xyzt Posted November 27, 2013 Share Posted November 27, 2013 (edited) . So here is the scenario and a Minkowski diagram. The three ships start out at rest in the inertial frame, rear at x=0.5, center at x=0.75, and front at x=1.0. Simultaneously, center sends flashes rearward and forward, and the ships begin constant proper acceleration of rear a=2, center a=1.33, and front a=1. This maintains their proper distance in their own accelerating reference frame. Their proper distance wrt to what? Each other? It doesn't appear so since [math]x_i(\tau)=x_{i0}+\frac{c}{a_i^2}(ch \frac{a_i \tau}{c}-1)[/math] so [math]x_2-x_1 \ne x_3-x_2[/math] . can you explain what you mean? In the example I worked for you, the distance from the rocket center to the aft and fore point were equal. The other thing I'd like you to explain is how all your scenarios relate to the title of this thread. The coordinate - independent light speed is not affected by the gravitational field, neither is it affected by accelerated frames, so what is with all these scenarios? Edited November 27, 2013 by xyzt Link to comment Share on other sites More sharing options...

JVNY Posted November 27, 2013 Author Share Posted November 27, 2013 Yes, the center ship is equidistant from the front and rear ships at the start. All three are at rest in the inertial frame, with rear at x=0.5, center at x=0.75, and front at x=1.0. Then, when they accelerate, they maintain their proper distance in their accelerating frame. In their frame, there is always 0.25 distance between rear and center, and 0.25 distance between center and front. In the inertial frame the distances length contract, however. The scenarios relate to the title because it seems that the acceleration pattern causes the rearward flash to return to the center after the forward flash returns to the center in the ships' frame, even though they travel the same spatial distance in the ships' frame. Analogize to gravity, and the same would occur. This seems to be the first step in a deeper analysis that may well end with your result -- that light is still traveling at c, but there is a spacetime effect that explains why the rearward or downward flash returns later. Link to comment Share on other sites More sharing options...

xyzt Posted November 27, 2013 Share Posted November 27, 2013 Yes, the center ship is equidistant from the front and rear ships at the start. All three are at rest in the inertial frame, with rear at x=0.5, center at x=0.75, and front at x=1.0. Then, when they accelerate, they maintain their proper distance in their accelerating frame. In their frame, there is always 0.25 distance between rear and center, and 0.25 distance between center and front. In the inertial frame the distances length contract, however. But the math of hyperbolic motion does not support your claim, it actually contradicts your claim. How do you explain that? The scenarios relate to the title because it seems that the acceleration pattern causes the rearward flash to return to the center after the forward flash returns to the center in the ships' frame, even though they travel the same spatial distance in the ships' frame. The difference in elapsed time is due to closing speeds, as I already explained to you in a couple of posts, not due to ANY variability in the speed of light, as the title of the thread claims. Analogize to gravity, and the same would occur. There is no analogy with gravity. I explained that as well. You are mixing up closing speeds in an accelerated frame with the calculation of the coordinate speed of light in the presence of a gravitational field. This seems to be the first step in a deeper analysis that may well end with your result -- that light is still traveling at c, but there is a spacetime effect that explains why the rearward or downward flash returns later There is no "deeper" meaning, this is what closing speeds to, the travel time is bigger when you chase something going away from you. Can anyone help explain whether gravity slows light Gravity does NOT "slow" light. Link to comment Share on other sites More sharing options...

John Cuthber Posted November 27, 2013 Share Posted November 27, 2013 Just a thought: after 3 pages discussing "Does Gravity Slow Light Moving Vertically?" is the speed of light (in vacuo) still exactly 299,792,458 metres per second? Link to comment Share on other sites More sharing options...

xyzt Posted November 27, 2013 Share Posted November 27, 2013 Just a thought: after 3 pages discussing "Does Gravity Slow Light Moving Vertically?" is the speed of light (in vacuo) still exactly 299,792,458 metres per second? Yes, it is. And no, gravity does not "slow down the speed of light". 1 Link to comment Share on other sites More sharing options...

md65536 Posted November 27, 2013 Share Posted November 27, 2013 Just a thought: after 3 pages discussing "Does Gravity Slow Light Moving Vertically?" is the speed of light (in vacuo) still exactly 299,792,458 metres per second? Yes, when measured locally. If you're measuring from a different gravity potential---different coordinates---you get different results. You can still calculate, as JVNY and Iggy have been doing. For example, from a distant observer's perspective, light near a black hole's event horizon is not moving very fast in your coordinates. You can say that it is slowed in the lower gravitational potential. However, in local coordinates it is still moving at c. 1 Link to comment Share on other sites More sharing options...

xyzt Posted November 27, 2013 Share Posted November 27, 2013 You can still calculate, as JVNY and Iggy have been doing. Neither of them "calculated" any such thing. For a basic answer, see here. Link to comment Share on other sites More sharing options...

