re - a short version of the argument presented in this recent article to GSJ

http://www.wbabin.net/physics/waterman12.pdf

Has anyone read it through ?

What if we throw some numbers into the equation x' = v - vt

given -

Coincident frames have a point at 11,0.
One frame is moved to the right by 7 while the other remains in place.
What is the coordinate of the point now, in reference to the unmoved frame ?

Voigt and then accepted by Einstein x' = x - vt
x = 11,0
vt = 7
x' = 4,0

reality x' = x + vt
x = 11,0
vt = 7
x' = 18,0

Houston we have a problem -
Relativity says 4 and reality says 18.

steve waterman

AAF,

Yes, please do remove all those comments but let the post below go through. thank you AAF.

....................................................................

I am hoping to find out those who agree and those whose do not.
and why not from any that have a mathematical objection.

i think things are summed up well in the questions on the first post above -
is it 4 or is it 18 ?

Granted, 4 = 11 -7, but 18 = 11 + 7 too,

As well, these is just numerical values,
and we are talking about the distance from 0,0 in the stationary frame.
Not all line segments are coordinates. His 4 distance does not get
measured from 0,0 in the stationary frame and that is wrong.

VT is not acceptable to count from to get 4, and neither can we get 4 by
counting backwards from 11. These are not coordinates lengths in the
stationary frame, only those commencing from 0,0 are.

steve

I'm going to weigh in!


Firstly, the [x' = x - vt] is one of the Galilean Equations:
http://psi.phys.wits.ac.za/teaching/Connell/phys284/2005/lecture-01/lecture_01/node5.html

And this equation is intuitively clear and is correct.


Secondly, Lorentz and Einstein and their followers multiplied this simple Galilean Equation by their Gamma Factor.

And their new equation is neither intuitively obvious nor physically meaningful nor correct.

To me, it is intuitively correct for the placing of the second origin toward the left. It looks to me as if all these years, the fact that the abscissa runs right and left has been ignored. You cannot measure backward (toward the origin) and have a meaningful measurement for the transform. The conventional diagram is too complicated, and I believe most people quit trying to figure it out, just accept it, and move on.

To say that (as you do, AAF): "Firstly, the [x' = x - vt] is one of the Galilean Equations"

is to over look the fact that the quote:

"S' is moving with respect to S with velocity "v" (as measured in S) in the "x" direction," is to overlook the fact that:

The "x" axis has two directions. TWO DIRECTIONS. GET IT????(sorry to be so loud in my effort to get you to see the problem!)

". . .in the "x" direction." IS AMBIGUOUS! the "x" axis proceeds to infinity to the right in POSITIVE numbers. It (unacknowledged) ALSO proceeds to the lEFT IN NEGATIVE NUMBERS.

What Mr. Waterman is declaring is: the statement: "Firstly, the [x' = x - vt] is one of the Galilean Equations" is wrong!!!!! It is NOT intuitive.


The gullible acceptance of this statement is the main problem with all the silly deductions of relativity.

I know: How can something so simple be overlooked by all the great minds of the last hundred years????

Well, smart guys like you and "Anonymous" don't see it. No maliciousness intended, be among the first to realize this major problem!!!!

Waterman is trying to get an answer about the math and logic of the basic transform. You and I always want to jump to another subject. Stare at his question, and only answer the question. You didn't answer his question!

AAF,

Thank you for weighing in.

Merely saying that x' = x - vt is intuitively right, is by no means a proof.

I am clearly saying that the article shows your intuition to be incorrect,
as is pertains to the equation. My article supplies a mathematical proof
of its hidden-for-a-hundred-year-old error.

Did you read the article ? All the way to the end ? Surely, then...if my article is wrong...one should be able to drive a Mac truck through my logic.
No theories, no feelings, no favoritism, no intuition, no gut feeling....just that good old math. - which must be either 100 percent right or 100 percent wrong.

Do you not see that he moves the second frame and does NOT move the point along with it as he goes ? etc etc I can only suspect that you may not
have read the whole article....

So, as feedback then...all I am hearing from you is, it is wrong, because you know its wrong. That AAF, is really not the feedback quality I was going for.

The question again is....if you disagree - what is your MATHEMATICAL objection. At this point AAF, you will perhaps be surprised that there has been no mathematical objection ever put forth to me. Indeed, only the occasional personal opinion...and then few enough times to easily count on my fingers.

Historically, others have initially mentally dismissed my premise as not even remotely possible...and yet not a single person has shown me why a word of it that is wrong. Pretty difficult to fight a phantom.

Unless there is a mathematical objection, then from that stand-point, I am right until proven wrong. Prove me wrong - I dare say, you cannot, nor can any one else. So placate me for a moment and tell me you finished reading all of the article. Please comments about the math of the article itself...

"just the math please"

steve

AAF,

AAF - Secondly, Lorentz and Einstein and their followers multiplied this simple Galilean Equation by their Gamma Factor.

Yes...however, they derived that equation using x' = x - vt. The damage was already done.

AAF -And their new equation is neither intuitively obvious nor physically meaningful nor correct.

Indeed...garbage in, garbage out. So we agree on this; the Gamma Factor
is wrong. Please drop the intuition comments...this is math and not in any way, shape or form, or condition, subjective. Hence, I am saying that the CAUSE of the Gamma Factor being incorrect IS that x' = x - vt is incorrect.

Some actual feedback...still not about the article...but we are getting there...
thank you AAF.

steve

woldemar+AAF+JR=bunch of idiots congratulating each other for their stupidity

And you claim others do nothing but call out names! You are the juvenile ad hominem baiter, the Latin scholar who doesn't know Latin from French. The child who plays with herself.

AAF,

Talk to me, can you block annon the mouse
on this one thread or not ?

If I must inform you every time this happens,
it only makes sense to try this at a different forum.

The response here so far, is pretty dismal. How do I
fight that to which the only objection so far, is that
x' = x - vt is right and i am wrong ?

steve

"Merely saying that x' = x - vt is intuitively right, is by no means a proof. "

The same applies to the Lorentz transformation.
Yet, the Galilean transform is so easy to proof and understand that it is difficult to believe this is not a joke.

If you don't understand that, consider writing with some chalk coordinates on a train and coordinates on the platform. Then think for a moment. You need of course to decide what the symbols in the equation relate to and particularly the speed. It may well be that you come up with a + instead of a -, depending on what you are talking about.

Now for AFF: just do the same but use clock synchronisation in each frame.
You will get the result you consider as stupid, unintuitive, ...
The perform some introspection to know why you find it stupid.

I will give this one more shot on this thread.

if 4 is your answer...

is it counting backwards from 11 - 7 ?

obviously this cannot be...
as we are not counting from 0,0 in the unmoved frame.

is it counting forwards from 7 - 11 ?

note - in the unmoved frame all coordinates start from 0,0.

if 18 is your answer, then you agree ( albeit likely unknowingly )
that although 4 = 11 - 7, the equation x' = x - vt is wrong.

Indeed, NO COORDINATE transformation ever occurred. When the length of 4 was taken, it WAS a coordinate in the moved frame. When it got placed
starting at 7 and not from 0,0, ...it is now NOT a coordinate in the unmoved frame....it is just a mere line segment.

steve

Anonymous (the one confused by French/Latin) said: "If you don't understand that, consider writing with some chalk coordinates on a train and coordinates on the platform. Then think for a moment. You need of course to decide what the symbols in the equation relate to and particularly the speed. It may well be that you come up with a + instead of a -, depending on what you are talking about."

Jose says: Good idea! Why don't you try it? You will find that Steve knows what he is talking about. You have never tried this yourself, otherwise you would understand. (you can do it with two pieces of paper if you want to try it right now.)

(His question has nothing to do with motion, only the alignment of the two origins to start with, then move one in the plus direction to the right.

Then realign the two origins and move one in the minus direction, to the left.)

Woldemar: AAF, Talk to me, can you block annon the mouse on this one thread or not ? If I must inform you every time this happens, it only makes sense to try this at a different forum. The response here so far, is pretty dismal. How do I fight that to which the only objection so far, is that x' = x - vt is right and i am wrong?





AAF Woldermar; first of all, don't mistake the responsible Anon of Belgium for the kid-like Anon (aka Ted) of Santa Clara! Of course, the IP can be blocked. But the problem here is that this 'son o' bitch' lives very close to the Silicon Valley and has access to many different IPs! In addition, his offense is quite trivial and consisting of little else beside his mantra: "woldemar+AAF+JR=bunch of idiots congratulating each other for their stupidity". I could be way wrong, but I like it! Why? Because, upon posting some argument, the most dreaded objections to it are the real and the fatal ones. Ted's objections are mere silly crap in this regard. Is this little childish comment of his very offensive? Well, I guess following around this moron across the Silicon Valley and blocking the IP of every Internet café he used is like using a tank to block a mouse hole; right Jose? Now, Woldermar, if the comments of this well-known idiot are really bothering you, then you should know that there is an other way to punish him. This retarded person, is somewhat of a notable in his area! He's also some sort of a published scientist and patent holder and company manager. And this means you can attach his real name to a lot of smear and Google bombs; is this right;; Bob? And not just his real name and his resume, his picture is on the NET too! Do you need any help, Woldemar? Just ask! However, the principles of justice require that you use this method only in defense of your real name.

