yknot Posted November 13, 2013 Posted November 13, 2013 To keep it as simple as possible, we will use only the right half of E’s train example. O1 -----------------------------------------------<~~~~~light O2--> A light ray starts equidistant from inertial observers O1 and O2, as shown. And yet O2 sees the ray before O1. (And this is absolutely before because these are light-like events.)Why this difference? I’m guessing that is has nothing to do with either the event or light’s speed, but that it has everything to do with the fact that the observers moved differently in relation to the approaching light ray. But this does not show that simultaneity is relative; it merely shows that observers moving differently will see events differently. It is wrong to ignore these different motions, and to then go on and say that the events occurred differently for the two train example observers.
DimaMazin Posted November 13, 2013 Posted November 13, 2013 The distance has less meters in O2 frame.The distance has more meters in O1 frame.
swansont Posted November 13, 2013 Posted November 13, 2013 You only show a single event, so where are you thinking simultaneity would come into play?
DimaMazin Posted November 13, 2013 Posted November 13, 2013 You only show a single event, so where are you thinking simultaneity would come into play? I see more events.
swansont Posted November 13, 2013 Posted November 13, 2013 I see more events. Saying what they are would be useful. There's a single flash of light. What else?
yknot Posted November 13, 2013 Author Posted November 13, 2013 The distance has less meters in O2 frame.The distance has more meters in O1 frame. This was 1/2 of E's train ex. & he had his observers equidistant from both light-emitting events. (The lightning strikes could leave burn marks in each frame to verify this, as many have said.)
swansont Posted November 13, 2013 Posted November 13, 2013 This was 1/2 of E's train ex. & he had his observers equidistant from both light-emitting events. (The lightning strikes could leave burn marks in each frame to verify this, as many have said.) I don't think that wording is used regarding the two observers' measurements— the setup is usually that they are co-located when the flash occurs. Saying they are equidistant has implications regarding length contraction. (If equidistant is used in the Einstein train example it would likely refer to the second flash, on the other side of the observers, which you are omitting.)
yknot Posted November 13, 2013 Author Posted November 13, 2013 Saying what they are would be useful. There's a single flash of light. What else? I started (in order to simplify) with a single flash to show that the difference between the two observers in the single-flash case has everything to do with their different motions, and really nothing at all to do with event times. Then this conclusion can easily be extended to cover the entire train example by simply adding a second lite flash, but this is not really necessary is it? (I'm just saying that E's train ex. does not show that sim. is rel. but merely shows that observers moving differently will see events differently if they judge event times by light-rays-from-the-events and totally ignore the fact that they moved differently during the experiment, which is what really caused them to see the events diff'ly.)
Strange Posted November 13, 2013 Posted November 13, 2013 I'm just saying that E's train ex. does not show that sim. is rel. but merely shows that observers moving differently will see events differently They will see things differently because of their relative motion. One of the things they will see differently is whether things are simultaneous or not. You can't analyse this with half the experiment because what the example shows is that the two lightning strikes will be seen as either simultaneous or not simultaneous by the two observers.
swansont Posted November 13, 2013 Posted November 13, 2013 I started (in order to simplify) with a single flash to show that the difference between the two observers in the single-flash case has everything to do with their different motions, and really nothing at all to do with event times. Then this conclusion can easily be extended to cover the entire train example by simply adding a second lite flash, but this is not really necessary is it? (I'm just saying that E's train ex. does not show that sim. is rel. but merely shows that observers moving differently will see events differently if they judge event times by light-rays-from-the-events and totally ignore the fact that they moved differently during the experiment, which is what really caused them to see the events diff'ly.) That's the whole point of relativity, though: what you observe depends on your frame of reference. E's version does in fact show that simultaneity is relative, precisely because the observers are not in the same frame and see things differently. Again, that's the whole point.
