Re: THE GENERAL RELATIVITY.

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)

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)

or:

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)

Right.

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

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

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

I will rather go back to my previous post.

Paul B. Andersen wrote:
| 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 - v2/c2) = f/sqrt(1 - v2/c2)
| 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 - v2/c2)
| f' = f*(1 - (v/c)2)/sqrt(1 - v2/c2)
| f' = f*sqrt(1 - v2/c2)
| This is a red shift.

You called this gibberish.
Now you have confirmed that it is correct.

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?

Paul


.

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. ... 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. ... If v and p are transverse in the observer frame S' ...
    (sci.physics.relativity)
  • Re: THE GENERAL RELATIVITY.
    ... The angle phi is thus pi/2 ... So we can conclude that if v and p are transverse in S, ... The Lorentz transform from S to S' is: ...
    (sci.physics.relativity)
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