Lowemack Posted February 17, 2012 Author Share Posted February 17, 2012 I think I get it:- Maxwell calculated that the speed of light is invariant. Michleson Morley proved it. This meant that if nothing can travel faster than light then we need to use Lorenze transformation formulae for relativistic speed additions. C is constant because of how we add up relative speeds. Link to comment Share on other sites More sharing options...

IM Egdall Posted February 17, 2012 Share Posted February 17, 2012 (edited) I think I get it:- Maxwell calculated that the speed of light is invariant. Michleson Morley proved it. This meant that if nothing can travel faster than light then we need to use Lorenze transformation formulae for relativistic speed additions. C is constant because of how we add up relative speeds. Not quite: Yes, Maxwell's equations say the speed of light is invariant. But physicists had difficulty interpreting the meaning of this. In 1905, Einstein proposed his light postulate - which in effect says no matter what (uniform) speed you are traveling at, no matter what speed the source of the light is going at, you measure the speed of the light beam as the same value (i.e. invariant). No, Michelson Morley did NOT prove the invariance of the speed of light. MM just found no evidence for the ether. Their experiment said nothing about the invariance of the speed of light. (A number of later experiments, however, did verify the invariance of the speed of light). C is constant because the speed of light, c is invariant. Einstein's formula on how to combine relative speeds obeys this principle. Edited February 17, 2012 by IM Egdall Link to comment Share on other sites More sharing options...

DrRocket Posted February 17, 2012 Share Posted February 17, 2012 Yes, MM failed to detect the ether. But it said nothing about the absolute speed of light. Not quite. Given that the invariance of the speed of light means that the speed is the same in ALL inertial reference frames, no experiment can prove that. There are too many invariant reference frames. There are in fact uncountably many inertial reference frames and one simply cannot check all of them directly by experiment. However, the alternative, assuming that Maxwell's equations are correct, is that either Maxwell's equations apply in only a single preferred reference frame, by definition the ether frame, or else the speed of light is indeed invariant. The former was the consensus prior to the Michelson Morely experiment. The experiment supported the latter alternative. If one wants to be picky, and you are being picky, then strictly speaking the existence of the ether has not been disproved. In fact there is a perfectly valid theory, the Lorentz Ether Theory, that includes an ether and is experimentally indistinguishable from special relativity -- it makes exactly the same predictions as does special relativity. So, again strictly speaking, the Michelson Morely experiment and special relativity did not and do not disprove the existence of an ether, but rather simply make an ether irrelevant in classical electrodynamics. 1 Link to comment Share on other sites More sharing options...

PaulS1950 Posted February 18, 2012 Share Posted February 18, 2012 The speed of light appears to be constant because we allow time and space to change to keep it at that speed. If we were to say that time or space was constant then we could manipulate velocity to make it seem so. These are our perceptions of an existance that we are still learning about. The major stumbling block that I see is that we have to learn that matter is only a perception and has no real for other than energy. Paul Link to comment Share on other sites More sharing options...

JohnStu Posted February 29, 2012 Share Posted February 29, 2012 I don't get why C is constant either. Someone enlighten me, please! Link to comment Share on other sites More sharing options...

swansont Posted March 1, 2012 Share Posted March 1, 2012 I don't get why C is constant either. Someone enlighten me, please! Did you read the thread? How about a question that's a little more specific? Link to comment Share on other sites More sharing options...