Iggy Posted November 28, 2013 Share Posted November 28, 2013 (edited) Just a thought: after 3 pages discussing "Does Gravity Slow Light Moving Vertically?" is the speed of light (in vacuo) still exactly 299,792,458 metres per second? It is not such without qualification. Take, for example, the following quotes by xyzt and Einstein: The speed of light is constant, c... Light speed, contrary to Iggy's crank claims, does NOT VARY in relativity, no matter if the motion is accelerated or not, it is a constant. In the second place our result shows that, according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity and to which we have already frequently referred, cannot claim any unlimited validity. (1920) Both of those quotes answer your question directly. One says, "yes, the velocity of light is c regardless". The other says "no, the velocity is not c regardless". You're right that three pages of posts seems like a lot, so consider also the following two quotes: The speed of light is constant... [math]xg[/math] is not constant for the mere reason that x is variable. [i.e. the speed of light doesn't change with position] Given ... that the speed of light is a function of position, it is easily deduced from Huygens's principle that light rays propagating at right angles to the gravity field must undergo deflection. (1911) [i.e. the speed of light changes with position] CaptainPanic was honest enough to offer earlier in the thread that he didn't know who was right or wrong, but I personally think that this subject has been settled for more than 100 years. Einstein is correct. You are now honest enough to say "is the speed of light (in vacuo) still exactly 299,792,458 metres per second?" implying that you can't imagine it being anything different. That the speed of light is "exactly 299,792,458 metres per second" is true only in inertial frames. Introducing gravity or acceleration opens nature's domain beyond that postulate. [math]c[/math] is a universal constant. The speed of a ray of light is not. Special relativity postulates that c and the speed of light are the same in inertial frames, but beyond that limited domain they are not. It is a common misconception to think they are equivalent which is why one frequently gets the question "if nothing can travel faster than light then why are galaxies receding faster than c?" This isn't a zero sum debate. The answers are mutually exclusive. Either xyzt or Einstein is correct. Either the speed of light varies or it doesn't, and if the last 3 pages of discussion, and the last 100 years of relativity, have failed to illuminate the answer then we have some ways to go. Edited November 28, 2013 by Iggy Link to comment Share on other sites More sharing options...

xyzt Posted November 28, 2013 Share Posted November 28, 2013 (edited) It is not such without qualification. Take, for example, the following quotes by xyzt and Einstein: Both of those quotes answer your question directly. One says, "yes, the velocity of light is c regardless". The other says "no, the velocity is not c regardless". Velocity is a vector, so it cannot "be c". The redlined sentence shows that you do not know the difference. I explained to you several times that while the velocity (a vector) changes (as in the Shapiro delay, gravitational lensing, etc), the speed of light (i.e. the scalar) does NOT vary. If you want to keep pushing your fringe ideas, you should try an anti-relativity forum, that is where cranks routinely use the quote from Einstein (1920) to support their claims that the "speed of light" (in vacuum) varies. I asked you before and I am asking you again: prove that the speed of light varies in the Shapiro delay, If you cannot prove that, have the integrity to admit that you were wrong all along and stop pushing fringe misconceptions as if they were mainstream concepts. Exactly. Shapiro delay confirms it. Light passing a mass is slowed on approach and slowed likewise on departure. yep yep The above claim is not only grossly incorrect but you repeated it several times. Contrary to these repeated claims, textbooks teach you that "light passing a mass is NOT slowed on approach and slowed likewise on departure". It is the velocity that changes, more exactly the direction of the vector, not the scalar (the speed). The speed stays the same,c. For your education , I am including a diagram that explains the Shapiro delay correctly, just to debunk your fringe "explanation". You can see that the velocity is what changes, this is what Einstein tried to teach you , so what I have been telling you is exactly what Einstein has been telling you, no disagreement. The only disagreement is between your claims and mainstream GR. Edited November 28, 2013 by xyzt Link to comment Share on other sites More sharing options...

JVNY Posted November 28, 2013 Author Share Posted November 28, 2013 Here is part of the Physics FAQ answer on the quotation and point: "In the 1920 book "Relativity: the special and general theory" he wrote: . . . according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity [. . .] cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity of propagation of light varies with position. Since Einstein talks of velocity (a vector quantity: speed with direction) rather than speed alone, it is not clear that he meant the speed will change, but the reference to special relativity suggests that he did mean so. This interpretation is perfectly valid and makes good physical sense, but a more modern interpretation is that the speed of light is constant in general relativity. The problem here comes from the fact that speed is a coordinate-dependent quantity, and is therefore somewhat ambiguous. To determine speed (distance moved/time taken) you must first choose some standards of distance and time, and different choices can give different answers. This is already true in special relativity: if you measure the speed of light in an accelerating reference frame, the answer will, in general, differ from c." http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html So even that careful web site seems to go back and forth on the answer. This is in part why I asked this question here, to get everyone's views. 1 Link to comment Share on other sites More sharing options...

