, because in SR both observers believe themselves stationary.

cinci: It doesn't make any difference what the observers think, they aren't doing this analysis, YOU are. YOU are analyzing the problem and YOU get to pick which one YOU consider stationary. So the terminology is perfectly valid.

No you don't pick who is stationary, according to relativity they both are. Relativity is part of the analysis.


**************************************


cinci: What the theory says is that you can't perform an experiment that proves you are moving.

Which means by relativity they are both treated as stationary.

cinci:Both observers can know that they have relative motion.

yes.

cinci:But they are free to assume that they are moving or stationary.

No, they don't know if they are moving or not, so can assume that they are stationary.

It's you that very confused about this.
*******************************


cinci: Here is the prrof of your confusion. First you say this observer sees the light travel along the hypotenuse (at c I assume, you don't say) abd then you say "The velocity....along the vertical is c'"

Yes

cinci:The velocity of what....you just said the light was along the hypotenuse. what's going along the vertical?

projection of c


******************************

cinci: I used trains to try to clairy your confusion about statioanry and moving observers......apparently it didn't work.

I think you are confused by it.

****************************

cinci: This observer see the length L there, nothing more.

The observer in the O frame measures L and that is the hyp of the observer in the O' frame.

***************************

cinci: the sides are ct, vt, and L. There is nothing moving along L for this oberver and therefore no velocity c' .


No the sides of the triangle for O' observer are ct, vt, (and third side let us call) L'. The L is what the O observer has. And L does not equal L'.


**********************************



Roger: Some texts miss out using c' and use other letters.

cinci: The texts don't use c' because that would indicate a velocity and nothing is moving along that leg. They use L or some other letter implying a distance.

They do use c', but can use the length L' instead.

So then-

...... B
......./|
....../.|
...../..|... L'
..../___|
...A.....

horizontal = 1/2 v delta t'
hyp = 1/2 c delta t'


Standard SR maths equates L to L' where L is 1/2 c delta t

And that is wrong.

You say : "... nothing is moving along that leg" --- that is the leg L'

And that is precisely my point-- In O' frame the light moves from mirror A to mirror B along the hyp of the triangle.
In the O frame the light moves from mirror A to mirror B along L.

The light does not move along L'.

So the SR texts stating L = L' are in error.

Light does not travel along L'

Yes, smoke could be used to divert some of the beam into diffuse light that could be seen. What could be seen is that the "light path" is perpendicular to the motion, if that's the direction it was directed. To directly sense the light from the emission or mirrors, some of it must pass outside the mirrors to be sensed in the other frame.

cinci: Roger, there is nothing travelling at c'. You're mixing up what one observer sees with what the other on sees. The observer with the clock sees light travelling along the lenght L. The other observer does not and this is where you define the c'.

Roger: Yes, I suppose you could look on it as nothing travelling at c', it is a projection of c along the vertical part of the triangle as seen by the person in the O' frame. It is not me mixing this c' up with c it is the usual maths presented in SR texts doing that; they mix up the length that c' acts on with the length that c acts on, and that is what I am saying they do wrong.

cinci: They aren't mixing up anything. Why would they want to deal witht the porjectin of a velocity? Why do you?
*************************************


cinci: Yes they each may consider themselves stationary, but I picked one situation here to analyze and the one I picked has the clock moving. Throwing in this comment is just to cause unnecessary confusion.

Roger: No it wasn't. It was about the relativity of observers, and it was important that you should be aware of it. Your way of talking as if there was only one stationary observer was implying you did not know relativity.

cinci: For any given analysis, there IS only one stationary observer. That sdoesn't mean I can[t analyze it making the other assumption, but trying to deal with them at the same time is bound to be confusing and it's why you're confused.
********************************


cinci: Yes, they do. Usually, people set this up with an observer perhaps standing by a railroad track and call him stationary. The light clock would be going by on a railroad car and would be moving relative to the statioanry observer.

Roger: But that would be bad terminology, because in SR both observers believe themselves stationary.

cinci: It doesn't make any difference what the observers think, they aren't doing this analysis, YOU are. YOU are analyzing the problem and YOU get to pick which one YOU consider stationary. So the terminology is perfectly valid.

Roger: No you don't pick who is stationary, according to relativity they both are. Relativity is part of the analysis.

cinci: No, you can't say the both ARE. You may consider either stationary but whn you have done that, the other is moving. Observers can't be moving and stationary at the same time.
**************************************


cinci: What the theory says is that you can't perform an experiment that proves you are moving.

Roger: Which means by relativity they are both treated as stationary.

cinci: No, it means that you can consider either stationary. It does not mean you can consider both stationary at the same time....that would violate the set up of the problem.
*********************

cinci: Both observers can know that they have relative motion.

Roger: yes.

cinci: But they are free to assume that they are moving or stationary.

Roger: No, they don't know if they are moving or not, so can assume that they are stationary.

It's you that very confused about this.

cinci: Yes, they can assume they are staionary, but they can't assume that both are stationary at the same time.
*******************************

cinci: Here is the proof of your confusion. First you say this observer sees the light travel along the hypotenuse (at c I assume, you don't say) and then you say "The velocity....along the vertical is c'"

Roger: Yes

cinci: The velocity of what....you just said the light was along the hypotenuse. what's going along the vertical?

Roger: projection of c

cinci: Why is that important? You could draw a 45 degree angle line and take the projection along it as well. Why don't you do that?
******************************

cinci: I used trains to try to clairy your confusion about statioanry and moving observers......apparently it didn't work.

Roger: I think you are confused by it.

cinci: You are wrong.
****************************

cinci: This observer see the length L there, nothing more.

Roger: The observer in the O frame measures L and that is the hyp of the observer in the O' frame.

cinci: No comment.
***************************

cinci: the sides are ct, vt, and L. There is nothing moving along L for this oberver and therefore no velocity c' .

Roger: No the sides of the triangle for O' observer are ct, vt, (and third side let us call) L'. The L is what the O observer has. And L does not equal L'.


**********************************



Roger: Some texts miss out using c' and use other letters.

cinci: The texts don't use c' because that would indicate a velocity and nothing is moving along that leg. They use L or some other letter implying a distance.

They do use c', but can use the length L' instead.

So then-

...... B
......./|
....../.|
...../..|... L'
..../___|
...A.....

horizontal = 1/2 v delta t'
hyp = 1/2 c delta t'


Standard SR maths equates L to L' where L is 1/2 c delta t

And that is wrong.

You say : "... nothing is moving along that leg" --- that is the leg L'

And that is precisely my point-- In O' frame the light moves from mirror A to mirror B along the hyp of the triangle.
In the O frame the light moves from mirror A to mirror B along L.

The light does not move along L'.

So the SR texts stating L = L' are in error.

Light does not travel along L'

cinci: The texts don't say L=L'. They say that delta(t) is the time ths stationary observer measures for the light to traverse L and that delta(t') is the time that the moving observer measures for the light to traverse L'.
**********************

cinci: The texts don't say L=L'. They say that delta(t) is the time ths stationary observer measures for the light to traverse L and that delta(t') is the time that the moving observer measures for the light to traverse L'.

Some texts use L = L'

Re: Special Relativity is Galilean Relativity January 5 2009, 8:12 PM

cinci: Roger, there is nothing travelling at c'. You're mixing up what one observer sees with what the other on sees. The observer with the clock sees light travelling along the lenght L. The other observer does not and this is where you define the c'.
Roger: Yes, I suppose you could look on it as nothing travelling at c', it is a projection of c along the vertical part of the triangle as seen by the person in the O' frame. It is not me mixing this c' up with c it is the usual maths presented in SR texts doing that; they mix up the length that c' acts on with the length that c acts on, and that is what I am saying they do wrong.
cinci: Why would they want to deal witht the porjectin of a velocity? Why do you?

Its one method available.


*************************************


cinci: Yes they each may consider themselves stationary, but I picked one situation here to analyze and the one I picked has the clock moving. Throwing in this comment is just to cause unnecessary confusion.
Roger: No it wasn't. It was about the relativity of observers, and it was important that you should be aware of it. Your way of talking as if there was only one stationary observer was implying you did not know relativity.
cinci: For any given analysis, there IS only one stationary observer. That sdoesn't mean I can[t analyze it making the other assumption, but trying to deal with them at the same time is bound to be confusing.

Both observers have their own stationary frame.

********************************


cinci: Yes, they do. Usually, people set this up with an observer perhaps standing by a railroad track and call him stationary. The light clock would be going by on a railroad car and would be moving relative to the statioanry observer.
Roger: But that would be bad terminology, because in SR both observers believe themselves stationary.
cinci: It doesn't make any difference what the observers think, they aren't doing this analysis, YOU are. YOU are analyzing the problem and YOU get to pick which one YOU consider stationary. So the terminology is perfectly valid.
Roger: No you don't pick who is stationary, according to relativity they both are. Relativity is part of the analysis.
cinci: No, you can't say the both ARE. You may consider either stationary but whn you have done that, the other is moving. Observers can't be moving and stationary at the same time.

They can by relativity theory.

**************************************


cinci: What the theory says is that you can't perform an experiment that proves you are moving.
Roger: Which means by relativity they are both treated as stationary.
cinci: No, it means that you can consider either stationary. It does not mean you can consider both stationary at the same time....

They can be stationary same time if they are stationary with respect to each other.



*********************

cinci: Here is the proof of your confusion. First you say this observer sees the light travel along the hypotenuse (at c I assume, you don't say) and then you say "The velocity....along the vertical is c'"
Roger: Yes
cinci: The velocity of what....you just said the light was along the hypotenuse. what's going along the vertical?
Roger: projection of c
cinci: You could draw a 45 degree angle line and take the projection along it as well. Why don't you do that?

I am doing for the c case at the moment.
******************************

cinci: I used trains to try to clairy your confusion about statioanry and moving observers......apparently it didn't work.
Roger: I think you are confused by it.
cinci: You are wrong.

No its you.

****************************


Roger: Some texts miss out using c' and use other letters.

cinci: The texts don't use c' because that would indicate a velocity and nothing is moving along that leg. They use L or some other letter implying a distance.

They do use c', but can use the length L' instead.

So then-

...... B
......./|
....../.|
...../..|... L'
..../___|
...A.....

horizontal = 1/2 v delta t'
hyp = 1/2 c delta t'


Standard SR maths equates L to L' where L is 1/2 c delta t

And that is wrong.

You say : "... nothing is moving along that leg" --- that is the leg L'

And that is precisely my point-- In O' frame the light moves from mirror A to mirror B along the hyp of the triangle.
In the O frame the light moves from mirror A to mirror B along L.

The light does not move along L'.

So the SR texts stating L = L' are in error.

