robinpike Posted June 30, 2014 Share Posted June 30, 2014 When photons move through empty space, they are found to always move at the same speed, i.e. that of the speed of light. This property of 'a single speed' can be demonstrated by the situation of several photons moving parallel to each other, side-by-side, always staying abreast to each other. This is regardless as to how the photons were created, or when they were created. In order for this to be the case, does this mean that light - no matter where it is in the universe and when it was created - is moving to an absolute frame of reference? Link to comment Share on other sites More sharing options...
swansont Posted June 30, 2014 Share Posted June 30, 2014 When photons move through empty space, they are found to always move at the same speed, i.e. that of the speed of light. This property of 'a single speed' can be demonstrated by the situation of several photons moving parallel to each other, side-by-side, always staying abreast to each other. This is regardless as to how the photons were created, or when they were created. In order for this to be the case, does this mean that light - no matter where it is in the universe and when it was created - is moving to an absolute frame of reference? That you can observe this measurement to be c in any inertial frame means there is no absolute frame. By definition. The frame of a photon is not an inertial frame. Link to comment Share on other sites More sharing options...
robinpike Posted June 30, 2014 Author Share Posted June 30, 2014 That you can observe this measurement to be c in any inertial frame means there is no absolute frame. By definition. The frame of a photon is not an inertial frame. Our measurement of the speed of a photon as 'c' in any inertial frame is not needed for the deduction. We only need to observe that a group of photons stay together. I don't understand how the observation that 'all photons move at the same speed' - gives rise to the deduction that photons do not move to an absolute frame of reference. Using the observation that photons move at the same speed, what are the steps in logic that break the deduction that they move to an absolute frame of reference? Link to comment Share on other sites More sharing options...
swansont Posted June 30, 2014 Share Posted June 30, 2014 Our measurement of the speed of a photon as 'c' in any inertial frame is not needed for the deduction. But it's part of your premise — that we observe the speed to be c — and it requires the conclusion be that there is no absolute frame. Ignoring this doesn't make it untrue. We only need to observe that a group of photons stay together. I don't understand how the observation that 'all photons move at the same speed' - gives rise to the deduction that photons do not move to an absolute frame of reference. Using the observation that photons move at the same speed, what are the steps in logic that break the deduction that they move to an absolute frame of reference? I don't see any. AFAICT it's not the argument that one would use to get to that conclusion. This is a non sequitur. Link to comment Share on other sites More sharing options...
robinpike Posted June 30, 2014 Author Share Posted June 30, 2014 But it's part of your premise — that we observe the speed to be c — and it requires the conclusion be that there is no absolute frame. Ignoring this doesn't make it untrue. Our observation of the speed to be 'c' is not part of the premise at all - the observation is that photons move at the same speed - there is no need for us to measure what that speed is, in that observation. Please explain in more detail how the observation that photons move at the same speed, can lead to a conclusion other than they move to an absolute frame of reference? Link to comment Share on other sites More sharing options...
xyzt Posted June 30, 2014 Share Posted June 30, 2014 Please explain in more detail how the observation that photons move at the same speed, can lead to a conclusion other than they move to an absolute frame of reference? The correct statement is: IN VACUUM ONLY, photons move AT THE SAME SPEED © with respect TO ALL REFERENCE FRAMES. Now , Robin, what would that "absolute frame of reference" be? Link to comment Share on other sites More sharing options...
swansont Posted June 30, 2014 Share Posted June 30, 2014 Our observation of the speed to be 'c' is not part of the premise at all - the observation is that photons move at the same speed - there is no need for us to measure what that speed is, in that observation. You said that "When photons move through empty space, they are found to always move at the same speed, i.e. that of the speed of light." So, silly me, I thought that this was part of the discussion. No matter. Please explain in more detail how the observation that photons move at the same speed, can lead to a conclusion other than they move to an absolute frame of reference? One possibility is that they all move at c regardless of reference frame. i.e. the lack of dispersion does not lead one to the conclusion of an absolute frame. That's only one possibility consistent with the observation. Link to comment Share on other sites More sharing options...
robinpike Posted June 30, 2014 Author Share Posted June 30, 2014 The correct statement is: IN VACUUM ONLY, photons move AT THE SAME SPEED © with respect TO ALL REFERENCE FRAMES. Now , Robin, what would that "absolute frame of reference" be? It is not necessary to add "with respect TO ALL REFERENCE FRAMES" (even though that is an observation). For the discussion on whether photons move to an absolute frame of reference, or not, all that is needed is the observation that a group of photons stay together, regardless as to which photons are picked to be in the group. If the point of view is that the photons do not move to a single reference frame, then what is the explanation as to how the group of photons stay together? Link to comment Share on other sites More sharing options...