xyzt Posted November 28, 2013 Share Posted November 28, 2013 (edited) Here is part of the Physics FAQ answer on the quotation and point: "In the 1920 book "Relativity: the special and general theory" he wrote: . . . according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity [. . .] cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity of propagation of light varies with position. Since Einstein talks of velocity (a vector quantity: speed with direction) rather than speed alone, it is not clear that he meant the speed will change, but the reference to special relativity suggests that he did mean so. This interpretation is perfectly valid and makes good physical sense, but a more modern interpretation is that the speed of light is constant in general relativity. The modern, mainstream view is quite clear on this issue, end of story. Did you read the explanation I wrote for you on the Shapiro delay? What did you understand from it? I also twice questioned your claim about the three rockets maintaining equal distances, can you answer those questions for me, please? Edited November 28, 2013 by xyzt Link to comment Share on other sites More sharing options...

JVNY Posted November 28, 2013 Author Share Posted November 28, 2013 On the Shapiro delay, the picture does not show purely radial travel that we are discussing here. Consider instead a mirror orbiting the gravitational mass between that mass and earth. One sends a signal directly toward the mass. The signal travels radially directly toward the mass, then reflects off the mirror and returns to earth on the same radial path. There is no deflection like the curved path under the mass in the picture. There is no change of direction other than the 180 degree reflection. The light will take longer to make that trip than it would absent the mass. There is no curving in space. But there seems to be curvature in spacetime, and some people say that it is due to gravitational time dilation. On the proper distance, by setting the proper accelerations at 2, 1.33, and 1, one ensures that the ships always stay 0.25 apart in their own reference frame. This is explained more fully in http://www.mathpages.com/home/kmath422/kmath422.htm . Link to comment Share on other sites More sharing options...

Iggy Posted November 28, 2013 Share Posted November 28, 2013 (edited) Here is part of the Physics FAQ answer on the quotation and point: "In the 1920 book "Relativity: the special and general theory" he wrote: . . . according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity [. . .] cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity of propagation of light varies with position. Since Einstein talks of velocity (a vector quantity: speed with direction) rather than speed alone, it is not clear that he meant the speed will change, but the reference to special relativity suggests that he did mean so. This interpretation is perfectly valid and makes good physical sense, but a more modern interpretation is that the speed of light is constant in general relativity. The problem here comes from the fact that speed is a coordinate-dependent quantity, and is therefore somewhat ambiguous. To determine speed (distance moved/time taken) you must first choose some standards of distance and time, and different choices can give different answers. This is already true in special relativity: if you measure the speed of light in an accelerating reference frame, the answer will, in general, differ from c." http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html So even that careful web site seems to go back and forth on the answer. This is in part why I asked this question here, to get everyone's views. That site is mistaken. The paper I quoted was originally written by Einstein in 1916. It was translated into English in 1920. The German language does not have different words for the English: 'velocity' and 'speed', so any notion that Einstein meant one and not the other is mistaken. I should find a source for that... If we look at the lightspeed issue in Einstein’s 1905 paper on Special relativity: “On the Electrodynamics of moving bodies,” a good thing to note is there has been a translation problem from Einstein’s original German into English. The German language does not differentiate between the English words “speed” and “velocity”. (Others have noticed this, so I cannot take credit for discovering it. Einstein’s 1905 Speed/velocity of Light errors -- Roger J. Anderton For example, the German word for speed/velocity was translated to speed in the following two quotes: "Aus dem . . ., dass die Lichtgeschwindigkeit im Schwerefelde eine Funktion des Ortes ist, lasst sich leicht mittels des Huygensschen Prinzipes schliessen, dass quer zum Schwerefeld sich fortpflanzende Lichtstrahlen eine Krummung erfahren mussen." ("Given ... that the speed of light is a function of position, it is easily deduced from Huygens's principle that light rays propagating at right angles to the gravity field must undergo deflection." “Das Prinzip der Konstanz der Lichtgeschwindigkeit kann nur insofern aufrechterhalten werden, als man sich auf für Raum-Zeitliche-Gebiete mit konstantem Gravitationspotential beschränkt.“ (“The principle of the constancy of the speed of light can be kept only when one restricts oneself to space-time regions of constant gravitational potential.”) wiki -- variable speed of light Xyzt's earlier notion that the speed of light is constant while the coordinate speed of light is not -- that idea isn't really meaningful. A coordinate speed is a speed. If one insists that the coordinate speed is not constant then by necessity 'the speed' is not constant. No doubt, any non-local speed in GR is ambiguous, but an ambiguous thing does not a constant and invariant thing make. Edited November 28, 2013 by Iggy Link to comment Share on other sites More sharing options...