Light does not travel along L'

cinci: The texts don't say L=L'. They say that delta(t) is the time ths stationary observer measures for the light to traverse L and that delta(t') is the time that the moving observer measures for the light to traverse L'.

Some texts say L = L'
**********************

Roger, all Cincirob is trying to say, is that the observer stationed on the moving train can consider himself to be at rest, while the observer who is stationary is assumed to be moving- in reference of the train observer.

Re: Special Relativity is Galilean Relativity January 5 2009, 8:12 PM

cinci: Roger, there is nothing travelling at c'. You're mixing up what one observer sees with what the other on sees. The observer with the clock sees light travelling along the lenght L. The other observer does not and this is where you define the c'.
Roger: Yes, I suppose you could look on it as nothing travelling at c', it is a projection of c along the vertical part of the triangle as seen by the person in the O' frame. It is not me mixing this c' up with c it is the usual maths presented in SR texts doing that; they mix up the length that c' acts on with the length that c acts on, and that is what I am saying they do wrong.

cinci: Why would they want to deal witht the porjection of a velocity? Why do you?

Roger: Its one method available.

cinci: A method for what?
*************************************


cinci: Yes they each may consider themselves stationary, but I picked one situation here to analyze and the one I picked has the clock moving. Throwing in this comment is just to cause unnecessary confusion.

Roger: No it wasn't. It was about the relativity of observers, and it was important that you should be aware of it. Your way of talking as if there was only one stationary observer was implying you did not know relativity.

cinci: For any given analysis, there IS only one stationary observer. That doesn't mean I can't analyze it making the other assumption, but trying to deal with them at the same time is bound to be confusing.

Roger: Both observers have their own stationary frame.

cinci: You can't analyze it that way. If one observer is considered statioanry the other must be moving.
********************************


cinci: Yes, they do. Usually, people set this up with an observer perhaps standing by a railroad track and call him stationary. The light clock would be going by on a railroad car and would be moving relative to the statioanry observer.

Roger: But that would be bad terminology, because in SR both observers believe themselves stationary.

cinci: It doesn't make any difference what the observers think, they aren't doing this analysis, YOU are. YOU are analyzing the problem and YOU get to pick which one YOU consider stationary. So the terminology is perfectly valid.

Roger: No you don't pick who is stationary, according to relativity they both are. Relativity is part of the analysis.

cinci: No, you can't say the both ARE. You may consider either stationary but when you have done that, the other is moving. Observers can't be moving and stationary at the same time.

Roger: They can by relativity theory.

cinci: No, either one or the other can be considered stationary. If you think relativity says otherwise then look in Alberts 1905 paper and show me where:

http://www.phys.lsu.edu/mog/100/elecmovbodeng.pdf
**************************************

cinci: What the theory says is that you can't perform an experiment that proves you are moving.

Roger: Which means by relativity they are both treated as stationary.

cinci: No, it means that you can consider either stationary. It does not mean you can consider both stationary at the same time....

Roger: They can be stationary same time if they are stationary with respect to each other.

cinci: In that case there is no relative motion and relativity doesn't apply. You have thrwon in another red herring for the sake of confusion.
*********************

cinci: Here is the proof of your confusion. First you say this observer sees the light travel along the hypotenuse (at c I assume, you don't say) and then you say "The velocity....along the vertical is c'"

Roger: Yes

cinci: The velocity of what....you just said the light was along the hypotenuse. what's going along the vertical?

Roger: projection of c

cinci: You could draw a 45 degree angle line and take the projection along it as well. Why don't you do that?

Roger: I am doing for the c case at the moment.

cinci: That would be a case for c also and your c' would be a projection along that line. This is nonsense. Ther is no velocity c' and there is no reason to measure the projection of c.
******************************

cinci: I used trains to try to clariy your confusion about statioanry and moving observers......apparently it didn't work.

Roger: I think you are confused by it.

cinci: You are wrong.

Roger: No its you.

cinci: I guess you're out of real arguments. Of course you have been all along.
****************************

Roger: Some texts miss out using c' and use other letters.

cinci: The texts don't use c' because that would indicate a velocity and nothing is moving along that leg. They use L or some other letter implying a distance.

They do use c', but can use the length L' instead.

So then-

...... B
......./|
....../.|
...../..|... L'
..../___|
...A.....

horizontal = 1/2 v delta t'
hyp = 1/2 c delta t'


Standard SR maths equates L to L' where L is 1/2 c delta t

And that is wrong.

You say : "... nothing is moving along that leg" --- that is the leg L'

And that is precisely my point-- In O' frame the light moves from mirror A to mirror B along the hyp of the triangle.

In the O frame the light moves from mirror A to mirror B along L.

The light does not move along L'.

So the SR texts stating L = L' are in error.

Light does not travel along L'

cinci: The texts don't say L=L'. They say that delta(t) is the time ths stationary observer measures for the light to traverse L and that delta(t') is the time that the moving observer measures for the light to traverse L'.

Roger: Some texts say L = L'

cinci: If you have a text that shows the leg of right triangle as L' and the hypotenuse of the same triangle as L, then throw it away.

Bottom line here is that I showed your analysis is wrong and you're just stalling around now hoping I'll drop it. OK, now that I've shown the error of what you've done and you are simply in denial, I will drop the subject out of pity. Have a nice day.
**********************

Roger, all Cincirob is trying to say, is that the observer stationed on the moving train can consider himself to be at rest, while the observer who is stationary is assumed to be moving- in reference of the train observer.

That's fair enough, but at the same time cincirob if saying that, is trying to insult me.

cinci: Roger, there is nothing travelling at c'. You're mixing up what one observer sees with what the other on sees. The observer with the clock sees light travelling along the lenght L. The other observer does not and this is where you define the c'.
Roger: Yes, I suppose you could look on it as nothing travelling at c', it is a projection of c along the vertical part of the triangle as seen by the person in the O' frame. It is not me mixing this c' up with c it is the usual maths presented in SR texts doing that; they mix up the length that c' acts on with the length that c acts on, and that is what I am saying they do wrong.
cinci: Why would they want to deal witht the porjection of a velocity? Why do you?
Roger: Its one method available.
cinci: A method for what?

Doing the maths.



*************************************


cinci: Yes they each may consider themselves stationary, but I picked one situation here to analyze and the one I picked has the clock moving. Throwing in this comment is just to cause unnecessary confusion.
Roger: No it wasn't. It was about the relativity of observers, and it was important that you should be aware of it. Your way of talking as if there was only one stationary observer was implying you did not know relativity.
cinci: For any given analysis, there IS only one stationary observer. That doesn't mean I can't analyze it making the other assumption, but trying to deal with them at the same time is bound to be confusing.
Roger: Both observers have their own stationary frame.
cinci: You can't analyze it that way. If one observer is considered statioanry the other must be moving.

You can; you can analyse from each stationary frame in turn.



********************************


cinci: Yes, they do. Usually, people set this up with an observer perhaps standing by a railroad track and call him stationary. The light clock would be going by on a railroad car and would be moving relative to the statioanry observer.
Roger: But that would be bad terminology, because in SR both observers believe themselves stationary.
cinci: It doesn't make any difference what the observers think, they aren't doing this analysis, YOU are. YOU are analyzing the problem and YOU get to pick which one YOU consider stationary. So the terminology is perfectly valid.
Roger: No you don't pick who is stationary, according to relativity they both are. Relativity is part of the analysis.
cinci: No, you can't say the both ARE. You may consider either stationary but when you have done that, the other is moving. Observers can't be moving and stationary at the same time.
Roger: They can by relativity theory.
cinci: No, either one or the other can be considered stationary. If you think relativity says otherwise then look in Alberts 1905 paper and show me where:

http://www.phys.lsu.edu/mog/100/elecmovbodeng.pdf

Einstein says: "We will raise
this conjecture (the purport of which will hereafter be called the
``Principle of Relativity'') to the status of a postulate,"

**************************************

cinci: What the theory says is that you can't perform an experiment that proves you are moving.
Roger: Which means by relativity they are both treated as stationary.
cinci: No, it means that you can consider either stationary. It does not mean you can consider both stationary at the same time....
Roger: They can be stationary same time if they are stationary with respect to each other.
cinci: In that case there is no relative motion and relativity doesn't apply. You have thrwon in another red herring for the sake of confusion.

I am telling you relativity applies, hence relative motion; you are confused.


*********************

cinci: Here is the proof of your confusion. First you say this observer sees the light travel along the hypotenuse (at c I assume, you don't say) and then you say "The velocity....along the vertical is c'"
Roger: Yes
cinci: The velocity of what....you just said the light was along the hypotenuse. what's going along the vertical?
Roger: projection of c
cinci: You could draw a 45 degree angle line and take the projection along it as well. Why don't you do that?
Roger: I am doing for the c case at the moment.

cinci: That would be a case for c also and your c' would be a projection along that line. This is nonsense. Ther is no velocity c' and there is no reason to measure the projection of c.

No I am forming a triangle within the radius of an expanding circle where radius = c, and c' the side of a triangle smaller than radius.



******************************

cinci: I used trains to try to clariy your confusion about statioanry and moving observers......apparently it didn't work.
Roger: I think you are confused by it.
cinci: You are wrong.
Roger: No its you.
cinci: I guess you're out of real arguments. Of course you have been all along.

No you are out of real arguments.

****************************

Roger: Some texts miss out using c' and use other letters.
cinci: The texts don't use c' because that would indicate a velocity and nothing is moving along that leg. They use L or some other letter implying a distance.
Roger:They do use c', but can use the length L' instead.

So then-

...... B
......./|
....../.|
...../..|... L'
..../___|
...A.....

horizontal = 1/2 v delta t'
hyp = 1/2 c delta t'


Standard SR maths equates L to L' where L is 1/2 c delta t

And that is wrong.

You say : "... nothing is moving along that leg" --- that is the leg L'

And that is precisely my point-- In O' frame the light moves from mirror A to mirror B along the hyp of the triangle.

In the O frame the light moves from mirror A to mirror B along L.

The light does not move along L'.

So the SR texts stating L = L' are in error.

Light does not travel along L'

cinci: The texts don't say L=L'. They say that delta(t) is the time ths stationary observer measures for the light to traverse L and that delta(t') is the time that the moving observer measures for the light to traverse L'.

Roger: Some texts say L = L'

cinci: If you have a text that shows the leg of right triangle as L' and the hypotenuse of the same triangle as L, then throw it away.

cinci:Bottom line here is that I showed your analysis is wrong and you're just stalling around now hoping I'll drop it. OK, now that I've shown the error of what you've done and you are simply in denial, I will drop the subject out of pity. Have a nice day.

You haven't shown anything, other than go into denial about relativity.





**********************

cinci: If you have a text that shows the leg of right triangle as L' and the hypotenuse of the same triangle as L, then throw it away.

