md65536 383 Posted December 31, 2013 Share Posted December 31, 2013 Read post 3, it is made clear from the first lines.It is not made any clearer. Also I get confused here: 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].What do you mean by that the speed of light measured by a distant observer approaches zero? Doesn't the speed of light remain c, and only the coordinate speed of light approach zero? Link to post Share on other sites

xyzt 81 Posted December 31, 2013 Share Posted December 31, 2013 (edited) It is not made any clearer. Also I get confused here: What do you mean by that the speed of light measured by a distant observer approaches zero? Doesn't the speed of light remain c, and only the coordinate speed of light approach zero? 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. It is clear that this is what's been discussed all along. It's been pointed out several times (including by you) that it's coordinate speed of light that is being discussed. If it's confusing or bothersome perhaps everyone could repeat "coordinate speed" every time it's used. 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. I apologize profusely. Is the variable speed of light the unphysical entity, or the "standard answer", both of which you'd said. 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? They are physically dielectric. This sentence is meaningless. Edited December 31, 2013 by xyzt -1 Link to post Share on other sites

Iggy 344 Posted December 31, 2013 Share Posted December 31, 2013 They are physically dielectric.This sentence is meaningless. Dielectric means that you can't get from one to the other. My hope for you faded fast. We don't understand each other. I'll welcome your solution to Shapiro delay as proffered in post 3, or I'll welcome your insistence that nothing non-local is physical. I'll welcome either thing you've said, but they are fiercely dielectric. You can't get from one to the other, you know? The forces of the universe will keep those assertions apart despite your best efforts. Link to post Share on other sites

xyzt 81 Posted December 31, 2013 Share Posted December 31, 2013 (edited) Dielectric means that you can't get from one to the other. 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? I'll welcome your solution to Shapiro delay as proffered in post 3, 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. then the radial coordinate velocity of light is, [math]v_c = \pm \left(1- \frac{r_s}{r} \right)[/math] and in terms of acceleration of gravity, g, and two radial points, A and B... [math]v_{BA} = \left(1- \frac{gr}{2c^2} \right)[/math] (where A is further than B and [math]v_{BA}[/math] is the average velocity of light between the two as measured by A) If true then we get Shapiro delay for a round trip A->B->A... [math]\Delta \tau = \Delta t_{flat} \left(1 + \frac{gr}{2c^2} \right)[/math] and then light is slowed relative to flat spacetime, and many experiments have confirmed this, therefore, 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? Edited December 31, 2013 by xyzt Link to post Share on other sites

JVNY 8 Posted December 31, 2013 Author Share Posted December 31, 2013 It seems that the biggest issues are (1) whether slower coordinate speed of light has any physical meaning, (2) if one believes so whether one is mainstream or fringe, and (3) whether the coordinate speed of light (the speed as measured by a distant observer) is a valid explanation of the Shapiro delay. Taylor and Wheeler are certainly mainstream, and here are excerpts from their book "Exploring Black Holes: Introduction to General Relativity." The text answers the questions (1) yes, (2) mainstream, and (3) yes. Project (i.e., chapter) E is titled "Light Slowed Near Sun." That pretty much answers all of the questions implicitly from the start. Here goes with excerpts. The text sets forth the initial questions: * What happened to "Light always moves with the same speed"? * Who says light slows down near Sun? * How much does light slow down near Sun? * Does observation verify the predicted value of the slow-down? [from page facing E-1] The Schwarzschild bookkeeper records a "smaller speed of light" than do . . . free-float observers . . . That is the prediction of our analysis . . . Does this prediction have a physical meaning at all? Is there any way to measure this "slowing down of the speed of light" as reckoned by remote observers? "Yes" and "yes" were the answers predicted by Irwin Shapiro and demonstrated by him and coworkers. [from page E-1] So, Taylor and Wheeler literally ask the question whether the slower speed of light as reckoned by remote observers has a physical meaning, and they state that Shapiro answers "yes" it does. Question (1) above is answered "yes." This answer comes from Shapiro himself (according to Taylor and Wheeler) and it is included in Taylor and Wheeler's book, so it seems pretty mainstream. Question (2) above is answered "mainstream." Taylor and Wheeler then analyze the Shapiro delay. They state that it is too awkward to integrate the equation for the "skimming" path of light next to Sun that the Shapiro experiments actually use [page E-2]. So they state that: Instead we obtain an estimate of the delay time by assuming a slightly different path . . . that comes in radially from Earth, circles halfway around Sun in a tangential semicircular path, then moves out radially . . . This alternative path . . . yields an estimate much closer to a precise calculation than we might have expected. [page E-2] They use an illustration like the attached. They then calculate the delay as a combination of the radial and semicircular segments. The calculation uses the radial Shapiro delay formula, as does Iggy. Taylor and Wheeler's slightly different path generates a delay very close to the actual delay determined by experiments (approximately 250 ms modeled delay for a roundtrip to Mars). This is a delay "reckoned by remote observers" [per Taylor and Wheeler, page E-1], so it is in fact the coordinate speed of light measured by distant observers [as phrased by xyzt]. So, the coordinate speed of light is the valid explanation of the Shapiro delay. The answer to question (3) above is "yes." Going back to Taylor and Wheeler's introductory questions to the project, and answering them: * What happened to "Light always moves with the same speed"? [light travels slower near gravity as measured by distant observers, which has real physical meaning] * Who says light slows down near Sun? [shapiro, for one, according to Taylor and Wheeler] * How much does light slow down near Sun? [approximately 250 ms for the Earth-Mars round trip] * Does observation verify the predicted value of the slow-down? [yes] 2 Link to post Share on other sites