AAF,

Too bad. The mouse wins, I give up. I am gone again,
and this time, most likely for good. Sorry to
have to pick up my football and leave.

No one ever emailed me about by posted stuff for a year...
so, it is time for me to leave this non-discussion arena.

I do not care who the mouse is any more - let him stay
anonymous. You said you enjoy the banter and name calling.
Each to their own.

steve

AAF,

OH ...I forgot to answer this....

NO, NO, NO...I DO NOT WANT TO DO ANYTHING.

I want to be left alone and I suggest you leave
anonymous alone too. He does have the right to be
contrary. I have the right to leave the forum.

Let's all just move on.

steve

Woldemar; it's nice to take the high road!

Too bad; the mouse won! ---->>>>>>

But before you say farewell, let me tell you something:

When you set out to criticize something as widespread and well-rooted as the theory of Relativity, you have to be prepared to withstand the heat and to take in stride the name-calling and all the crap that inevitably to be thrown on you. And you have to be thick-skinned and to take any criticism whatever it is and let others pay for it in kind. Otherwise, it would be better and much safer to stay silent and say nothing.

Finally, I have composed the answer to your previous reply to my objections; and you may choose to wait a bit longer to take a look at it.

AAF - When you set out to criticize something as widespread and well-rooted as the theory of Relativity, you have to be prepared to withstand the heat and to take in stride the name-calling and all the crap that inevitably to be thrown on you. And you have to be thick-skinned and to take any criticism whatever it is and let others pay for it in kind. Otherwise, it would be better and much safer to stay silent and say nothing.

Already have the complete set of t-shirts.

So, either i shut up or i have to accept personal attacks from anonymous voices who can defame and slander and whatever they want ? Woldemar has left the building.


AAF - Finally, I have composed the answer to your previous reply to my objections;

What objections....did you object to something i said ?

AAF - and you may choose to wait a bit longer to take a look at it.

Bring it on please....but not here...try me through my site if you are interested in a discussion.

steve

AAF: Be patient, Woldemar; let me first listen very carefully to what our colleague, Jose wants to tell me!





Jose: To me, it is intuitively correct for the placing of the second origin toward the left. It looks to me as if all these years, the fact that the abscissa runs right and left has been ignored. You cannot measure backward (toward the origin) and have a meaningful measurement for the transform. The conventional diagram is too complicated, and I believe most people quit trying to figure it out, just accept it, and move on. To say that (as you do, AAF): "Firstly, the [x' = x - vt] is one of the Galilean Equations" is to overlook the fact that the quote: "S' is moving with respect to S with velocity "v" (as measured in S) in the "x" direction," is to overlook the fact that: The "x" axis has two directions. TWO DIRECTIONS. GET IT????(sorry to be so loud in my effort to get you to see the problem!) ". . .in the "x" direction." IS AMBIGUOUS! the "x" axis proceeds to infinity to the right in POSITIVE numbers. It (unacknowledged) ALSO proceeds to the lEFT IN NEGATIVE NUMBERS. What Mr. Waterman is declaring is: the statement: "Firstly, the [x' = x - vt] is one of the Galilean Equations" is wrong!!!!! It is NOT intuitive. The gullible acceptance of this statement is the main problem with all the silly deductions of relativity. I know: How can something so simple be overlooked by all the great minds of the last hundred years???? Well, smart guys like you and "Anonymous" don't see it. No maliciousness intended, be among the first to realize this major problem!!!! Waterman is trying to get an answer about the math and logic of the basic transform. You and I always want to jump to another subject. Stare at his question, and only answer the question. You didn't answer his question!





AAF: Look, Jose; it's really simple! The co-ordinate x is cut to two equal halves by the origin (0, 0, 0): One half extends to the right all the way to the right-hand infinity. And the other half extends to the left all the way to the left-hand infinity. Now, if the motion is along the negative direction of the co-ordinate x, then it must be expressed explicitly that it is in the direction of (-x). But if the motion is along the positive direction of the co-ordinate x, then you can say it's in the direction of (x), which by convention means the direction of (+x). Do you like Woldemar's redundant notion of writing over and over again Cartesian points like these: (+1, +4, +3), (+10, +14, +33), (+111, +444, +333)? I guess not! Time is money; and doing little aesthetic games of this sort can hurt your chances of passing the math exam; right, Jose? It follows, therefore, that Woldemar's objection to the validity of the Galilean transformations is really a kind of personal dislike to the Cartesian convention and little else.

Woldemar: AAF, Thank you for weighing in. Merely saying that x' = x - vt is intuitively right, is by no means a proof. I am clearly saying that the article shows your intuition to be incorrect, as is pertains to the equation. My article supplies a mathematical proof of its hidden-for-a-hundred-year-old error. Did you read the article ? All the way to the end ? Surely, then...if my article is wrong...one should be able to drive a Mac truck through my logic. No theories, no feelings, no favoritism, no intuition, no gut feeling....just that good old math. - which must be either 100 percent right or 100 percent wrong. Do you not see that he moves the second frame and does NOT move the point along with it as he goes ? etc etc I can only suspect that you may not have read the whole article.... So, as feedback then...all I am hearing from you is, it is wrong, because you know its wrong. That AAF, is really not the feedback quality I was going for. The question again is....if you disagree - what is your MATHEMATICAL objection. At this point AAF, you will perhaps be surprised that there has been no mathematical objection ever put forth to me. Indeed, only the occasional personal opinion...and then few enough times to easily count on my fingers. Historically, others have initially mentally dismissed my premise as not even remotely possible...and yet not a single person has shown me why a word of it that is wrong. Pretty difficult to fight a phantom. Unless there is a mathematical objection, then from that stand-point, I am right until proven wrong. Prove me wrong - I dare say, you cannot, nor can any one else. So placate me for a moment and tell me you finished reading all of the article. Please comments about the math of the article itself... "just the math please"





AAF: As I said before, it appears on the face of it that you've based the principal argument of your fore-mentioned paper upon overlooking the implicit meaning of (x) as (+x) in the current Cartesian convention. If this is the only objection you have against the Galilean transformation, then that paper can't go far towards its intended goal; i.e. uncovering a fatal error and blowing to smithereens the equa­t­ions of Galileo and Newton and the weakling of physics called 'Lorentz & Einsteins'! I must, however, salute your daring spirit and clear exposition; and I dare to say that this work of yours is an important stepping stone and ought to lead you towards developing and refining more your arguments in the near future. So long & farewell, Woldemar; and we wish you success....



A Farewell to Arms

AAF - that you've based the principal argument of your fore-mentioned paper upon overlooking the implicit meaning of (x) as (+x) in the current Cartesian convention.

What in hell are you talking about ? I never said that or even anything remotely similar. You do not speak for me...please do not try.

AAF -The implicit meaning of (x) as (+x)
...this is really insulting AAF.

I said x' = x - vt is wrong - it should be x' = x + vt

You do not seem to even grasp what I proposed, as this seems apparent
to me, when you attempt to summarize the article with such an incredible
statement. Some discussion...now I am being told that my paper boils down to me saying, well according to you that is.... the implicit meaning of (x) as (+x). I am almost speechless AAF. I do not have the words to express how offended I am by this. After the recent, and still "anonymous" fiasco, this be the cherry on the cake.

Even the ghost of woldemar has hit the road this time.

Well, there is no doubt that each inventor should have been given back 13.333 bucks, not ten bucks. It seems that the thief who decided to take twenty bucks, and returning thirty, was the logical solution to his predicament. Each inventor, eventually receives ninety bucks, and three times ninety equals two- seventy. So, where is the missing ten bucks? This is mind boggling! Is there a solution to this paradox? It seems that if each inventor received 13.333 bucks, this would equal forty bucks, plus the twenty that he stole would equal three hundred and ten bucks. Here, where is the excess of ten bucks? This is truly perplexing.

The only solution to this problem must be in dimensional analysis. We must be using our scales differently. Certainly, if you lay three hundred dollars down on a table, no matter how you mix the bills up, in the end, everybody involved has to feel that they have received their money back. This goes without saying.

So, you have three groups of ten and you need to take five units from this group of ten- from the fifty bucks from the three hundred. This leaves two groups of ten with only eight and one group with nine. If you give them each ten bucks back, two groups will eventually receive ninety bucks and one group receive one hundred bucks.

http://www.wbabin.net/physics/waterman12.pdf

Waterman: "The solution is to find the flaw in his statements...not in the stating of truths.

The deception is in the phrase "plus the twenty which the clerk tipped himself". This twenty bucks should not be added to the two seventy; it should be subtracted from the two seventy.

The twenty the clerk got is part of the two seventy the three inventors spent collectively. If you subtract the twenty from the two seventy you get the two fifty that the
patent office received..

This exact same trick is employed in the Galilean equation x' = x - vt... when the correct answer is x' = x + vt. Actually...the Galilean transformation does the opposite by subtracting when he should be adding."

The other 13 1/2 pages is just over kill!

Thanks for your feedback Bob.

I agree...the rest is overkill. However, without it,
there is no proof, just contention of the formula.
My writings skills well, I do the best I can, as i did with
the graphics for it.

