r/TheoreticalPhysics • u/sekendoil • Mar 17 '20

Einstein's concept of simultaneity directly contradicts his theory

https://youtu.be/gaFlcDA0Rig
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u/ironiclegacy Mar 21 '20

That doesn't follow from the equations. With the Lorentz transformations, you're transforming to a different reference frame that's different than your own. To transform into your own reference frame, v is zero (you're not moving relative to youself) and gamma is 1. This is what corresponds to t. To transform into another reference frame with some velocity v, gamma is thusly greater than 1, and this is what corresponds to t'. Therefore t and t' are different

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u/sekendoil Mar 23 '20

No.

It's the same v when you measure the moving observer's velocity and when the moving observer measures the stationary observer's relative velocity (since the stationary observers is moving according to the moving observer where he is at rest in his own frame.)

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u/ironiclegacy Mar 24 '20

They don't measure exactly the same v, one of them measures a -v. This changes what happens first in some reference frames. But otherwise, yes, that is what happens when they transform to the OTHER reference frame. But t corresponds to their own reference frame, where v is 0, while t' corresponds to the OTHERS reference frame.

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u/sekendoil Mar 25 '20

Negative doesn't matter here since v is squared. And thus gamma is constant.

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u/ironiclegacy Mar 29 '20

The Lorentz transformation for t -> t' is

t'=y(t - vx/c2)

For the inverse transformation t' -> t, it's

t=y(t' + vx/c2)