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You are thinking Gravitoelectromagnetism

Hint: You know that [math] b(x)=b(l) + ((l-x)/l)* (b(0)-b(l))[/math] Then: [latex]b(x+dx)=b(l) + ((l-x-dx)/l)* (b(0)-b(l))=b(x)-dx/l (b(0)-b(l)) [/latex] Then, the volume element is: [latex]dV=\frac{1}{3} (b^2(x)+b^2(x+dx)+b(x)b(x+dx))dx[/latex] You are left with integrating [latex]x^2 dV[/latex] from 0 to l. (you will need to ignore the powers of dx higher than 2)

Actually, it DOES. It gives you the formulas for [latex]A[R],B[R][/latex]. It teaches you how it replaces the two bodies of masses [latex]m_1,m_2[/latex] with a single body of mass [latex]\frac{m_1m_2}{m_1+m_2}[/latex]..... You will need to find the extremum of the function [latex]-A[R]+\frac{R^2 \omega^2}{c^2}[/latex]. where [latex]A[R]=1+\frac{a_1}{c^2R}[/latex] (see page 3) This is beyond pale, after teaching you how this type of problem is being solved.

Thank you, it is correct. This is a very famous solution to a very difficult problem.

Damour, in a series of famous papers, has reduced the two-body problem to an equivalent one-body problem. The solution to the EFE's is the metric: [latex](cd \tau)^2=-A[r](cdt)^2+B[r]dr^2+(r d \theta)^2 [/latex] (see page 4 for definitions of terms and coefficients). PS: Latex doesn't seem to want to work this morning.

For good reason, the frequencies are equal.

Yes, I can. You should be able to , after I showed you.

Yes, I did.

Here is the copy of the paper, contrary to your claims it is readily available. You need to stop spewing crankeries.

That would be difficult for the general case. See here the metric for particular cases. You already have the approach to the solution , in post 22. You know what to do.

...except that the above is just a simplification of the actual solution. It uses the Schwarzschild solution applicable to a single gravitating body. What is needed is the two-body EFE solution.

Cahill is a crank. Stop rubbishing the forum with your own crankreries.

Actually, this is nullified by TWO reasons: 1. In MMX the source ant the receiver are in the same place, as explained. 2. Even if you assumed that the source and the mirror have different velocities, their distance remains constant, hence no Doppler effect . You may want to study the Mossbauer experiment where the source and receptor are placed diametrically opposite (hence they have different velocities), yet there is absolutely no Doppler effect. You are co-mingling Doppler with gravitational redshift. These are different effects.

Doppler effect can only exist in the case of relative motion between source and detector, in MMX the source and the detector are at the SAME location. So, there is no way that there is any Doppler in MMX, the acceleration is irrelevant.

This is a mainstream forum, not one for pushing crackpot theories.

The equation is central to all experiments involving relativistic particle collisions, Compton effect, inverse Compton effect, etc.

Yes, the standard is given in post 23. Doesn't change the answer since you need to take the derivative wrt x.

You need to make the effort to LEARN the Standard Model. This IS the explanation for your questions, there is no shortcut or dumbed down version,.

Yes, check Standard Model

R is the Earth radius. The formula (derived per your earlier request) gives the ratio of frequencies of a clock placed on the Earth surface vs a clock situated at distance d-x from the Earth center moving with angular speed [latex]\omega[/latex]

Not necessarily, you will need to calculate instead the extremum of the function: [latex]\frac{f®}{f(x)}=\sqrt{\frac{1+2\Phi(x)/c^2-((d-x) \omega/c)^2}{1+2\Phi®/c^2}}[/latex]

No, the gravitational forces do not cancel in the above point, the gravitational potential has an extremum. This is different. The gravitational forces cancel each other for : [latex]\frac{GMm_p}{(R-x)^2}=\frac{Gmm_p}{x^2}[/latex] It so happens that the two equations have the same solution.This is because the force is the gradient (derivative wrt x) of the potential. At the distance x calculated above. I do not think that there is any experiment confirming it but it would be fairly straightforward.

Correction: [latex]\frac{d \tau_A}{d \tau_B}=\sqrt{\frac{1+2 \frac{\Phi_A}{c^2}-(r_A \omega_A/c)^2}{1+2 \frac{\Phi_B}{c^2}-(r_B \omega_B/c)^2}}[/latex]

It comes from the following: The Schwarzschild solution to the EFE's is (see Rindler "Relativity, Special, General and Cosmological"): [latex](c d \tau)^2=(1+2 \frac{\Phi}{c^2})(cdt)^2-dr^2-(rd \theta)^2[/latex] If [latex]dr=0[/latex] (no radial motion) and [latex]d \theta=0[/latex] (no angular motion) then: [latex]\frac{d \tau}{dt}=\sqrt{1+2 \frac{\Phi}{c^2}}[/latex] For the case of two clocks A and B situated in gravitational potentials [latex]\Phi_A, \Phi_B[/latex] : [latex]\frac{d \tau_A}{dt}=\sqrt{1+2 \frac{\Phi_A}{c^2}}[/latex] [latex]\frac{d \tau_B}{dt}=\sqrt{1+2 \frac{\Phi_B}{c^2}}[/latex] So, the clock periods are in the relation: [latex]\frac{d \tau_A}{d \tau_B}=\sqrt{\frac{1+2 \frac{\Phi_A}{c^2}}{1+2 \frac{\Phi_B}{c^2}}}[/latex] The frequencies are in the inverse relation: [latex]\frac{f_B}{f_A}=\sqrt{\frac{1+2 \frac{\Phi_A}{c^2}}{1+2 \frac{\Phi_B}{c^2}}}[/latex] So, what I should have written is: [latex]\frac{f®}{f(x)}=\sqrt{\frac{1+2\Phi(x)/c^2}{1+2\Phi®/c^2}}[/latex] The extremum calculation shown earlier is not affected. If there is angular motion, things get more complicated: [latex]\frac{d \tau_A}{d \tau_B}=\sqrt{\frac{1+2 \frac{\Phi_A}{c^2}-(r_A \omega_A)^2}{1+2 \frac{\Phi_B}{c^2}-(r_B \omega_B)^2}}[/latex]

Hard to tell, Lagrange points are determined by forces , the clock frequencies are a function of potentials.... ETA: the Lagrange points are much farther than the point calculated above