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Yeah, exactly as expected, you are using an incomplete formula (and your numerical accuracy is wrong, to boot). The oversimplified formula you are using does not account for the difference in angular speed (the satellites are NOT geostationary, they MOVE wrt. the Earth surface) and the trajectories are NOT circular, they are elliptical. Yet another attempt by Bart to "I can prove relativity wrong" that falls flat on its face.

Sidereal. This is just another Bart "I can prove relativity wrong" thread that belongs in the Trash Can. Does not belong in "Science".

To be exact, the difference in clock frequency is derived directly from the Schwarzschild solution to the EFE: [math]\displaystyle \frac{f_1}{f_2}= \sqrt{\frac{1-r_s/r_2-\frac{(v_2/c)^2}{1-r_s/r_2}-(\frac{r_2 \omega_2}{c})^2}{1-r_s/r_1-\frac{(v_1/c)^2}{1-r_s/r_1}-(\frac{r_1 \omega_1}{c})^2}}[/math] where the difference in frequency comes from: 1. gravitational potential difference , via the term in [math]\frac{r_s}{r_i}[/math] 2. radial speed difference (if it exists), via the term in [math]v_i[/math] (the sat trajectories are not circular) 3. angular speed difference, via the term in [math]r_i \omega_i[/math] Traditionally, people say that there is an "SR component" and a "GR component" to the explanation. In reality, the whole explanation is purely GR.

You mean another one of "Einstein is wrong" ? Yes, you should try another forum, this one is, at least, trying to maintain a semblance of a science forum.

Actually this is not correct: 1. In math, the proofs are 100%. There is no such thing as a 93% proof. It is either proved or not. 2. In physics, theories are not proven at all, they can only be disproven. Experimentally.

Yes. No.

You don't get it, do you? The speed of the photon is "c".

1. Time and length don't "know" anything. They aren't endowed with knowledge. 2. The laws of nature don't care about our presence or absence. 3. No one knows, this is why it is an axiom. 4. Same answer as 3. 5. Same answer as 1.

Yet, according to mainstream physics, the Earth's orbit would take a little over 8 minutes to change. Gravitational waves propagate at the speed of em waves, which is "c". Contrary to your misconceptions, they don't propagate with infinite speed. This is why the Sun disappearance does not have an instantaneous effect on the Earth. Nope, when you push a pen of length [math]L[/math] at one end, the other end will start moving after a delay of [math]\Delta t=\frac{L}{s}[/math] where [math]s[/math] is the speed of sound in the material that the pen is made from. Repeating the same crank ideas over and over doesn't make them true.

Energy doesn't move.

I have posted the exact equations of motion. From these equations it is not possible to decide whether the two test probes fall together or separate. Since none of the "doubters" posted any math, none of them proved anything one way or another so far. Nevertheless, the fact that when dropped "alone", the heavier test probe hits in a shorter time than the lighter test probe is undisputable and undisputed. I established this very clearly. Until someone manages to integrate (wrt z) the equations of motion I have posted, the other case will remain unresolved. "Prose" arguments don't count. The "angular separation" is zero AT START. You have no proof that the angular separation is zero throughout the fall. You are simply stating your conclusion as "proof". Needless to say, this doesn't form a valid proof. I spent quite a bit of time with swansont explaining to him this failure, it seems he understood and abandoned this flawed line of argumentation. We seem to have found a more rigorous way of finding out the truth, based on the analysis of the differential equations describing the equations of motion. can you put the above in mathematical language? Please.

Working out the solution through GR (the Schwarzschild solution) one obtains : [math]t_i=\frac{D^{3/2}}{\sqrt{2GM}}(arctan \sqrt{\frac{R+r_i}{D-(R+r_i)}}-\frac{\sqrt{(R+r_i)(D-(R+r_i)}}{D})[/math] After the initial shock, one needs to remember that the Schwarzschild solution is obtained for a testprobe of zero mass, the "disagreement" is due to GR using [math]m=0[/math].

Glad that I managed to get you all squared.

This is basic introductory physics. It simply says [math]m \frac{d^2x}{dt^2}=NetForce[/math] You are rolling multiple basic errors in your post. To start with, [latex]f=\frac{G(m1+m2)}{r^2}[/latex] doesn't have the dimension of a force. Would you please let me continue the discussion with swansont undisturbed?