It is sloppy perhaps, but the issue is driven home through
all the graphics. Again, thanks for weighing in.

steve

Re: Waterman, an ember yet glows August 24 2009, 2:08 PM

"[A]n ember yet glows"? Was there some other part of the paper you wanted an opinion on?

Bob,

No. I am just looking for those that agree and for those that are contrary,
to say why. So I am fine with your response.

The ember remark had to do woldemar leaving, yet surprisingly, a spark of your comment still came from the barely smoldering fires of GSJ regarding the article.

Woldy

AAF: As I said before, you've based the principal argument of your fore-mentioned paper upon overlooking the implicit meaning of (x) as (+x) in the current Cartesian convention.





The Ghost of Woldemar: huh? What in hell are you talking about? I never said that or even anything remotely similar. You do not speak for me...please do not try. I said x' = x - vt is wrong - it should be x' = x + vt You do not seem to even grasp what I proposed, as this seems apparent to me, when you attempt to summarize the article with such an incredible statement. Some discussion...now I am being told that my paper boils down to me saying, well according to you that is.... the implicit meaning of (x) as (+x). I am almost speechless AAF. I do not have the words to express how offended I am by this. After the recent, and still "anonymous" fiasco, this be the cherry on the cake. Even the ghost of woldemar has hit the road this time.






AAF: Sorry, GW; that is only your side of the story! And here's my side of it: [A] You started the day of last Sunday by whining and complaining bitterly about the small trivialities of Moron Ted. [B] In good faith I offered my help and was about to introduce you to Gregory9, the most famous serial winner over the Idiot of Santa Clara (aka Ted); but you chose to play monk instead and to turn down my good offer! [C] And finally, I got inspired and fired up by Jose's excellent summary of your thesis and I dared to explain to you that your entire argumentation against those transformations is based upon the implicit assumption of the supposed ambiguity of the symbol x; and that is glaringly an oversight on your part. And now, your ghost is telling me that you've been offended and can't take it anymore! Well, GW, go back and tell Woldemar that (v) in Galileo's first equa­t­ion is a vector too; and when it points to the direction of (-x), it must be treated as (-v) as well.

AAF -

[B] In good faith I offered my help and was about to introduce you to Gregory9, the most famous serial winner over the Idiot of Santa Clara (aka Ted); but you chose to play monk

W - Oh really ...you want to point me to that offer ?


AAF - instead and to turn down my good offer!

S - The offer that you never bothered to mention to me...that one ?


AAF - [C] And finally, I got inspired and fired up by Jose's excellent summary of your thesis and I dared to explain to you that

"your entire argumentation against those transformations is based upon the implicit assumption of the supposed ambiguity of the symbol x;"

S - I never said that. Did you forget that I mentioned this that already ?
It is certainly NOT my entire argument...
You use "implicit assumption of the supposed ambiguity
of the symbol x:" to characterize the entire article.
The article has absolutely nothing at all even related to that.

AAF - and that is glaringly an oversight on your part.

S - What...the thing I never said ?

AAF - And now, your ghost is telling me that you've been offended and can't take it anymore! Well, GW, go back and tell Woldemar that


(v) in Galileo's first equa­t­ion is a vector too; and when it points to the direction of (-x), it must be treated as (-v) as well.

S - So now you wind it all up with a simple true statement of your own, implying that I would disagree...give me a break. I wonder why you mention it ? Is this your proof that I am wrong ? And off we would go, on a another side-tracking. [ So this is not pertinent to the math presented either.

We disagree, fine. So be it. I will accept that you were not trying to be offensive with your comments, and that you were just trying to supply some honest feedback.

What I objected to, was the feeling that I was being spoken for with words and thoughts that were not mine, and thus being misrepresented. All of this could have been avoided if you had first asked me if that is what i was saying - or just supplied the appropriate quote....

In the end AAF, it turns out some get it and some do not. I have to place you into the did not get it box. Win some lose some. Its all okay.

Interesting,

My last post was a few minutes ago...

now has it the anonymous name on it as if he had posted it.
Can that even be done ? I assure you it was me writing his name.

This was my response to you AAF.

I guess I will just have to stop looking into the Forum.

Woldemar; it's OK!

On the Network54 forums, like this one, when you don't fill in the first field of the forum editor,
you automatically get the alias 'anonymous' by default.

AAF: Well, GW, go back and tell Woldemar that (v) in Galileo's first equa­t­ion is a vector too; and when it points to the direction of (-x), it must be treated as (-v) as well





Steve: So now you wind it all up with a simple true statement of your own, implying that I would disagree...give me a break. I wonder why you mention it ? Is this your proof that I am wrong ? And off we would go, on a another side-tracking. [ So this is not pertinent to the math presented either. We disagree, fine. So be it. I will accept that you were not trying to be offensive with your comments, and that you were just trying to supply some honest feedback. What I objected to, was the feeling that I was being spoken for with words and thoughts that were not mine, and thus being misrepresented. All of this could have been avoided if you had first asked me if that is what i was saying - or just supplied the appropriate quote.... In the end AAF, it turns out some get it and some do not. I have to place you into the did not get it box. Win some lose some. Its all okay.





AAF: Well, then, would you please summarize your main argument for us to discuss here?





Steve: A short version of the argument presented in this recent article to GSJ http://www.wbabin.net/physics/waterman12.pdf Has anyone read it through? What if we throw some numbers into the equation x' = v - vt given - Coincident frames have a point at 11,0. One frame is moved to the right by 7 while the other remains in place. What is the coordinate of the point now, in reference to the unmoved frame? Voigt and then accepted by Einstein x' = x - vt x = 11,0 vt = 7 x' = 4,0 reality x' = x + vt
x = 11,0 vt = 7 x' = 18,0 Houston we have a problem - Relativity says 4 and reality says 18.






AAF: Thank you Steve... For the time being let's put aside the published article, and concentrate instead on the numbers and the scenario above! We have two coincident Cartesian co-ordinate systems. The term 'coincident', within this context, implies that the points of origin for the two co-ordinate systems are exactly at the same geometrical point of Newton's immobile and absolute space at this specific moment of time. And of course, it should be self-evident that when the points of origin of two co-ordinate systems coincide, all other corresponding points coincide by logical necessity as well. So far so good! Now, let, as specified above by Steve's scenario, the origin of one of the two coordinate systems move with a constant velocity (v) along the positive direction of the co-ordinate x relative to the point of origin of the other co-ordinate system. Steve, please pay close attention to the continuous changing displacement of the origin of the moving co-ordinate system with time! The motion of the point of origin is by the far the most import point of co-ordinates in dealing with any moving co-ordinate system. That is because the precise determination of the instantaneous position of the point of origin of the moving system with respect to the stationary origin of the other system automatically means and necessarily implies the precise determination of every other point of the moving system. Likewise, the failure to determine the instantaneous position of the point of origin of the moving co-ordinate system necessarily leads to the inescapable failure to find out the instantaneous position of every other point of that system. For this important reason, one of the main uses of this great equa­t­ion [x' = x - vt] is to compute and to track the instantaneous position of the point of origin of the moving co-ordinate system with time at each and every instant and from start to finish. Now, once again, pay attention, Steve! Whenever you perform measurements or transformation of measurement from the standpoint of some co-ordinate system, you have to assume all the co-ordinates for the origin of that system to be all zeros; i.e (0, 0, 0). So, here we go! Let's suppose you performed some measurements while the two systems were coincidental and wrote down in your notebook the co-ordinates of three Cartesian points: (5, 0, 0), (11, 0, 0), and (25, 0, 0). Then you decided to move along with one of the two co-ordinate systems in the positive direction of the co-ordinate x at constant speed of 10 co-ordinate units per second. And after one second, you decided once again to freeze everything and take measurements of the same three points in your notebook. What would you find? Well, if everything was done correctly, then you should obtain these new co-ordinates of the three points respectively: (-5, 0, 0), (1, 0, 0), and (15, 0, 0). In other words, the value of (vt), in this case, is equal to 10 co-ordinate units. And all what you have to do to transform these points from the old system to the new system is to subtract 10 co-ordinate units from the values of their co-ordinate x. And that is it; just computing... no measurement.... no headache... Galileo is great... Long live, Equa­t­ion #1 of Galileo....

Just my opinion AAF, you are going from the new coordinate backward, while Steve was going from the old coordinate forward.

AAF: Thank you Steve... For the time being let's put aside the published article, and concentrate instead on the numbers and the scenario above! We have two coincident Cartesian co-ordinate systems. The term 'coincident', within this context, implies that the points of origin for the two co-ordinate systems are exactly at the same geometrical point of Newton's immobile and absolute space at this specific moment of time. And of course, it should be self-evident that when the points of origin of two co-ordinate systems coincide, all other corresponding points coincide by logical necessity as well. So far so good!

S - Yes.

AAF - Now, let, as specified above by Steve's scenario,

S = Actually this is Voigt's scenario, but yes I use it too.