A lot of relativity texts have one error or another, if we start throwing them away it makes a very large pile for a bonfire.

cinci: If you have a text that shows the leg of right triangle as L' and the hypotenuse of the same triangle as L, then throw it away.

Roger: A lot of relativity texts have one error or another, if we start throwing them away it makes a very large pile for a bonfire.

cinci: If you don't understand that saying the leg of a right triangle equals the hypotenuse of the triangle violates a theorem of geometry as old as Euclid, then stop saying you know enough to say relativity is wrong.
************************

cinci: If you don't understand that saying the leg of a right triangle equals the hypotenuse of the triangle violates a theorem of geometry as old as Euclid,

Geometry has now moved onto having the option of being NonEuclidean; presumably you are thousands of years out of date as regards maths.

cinci: If you don't understand that saying the leg of a right triangle equals the hypotenuse of the triangle violates a theorem of geometry as old as Euclid,

Roger: Geometry has now moved onto having the option of being NonEuclidean; presumably you are thousands of years out of date as regards maths.

cinci: Yes, there is that option, but your problem doesn't take that option. You've simply made a mistake and noneuclidiean geometry won't fix it.
*****************************

cinci: If you don't understand that saying the leg of a right triangle equals the hypotenuse of the triangle violates a theorem of geometry as old as Euclid,
Roger: Geometry has now moved onto having the option of being NonEuclidean; presumably you are thousands of years out of date as regards maths.
cinci: Yes, there is that option,

Yes

cinci: but your problem doesn't take that option.

don't need to.

cinci:You've simply made a mistake and noneuclidiean geometry won't fix it.

You are deluded, you invent this "fictious claim" of "the leg of a right triangle equals the hypotenuse of the triangle" amd pretend I made it, then waffle on about it.






*****************************

cinci: If you don't understand that saying the leg of a right triangle equals the hypotenuse of the triangle violates a theorem of geometry as old as Euclid.

Roger: Geometry has now moved onto having the option of being NonEuclidean; presumably you are thousands of years out of date as regards maths.

cinci: Yes, there is that option,

Roger: Yes

cinci: but your problem doesn't take that option.

Roger: don't need to.

cinci: You've simply made a mistake and noneuclidiean geometry won't fix it.

Roger: You are deluded, you invent this "fictious claim" of "the leg of a right triangle equals the hypotenuse of the triangle" amd pretend I made it, then waffle on about it.

cinci: Aw Roger, you even said you knew of text books that say L=L'. Are you now trying to say you were defending me? Give us a break.


Bottom line is your analysis is wrong and I've shown you why...why don't you just move on.
**************************

cinci: If you don't understand that saying the leg of a right triangle equals the hypotenuse of the triangle violates a theorem of geometry as old as Euclid.

[]

Roger: You are deluded, you invent this "fictious claim" of "the leg of a right triangle equals the hypotenuse of the triangle" amd pretend I made it, then waffle on about it.

cinci: Aw Roger, you even said you knew of text books that say L=L'. Are you now trying to say you were defending me? Give us a break.


___
.|
.|
.|...L
_|_


........./.|
......../..|...L'
......./...|
....../____|


nothing about legs of triangle

Roger, Cincirob cannot fathom the contradiction. He says "the observers aren't making the analysis, you are." He can't imagine that both observers are imagining themselves to be at rest at the same time, or imagining themselves both to be moving in relation to the other, at the same time. The Gedankin does not have to run separately for each observer. Each sees the other's clock running slower. No way around it.

Curt: Roger, Cincirob cannot fathom the contradiction. He says "the observers aren't making the analysis, you are." He can't imagine that both observers are imagining themselves to be at rest at the same time, or imagining themselves both to be moving in relation to the other, at the same time. The Gedankin does not have to run separately for each observer. Each sees the other's clock running slower. No way around it.


cinci: I didn't say each observer couldn't imagine themselves stationary at the same time, I said you can't do the analysis that way. For a guy who can't figure out how each can see the other's clock running slow you very easily accept that two observers moving relative to each other can be stationary at the same time.

And you are wrong as far as relativity is concerned, the Gedanken does have to run separately for each observer. There is no mathematics in the theory for the case where both realtively moving observers are considered stationary since that situation is a contradiction. If you think such math exists, then show it.
*************************

Curt: Roger, Cincirob cannot fathom the contradiction. He says "the observers aren't making the analysis, you are." He can't imagine that both observers are imagining themselves to be at rest at the same time, or imagining themselves both to be moving in relation to the other, at the same time. The Gedankin does not have to run separately for each observer. Each sees the other's clock running slower. No way around it.

Hi Curt

Yeah Cinci seems another one those people that can't understand relativity, and then throws up masses of confusion on everything.

When Cinci says "I didn't say each observer couldn't imagine themselves stationary at the same time, I said you can't do the analysis that way." -- He's making contradictory statements; so when we try to understand what he's saying he will contradict himself. Somehow to such people saying contradictory statements makes sense to them; is he what is called an "einstein zombie"?

I think I got around the time dilation issue, instead of Lorentz factor on t and x (x the direction of motion), its on y and z instead.

Roger

Roger: Yeah Cinci seems another one those people that can't understand relativity, and then throws up masses of confusion on everything.

cinci: So what you're saying Roger, is that it's only people who think relativity is wrong, like you, that understand it.
**************************

Roger: When Cinci says "I didn't say each observer couldn't imagine themselves stationary at the same time, I said you can't do the analysis that way." -- He's making contradictory statements; so when we try to understand what he's saying he will contradict himself. Somehow to such people saying contradictory statements makes sense to them; is he what is called an "einstein zombie"?

cinci: Well gee Roger, now all you have to do is show me the analysis where both observers are considered stationary at the same time. You don't havce to work it all out, just show me the equations you're going to use. You can do that, can't you? Or are you just blowing smoke?
***************************

Roger: I think I got around the time dilation issue, instead of Lorentz factor on t and x (x the direction of motion), its on y and z instead.

cinci: Got around it? You mean you just woke up to the fact that time dilation is the same throughout the whole x'y'z' frame.....that shouldn't be news to someone who thinks he knows more about relativity than me.
**********************

Roger: Yeah Cinci seems another one those people that can't understand relativity, and then throws up masses of confusion on everything.
cinci: So what you're saying Roger, is that it's only people who think relativity is wrong, like you, that understand it.
.
Well you are wrong, because I am keeping relativity; just saying the equations from it are derived wrong.

**************************

Roger: When Cinci says "I didn't say each observer couldn't imagine themselves stationary at the same time, I said you can't do the analysis that way." -- He's making contradictory statements; so when we try to understand what he's saying he will contradict himself. Somehow to such people saying contradictory statements makes sense to them; is he what is called an "einstein zombie"?
cinci: Well gee Roger, now all you have to do is show me the analysis where both observers are considered stationary at the same time.
.
It is not in the ability for zombies to understand such things; relativity is about relative motion, hence each inertial observer treats themselves as stationary.


***************************

Roger: I think I got around the time dilation issue, instead of Lorentz factor on t and x (x the direction of motion), its on y and z instead.
cinci: Got around it? You mean you just woke up to the fact that time dilation is the same throughout the whole x'y'z' frame.....that shouldn't be news to someone who thinks he knows more about relativity than me.
.
It is to me, because I didn't know the zombies had been at that specific equation messing it up.
**********************

Roger: Yeah Cinci seems another one those people that can't understand relativity, and then throws up masses of confusion on everything.

cinci: So what you're saying Roger, is that it's only people who think relativity is wrong, like you, that understand it.
.
Roger: Well you are wrong, because I am keeping relativity; just saying the equations from it are derived wrong.

cinci: ROTFLMAO!
**************************

Roger: When Cinci says "I didn't say each observer couldn't imagine themselves stationary at the same time, I said you can't do the analysis that way." -- He's making contradictory statements; so when we try to understand what he's saying he will contradict himself. Somehow to such people saying contradictory statements makes sense to them; is he what is called an "einstein zombie"?

cinci: Well gee Roger, now all you have to do is show me the analysis where both observers are considered stationary at the same time.
.
Roger: It is not in the ability for zombies to understand such things; relativity is about relative motion, hence each inertial observer treats themselves as stationary.

cinci: Yes they do, but I notice you dropped the phrase "at the same time".
***************************

Roger: I think I got around the time dilation issue, instead of Lorentz factor on t and x (x the direction of motion), its on y and z instead.

cinci: Got around it? You mean you just woke up to the fact that time dilation is the same throughout the whole x'y'z' frame.....that shouldn't be news to someone who thinks he knows more about relativity than me.
.
Roger: It is to me, because I didn't know the zombies had been at that specific equation messing it up.

cinci: I see you are adopting Pentcho's habit of calling relativsts zombies. If you can't be any more creative than Pentcho, give it up....you lost the argument.
**********************

Roger: When Cinci says "I didn't say each observer couldn't imagine themselves stationary at the same time, I said you can't do the analysis that way." -- He's making contradictory statements; so when we try to understand what he's saying he will contradict himself. Somehow to such people saying contradictory statements makes sense to them; is he what is called an "einstein zombie"?

cinci: Well gee Roger, now all you have to do is show me the analysis where both observers are considered stationary at the same time.
.
Roger: It is not in the ability for zombies to understand such things; relativity is about relative motion, hence each inertial observer treats themselves as stationary.

cinci: Yes they do, but I notice you dropped the phrase "at the same time".
.
Roger: Do you want that phrase back?
***************************



cinci: I see you are adopting Pentcho's habit of calling relativsts zombies.

Poor zombies; they used to be human once and able to think.

Roger: When Cinci says "I didn't say each observer couldn't imagine themselves stationary at the same time, I said you can't do the analysis that way." -- He's making contradictory statements; so when we try to understand what he's saying he will contradict himself. Somehow to such people saying contradictory statements makes sense to them; is he what is called an "einstein zombie"?

cinci: Well gee Roger, now all you have to do is show me the analysis where both observers are considered stationary at the same time.
.
Roger: It is not in the ability for zombies to understand such things; relativity is about relative motion, hence each inertial observer treats themselves as stationary.

cinci: Yes they do, but I notice you dropped the phrase "at the same time".
.
Roger: Do you want that phrase back?

cinci: It's your phrase. Why did you drop it?
***************************

cinci: I see you are adopting Pentcho's habit of calling relativsts zombies.

Roger: Poor zombies; they used to be human once and able to think.

cinci: Roger, there aren't any real zombies, just like there aren't any real theories that can replace relativity. You need to stop believing in all these fantasies.
**************************************

cinci: Roger, while its easy to just keep on making you look silly, it is tiresome. So Im going to explain the so-called twin paradox to you with analysis ala relativity to see it that convinces you.