D H Posted March 1, 2012 Share Posted March 1, 2012 The speed of light appears to be constant because we allow time and space to change to keep it at that speed. If we were to say that time or space was constant then we could manipulate velocity to make it seem so. These are our perceptions of an existance that we are still learning about. The major stumbling block that I see is that we have to learn that matter is only a perception and has no real for other than energy. No. I don't get why C is constant either. Someone enlighten me, please! I. Special Relativity As a starting point, the speed of light is constant because that is exactly what falls out of Maxwell's equations. Neither the velocity of the emitter nor that of the receiver appears in the calculation of the velocity of Maxwell's electromagnetic waves. There were two big problems here: Every wave phenomenon known to the physicists of that time required some kind of medium as a transport mechanism for the waves, and This was very much at odds with Newtonian mechanics. This conflict between electromagnetism and Newtonian mechanics was one of the biggest problems of latter 19th century physics. There were two ways around this. One is to say that there is some special frame, the ether frame, in which the one-way speed of light is indeed c. We can't measure the one-way speed of light. All experiments are ultimately measurements of the two-way (round trip) speed of light. We can't see this ether frame because the Lorentz transformation act to hide it from us. This is Lorentz Ether Theory, the main proponents of which were Poincare and Lorentz. Another way around this dilemma is to simply accept Maxwell's equations at face value: The speed of light is the same to all inertial observers. This requires some fundamental changes in how one views space and time. This was Einstein's insight. These two theories, Lorentz Ether Theory and Special Relativity, are experimentally indistinguishable from one another. So why choose one over the other? Physicists obviously have done so; Lorentz Ether Theory nowadays lives on only at crackpot websites. The reasons are many. Lorentz Ether Theory has some rather goofy axioms. Time dilation and length contraction are axiomatic in Lorentz Ether Theory (watch me pull this rabbit out of a hat!). The existence of an ether frame (also axiomatic) is even worse. This ether frame that lies at the heart of Lorentz Ether Theory is an untestable axiom. It's metaphysics, not physics. Compare to the axioms of special relativity: (1) The speed of light is the same to all inertial observers, which is bizarre but is consistent with theory and experimental results, and (2) The laws of physics are the same to all inertial observers, which is anything but bizarre. This axiom goes back to Galileo (Dialogue Concerning the Two Chief World Systems). The final nail in the coffin for Lorentz Ether Theory was quantum mechanics. There is no need for this metaphysical ether frame that contains some ether that acts as the transport mechanism for electromagnetic phenomena. Electromagnetic phenomena do not need a medium. Electromagnetism is carried by photons. So one answer to your question, "why is C constant?" is that it is axiomatic. If you pester your parents or teachers with "why" question after "why" question, you will inevitably get the non-answer "because we said so." Saying that a constant speed of light is axiomatic is a scientific way of saying "because we said so." II. A bit more philosophical. Start of with a couple of simple observations. (1) There is this thing that we call "space" that appears to be described by Euclidean three dimensional geometry, and (2) there is this thing called "time" that appears to be the independent variable with which we can describe how things interact. Now add one simple hypothesis: There exists one speed such if any one observer sees a particle moving at this speed, all observers see that particle moving at that same speed. In other words, there exists some invariant speed. There are two basic choices for this universally-agreed upon speed. One choice is that this universally-agreed upon speed is infinite. This choice leads to the Newtonian universe. Time is absolute, and space is Euclidean everywhere. Should one observer see something moving at a finite velocity, there will always be some other observer who will see that thing moving at a different finite velocity. The only universally-agreed upon speed is that of a particle that is moving infinitely fast. The other choice is that this universally-agreed upon speed is finite. This requires recasting those simple observations of time and space in a new light. Just because those observations appear to be true locally does not mean that they are universal. All of the experiments that led to the conclusions that time is the independent variable that describes motion through Euclidean three dimensional space are ultimately local experiments. A physical theory must agree with these local experiments. This choice of a finite universal speed coupled with this local need for space to appear to be Euclidean three space and time to appear to be the independent variable leads to a radically different geometry of space and time. Couple this finite speed with some basic laws of physics results in massive particles always moving at a speed less than this universal speed while massless particles can only move at this universal speed. So now we have two competing hypotheses of the universe: The Newtonian hypothesis that the universe is globally Euclidean and time is absolute versus this non-Euclidean universe. There's a simple way to test one against the other: Is there a universally-agreed upon speed that is finite? Experiment after experiment says there is such a finite universally-agreed upon speed, and it is the speed of light. So why is the speed of light constant? Once again, the answer right now is that it is axiomatic ("because we said so"). Of course, those axioms have to agree with reality -- and they do. 2 Link to comment Share on other sites More sharing options...

Lowemack Posted March 6, 2012 Author Share Posted March 6, 2012 So if on my lab, on my space ship, I have a closed long tube with a photo sensor at both ends, connected to an accurate clock. If I travel towards the sun @0.5C, open one end of the tube, the clock starts when the light from the sun hits the first sensor and stops on the second sensor, then I get a measurement of time. If I repeat the experiment moving away from the sun, will I read the same time on my clock. If yes, will the distance between the sensors, as measured on the ship divided by the time on the clock = C. Obviously time dialation/ length contraction can't explain why both experiments give the answer C, I think the answer must be "C is invariant" because it is(as stated above), not because time slows down, or length contracts. But if that is the case, does time really slow down for moving objects and do lengths contract, because surely they would have opposite effects for both experiments, yet the answer is C in both cases. If the photo sensors had lights on them that lit when they were activated, would a stationary observer (relative to the sun) measure the period between pulses as the same for both experiments? His measurement for distance that light travelled, would be different. So how can this be reconciled? Link to comment Share on other sites More sharing options...