AAF - the origin of one of the two coordinate systems move with a constant velocity (v) along the positive direction of the co-ordinate x relative to the point of origin of the other co-ordinate system. Steve, please pay close attention to the continuous changing displacement of the origin of the moving co-ordinate system with time! The motion of the point of origin is by the far the most import point of co-ordinates in dealing with any moving co-ordinate system. That is because the precise determination of the instantaneous position of the point of origin of the moving system with respect to the stationary origin of the other system automatically means and necessarily implies the precise determination of every other point of the moving system. Likewise, the failure to determine the instantaneous position of the point of origin of the moving co-ordinate system necessarily leads to the inescapable failure to find out the instantaneous position of every other point of that system. For this important reason, one of the main uses of this great equa­t­ion [x' = x - vt] is to compute and to track the instantaneous position of the point of origin of the moving co-ordinate system with time at each and every instant and from start to finish. Now, once again, pay attention, Steve! Whenever you perform measurements or transformation of measurement from the standpoint of some co-ordinate system, you have to assume all the co-ordinates for the origin of that system to be all zeros; i.e (0, 0, 0). So, here we go!

S - We are still good.

AAF - Let's suppose you performed some measurements while the two systems were coincidental and wrote down in your notebook the co-ordinates of three Cartesian points: (5, 0, 0), (11, 0, 0), and (25, 0, 0). Then you decided to move along with one of the two co-ordinate systems in the positive direction

S -Still agree.

AAF -of the co-ordinate x at constant speed of 10 co-ordinate units per second. And after one second, you decided once again to freeze everything and take measurements of the same three points in your notebook. What would you find? Well, if everything was done correctly, then you should obtain these new co-ordinates of the three points respectively: (-5, 0, 0), (1, 0, 0), and (15, 0, 0).

S -Here is where we disagree. You are gonna move right and yet subtract.
Those points would now be wrt the unmoved frame
(15, 0, 0), (21, 0, 0), and (35, 0, 0)

I an gonna move right and add !

AAF - In other words, the value of (vt), in this case, is equal to 10 co-ordinate units. And all what you have to do to transform these points from the old system to the new system is to subtract

S - Not if you went in the positive direction initially.

AAF - 10 co-ordinate units from the values of their co-ordinate x. And that is it; just computing... no measurement.... no headache... Galileo is great... Long live, Equa­t­ion #1 of Galileo.... [linked image]

S - Galileo is great, but my understanding is that it is the Galilean out of tribute and he had nothing to do with the transformations.

Your example and this response may make my contention clearer...?

RECAP When moving right, x' = x - vt is mathematically incorrect,
the correct equation when moving right is x' = x + vt.
Noting that this still applies if distance is substituted for vt.

You are saying what the masses would say, I am saying he zigs when
he should zag.

btw, Thanks for putting the effort and thought into this post of yours, for this discussion.

Bob S: Just my opinion AAF, you are going from the new coordinate backward, while Steve was going from the old coordinate forward.





AAF: If you mean by that the initial Cartesian points (5, 0, 0), (11, 0, 0), & (25, 0, 0) of the stationary system have drifted in the backward direction and transformed to (-5, 0, 0), (1, 0, 0), & (15, 0, 0) of the moving system, then you're right on target, Bob. And the same applies to all the points of the first system upon the transformation to the second system. And that is to be expected and very reasonable, Bob!





Casper: Here is where we disagree. You are gonna move right and yet subtract. Those points would now be wrt the unmoved frame (15, 0, 0), (21, 0, 0), and (35, 0, 0) I an gonna move right and add ! RECAP When moving right, x' = x - vt is mathematically incorrect, the correct equation when moving right is x' = x + vt. Noting that this still applies if distance is substituted for vt.





AAF: Casper; when the velocity (v) of the moving system is along the co-ordinate axis of ( x), only four cases are possible. Let's take a closer look and see where your beloved (x' = x + vt) would apply:

(1) The vector (v) is along the positive direction of the axis of (x):

[A] The case of transforming measurements made in the stationary system to the moving system:

Here, due to the forward motion of the moving system, the Cartesian points must recede and drift backward towards (-x). And hence, this case belongs to the domain of (x' = x - vt).

[B] The case of transforming measurements made in the moving system to the stationary system:

Since this case is the reverse of Case [A], your beloved equation, (x' = x + vt), is obviously the ruler of this domain!


(2) The vector (v) is along the negative direction of the axis of (x):

[A] The case of transforming measurements made in the stationary system to the moving system:

Here, due to the fact that the 'negative of the negative is positive', we just insert the negative velocity (-v) into (x' = x - vt); and voila, your pet equation (x' = x + vt) is born! And it's clearly the ruler of this domain.

[B] The case of transforming measurements made in the moving system to the stationary system:

Since this case is the reverse of Case [A], it must belong to the domain of the equation, (x' = x - vt).


So, Casper; in the end, it's just like the Americas; one half of 'em belongs to Jose; and one half of 'em belongs to Bob!

AAF - due to the forward motion of the moving system, the Cartesian points must recede and drift backward

S - due to the positive direction, the x' related Cartesian points must drift forward.

What you are saying, is the EXACT thing that I am mathematically challenging.

steve

Re: AAF, Anyone dare to weigh in with your opinion ? August 26 2009, 4:31 PM

Bob S: "Just my opinion AAF, you are going from the new coordinate backward, while Steve was going from the old coordinate forward."

AAF: "If you mean by that the initial Cartesian points (5, 0, 0), (11, 0, 0), & (25, 0, 0) of the stationary system have drifted in the backward..."

That is not at all what I meant AAF! The stationary system "x" does not drift backwards it remains stationary, that's why, I'll go out on a limb here and say, it's called STATIONARY!

Steve's point was simple enough even Max could have understood it.

Firstly, if the origin of the "Moving System" moves to the right, so does everything within it. If point P is in the moving system, and to the right of the origin in the "Moving System" when both origins are aligned, it will still be to the right of the origin when the "Moving System" has moved. It will also be further to the right in the "Stationary System" by the same amount as the origin has moved.

Secondly, if point P is in the "Stationary System" the same distance from the origin as the above point, The "Moving System" will find the point to the left of where the point was when the origins were aligned, after the move.

The conventional story, as told by Woldemar Voigt, changes horses in the middle of the stream. He starts off the story as in the first scenario (above), but tells the aftermath as if the point under discussion were in the second scenario.

What say you?

Re: Jose,Anyone dare to weigh in with your opinion ? August 27 2009, 1:50 AM

Jose: "The conventional story, as told by Woldemar Voigt, changes horses in the middle of the stream. He starts off the story as in the first scenario (above), but tells the aftermath as if the point under discussion were in the second scenario.

What say you?"

In spite of what Voight said, Steve asked, "What is the coordinate of the point now, in reference to the unmoved frame?", which clearly, and without any doubt, can only be solved with

x' = x + vt

So say I.

Jose,

There is no point P.

When the two frames are coincident,
there is point x at 11,0
there is point x' at 11,0.

Move the movable grid to the right by 7,
keep the stationary grid, stationary.

X = 11
X' = 18

Jose can you see ?

steve

I can take a P if I want. My point, in your scenario, is that from the alternate viewpoint of "the movable grid," the stationary grid "moves" the opposite way (to the left) and in that situation, 18-7=11. Which is probably why AAF is confused. He must have switched his viewpoint before the dust settled.

I think that Bob understands, but please do not accuse me of putting words in others mouths, pigflew and all, you know.

J -I can take a P if I want.

Not in my sandbox if you please. Just adds confusion.


My point, in your scenario, is that from the alternate viewpoint of "the movable grid," the stationary grid "moves" the opposite way (to the left) and in that situation, 18-7=11.

W -You want to consider the stationary as movable and the movable as stationary? Talk about confusing the thread....please stick with the given.


J - Which is probably why AAF is confused. He must have switched his viewpoint before the dust settled.

W - Hell, your confusing me too....the stationary is the movable and from that viewpoint....that does nothing to clarify at all...and is quite counter-production to any understanding.

J - I think that Bob understands,

S - He does. I do not know if you agree with me or not yet...be clear.
18-7=11. Does this mean you get that the correct answer is 18 and not 4 ?
Does that mean you agree that x' = x - vt is wrong, and that
x' = x + vt is right ?

J - but please do not accuse me of putting words in others mouths,

S -whoa, whoa....you are gonna have to clue me in here...
What words have I put in what other people's mouth please ?
I cannot even imagine...as that is something I just would not do.

J -pigflew and all, you know.

No, I am not accusing you of doing anything.

Jose,

Oh sorry, you were saying that YOU
could not speak for Bob. I get it now.

So....are in with x' = x - vt or are you with the x' = x + vt club ?

steve

AAF: Casper; due to the forward motion of the moving system, the Cartesian points must recede and drift backward towards (-x). And hence, this case belongs to the domain of (x' = x - vt). due to the forward motion of the moving system, the Cartesian points must recede and drift backward




Casper: Due to the positive direction, the x' related Cartesian points must drift forward. What you are saying, is the EXACT thing that I am mathematically challenging.




AAF: Of course, it's the exact thing that you're challenging; Casper!




Bob S: Just my opinion AAF, you are going from the new coordinate backward, while Steve was going from the old coordinate forward.