The main problem that antirelativists have with this problem is that they think only two clocks are involved in the analysis; there are at least three and if you analyze what both observers see, then there are four. Here is the Lorentz time transformation:

(1) t = (t vx/cc)/(1 (v/c)^2)^.5

To simplify the algebra, lets call the twins A and B and let A and B be the time on their respective clocks. Thats two clocks, one located with each observer and the equation above becomes.

(2) B = (A vx/cc)/(1 (v/c)^2)^.5

Now lets see where what the other two clocks are. A (or t) represents not just the time at the location of its observer. It is the time for all points in the frame of its observer; that is, at every point in that frame you can assume there is a clock that is synchronous with the clock located with the observer. B (or t) in the equation is the time on the clock Bs frame located at x in the frame of A. This can be shown by noting that the Lorentz time transformations is structured so that x=0 at t = t = 0. This is made clear in Einsteins 1905 paper.
So the location of B is always

(3) x = vt.

Substituting this into (2) yields

(4) B = A (1 (v/c)^2)^.5

Which is the time dilation formula.

What you get if you use a different x is the time on a clock in the frame of B that is synchronized with B. So, there is a clock in Bs frame located adjacent to clock A and A is at x = 0. Lets call that clock B(at A).

Obviously B(at A) will not equal B in general because clocks that are synchronized in one frame will not be synchronized when observed from a relatively moving frame. Notice, by this method of defining clocks, the A equation (4) becomes A(at B). So lets rewrite it.

(5) B = A(at B)( 1 (v/c)^2)^.5

This is a lttle trivial because that equation is based on the clock As frame where all As clock read the same.

Now lets write the formula for B(at A).

(6) B(at A) = (A v*0/cc)/( 1 (v/c)^2)^.5 = A/(1 (v/c)^2)^.5

Or,

(7) A = B(at A)( 1 (v/c)^2)^.5

Now we have our four clocks: A, B, A(at B), and B(at A)
Comparing equations (5) and (7), we see that from (5)

a. B is time dilated relative to A(at B)

And from (7),

b. A is time dilated relative to B (at A).

So, symmetry is preserved and your statement that each clock runs slower than the other is a misunderstanding of simultaneity and that fact that you are not comparing A and B when you say it. You are comparing B with a clock synchronized with A but separated from A by the distance vt, A (at B).
*********************************

Roger: It is not in the ability for zombies to understand such things; relativity is about relative motion, hence each inertial observer treats themselves as stationary.

cinci: Yes they do, but I notice you dropped the phrase "at the same time".
.
Roger: Do you want that phrase back?

cinci: It's your phrase. Why did you drop it?
.
What repeat myself like a zombie; you must be happier talking to zombies.


**************************************

cinci: Roger, while its easy to just keep on making you look silly, it is tiresome. So Im going to explain the so-called twin paradox to you with analysis ala relativity to see it that convinces you.

Well its tiresome. I can point out that you are wrong in such an explanation, but you are just going to ignore that, and carry on in a zombielike trance.

cinci:The main problem that antirelativists have with this problem is that they think only two clocks are involved in the analysis; there are at least three and if you analyze what both observers see, then there are four.

Jeez already you deviate from standard relativity texts. You should be telling the writers of those relativity texts that they got it wrong, not me.


cinci: Here is the Lorentz time transformation:
(1) t = (t vx/cc)/(1 (v/c)^2)^.5

If you read any of my postings, you will find that I dispute that already. I think the Lorentz transforms were derived wrong.

Since I disagree with that, I disagree with everything derived using it.

Show me how you derive the Lorentz transforms and I will say its all based on maths mistakes.

Roger: Show me how you derive the Lorentz transforms and I will say its all based on maths mistakes.

cinci: Well anybody could SAY it's all based on maths mistakes. If you really know anything about it, you could point out the mistakes. I think you're whistling past the graveyard hoping nobody calls your bluff. Here's my derivation, take your best shot.


Einstein postulates that c is the same for all reference frames. I can define one inertial frame as x',y',z',t' and assume this frame's axes are all parallel to their counterparts in x,y,z, and that x and x are coincident. For the same expanding sphere of light in each frame of reference, an inertial observer in each of these frames should see light from a point source expand as a spherical wave front. Setting t=0 when t'=0, I can write:

(1) x^2 + y^2 + z^2 = c^2t^2

and

(2) x'^2 + y'^2 + z'^2 = c^2t'^2 .

Assume the transformations from inertial coordinate system x,y,z,t moving a v relative to coordinate system x,y,z,t where x ans x are coincident, y is parallel to y, z is parallel to z, and x=x=0 at t=t=0 are of the form:

(3) x = A(x-vt)
(4) y = y
(5) z = z
(6) t = D(t-Bx).

Substituting (3), (4), (5), and (6) into (2) yields

(8) (A(x-vt))^2 + y^2 + z^2 = cD(t-Bx))^2

(9) A^2x^2 2Avxt + A^2v^2t^2 + y^2 + z^2 = c^2D^2t^2 2c^2D^2Bxt + c^2D^2B2^x^2

(10) x^2(A^2c^2D^2B^2) + y^2 + z^2 = (c^2t^2)(D^2-A^2v^2/c^2) + (2A^2v2c^2D^2B)xt

Because x,y,z,t and xt are independent variables, comparing (10) with (1) and setting the coefficients of like terms equal yields

(11) A^2 c^2D^2B^2 = 1

(12) D^2 A^2v^2/c^2 = 1

(13) 2A^2v 2c^2D2^2B = 0

These equations yield

(14) A = D = 1/(1-(v/c)^2)1/2

(15) B = v/c^2

Therefore
(16) x = (x-vt)/(1-(v/c)^2)1/2

(17) y = y

(18) z = z

(19) t = (t vx/c^2) / (1-(v/c)^2)1/2
*****************************

cinci: Assume the transformations from inertial coordinate system x,y,z,t moving a v relative to coordinate system x,y,z,t where x ans x are coincident, y is parallel to y, z is parallel to z, and x=x=0 at t=t=0 are of the form:

(3) x = A(x-vt)
(4) y = y
(5) z = z
(6) t = D(t-Bx).

roger: The dashes are missed out; the computer editor of this forum sometimes deletes them

Putting the dashes in:

cinci: Assume the transformations from inertial coordinate system x,y,z,t moving a v relative to coordinate system x,y,z,t where x ans x are coincident, y is parallel to y, z is parallel to z, and x=x=0 at t=t=0 are of the form:

(3) x = A(x'-vt')
(4) y = y
(5) z = z
(6) t = D(t'-Bx').

roger: Presume that was what was meant.

instead of that scenario, consider the following scenario:
x = x vt
y= Ey
z=Ez
t= t

then:

x^2 + v^2 t^2 + E^2y^2 + E^2z^2 c^2t^2 = 0 (a)

and
x^2 +y^2 + z^2 c^2t^2 = 0 (b)

from (b) gives us:

y^2 + z^2 = -(x^2 c^2t^2) (c)

(a) gives
x^2 + (v^2 c^2) t^2 + E^2 (y^2 +z^2) = 0

-y^2 - z^2 +v^2t^2 + E^2 (y^2 +z^2) =0

(E^2 -1) (y^2 +z^2) = - v^2t^2



take (y^2 +z^2) = r^2 = c^2 t^2

then

(E^2 1)c^2 = - v^2


(E^2 1) = - v^2/c^2


E^2 = -v^2/c^2 +1

E = sqrt(1 - v^2/c^2)


i.e.

x = x vt
y= sqrt(1 -v^2/c^2) y
z= sqrt(1 -v^2/c^2) z
t= t

is a solution!

Since that is a possibility and the setup cinci gave as:


(3) x = A(x'-vt')
(4) y = y
(5) z = z
(6) t = D(t'-Bx').

does not consider all possibilities for solutions, it is therefore in error by NOT considering all possibilities.

Putting the dashes in:

cinci: Assume the transformations from inertial coordinate system x,y,z,t moving a v relative to coordinate system x,y,z,t where x ans x are coincident, y is parallel to y, z is parallel to z, and x=x=0 at t=t=0 are of the form:

(3) x = A(x'-vt')
(4) y = y
(5) z = z
(6) t = D(t'-Bx').

roger: Presume that was what was meant.

instead of that scenario, consider the following scenario:
x = x' vt'
y= Ey'
z=Ez'
t= t'

then:

x'^2 + v^2 t'^2 + E^2y'^2 + E^2z'^2 c^2t'^2 = 0 (a)

and
x^2 +y^2 + z^2 c^2t^2 = 0 (b)

from (b) gives us:

y^2 + z^2 = -(x^2 c^2t^2) (c)

(a) gives
x'^2 + (v^2 c^2) t'^2 + E^2 (y'^2 +z'^2) = 0

-y^2 - z^2 +v^2t^2 + E^2 (y'^2 +z'^2) =0

(E^2 -1) (y'^2 +z^2) = - v^2t^2



take (y'^2 +z'^2) = r'^2 = c^2 t^2

then

(E^2 1)c^2 = - v^2


(E^2 1) = - v^2/c^2


E^2 = -v^2/c^2 +1

E = sqrt(1 - v^2/c^2)


i.e.

x = x' vt'
y= sqrt(1 -v^2/c^2) y'
z= sqrt(1 -v^2/c^2) z'
t= t'

is a solution!

Since that is a possibility and the setup cinci gave as:


(3) x = A(x'-vt')
(4) y = y
(5) z = z
(6) t = D(t'-Bx').

does not consider all possibilities for solutions, it is therefore in error by NOT considering all possibilities.

cinci: Assume the transformations from inertial coordinate system x,y,z,t moving a v relative to coordinate system x,y,z,t where x ans x are coincident, y is parallel to y, z is parallel to z, and x=x=0 at t=t=0 are of the form:

(3) x = A(x-vt)
(4) y = y
(5) z = z
(6) t = D(t-Bx).

roger: The dashes are missed out; the computer editor of this forum sometimes deletes them

cinci: Do you mean the primes? That m,ay have been my mistake, these equaiotns should be:

(3) x' = A(x-vt)
(4) y' = y
(5) z' = z
(6) t' = D(t-Bx).

I'll have to post it to see if it works, but the left side of the equations should be primed.
******************************

Putting the dashes in:

cinci: Assume the transformations from inertial coordinate system x,y,z,t moving a v relative to coordinate system x,y,z,t where x ans x are coincident, y is parallel to y, z is parallel to z, and x=x=0 at t=t=0 are of the form:

(3) x = A(x'-vt')
(4) y = y
(5) z = z
(6) t = D(t'-Bx').


*******************
cinci: No, it should be

(3) x' = A(x-vt)
(4) y' = y
(5) z' = z
(6) t' = D(t-Bx).