Iggy 344 Posted December 31, 2013 Share Posted December 31, 2013 (edited) 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? So... you aren't just mistaking my meaning, you are literally mistaking my words? How fun! If we were involved in something dialectic then it would suggest that there were a difference of opinion, but I assure you that is not the case. It's just that you said two wildly different things, and that is *dielectric*. Perhaps you could look the word up, and consider your position, and get back to me, because either "Schwarzschild coordinates are the standard", or "the coordinate speed of light isn't physical". You'll have to pick one or the other, and I guarantee you the dictionary says that the two are dielectric. As explained several times already, this is clearly incorrect. Using your notation: It's good to terms of -c^2. If you're good to salt then you know exactly from whence it came, and I'm a little embarrassed you ask. Edited December 31, 2013 by Iggy Link to post Share on other sites

xyzt 81 Posted December 31, 2013 Share Posted December 31, 2013 (edited) So... you aren't just mistaking my meaning, you are literally mistaking my words? Neither, I am simply pointing out your mistakes. Taylor and Wheeler then analyze the Shapiro delay. They state that it is too awkward to integrate the equation for the "skimming" path of light next to Sun that the Shapiro experiments actually use [page E-2]. So they state that: Instead we obtain an estimate of the delay time by assuming a slightly different path . . . that comes in radially from Earth, circles halfway around Sun in a tangential semicircular path, then moves out radially . . . This alternative path . . . yields an estimate much closer to a precise calculation than we might have expected. [page E-2] 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. They then calculate the delay as a combination of the radial and semicircular segments. The calculation uses the radial Shapiro delay formula, as does Iggy. Taylor and Wheeler's slightly different path generates a delay very close to the actual delay determined by experiments (approximately 250 ms modeled delay for a roundtrip to Mars). This is a delay "reckoned by remote observers" [per Taylor and Wheeler, page E-1], so it is in fact the coordinate speed of light measured by distant observers [as phrased by xyzt]. So, the coordinate speed of light is the valid explanation of the Shapiro delay. The answer to question (3) above is "yes." 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. Edited December 31, 2013 by xyzt 1 Link to post Share on other sites

Iggy 344 Posted December 31, 2013 Share Posted December 31, 2013 though he claims that it agrees up to powers of [math]\frac{1}{c^2}[/math]). I do. It's quite approximate and weak-field-limit of me, but that's just the way I talk. Who can't talk in terms over c squared? Quite dielectric. Link to post Share on other sites

xyzt 81 Posted December 31, 2013 Share Posted December 31, 2013 (edited) I do. It's quite approximate and weak-field-limit of me, but that's just the way I talk. 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. Who can't talk in terms over c squared? Quite dielectric. 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. Edited December 31, 2013 by xyzt Link to post Share on other sites

JVNY 8 Posted December 31, 2013 Author Share Posted December 31, 2013 . . . Do Taylor and Wheeler use the coordinate speed of light in their calculations? Can you post them? Sure. Although I do not understand them, I can post them for you. For radial path segments they use: dr/dt = 1 - (2M / r) [note that for c=1, I believe that this is the same as your post 3 formula: ] and then integrate: dt = dr / (1 - (2M / r)) They define r as the "coordinate" radius (or reduced circumference, or r-coordinate) and say that it is called a radius "despite its being no true radius" (page 2-9). They use an approximation for 2M/r: (1 + d)^n = 1 + nd And from this they get 53 microseconds delay for the radial segment Earth to Sun. Then, for tangential semicircular path they use: ds/dt = (1 - (2M / r))^(1/2) They assume that r = R for the entire semicircle, so they solve that equation for dt and sum it along the entire semicircular path of length πR, being: πR / (1 - (2M /r))^(1/2) And they use the approximation again and get a delay of approximately 15 microseconds. They don't do the Sun to Mars segment in detail, but they instruct the reader to use r_{Morbit} = 2.28 E +11 meters, and they say that this generates a time delay for that segment of 57 microseconds. So their total time delay is 53 + 15 + 57 = 125 microseconds each way, or 250 microseconds for the round trip. The chapter on the time delay is a "project," so Taylor and Wheeler do not show each step of their calculation. They show what I posted, generally saying to the reader to use certain formulas and then ask the reader to show that the results are the numbers above. 2 Link to post Share on other sites

