Re: THE GENERAL RELATIVITY.

Koobee Wublee wrote:
Paul B. Andersen wrote:
| Paul B. Anderson bragged about his own mistake again:
Koobee Wublee wrote:

| I challenge you to prove that you understand
| the very mathematics of the Lorentz transform
| by solving the following problem:
|
| In a frame of reference S, a photon with momentum p is moving
| along the y -axis. The angle phi is thus pi/2
|
| p
| ^
| | phi
| -----|---------------------------> x
|
| A frame of reference S' is moving in the x direction of S
| with the speed v.
|
| p
| ^
| | phi'
| -----|---------------------------> x'
|
| What is the angle phi' ?
|
| Put up, or shut up.

| [...]
|
phi' = - sin[^]-1(v / c)
This is correct from your definition of the angle.
With my definition of the angle,
the correct answer is phi' = arccos(-v/c)

[...]

\
\ phi'
----\-----------------------------------> x'

So we can conclude that if v and p are transverse in S,
they are NOT transverse in S', because the angle between
v and p is arccos(-v/c) in S'.

Yes. For reference, this is the scenario of S1 to S'.

Conversely we can conclude that if v and p are transverse in S',
they are NOT transverse in S, because the angle between
v and p is arccos(v/c) in S.

No. For reference, this is the scenario of S' to S2.

I can only interpret your "no" to mean that you claim
my statement in front of that "no" to be wrong.

Let's settle this before we continue.

The axes of S' and S are aligned, that is the x'-axis of S'
is parallel to the x-axis of S, ditto for the y and z axes.
The origo of S' is moving in the positive x-direction in S
with the speed v, and consequently the origo of S is moving
in the negative x'-direction in S' with the speed v.

The Lorentz transform from S to S' is:
t' = (t - vx/c^2)/sqrt(1-v^2/c^2)
x' = (x - vt)/sqrt(1-v^2/c^2)
y' = y
z' = z

The Lorentz transform from S' to S is:
t = (t' + vx'/c^2)/sqrt(1-v^2/c^2)
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'

According to this transformation, the direction of the velocity
of a light beam (or the velocity/momentum of a photon)
transforms like this:

From S to S':
-------------


^
| phi
-----|---------------------------> x


\
\ phi'
----\-----------------------------------> x'

The general equation relating phi and phi' is:
cos(phi') = (cos(phi) - v/c)/(1 - (v/c)*cos(phi))

If phi = pi/2, phi' = arccos(-v/c)

So we can conclude that if v and p are transverse in S,
they are NOT transverse in S', because the angle between
v and p is arccos(-v/c) in S'.

This far, we agree.
You do however NOT agree to the following:

From S' to S
============

^
| phi'
-----|---------------------------> x'


/
/ phi
----/----------------------------> x

The general equation relating phi' and phi is:
cos(phi) = (cos(phi') + v/c)/(1 + (v/c)*cos(phi'))

If phi' = pi/2, phi = arccos(v/c)

So we can conclude that if v and p are transverse in S',
they are NOT transverse in S, because the angle between
v and p is arccos(v/c) in S.

You claimed this to be wrong.
Please specify exactly what's wrong.


We will go on when this is settled.

There is no point in applying the LT on Thims's experiment
before we agree how the velocity of light transforms
according to SR.


Paul
.

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