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The "two" are one and the same. What is the (simple) mathematical formalism that describes it? We have already been over your claim, the "paper" is not consistent with SR. Have you submitted it to either journal I suggested? If yes, have they answered? If yes, what was the answer? Nope, you are clearly confused on the basic concept of what RoS means. Write down the mathematical formulation and we can talk. Principle of Incertitude.

Sure they do, this is how they were derived. What is "measured" and what is "seen" are one and the same thing. This is not how the relativistic Doppler effect is derived. Based on what I seen in your paper, the answer is no. Then, what you are speaking of is not mainstream. There is a very simple, mainstream definition of RoS. It has a very clean cut mathematical definition. This is what I am talking about. This is false , as well. SR permits (under certain circumstances) that two events simultaneous in one frame are simultaneous in all frames. Yep, there isn't such test. Which makes your paper DoA. Not measurable means no way of telling the difference in practice. This is not what RoS means, so you are mixing concepts. Not. The PoI tells you something completely different about simultaneity but it has nothing to do with RoS.

SR lets you do that by using the Lorentz transforms that encapsulate all the effects. So far, the Lorentz transforms have proven totally adequate for the task. This is quite obvious given the known fact that the delay between two remote events can be measured through "probing" the events via light signals. This is not novel. Nor is it very useful since relativity of simultaneity has little to no application in practice (actually, it is the only prediction of relativity that isn't even testable).

How do you know when the pulse of light was sent from the perspective of the moving observer? You don't and you can't, nature "conspires" against all your attempts at finding out. .

Yes, pretty bad, especially coming from two physicists of their stature.

It is really a terrible hack, then. Especially when the correct treatment is not that complicated.

Correct, on both accounts. Taylor and Wheeler (if that is their drawing) are wrong on both accounts. Microsecond.

We see a relativistic Doppler effect , so 1 is incorrect. See here.

No, they aren't, they are made by nature.

You can't change the laws of physics, only nature can do that.

OK, This is not as bad as I thought it would be. The takeaway is that they don't make any reference to the (variable) coordinate - dependent light speed. What they do is pretty straightforward: -Start with the Schwarzschild solution: [math]ds^2=(1-r_s/r)(cdt)^2-\frac{dr^2}{1-r_s/r}-(r d \theta)^2-(r sin \theta d \phi)^2[/math] -Neglect the rotational terms since the trajectory is very close to being a straight line: [math]ds^2=(1-r_s/r)(cdt)^2-\frac{dr^2}{1-r_s/r}[/math] -Light follows null geodesics, so: [math]0=(1-r_s/r)(cdt)^2-\frac{dr^2}{1-r_s/r}[/math] -This gives: [math]cdt=\pm \frac{dr}{1-r_s/r}[/math] -Use the fact that [math]r_s<<r[/math] in order to further simplify the ODE to: [math]cdt=\pm (1+r_s/r)dr[/math] From this point they have an ordinary differential equation that they can integrate wrt [math]r[/math]. Actually, this is not quite right, so they insert another hack, making the path look like a line with a bump in the middle, around the gravitational body (when, in reality, the path is a parabola). Not very elegant, nor exact but it works. Nowhere is there any mention of using the coordinate-dependent light speed. I think that the correct answer is [math] \approx 200 us [/math]

Yet, your formula is clearly incorrect. I gave you the correct formula and I also gave you the page and paragraph in the Rindler book. Rather than continuing to push fringe stuff, may I suggest that you study how things are done correctly? Your formula doesn't even contain the correct variables ([math]X_1,X_2,R[/math]) that describe the geometry of the problem. That's not the problem, the problem is that your formula, wherever you got it from , is not the right one. Since you show no derivation, it is hard to point out all your mistakes, just that it disagrees with the mainstream result.

Neither, I am simply pointing out your mistakes. It is not "awkward' at all to calculate the integral. It is embarrassing that Taylor and Wheeler feel the need to resort to a hack in order to do such a simple calculation. To make matters worse, the geodesic followed by light doesn't look anything like the ridiculous drawing you have attached. Given that their book is a popularization book, it is less surprising. Can you post their calculations? I have no intentions spending 70$ on their book. Iggy did NOT do any calculations, he simply posted the radial speed (that I already derived in post 3) and the final formula (that , as I pointed out) disagrees with the correct formula. Actually, it has nothing to do with the correct formula (though he claims that it agrees up to powers of [math]\frac{1}{c^2}[/math]). Given his predilection to post hacks that are later proven to be wrong, I have no reason to believe any of his "stuff". Do Taylor and Wheeler use the coordinate speed of light in their calculations? Can you post them? For a rigorous derivation that does not resort to embarrassing "variable light speed", please look up Rindler page 237 or D'Inverno's excellent chapter 15.6 , free, here. I can also provide you with my own personal derivation.

