Re: SR fundamental contradiction
- From: Brian Kennelly <bwkennelly@xxxxxxx>
- Date: Wed, 11 Oct 2006 09:41:05 -0700
mluttgens@xxxxxxxxxx wrote:
No. It only implies that I am using the S' coordinates to use calculate the length. Using the S' coordinates allows me to work with x' and t' without the need to calculate any t values.
Brian:
"If you wish to calculate the S' length of a stick at rest in S,
with endpoints at vT and cT [so that L=(c-v)T], we can use the
LT as follows:
vT=g(x1'+vt')
cT=g(x2'+vt')
Subtracting we get:
(c-v)T=g(x2'-x1')
The length is:
x2'-x1'=(c-v)T/g
= L/g
It is contracted."
By using the 'inverse' LT x = g(x'+vt), you imply that the stick
is at rest in S', not in S.
Indeed, you first calculate x1 and x2 from x1' and x2', which isNo. <vT> and <cT> are constants, so those equations allow us to calculate the endpoints in S' from the fixed values in S, for any time t'.
possible only if you know the endpoints of the stick in S':
x1 = vT = g(x1'+vt')
x2 = cT = g(x2'+vt')
If it is easier, you can express them as:
x1'=vT/g-vt'
x2'=cT/g-vt'
Then, by substrating, you calculate the length of the stick in S:Where is the circularity?
x2 - x1 = g(x2'-x1'), or
L = gL'
Don't you realize that you reasoning is circular?From this, you calculate back the length in S' as L'= L/g !
I used the fixed endpoints in S and the LT to calculate the length in S'.
If you prefer, we can do it the long way:
x1'=g(vT-vt1)
x2'=g(cT-vt2)
t1'=g(t1-v^2T/c^2) (x1=vT)
t2'=g(t2-vT/c) (x2=cT)
Setting t1'=t2', we get
t2=t1-v^2T/c^2+vT/c
Choosing t1=T, so that x1'=0
t2=T(1-v^2/c^2+v/c)
So:
x2'=g(cT-vT(1-v^2/c^+v/c))
=g(cT-v^2T/c-vT(1-v^2/c^2))
= g(cT(1-v^2/c^2)-vT(1-v^2/c^2))
=g((c-v)T(1/g^2)
=(c-v)T/g
Again, we find that the length is contracted in S'.
Yes, the stick is at rest in the S' frame.
Luttgens:
Here is an exemple given by the tracking guru, which
indirectly proves the dilation:
"Now imagine a stick with this particular length
x' = g (c-v) t
at rest in the S' frame.
What is the length of such a stick in the S-frame?
If you apply length contraction, you find that this length
would be
x' / g = (c - v) t
in the S frame."
The length (c - v)t represents the distance in S between the
endpoints ct and vt (for a given value of t).
The particular length x' = g (c-v)t corresponds of course
to the dilated length obtained by applying the LT x' = g(x - vt)
to that case where x = ct in S.
Brian:
"So, when the stick is at rest in the S' frame, it is contracted
in the S frame."
So, you are claiming that L = L'/g.
Yes, when the stick is at rest in the S frame.
Brian:
"When the stick is at rest in the S frame, it is
contracted in the S' frame, as I showed above."
Now, you claim that L' = L/g, hence, according to you,
L = gL'. Or you also claimed that L = L'/g.
Don't you see the contradiction?No, because the stick cannot be at rest in both frames, unless v=0, in which case g=1, and L=L'
You took the statement out of context. I said that, given the LT, there is no single value of t' that allows both <x1'=0> and <x2'=g(c-v)T> to both be true.
Brian:
"Okay, then from the LT, we have:
vT=g(x1'+vt1')
cT=g(x2'+vt2')
This allows us to find x2' and x1' at the same t'
There is no value of t' that makes both of these equations true.
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."
Another contradiction of you:
- There is no value of t' that makes both of these equations true.
- Setting t1'=t2' in the LT equations above ..., the length is
contracted.
IOW, this system of equations is inconsistent:
vT=g(x1'+vt1')
cT=g(x2'+vt2')
x1'=0
x2'=g(c-v)T
t1'=t2'
The first two equations are the LT, and the last is a requirement for measuring length in S'. You must drop either the third or fourth equation. In either case, you find that the length is contracted.
E.g., if we drop the fourth equation, then we get
vT=g(vt1')
cT=g(x2'+vt1')
t1'=T/g
cT=gx2'+vT
x2'=(c-v)T/g
Earlier, I showed the calculation obtained by dropping the third equation.
The inverse transformation has the same content as the direct equations.
On the other side, you use again the 'inverse' transform, making
the same circular reasoning as above.
So the stick is at rest in S', and moving in S
Luttgens:
Notice that x1' is always zero, its value is independent of time.
Brian:
"The 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':
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 "
The end vT of the stick in S always coincide with the S' origin,
so in S', such end is independent of T, thus its coordinate x1'
is always 0.
By applying the 'direct' LT, which is the correct procedure, one getsBecause you stated above that the stick is at rest in S', we conclude that it is contracted in S.
L' = x2' - 0 = g(c-v)T = gL
The length is thus dilated in S', according to the Einsteinian LT.
The LT never implies that a stick will have a greater length than it does in its rest system.
.
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