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Correct, there is no curving in the case of radial trajectory. Let's look at the formula I derived: [math]\frac{dr}{dt}=c(1-r_s/r)[/math] For the observer at infinity ([math]r=\infty[/math]) [math]\frac{dr}{dt}=c[/math], i.e. such an observer measures light speed to be ...exactly [math]c[/math]. For large values of [math]r[/math], the observer at infinity still measures light pulses to arrive to him/her at close the speed of light. As the light source approaches the event horizon (i.e. [math]r \to r_s[/math]), the light pulses appear to slow more and more, as [math]r[/math] decreases. When the light source reaches the EH, light never appears to arrive to the distant observer. It is as if the BH exerts and ever increasing "attraction" on the photons. Indeed, photons emitted from behind the EH never "escape", they can never be seen by the distant observer. Does this help?

It is not "we" who have a problem, it is YOU who has a problem. Experiments contradict your claim above.

This is what I get for quoting from memory, I opened my Landau and Lifschitz and, sure enough, charge denisty transforms under the Lorentz transform as : [math]\rho=\gamma(\rho'+j' \frac{v}{c^2})[/math]

This is an example of an ill posed question resulting into a lot of confusing/meaningless answers. The issue is that indeed, "relativistic" mass increases with speed by [math]\gamma[/math] while volume decreases by [math]1/\gamma[/math], so it would appear that COORDINATE-DEPENDENT density would increase by [math]\gamma^2[/math]. But, this is a meaningless exercise, since COORDINATE-DEPENDENT density is not a meaningful physics quantity.

This is a complicated question, the answer has two parts: 1. Vesselin Petkov is a crank, this is known in the physics community. 2. Your question is answered trivially in most textbooks: 2.1 LOCALLY, the speed of light in a gravitational field is CONSTANT, if it weren't the whole GR theory would fall apart. 2.2. Nevertheless, the COORDINATE speed of light is variable. Indeed, start with the reduced Schwarzschild metric: [math]ds^2=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)[/math] Light follows null geodesics, therefore: [math]0=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)[/math] giving: [math]\frac{dr}{dt}=c(1-r_s/r)[/math] Th LHS represents the COORDINATE speed of light, i.e., the speed of light measured by a distant observer. As one can easily see, it approaches 0 when the light approaches the event horizon , i.e. [math]r \to \infty[/math]. 2.3. When starting from the complete metric, things get even more complicated, resulting into what is taught as "photon orbits" in GR classes. For example: [math]ds^2=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)-r^2 (d\theta)^2[/math] Since light follows null geodesics: [math]0=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)-r^2 (d\theta)^2[/math] 2.3,1. If there is no radial motion (i.e [math]dr=0[/math]) then we obtain the so-called spherical orbits": [math]0=(1-r_s/r)(cdt)^2-r^2 (d\theta)^2[/math] The tangential speed is: [math]r \omega=\pm c \sqrt{1-r_s/r}[/math] 2.3.2 If there is both radial and tangential motion, then: [math]0=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)-r^2 (d\theta)^2[/math] so: [math]\frac{dr}{dt}= \sqrt {(1-r_s/r)^2 c^2-(1-r_s/r) (r \omega )^2}[/math]

This false, when you take a series of timestamped snapshots of a batted baseball or of a football thrown by a quarterback you get a set {x,y,z,t} of "points" in 4D. Contrary to your claims, "t", as a coordinate, has equal footing with the rest , i.e. with "x,y,z". The quartet (x,y,z,t) is the foundation of the Minkowski space.

False, think about the trajectory described by an object moving with respect to the observer.

Because experiment tells us that this is the case.

Let me re-write Janus LatEx formulas for you: [math] \frac{0.866c+1.11e^{-6}c}{1+\frac{0.886c(1.11e^{-6}c)}{c^2}}- 0.866c = ~2.78e^{-6}c = 0.0832646 km/s. [/math] If L' = 0.08325 km, then the trip time for the forward leg is ~1 sec. For the return trip, we get a speed of [math] \frac{0.866c-1.11e^{-6}c}{1-\frac{0.886c(1.11e^{-6}c)}{c^2}}- 0.866c = 0.0832647 km/s. [/math] Which gives a return trip time of ~1 sec He'll explain the formulas for you (they are simply the relativistic speed composition formulas).

OK, makes sense

The units seem messed up.