No, it doesn't, I think the term you are looking for is "dialectic", "dielectric" is something totally different and makes no sense in the context. Besides, I explained to you repeatedly that the two sentences are not connected logically so, why do you insist in beating up the strawman you have created? My post 3 explains the variability of coordinate light speed, as such, it has nothing to do with Shapiro delay. This must be the sixth time I explained that to you. As explained several times already, this is clearly incorrect. Using your notation: [math]\Delta t_{flat}=2 \frac{X_1+X_2}{c}[/math] where [math]X_1,X_2[/math] are the distances of the gravitating body that bends the light ray to the source and to the reflector. [math]R[/math] is the distance of minimal approach. I already referred you to the appropriate textbook on this subject. [math]\Delta \tau =2(\frac{X_1+X_2}{c}+\frac{r_s ln 4X_1 X_2/R^2}{c})[/math] So, [math]\frac{\Delta \tau}{\Delta t_{flat}}=1+\frac{r_s ln4 X_1 X_2/R^2}{X_1+X_2}[/math]. The above correct result contradicts your formula pulled out of nowhere mainstream. Out of curiosity, how did you come up with it?

My sentence you just quoted tells you that the speed of light measured by a distant observer is the same thing as the coordinate speed of light. Yes, it is the coordinate speed of light that is being discussed. And, no, the coordinate speed of light (or its variability), is NOT the valid explanation of the Shapiro delay, contrary to the fringe ideas pushed by Iggy. No, I didn't, I said that the coordinate speed of light is unphysical and that the Schwarzschild solution is the standard answer for finding the Shapiro delay. Are you trying to be deliberately obtuse in order to cover your errors? This sentence is meaningless.

Read post 3, it is made clear from the first lines. Do you really not understand the "variable" speed of light is not the mainstream explanation of the Shapiro delay ? That you are pushing your personal fringe idea? That your "solution" is negated by the mainstream solutions? Let's try again: 1. The Schwarzschild solution is used for deriving the Shapiro delay. 2. The Schwarschild/Kerr/Riessner/etc. solutions happen to predict variable coordinate light speed. 3. This does NOT mean that variable light speed is the valid/mainstream explanation of the Shapiro delay, contrary to your fringe ideas pushed in your posts. There is no physical connection between statements 1 and 2. Both are valid but they aren't connected.

Sure it is, it points out that the variability is applied to a unphysical entity, to the coordinate speed of light , not to the local speed of light. This has been explained early on, in post 3, I recommend that you re-read it and try to understand it. This demonstrates that you STILL do not understand the issue: using the Schwarzschild solution doesn't mean that one uses the "variable" speed of (coordinate) light as a means of explaining the Shapiro delay. I did. It is right there, in post 3, has been there from the beginning. I suggest that you re-read it and try to understand it.

Correction: the coordinate speed of light is variable in Schwarzschild coordinates. As it is in Riessner coordinates, as it is in Rindler coordinates, as it is in Kerr coordinates. So what? This unphysical effect is not used in the derivation of the Shapiro delay. Why do you persist in pushing your fringe ideas? This is supposed to be a science forum. Both my statements are correct, so there isn't going to be any retraction. As I explained to you (repeatedly), the Schwarzschild solution is used in deriving the correct prediction for the Shapiro delay (not the incorrect one that you posted). The Schwarzschild solution also predicts a variable coordinate speed of light. You cannot infer from this that a variable speed of light explains the Shapiro delay, you are committing a basic fallacy: A->B and A->C doesn't mean C->B.

This is because the sources (binary stars) are very weak. If the Sun "disappeared", the effects would be detectable : the Earth would continue on a tangent to the current elliptical trajectory, never to return to the same point, the passage of seasons would disappear. SR is instrumental in the design of particle accelerators. SR is instrumental in determining the amount of energy released in a nuclear reaction. SR is instrumental in ....

The Earth gravitational field is created by the Earth, so what happens to the Sun is irrelevant, your question makes no sense. Moreover, if the Sun "exploded", the gravitational field of the resulting bits and pieces would still be there, so the gravitational field of the "exploded Sun" would be present. If the Sun magically "disappeared", then , it would take about 8.3 minutes for the effect to be felt on Earth, since the gravitational waves propagate at the speed of light and it takes light about 8.3 minutes to propagate from the Sun to the Earth.

Minkowski metric applies only to inertial observers, the observers in JVNY's exercise are accelerated, so your attempt at applying the Minkowski metric in order to calculate distances is clearly incorrect. There are multiple mistakes in the above answer: 1. The Shapiro delay is not calculated using the radial speed of light. 2. The light ray in the Shapiro delay doesn't move radially, it moves on a geodesic that grazes the gravitating object 3. The formula that you posted for the Shapiro delay is incorrect, the correct formula has been given early in this thread. The Shapiro delay requires a simple understanding of how to calculate the elapsed time of the photon traveling along a null geodesic. Nowhere in the calculation one uses "variable speed of light". The use of the Schwarzschild metric in figuring out the delay does not validate in any form the fringe concept of "variable light speed". Post 3 is textbook stuff, therefore it is correct. Your posts, on the other hand, are not.