You can tell because I substituted (3)-(6) into (2).
**************************

roger: Presume that was what was meant.

instead of that scenario, consider the following scenario:
x = x' vt'
y= Ey'
z=Ez'
t= t'

then:

x'^2 + v^2 t'^2 + E^2y'^2 + E^2z'^2 c^2t'^2 = 0 (a)

and
x^2 +y^2 + z^2 c^2t^2 = 0 (b)

from (b) gives us:

y^2 + z^2 = -(x^2 c^2t^2) (c)

(a) gives
x'^2 + (v^2 c^2) t'^2 + E^2 (y'^2 +z'^2) = 0

-y^2 - z^2 +v^2t^2 + E^2 (y'^2 +z'^2) =0

(E^2 -1) (y'^2 +z^2) = - v^2t^2



take (y'^2 +z'^2) = r'^2 = c^2 t^2

then

(E^2 1)c^2 = - v^2


(E^2 1) = - v^2/c^2


E^2 = -v^2/c^2 +1

E = sqrt(1 - v^2/c^2)


i.e.

x = x' vt'
y= sqrt(1 -v^2/c^2) y'
z= sqrt(1 -v^2/c^2) z'
t= t'

is a solution!

Since that is a possibility and the setup cinci gave as:

(3) x = A(x'-vt')
(4) y = y
(5) z = z
(6) t = D(t'-Bx').

does not consider all possibilities for solutions, it is therefore in error by NOT considering all possibilities.


cinc: There is no rule in mathematics that says you have to consider an infinite number of possibilities. This is an algebraic derivation and will only be correct if you make your initial assumptions sufficiently general. So it's OK to try yours, but after you try mine you know yours is wrong. If your solution were correct, the mine would have produced

D = 1 and B = 0.

If you introduce any more undefined constants that I did you cannot achieve a solution.

But the real reason that we know mine is correct is from Einstein's 1905 paper. In that derivation he derived the equations from a differential equation and that equation determines the form of the solution mathematically and does not permit other forms. Your assumption above does not follow that form, mine does.
***********************************

cinc: There is no rule in mathematics that says you have to consider an infinite number of possibilities.

When modelling the general case should be considered.

add on:

Einstein's multiple unequal definitions of the symbol x' in his equation (2) invalidate that equation. Since the remainder of his analysis is predicated on equation (2), his derivation of his Transformation of Coordinates and Times is meaningless. http://philica.com/display_article.php?article_id=16

Roger: add on:

Einstein's multiple unequal definitions of the symbol x' in his equation (2) invalidate that equation. Since the remainder of his analysis is predicated on equation (2), his derivation of his Transformation of Coordinates and Times is meaningless. http://philica.com/display_article.php?article_id=16

cinci: (2) x'^2 + y'^2 + z'^2 = c^2t'^2 is the equation of a sphere in the coordinates x',y',z',t'. It comes from Euclid. There aren't any "multiple unequal definitions of the symbol x'" whatever that means. You can't just make stuff up.

And http://philica.com/display_article.php?article_id=16 is written by RJ Hannon, whom I have been discussing relativity with for years. He doesn't have a clue.

The article above is not about the derivation I presented, it's about Einstein's 1905 paper. If you want to argue about RJ erroneous claims about that derivation, bring it on. I've heard 'em all and they're all wrong. And the x' in that derivation is a different parameter. In Einstein's paper the x',y',z',t' parameters are greek letters. You're just guessing now.

The thing you should note is that the model for my derivation is relatively moving observers viewing a single, expanding sphere of light. The 1905 paper's model is basically the leg of the Michelson-Morley experiment that is parallel to the direction of motion and despite these differences, the same equations arise.

You're clutching at straws now.
**********************

Einstein's multiple unequal definitions of the symbol x' in his equation (2) invalidate that equation. Since the remainder of his analysis is predicated on equation (2), his derivation of his Transformation of Coordinates and Times is meaningless. http://philica.com/display_article.php?article_id=16
cinci: (2) x'^2 + y'^2 + z'^2 = c^2t'^2 is the equation of a sphere in the coordinates x',y',z',t'. It comes from Euclid. There aren't any "multiple unequal definitions of the symbol x'" whatever that means. You can't just make stuff up.

What it means is Einstein does not use x' appropriately.

cinci:And http://philica.com/display_article.php?article_id=16 is written by RJ Hannon, whom I have been discussing relativity with for years. He doesn't have a clue.

So you are asking me who I should believe him or you. Anyway -I have pointed out the error in your derivation; whether you want to acceept that or not is no concern of mine.

Roger: Einstein's multiple unequal definitions of the symbol x' in his equation (2) invalidate that equation. Since the remainder of his analysis is predicated on equation (2), his derivation of his Transformation of Coordinates and Times is meaningless. http://philica.com/display_article.php?article_id=16

cinci: (2) x'^2 + y'^2 + z'^2 = c^2t'^2 is the equation of a sphere in the coordinates x',y',z',t'. It comes from Euclid. There aren't any "multiple unequal definitions of the symbol x'" whatever that means. You can't just make stuff up.

Roger: What it means is Einstein does not use x' appropriately.

cinci: No, Roger what it means is you've never looked at Einstein's 1905 paper. Most references today use x,y,z,t and x',y,z,t, because it's difficult to use Greek letters. In his 1905 paper Einstein used Greek letters for the primed variables. He used x' for a distance defined in the moving frame but measured in the stationary frame in that paper. Here I used it as a Cartesian coordinate. There is no mathematical error. Here is the paper:

http://www.phys.lsu.edu/mog/100/elecmovbodeng.pdf

As far as I know Einstein never did the derivation that I gave so you're quibbling about what I mnamed parameters, not about math errors. He did suggest this method on page 10 but he didn't use x', he used a Greek letter, as I have said.

Does my derivation change if I use a,b,c,t instead of x',y',z',t'?

Give it up Roger, there is no error.
*****************************

cinci:And http://philica.com/display_article.php?article_id=16 is written by RJ Hannon, whom I have been discussing relativity with for years. He doesn't have a clue.

Roger: So you are asking me who I should believe him or you. Anyway -I have pointed out the error in your derivation; whether you want to acceept that or not is no concern of mine.

cinci: No, you did not point out an error in my paper. You just directed me to a poorly thought out paper about a different derivation than mine. Even Bob S wouldn't buy your argument.

As far as who you should believe, you claim to know all about relativity. So why do you have to bring somebody else into it?

The only honest thing you've said in all this is: "Roger: Show me how you derive the Lorentz transforms and I will say its all based on maths mistakes." And you're still just "saying" it. You have found no error.
**************************

Re: Cincirob, Special Relativity is Galilean Relativity January 10 2009, 10:28 AM

"Even Bob S wouldn't buy your argument."!!!
Cinci, I have been staring at that statement for a good 1/2 hour and can't, for the life of me, think of a reason for you to say that. (bear with me here) For you to make that statement there can be two conclusions that I know of, 1.) That I would buy your argument or 2.) That I would not buy either argument. I could reject your statement and buy Roger's argument but that is an option and Im working with your assumption, not my choices, because I could also choose to ignore what you said but I can't do that because I am thoroughly intrigued as to why you would make such a statement and I do have an opinion.

I have been reading the exchanges between you and Roger for some time going back over different strings. So I'll tell you a story. Some years back, mid 70s, I was enjoying some beers at my local tavern and there were two guys having a rather heated discussion about Chevy v. Ford, blah blah blah. At one point (I knew both of them) one guy turned to me and said "you agree with me, right Bob?". Now you have to understand, I was not a party to their discussion, so I turned to them both and said, do you know that the firing sequence on a Ford 289 is the same as it is on a Chevy 327!

Roger: Einstein's multiple unequal definitions of the symbol x' in his equation (2) invalidate that equation. Since the remainder of his analysis is predicated on equation (2), his derivation of his Transformation of Coordinates and Times is meaningless. http://philica.com/display_article.php?article_id=16

cinci: (2) x'^2 + y'^2 + z'^2 = c^2t'^2 is the equation of a sphere in the coordinates x',y',z',t'. It comes from Euclid. There aren't any "multiple unequal definitions of the symbol x'" whatever that means. You can't just make stuff up.

Roger: What it means is Einstein does not use x' appropriately.

cinci: No, Roger what it means is you've never looked at Einstein's 1905 paper. Most references today use x,y,z,t and x',y,z,t, because it's difficult to use Greek letters. In his 1905 paper Einstein used Greek letters for the primed variables. He used x' for a distance defined in the moving frame but measured in the stationary frame in that paper. Here I used it as a Cartesian coordinate. There is no mathematical error. Here is the paper:

http://www.phys.lsu.edu/mog/100/elecmovbodeng.pdf

As far as I know Einstein never did the derivation that I gave so you're quibbling about what I mnamed parameters, not about math errors. He did suggest this method on page 10 but he didn't use x', he used a Greek letter, as I have said.

Does my derivation change if I use a,b,c,t instead of x',y',z',t'?

Give it up Roger, there is no error.

roger: I see what you are doing, you just want to muddle things up. I refer to an article saying about Einstein's 1905 paper being wrong, and you want to believe the 1905 paper so you make up some nonsense about the use of Greek letters which is non sequitur. Believe whatever nonsense you like, it is no concern of mine.





*****************************

cinci:And http://philica.com/display_article.php?article_id=16 is written by RJ Hannon, whom I have been discussing relativity with for years. He doesn't have a clue.

Roger: So you are asking me who I should believe him or you. Anyway -I have pointed out the error in your derivation; whether you want to acceept that or not is no concern of mine.

cinci: No, you did not point out an error in my paper. You just directed me to a poorly thought out paper about a different derivation than mine. Even Bob S wouldn't buy your argument.

Roger: Bob S does not appear concerned. I am not concerned. I pointed out your error; believe whatever you like.


**************************

Roger: Einstein's multiple unequal definitions of the symbol x' in his equation (2) invalidate that equation. Since the remainder of his analysis is predicated on equation (2), his derivation of his Transformation of Coordinates and Times is meaningless. http://philica.com/display_article.php?article_id=16

cinci: (2) x'^2 + y'^2 + z'^2 = c^2t'^2 is the equation of a sphere in the coordinates x',y',z',t'. It comes from Euclid. There aren't any "multiple unequal definitions of the symbol x'" whatever that means. You can't just make stuff up.

Roger: What it means is Einstein does not use x' appropriately.

cinci: No, Roger what it means is you've never looked at Einstein's 1905 paper. Most references today use x,y,z,t and x',y,z,t, because it's difficult to use Greek letters. In his 1905 paper Einstein used Greek letters for the primed variables. He used x' for a distance defined in the moving frame but measured in the stationary frame in that paper. Here I used it as a Cartesian coordinate. There is no mathematical error. Here is the paper:

http://www.phys.lsu.edu/mog/100/elecmovbodeng.pdf

As far as I know Einstein never did the derivation that I gave so you're quibbling about what I named parameters, not about math errors. He did suggest this method on page 10 but he didn't use x', he used a Greek letter, as I have said.

