Re: THE GENERAL RELATIVITY.

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.

The scenarios of S1 to S' and of S' back to S2 are completely
independent of each other.

The situation when S1 and S' are traveling in parallel to each other
is the same as when S' and S2 are traveling in parallel to each
other. This is not the scenario of Mr. Thim's setup.

In the scenario of S1 to S', we have

E1' = (E1 - v * p1) / sqrt(1 - v^2 / c^2)

Where

** E1' = h f'
** E1 = h f
** p1 = Momentum vector of the photon as observed by S
** v = Velocity vector of S' as observed by S1

Further more,

** | p1 | = h f1 / c

In a transverse case, we have

f1' = f1 / sqrt(1 - v^2 / c^2)

Where

** v * p1 = 0

Notice we don't give a damn about the angle of the dot product, v'
* p1', which is related to your phi'.

Now, in the scenario of S' to S2, we have

E2 = (E2' - v' * p2') / sqrt(1 - v'^2 / c^2)

Where

** E2 = h f2
** E2' = h f2'
** p2' = Momentum vector of the photon as observed by S'
** v' = Velocity vector of S2 as observed by S'

Further more,

** | p2' | = h f2' / c

S1 and S2 are at rest relative to each other. Thus, more
simplification,

** v' = - v


In a transverse case, we have

f2 = f2' / sqrt(1 - v'^2 / c^2) = f2' / sqrt(1 - v^2 / c^2)

Where

** v' * p2' = 0

Again, notice we don't give a damn about the angle of the dot
product, v * p2, which is related to your phi.

In a nutshell, we have the frequency of the photon emitted by S1 and
observed by S' as follows.

f1' = f1 / sqrt(1 - v^2 / c^2)

This indicates a blue shift.

Continue with the situation from S' to S2, we have the frequency of
the photon emitted by S' as f2' and received by S2 as f2.

f2 = f2' / sqrt(1 - v^2 / c^2)

This also indicates a blue shift.

Although f1' does not have to be f2', you probably would demand
(f1' = f2'). This is OK. This is a special case. In such as
case, we have

f2 = f1' / sqrt(1 - v^2 / c^2) = f1 / (1 - v^2 / c^2)

So, as a photon is emitted in S1, absorbed by S', re-emitted by S',
and at last absorbed by S2 (S1 and S2 are at rest relative to each
other), the observed photon by S2 will be blue shifted from the
original S1. The photon gains energy by doing so. The velocity of
S' will decrease and change direction to compensate for the energy
gain in the photon.

This is the prediction of the Lorentz transform.

[...]

The rest of garbage snipped due to the fundamental blunder by the
professor as pointed out in the above derivation.

In Mr. Thim's experimental setup, the rotating apparatus (S' in our
terminology) is actually a conductor. Thus, special relativity does
not apply. However, if he were to actually add sensors all along the
edge of S' and transmitter transmitting exactly frequency of the
received photon, then the situation of S' going to S2 is very
different from what I have derived above. Instead, we have

v' * p2' = E2' v'^2 / c^2

Where

** E2 = (E2' - v' * p2') / sqrt(1 - v'^2 / c^2)

This is because only at that certain angle, the emitted photon from the
rotating S' can reach S2. Then, we have

f2 = f2' sqrt(1 - v'^2 / c^2)

The overall observed frequency by S2 as emitted by S1 is

f2 = f1

Just as you have said, but with a very different reason. This is
another one of the situations of the Sagnac non-effect.

Now, put up or shut up.

.

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