xyzt Posted November 28, 2013 Share Posted November 28, 2013 (edited) AS far as the site's insistence that the speed of light is constant while the coordinate speed of light is not -- that statement is just meaningless. A coordinate speed is a speed. If one insists that the coordinate speed is not constant then by necessity 'the speed' is not constant. Frankly, I do not understand why you persist in this embarrassing charade. I should find a source for that... If we look at the lightspeed issue in Einstein’s 1905 paper on Special relativity: “On the Electrodynamics of moving bodies,” a good thing to note is there has been a translation problem from Einstein’s original German into English. The German language does not differentiate between the English words “speed” and “velocity”. (Others have noticed this, so I cannot take credit for discovering it. Einstein’s 1905 Speed/velocity of Light errors -- Roger J. Anderton Roger Anderton is a well known antirelativistic crackpot, thank you for revealing your source of information. Now, it all becomes clear. On the Shapiro delay, the picture does not show purely radial travel that we are discussing here. Consider instead a mirror orbiting the gravitational mass between that mass and earth. One sends a signal directly toward the mass. The signal travels radially directly toward the mass, then reflects off the mirror and returns to earth on the same radial path. There is no deflection like the curved path under the mass in the picture. There is no change of direction other than the 180 degree reflection. The light will take longer to make that trip than it would absent the mass. There is no curving in space. But there seems to be curvature in spacetime, and some people say that it is due to gravitational time dilation. Yes, so what? The point is that what causes the delay is the change in velocity direction , not magnitude, as you and Iggy incorrectly claim. On the proper distance, by setting the proper accelerations at 2, 1.33, and 1, one ensures that the ships always stay 0.25 apart in their own reference frame. This is explained more fully in http://www.mathpages.com/home/kmath422/kmath422.htm . I don't see anything supporting your claim, why don't you explain, with math, in your own words , your claim. Please do so. Edited November 28, 2013 by xyzt Link to comment Share on other sites More sharing options...

Iggy Posted November 28, 2013 Share Posted November 28, 2013 (edited) Roger Anderton is a well known antirelativistic crackpot, thank you for revealing your source of information. Now, it all becomes clear. Fair enough, here is another source: 4.2 Dual Equivalence As soon as terms like Geschwindigkeit (speed, velocity) or Spannung (stress, tension) appear in a text the translator's problems begin. Two approaches are adopted... First, the Polyseme Dictionary lists samples of adjacent compounds at entries like Geschwindigkeit, Spannung so that the translator realises the terms are polysemous... A Basis for Scientific and Engineering Translation - Michael Hann - 2004 I don't see anything supporting your claim, why don't you explain, with math, in your own words , your claim. Please do so. He has done that repeatedly. Edited November 28, 2013 by Iggy -1 Link to comment Share on other sites More sharing options...

md65536 Posted November 28, 2013 Share Posted November 28, 2013 This interpretation is perfectly valid and makes good physical sense, but a more modern interpretation is that the speed of light is constant in general relativity.The modern, mainstream view is quite clear on this issue, end of story. When there are multiple "perfectly valid" interpretations, one alone is never the "end of story", and I don't think it's a good reason to try to kill a discussion that others are having. Others are calculating things using a perfectly valid interpretation. Do you have a problem with the results, or their interpretation? Link to comment Share on other sites More sharing options...

xyzt Posted November 28, 2013 Share Posted November 28, 2013 (edited) Fair enough, here is another source: 4.2 Dual Equivalence As soon as terms like Geschwindigkeit (speed, velocity) or Spannung (stress, tension) appear in a text the translator's problems begin. Two approaches are adopted... First, the Polyseme Dictionary lists samples of adjacent compounds at entries like Geschwindigkeit, Spannung so that the translator realises the terms are polysemous... A Basis for Scientific and Engineering Translation - Michael Hann - 2004 You are trying to turn the attention from your crank claims into a debate over terminology. When there are multiple "perfectly valid" interpretations, one alone is never the "end of story", and I don't think it's a good reason to try to kill a discussion that others are having. There are no "multiple interpretations", Iggy has been very proactive in pushing his fringe ideas. Others are calculating things using a perfectly valid interpretation. Do you have a problem with the results, or their interpretation? I see the errors that other (Iggy, JVNY ) are posting, I do not see any valid calculations. The one that is the most glaring is the claim that Shapiro delay is a result of "light speed slowing down". Edited November 28, 2013 by xyzt Link to comment Share on other sites More sharing options...

CaptainPanic Posted November 28, 2013 Share Posted November 28, 2013 ! Moderator Note xyzt, I have had it up to here (*points somewhere pretty high up*) with you. You seem incapable of discussing in a polite manner. Perhaps you do not see the difference between a person who attacks your idea, and a person who attacks your person. You certainly don't seem to see the difference between a Big Red Moderator Note and a normal post, or you would have taken my previous post more seriously. You do not seem to realize who you are dealing with here. I am a moderator. I don't care who is wrong or right. I don't care about the topic of this thread. In my role of moderator, I don't care about the contents at all. But I do care about people behaving on this forum, and you don't behave. You are trying to 'win' this discussion by calling names and by discrediting your opponent, while the other members in this thread are just discussing the science. I enforce the rules here, and at this point you are the only one in this thread who is violating the rules, specifically section 2, rules 1, 4 and 5. Please note that this was your final warning. Any further violations of the rules of our forum will get you suspended from this forum (even if you convinced you are right, and the other is wrong). Do not respond to this mod note in this thread. If you have any issues with this mod note, use the report function and (other) moderators will have a look at it. Link to comment Share on other sites More sharing options...