Does my derivation change if I use a,b,c,t instead of x',y',z',t'?

Give it up Roger, there is no error.

roger: I see what you are doing, you just want to muddle things up. I refer to an article saying about Einstein's 1905 paper being wrong, and you want to believe the 1905 paper so you make up some nonsense about the use of Greek letters which is non sequitur. Believe whatever nonsense you like, it is no concern of mine.

cinci: No Roger, you are the great muddler. I give you a very clear derivation of the Lorentz transformations and you try to tell me it's wrong because of a completely different derivation. And you didn't even look at the other derivation to find out that the x' in that derivation is different parameter than the one in mine.

I thought you were supposed to be the guy who could expose mathematical errors in analyses and you can't even tell two very distinct analyses apart. And apparently don't recognize what Greek letters are. And you think you're going to bring down relativity? I am ROTFLMAO.
*****************************

cinci:And http://philica.com/display_article.php?article_id=16 is written by RJ Hannon, whom I have been discussing relativity with for years. He doesn't have a clue.

Roger: So you are asking me who I should believe him or you. Anyway -I have pointed out the error in your derivation; whether you want to acceept that or not is no concern of mine.

cinci: No, you did not point out an error in my paper. You just directed me to a poorly thought out paper about a different derivation than mine. Even Bob S wouldn't buy your argument.

Roger: Bob S does not appear concerned. I am not concerned. I pointed out your error; believe whatever you like.

cinci: No, you didn't even point out an error in my derivation. You simply said something about Einstein's "multiple unequal definitions" any made no comment at all about my derivation.

Even AAF wouldn't stick to this story.
*************************************

Re: Special Relativity is Galilean Relativity January 10 2009, 4:58 PM

Roger: Einstein's multiple unequal definitions of the symbol x' in his equation (2) invalidate that equation. Since the remainder of his analysis is predicated on equation (2), his derivation of his Transformation of Coordinates and Times is meaningless. http://philica.com/display_article.php?article_id=16

cinci: (2) x'^2 + y'^2 + z'^2 = c^2t'^2 is the equation of a sphere in the coordinates x',y',z',t'. It comes from Euclid. There aren't any "multiple unequal definitions of the symbol x'" whatever that means. You can't just make stuff up.

Roger: What it means is Einstein does not use x' appropriately.

cinci: No, Roger what it means is you've never looked at Einstein's 1905 paper. Most references today use x,y,z,t and x',y,z,t, because it's difficult to use Greek letters. In his 1905 paper Einstein used Greek letters for the primed variables. He used x' for a distance defined in the moving frame but measured in the stationary frame in that paper. Here I used it as a Cartesian coordinate. There is no mathematical error. Here is the paper:

http://www.phys.lsu.edu/mog/100/elecmovbodeng.pdf

As far as I know Einstein never did the derivation that I gave so you're quibbling about what I named parameters, not about math errors. He did suggest this method on page 10 but he didn't use x', he used a Greek letter, as I have said.

Does my derivation change if I use a,b,c,t instead of x',y',z',t'?

Give it up Roger, there is no error.

roger: I see what you are doing, you just want to muddle things up. I refer to an article saying about Einstein's 1905 paper being wrong, and you want to believe the 1905 paper so you make up some nonsense about the use of Greek letters which is non sequitur. Believe whatever nonsense you like, it is no concern of mine.

cinci: No Roger, you are the great muddler. I give you a very clear derivation of the Lorentz transformations and you try to tell me it's wrong because of a completely different derivation.

Yes, because your derivation was not general enough to consider all relevant possibilities.

cinci:And you didn't even look at the other derivation to find out that the x' in that derivation is different parameter than the one in mine.

Not true, I worked through one possibility that you had not considered in your false derivation.

cinci:I thought you were supposed to be the guy who could expose mathematical errors in analyses

I can and I have done, but whether you are prepared to accept your error is not my concern.



*****************************



Roger: Bob S does not appear concerned. I am not concerned. I pointed out your error; believe whatever you like.

cinci: No, you didn't even point out an error in my derivation. You simply said something about Einstein's "multiple unequal definitions" any made no comment at all about my derivation.

I pointed out the error in your derivation, and you tried to invoke Einstein's 1805 paper for help, and I pointed out a paper then criticising Einstein's paper. Whether you accept that or not, is as I have pointed out not my concern.

*************************************

correction-

Einstein's 1905 paper not his 1805 paper; unless he was a time traveller LOL

Roger: correction-

Einstein's 1905 paper not his 1805 paper; unless he was a time traveller LOL.

cinci: Well at least you found an error. Now read the paper.
************************

cinci: Now read the paper.

Yes loads of mistakes as RJ Hannon has pointed out. For instance Einstein declares speed of light constant and in next breadth talks of light having speed sqrt(c^2 - v^2) along YZ. He just says one thing followed later by saying the opposite. i.e contradictory.

************************

Roger: Yes loads of mistakes as RJ Hannon has pointed out. For instance Einstein declares speed of light constant and in next breadth talks of light having speed sqrt(c^2 - v^2) along YZ. He just says one thing followed later by saying the opposite. i.e contradictory.

cinci: You're not reading Einstein's paper, you're reading Hannon's.

And yes he writes

(1) y/(c^2 - v^2)^.5 = t

Which is comes from the relationship

(2) y^2 = (ct)^2 - (vt)^2

or

(3) (ct)^2 = y^2 + (vt)^2.

This is what the stationary observer sees when a ray of light traverses the veritcal axis of the moving frame. It's the usual zigzag path you get in the light clock.

ct is the diagonal, vt is the distance the moving frame moves and, y is the vertical distance as seen bythe stationary observer.

Y and vt are the legs of a right triangle and ct is the hypotenuse. Equation (3) is simply the Pythagorean theorem. Einstein doesn't show all his steps, maybe because he was genius enough to see these answers without going through them. Years of discussion with Mr. Hannon has shown me that he is either unable or unwilling to figure any of this out. But since you already screwed up the light clock, you probably won't get it here either.

But it is not an error.
*********************

Roger: Yes loads of mistakes as RJ Hannon has pointed out. For instance Einstein declares speed of light constant and in next breadth talks of light having speed sqrt(c^2 - v^2) along YZ. He just says one thing followed later by saying the opposite. i.e contradictory.

cinci: You're not reading Einstein's paper, you're reading Hannon's.

And yes he writes

(1) y/(c^2 - v^2)^.5 = t

Which is comes from the relationship

(2) y^2 = (ct)^2 - (vt)^2

or

(3) (ct)^2 = y^2 + (vt)^2.

This is what the stationary observer sees when a ray of light traverses the veritcal axis of the moving frame. It's the usual zigzag path you get in the light clock.

ct is the diagonal, vt is the distance the moving frame moves and, y is the vertical distance as seen bythe stationary observer.

Y and vt are the legs of a right triangle and ct is the hypotenuse. Equation (3) is simply the Pythagorean theorem. Einstein doesn't show all his steps, maybe because he was genius enough to see these answers without going through them. Years of discussion with Mr. Hannon has shown me that he is either unable or unwilling to figure any of this out. But since you already screwed up the light clock, you probably won't get it here either.

But it is not an error.


roger: what I was referring to was- Einstein declares speed of light constant and in next breadth talks of light having speed sqrt(c^2 - v^2) along YZ.- none of your stuff about equation (1), (2), (3); you just go off on another diversion. Einstein makes two statements that contradict each other, that is an error. And it leads to ambiguity; since one statement is wrong we have to correct one of the statements, different people then pursue different paths in which statement they correct. This leads to different theories not just on that example but on the numerous such mistakes that Einstein makes.

Roger Rumberton: what I was referring to was- Einstein declares speed of light constant and in next breadth talks of light having speed sqrt(c^2 - v^2) along YZ.-

Ted: no, he doesn't. But you are too thoroughly stooopid to understand :lol:

Roger: what I was referring to was- Einstein declares speed of light constant and in next breadth talks of light having speed sqrt(c^2 - v^2) along YZ.-

Ted: no, he doesn't.

roger: yes he does.

Roger: Yes loads of mistakes as RJ Hannon has pointed out. For instance Einstein declares speed of light constant and in next breadth talks of light having speed sqrt(c^2 - v^2) along YZ. He just says one thing followed later by saying the opposite. i.e contradictory.

cinci: You're not reading Einstein's paper, you're reading Hannon's.

And yes he writes

(1) y/(c^2 - v^2)^.5 = t

Which is comes from the relationship

(2) y^2 = (ct)^2 - (vt)^2

or

(3) (ct)^2 = y^2 + (vt)^2.

This is what the stationary observer sees when a ray of light traverses the veritcal axis of the moving frame. It's the usual zigzag path you get in the light clock.

ct is the diagonal, vt is the distance the moving frame moves and, y is the vertical distance as seen bythe stationary observer.

Y and vt are the legs of a right triangle and ct is the hypotenuse. Equation (3) is simply the Pythagorean theorem. Einstein doesn't show all his steps, maybe because he was genius enough to see these answers without going through them. Years of discussion with Mr. Hannon has shown me that he is either unable or unwilling to figure any of this out. But since you already screwed up the light clock, you probably won't get it here either.

