Re: THE GENERAL RELATIVITY.

Paul B. Andersen wrote:
Koobee Wublee wrote:
Paul B. Andersen wrote:
Koobee Wublee wrote:
Paul B. Andersen wrote:
Koobee Wublee wrote:
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
For reference, the above set of equations is equation 1.

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'
For reference, the above set of equations is equation 2.

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))
For reference, the above is equation 3.

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.
Yes, but you have not derived equation 3 from either or both equations
1 and 2. Your equations 1 and 2 are not setup to give you equation 3.
Do you really understand the Lorentz transform, the equations 1 and 2?
It is trivially simple to derive equation 3 from either one of 1 and 2.

You then, wrote down the energy transformation equation as equation 4
below.

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

Since (phi = pi / 2), the above equation becomes

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

The above equation is how S' observes a photon sent by S which is a
blue shift.

So if the source is stationary in S and the observer in S',
we can say:
If v and p are transverse in the source frame S (phi = pi/2)
then the observer will see:
f' = f / sqrt(1 - v^2 / c^2)
which is a blue shift.


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.
I actually don't see you are wrong here. What I meant previously was
your the next step in which you are about to take is wrong --- the step
on transformation of energy. Your mistake was to recycle equation 4.
The equation of you mistake is

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

It should be the following related to equation 2.

f = f'(1 + v cos(phi') / c) / sqrt(1 - v^2 / c^2)
Right.

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

Since (phi' = pi / 2), the above equation becomes

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

The above equation is how S observes a photon sent by S' which is
also a blue shift.

Right.

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

The above equation is what frequency of the photon sent by S' as it
is observed by S which indicates a red shift in the emitted frequency.

So if the source is stationary in S and the observer in S',

No, it does not matter if S or S' is moving. All transverse
observations are blue shifted.

we can say:
If v and p are transverse in the observer frame S' (phi' = pi/2)
then the observer will see:
f' = f * sqrt(1 - v^2 / c^2)
which is a red shift.

You are grossly confused. The above equation indicates the emitted
frequency not the observed frequency. The emitted frequency relative
to the observed frequency is red shifted. The observed frequency
relative to the emitted frequency is again also a blue shift.

Now, you need to go back to address my previous post.
I will rather go back to my previous post.

Paul B. Andersen wrote previously:
| 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.

You called this gibberish.

Yes, a magic trick, sign of a swindler or a con-artist. You can fool
harry but not me.

Now you have confirmed that it is correct.

No, I have confirmed you are wrong.

Can we now consider this for settled, and apply it
on Thim's paper, or are you yet again going to say that
what is correct, is wrong, and then demonstate that it is correct?

Yes, it is settled that you are wrong.

How the hell is it possible to calculate the correct
equation and still claim them to be wrong?

No. Your equation relating the frequency is wrong.

Look. Let us repeat what you have stated is correct.

This is the equation for the Doppler shift of a signal
emitted from a source in S, and observed by an observer in S.

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

That is, when p and v are transverse in the source frame S,
then phi = pi/2 and the equation above yields:
f' = f/sqrt(1 - v^2/c^2)

But if v and p are transverse in the observer frame S',
then phi' = pi/2 and cos(phi) = v/c

You should write p' instead of p to indicate the momentum of the
photon as observed by S'.

Paul B. Andersen wrote;
| 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.

Koobee Wublee responded:
| I actually don't see you are wrong here.

That is because it is correct!
So get this into your head, and don't repeat that
it is wrong in every second posting!

The issue is not the above.

Read carefully:
| If v and p are transverse in the observer frame S',
| they are NOT transverse in the source frame S,
| because the angle between v and p is arccos(v/c) in S.

You have admitted that this is correct!

And when cos(phi) = v/c, the Doppler equation:
f' = f*(1 - (v/c)*cos(phi))/sqrt(1 - v^2/c^2)

The equation above is wrong. It should be

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

yields:
f' = f*sqrt(1 - v^2/c^2)
which is a red shift

Wrong here. The observed frequency by S in this case should be

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

Which is always a blue shift.

========================================================
So when v and p are transverse in the observer frame,
a red shift is observed!
========================================================

========================================================
No, any transverse Doppler shift according to the Lorentz transform is
observed to be blue shifted as the mathematics indicates.
========================================================

I gave you a most thorough and professional derivation over yours, and
you have not even read it.

http://groups.google.com/group/sci.physics.relativity/msg/de06b6a03ec2e430?hl=en&;

You cannot keep contradicting yourself in every second posting,
so unless you admit that this is correct, you are a crank!

I have taken your challenge and pointed out exactly where you have
failed. If you insist on these trollish behaviors, I have no time to
discuss with a professor such as yourself who blunders at the most
basic math and physics.

.

Relevant Pages

  • Re: THE GENERAL RELATIVITY.
    ... So we can conclude that if v and p are transverse in S, ... which is a blue shift. ... If v and p are transverse in the observer frame S' ...
    (sci.physics.relativity)
  • Re: THE GENERAL RELATIVITY.
    ... So we can conclude that if v and p are transverse in S, ... which is a blue shift. ... So if the source is stationary in S and the observer in S', ...
    (sci.physics.relativity)
  • Re: THE GENERAL RELATIVITY.
    ... So we can conclude that if v and p are transverse in S, ... which is a blue shift. ... So if the source is stationary in S and the observer in S', ...
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  • Re: THE GENERAL RELATIVITY.
    ... So we can conclude that if v and p are transverse in S, ... which is a blue shift. ... If v and p are transverse in the observer frame S' ...
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