Iggy Posted November 28, 2013 Share Posted November 28, 2013 (edited) On the proper distance, by setting the proper accelerations at 2, 1.33, and 1, one ensures that the ships always stay 0.25 apart in their own reference frame. This is explained more fully in http://www.mathpages.com/home/kmath422/kmath422.htm . I don't see anything supporting your claim, why don't you explain, with math, in your own words , your claim. Please do so. I do not see any valid calculations. The part for which you asked -- the part about maintaining a constant proper distance -- can be quickly proved. Given JVNY's derivation (and illustration) of his thought experiment: The proper distance at t=0 is given along the x axis as: [math]\sigma = 1 - 0.5 = 0.5[/math] The proper distance at the dotted line between the accelerated hyperbolic worldlines is given from the numbers in this frame by length contraction: [math]\sigma = L_{0}\sqrt{1-v^{2}} = (1.25-0.625)(\sqrt{1-0.6^{2}}) = 0.5[/math] So, the rest frame (the inertial frame, or lab frame) calculates that the proper distance between the front and rear rockets remains equal (always 0.5 in their frame) while the front accelerates at a=1 and the aft at a=2. This is further verified by the definitional meaning of Rindler coordinates in which I've already run these numbers. All x coordinates maintain constant proper distance in Rindler (accelerated) coordinates, and these numbers work out in Rindler coordinates (in which all observers are taken to be at rest in their accelerated frame. It is further cited in the link JVNY gave (2.9 Accelerated Travels) And, here is another well-written article on Born rigidity reporting the same: Likewise every other hyperbolic path asymptotic to the same light lines (i.e., sharing the same pivot event) maintains a constant proper distance from that event and, therefore, from every other path in this class. Hence, in order for our set of particles distributed along the x axis to accelerate while maintaining constant proper distances from each other, the magnitude of the constant proper acceleration of each particle must be inversely proportional to it's (signed) distance from the pivot event x_{p}. In other words, if x_{j} is the location of the jth particle at time t = 0, then this particle's constant proper acceleration must be a_{j} = 1/(x_{j} - x_{p}) Born Rigidity, Acceleration, and Inertia [edit]... I apologize. JVNY already gave this link in addition to the one directly above it. It continually amazes me how quick he finds these relevant links, reads them, and correctly reports and understands what they say. I'm struggling to keep up. [/edit] You are asking how acceleration of 2, 1.33, and 0.5 can maintain constant proper distance at a = 2, x = 0.5; a = 1.33, x = 0.75; and a = 1, x = 1. The bolded portion above answers that they do because they are inversely proportional. In other words, all three of those sets of numbers follows x = 1/a. Cited by two good sources and solved by at least two independent methods, I don't understand how you can continue to object with nothing more than "explain, with math, in your own words". Edited November 28, 2013 by Iggy Link to comment Share on other sites More sharing options...

xyzt Posted November 28, 2013 Share Posted November 28, 2013 (edited) The part for which you asked -- the part about maintaining a constant proper distance -- can be quickly proved. Given JVNY's derivation (and illustration) of his thought experiment: The proper distance at t=0 is given along the x axis as: [math]\sigma = 1 - 0.5 = 0.5[/math] The proper distance at the dotted line between the accelerated hyperbolic worldlines is given from the numbers in this frame by length contraction: [math]\sigma = L_{0}\sqrt{1-v^{2}} = (1.25-0.625)(\sqrt{1-0.6^{2}}) = 0.5[/math] I asked IVNY to explain how he arrived to his conclusions, I asked you to justify your claims that velocity and speed are interchangeable and that "light speed slowdown explains the Shapiro delay". You are trying to answer the question posed to IVNY but your approach using length contraction is incorrect. Let's try again: the trajectories of the three rockets are [math]x_i(\tau)=x_{i0}+\frac{c^2}{a_i}(ch \frac{a_i \tau}{c}-1)[/math] with [math]i=1,3[/math] and [math]x_{i0}=0.5, 0.75,1.0[/math]. It is obvious that for any given [math]\tau >0[/math] [math]x_2(\tau)-x_1(\tau) \ne x_3(\tau)-x_2(\tau)[/math], so a naive solution based on length contraction fails miserably. You may not realize that what you tried to prove is that [math]x_2(t)-x_1(t) = x_3(t)-x_2(t)[/math], i.e. the distance between the rockets stays constant in the frame of the launch pad. So, the rest frame (the inertial frame, or lab frame) calculates that the proper distance between the front and rear rockets remains equal (always 0.5 in their frame) while the front accelerates at a=1 and the aft at a=2. The lab frame cannot "calculate that the proper distance between the front and rear rockets remains...", this is a contradiction of terms. The proper distance between the rockets is [math]x_2(\tau)-x_1(\tau) , x_3(\tau)-x_2(\tau)[/math] . What the "lab frame calculates..." is the coordinate , not the proper distance. Do you know the difference? Edited November 28, 2013 by xyzt Link to comment Share on other sites More sharing options...