But it is not an error.


roger: what I was referring to was- Einstein declares speed of light constant and in next breadth talks of light having speed sqrt(c^2 - v^2) along YZ.- none of your stuff about equation (1), (2), (3); you just go off on another diversion. Einstein makes two statements that contradict each other, that is an error. And it leads to ambiguity; since one statement is wrong we have to correct one of the statements, different people then pursue different paths in which statement they correct. This leads to different theories not just on that example but on the numerous such mistakes that Einstein makes.


cinci: Are you just trying to be obtuse. Here he is talking about the projection of the velocity of the ray of light, not the ray of light itself. Equation (3) says that light is travelling at c, see the term ct. It says the farme is travelling at v, see the term vt. Equation (1) comes from equation (3). Equation (10 is just an algebraic manipulation of equation (3), it doesn't imply anything physical. Ther is no diversion here, just you trying to weasel out of your mistaken statement. Get over it Roger, you blew it. Einstein did not make a mistake.
*************************

roger: what I was referring to was- Einstein declares speed of light constant and in next breadth talks of light having speed sqrt(c^2 - v^2) along YZ.- none of your stuff about equation (1), (2), (3); you just go off on another diversion. Einstein makes two statements that contradict each other, that is an error. And it leads to ambiguity; since one statement is wrong we have to correct one of the statements, different people then pursue different paths in which statement they correct. This leads to different theories not just on that example but on the numerous such mistakes that Einstein makes.


cinci: Are you just trying to be obtuse. Here he is talking about the projection of the velocity of the ray of light, not the ray of light itself.

roger: I have no idea what you are talking about. Einstein says and I quote: "to the axes of Y and Z--it being borne in mind that light is always propagated along these axes, when viewed from the stationary system, with the velocity sqrt (c^2 - v^2). whatever you have decided to talk about bears no relation.

roger: what I was referring to was- Einstein declares speed of light constant and in next breadth talks of light having speed sqrt(c^2 - v^2) along YZ.- none of your stuff about equation (1), (2), (3); you just go off on another diversion. Einstein makes two statements that contradict each other, that is an error. And it leads to ambiguity; since one statement is wrong we have to correct one of the statements, different people then pursue different paths in which statement they correct. This leads to different theories not just on that example but on the numerous such mistakes that Einstein makes.


cinci: Are you just trying to be obtuse. Here he is talking about the projection of the velocity of the ray of light, not the ray of light itself.

roger: I have no idea what you are talking about.

cinci: Yes, I can see that.
*********************

Roger: Einstein says and I quote: "to the axes of Y and Z--it being borne in mind that light is always propagated along these axes, when viewed from the stationary system, with the velocity sqrt (c^2 - v^2). whatever you have decided to talk about bears no relation.

cinci: OK, let's do it baby style:

1. You have to read the paper to know he's talking about the axes in the moving frame when he says "these axes".

2. If the leading edge of a ray (and that's what he's describing) moves along a vertical line and taht line is moving, then the point that represnts that ray travels verically the distance y and horizontally the distance vt.

3. The ray travels at c and the distance the ray travels is the hypotenuse of the triangle formed by the horizontal distance vt and the vertical distance y,

So (ct)^2 + (vt)^2 + y^2 is the equation you write to find out what t is. If you don't understand that you're not good enough at math to find errors in Eisntein's work.

4. Solving that equation for t yields

y = t(c^2-v^2)^.5.

(distance) = (duration)(speed) so this equation has this form and (c^2-v^2)^.5 can be regarded of a speed.

5. To understand what (c^2-v^2)^.5 is, you have to pay attention to the part of Einstein's statement that says "when viewed from the stationary system". It's the velocity in the vertical direction of the ray, but from thepoint of view of the stationary observer, the ray doesn't go in the vertical direction, it goes along the hypotenuse as I described above. In this reagrd it is the projection of the velocity of light on the vertical axis.

6. If you read the rest of the paper, you find that in the moving frame he says Y = cT where Y and T are in Greek letters.

There is no error in all this. Further, none of this has anything to do with my analysis where there is also no error.

If youwant to go on ignoring good science and clinging to the nonsense of RJHannon, that's your choice. I am continually amazed by how many people choose nonsense over science.
*************************

Roger: Einstein says and I quote: "to the axes of Y and Z--it being borne in mind that light is always propagated along these axes, when viewed from the stationary system, with the velocity sqrt (c^2 - v^2)." whatever you have decided to talk about bears no relation.

cinci: OK, let's do it baby style:

1. You have to read the paper to know he's talking about the axes in the moving frame when he says "these axes".

2. If the leading edge of a ray (and that's what he's describing) moves along a vertical line and taht line is moving, then the point that represnts that ray travels verically the distance y and horizontally the distance vt.

3. The ray travels at c and the distance the ray travels is the hypotenuse of the triangle formed by the horizontal distance vt and the vertical distance y,

So (ct)^2 + (vt)^2 + y^2 is the equation you write to find out what t is. If you don't understand that you're not good enough at math to find errors in Eisntein's work.

4. Solving that equation for t yields

y = t(c^2-v^2)^.5.


roger: the equation you give of (ct)^2 + (vt)^2 + y^2 is probably(ct)^2 = (vt)^2 + y^2 ; you made a typo slip with "+" and "=". Anyway, still bears no relation with what Einstein said in his quote of lightspeed not being constant c.

Roger: Einstein says and I quote: "to the axes of Y and Z--it being borne in mind that light is always propagated along these axes, when viewed from the stationary system, with the velocity sqrt (c^2 - v^2)." whatever you have decided to talk about bears no relation.

cinci: OK, let's do it baby style:

1. You have to read the paper to know he's talking about the axes in the moving frame when he says "these axes".

2. If the leading edge of a ray (and that's what he's describing) moves along a vertical line and taht line is moving, then the point that represnts that ray travels verically the distance y and horizontally the distance vt.

3. The ray travels at c and the distance the ray travels is the hypotenuse of the triangle formed by the horizontal distance vt and the vertical distance y,

So (ct)^2 + (vt)^2 + y^2 is the equation you write to find out what t is. If you don't understand that you're not good enough at math to find errors in Eisntein's work.

4. Solving that equation for y yields

y = t(c^2-v^2)^.5.


roger: the equation you give of (ct)^2 = (vt)^2 + y^2 is probably(ct)^2 = (vt)^2 + y^2 ; you made a typo slip with "+" and "=". Anyway, still bears no relation with what Einstein said in his quote of lightspeed not being constant c.

cinci: Thanks for catching the typo.

As for the rest of your comment, you still don't realize that from the perspective of the stationary observer there is nothing travelling along y. So while the motion along the hypotenuse is ct, for someone moving with that axis it would look like it was moving at (c^2 - v^2)^.5 along the axis. But we know that the observer moving with the y-axis sees light travel at c. For this to make sense, it has to be that time is different for the moving observer by the ratio (c^2 - v^2)^.5/c which is (1 - (v/c)^2)^.5. So t'(the moving observer's time) = t(1 - (v/c)^2)^.5.

This is one of the basic equations in relativity. If you're ever going to find a real flaw in relativity, which you are apparently keen to do, you will have to learn what it says and why it says it. Of course, if you do this you may wind up believing it....is that why you cling to your ignorance here...are you afraid you might have to change your mind?
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Re: Special Relativity is Galilean Relativity January 14 2009, 10:37 AM

Roger: Einstein says and I quote: "to the axes of Y and Z--it being borne in mind that light is always propagated along these axes, when viewed from the stationary system, with the velocity sqrt (c^2 - v^2)." whatever you have decided to talk about bears no relation.

[snip]

cinci: Thanks for catching the typo.

ok

cinci:As for the rest of your comment, you still don't realize that from the perspective of the stationary observer there is nothing travelling along y.

Einstein refers to it as "light" not as you refer to it as "nothing."

cinci: So while the motion along the hypotenuse is ct, for someone moving with that axis it would look like it was moving at (c^2 - v^2)^.5 along the axis. But we know that the observer moving with the y-axis sees light travel at c. For this to make sense, it has to be that time is different for the moving observer by the ratio (c^2 - v^2)^.5/c which is (1 - (v/c)^2)^.5. So t'(the moving observer's time) = t(1 - (v/c)^2)^.5.

I think you have things confused. But anyway back to criticising Einstein he refers to light travelling at velocity sqrt (c^2 - v^2), when in other places he says it must always be c. So he contradicts himself.
Einstein says and I quote: "to the axes of Y and Z--it being borne in mind that light is always propagated along these axes, when viewed from the stationary system, with the velocity sqrt (c^2 - v^2)."

***************************

cinci:As for the rest of your comment, you still don't realize that from the perspective of the stationary observer there is nothing travelling along y.

Roger: Einstein refers to it as "light" not as you refer to it as "nothing."

cinci: Einstein refers to it as light moving along the moving axis as seen by the moving observer. I said nothing was travelling along the moving axis from the perspsctive of the stationary observer. So Einstein's statement doesn't conflict with mine.
***********************

cinci: So while the motion along the hypotenuse is ct, for someone moving with that axis it would look like it was moving at (c^2 - v^2)^.5 along the axis. But we know that the observer moving with the y-axis sees light travel at c. For this to make sense, it has to be that time is different for the moving observer by the ratio (c^2 - v^2)^.5/c which is (1 - (v/c)^2)^.5. So t'(the moving observer's time) = t(1 - (v/c)^2)^.5.

Roger: I think you have things confused.

cinci: I don't.
*******************

Roger: But anyway back to criticising Einstein he refers to light travelling at velocity sqrt (c^2 - v^2), when in other places he says it must always be c. So he contradicts himself.
Einstein says and I quote: "to the axes of Y and Z--it being borne in mind that light is always propagated along these axes, when viewed from the stationary system, with the velocity sqrt (c^2 - v^2)."

cinci: You're not paying attention to the phrase "when viewed from the stationary system". It's the same problem with your first statement above. You are simply not viualizing what is going on correctly.

Here's a nice graphic. You can vary the speed, v. If you don't see it after this, I oficcialy give up on you.
*********************
********************

Re: Special Relativity is Galilean Relativity January 14 2009, 4:20 PM

cinci:As for the rest of your comment, you still don't realize that from the perspective of the stationary observer there is nothing travelling along y.

Roger: Einstein refers to it as "light" not as you refer to it as "nothing."

cinci: Einstein refers to it as light moving along the moving axis as seen by the moving observer. I said nothing was travelling along the moving axis from the perspsctive of the stationary observer. So Einstein's statement doesn't conflict with mine.

roger: so now you are trying to make out that what observer sees as "light" the other sees "nothing".

***********************



Roger: But anyway back to criticising Einstein he refers to light travelling at velocity sqrt (c^2 - v^2), when in other places he says it must always be c. So he contradicts himself.
Einstein says and I quote: "to the axes of Y and Z--it being borne in mind that light is always propagated along these axes, when viewed from the stationary system, with the velocity sqrt (c^2 - v^2)."

cinci: You're not paying attention to the phrase "when viewed from the stationary system". It's the same problem with your first statement above. You are simply not viualizing what is going on correctly.

Here's a nice graphic. You can vary the speed, v. If you don't see it after this, I oficcialy give up on you.

roger: there is no graphic; if you posted it; it could not have come through.

Roger Dumberton: But anyway back to criticising Einstein he refers to light travelling at velocity sqrt (c^2 - v^2), when in other places he says it must always be c. So he contradicts himself.

Ted: No, he doesn't. But your dementia is too far advanced for you to get it.

yes he does

Here's a nice graphic. You can vary the speed, v. If you don't see it after this, I oficcialy give up on you.

roger: there is no graphic; if you posted it; it could not have come through.

cinci: Sorry, here it is: http://galileoandeinstein.physics.virginia.edu/more_stuff/flashlets/lightclock.swf

http://galileoandeinstein.physics.virginia.edu/more_stuff/flashlets/lightclock.swf

Ha..ha.hahahahah That has to be the second stupidest demonstration of Relativity I have ever seen, allowing for the inevitability of an even stupider one that will, no doubt, coma along.

