Lets start a fresh looking at the maths of SR:

Standard SR maths is as follows-

SR has it that the transformation between the two equations

x^2 - c^2t^2 = 0 (1)
x' ^2 - c^2 t' ^2 = 0 (2)

is the Lorentz transformation

(this is considering the case along x and x' direction and treating y= y' =z = z' =0)

Lorentz transformations is:

t' = gamma (t - vx/c^2) (3)
x' = gamma (x - vt) (4)

Let us accept that for the moment, and look at the issue of discovering what gamma equals.

if we subst (3) and (4) into (2) we get equation (1) as required.


Now let us try using 1/gamma instead of gamma, so that we have:

t' = (1/gamma) (t - vx/c^2) (3b)
x' = (1/gamma) (x - vt) (4b)

i.e. subst (3b) and (4b) into (2) we also get equation (1).


So we can conclude gamma = 1/gamma therefore equals "1."


Its strange that most SR texts do their maths wrong and get gamma equal to something other than "1." But SRists were never known for being any good at maths.


Roger

c.RJAnderton2009

You call that a fresh look?
You call that standard SR maths?
Probably you need a refresh in maths?

Let's start: assume x=2 and t=2, clearly you have:

x²-t²=0

but you also have:

(13x)² - (13t)² = 0

as well as

(17x)² - (17t)² = 0

and -of course- this does not at all implies that 13=17 as you would pretend.

So take the time to refresh your maths and see you again in a few weeks for your next rambling.

Scientia ac Labore

Death of laughter

Let's start: assume x=2 and t=2, clearly you have:

x²-t²=0



**** thats taking c = 1
the Lorentz then are t' = gamma(t-vx)
x' = gamma (x - vt)


but you also have:

(13x)² - (13t)² = 0

**** you missed out the dashes that would be x' = 13x, t' = 13t

Substituting (3) and (4) into equation (2) yields:

gamma²(1-v²/c²) (x²-c²t²) = 0

Doing that for gamma=13 and for gamma=17 leads to:

13²(1-v²/c²) (x²-c²t²) = 0
17²(1-v²/c²) (x²-c²t²) = 0

and this does not imply that 13=17, as you seem to believe.

The transformation (3-4) just implies that (x²-c²t²)=0 since (x'²-c²t'²)=0.
And this is just what the transformation is designed to.

You need to understand that your restricted constraints just leave gamma undertermined.
However, assuming that the same transformation must apply forward and backward by just substituting v by -v will lead you the century-old gamma factor. Indeed, this implies that:

gamma²(1-v²/c²) = 1

whence the final result.

Substituting (3) and (4) into equation (2) yields:

gamma²(1-v²/c²) (x²-c²t²) = 0

Doing that for gamma=13 and for gamma=17 leads to:

13²(1-v²/c²) (x²-c²t²) = 0
17²(1-v²/c²) (x²-c²t²) = 0

*** you did the rest wrong

for 13²(1-v²/c²) (x²-c²t²) = 0

divide through by 13²(1-v²/c²) gives (x²-c²t²) = 0

and for 17²(1-v²/c²) (x²-c²t²) = 0

divide through by 17²(1-v²/c²) gives (x²-c²t²) = 0

nothing about equating 13 to 17

"nothing about equating 13 to 17 "

And similarly nothing about gamma = 1/gamma .

that last bit was wrong; go back and learn maths instead of drivel

that's how you answer when you are trapped in your muddy logic

To illustrate the stupidity of Roger argument, let us use logic in the same way on another problem:

assume w² - 3 w + 2 = 0 (1)

Observe that w = 1 satisfies equation (1)
observe also that w = 2 statisfies equation (1)

therefore, according to roger, 1 = 2

_________________________________________________

Roger has not yet understood that his equations (1,2) simply do not determine any choice of gamma.
If only equation (1,2) are taken into account (compatibility), then any factor gamma satisfies the compatibility condition.

He has of course no hope to understanding what the other condition is and why it is needed.
Therefore his next argument will be that this additional condition is arbitrary.

But let's wait first till he understands the first step.

no, you are trapped in nonsense.


you have:


(13x)² - (13t)² = 0

as well as

(17x)² - (17t)² = 0


and so you are asking for the dashed frame to have its time and length sqrt (13) and sqrt (17) times greater than the undashed frame, which is nonsense because 13 does not equal 17.

Whereas I have:


t' = gamma (t - vx/c^2) (3)
x' = gamma (x - vt) (4)

and

t' = (1/gamma) (t - vx/c^2) (3b)
x' = (1/gamma) (x - vt) (4b)

both as solution to (1) - (2).

By the first we have the dashed frame gamma times greater than the time and length of the undashed frame, and by the second we have it 1/gamma greater.