Your spreadsheet is false. The correct information is that, no matter what type of clock you use (light, atomic, sound, ping-pong balls), the periods are as follows: 1. In the frame co-moving with the clock : [math]T=\frac{2L}{s}[/math] 2. In the frame wrt. the clock moves at speed [math]v[/math]: [math]T=\frac{2L}{s \sqrt{1-(v/c)^2}}[/math] where [math]s[/math] is the speed of the signal ([math]s=c[/math] for light, [math]s=v_{sound}[/math] for sound, [math]s=v_{pingpong}[/math] for ping-pong balls, etc). The proof is non-trivial, it requires serious understanding of SR.

This is not quite correct. The principle of equivalence establishes the equivalence between a uniform gravitational field and uniform acceleration. The non-uniform gravitational fields exhibit tidal forces and such fields cannot be made equivalent to (uniform) acceleration, so EEP does not apply to them.

This is why I gave the derivation from first principles, to show the rigorous answer, rather than its components and to show the conditions under which the total effect is the sum of the components. These conditions are restricted to both weak gravitational fields and to low speeds wrt. light speed. I am glad that we are in clear now. Do you understand the other issues with your answer? You never acknowledged the issues, this is why I am asking.

The argument that the math is the same. The fact that you don't recognize this simple fact is irrelevant. Because correct application of GR teaches you so.

Actually, it does, since I wrote it as well. The only difference is that it shows how the effects are calculated on each segment of the trip in the experiment (the plane took off, reached altitude, slowed down, landed a few times, etc).

The effects are additive under certain conditions: -low radial and tangential speeds -low gravitational fields You cannot claim that , in general, the effects are additive.

1. The halving of the frequency happens when: [math]\sqrt{\frac{1-r_s/r_1}{1-r_s/r_2}}=\frac{1}{2}[/math] i.e. [math]\frac{4}{r_1}-\frac{1}{r_2}=\frac{3}{r_s}[/math] 2. In GR, the clock would "halt" if : [math]1-\frac{r_s}{r}-\frac{(v/c)^2}{1-r_s/r}=0[/math] This means : [math]v=c(1-r_s/r)[/math] . This is way off from your [math]v=0.5c[/math]. Using intuition is a very bad way of tackling problems. 1. You got the "kinematic" term wrong, you are off by [math]\frac{1}{1-r_s/r}[/math]. In strong gravitational fields, the error is significant. 2. The terms do not really "combine" since the relationship is , in general, non-linear. 3. Only in very special cases (low speeds and weak gravitational fields) one can claim that the terms "combine". 4. Either way, this"combination" is not a combination of GR and SR, as claimed in the OP, the effect is purely GR since it is all derived from the GR metric.

The first term is correct, the second term isn't quite correct. It is missing a [math]\frac{1}{1-r_s/r}[/math] (and so does the wiki page) . You are missing the third term, the one that quantifies the rotational motion, [math]\frac{r^2\omega^2}{c^2}[/math]. Technically, there can be even a fourth term, if the motion is not planar. My claim is that the solution doesn't "combine GR with SR", the whole solution is GR (as I have just shown). Please read carefully what I wrote.

1. Have you looked at the formulas he's posted? 2. You do realize that all the components are derived from GR, not from a combination of GR and SR, right? There is no SR involved in the derivation.

Multiplying the effects is a common mistake. It turns out that the effect is (heavily) non-linear , so you cannot resort to multiplication. The Schwarzschild solution provides you all the information necessary for performing the correct calculation. The Hafele-Keating experiment has been reprised with a much higher precision. Incidentally, the additive effect is also at the foundation of the GPS relativistic corrections. The overall effect is derived from the Schwarzschild solution: [math](c d \tau)^2=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)-r^2(d\theta)^2[/math] giving: [math]\tau=\int{\sqrt{1-r_s/r-(v/c)^2/(1-r_s/r)-(r/c)^2 \omega^2} dt}[/math] For small values of [math]r_s, v/c, r \omega /c[/math] the above can be simplified to: [math]\tau=\int{(1-0.5(r_s/r+(v/c)^2/(1-r_s/r)+(r/c)^2 \omega^2)) dt}[/math] ...recovering the explanation from the wiki article on Hafele-Keating.

GR makes a different prediction than the one you have, the changes are much more complicated than you think. A good synopsis can be found in the description of the Hafele-Keating experiment. The effects are additive, not multiplicative.

I didn't give you any negs.

First take your gratuitous negs back. BTW: it is "Schwarzschild", not "swartzchild".

You quoted (incorrectly) separate sources for separate effects. You cannot combine them in order to get a correct answer. Like I said, the Schwarzschild solution produces all the effects in one equation. You can give me all the negatives you have available, you will still not have learned anything.