swansont Posted June 30, 2014 Share Posted June 30, 2014 If the point of view is that the photons do not move to a single reference frame, then what is the explanation as to how the group of photons stay together? Because they always move at c with respect to any observer. Link to comment Share on other sites More sharing options...
robinpike Posted June 30, 2014 Author Share Posted June 30, 2014 You said that "When photons move through empty space, they are found to always move at the same speed, i.e. that of the speed of light." So, silly me, I thought that this was part of the discussion. No matter. True - my mistake, the introduction should not have mentioned our measurement of the speed of light - that is not the important part: the important part is that a group of photons , however picked, stay together in space. Because they always move at c with respect to any observer. But it is not necessary to include "with respect to any observer". If the deduction that photons move to a single reference frame is being considered as false, then what are the steps that refute that deduction? Please only consider how the group of photons stay together - not what different observers measure the speed of the photons. Link to comment Share on other sites More sharing options...
xyzt Posted June 30, 2014 Share Posted June 30, 2014 But it is not necessary to include "with respect to any observer". Actually, it is. If the deduction that photons move to a single reference frame is being considered as false, then what are the steps that refute that deduction? From a wrong premise you can draw any conclusion. This is known under "GiGo" (garbage in, garbage out). Link to comment Share on other sites More sharing options...
robinpike Posted June 30, 2014 Author Share Posted June 30, 2014 robinpike, on 30 Jun 2014 - 5:28 PM, said: But it is not necessary to include "with respect to any observer". Actually, it is. Quote robinpike, on 30 Jun 2014 - 5:28 PM, said:: If the deduction that photons move to a single reference frame is being considered as false, then what are the steps that refute that deduction? From a wrong premise you can draw any conclusion. This is known under "GiGo" (garbage in, garbage out).Actually, it is. The premise is that a group of photons moving together in space stay together - the deduction is that are moving to a single reference frame. Please explain how the premise is garbage? Link to comment Share on other sites More sharing options...
xyzt Posted June 30, 2014 Share Posted June 30, 2014 (edited) The premise is that a group of photons moving together in space stay together Correct - the deduction is that are moving to a single reference frame.Please explain how the premise is garbage? Because it doesn't follow, the photons "move together" in ALL reference frames. Hence GiGo. Edited June 30, 2014 by xyzt 1 Link to comment Share on other sites More sharing options...
robinpike Posted June 30, 2014 Author Share Posted June 30, 2014 xyzt, thank you for taking part in this discussion, but please think more carefully before you start posting: garbage in, garbage out. The premise is that photons moving together, stay together - that is the premise. The deduction is that they are moving to a single reference frame. Please can you explain how the deduction doesn't follow? Thank you. Okay, here is a different way of expressing the premise and deduction. Two spaceships are next to each other moving through space - the observation is that they always stay abreast of each other. The deduction is that the two spaceships are moving in the same reference frame. Before we go back to the original premise and deduction, does anyone disagree with this deduction from the observation? Link to comment Share on other sites More sharing options...
xyzt Posted June 30, 2014 Share Posted June 30, 2014 (edited) xyzt, thank you for taking part in this discussion, but please think more carefully before you start posting: garbage in, garbage out. Because this is what you have been doing <shrug> Two spaceships are next to each other moving through space - the observation is that they always stay abreast of each other. The deduction is that the two spaceships are moving in the same reference frame. You mean that they share the same reference frame, you don't even have your basic terminology right. This doesn't mean anything about the existence of an "absolute" reference frame. Edited June 30, 2014 by xyzt Link to comment Share on other sites More sharing options...
