Thread: Relativity of Simultaneity View Single Post
P: 359
 Quote by DaleSpam This does not make sense. The number of reference frames is not a property of a universe. Universes don't "have" reference frames, they are mathematical devices for analyzing physics, not physical features themselves. You could use an infinite number of reference frames to describe even the simplest possible universe. If you want to talk about something "having" reference frames, then it would be an analysis which has reference frames.
Apologies, I am aware of that, but don't often use terminology that makes that clear.

 Quote by DaleSpam No. Even if everything were at rest to each other you could still analyze it in different reference frames, and if the transformation of the time coordinate included a spatial term then there would be relativity of simultaneity.
That would be another thing that I don't understand, namely how, or why, the time co-ordinate would include a spatial term.

 Quote by DaleSpam OK, let's look at the equations $t'=t-vx$ and $x'=x-vt$ in a little more detail. Suppose we have three events with coordinates $(t_A,x_A)=(0,0)$, $(t_B,x_B)=(0,1)$, and $(t_C,x_C)=(1,0)$. A and B are simultaneous, since $t_A=t_B$, and the time between A and C is 1. Now, transforming to the primed coordinates using the above formulas (v=0.5) gives $(t'_A,x'_A)=(0,0)$, $(t'_B,x'_B)=(-.5,1)$, and $(t'_C,x'_C)=(1,-.5)$. So we see that $t_A \ne t_B$ meaning that simultaneity is relative, and the time between A and C is still 1 meaning that time does not dilate. Therefore, the relativity of simultaneity is possible without time dilation. It doesn't have to do with motion, but with the transformation between different reference frames. Yes to the above, although again mathematically I don't think that it is possible for only one event to be non-simultaneous.
Thanks for going through the above; the part I don't understand is the initial equations; I read $t'=t-vx$ as meaning $t'$ equals t minus the velocity along the X-axis, but I don't understand why the velocity comes into it.

and $x'=x-vt$ I read as $x'$ equals x minus the velocity multiplied by the time - which makes a bit more sense to me.

My interpretation of it would be that, if the clocks which give the time co-ordinates all ran at the same rate, then absolute simultaneity should prevail; and in order for RoS to prevail clocks would have to give different times (co-ordinates).

I suppose, essentially, where I have trouble is how we can go from the scenario where an event (or all events) are absolutely simultaneous across all reference frames, to a scenario where there is RoS. Presumably the initial scenario of absolute simultaneity would involve a transform (to affirm absolute simultaneity); I don't understand where a different transform could result in [the conclusion of] RoS if the initial transform leads to the conclusion of absolute simultaneity.

Hopefully that makes some bit of sense.