swansont Posted March 6, 2012 Share Posted March 6, 2012 So if on my lab, on my space ship, I have a closed long tube with a photo sensor at both ends, connected to an accurate clock. If I travel towards the sun @0.5C, open one end of the tube, the clock starts when the light from the sun hits the first sensor and stops on the second sensor, then I get a measurement of time. If I repeat the experiment moving away from the sun, will I read the same time on my clock. If yes, will the distance between the sensors, as measured on the ship divided by the time on the clock = C. Obviously time dialation/ length contraction can't explain why both experiments give the answer C, I think the answer must be "C is invariant" because it is(as stated above), not because time slows down, or length contracts. But if that is the case, does time really slow down for moving objects and do lengths contract, because surely they would have opposite effects for both experiments, yet the answer is C in both cases. If the photo sensors had lights on them that lit when they were activated, would a stationary observer (relative to the sun) measure the period between pulses as the same for both experiments? His measurement for distance that light travelled, would be different. So how can this be reconciled? You get an answer of c. If I am an observer in another frame, I see that your tube has length contracted and your time has slowed, each by the same factor, so there's no reason to expect a contradiction. Link to comment Share on other sites More sharing options...

DrRocket Posted March 6, 2012 Share Posted March 6, 2012 Obviously time dialation/ length contraction can't explain why both experiments give the answer C, I think the answer must be "C is invariant" because it is(as stated above), not because time slows down, or length contracts. But if that is the case, does time really slow down for moving objects and do lengths contract, because surely they would have opposite effects for both experiments, yet the answer is C in both cases. If you start with "c is invariant", plus the laws of physics are the same in all inertial reference frames, then you can logically deduce the Lorentz transformations that describe length contraction and time dilation. This is the historical train of logic followed by Einstein in the original discovery of special relativity. On the other hand if you start with the Lorentz transformations, i.e. with length contraction and time dilation, then you can deduce that c is invariant. So, yes, time dilation and lenght contraction can indeed explain why "the answer is C in both cases." The two assumptions are logically equivalent. As noted earlier the invariance of c also comes from Maxwell's equations of classical electrodynamics, so on e might say that invariance of c has an established foundation in the physics that was known prior to the special theory of relativity. However, historically the interpretation prior to Einstein was that electromagnetic waves propagated through a universal "aether" that provided an absolute reference frame for Newtonian mechanics. Now we know better. Link to comment Share on other sites More sharing options...

Schrödinger's hat Posted March 7, 2012 Share Posted March 7, 2012 So, yes, time dilation and lenght contraction can indeed explain why To be nit-picky, time dilation+length contraction (at least as they are understood by most who read those words) are not a sufficient explanation on their own. They are, however, a necessary consequence of a constant speed or of the lorentz transforms (both of which are). I suspect that the part Lowemack is missing is the idea of how simultanaety works in a relativistic universe. Not only do durations measured by one clock change between reference frames, but distant, originally synchronised clocks will not show the same time. Link to comment Share on other sites More sharing options...

michel123456 Posted March 7, 2012 Share Posted March 7, 2012 To be nit-picky, time dilation+length contraction (at least as they are understood by most who read those words) are not a sufficient explanation on their own. They are, however, a necessary consequence of a constant speed or of the lorentz transforms (both of which are). I suspect that the part Lowemack is missing is the idea of how simultanaety works in a relativistic universe. Not only do durations measured by one clock change between reference frames, but distant, originally synchronised clocks will not show the same time. Exactly. And what will happen in an experiment in which scientists will manage to make the clocks show the same time (after a fight against Relativity)? Link to comment Share on other sites More sharing options...