Iggy Posted November 28, 2013 Share Posted November 28, 2013 (edited) I asked IVNY to explain how he arrived to his conclusions, I asked you to justify your claims that velocity and speed are interchangeable. Velocity and speed are equivalent in German because the same word in German means both. The site JVNY quoted said "Since Einstein talks of velocity (a vector quantity: speed with direction) rather than speed alone...", but like I said, that site doesn't realize that the quote was originally written in German in 1916, and translated to English in 1920. Einstein's quote says "the law of the constancy of the velocity of light in vacuo... cannot claim any unlimited validity", and the word "velocity" there is a translator's choice. It could have just as easily and correctly been translated as "speed" (just like it was in the other two quotes I gave). You are trying to answer the question posed to IVNY but your approach using length contraction is incorrect. Let's try again: the trajectories of the three rockets are [math]x_i(\tau)=x_{i0}+\frac{c}{a_i^2}(ch \frac{a_i \tau}{c}-1)[/math] with [math]i=1,3[/math] and [math]x_{i0}=0.5, 0.75,1.0[/math]. It is obvious that for any given [math]\tau >0[/math] [math]x_2(\tau)-x_1(\tau) \ne x_3(\tau)-x_2(\tau)[/math], There are a couple of problems here. I'll address them at the end of the post as they require some latex so a naive solution based on length contraction fails miserably. Length contraction is what you get when one frame solves the distance of a length in another frame. It is an application of the Lorentz transformations (the transformation from one frame to another). You may not realize that what you tried to prove is that [math]x_2(t)-x_1(t) = x_3(t)-x_2(t)[/math], i.e. the distance between the rockets stays constant in the frame of the launch pad. Certainly not. Notice what I said in my last post: So, the rest frame (the inertial frame, or lab frame) calculates that the proper distance between the front and rear rockets remains equal (always 0.5 in their frame)... these numbers work out in Rindler coordinates (in which all observers are taken to be at rest in their accelerated frame. I'm calculating "their frame" (the accelerated observers). I'm not calculating distance in the lab frame. Their proper distance is measured in their frame and that is what I calculated (which is to say... not the person on the ground doing the calculating, but them whom are accelerating doing the measuring) JVNY said the same earlier: In their frame, there is always 0.25 distance between rear and center, and 0.25 distance between center and front. In the inertial frame the distances length contract, however In their frame the distance remains constant. In the lab frame it shortens. Returning to the thing before... it looks like you mistyped/miscopied the equation for proper distance of uniform accelerated motion. The equation as given here (second one down) is: [math](\frac{c^2}{a}) [\cosh(\frac{aT}{c}) - 1][/math] and you wrote: [math]\frac{c}{a_i^2}(\cosh \frac{a_i \tau}{c}-1)[/math] so it looks like you squared the denominator of the first term rather than the numerator. However, even if you had gotten that portion correct, this still wouldn't work: [math]x_i(\tau)=x_{i0}+\frac{c}{a_i^2}(ch \frac{a_i \tau}{c}-1)[/math] Because [math]x_{i0}[/math] gives part of the distance in the lab frame and [math]\frac{c^2}{a_i}(\cosh \frac{a_i \tau}{c}-1)[/math] gives the other part in the accelerated frame. All distances must be measured in the accelerated frame in order to calculate the distance in the accelerated frame. One can't simply add lab frame proper distances to accelerated frame proper distances when the claim to which you disagree is that distance remains constant for observers undergoing these particular accelerations in their (accelerated) frames Edited November 28, 2013 by Iggy 1 Link to comment Share on other sites More sharing options...

