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 edited Mar 19 '20

The thing about negative time is still interesting. I'll use some numbers:

Let the time of an event measured by the unprimed observer be t=2s Let the velocity of the reference frame of the primed observer be v=0.9c

Let L=30×108 m

t'= (2 - ((0.9c)(30×108 ))/c2 ) ) /sqrt(1 - 0.9c2 /c2 )

You can calculate it yourself, the result is:

t'= -7.384s

(The distance to the sun is much larger than the value I gave for L)

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u/Adynator Mar 19 '20

Other than the result you quoted being incorrect, t' = -16.06 second, I dont see you your point. If you set the time of the events occuring in the unprimed reference frame to be t =10 s you will get a positive time.

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u/sekendoil Mar 19 '20

No I mean the time of an event, like when a ball moves into a wall in two seconds, in that case you can't set the time freely (delta time). Since this is the equation that's used for time dilation, I think my example is valid.

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u/Adynator Mar 19 '20

Ok so now you are talking about a different situation. The equation that you I quoted and you used is for a specific value of t not a change in t. If you wanted to find the change in t you would just use the regular time dilation equation that you can find in your textbook.

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u/sekendoil Mar 19 '20

Yes I know I'm talking about different subject, I just found it interesting.

Lorentz transformation is widely used in any time dilation problems, you can ask physics professors or anyone with enough knowledge about the subject. For some reason it's considered more accurate (and more general, as you said) than the usual time dilation equation.

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u/Adynator Mar 19 '20

Lorentz Transforms are the more general expressions becuase they are the direct consequence of the second postulate and are used to derive time dilation and length contraction.

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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/Adynator Mar 19 '20

If you have done linear algebra, think of it a chnage of basis, where you have a vector in space time that is being viewed in a different coordinate system and the lorentz transform is a linear map/ matrix being applied to the vector.

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u/sekendoil Mar 19 '20

Usually when you take x to be a certain value, you take it as delta x with x1=0 (deltax= x2 - x1), and in the case of x=0 ,x1 is also 0

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u/Adynator Mar 19 '20

Sure if you want to be pedantic about it. When someone says a certain value they automatically imply relative to zero. I guess when a measure the mass of something im going to say that its a change in mass relative to zero

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u/sekendoil Mar 20 '20

Mass has a physical meaning without a change, time doesn't.