Schrödinger's hat Posted March 7, 2012 Share Posted March 7, 2012 Exactly. And what will happen in an experiment in which scientists will manage to make the clocks show the same time (after a fight against Relativity)? Not quite sure what you're getting at/asking here, but if I interpret it correctly, the answer is that they will not show the same time in the original frame. Note that I am taking the conventional definition of present (half-way between past and future in flat space) here, and assuming that 'show' includes correcting for light delay. Link to comment Share on other sites More sharing options...

michel123456 Posted March 7, 2012 Share Posted March 7, 2012 Not quite sure what you're getting at/asking here, but if I interpret it correctly, the answer is that they will not show the same time in the original frame. Note that I am taking the conventional definition of present (half-way between past and future in flat space) here, and assuming that 'show' includes correcting for light delay. I am trying to say that if those 2 clocks are used to measure a speed (knowing also the exact distance between the 2 clocks), the result will be wrong. Link to comment Share on other sites More sharing options...

DrRocket Posted March 7, 2012 Share Posted March 7, 2012 To be nit-picky, time dilation+length contraction (at least as they are understood by most who read those words) are not a sufficient explanation on their own. They are, however, a necessary consequence of a constant speed or of the lorentz transforms (both of which are). To be correct, go read my post. You can derive the Lorentz transformation from the constancy of the speed of light. You can derive the constancy of the speed of light from the Lorentz transformations. It is a matter of taste which proposition you take as being fundamental. They are logically equivalent. Link to comment Share on other sites More sharing options...

Schrödinger's hat Posted March 7, 2012 Share Posted March 7, 2012 To be correct, go read my post. Yes, constant c and Lorentz transform are equivalent. I was trying to point out that 'length contraction and time dilation' is not a sufficient description of the Lorentz transform. Upon re-reading your post, I see you implied are synonymous, which I somehow missed the first time. Anyone familiar with relativity would know what you mean immediately. However, without some mention of simultanaety it is unclear and/or confusing to someone without this knowledge. TL;DR [math] \mbox{Lorentz transform}\leftrightarrow\mbox{constancy of c}\rightarrow \mbox{Time dilation+length contraction} [/math] NB: implication only runs one way on that last. Such are the perils of not using maths to talk about precise concepts. Link to comment Share on other sites More sharing options...

DrRocket Posted March 7, 2012 Share Posted March 7, 2012 (edited) Yes, constant c and Lorentz transform are equivalent. I was trying to point out that 'length contraction and time dilation' is not a sufficient description of the Lorentz transform. Upon re-reading your post, I see you implied are synonymous, which I somehow missed the first time. Anyone familiar with relativity would know what you mean immediately. However, without some mention of simultanaety it is unclear and/or confusing to someone without this knowledge. TL;DR [math] \mbox{Lorentz transform}\leftrightarrow\mbox{constancy of c}\rightarrow \mbox{Time dilation+length contraction} [/math] NB: implication only runs one way on that last. Such are the perils of not using maths to talk about precise concepts. The "relativity of simultaneity" is a statement regarding the surface of constant time in coordinatizations of spacetime corresponding to observers in relative motion. It is usually somewhat loosely stated in textbooks. But it is something that is derived from the Lorentz transformations rather than something that implies those transformations. I know of 3 ways to arrive at special relativity or its equivalent. 1) The traditional way seen in introductory textbooks, which is to postulate that the speed of light is constant in all inertial frames and the equations are the same in such frames. One then goes through some physical reasoning to arrive at the usual Lorentz transformations. This has the advantage of presenting the physical reasoning that was the historic path to the special theory of relativity. 2) Simply choose a reference frame and by fiat demand that all other reference frames are related to it by the usual Lorentz transformations. If you call the chosen frame the "ether frame" this is the Lorentz Ether Theory. It is not usually taught, but it is equivalent to special relativity as it provides EXACTLY the same predictions. The reason that it is not taught is that there is no particular point to teaching it. It does nothing that special relativity in its usual form does not do, and is philosophically less satisfying. It does make clear that special relativity did not really prove the non-existence of an ether, but rather made its existence or non-existence irrelevant. 3) Start with real 4-space and the Minkowski metric. Then determine those transformations that preserve the metric. One notes that these transformations also preserve "light cones" and one can define a positive direction for time (arbitrarily chosen from two possibilities). The metric preserving transformations that also preserve the positive direction of time (orthochronic transformations) are the Lorentz transformations used in physics. This path is rather formal, but has the advantage of revealing deep connections between geometry and the physical theory. It also has the advantage that it is clearly related to the underlying mathematics and makes the mathematical consistency of the special theory of relativity clear (assuming the consistency of ordinary mathematics). There is a large advantage to understanding all three approaches, as it can make clear what is fundamental and what is not, and because for any given issue one perspective may be more enlightening than the others. It is also worthwhile noting that it is sufficient to find any phenomena that propagates at a fixed speed in all inertial frames. That speed, acall it "x" then plays the role of "c" and one gets the usual Lorentz transformations with "x" in place of "c". This shows that there can be only one invariant speed --- a fact that can be derived in reverse from the Lorentz transformations. The fact that the speed of light in a vacuum fulfills that requirement results from experiment or from Maxwell's equations. Edited March 7, 2012 by DrRocket Link to comment Share on other sites More sharing options...

