Graduate Lorentz Transformations: Why We Need the Same $\gamma$

[LagrangeEuler]

If we have motion of system $S'$ relative to system $S$ in direction of $x,x'$ axes, Lorentz transformation suppose that observers in the two system measure different times $t$ and $t'$.
x'=\gamma(x-ut)
x=\gamma(x'+ut')
Why we need to use the same $\gamma$ in both relations? Why not
x'=\gamma'(x-ut)
x=\gamma(x'+ut')

 

[Ibix]

I'd suggest looking up the derivation of the Lorentz transformations from symmetry considerations. I think there's a section on Wikipedia, and Palash Pal's article "Nothing but relativity" has another treatment.

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.

 

[PeroK]

If we have motion of system $S'$ relative to system $S$ in direction of $x,x'$ axes, Lorentz transformation suppose that observers in the two system measure different times $t$ and $t'$.
x'=\gamma(x-ut)
x=\gamma(x'+ut')
Why we need to use the same $\gamma$ in both relations? Why not
x'=\gamma'(x-ut)
x=\gamma(x'+ut')

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

 

[PeroK]

I'd suggest looking up the derivation of the Lorentz transformations from symmetry considerations. I think there's a section on Wikipedia, and Palash Pal's article "Nothing but relativity" has another treatment.

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.

$\gamma$ 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!

 

[Tendex]

The argument is basically the relativity of simultaneity for the difference between t and t' mentioned by the OP, and Einstein's simultaneity convention(a galilean notion) for the symmetry of $u$ from S to S' and back.

 

[Erland]

If you invert the Lorentz transformation, you will see that it is the same $\gamma$ in the inverse transformation.

 

[GrayGhost]

By convention, the unprimed frame records the primed system moving at +v, and so the primed frame records the unprimed system moving at -v.

As Erland said above, you can just substitute -v for v in the LTs, and you'll find that the inverse LTs are attained. Gamma is found to be the same for the primed and unprimed frames.

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

 

[Herman Trivilino]

Why we need to use the same $\gamma$ in both relations?

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.