There is no "extra M" term. M is the mass of the Earth, [math]M\frac{d^2X}{dt^2}[/math] is the force exerted by the two test probes onto the Earth. That is obviously equal to [latex]\frac{GMm_1}{(X-x_1)^2}+\frac{GMm_2}{(X-x_2)^2}[/latex]

You are confused, the EP establishes the equivalence between bodies stationary in a uniform gravitational field and uniformly accelerated bodies. This is the EEP. There is another form, the WEP , that establishes the equivalence between the inertial and gravitational mass. Because the problem being discussed has nothing to do with "Kepler's laws". I explained to you that your attempt at doing physics via quote-mining doesn't work, this is not how physics is done. Simple, this is how the equations of motion get formed, the LHS contains the mass X acceleration, the RHS contains the expression of the force, be it gravitational (in the case of this thread), Coulomb, Lorentz, etc. Nope, you are much mistaken, about most everything you posted.

No, but the approach through studying the properties of the equations of motion is a better, more rigorous approach to solving the problem, one that I have tried to push you all along, away from the hacks based on looking at relative acceleration. I just realized though that there is no theorem of existence and uniqueness for the type of non-linear equations we are examining, so you cannot conclude that the two solutions are identical. For starters, there is nothing to prove that the solution even exists.So, this approach needs more work (but it is more promising).

Zero. Totally irrelevant, you keep insisting on working off acceleration , when the rigorous way is to work with the complete equations of motion. This is exactly like the people that try to solve SR problems using length contraction instead of using the full Lorentz transforms.

You do not know that. The two test probes have the same acceleration only when they freefall by themselves, not when there are two probes present, you need to prove your claim You keep trying to extrapolate assertions without any basis. As a matter of fact , if you wanted to do things in a rigorous way, you shouldn't even operate off acceleration, you need to write down the complete equations of motion. Actually, the equations of motion for the test probes have identical form and have identical boundary conditions. Therefore, the two equations have solutions that can only differ by a constant. Since the initial conditions are the same, the constant is zero. This might be the beginning of a valid proof.

Correction, [math]\alpha=\alpha(t)[/math]. All you know is that [math]\alpha(0)=\frac{\pi}{2}[/math]. What you are doing is just another way of your asserting with no proof that [math]z_1(t)-z_2(t)=0[/math]. You need to prove that. Proof by assertion is not a valid form of proof.

The equations of motion along the Z (radial) axis are: [math]M\frac{d^2Z}{dt^2}=+(\frac{GMm_1}{(Z-z_1)^2}+\frac{GMm_2}{(Z-z_2)^2})[/math] [math]m_1\frac{d^2z_1}{dt^2}=-(\frac{Gm_1M}{(Z-z_1)^2}+\frac{Gm_1m_2cos \alpha}{(z_1-z_2)^2+(r_1+r_2)^2})[/math] [math]m_2\frac{d^2z_2}{dt^2}=-(\frac{Gm_2M}{(Z-z_2)^2}+\frac{Gm_1m_2cos \alpha}{(z_1-z_2)^2+(r_1+r_2)^2})[/math] [math]cos \alpha=\frac{z_1-z_2}{\sqrt{(z_1-z_2)^2+(r_1+r_2)^2}}[/math] with the initial conditions: [math]z_1(0)=z_2(0)=D[/math] [math]Z(0)=0[/math] [math]\frac{dZ}{dt}=\frac{dz_1}{dt}=\frac{dz_2}{dt}=0[/math] at [math]t=0[/math] The "times to collision" are obtained from the conditions [math]Z(t)-z_i(t)=R+r_i[/math], [math]i=1,2[/math] Please solve the above and prove your claim that [math]z_1(t)=z_2(t)[/math]. You have all the information. Personally, I do not think that the system of equations is solvable, at least symbolically.

The "time to collision" in VOID is given by: [math]t_i=\frac{D^{3/2}}{\sqrt{2G(M+m_i)}}(arctan \sqrt{\frac{R+r_i}{D-(R+r_i)}}-\frac{\sqrt{(R+r_i)(D-(R+r_i)}}{D})[/math] where [math]R[/math] is the radius of the massive body and [math]r_i[/math] is the radius of the test probe. If the two test probes have the same radius [math]r_1=r_2[/math], then the "time to collision" depends only on the mass of the respective test probes, the larger the mass, the shorter the time. The reason is that the object attracts the Earth, just as the Earth attracts the object. More massive bodies make the Earth close the distance faster.

[math]z_1=z_2[/math] is your claim, not mine. I pointed out that I am skeptical of it, you presented nothing in terms of supporting your claim. You have not presented any proof, so there is very little to rebut.

No, according to you [math]z_1(t)=z_2(t)[/math] (the two bodies fall as one), so [math]z_1-z_2=0[/math] is NOT (only) a boundary condition, it is the permanent condition. It is actually your ongoing, not supported mathematically, claim. Once again, I would love to see a mathematical proof of your claim , all you have been doing is offering unsupported claims. Back up your claims with math, would you?

It is an excellent book. Unfortunately, none of the stuff applies directly to the radial fall discussed in this thread. You would need to do your own calculations, as I did.