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lorentz x' = γ(x - vt) We know the time that has passed from each references are different. And it influences the speed of things in each station and bus, but I think that has nothing to do with the position. For example we still can say that x' = x - vt, then in the station 1 second has passed and yet in the bus time has passed in less than 1 second. Isn't it supposed to be like that?

Or let's say, if this world just works like that lorentz-equation. Then based on that equation. Let's say we have 3 objects (A, B, C) which the velocity of A is 0.1c, velocity of B is 0.8c and velocity of C is 0.9c. And imagine if we stop all the time, so there won't be any transformation of energies. Then let's say we are at A and we see the distance of B to C is 1 million years of c. Then while the time still freezing if we change the reference frame to B, based on that equation x'= γ(x-vt) we need to multiply the result by γ. So the distance of B to C won't be 1 million years of c. But γ*(1 millions years of c), whether 0.8 million.. or etc. Does it really work that way? or I misinterpreted something?

nitori
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1 Answers1

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Welcome to the site,

The equation I think you are either misreading or was mistyped is probably the special relativity expression for transforming between inertial frames: $$ x' = \gamma(x-vt) $$ where $\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$ is a dimensionless factor (In particular, it does not represent length/distance).

The role of $\gamma$ in the equation is to correctly relate the coordinates of the two inertial frames. Note that in particular for $v \lt\lt c$ , we have $\gamma \approx 1$, and the equation reproduces the Galilean transformation you mentioned $x'=x-vt$. This may give you a feel for why this is a completely sensible expression.

I don't know what is your source, but perhaps if you give it a second look this will make better sense in the context of what you're reading / looking at. Good luck.

Amit
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  • Okay, maybe I make you or other people confused. Thanks anyway, I have edited my question. But the problem isn't that, I don't think we need to multiply it by something, which makes the equation rather nonsense. – nitori Jan 31 '23 at 12:54
  • The factor $\gamma$ is dimensionless and it doesn't represent any multiplication by distance, which I think is what you're confused about. – Amit Jan 31 '23 at 12:56
  • hmm.. so from what you're saying and here https://physics.stackexchange.com/a/206724/354148 I got my conclude that is: the x and x' here is just the thing's inertial frame coordinat and doesn't mean the real distance of one's distance, like that? – nitori Jan 31 '23 at 15:18
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    That's right. In fact the word "real" will get you into a lot of trouble if you aren't careful with it about these matters. It is better to use language here as follows: The distances that observers in different frames measure between points in spacetime may be different due to their relative velocity. No need to bring "reality" into the picture :) – Amit Jan 31 '23 at 15:21
  • ah I see that, really thank you. That helps me so much. – nitori Jan 31 '23 at 15:31