Clearly it cannot be both gamma times and 1/gamma else its like saying 13 = 17, so the requirement is that to make sense gamma must equal 1/gamma.

So I am looking for a gamma that has the property gamma = 1/gamma

and find it is gamma = 1,

You said it yourself:

"both as solution to (1) - (2)" (you)

Therefore you are in fact introducing an extra hypothesis: gamma = 1.
instead of proving its need based on equations (1) and (2)

You are right that an extra hypothesis is needed.
You are wrong to believe that "obvioulsy it must be gamma = 1".
And you are doubly wrong to believe you have deduced that from a mathematical contradiction.
You have actually contradicted yourself since you admitted:

"both as solution to (1) - (2)" (you)

But a little thinking should prove you easily that assuming gamma = 1 is untenable.
I have already explained you that at length:

Transforming from frame R to frame R' with a velocity v,
and then transforming back from R' to R with velocity -v,
must bring everything back.
Assuming gamma = 1 does not have this property.
Here is the calculation:

t' = t - v x / c^2 . . . . . . . . . from R to R' velocity v
x' = x - v t

t'' = t' + v x' / c^2 . . . . . . . . from R' to R velocity -v
x'' = x' + v t'

yields

t'' = t (1 - v²/c²)
x'' = x (1 - v²/c²)

Therefore, with your assumption that gamma = 1, a forth and back transformation leaves you with a residual time and space compaction.
Therefore only the choice gamma = 1/sqrt(1 - v²/c²) is consistent.
____________________________________________________________
How much time have you now lost on this elementary question?

you are wrong, the way you do it requires that 13 =17

we have to introduce the requirement that gamma = 1/gamma or else the dashed frame is 13 times more than the undashed frame and 17 times or any number of times, so the requirement is that gamma = 1/gamma

1/sqrt(1 - v²/c²) does not have the required property, so is inconsistent; this is elementary maths that you are confused about.

whereas with this supposed claim---

t' = t - v x / c^2 . . . . . . . . . from R to R' velocity v
x' = x - v t

t'' = t' + v x' / c^2 . . . . . . . . from R' to R velocity -v
x'' = x' + v t'


you know it is false because you introduced extra dashes for the second set of equations so it is not a transformation from a frame with one dash as you claim, but from two dashes. So what you are trying to do is add extra errors.

whereas with this supposed claim---

t' = t - v x / c^2 . . . . . . . . . from R to R' velocity v
x' = x - v t

t'' = t' + v x' / c^2 . . . . . . . . from R' to R velocity -v
x'' = x' + v t'


you know it is false because you introduced extra dashes for the second set of equations so it is not a transformation from a frame with one dash as you claim, but from two dashes. So what you are trying to do is add extra errors.

_________________________________________________________________________________________________

I introduced the double dashes simply because I performed two transformations subsequently to illustrate where you fail. This is simply a counting notation. I did so for the clarity.
I considered then a special case: I transformed first from R to R' and subsequently from R' to R.
The combination of these two transformations should bring back the original coordinates.
Therefore we should have:

t'' = t
x'' = x

Your transformation is not able to satisfy this simple requirement.
Instead you get this result:

t'' = t (1 - v²/c²)
x'' = x (1 - v²/c²)

This result is obviously an absurdity.
This is like translating "chaise" in french to "chair" in english and then translating back "chair" from english to "table" in french.

Roger, you should understand that the two sets of coordinates (x,t) and (x',t') are simply the vocabulary used by observer in two different frames of reference. The translation process should be reliable and therefore you should at least check the forth and back translation process.

Additonal comment:

As english is not my native language, I sometimes do perform this forth and back translation process using online dictionaries. This is a useful check that there is no misunderstanding in the choice of a translation.
Not doing so could sometimes leak very funny translations.

Cleary your dictionary does not pass the test as it introduces a distortion during the process.
The distortion appears in a factor (1-v²/c²) that I have explained you already.

moron-As english is not my native language, I sometimes do perform this forth and back translation process using online dictionaries.

And it shows, not do you talk bull**** in the language of maths, you talk it in English. Learn English and maths you moronic turd.

moron -that the two sets of coordinates (x,t) and (x',t') are simply the vocabulary used by observer in two different frames of reference. The translation

No, they are not you are just be absurd. There are two observers A and B; one observer has undashed coordinates and observes dashed coordinates of the other.

yes I am a turd because I am being absurd

Glad to hear you admit it. So you are just messing around by talking nonsense. It makes a bit of a mockery of this forum though for us others trying to be serious.

"There are two observers A and B; one observer has undashed coordinates and observes dashed coordinates of the other. " (roger, JR, Anderton, and the like)
_____________________________

The coordinates (x,t) and (x',t') are how these observers label the events they observe in their own frame of reference.
For example, a spark located at (x,t) by observer A will be located at (x',t') by observer B.
Both observer agree that they observed the same spark but they gave it different coordinates.

