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What makes you think that the two test probes do not separate during the fall? You are making an assumption that is not backed up by any math. Actually, the more I think about it, the more I become convinced that your assumption has no basis.

The solution posted is for one test probe being dropped at a time. I think that dropping both test probes simultaneously is a much more complicated problem (one test probe at a time is already very complicated) , I do not know the solution (I only know that this would be a variant of the "Three-body problem") and I do not know if your claim is true or false, can you post the math supporting your claim? I would be very interested in learning.

If two test probes of massses [math]m_1,m_2[/math] are dropped one at a time from the same altitude [math]D[/math] above a massive body of mass [math]M>>m_{1,2}[/math] then the test probe of larger mass will hit first. This is due to the fact that the "time to collision" 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 bode 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 derivation of the above formula is not trivial but I could post it if there is enough interest.

[math]a=\frac{GM}{d^2}[/math] where [math]d[/math] is the distance between the centers of the two attracting bodies. 1. For a test probe mass (negligible radius) situated on the surface of the Earth [math]d=R[/math] so: [math]a_1=\frac{GM}{R^2}[/math] 2. For two "Earths" touching each other [math]d=2R[/math] so: [math]a_2=\frac{GM}{(2R)^2}=\frac{a_1}{4}[/math] Spyman has ninja'd me.

GR says that it can. Not only that it can, but, upon crossing the event horizon, it will exceed the speed of light. For a rigorous derivation, see here.

[math]r_s[/math] is the Schwarzschild radius.

Your calculations are based on Newtonian mechanics, as such they are invalid since Newtonian mechanics cannot deal with the issue correctly. In order to get the correct answer you need to apply GR. When you do that you find that the speed is [math]c\sqrt{\frac{r_s}{r}}[/math]. At the singularity , [math]r=0[/math] so.... On a different note, at the even horizon [math]r=r_s[/math], so the infalling particle attains light speed already there.

Nothing can escape from INSIDE the event horizon, g-waves are not coming from within the even horizon.

No, in GR geodesic motion is not accelerated motion, you are trying to force fit your notions of Newtonian mechanics into GR. This seems to be a fringe misconception generated by your lack of understanding the mainstream physics. No.

You asked if I have seen "papers showing that White's energy calculations are wrong". I answered.

No, because White hasn't published in a peer reviewed journal in the first place.

Thank you, I suspected this must be the writeup, this isn't new, it has not been published in a peer reviewed journal and it has met with a lot of skepticism.

Could you tell me the author, title and publication (if it was published)? Thank you.

The link is dead, it is most likely some practical joke or a hoax.

You are welcome.

You are wrong, acceleration is change in velocity. This includes change in direction.

No, it isn't. You can prove that to yourself by transforming (x,y,z,t) into (x',y',z',t'). It is a simple algebra exercise.

The explanation has nothing to do with relativity of simultaneity. To get a correct grasp of what is going on, one needs to apply the equations of accelerated motion in SR. The difference between the elapsed times is due to the presence of acceleration in the case of the "traveling twin". See the detailed explanation here. The traveling twin measures an elapsed time of [math]\Delta \tau = 2 T_c / \sqrt{ 1 + (a \ T_a/c)^2 } + 4 c / a \ \text{arsinh}( a \ T_a/c )[/math]. The "inertial twin", measures an elapsed time of [math]\Delta t = 2 T_c + 4 T_a\,[/math]. A little bit of calculus shows that : 1. [math]\Delta t > \Delta \tau\,[/math] for [math] a \ne 0[/math] 2. [math]\Delta t = \Delta \tau\,[/math] for [math] a = 0[/math]

In answer to "caledonia" question(s): -if [math]v->c[/math] then, when you move away from the distant clock at speed [math]v[/math] you will measure the frequency to decrease to: [math]\displaystyle{f_o=f_s \sqrt{\frac{1-\beta}{1+\beta}}}->0[/math] because [math]\beta->1[/math]. In other words, it would appear that the frequency of the clock under observation goes to zero when it "separates" from the observer at a speed approaching the speed of light. On the other hand, when the observer moves towards the distant clock at speed [math]v->c[/math] then the measured frequency goes to infinity: [math]f_o=f_s \sqrt{\frac{1+\beta}{1-\beta}}->\infty[/math] because [math]\beta->1[/math] Makes sense? Not "double" , nor "infinite" speed. The clock and the observer are moving towards each other at [math]v->c[/math]

These are the basic relativistic Doppler formulas. Do you want a derivation from base principles?

Let's say that in the absence of relative motion you observe the distant clock to tick at the frequency [math]f_s[/math]. When you move towards the distant clock at speed [math]v[/math] you will measure the frequency to increase to: [math]f_o=f_s \sqrt{\frac{1+\beta}{1-\beta}}[/math] where [math]\beta=\frac{v}{c}[/math] When you move away from the distant clock at speed [math]v[/math] you will measure the frequency to decrease to: [math]\displaystyle{f_o=f_s \sqrt{\frac{1-\beta}{1+\beta}}}[/math]

correct

Well , causality NECESSARILY applies to events that are separated in the time dimension. On the other hand, causality may or may not apply to events separated in the spatial dimension. When the spatially separated events are not causally linked in one frame of reference, their order will be measured differently in other reference frames. So, essentially, you got the gist.

Correct, Causally linked effects are measured to be in the same order in all frames of reference. Space-like separated effects may or may not be causally linked so they can be measured in contradicting orders in different frames of reference. I gave a rigorous mathematical explanation at the beginning of the thread.

Accelerated motion, unlike uniform rectilinear motion, is absolute. You can tell which twin is accelerating from the one who isn't by using accelerometers attached to each twin. So, your scenario is flawed.