Klaynos Posted June 30, 2014 Share Posted June 30, 2014 xyzt, thank you for taking part in this discussion, but please think more carefully before you start posting: garbage in, garbage out. The premise is that photons moving together, stay together - that is the premise. The deduction is that they are moving to a single reference frame. Please can you explain how the deduction doesn't follow? Thank you. Okay, here is a different way of expressing the premise and deduction. Two spaceships are next to each other moving through space - the observation is that they always stay abreast of each other. The deduction is that the two spaceships are moving in the same reference frame. Before we go back to the original premise and deduction, does anyone disagree with this deduction from the observation? The deduction doesn't follow as they are moving together in all frames, not just one. Link to comment Share on other sites More sharing options...
pzkpfw Posted June 30, 2014 Share Posted June 30, 2014 (edited) If you want to use light as some absolute reference frame then you could try to turn it around ... Take three spaceships moving such that they can all measure off each other and know they are moving at different speeds (well, velocity, but ignore direction). "Aha", they say, "we'll measure our 'real' speed off the absolute reference frame". So they all measure the speed of light. Oops. It's c, for all of them. Edited June 30, 2014 by pzkpfw Link to comment Share on other sites More sharing options...
swansont Posted June 30, 2014 Share Posted June 30, 2014 True - my mistake, the introduction should not have mentioned our measurement of the speed of light - that is not the important part: the important part is that a group of photons , however picked, stay together in space. But it is not necessary to include "with respect to any observer". If the deduction that photons move to a single reference frame is being considered as false, then what are the steps that refute that deduction? Please only consider how the group of photons stay together - not what different observers measure the speed of the photons. You still can't draw your conclusion, as more than one explanation works — more evidence is needed. A may imply B, but the observation of B does not imply A. That's basic logic. Link to comment Share on other sites More sharing options...
Janus Posted June 30, 2014 Share Posted June 30, 2014 xyzt, thank you for taking part in this discussion, but please think more carefully before you start posting: garbage in, garbage out. The premise is that photons moving together, stay together - that is the premise. The deduction is that they are moving to a single reference frame. Please can you explain how the deduction doesn't follow? Thank you. Okay, here is a different way of expressing the premise and deduction. Two spaceships are next to each other moving through space - the observation is that they always stay abreast of each other. The deduction is that the two spaceships are moving in the same reference frame. Before we go back to the original premise and deduction, does anyone disagree with this deduction from the observation? xyzt, thank you for taking part in this discussion, but please think more carefully before you start posting: garbage in, garbage out. The premise is that photons moving together, stay together - that is the premise. The deduction is that they are moving to a single reference frame. Please can you explain how the deduction doesn't follow? Thank you. Okay, here is a different way of expressing the premise and deduction. Two spaceships are next to each other moving through space - the observation is that they always stay abreast of each other. The deduction is that the two spaceships are moving in the same reference frame. Before we go back to the original premise and deduction, does anyone disagree with this deduction from the observation? Actually, the deduction is that they are at rest with respect to one particular frame. There are an infinite number of reference frames in which they are moving. They stay together in all of these frames, so this fact does not give any of the frames priority or preference over the others. To claim a absolute frame of reference you have to show that there is one frame that is unique or special. This is where the consistency in the speed of light comes in. If you could show that the speed of light was somehow tied to one particular frame, then you can say that that an absolute frame exists. For example, you and someone else are moving relative to each other. At the instant you pass each other you both briefly flash two flashlights in opposite directions and along the line of your relative motion. Now according to both of you, the individual photons of each flash stay together as a group. In addition, the two groups of photons from the flashlights pointing in the same direction stay side by side. Neither of these facts of themselves are an indication of there being an special or unique frame of reference. Instead we need to look at how those groups behave as measured by by the two of you. Now, let's say that you measure that all the photons maintain an equal distance from you all all times and the other person measured the distance between one pair of photon and himself as being and the distance between himself and the other pair as being unequal. Thus for you, the speed of all the groups are the same relative to you, while from him they are not. Now this would be evidence of a unique absolute frame of reference. It would be the in which light always travels at the same speed with respect to. However, in the real universe, this is not what happens. both of you measure the groups of light as all moving at the same speed with respect to yourself. There is no uniqueness between you two reference frames and they cannot be told apart from each other. (if the two of you suddenly switched reference frames, neither of you would be able to tell it from observing the behavior of the light groups.) Since all reference frames will measure this same thing, there is no unique or preferred frame that can be called absolute. 4 Link to comment Share on other sites More sharing options...
md65536 Posted July 3, 2014 Share Posted July 3, 2014 The premise is that a group of photons moving together in space stay together - the deduction is that are moving to a single reference frame. Please explain how the premise is garbage? Because they are only together in some reference frames. If they are "side-by-side" implying separated by some distance and emitted simultaneously, you must realize that in some other frames they are not emitted simultaneously and thus not "together" in all frames. If they are not separated by any distance then the argument is vacuous. Otherwise, you have to falsely assume absolute simultaneity in order to conclude that they're "together" for all observers. Link to comment Share on other sites More sharing options...