xyzt Posted November 28, 2013 Share Posted November 28, 2013 (edited) Velocity and speed are equivalent in German because the same word in German means both. You are playing word games, so what would velocity/speed "slowdown" mean in your post on the explanation of the Shapiro delay be? Contrary to your claims, there is no "speed slowdown" in the mainstream explanation of the Shapiro delay. I'd like you to answer the challenge to your claim that : Exactly. Shapiro delay confirms it. Light passing a mass is slowed on approach and slowed likewise on departure. yep yep Or you may want to retract it. In their frame the distance remains constant. In the lab frame it shortens. But your attempt at proving that the "distance is constant" used length contraction, so you are contradicting yourself or you do not realize that you have actually (as you admitted at the end of your post) actually calculated the length in the lab frame.See your [math]L=L_0 \sqrt {1-v^2}[/math]? As an aside, how did you get [math]v=0.6c[/math] in the above? The rockets have different proper accelerations, they have different speeds, so how did you get the value you plugged in? Returning to the thing before... it looks like you mistyped/miscopied the equation for proper distance of uniform accelerated motion. The equation as given here (second one down) is: [math]\frac{c^2}{a} (cosh(\frac{aT}{c}) - 1)[/math] and you wrote: [math]\frac{c}{a_i^2}(\cosh \frac{a_i \tau}{c}-1)[/math] Yes, it was a typo, I corrected it. Doesn't change the fact that your "proof" failed. Try proving that [math]\frac{c^2}{a_3} (cosh(\frac{a_3T}{c}) - 1)-\frac{c^2}{a_2}(cosh(\frac{a_2T}{c}) - 1)=constant[/math], please. One last thing, at [math]t=0[/math], [math]T=0[/math] so [math]x_{i0}=X_{i0}[/math] because the rockets are at rest, so their instantaneous frames of reference coincide with the launcher frame of reference. Therefore, contrary to your objection, [math]x_i(T)=x_{i0}+\frac{c^2}{a_i} (cosh(\frac{a_iT}{c}) - 1)[/math] is the correct general formula since [math]x_i(T=0)=x_i(t=0)=x_{i0}[/math]. But this is just a set of secondary issues, a sideshow to our MAIN disagreement. I'd like you to answer the challenge to your claim that : Exactly. Shapiro delay confirms it. Light passing a mass is slowed on approach and slowed likewise on departure. yep yep Or you may want to retract it. Either way, I'd like you to stop evading the main topic of our disagreement and address it. Can anyone help explain whether gravity slows light going directly up or down a gravity well over an extended (not infinitesimally small) path, Mainstream view is that there is no such thing as a "light slowdown". Edited November 28, 2013 by xyzt Link to comment Share on other sites More sharing options...

Iggy Posted November 29, 2013 Share Posted November 29, 2013 You are playing word games A site that says Einstein meant "velocity" rather than "speed" when he said "geschwindigkeit" is, itself, playing word games. I just corrected the site. Whenever Einstein says "velocity" in his early publications, one should never assume that he must be referring to a direction-dependent quantity. Plus I gave two other quotes saying the same thing that use the english word "speed". I'm sorry this appears to you as a word game. so what would velocity/speed "slowdown" mean in your post on the explanation of the Shapiro delay be? Contrary to your claims, there is no "speed slowdown" in the mainstream explanation of the Shapiro delay. I didn't explain Shapiro delay -- I said that it verifies a variable speed of light. It is shown to do that because the speed of light approaches zero as light approaches an event horizon (either the horizon of a black hole or the horizon of rindler coordinates) My claim also means that there are many sources saying the same: The rule about the speed of light being constant only applies locally to patches of spacetime small enough to be effectively flat, i.e. ones which can be described by special relativity... Close to the sun... from our point of view the light is going slower. University of Illinois - Department of Physics On September 8, 2002, Jupiter passed almost in front of a quasar, and Kopeikin and Fomalont made precise measurements of the Shapiro delay with picosecond timing accuracy... ...several authors pointed out that this 1.5PN effect does not depend on the speed of propagation of gravity, but rather only depends on the speed of light [14, 288Jump To The Next Citation Point, 232, 49, 233] - The Confrontation between General Relativity and Experiment - 3.43 - Clifford M. Will since speed of light is reduced in the gravitational field, [math]c` < c[/math], travel time is increased... this is called the Shapiro delay Gravitational Lensing - p.44 - Matthias Bartelmann (University of Heidelberg, Germany) These are the citations, logic, and science I'm willing to devote to the subject. I would rather focus on the other thing we're talking about since it came up first and it's closer to the topic, so please forgive me if you keep talking about Sapiro delay and I keep failing to respond to it. But your attempt at proving that the "distance is constant" used length contraction, so you are contradicting yourself or you do not realize that you have actually (as you admitted at the end of your post) actually calculated the length in the lab frame.See your [math]L=L_0 \sqrt {1-v^2}[/math]? Distances can be constant in a length-contracted frame. I feel like I'm befuddling you. As an aside, how did you get [math]v=0.6c[/math] in the above? The rockets have different proper accelerations, they have different speeds, so how did you get the value you plugged in? It can be done with two equations of which I can think (both given on this page). I'll do a=1 and a=2 (the velocity should be 0.6 with both equations for both accelerations) like you ask me to demonstrate... a=1: [math]v = c \cdot \tanh{ \left( \frac{a \tau}{c} \right)} = \tanh{(0.6931)} = 0.6[/math] [math]v = \frac{at}{\sqrt{1+ \left( \frac{at}{c} \right)^2}} = \frac{2 \cdot 0.75}{\sqrt{1 + (2 \cdot 0.75)^2}} = 0.6[/math] a = 2: [math]v = c\cdot\tanh{ \left( \frac{a \tau}{c} \right)} = \tanh{(2 \cdot 0.34665)} = 0.6[/math] [math]v = \frac{at}{\sqrt{1+ \left( \frac{at}{c} \right)^2}} = \frac{2 \cdot 0.375}{\sqrt{1 + (2 \cdot 0.375)^2}} = 0.6[/math] So... that's how it can be done... how one can get v = 0.6 JVNY also did the calculation earlier, but I don't remember what method he used Yes, it was a typo, I corrected it. Doesn't change the fact that your "proof" failed. Try proving that [math]\frac{c^2}{a_3} (cosh(\frac{a_3T}{c}) - 1)-\frac{c^2}{a_2}(cosh(\frac{a_2T}{c}) - 1)=constant[/math], please. Why would the proper distance traveled from launch pad to finish of the lead ship minus the proper distance of the middle ship be constant? Why would I attempt to prove that? Nobody claimed those things were constant and it isn't logical or relevant to ask for that proof. The claim is that the distance between ships is constant, not that the distance traveled by any one ship is constant with the distance traveled by another. That was really off base... let me draw a diagram and see if we can get on the same page... Given the blue line is 0.5 in the lab frame, can you find the length of the green line in a frame with a relative leftward velocity of 0.6c (relative to the lab frame)? I'm asking because it is necessary to answer the above question in order to calculate what we're doing -- the proper distance between them -- and it is quite literally the easiest problem relativity has to offer. If you're familiar with the Minkowski metric then it can be done in your head. A simple number for the length of the green line in the v=0.6 frame would delight me beyond telling. Please don't answer that you've said previously that you don't type numbers, and I know your history shows a pattern such that you are inclined to point out an error in the last three paragraphs rather than answering, but I'm asking, please: you disagree with my number so... what is the length of the green line in a frame v=0.6c relative to the frame showed? Link to comment Share on other sites More sharing options...