The transformation from one frame to another is -literally speaking- a dictionary that translates the labels (or coordinates) in the frame of one observer to the labels in the frame of the other observer.

Therefore, translating from the language of A to the language of B and then from the language of B to the language of A must restitute the original labels given by A. (a "chair" should not become a "table")

In this respect, I repeat, your transformation fails to work properly as this double translation from R to R' and back from R' to R does not restitute the original labels in R. Instead you get at the end of this process:

t'' = t (1 - v²/c²)
x'' = x (1 - v²/c²)

and therefore

t'' =/= t
x'' =/= x

which is a very bad defect of your translation formula.

It goes without saying that the Lorentz transformation does not suffer from such a flaw.

ah you are back pretending again, the Lorentz transformations is:

t' = gamma (t - vx/c^2) (3)
x' = gamma (x - vt) (4)

and not as you try to pretend it to be.

I must admit that I cannot object anymore: gamma must be given by the usual Lorentz formula.
Thanks Anonymous, despite your strong words for me, I conceide that you could explain that clearly.
The comparison with a dictionary was really great.

roger

so for

Lorentz transformations is:

t' = gamma (t - vx/c^2) (3)
x' = gamma (x - vt) (4)

if we were to have gamma as 1

then

t' = (t - vx/c^2)
x' = (x - vt)

and using x = ct this becomes:

t' = (c - v)t/c
x' = (c - v)t

with inverse as

t = (c + v)t'/c
x = (c - v)t'

oops missed out "+" with inverse as

t = (c + v)t'/c
x = (c + v)t'

"so for

Lorentz transformations is:

t' = gamma (t - vx/c^2) (3)
x' = gamma (x - vt) (4)

if we were to have gamma as 1

then

t' = (t - vx/c^2)
x' = (x - vt)

and using x = ct this becomes:

t' = (c - v)t/c
x' = (c - v)t

with inverse as

t = (c + v)t'/c
x = (c + v)t'
"
_______________________________
You can check that indeed x=ct implies x'=ct'.
That's an elementary property of the Lorentz transformation.

and v = 0 from both frames

And that's again where you see the inconsistency of assuming gamma = 1 .
Let's redo the same but keeping the factor gamma from SR: gamma = 1/Sqrt(1-v²/c²)
__________________________________________________________________________________
Lorentz transformations is:

t' = gamma (t - vx/c²) (3)
x' = gamma (x - vt) (4)

let us now assume gamma = 1/sqrt(1-v²/c²)

then the inverse transformation is

t = gamma (t + vx/c²)
x = gamma (x + vt)

and using x = ct this becomes:

t' = gamma (c - v)t/c
x' = gamma (c - v)t

with inverse as

t = gamma (c + v)t'/c
x = gamma (c + v)t'
__________________________________________________________________________________

For the two expressions to be compatible you need to have:

t' = gamma (c - v)t/c = t' = gamma (c - v) gamma (c + v)t'/c /c

or

t' = gamma² (c²-v²)/c² t'

or

gamma² (c²-v²)/c² = 1

or

gamma² = 1/(1-v²/c²)

which is satisfied by the SR choice for gamma = 1/Sqrt(1-v²/c²)
__________________________________________________________________________________

This shows again that the SR choice for the gamma factor is the only consistent choice.

it does not such thing, we already showed it false

Lets start a fresh looking at the maths of SR:

Standard SR maths is as follows-

SR has it that the transformation between the two equations

x^2 - c^2t^2 = 0 (1)
x' ^2 - c^2 t' ^2 = 0 (2)

is the Lorentz transformation

(this is considering the case along x and x' direction and treating y= y' =z = z' =0)

Lorentz transformations is:

t' = gamma (t - vx/c^2) (3)
x' = gamma (x - vt) (4)

Let us accept that for the moment, and look at the issue of discovering what gamma equals.

if we subst (3) and (4) into (2) we get equation (1) as required.


Now let us try using 1/gamma instead of gamma, so that we have:

t' = (1/gamma) (t - vx/c^2) (3b)
x' = (1/gamma) (x - vt) (4b)

i.e. subst (3b) and (4b) into (2) we also get equation (1).


So we can conclude gamma = 1/gamma therefore equals "1."


Its strange that most SR texts do their maths wrong and get gamma equal to something other than "1." But SRists were never known for being any good at maths.

http://www.haltonarp.com/articles/the_observational_impetus_for_le_sage_gravity

Quote:

"Now that we reference the primary reference frame we are reminded that this is yet another strike against the hallowed relativity theory which is supposed to have no primary reference frame. But the existence of the microwave background certainly reminds us that an average over the detectable universe certainly represents an obvious, primary reference frame. Moreover laboratory experiments like the Sagnac effect by Selleri and others reveals the presence of such a frame."