Re: SR fundamental contradiction
- From: Brian Kennelly <bwkennelly@xxxxxxx>
- Date: Tue, 10 Oct 2006 07:35:37 -0700
mluttgens@xxxxxxxxxx wrote:
Brian Kennelly wrote:Okay, then from the LT, we have:mluttgens@xxxxxxxxxx wrote:If you consider a length g (c-v)T at rest in the S'-frame, you
get a length (c-v)T in the S-frame, by applying length contraction
(this is what the tracking guru did).
If you consider a stick of length (c-v)T at rest in the S-frame,
you get a length g(c-v)T in the S'-frame according to the LT.
It is dilated in the S'-frame. Indeed, if the ends of the stick
are x1 = vT and x2 = cT in the S-frame,
vT=g(x1'+vt1')
cT=g(x2'+vt2')
This allows us to find x2' and x1' at the same t'
x1' = 0 and x2' = g(c-v)T,There is no value of t' that makes both of these equations true.
When x1'=0, then (from the first equation):
vT=g(vt1')
So t1' = T/g
When x2'=g(c-v)T, then
cT=g(g(c-v)T+vt2')
So t2'=cT/vg - g(c-v)T/v
\=t1'
To find the length in S', we must measure the distance between the endpoints at the same time in S' (the same t')
Setting t1'=t2' in the LT equations above and subtracting we get
cT-vT=g(x2'-x1')
x2'-x1'=(c-v)T/g
The length is contracted.
hence x2' - x1' = g(c-v)T. Notice that x1' is always zero, itsThe only way that x1' can be always zero is if it is at rest in S'. Because you stated that the stick is at rest in S, it, and therefore its endpoints, will be moving in S':
value is independent of time.
vT=g(x1'+vt')
x1'=vT/g-vt'
In particular, when x2'=g(c-v)T, then t'=cT/vg - g(c-v)T/v and x1' is:
x1' = vT/g-v(cT/vg - g(c-v)T/v)
= vT/g- cT/g +g(c-v)T
=g(c-v)T - (c-v)T/g
Subtracting this from x2' again gives the result derived above:
x2'-x1'=(c-v)T/g
No, L=L'/g and L'=L/g apply to two different scenarios. In the first, the stick is at rest in S', in the second, it is at rest in S.
This is obvious: if L = L'/g (as in the guru exemple), L' is
necessarily gL. The one who got L = L'/g AND L' = L/g made a logical
mistake somewhere.
.
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