xyzt Posted November 29, 2013 Share Posted November 29, 2013 (edited) I didn't explain Shapiro delay -- I said that it verifies a variable speed of light. It is shown to do that because the speed of light approaches zero as light approaches an event horizon (either the horizon of a black hole or the horizon of rindler coordinates) Despite of your repeated claims, there is no "variable speed of light" and Shapiro delay doesn't "verify" any such thing, the Shapiro delay mainstream explanation (you can try MTW, Rindler, Moller, to name just afew), shows that the delay is due to the deviation of the geodesic from a straight line. I gave the mainstream explanation in this post, the one that you chose to ignore. For further information, there is no light approaching the event horizon in the Shapiro delay. a=1: [math]v = c \cdot \tanh{ \left( \frac{a \tau}{c} \right)} = \tanh{(0.6931)} = 0.6[/math] [math]v = \frac{at}{\sqrt{1+ \left( \frac{at}{c} \right)^2}} = \frac{2 \cdot 0.75}{\sqrt{1 + (2 \cdot 0.75)^2}} = 0.6[/math] a = 2: [math]v = c\cdot\tanh{ \left( \frac{a \tau}{c} \right)} = \tanh{(2 \cdot 0.34665)} = 0.6[/math] [math]v = \frac{at}{\sqrt{1+ \left( \frac{at}{c} \right)^2}} = \frac{2 \cdot 0.375}{\sqrt{1 + (2 \cdot 0.375)^2}} = 0.6[/math] So... that's how it can be done... how one can get v = 0.6 Err, what did you do with [math]\tau[/math]? How did it "disappear" from the formulas? [math]v=v(\tau)[/math] , remember? You just calculated the value of the speed at a particular instant in time , yet you extended the claims to the whole trajectory. This is clearly incorrect. Why would the proper distance traveled from launch pad to finish of the lead ship minus the proper distance of the middle ship be constant? Why would I attempt to prove that? ....because that is the proper distance between the two ships. And you claimed that that distance is constant for ant value [math]\tau[/math]. So, I would like to see your calculations. Please refrain from plugging in numbers and from pointing to colored lines on drawings. You appear to be thinking that the proper distance between ships is to be measured at the points (in time) where the three ships have the same proper distance, this means different values of the proper time [math]\tau[/math] for each of them (due to the differences in proper acceleration, the ships do not achieve the same speed at the same proper time). Proper distance is measured by marking the points at the same time, this is why your approach is incorrect. Edited November 29, 2013 by xyzt Link to comment Share on other sites More sharing options...

Iggy Posted November 29, 2013 Share Posted November 29, 2013 Despite of your repeated claims, there is no "variable speed of light" and Shapiro delay doesn't "verify" any such thing, the Shapiro delay mainstream explanation (you can try MTW, Rindler, Moller, to name just afew), shows that the delay is due to the deviation of the geodesic from a straight line. I gave the mainstream explanation in this post, the one that you chose to ignore. The blue line is 0.5 in the rest frame Since you disagree with my calculation of the proper distance between the first and the third worldlines, could you please calculate the distance of the green line in a v=0.6 frame? Thank you. I should also mention -- I appreciate your persistence and dedication in neg-reppring all of my posts. Occasionally, I think you might forget one, but you are very meticulous and I can appreciate that quality. Any case... the length of the green line? Link to comment Share on other sites More sharing options...

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