Re: SR fundamental contradiction



mluttgens@xxxxxxxxxx wrote:
Yes, that was intended in the scenario, reread the first posts of the
sadomaso,
even *he* had no doubt about this. He was the first to specify "at some
specific
time", which means, of course, t = T.
Reading the early posts, I got the impression that you were describing the distance between the light signal and another point moving at a constant velocity.

Of course, in that case, the length is increasing with time, so it is not straightforward to compare the lengths between systems.

The answer is that the distance at any given time in S is (c-v)t, and the distance at any given time in S' is ct'. *But*, the two distances do not correspond to the same fixed length, because the times cannot be directly compared.

If we pick time t'=T', and calculate the locations of the origin of S' and the light signal in the S system, we get:
x1=g(vT')
x2=g(cT'+vT')

If these points correspond to the ends of an interval at rest in S, the length is x2-x1=gcT' or cT'=(x2-x1)/g.

So, the length in S', cT', is contracted when compared to a fixed length in S whose endpoints correspond to the ends of the interval at time T'.

If you measure the distance in S, at time T, and compare it to an interval at rest in S', you get:
x1'=g(vT-vT)
=0
x2'=g(cT-vT)
=g(c-v)T

So
x2'-x1'=g(c-v)T
or (c-v)T=(x2'-x1')/g
So, the length in S is contracted when compared to a fixed length in S' whose endpoints correspond to the ends of the interval at time T.

The situation is entirely symmetrical, and in both cases, the length is contracted when compared to a fixed length at rest in the other system.


>From x' = g(x-vt), x'1 = g(x1-vt), hence x' =0 as x1=vt..
Aftrwards, you complicated the problem by introducing t1 and t2.

No, you complicated the problem by ignoring them, and expecting
S' to mark the ends at two different times.

S and S' measure time differently.


Of course, in the general case, but not in the scenario.
Please explain.
.



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