AAF: If you mean by that the initial Cartesian points (5, 0, 0), (11, 0, 0), & (25, 0, 0) of the stationary system have drifted in the backward direction and transformed to (-5, 0, 0), (1, 0, 0), & (15, 0, 0) of the moving system, then you're right on target, Bob. And the same applies to all the points of the first system upon the transformation to the second system. And that is to be expected and very reasonable, Bob!




Bob S: That is not at all what I meant AAF! The stationary system "x" does not drift backwards it remains stationary, that's why, I'll go out on a limb here and say, it's called STATIONARY! Steve's point was simple enough even Max could have understood it.





AAF: The "stationary system "x"" does drift backwards with respect to the moving co-ordinate system. There can be not doubt about that. Let me give you a concrete example! Suppose that the origin of the stationary system is the city center of Paris. And the Cartesian point is the co-ordinate x of the city center of Rome with respect to the center of Paris; and let's say it's (444, 0, 0). And finally, the center of your car is the origin point of the moving co-ordinate system. Now, Bob, this is the question: As you drive your car from the downtown of Paris towards the downtown of Rome, is the value of the x co-ordinate of Rome decreasing or increasing with respect of the center of your car? Certainly, the value of the x co-ordinate of Rome is decreasing by the minute relative to the moving co-ordinate system of your car. And therefore, you really should use the equa­t­ion (x' = x - vt) in your computations in this case; and just drop very quietly Casper's equa­t­ion on the left-hand side of the highway!

Re: AAF, Anyone dare to weigh in with your opinion ? August 27 2009, 5:29 PM

AAF: "The "stationary system "x"" does drift backwards with respect to the moving co-ordinate system."

No...it does not, x' drifts positive.

AAF: "There can be not doubt about that."

If there were any drift x would not be called "stationary!

AAF: "Let me give you a concrete example!"

Instead of belaboring your examples why not just answer the question Steve asked.

AAF: "Suppose that the origin of the stationary system is the city center of Paris. And the Cartesian point is the co-ordinate x of the city center of Rome with respect to the center of Paris; and let's say it's (444, 0, 0). And finally, the center of your car is the origin point of the moving co-ordinate system. Now, Bob, this is the question: As you drive your car from the downtown of Paris towards the downtown of Rome, is the value of the x co-ordinate of Rome decreasing or increasing with respect of the center of your car? Certainly, the value of the x co-ordinate of Rome is decreasing by the minute relative to the moving co-ordinate system of your car. And therefore, you really should use the equa­t­ion (x' = x - vt) in your computations in this case;..."

Would it not have been easier to just answer the question Steve asked!

AAF: "...and just drop very quietly Casper's equa­t­ion on the left-hand side of the highway!"

Why not drop the equation on the right hand side of the highway as Steve asked.

It seems that AAF's picture now has Three FOR, (Frames of Reference) unless Paris and Rome are in the same frame and immovable wrt (with respect to) each other.

I remember discussing this with minimax in the misty past. I could never get him to understand that within a frame of reference, nothing can move. Only the entire frame can change position, wrt something outside the frame. I couldn't get him to agree that frames are imaginary.

It is going to be easier to have only one origin per frame. I suppose there could be more origins in a frame, but they would have to all be stationary wrt some master origin within the same frame.

Just to please Anonymous, let's say the back of the train corresponds to a mark on the platform. Let's say the back of the train, and this mark on the platform are the origins of the two ordinate systems. Let's say the front of the train and a corresponding mark on the platform are the positions x' and x respectively.

Now, let's say the back of the train has moved to the platform mark that corresponds with where the front of the train was earlier. Now the front of the train has moved further down the platform, and the x' now corresponds to x plus the length of the train. This seems trivial, but that's all there is to it.

This is all Woldemar, or his ghost, Casper, is seeking agreement upon. I have to agree with that.

AAF, the example is that initially, both coordinate systems overlay each other, including x and x'. In your example, to fit the original scheme, the car would have an imaginary extension reaching Rome, at the initial condition.

We can argue that the people on the train see themselves as stationary, and it is the platform that has moved. We can argue that the people on the train see the platform has moved to the left. But this is an immaterial straw argument, since the people there can position themselves so that they consider the platform to have moved to either the right or the left. So, the formula is still x=x'+d for the platform, wrt the train.

AAF: Just drop very quietly Casper's equa­t­ion on the left-hand side of the highway!





Bob S: Why not drop the equation on the right hand side of the highway as Steve asked.




AAF: Why? Because that equa­t­ion is very useful to the people who are driving in the opposite direction towards the city of Paris!

Jose wrote -


It seems that AAF's picture now has Three FOR, (Frames of Reference)

Anyone remember the given ?

Coincident frames have a point at 11,0.
One frame is moved to the right by 7 while the other remains in place.
What is the coordinate of the point now, in reference to the unmoved frame ?

The idea was to say if the answer is 4 or 18.

AAF is it 4 or 18 ?

Just answer MY question please as presented. Pick one,
unless you have another number you think is correct.

How hard can this be ?

Boys, have you come to a conclusion?

Casper: Jose wrote - It seems that AAF's picture now has Three FOR, (Frames of Reference) Anyone remember the given? Coincident frames have a point at 11,0. One frame is moved to the right by 7 while the other remains in place. What is the coordinate of the point now, in reference to the unmoved frame? The idea was to say if the answer is 4 or 18. AAF is it 4 or 18? Just answer MY question please as presented. Pick one, unless you have another number you think is correct. How hard can this be?






AAF: It is not hard at all to answer your question; Casper! The point of coincidence remains in exactly the same place and will remain forever in exactly the same place regardless of what will happen to our two co-ordinate systems in the future. And so, the co-ordinate of the point of coincidence in reference to the unmoved frame is still (11, 0); and in reference to the moved frame is now (4, 0). And that is the answer to your question.

However, I can see clearly your idea; but please don't say I'm putting words in your mouth again; because what I gonna say is not your words, but my words to express how I understand your idea.

Your main idea (as I understand) is to move one of the two systems 7 co-ordinate units to the right. And then we go to the moved system and ask ourselves: Where is the point whose co-ordinate is (11, 0)? And for sure, we shall find one. Every Cartesian co-ordinate system in the entire universe has one. After finding the point whose co-ordinate is (11, 0) in reference to the moved frame, we ask once again: What is the co-ordinate of this point in reference to the unmoved frame? And since, by ask this question, we are actually doing transformations from the moved system to the unmoved system, we must use Casper's equa­t­ion (x' = x + vt). By using this equa­t­ion, we obtain the co-ordinate value (18, 0), just as Woldemar did before us! And that is it... Fair and square.... everybody is happy... Woldemar is happy because his calculations are correct... And we are happy because our Galilean transformations remain intact and still king! Does that answer your question; Casper?

AAF - Your main idea (as I understand) is to move one of the two systems 7 co-ordinate units to the right. And then we go to the moved system and ask ourselves: Where is the point whose co-ordinate is (11, 0)? And for sure, we shall find one. Every Cartesian co-ordinate system in the entire universe has one. After finding the point whose co-ordinate is (11, 0) in reference to the moved frame, we ask once again: What is the co-ordinate of this point in reference to the unmoved frame? And since, by ask this question, we are actually doing transformations from the moved system to the unmoved system, we must use Casper's equa­t­ion (x' = x + vt). By using this equa­t­ion, we obtain the co-ordinate value (18, 0),

S -Good, we are perfect up to right here.
I am happy that you see that I do obtain 18 and not 4, moving to the right.

AAF -just as Woldemar did before us!

S- disagree - Woldemar Voigt would say it is 4, being x - vt.
Confusing, as I was using the name Woldemar here, as you know.
Was it voigt or me that you refer to ?

AAF - And that is it... Fair and square.... everybody is happy... Woldemar is happy because his calculations are correct

S - I am happy when agree with 18, HGowever I am stll not sure if you do, although it appears you do in what you have written above in your last response.

AAF - ... And we are happy because our Galilean transformations remain intact and still king!

S - Not so. How can you say that ? Do you agree, AAF, with what you said
up to point where I say "Good we are perfect up to right here".????

AAF - As in that case, the Galilean transformations do not remain intact.

Does that answer your question; Casper?

SA - Some of it,

AAF - "we are actually doing transformations from the moved system to the unmoved system, we must use Casper's equa­t­ion (x' = x + vt)."

Is that not the essence of what a coordinate transformation should do ?
"doing transformations from the moved system to the unmoved system"

S - It sounds to me that you agree with me here ? (x' = x + vt)

AAF - And we are happy because our Galilean transformations remain intact and still king

S - Yet we still disagree on this. (x' = x - vt)

Ooops ...I misread your statement,

"doing transformations from the moved system to the unmoved system"

I thought you said....
"doing transformations from the unmoved system to the moved system"

and so, I can only imagine we are back at square one.

The transformation is from the unmoved frame, when x' is 11,
to the moved frame, where x' is 18.

In fact, I think this part of why we disagree. Do you think I am
doing transformations from the moved system to the unmoved system,
because I am not.

Indeed, I believe that you and Voigt are
doing transformations from the moved system to the unmoved system.

Which I see as reversed. Now if X' went from 11 to 4, that would be
doing transformations from the moved system to the unmoved system.

Again, sorry for not reading carefully enough.

