- #1

- 587

- 10

[tex]x'=\gamma(x-ut)[/tex]

[tex]x=\gamma(x'+ut')[/tex]

Why we need to use the same ##\gamma## in both relations? Why not

[tex]x'=\gamma'(x-ut)[/tex]

[tex]x=\gamma(x'+ut')[/tex]

- A
- Thread starter LagrangeEuler
- Start date

- #1

- 587

- 10

[tex]x'=\gamma(x-ut)[/tex]

[tex]x=\gamma(x'+ut')[/tex]

Why we need to use the same ##\gamma## in both relations? Why not

[tex]x'=\gamma'(x-ut)[/tex]

[tex]x=\gamma(x'+ut')[/tex]

- #2

Ibix

Science Advisor

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Alternatively, the Lorentz transformations are the ones that work in this universe to the limits of our experimental knowledge - all the justification you need in science.

- #3

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The gamma factors must be the same by symmetry. Can you prove or justify this yourself?

[tex]x'=\gamma(x-ut)[/tex]

[tex]x=\gamma(x'+ut')[/tex]

Why we need to use the same ##\gamma## in both relations? Why not

[tex]x'=\gamma'(x-ut)[/tex]

[tex]x=\gamma(x'+ut')[/tex]

- #4

- 14,711

- 6,955

##\gamma## must be the same function of ##u## in both cases. A symmetry argument is needed, perhaps, for why ##u## is the same in both cases!

Alternatively, the Lorentz transformations are the ones that work in this universe to the limits of our experimental knowledge - all the justification you need in science.

- #5

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- #6

Erland

Science Advisor

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- #7

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As

As far as gamma itself goes, you could also just substitute -v for v into the gamma function, and you'll find that the same gamma function is attained. As such, gamma is not dependent on direction, only the relative speed.

Best Regards,

GrayGhost

- #8

Mister T

Science Advisor

Gold Member

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I would say that there are two important reasons. One, we want the theory to be self-consistent, and two, we want the theory's predictions to match observation.Why we need to use the same ##\gamma## in both relations?

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