Re: SR false?

Paul B. Andersen wrote:
Koobee Wublee wrote:

OK, since you insist. However, I do reserve the right to make my life
similar that is for me to set the velocity of x as observed by x'
parallel to both x and x' axes. The magnitude of the velocity is
positive going from left to right. I am also defining this angle as 0
when your angle = pi / 2. Also defining clockwise phi as positive, we
have

phi' = - cos^-1(v / c)

Almost right.
The correct answer is phi' = arccos(-v/c) (or phi' = cos^-1(-v/c))
I will assume this is what you meant.

From my definition of phi', I actually meant

phi' = - sin^-1(v / c)

In your definition assuming clockwise, it should be

phi' = pi / 2 - cos^-1(v / c)

In your definition assuming counter-clockwise, it should be

phi' = pi / 2 + cos^-1(v / c)

Now, if the source is stationary in S, and the observer
is stationary in S' we can conclude:

If the velocity of the observer is transverse to the wave vector
in the source frame S, then phi = pi/2, and the observed Doppler shift
will be:
f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v^2/c^2) = f/sqrt(1 - v^2/c^2)
This is a blue shift

If the velocity of the source is transverse to the wave vector
in the observer frame, then phi' = pi/2 and cos(phi) = v/c,
and the observed Doppler shift will be:
f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v^2/c^2)
f' = f*(1 - (v/c)^2)/sqrt(1 - v^2/c^2)
f' = f*sqrt(1 - v^2/c^2)
This is a red shift.

This is gibberish. See my mistake and your mistake above.

Since this follows from the very mathematics of
the Lorentz transform which you claim to understand,
I will assume you know this is correct.

In the case you don't agree, I challenge you to show that
the above is NOT what the Lorentz transform predicts.
General talk about symmetry won't do.
You will have to do the math.

f' = (f - v * p / h) / sqrt(1 - v^2 / c^2)

In transverse Doppler case, we always have

v * p = 0

So, we always have

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

The scenario bellow satisfies the above description:

A B -> v C

If f is the frequency of the signal emitted from A,
B will according to the Lorentz transform measure
the frequency f' = f*sqrt((c-v)/(c+v)) and thus
transmit the same frequency f'.
C will according to the Lorentz transform measure
the frequency f" = f'*sqrt((c+v)/(c-v)) = f

Thus the Lorentz transform predicts that C measures
the same frequency as is emitted from A.
And that is also what a real experiment would show,
if it didn't, we would have a causality violation.

Yes, this is true. However, this is not transverse Doppler effect.

But nevertheless it falsifies your claim:
"Because the Lorentz transform is symmetrical, it predicts
that C must must see the signal from B Doppler shifted by
the same amount as B sees the signal from A Doppler shifted
with."

In this scenario, we have

** f_B = f_A (1 - B) / sqrt(1 - B^2)
** f_C = f_B (1 + B) / sqrt(1 - B^2)

f_C = f_A (1 - B) (1 + B) / (1 - B^2) = f_A

You just don't understand how to apply the Lorentz transform.

In the case of Mr. Thim's experiment, we have

** f_B = f_A / sqrt(1 - B^2)
** f_C = f_B / sqrt(1 - B^2)

Thus,

f_C = f_A / (1 - B^2)

No.

In the case of Mr. Thim's experiment, we have:

v
^
|
A ->p B ->p C

Here the velocity of B is transverse to the wave vector
(or momentum) of the radiation in the stationary frame.

Case #1:
A is the source, B is the observer.
The stationary frame is the source frame.
The velocity of the observer is transverse to the wave vector
in the source frame, and the observed Doppler shift is thus:
f' = f/sqrt(1 - v^2/c^2)

Yes.

Case #2:
B is the source, C is the observer.
The stationary frame is the observer frame.
The velocity of the source is transverse to the wave vector
in the observer frame, and the observed Doppler shift is thus:
f' = f*sqrt(1 - v^2/c^2)

Wrong. In SR, there is no distinction between who is dong the moving.
So, your Case #2 should be the same as Case #1. The correct math is
also

f' = f / sqrt(1 - B^2)

The Lorentz transform predicts that the two Doppler shifts
will cancel each other, and C will detect the same frequency
as is emitted by A.

You must have a different version of the Lorentz transform.

Your cancellation method violates the very mathematics of the Lorentz
transform. You are totally lost.

I challenge you to point out exactly where I got lost.

Put up, or shut up.

I just did point out your error. Are you going to shut up now?

.

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