Casper: Ooops ...I misread your statement, "doing transformations from the moved system to the unmoved system" I thought you said.... "doing transformations from the unmoved system to the moved system" and so, I can only imagine we are back at square one. The transformation is from the unmoved frame, when x' is 11, to the moved frame, where x' is 18. In fact, I think this part of why we disagree. Do you think I am doing transformations from the moved system to the unmoved system, because I am not. Indeed, I believe that you and Voigt are doing transformations from the moved system to the unmoved system. Which I see as reversed. Now if X' went from 11 to 4, that would be doing transformations from the moved system to the unmoved system. Again, sorry for not reading carefully enough.





AAF: Yes, Casper; you're doing transformations from the moved system to the unmoved system. Even Jose & Bob can hardly disagree with this! Look one more time at this equa­t­ion: [x' = x - vt]. Written this way, this equa­t­ion transforms any Cartesian point in the unmoved system to the moved system. But when we rearrange this equa­t­ion and solve for (x), we obtain this equa­t­ion: [x = x' + vt]. This equa­t­ion transforms any Cartesian point in the moved system to the unmoved system. Furthermore this last equa­t­ion is exactly the same as this [[x' = x + vt]. Why is the notation different? It's because contemporary physicists have to rewrite [[x = x' + vt] and to change it to [x' = x + vt] to satisfy the Einsteinian Condition that we must always treat the system we transform from as stationary and the system we transform to as moving. If we changed the notation this way to comply with such a silly condition during an exam of pure mathematics, we would probably get a big red 'X'! So, Casper, your [x' = x + vt] gives exactly the same results as this [x = x' + vt]. and the two are the one and the same equa­t­ion.

Re: AAF, Anyone dare to weigh in with your opinion ? August 28 2009, 10:06 PM

AAF: "Yes, Casper; you're doing transformations from the moved system to the unmoved system. Even Jose & Bob can hardly disagree with this!"

Well, I do not agree AAF, and, unless you are addressing me or asking me I wish you would stop assuming what I do or do not agree with, thank you!

I said you can hardly disagree; Bob!


Anyway, Woldemar seems to be satisfied by my ANWSER to his question.


Do you think this ANSWER of mine is satisfactory, Bob?

Re: AAF, Anyone dare to weigh in with your opinion ? August 29 2009, 2:50 PM

AAF: "I said you can hardly disagree; Bob!"

And I said I did not agree with the quoted sentence, so what is your point!

AAF: "Anyway, Woldemar seems to be satisfied by my ANWSER to his question."

So it would seem!

AAF: "Do you think this ANSWER of mine is satisfactory, Bob?"

Without knowing what the question was that you are referring to I can not answer your question.

AAF -

To avoid confusion AAF,

S - Please only call me steve in the future( aka casper aka woldemar .
Since I am opposing Woldemar's Voigt's math...when you just refer to
woldemar...I have not a clue if you are now saying x' = x - vt or x + vt.


You did not answer me at all. I am certainly NOT NOT NOT
in agreement with this key statement below. I object to you saying that I agree carte blanche, and then cloaking your own conclusions. This is crap. You are doing it again....speaking falsely for others. You implied that I agreed with the following statement....when you said

""Anyway, Woldemar seems to be satisfied by my ANWSER to his question."

Are you f**king kidding me....knock it off. I am getting annoyed by this ploy of yours.

AAF - And we are happy because our Galilean transformations remain intact and still king


S - Please answer the three questions below. ( as asked previously )


1 Is that not the essence of what a coordinate transformation should do ?
"doing transformations from the moved system to the unmoved system"

answer ? yes or no please

2 It sounds to me that you agree with me here ? (x' = x + vt)

answer ? yes or no please

AAF - And we are happy because our Galilean transformations remain intact and still king

S -
3 Yet we still disagree on this ? (x' = x - vt)

answer? yes or no please

We continue to go round in circles. You seem to agree yet disagree at the same time. That is why I asked you these 3 questions.
Can you answer them or not ?

AAF

You have just pushed me over the edge, just as Anonymous had done before.

Bantering with you is too frustrating. There are those that agree with me here, and those that do not. I have already explained my ass off. There
is no more for me to say here on this subject. Sorry we cannot come to agreement AAF. So, I am afraid it is time for my stay at GSJ Forum to be over. You appear to be the only hold-out left to even attempt to debug me,
in this thread.

No hard feelings...it is just the appropriate time now, for me to move on.
This will be my last post here. Noting that I will continue to read the thread from time to time. Thus you may have the last word, as many times as you wish...assuming then, that only my math work is under scrutiny.

steve

Steve: To avoid confusion AAF, Please only call me steve in the future( aka casper aka woldemar . Since I am opposing Woldemar's Voigt's math...when you just refer to woldemar...I have not a clue if you are now saying x' = x - vt or x + vt. You did not answer me at all. I am certainly NOT NOT NOT in agreement with this key statement below. I object to you saying that I agree carte blanche, and then cloaking your own conclusions. This is crap. You are doing it again....speaking falsely for others. You implied that I agreed with the following statement....when you said





AAF: What I've posted so far about the [x' = x - vt & x' = x + vt] is crystal-clear and can't be misunderstand by anybody; Steve! I understand those equations are at the heart of your thesis and that may make it difficult for you to see or comprehend my criticism in this regard; but eventually you'll see the truth of it. Furthermore, you have no claim whatsoever for banning people from interpreting your ideas or predicting your future behavior based on what you say. Bob; do you think this ANSWER of mine is satisfactory, Bob?





Bob S: Without knowing what the question was that you are referring to I can not answer your question.





AAF: Is Steve satisfied by my previous ANSWER to his question?





Steve: Are you f**king kidding me....knock it off. I am getting annoyed by this ploy of yours.





AAF I thought you don't like this kind of talk, Steve; old Ted wasn't so bad after all! But we should be happy that our Galilean transformations remain intact and still king.





Steve: Please answer the three questions below. ( as asked previously ) 1 Is that not the essence of what a coordinate transformation should do? "doing transformations from the moved system to the unmoved system" answer ? yes or no please





AAF: The answer is no. The Galilean equa­t­ion that you're opposing is only about the transformation from the unmoved system to the moved system. You can't just change the domain of its applicability on your own, and then claim that the equa­t­ion is wrong. Stick to the domain of the equa­t­ion; Steve!




Steve: 2 It sounds to me that you agree with me here? (x' = x + vt) answer? yes or no please





AAF: I explained to you many times that this equa­t­ion (x' = x + vt) is the Galilean equation for the transformation from the moved system to the unmoved system.





Steve: 3 Yet we still disagree on this ? (x' = x - vt) answer? yes or no please





AAF: I also pointed out to you many times that this equa­t­ion (x' = x - vt) is the Galilean equation for the transformation from the unmoved system to the moved system.





Steve: We continue to go round in circles. You seem to agree yet disagree at the same time. That is why I asked you these 3 questions. Can you answer them or not?





AAF: No, Steve; we are not going in circles; we are going forwards; and all your questions have been answered! In short: Equa­t­ion (x' = x - vt) and Equa­t­ion (x' = x + vt) don't contradict each other. They complement each other perfectly.. . They complement each other; Steve.. . They complement each other perfectly.. .

Re: AAF, Anyone dare to weigh in with your opinion? August 30 2009, 12:00 AM

AAF: "Is Steve satisfied by my previous ANSWER to his question?"

For me, satisfaction is a work in progress, as to Steve's satisfaction with your answers, I am not concerned.

Steve: AAF; You have just pushed me over the edge, just as Anonymous had done before. Bantering with you is too frustrating. There are those that agree with me here, and those that do not. I have already explained my ass off. There is no more for me to say here on this subject. Sorry we cannot come to agreement AAF. So, I am afraid it is time for my stay at GSJ Forum to be over. You appear to be the only hold-out left to even attempt to debug me, in this thread. No hard feelings...it is just the appropriate time now, for me to move on. This will be my last post here. Noting that I will continue to read the thread from time to time. Thus you may have the last word, as many times as you wish...assuming then, that only my math work is under scrutiny.







AAF: Steve; you've been told right from the beginning that real and fatal objections to an argument whatever it's are the most dreaded, but you wouldn't listen! And yet no matter how painful and frustrating the experience can be, it is always better to discover the weakness of your argument and the flaws of your theory sooner than latter. And in the end, the truth and only the truth should matter. I wish you success and more robust argumentation in the future... God bless...

Jose: It seems that AAF's picture now has Three FOR, (Frames of Reference) unless Paris and Rome are in the same frame and immovable wrt (with respect to) each other. I remember discussing this with minimax in the misty past. I could never get him to understand that within a frame of reference, nothing can move. Only the entire frame can change position, wrt something outside the frame. I couldn't get him to agree that frames are imaginary. It is going to be easier to have only one origin per frame. I suppose there could be more origins in a frame, but they would have to all be stationary wrt some master origin within the same frame. Just to please Anonymous, let's say the back of the train corresponds to a mark on the platform. Let's say the back of the train, and this mark on the platform are the origins of the two ordinate systems. Let's say the front of the train and a corresponding mark on the platform are the positions x' and x respectively. Now, let's say the back of the train has moved to the platform mark that corresponds with where the front of the train was earlier. Now the front of the train has moved further down the platform, and the x' now corresponds to x plus the length of the train. This seems trivial, but that's all there is to it. This is all Woldemar, or his ghost, Casper, is seeking agreement upon. I have to agree with that.