The point was that you used the MINKOWSKI metric in your attempt to show that the distance between rockets is constant. When do you plan to retract your false statement that the Shapiro delay somehow proves the "slowing" of light speed?

Of course we are not any closer until you retract yur erroneous claim thath the Shapiro delay would somehow confirm the "slowing of light speed". The point I made was that your proof that the proper distance between the rockets is constant is incorrect. The thing with the linear dependence on distance for the coordinate speed is not the point of disagreement, the validity of your proof of constant distance is.You did not use the Rindler metric in computing the distance, aas you should have, you used the Minkowski metric. You never addressed this criticsm.

JVNY, What Ashby and Weiss are telling you (without spelling out the exact formulas) is that from [math]0=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)-r^2 (d\theta)^2[/math] one gets: [math]\frac{dt}{dr}= \frac{1}{\sqrt {(1-r_s/r)^2 c^2-(1-r_s/r) (r \omega )^2}}[/math] so, the one way total delay time can be obtained via an integration wrt the radial coordinate: [math]T_{total}=\int_{r_1}^{r_2}{\frac{dr}{\sqrt {(1-r_s/r)^2 c^2-(1-r_s/r) (r \omega )^2}}}[/math] The above integral doesn't have a closed form (best I know), so, a series of approximations needs to be done. There is no "slowing down of light speed", the effect can be calculated rigorously by a simple integration. If you neglect the rotational motion of the Earth, the integral is even simpler: [math]c*T_{total}=\int_{r_1}^{r_2}{\frac{dr}{1-r_s/r}} \approx \int_{r_1}^{r_2}{dr}+r_s \int_{r_1}^{r_2}{\frac{dr}{r}}=r_2-r_1+r_s ln \frac{r_2}{r_1}[/math] where [math]r_s=\frac{2GM}{c^2}[/math] is the Schwarzschild radius of the Earth. Post 3 contains all the mathematical formalism that you needed in order to answer your questions. BTW, the second term is what Ashby calls "the Shapiro delay" on slide 3. The "two perspectives" is only one, GPS was designed based on the GR machinery.

The simple answer is that you are trying to calculate the proper length using the Minkowski metric [math]ds^2=dx^2-(cdt)^2[/math], when , in reality, the metric is Rindler: [math]ds^2=(d \zeta)^2-(1+\frac{g \zeta}{c^2})^2 (c d \tau)^2[/math] where [math]\tau, \zeta[/math] are respecively, the proper time, proper distance measured by one of the accelerated observers. I have pointed out your mistake repeatedly in this thread, in different ways, this is yet another way to point out why your claims are wrong. They are just as wrong as your even more serious claim that the Shapiro delay would somehow be explained by light "slowing down" in the presence of a gravitating body. I mentioned before that I will not even respond anymore to your calculations on the JVNY exercise until you retract the more egregious claim about the Shapiro delay but I broke my promise since you persist in the claim that the distance between the rockets is constant. Once again, this is the coordinate speed of light, an entity devoid of any physical meaning. As explained already in post 3, this entity can take any value (in GR, it is dependent on the coordinates [math]r, \theta[/math] ), in accelerated frames, is dependent on the acceleration and on the coordinate x. The elegant way of showing that is to remember that light follows null geodesics and to make [math]ds=0[/math] in the Rindler metric above. You would get [math]d \zeta=(1+g \zeta/c^2)(c d \tau)[/math] That is, the correct formula of the coordinate speed of light for the accelerated observer is [math]\frac{d \zeta}{d \tau}=c(1+g \zeta/c^2)[/math]. According to your above logic, the observer at [math]x=0[/math] will measure the speed of light to be [math]v_x=0[/math], which is clearly absurd. The reason is that your formula, wherever you picked it up from (since you show no derivation) is incorrect. The correct formula is, as derived above , [math]\frac{d \zeta}{d \tau}=c(1+g \zeta/c^2)[/math]. This means that an observer at [math]\zeta=0[/math] measures the speed of light to be [math]c[/math] , as one would expect. As the observer accelerates away from the light source, the coordinate speed of light decreases, following the rule [math]\frac{d \zeta}{d \tau}=c(1-g | \zeta |/c^2)[/math]. At the Rindler horizon [math]\zeta=\frac{c^2}{g}[/math] so [math]c=0[/math]. A pulse of light emitted when the accelerated observer is already past the Rindler horizon, will never "catch up" with the accelerated observer. On the other hand, if the observer accelerates towards the light source, then [math]\frac{d \zeta}{d \tau}=c(1+g \zeta/c^2)[/math] teaches you that the coordinate speed of light "increases" linearly, as the observer "speeds up" towards the light source (and away from the launch pad). Notice how similar are these formulas with the ones derived from the Schwarzschild metric in post 3.