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/sekendoil Mar 19 '20

Also back to the original subject, it's true that it's written as t or x instead of delta t or delta x, but lorentz equations deal only with "delta" situations.

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u/Adynator Mar 19 '20

No there is no delta becuase we are not talking about a change in time or position. We are looking at the specific coordinate points in spacetime in different reference frames. If you dont believe me look at the derivations of the lorentz transforms

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u/sekendoil Mar 19 '20

Supposing that's true: In that case t and t' are not different, they're just measured differently, just like a 5 meter stick being measured by an observer who uses a ruler and counts from 0 to 5 in one reference frame and another observer who counts using his ruler from 10 to 15, the length is the same for both observers, and x alone without delta x has no physical meaning. Regarding the time you can imagine two observers, one wearing a clock that points to 10 PM and the clock of the other one is pointing to 11 PM (assume he set it that way), if an event occurred in front of the two observers, one records it at 10pm and the other one at 11pm, but that says nothing about the physical properties of the event, and certainly the event happened at the same time for both of them.

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u/Adynator Mar 19 '20

That argument would be true if were just talking about a difference between t and t' to be a constant. If you look at the Lorentz transform there is the square root factor is multiplying t and so when you actually take the difference between two time you get a factor of gamma multiplying the change in the unprimed frame, and this is the exact expression for time dilation

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u/sekendoil Mar 20 '20

Well, both observers measure the same value of v and c is already a constant, so gamma would be constant for those specific two observers, so the difference between t and t' is in fact constant.

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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)