AAF: Of course, Paris & Rome are in the same frame and immovable with respect to each other! Paris, in the fore-mentioned example, is the point of origin of the co-ordinate system with its three co-ordinates set to zero; i.e. (0, 0, 0). And Rome is just one Cartesian point whose co-ordinate values (444, 0, 0). It follows, therefore, that there is only one frame, here, not 'three FOR'; Jose. As for the rest of your post, it's already done with and disposed of; see the last three posts above.... QED.....

Re: AAF, Anyone dare to weigh in with your opinion? August 30 2009, 12:00 AM

AAF: "No, Steve; we are not going in circles; we are going forwards; and all your questions have been answered! In short: Equa­t­ion (x' = x - vt) and Equa­t­ion (x' = x + vt) don't contradict each other."

I don't see where Stave was trying to say the 2 equations "contradict" each other.

His question was; "What is the coordinate of the point now, in reference to the unmoved frame?"
The coordinate point now x'.
In reference to the original point x.
x' moved (past tense) at a given velocity v, for a given time t.
The correct, and proper, equation to find for x' is x'= x +vt, to verify the findings the equation would (and should) be x = x'-vt.
Addition is used to find.
Subtraction is used to verify.
Voigt and Einstein et al are wrong...QED QED!

AAF: No, Steve; we are not going in circles; we are going forwards; and all your questions have been answered! In short: Equa­t­ion (x' = x - vt) and Equa­t­ion (x' = x + vt) don't contradict each other. They complement each other perfectly.. . They complement each other; Steve.. . They complement each other perfectly.. .






Bob S: I don't see where Stave was trying to say the 2 equations "contradict" each other. His question was; "What is the coordinate of the point now, in reference to the unmoved frame?" The coordinate point now x'. In reference to the original point x. x' moved (past tense) at a given velocity v, for a given time t. The correct, and proper, equation to find for x' is x'= x +vt, to verify the findings the equation would (and should) be x = x'-vt. Addition is used to find. Subtraction is used to verify. Voigt and Einstein et al are wrong...QED QED!






AAF: Galileo's transformations are wrong? NO! Einstein's transformations are wrong? YES! Steve's analysis, therefore, is half-wrong & half-right... QED... QED....

"Henri Poincare discovered that Lorentz's equations could be reversed in a very unusual way, just using the plus sign instead of the minus sign, actually the same way as Descartes did for Galileo's Relativity principle.

Galileo: x' = x + v t

Reversed: x = x' - v t


Poincare: x' = gamma * (x + beta * t)

Reversed: x = gamma * (x' - beta * t')


Mathematically, x should rather be recovered like this, using t instead of t':

x = x' / gamma beta * t



Lorentz borrowed his equations from Woldemar Voigt, whose goal in 1887 was to cancel the Doppler effect on Maxwell's equations":





http://www.glafreniere.com/sa_Doppler.htm

"Houston we have a problem -
Relativity says 4 and reality says 18. "

Mind your own business.

Re: Anonym, Anyone dare to weigh in with your opinion? September 1 2009, 1:52 AM

Anonym:
""Houston we have a problem -
Relativity says 4 and reality says 18. "

Mind your own business."

Son...if it is posted on this board it is our business!

Here is the problem with this image from your link, AAF,



The "spherical" wave fronts pictured are not really circular. Can't you see that the foci are not in the center of the "circle?" In other words, the wave fronts are ellipsoid. They will be oblate if the speed of light is determined by the ether. If the speed is determined by particles traveling at c+v, the ellipsoid will be elongate.

Otherwise, the transverse signal will appear to be "Doppler shifted," when in reality it is not.

(I forgot to post my name to the above post.)

Re: AAF, Anyone dare to weigh in with your opinion ? August 31 2009, 4:09 PM

Bob S: "I don't see where Stave was trying to say the 2 equations "contradict" each other. His question was; "What is the coordinate of the point now, in reference to the unmoved frame?" The coordinate point now x'. In reference to the original point x. x' moved (past tense) at a given velocity v, for a given time t. The correct, and proper, equation to find for x' is x'= x +vt, to verify the findings the equation would (and should) be x = x'-vt. Addition is used to find. Subtraction is used to verify. Voigt and Einstein et al are wrong...QED QED!"

AAF: "Galileo's transformations are wrong? NO! Einstein's transformations are wrong? YES! Steve's analysis, therefore, is half-wrong & half-right... QED... QED...."

I didn't see where Stave was trying to say the 2 equations "contradict" each other because he did not make a contradictory statement.

If, as you say, Steve's analysis is half wrong and half right that would make your analysis half right and half wrong...right AAF? But more importantly, I am not wrong at all!

Jose: Here is the problem with this image from your link, AAF, The "spherical" wave fronts pictured are not really circular. Can't you see that the foci are not in the center of the "circle?" In other words, the wave fronts are ellipsoid. They will be oblate if the speed of light is determined by the ether. If the speed is determined by particles traveling at c+v, the ellipsoid will be elongate. Otherwise, the transverse signal will appear to be "Doppler shifted," when in reality it is not.







AAF: Below is the diagram, Jose! As you can see, the author wants to demonstrate that spherical waves of light become spheroid or ellipsoid because of the Doppler effect.


The Doppler effect explains Relativity:




The interesting historical fact, here, is that Woldemar Voigt derived his transformations and wrote his article in German in 1887. This implies that Einstein saw it and read it and was aware of the transformations in it. It follows, therefore, that wasn't quite miraculous year for him to come up with Special Relativity in 1905; right, Jose? Even Aaron could have come up with it, if he were to read Voigt's article in 1887; or even Cincirob could have done it; and it wasn't difficult!

"This implies that Einstein saw it and read it and was aware of the transformations in it. It follows, therefore, that wasn't quite miraculous year for him to come up with Special Relativity in 1905; right, Jose? Even Aaron could have come up with it, if he were to read Voigt's article in 1887; or even Cincirob could have done it; and it wasn't difficult! "

Yet so many people still don't understand Einstein's 1905 paper, specially in this forum!

In addition, the more I think to it, the more I consider Einstein's 1905 paper as a didactic paper.
What is explained is extremely simple but totally new: the idea that space and time (I weight my words!) has to be constructed in each frame of reference was completely new.
The consequences are described by Einstein with such a didactic care that today it is nearly unbearable. In the end it is all about explaining what a (2x2) linear transformation means as well as each coefficient involved. Yet still today, many people have not been able to understand that and remain trapped in dumb discussions and paradoxes, specially in this forum!

AAF: The interesting historical fact, here, that Woldemar Voigt derived his transformations and wrote his article in German in 1887. This implies that Einstein saw it and read it and was aware of the transformations in it. It follows, therefore, that wasn't quite miraculous year for him to come up with Special Relativity in 1905; right, Jose? Even Aaron could have come up with it, if he were to read Voigt's article in 1887; or even Cincirob could have done it; and it is not difficult!





Anon: Yet so many people still don't understand Einstein's 1905 paper, specially in this forum! In addition, the more I think to it, the more I consider Einstein's 1905 paper as a didactic paper. What is explained is extremely simple but totally new: the idea that space and time (I weight my words!) has to be constructed in each frame of reference was completely new. The consequences are described by Einstein with such a didactic care that today it is nearly unbearable. In the end it is all about explaining what a (2x2) linear transformation means as well as each coefficient involved. Yet still today, many people have not been able to understand that and remain trapped in dumb discussions and paradoxes, specially in this forum!






AAF: Yes, it's didactic and verbose; Anon! But why did Einstein have to make his 1905 paper tediously didactic and verbose? It's because he was trying to convince his fellow physicists to accept as scientific and logical something that is neither scientific nor logical. Einstein's didacticism, here, is very similar to Hegel's didacticism regarding his illogical Hegels Science of Logic . In a nutshell, Einstein's idea of elastic space & time is extremely illogical & absurd; and didacticism is the only way to convince others of its viability. That is the main reason; Anon. But there can be no good reason to explain away why very smart individuals, like yourself, are unable to spot the colossal logical defect at the heart of Einstein's theory of Relativity. That is the big mystery, Anon... Why... Why... Why....

The only defect aspect is your brain, AAF.
You are sooo stooopid.

Iggy liggety strike again.

AAF: Galileo's transformations are wrong? NO! Einstein's transformations are wrong? YES! Steve's analysis, therefore, is half-wrong & half-right....







Bob S: I didn't see where Stave was trying to say the 2 equations "contradict" each other because he did not make a contradictory statement. If, as you say, Steve's analysis is half wrong and half right that would make your analysis half right and half wrong...right AAF? But more importantly, I am not wrong at all!







AAF: Don't see where; Bob? It's right smack in his 'Hey-Houston' S.O.S post! There, Steve asserts that the correct sign is (+), and that the equa­t­ion with (-) is wrong....