Iggy Posted March 8, 2012 Share Posted March 8, 2012 To be nit-picky, time dilation+length contraction (at least as they are understood by most who read those words) are not a sufficient explanation on their own. They are, however, a necessary consequence of a constant speed or of the lorentz transforms (both of which are). Yes, constant c and Lorentz transform are equivalent. I was trying to point out that 'length contraction and time dilation' is not a sufficient description of the Lorentz transform. I might look at it a little differently in that time dilation or length contraction are no more a consequence of an invariant speed than the other way around. I mean... I'm not sure having a constant finite speed is any more fundamental than time dilation or any other relativistic observation in deriving the Lorentz transforms. Certainly, the Einstein-like way of doing it, Principle of relativity + finite invariant speed -> Lorentz transforms is more popular. But, I think it would be just as admissible to do... Principle of relativity + time dilation -> Lorentz transforms or any other relativistic effect, because if you derive a general set of equations from just the principle of relativity (and the usually unwritten postulates isotropy and homogeneity) you get an equation with one free parameter, usually written Kappa I think. Then you can either set it positive, negative, or zero and the resulting transforms are either Lorentzian, Galilean, or something that makes no physical sense (wikipedia references section 2.1 here as an example of that approach). Any relativistic observation (time dilation, length contraction, invariant speed, relativity of simultaneity) would serve as an additional postulate allowing one to chose the Lorentz transforms. In othe rwords, I don't think I would say that a constant speed is equivalent to the Lorentz transforms any more than I would say time dilation, for example, is equivalent. Link to comment Share on other sites More sharing options...

swansont Posted March 8, 2012 Share Posted March 8, 2012 I think it would be just as admissible to do... Principle of relativity + time dilation -> Lorentz transforms What happens if you assume length to be an invariant? Link to comment Share on other sites More sharing options...

Iggy Posted March 8, 2012 Share Posted March 8, 2012 What happens if you assume length to be an invariant? If you assume the principle of relativity (something you have to do in either case) then your options are only the Galilean or Lorentzian transformations. That was my meaning. In neither case would an invariant length with time dilation be sensible. Looked at another way... if we could for some reason not be able to reach the invariant speed. If light, for some reason, went half the invariant speed and we couldn't accelerate massive objects any faster than light... we still could easily derive the lorentz transforms by making just one time dilation observation. We could even solve what the invariant speed is. Link to comment Share on other sites More sharing options...

Schrödinger's hat Posted March 8, 2012 Share Posted March 8, 2012 (edited) If you assume the principle of relativity (something you have to do in either case) then your options are only the Galilean or Lorentzian transformations. That was my meaning. In neither case would an invariant length with time dilation be sensible. I could put whatever I like into [math] \alpha, \;\beta,\;\gamma,\mbox{ and }\delta[/math] here: [math] x' = \alpha x + \beta t[/math] [math] t' = \delta t + \gamma x [/math] As long as it gives the suitably mirrored result when reversing v and still have a consistent set of equations. [math] x' = \frac{1}{\gamma} x[/math] [math]t' = \gamma t [/math] (with the traditional definition of gamma) Is one example. This produces the correct time dilation and length contraction formulae, it gives the same result in both frames, but it produces nonsense when we try to apply it anywhere else. So Principle of relativity + time dilation -> Lorentz transforms Doesn't hold without further data/constraints/assumptions Principle of relativity + time dilation + transformation consistent with measurements-> Lorentz transforms Does, but then you can just go Principle of relativity + consistent with measurements-> Lorentz transforms Edited March 8, 2012 by Schrödinger's hat Link to comment Share on other sites More sharing options...

swansont Posted March 8, 2012 Share Posted March 8, 2012 If you assume the principle of relativity (something you have to do in either case) then your options are only the Galilean or Lorentzian transformations. That was my meaning. In neither case would an invariant length with time dilation be sensible. Looked at another way... if we could for some reason not be able to reach the invariant speed. If light, for some reason, went half the invariant speed and we couldn't accelerate massive objects any faster than light... we still could easily derive the lorentz transforms by making just one time dilation observation. We could even solve what the invariant speed is. But you haven't a priori assumed an invariant speed with this approach. I'm just wondering if you've underconstrained the problem. Link to comment Share on other sites More sharing options...