xyzt Posted July 4, 2014 Share Posted July 4, 2014 Because they are only together in some reference frames. If they are "side-by-side" implying separated by some distance and emitted simultaneously, you must realize that in some other frames they are not emitted simultaneously and thus not "together" in all frames. Any two particles that have a comoving frame, are comoving in all frames. The reason is simple, they have the same speed wrt of all frames. You are misapplying RoS. Link to comment Share on other sites More sharing options...
md65536 Posted July 4, 2014 Share Posted July 4, 2014 (edited) Any two particles that have a comoving frame, are comoving in all frames. The reason is simple, they have the same speed wrt of all frames. You are misapplying RoS.No, I'm not. You're right that they have the same speed, but if they're not emitted simultaneously then they won't be "several photons moving parallel to each other, side-by-side, always staying abreast to each other," as per OP. Unless their separation is negligible, they can't be emitted simultaneously in all frames. This is a correct application of RoS. For example, if there is separation and you allow observers arbitrarily close to the speed of light, you can shorten the travel time of light arbitrarily (due to length contraction), and contrive an observer for whom one photon reaches its destination before the other photon is emitted. In this case the photons don't travel together in any way. Edited July 4, 2014 by md65536 Link to comment Share on other sites More sharing options...
xyzt Posted July 4, 2014 Share Posted July 4, 2014 (edited) No, I'm not. You're right that they have the same speed, but if they're not emitted simultaneously then they won't be "several photons moving parallel to each other, side-by-side, always staying abreast to each other," as per OP. Unless their separation is negligible, they can't be emitted simultaneously in all frames. This is a correct application of RoS. For example, if there is separation and you allow observers arbitrarily close to the speed of light, you can shorten the travel time of light arbitrarily (due to length contraction), and contrive an observer for whom one photon reaches its destination before the other photon is emitted. In this case the photons don't travel together in any way. The speed of the two particles is 0 in their comoving frame. Their relative speed is zero. The speed of the two particles is V in any frame moving at speed V wrt their comoving frame. Their relative speed is....0! Nothing to do with RoS. Edited July 4, 2014 by xyzt Link to comment Share on other sites More sharing options...
pzkpfw Posted July 4, 2014 Share Posted July 4, 2014 I would expect their relative position may not be the same according to all observers in all frames (i.e. are they side by side according to everyone?) due to ROS, but their relative position will stay the same, according to an observer, if their speed (speed of the two particles) is the same. There's an experiment involving a falling "horizontal" bar. It turns out an observer moving sideways with respect to the bar, won't (due to ROS) necessarily consider the bar to be horizontal. But, they don't see the bar turning, the relative position of the ends stays the same. (The ends of the bar I am equating to the two particles in question.) Link to comment Share on other sites More sharing options...
md65536 Posted July 4, 2014 Share Posted July 4, 2014 (edited) The speed of the two particles is 0 in their comoving frame. Their relative speed is zero. The speed of the two particles is V in any frame moving at speed V wrt their comoving frame. Their relative speed is....0! Nothing to do with RoS. Are the photons emitted simultaneously in all frames? If no, are they side-by-side in all frames? If no, is OP's premise in the thought experiment flawed? Isn't that relevant? Do photons have a comoving frame? Edit: I guess you could put the photons together and thus have them emitted simultaneously in every frame, and still come to the same conclusions as OP did if you try to consider an invalid frame in which the photons' relative speed is 0. So RoS isn't a necessary part of the answer and it's possible to have the photons stay together in every frame. I guess this would only come into play if you used the distance between the photons in a calculation that derives a contradiction if absolute simultaneity is assumed, which I don't think has been done here, so you're right that RoS need not be considered. There's an experiment involving a falling "horizontal" bar. It turns out an observer moving sideways with respect to the bar, won't (due to ROS) necessarily consider the bar to be horizontal. But, they don't see the bar turning, the relative position of the ends stays the same. (The ends of the bar I am equating to the two particles in question.)That's a good point. No matter how distorted the bar is by relativistic effects, it remains "together" in any frame. OP could use that if "the photons remain side-by-side" was replaced with "the photons remain close to each other." However the bar is only an analogy, since the bar has a rest frame but photons don't. (The bar ends don't have an invariant "single speed" but the photons do.) Edited July 4, 2014 by md65536 Link to comment Share on other sites More sharing options...
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