Bob S: http://galileoandeinstein.physics.virginia.edu/more_stuff/flashlets/lightclock.swf

Ha..ha.hahahahah That has to be the second stupidest demonstration of Relativity I have ever seen, allowing for the inevitability of an even stupider one that will, no doubt, coma along.

cinci: You know Bob I'd say you were as dumb as a bag of hammers but I don't want to insult the hammers. So I'll just let it pass with this: There are none so blind as those who will not see.
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Re: Cincirob, Special Relativity is Galilean Relativity January 15 2009, 11:21 PM

cinci: "You know Bob I'd say you were as dumb as a bag of hammers but I don't want to insult the hammers. So I'll just let it pass with this: There are none so blind as those who will not see."
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Well I'm glad you didn't say it Cinci, because hammers can very vengeful and the last thing I want to do is try and stop a bag full of hammers from doing what they want to do. But my hammers serve a more useful purpose than that stupid video "demonstration". I'm gonna link that site for when I need a good, I mean a really good, laugh.

cinci: "You know Bob I'd say you were as dumb as a bag of hammers but I don't want to insult the hammers. So I'll just let it pass with this: There are none so blind as those who will not see."


Bob S: Well I'm glad you didn't say it Cinci, because hammers can very vengeful and the last thing I want to do is try and stop a bag full of hammers from doing what they want to do.

cinci: Why are you and AAF afraid of hammers?
************************

Bob S: But my hammers serve a more useful purpose than that stupid video "demonstration". I'm gonna link that site for when I need a good, I mean a really good, laugh.

cinci: Yes, I'm sure you will. It's a lot easier to laugh at something than to try to understand it.
**************************

Re:Cincirob, Special Relativity is Galilean Relativity January 16 2009, 4:59 PM

Bob S: "But my hammers serve a more useful purpose than that stupid video "demonstration". I'm gonna link that site for when I need a good, I mean a really good, laugh."

cinci: "Yes, I'm sure you will. It's a lot easier to laugh at something than to try to understand it."
**************************

I do understand the nature of motion Cinci, that's what makes that ridiculous video all the more funny. This explanation might help, considering the problem you have.

sA lliJ si gnivom latnoziroh morf tfel ot thgir ta )v(, eht thgil ecruos lliw osla evom latnoziroh morf tfel ot thgir ta )v(. ehT thgil maeb\eslup si dettime ta )c( ralucidneprep, lacitrev, ot eht thgil ecruos neht detcelfer kcab ot sti ecrous. ehT thgil eslup\maeb lliw naitnaim )v( dna )c( rof eht noitarud )t( fo sti htap. )t( sniamer lauqe rof lliJ dna kcaJ rof lla sesac fo )v(.

s bob

Bob S: I do understand the nature of motion Cinci, that's what makes that ridiculous video all the more funny. This explanation might help considering the problem you have.

sA lliJ si gnivom latnoziroh morf tfel ot thgir ta )v(, eht thgil ecruos lliw osla evom latnoziroh morf tfel ot thgir ta )v(. ehT thgil maeb\eslup si dettime ta )c( ralucidneprep, lacitrev, ot eht thgil ecruos neht detcelfer kcab ot sti ecrous. ehT thgil eslup\maeb lliw naitnaim )v( dna )c( rof eht noitarud )t( fo sti htap. )t( sniamer lauqe rof lliJ dna kcaJ rof lla sesac fo )v(.

cinci: You only put the letters in reverse order. I prefer the words to be reversed also.

By the way, I haven't seen anybody work this hard to misunderstand relativity since Dingle. Maybe somebody is giving a prize for it.

Did you figure out the calendar yet?
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MP 65:6 ,9002 61 yraunaJ ytivitaleR naelilaG si ytivitaleR laicepS ,boricniC :eR

cinci: "You only put the letters in reverse order. I prefer the words to be reversed also."

.od ekam ot evah tsuj lliw uoy ,icniC siht ertfA

latnoziroh evom osla lliw ecrous thgil eht ,)v( ta thgir ot tfel morf latnoziroh gnivom si lliJ sA
eht ot ,lacitrev ,ralucidneprep )c( ta dettime si maeb\eslup thgil ehT .)v( ta thgir ot tfel morf thgil
eht rof )c( dna )v( naitnaim lliw eslup\maeb thgil ehT .ecrous sti ot kcab detcelfer neht ecrous
.)v( fo sesac lla rof kcaJ dna lliJ rof lauqe sniamer )t( ,htap sti fo )t( noitarud

cinci: "You only put the letters in reverse order. I prefer the words to be reversed also."


Bob S" .od ekam ot evah tsuj lliw uoy ,icniC siht ertfA

latnoziroh evom osla lliw ecrous thgil eht ,)v( ta thgir ot tfel morf latnoziroh gnivom si lliJ sA
eht ot ,lacitrev ,ralucidneprep )c( ta dettime si maeb\eslup thgil ehT .)v( ta thgir ot tfel morf thgil
eht rof )c( dna )v( naitnaim lliw eslup\maeb thgil ehT .ecrous sti ot kcab detcelfer neht ecrous
.)v( fo sesac lla rof kcaJ dna lliJ rof lauqe sniamer )t( ,htap sti fo )t( noitarud


cinci: Thanks Bob, you just made 10 bucks for me. My son bet me that I couldn't get anybody to do anything this dumb.
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Re: Cincirob, Special Relativity is Galilean Relativity January 17 2009, 7:23 PM

cinci: Thanks Bob, you just made 10 bucks for me. My son bet me that I couldn't get anybody to do anything this dumb.
*******************************************

Youre quite welcome Cinci, you got 10 bucks and I got a challenge, it werent perfect but it was fun, but thats the kinda guy I am; a not perfect, fun loving guy. The best part was, is that it's true.

Re: Cincirob, Special Relativity is Galilean Relativity January 17 2009, 7:23 PM

cinci: "Thanks Bob, you just made 10 bucks for me. My son bet me that I couldn't get anybody to do anything this dumb."

Cinci, I was showing my boy, Gary, the exchange between us and how you won $10 bucks...my boy smiled and said "a kid should never bet against his dad!" Then, after a short pause, he said "I wonder how he feels knowing his boy doubts him". Good question Gary...I said, I'll ask him; so Cinci, how does it make you feel knowing that your boy doubts you?

cinci: "Thanks Bob, you just made 10 bucks for me. My son bet me that I couldn't get anybody to do anything this dumb."

Bob S: Cinci, I was showing my boy, Gary, the exchange between us and how you won $10 bucks...my boy smiled and said "a kid should never bet against his dad!" Then, after a short pause, he said "I wonder how he feels knowing his boy doubts him". Good question Gary...I said, I'll ask him; so Cinci, how does it make you feel knowing that your boy doubts you?

cinci: I think it's nice that your son humors you. His mother must have trained him well.
*****************

Re: Cincirob, Special Relativity is Galilean Relativity January 18 2009, 1:06 PM

cinci: "Thanks Bob, you just made 10 bucks for me. My son bet me that I couldn't get anybody to do anything this dumb."

Bob S: "Cinci, I was showing my boy, Gary, the exchange between us and how you won $10 bucks...my boy smiled and said "a kid should never bet against his dad!" Then, after a short pause, he said "I wonder how he feels knowing his boy doubts him". Good question Gary...I said, I'll ask him; so Cinci, how does it make you feel knowing that your boy doubts you?"

cinci: "I think it's nice that your son humors you. His mother must have trained him well."
*****************

Good call Cinci, no one, more than Gary's mom, knows better than to bet against me. However, since I can only imagine what it must be like to have a son doubt his father, I will just apply that as your answer...sigh!

cinci: "Thanks Bob, you just made 10 bucks for me. My son bet me that I couldn't get anybody to do anything this dumb."

Bob S: "Cinci, I was showing my boy, Gary, the exchange between us and how you won $10 bucks...my boy smiled and said "a kid should never bet against his dad!" Then, after a short pause, he said "I wonder how he feels knowing his boy doubts him". Good question Gary...I said, I'll ask him; so Cinci, how does it make you feel knowing that your boy doubts you?"

cinci: "I think it's nice that your son humors you. His mother must have trained him well."


BobS: Good call Cinci, no one, more than Gary's mom, knows better than to bet against me.
However, since I can only imagine what it must be like to have a son doubt his father, I will just apply that as your answer...sigh!

cinci: I just hope he doesn't make unfounded assuptions as you continue to do. And since you can apparently still afford a computer, it's clear you haven't been betting on your physics ideas.
****************

Roger,

Your latest scientific breakthrough (http://www.wbabin.net/science/anderton26.pdf) was counter productive. You should check your calculations since the composition of two "Anterton transformations" (AT) is not an AT anymore. Instead, the composition of two AT pops up additional terms bringing it closer to the LT.

I leave it to you to check that the composition of many small AT makes an LT actually. This is of course not surprising, as the LT for small velocities is an AT. This also known by any undergraduate.

Just read my recent thread "Anderton needs corrections":

---
In this paper, Andertons pretends to correct 100 years of mistakes in the derivation of the Lorentz transformation.
After "subtle corrections" he proposes the following replacement where he dropped the relativistic factor from the LT:

x' = x-Vt = (c-V)t (8)
t = (c-V)t/c = (t Vx/c²) (9)

There must be some problem with your derivation Dr Anderton !
Just write your transformation twice for subsequent changes of reference frames.
I do this below using 0,1 and 2 to identify the reference frame:

(A1) x1 = x0 - v0 t0
(A2) t1 = t0 - x0 v0 / c²

(B1) x2 = x1 - v1 t1
(B2) t2 = t1 - x1 v1 / c²

Now substitute equations A in equations B and you get:

(C1) x2 = x0 (1 + v0 v1 /c²) - (v0+v1) t0
(C2) t2 = t0 (1 + v0 v1 /c²) - (v0+v1)/c² x0

Anderton, doesn't it look like the relativistic factor you dropped out is trying to come back again from your own equation?

Anderton, check your calculation!

Roger,

Does your silence mean that you a looking for your mistake?

No its means that you are a moron, just talking nonsense and there is no pint talking to morons you thicko.

Roger,

There are two much better explanation to your silence,

1) either you don't understand what I am talking about,
2) or you don't have any answer to my simple argument

But you must admit that the Anderton transformation (AT) suffer from some problems.

Imagine the earth is moving around the sun, approximatively as an inertial frame.
Inmagine a boat moving with respect to the earth, also approximativel as an inertial frame.

Isn't it logical to assume that the AT should be applicable to:

1) transform from solar frame to earth frame (S->E)
2) transform from earth frame to boat frame (E->B)
3) transform from solar frame to boat frame (S->B)

Yet, by using the AT twice to perform the S->B transformation, you get a contradicting result.

Your theory fails this elementary test.
And, ironically, trying this you see a relativistic factor that pops up !!!

I can be a moron, if say it.