xyzt 81 Posted December 31, 2013 Share Posted December 31, 2013 (edited) Sure. Although I do not understand them, I can post them for you. For radial path segments they use: dr/dt = 1 - (2M / r) [note that for c=1, I believe that this is the same as your post 3 formula: ] and then integrate: dt = dr / (1 - (2M / r)) They define r as the "coordinate" radius (or reduced circumference, or r-coordinate) and say that it is called a radius "despite its being no true radius" (page 2-9). They use an approximation for 2M/r: (1 + d)^n = 1 + nd And from this they get 53 microseconds delay for the radial segment Earth to Sun. Then, for tangential semicircular path they use: ds/dt = (1 - (2M / r))^(1/2) They assume that r = R for the entire semicircle, so they solve that equation for dt and sum it along the entire semicircular path of length πR, being: πR / (1 - (2M /r))^(1/2) And they use the approximation again and get a delay of approximately 15 microseconds. They don't do the Sun to Mars segment in detail, but they instruct the reader to use r_{Morbit} = 2.28 E +11 meters, and they say that this generates a time delay for that segment of 57 microseconds. So their total time delay is 53 + 15 + 57 = 125 microseconds each way, or 250 microseconds for the round trip. The chapter on the time delay is a "project," so Taylor and Wheeler do not show each step of their calculation. They show what I posted, generally saying to the reader to use certain formulas and then ask the reader to show that the results are the numbers above. 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. So their total time delay is 53 + 15 + 57 = 125 microseconds each way, or 250 microseconds for the round trip. I think that the correct answer is [math] \approx 200 us [/math] Edited December 31, 2013 by xyzt 1 Link to post Share on other sites

DimaMazin 93 Posted January 6, 2014 Share Posted January 6, 2014 It http://www.scienceforums.net/uploads/monthly_12_2013/post-102023-0-11282600-1388504740.jpg The actual path and the modeled path are incorrect because they have no deflection angle there. I think that the correct answer is [math] \approx 200 us [/math] What is measurement unit "us"? Link to post Share on other sites

xyzt 81 Posted January 6, 2014 Share Posted January 6, 2014 The actual path and the modeled path are incorrect because they have no deflection angle there. Correct, on both accounts. Taylor and Wheeler (if that is their drawing) are wrong on both accounts. What is measurement unit "us"? Microsecond. Link to post Share on other sites

JVNY 8 Posted January 6, 2014 Author Share Posted January 6, 2014 Taylor and Wheeler (if that is their drawing) are wrong on both accounts. It is essentially their drawing (without copying it outright and violating any copyright they might have in it). But remember that they expressly state that they are modeling a different path than the actual experiment, in order to simplify the math. They acknowledge that their answer is therefore only an approximation of the actual experiment. Their method yields 250 microseconds; xyzt recalls that the actual is around 200 microseconds. So it is clearly only an approximation. -1 Link to post Share on other sites

xyzt 81 Posted January 6, 2014 Share Posted January 6, 2014 It is essentially their drawing (without copying it outright and violating any copyright they might have in it). But remember that they expressly state that they are modeling a different path than the actual experiment, in order to simplify the math. They acknowledge that their answer is therefore only an approximation of the actual experiment. Their method yields 250 microseconds; xyzt recalls that the actual is around 200 microseconds. So it is clearly only an approximation. It is really a terrible hack, then. Especially when the correct treatment is not that complicated. Link to post Share on other sites

JVNY 8 Posted January 6, 2014 Author Share Posted January 6, 2014 I think that they even state that the line for the actual path is straight (no aberration) because the aberration is so small that it cannot be represented in the simple drawing. -1 Link to post Share on other sites

xyzt 81 Posted January 6, 2014 Share Posted January 6, 2014 I think that they even state that the line for the actual path is straight (no aberration) because the aberration is so small that it cannot be represented in the simple drawing. Yes, pretty bad, especially coming from two physicists of their stature. Link to post Share on other sites

xyzt 81 Posted January 14, 2014 Share Posted January 14, 2014 I think that they even state that the line for the actual path is straight (no aberration) because the aberration is so small that it cannot be represented in the simple drawing. I am glad that you are getting independent confirmation of the things I've been telling you. Link to post Share on other sites

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