On Oct 26, 7:58 am, "Paul B. Andersen"
wrote:
Dr. Henri Wilson wrote:

I appreciate your efforts but it doesn't work as you say. You have ignored the
movement of the start point in the source frame. You are also ignoring the fact
that the 'intrinsic frequency' appears Doppler shifted at the detector because
the latter is moving wrt the startpoint. Even though the travel time of the
rays is the same, the number of cycles arriving at the detector differs for
each ray.
In short, you are confusing the start point frame with the source/detector
frame.
The fact is, the photon experiences one INTRINSIC cycle every wavelength
traveled.

The path lengths of the two rays are different, therefore the photons generally
end up out of phase.

Thankyou for helping me develop my model. It is all coming together nicely now.

So let's take one step at the time. So far, we have only stated what
the equation for the phase of your BaTh photon must be in the source frame.

Your talk about Doppler shift and motion relative some point in
another frame is thus utterly irrelevant.

Let's first agree on the equation describing the phase of your
BaTh photon in the source frame.
We can then take it from there later.

So read again, carefully this time:

You said:
1. A photon has an intrinsic oscillation of an unknown nature. During the absolute
time interval defined by one period of that oscillation, an identifiable point
in the photon body moves through a 'spatial interval' at c wrt the source.
The absolute distance it moves in the source frame is its 'wavelength'.
Like ALL lengths, that wavelength is the same in all frames.
The front of a BaTh photon oscillates once every absolute wavelength traveled.

This is YOUR oral description of your 'approach'.
All I do below is to express this description mathematically.
If you find an error in my math, please point out exactly what
it is, and show what the correct math should be.
Otherwise I will assume it is correct.

From your description, it follows that he phase at the front of any photon
must in the source frame fulfill the equation:
phi(t+T, x+cT) - phi(t,x) = 2pi
where T is the 'absolute time interval of one oscillation'
and cT = l is "the absolute distance it moves during T", that is the wavelength
T = l/c

If we assume that phi(t,x) is a linear function of x and t,
phi(t,x) = at + bx
we get:
(at + aT + bx + bcT)-(at + bx) = 2pi
aT+bcT = 2pi
b = (2pi+acT)/T = 2p/T + ac
Inserting T = l/c, we find:
phi(t,x) = at + (a/c + 2pi/l)x

Since the phase of any photon in a ray of photons must fulfill
this equation, it gives us the phase of the photon found at x at
the time t.

We know that the phase of the photon emitted from the source at x = x1
at the time t+T must be 2pi more than the photon emtted at the time t.
So:
phi(t+T,x1)-phi(t,x1) = 2pi
aT = 2pi
a = 2pi/T = 2pi.c/l (usually called the angular frequency w, of course)

So the equation becomes:
phi(t,x) = (2pi.c/l)t +((2pi.c/l)/c - 2pi/l)x = (2pi.c/l)t

================================================== =======
# So according to your BaTh:
#
# phi(t,x) = (2pi.c/l)t (in the source frame)
#
# The phase doesn't depend on x, all the photons in
# the ray have at any time the same phase.
================================================== =======

It's your model, Henri. You have now seen it expressed mathematically.
If your oral description of your 'approach' at the top is wrong,
please correct it, and I will express your changed description mathematically.

I understand that you are unable to do it, so I will have to help you.

I posted a brief response in the "Sagnac Threads United" thread.

Jerry