Iggy Posted March 8, 2012 Share Posted March 8, 2012 (edited) I could put whatever I like into [math] \alpha, \;\beta,\;\gamma,\mbox{ and }\delta[/math] here: [math] x' = \alpha x + \beta t[/math] [math] t' = \delta t + \gamma x [/math] As long as it gives the suitably mirrored result when reversing v and still have a consistent set of equations. [math] x' = \frac{1}{\gamma} x[/math] [math]t' = \gamma t [/math] (with the traditional definition of gamma) Is one example. velocity needs to be [math]-\beta / \alpha[/math] My conclusion that Lorentzian and Galilean transformations are the only two options with the postulates of the principle of relativity and the isotropy and homogeneity of space is well published. I gave a link in my last post. It gives a number of references. This one sets it out clearly in the abstract which might help: http://physics.sharif.ir/~sperel/paper1.pdf Wikipedia's derivation does the same thing under group postulates, but they forget to mention that the principle of relativity implies the group postulates. Bottom line, you can derive, [math] \begin{bmatrix} t' \\ z' \end{bmatrix} = \frac{1}{\sqrt{1 + \kappa v^2}} \begin{bmatrix} 1 & \kappa v \\ -v & 1 \end{bmatrix} \begin{bmatrix} t \\ z \end{bmatrix}. [/math] without assuming anything about an invariant speed. Setting [math]\kappa = 0[/math] gives the galilean transforms. [math]\kappa > 0[/math] gives a non-physical result. [math]\kappa < 0[/math] is Lorentzian and you set [math]c \, = \, \frac{1}{\sqrt{- \kappa}} [/math]. Once you have the equation above with kappa in it all you need is one relativistic observation and SR is derived. It wouldn't need to be an invariant speed. edit: at least, I don't think it would. Maybe I'm missing something But you haven't a priori assumed an invariant speed with this approach. I'm just wondering if you've underconstrained the problem. In a technical sense, either the invariant speed would be finite or infinite. In neither case would it make sense to have some relativistic effects without others. Edited March 8, 2012 by Iggy Link to comment Share on other sites More sharing options...

I-try Posted April 28, 2012 Share Posted April 28, 2012 I would suggest to Lowemack travelling towards the sun via a rocket travelling at .75C, that he would not be measuring the speed of light by referring to the condition of his clock. . If his rocket was equipped with a spectrometer and if it was still functioning due to its flattened state, it would be indicating that the H and K lines of calcium impacting from the sun would be displaced towards the ultraviolet, and those from earth would be equally displaced towards the read end of the light spectrum. Link to comment Share on other sites More sharing options...

Iggy Posted April 28, 2012 Share Posted April 28, 2012 But you haven't a priori assumed an invariant speed with this approach. I'm just wondering if you've underconstrained the problem. In a technical sense, either the invariant speed would be finite or infinite. In neither case would it make sense to have some relativistic effects without others. Revisiting this, I'm not sure how I thought that answered your point. I wish I could more naturally communicate better. I should have said that one can derive, as a conclusion, that there is an invariant speed when the principle of relativity and time dilation are postulates. Since the only options possible when one assumes the principle of relativity and the isotropy of spacetime are Galilean and Lorentzian, and time dilation is only consistent with the latter, if we assume time dilation then an invariant speed is a consequence along with the Lorentz transforms and the rest of the relativistic effects. In other words, these should both work with a sort of equal admissibility, 1)principle of relativity + 2)homogeneity and isotropy of space and isotropy of time + 3)invariant speed = ---------------------- Lorentz transforms (including time dilation and length contraction) 1)principle of relativity + 2)homogeneity and isotropy of space and isotropy of time + 3)time dilation = ---------------------- Lorentz transforms (including invariant speed and length contraction) I really do have trouble communicating complicated thoughts, so I hope that makes more sense. I would suggest to Lowemack travelling towards the sun via a rocket travelling at .75C, that he would not be measuring the speed of light by referring to the condition of his clock. But if he did use his clock light would still be traveling at c, right? Link to comment Share on other sites More sharing options...

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