Re: SR fundamental contradiction

mluttgens@xxxxxxxxxx wrote:
Brian Kennelly wrote:
mluttgens@xxxxxxxxxx wrote:
Brian Kennelly wrote:
mluttgens@xxxxxxxxxx wrote:
Brian Kennelly wrote:
Your solution (the LT solution) is correct, only if you
equate t1' with t2', but then t1 is different from t2. I call
this an incoherent solution.
The fact that simultaneity is not invariant is a feature of SR,
and probably the hardest to accept for many people. It follows
from the light postulate.

But, it is not incoherent.

You probably remember your following derivation:

"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'."

Choose t2 = T instead of t1 = T, and you will see that the LT
solution is incoherent.
No, if we choose t2=T, then we get
T=t1-v^2T/c^2+vT/c
t1=T+v^2T/c^2-vT/c

x1'=g(vT-vt1)
=g(vT-vT-v*v^2T/c^2+v*vT/c)
=g((c-v)(v^2T/c^2))
x2'=g(cT-vt2')
=g(cT-vT)
=g(c-v)T
Subtracting, we get
x2'-x1'=g(c-v)T(1-v^2/c^2)
=(c-v)T/g

This is the same result we got when t1=T, so the solution is still coherent.




Thank you for your pertinent analysis.
Contrary to those critics, who transformed this NG into a hornet's
nest,
you adopted a positive attitude, the only one that makes scientific
progress possible.
Anyhow, I probably had a big headache when I wrote my derivation.
Here is the revised version:

Derivation of the correct transformation:
----------------------------------------

Let's consider two frames of reference, S and S', each in uniform
translatory motion relative to the other, the velocity of S' relative
to S being v.

Basis relations:

x' = ax + bt
t' = ex + kt

At the origin of S', x' = 0 and x = vt.
Hence, 0 = (av+b)t, whence b = -av

The basis relations are now

(1) x' = a(x - vt)
(2) t' = ex + kt

Now, let's suppose that a light signal, starting from the coincident

origins of frames S and S' at t = t' = 0, travels toward positive x.
After a time t, it will be at
x = ct, and also at
x' = (c-v)t'
(Notice that Einstein postulated here that x' = ct')
It appears that, again, you are assuming a vector sum for the velocity vector.


Substituting these values of x and x' in relations
(1) and (2), one gets:

(c-v)t' = a(c-v)t, thus t' = at
t' = (ec+k)t, hence

(3) ec + k -a = 0

If the signal travels toward negative x,
x = -ct and x' = -(c+v)t'
(Here, Einstein postulated that x' = -ct')

Substituting these values of x and x' in relations
(1) and (2), one gets

-(c+v)t' = a(-ct-vt) = -a(c+v)t, thus t' = at
t' = (-ec+k)t, hence

(4) -ec + k - a = 0

From (3) and (4), one gets e = 0 and a = k

Hence, relations (1) and (2) become

(1) x' = k (x - vt)
(2) t' = k t

Now, a light signal follow the y' axis. Relatively to S,
it travels obliquely, for, while the signal goes
a distance ct, the y'-axis advances a distance x = vt.
Thus c^2t^2 = v^2t^2 + y^2, whence y = sqrt(c^2 - v^2) * t.
But, also, y' = ct' = c * k * t.
I don't see where you obtained y'=ct'. If your are using a vector sum for the velocity, you obtain y'=sqrt(c^2-v^2)t'


Equating y' to y, one gets k = sqrt(1 - v^2/c^2)
Again, using the vector sum for velocity, I get k=1.


The transform obtained without the *bold* Einstein's postulate that
the speed of light is the same in all frames is thus

(5) x' = sqrt(1 - v^2/c^2) * (x - vt) = (x - vt) / g
(6) t' = sqrt(1 - v^2/c^2) * t = t / g,

(g corresponds to Einstein's gamma).

The inverse transform is

(7) x = g(x'+vt')
(8) t = gt'



You cannot apply the transform twice.
t' = t/g gives the time in a moving frame wrt the time observed
in a frame at rest.
So, you are proposing that there is only one "rest frame"?

If you know t', you must use the inverse transform t = gt' to
calculate t, but not apply t/g twice.

More generally, your equations only represent a group when g=1.
I agree, but why should it represent a group?
Because the transformations should have the same form for any
value of v. In addition, the inverse of each transform should
have that form.

You automatically get a group when you use the Einstein's postulate
to derive the LT. Iow, the existence of such group has no objective
physical justification.

If there is not preferred frame, then the transformations between any two frames, including the inverse, will have the same form. This is the physical justification for the existence of a group.
.

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