Re: AAF, Anyone dare to weigh in with your opinion ? September 2 2009, 3:54 PM

AAF: "Don't see where; Bob? It's right smack in his 'Hey-Houston' S.O.S post! There, Steve asserts that the correct sign is (+), and that the equa­t­ion with (-) is wrong...."

Steve did not say the equation with (-) is wrong, he said the equation with (-) is being "used" wrong (read improperly) and he made that point quite clear in the opening paragraph of the paper he referencedhttp://www.wbabin.net/physics/waterman12.pdf you know, the one he asked if anyone had (read through), apparently you did not!

I tried explaining that to you in my post on August 31, 2009 @ 10:11 AM but then, you don't need to take the time to understand...do you AAF?

Steve was not wrong, you misunderstood!

AAF: The interesting historical fact, here, is that Woldemar Voigt derived his transformations and wrote his article in German in 1887. This implies that Einstein saw it and read it and was aware of the transformations in it. It follows, therefore, that wasn't quite miraculous year for him to come up with Special Relativity in 1905; right, Jose? Aaron could have come up with it, if he were to read Voigt's article in 1887; even Cincirob could have done it; and it is not difficult!





Anon: Yet so many people still don't understand Einstein's 1905 paper, specially in this forum! In addition, the more I think to it, the more I consider Einstein's 1905 paper as a didactic paper. What is explained is extremely simple but totally new: the idea that space and time (I weight my words!) has to be constructed in each frame of reference was completely new. The consequences are described by Einstein with such a didactic care that today it is nearly unbearable. In the end it is all about explaining what a (2x2) linear transformation means as well as each coefficient involved. Yet still today, many people have not been able to understand that and remain trapped in dumb discussions and paradoxes, specially in this forum!





AAF: Yes, it's didactic and verbose; Anon! But why did Einstein have to make his 1905 paper tediously didactic and verbose? It's because he was trying to convince his fellow physicists to accept as scientific and logical something that is neither scientific nor logical. Einstein's didacticism, here, is very similar to Hegel's didacticism regarding his illogical Hegels Science of Logic . In a nutshell, Einstein's idea of elastic space & time is extremely illogical; and didacticism is the only way to convince other of its viability. That is the main reason; Anon. But there can be no good reason to explain away very smart individuals, like yourself, are unable to spot the colossal logical defect at the heart of Einstein's theory of Relativity. That is the big mystery, Anon... Why... Why... Why....





Idiot Ted: The only defect aspect is your brain, AAF. You are sooo stooopid.




Jose: Iggy liggety strike again.




AAF: Jose; Moron Ted (aka the Idiot of Santa Clara) does not strike... Moron Ted can't strike... Imbecile Ted does not know how to strike.... Retarded Ted only lays eggs!

Ah, I,ve seen him before!

AAF: Galileo's transformations are wrong? NO! Einstein's transformations are wrong? YES! Steve's analysis, therefore, is half-wrong & half-right....





Bob S: I didn't see where Stave was trying to say the 2 equations "contradict" each other because he did not make a contradictory statement. If, as you say, Steve's analysis is half wrong and half right that would make your analysis half right and half wrong...right AAF? But more importantly, I am not wrong at all!




AAF: Don't see where; Bob? It's right smack in his 'Hey-Houston' S.O.S post! There, Steve asserts that the correct sign is (+), and that the equa­t­ion with (-) is wrong....




Bob S: Steve did not say the equation with (-) is wrong, he said the equation with (-) is being "used" wrong (read improperly) and he made that point quite clear in the opening paragraph of the paper he referencedhttp://www.wbabin.net/physics/waterman12.pdf you know, the one he asked if anyone had (read through), apparently you did not! I tried explaining that to you in my post on August 31, 2009 @ 10:11 AM but then, you don't need to take the time to understand...do you AAF? Steve was not wrong, you misunderstood!




AAF: So, Bob; do you think that Steve was making a lot of ado about absolutely nothing?

Re: AAF, Anyone dare to weigh in with your opinion ? September 3 2009, 4:09 PM

AAF: "So, Bob; do you think that Steve was making a lot of ado about absolutely nothing?"

You can't be serious AAF, your participation in the string was more doo...doo than Steve's work was ado!

Why could you not just admit to your misunderstanding and be done with it?

But, to answer your question AAF, as "I" always try and do; I gave Steve my opinion of his paper back on August 24, 2009 @ 1:43 PM, the 22nd response if memory serves me right! I would have responded sooner back then but I wanted to take the time to understand his paper.

AAF: So, Bob; do you think that Steve was making a lot of ado about absolutely nothing?







Bob S: You can't be serious AAF, your participation in the string was more doo...doo than Steve's work was ado! Why could you not just admit to your misunderstanding and be done with it? But, to answer your question AAF, as "I" always try and do; I gave Steve my opinion of his paper back on August 24, 2009 @ 1:43 PM, the 22nd response if memory serves me right! I would have responded sooner back then but I wanted to take the time to understand his paper.






AAF: I'm very serious, Bob! Surely, Steve knows very well that I understand his STRING very well. As a matter of fact, Steve has never encountered any critic of his THESIS more serious than I am... I'm very sure of that.... And you should not forget that I'm the one who has defeated the most talkative BAG of HUMBUG named 'Cinci' and forced him to run away forever.... But if you insist; Bob, would you explain to me Steve's IDEA as you understand it?

Re: AAF< Anyone dare to weigh in with your opinion ? September 5 2009, 6:26 PM <br>
AAF: "I'm very serious, Bob! ... Surely, Steve knows very well that I understand his STRING very well."

I, very seriously, doubt that AAF!

AAF: "As a matter of fact, Steve has never encountered any critic of his THESIS more serious than I am... I'm very sure of that...."

If you call doo...doo serious AAF, then maybe!

AAF: "And you should not forget that I'm the one who has defeated the most talkative BAG of HUMBUG named 'Cinci' and forced him to run away forever...."

No you did not AAF! You may have driven Steve off the board but not Cinci AAF! and, you didn't use reasoned argument to do it!, just doo...doo AAF! Now, it is you who is making much ado about doo...doo AAF!

AAF: "But if you insist; Bob, would you explain to me Steve's IDEA as you understand it?"

I did explain it to you once AAF, and, I even cited you back to that explanation. If you don't know how to use the scroll feature or the find feature that's your bad...AAF!

Stay tuned for the killer video!

AAF: I'm very serious, Bob! ... Surely, Steve knows very well that I understand his STRING very well.




Bob S: I, very seriously, doubt that AAF!




AAF: You're wrong, Bob! As a matter of fact, Steve has never encountered any critic of his THESIS more serious and more dangerous than I am... I'm very sure of that....




Bob S: If you call doo...doo serious AAF, then maybe!




AAF: 'Doo...doo' is much better than 'undoo...undoo'; Bob! And you should not forget that I'm the one who has defeated the most talkative BAG of HUMBUG named 'Cinci' and forced him to put his tale between his legs and to run away forever....




Bob S: No you did not AAF! You may have driven Steve off the board but not Cinci AAF! and, you didn't use reasoned argument to do it!, just doo...doo AAF! Now, it is you who is making much ado about doo...doo AAF!





AAF: I drove off & away Steve & Cincirob & Aaron. Let me summarize to you my annual achievements, Bob! I started this year by demolishing and re-demolishing Cincirob's derivation of the Lorentz transformations. Later on, I demolished Bob's thesis against Newton's Shell theorem. The demolition continues, and Aaron's fabulous Boson model was destroyed next. Then Steve came along and challenged everybody to bring it on and find something wrong with his anti-Galileo proposition. I got very serious and within few sunny days of summer I demolished his proposition. And I'm now more than ready to re-demolish it one more time, if you insist. Bob, would you explain to me Steve's IDEA as you understand it?




Bob S: I did explain it to you once AAF, and, I even cited you back to that explanation. If you don't know how to use the scroll feature or the find feature that's your bad...AAF!




AAF: Bob; that would not work! Be helpful & serious and explain to us Steve's idea according to your understanding of it....

Re: AAF, Anyone dare to weigh in with your opinion ? September 7 2009, 12:11 AM

AAF: "I drove off & away Steve & Cincirob & Aaron. Let me summarize to you my annual achievements, Bob! I started this year by demolishing and re-demolishing Cincirob's derivation of the Lorentz transformations. Later on, I demolished Bob's thesis against Newton's Shell theorem. The demolition continues, and Aaron's fabulous Boson model was destroyed next. Then Steve came along and challenged everybody to bring it on and find something wrong with his anti-Galileo proposition. I got very serious and within few sunny days of summer I demolished his proposition."

Your accomplishments speak for themselves, your logic is unassailable, what else can I say? nothing!

Watch your back, AAF, Steve Woldemar is about to rise again.

Jose; are you sure?

But Cinci would not....







Well, thank you, Bob!

Hi folks!


Im a bit busy cloning OS Hard Drives.

And I just want to say that I miss Cement Head ....
And also, I miss Bob ... Not to mention Steve ....

I guess Cinci and Cement Head couldn't take their own medicine. They must have decided to join Minimax's seminars. I think Bob is just taking a breather. Steve checks in once in a while, but doesn't see anything on which to comment.

Hi Jose;

I'm afraid CH has got the SWINE FLU!

There is absolutely no trace of him anywhere.