Janus Posted November 13, 2013 Posted November 13, 2013 I started (in order to simplify) with a single flash to show that the difference between the two observers in the single-flash case has everything to do with their different motions, and really nothing at all to do with event times. Then this conclusion can easily be extended to cover the entire train example by simply adding a second lite flash, but this is not really necessary is it? (I'm just saying that E's train ex. does not show that sim. is rel. but merely shows that observers moving differently will see events differently if they judge event times by light-rays-from-the-events and totally ignore the fact that they moved differently during the experiment, which is what really caused them to see the events diff'ly.) Here's an animation of the Train experiment, first shown from the embankment frame: Lightning strikes both train ends and red dots simultaneously. Observer on on embankment see's flashes simultaneously. Observer on train does does not. Here's the same experiment, but viewed from the Train's frame. A couple of things to note: The distance between red dots and the length of train is not the equal, as they were in the first animation. This is due to length contraction. From the embankment, the the train was moving so it undegoes length contraction, However, from the train, it is the embankment that it moving so it is length contracted. The light flashes from the strikes travel at c reative to the train (while in the first emebankment they travel at c relative to the embankment.) This is per the postulates of Relativity. Given that, one end of the train reaches a red dot before the other has. Since the lightning strikes the ends of the train when they are even with the red dots, and this must be true no matter which frame you are watching events from, as far as the train is concerned, the lightning must strike the ends of the trains at different times. Something else that must happen in both frames is that the light from each strike reach the emebankment observer at the same time, which this animation shows. In fact, if you compare the animations you will note that exactly the same part of the train is next to him in both when this happens. The same is true for the train observer, This is the basic argument of the train experiment, that if you apply the postulate that the speed of light is constant in all frames, and that any events that occur in one frame must occur in the other, then you must conclude that that the lightning strikes which are simultaneous in one frame are not so in the other. 3
yknot Posted November 14, 2013 Author Posted November 14, 2013 (edited) Let's worry about the very simplest part of my example. Forget about length contractions and invariant light speed (because no one is measuring either lengths or speeds). Forget about points of view where one person looking at another (because there are none). Let me just ask this one simple question: Why do my observers see the light ray differently? (This cannot be due to "different relative motions" because neither observer is observing or recording the motion of other.) (And I can just as well ask the same question about E's version - why did his two observers see the two rays arrive differently?) There has to be some physical reason why the observers see things differently. Edited November 14, 2013 by yknot
ACG52 Posted November 14, 2013 Posted November 14, 2013 Of course it has to do with different relative motion. You've got O2 moving towards the light flash, so he's going to see the flash first. the light ray has to travel a greater distance to get to observer O1.
yknot Posted November 14, 2013 Author Posted November 14, 2013 Of course it has to do with different relative motion. You've got O2 moving towards the light flash, so he's going to see the flash first. the light ray has to travel a greater distance to get to observer O1. To show that relative observer motion is not really involved, we can eliminate it by placing both observers in the same inertial frame as follows: --------------------O1------------------------------O2---------------------A-Frame x axis--------------------<~~~~~~~~~~light ray Just as in E's train ex. and as in my half-****d version,, O2 sees the ray before O1. Why do they see the ray arrive differently? (Catch you in the am)
ACG52 Posted November 14, 2013 Posted November 14, 2013 Again, because O2 is closer to the light source than O1. This has got to be simple trolling. This is too simple to misunderstand as repeatedly as you appear to.
md65536 Posted November 14, 2013 Posted November 14, 2013 Forget about length contractions and invariant light speed (because no one is measuring either lengths or speeds). Forget about points of view where one person looking at another (because there are none). Let me just ask this one simple question: Why do my observers see the light ray differently?Let me just play devil's advocate and admit: Yes, if you ignore relativistic effects and the postulates of SR, you're no longer speaking of SR, and you can describe it however you want. Perhaps you might speculate that light behaves differently for the different observers? Everyone is giving you answers according to SR. There's no point in asking others what the answer might be other than what SR says. If you think there's an answer, you'll have to provide that, but then you'll have a long road ahead showing that your answer is consistent with observations, the way that SR is.
swansont Posted November 14, 2013 Posted November 14, 2013 Let's worry about the very simplest part of my example. Forget about length contractions and invariant light speed (because no one is measuring either lengths or speeds). Forget about points of view where one person looking at another (because there are none). Let me just ask this one simple question: Why do my observers see the light ray differently? (This cannot be due to "different relative motions" because neither observer is observing or recording the motion of other.) (And I can just as well ask the same question about E's version - why did his two observers see the two rays arrive differently?) There has to be some physical reason why the observers see things differently. You can't just ignore length contraction and invariant light speed, since they are the very effects that are occurring and causing the effects. Whether the observers measure the other is irrelevant — all that is required is there is relative motion between them. The physical reason behind all of this is that the speed of light is invariant, and that the rules apply identically to all inertial frames. All of the effects stem from that.
DimaMazin Posted November 14, 2013 Posted November 14, 2013 Saying what they are would be useful. There's a single flash of light. What else? One event of radiation and two events of reception.Three events.They can't be simultaneous in two frames because then math of indication of clock will be wrong.
swansont Posted November 14, 2013 Posted November 14, 2013 One event of radiation and two events of reception.Three events.They can't be simultaneous in two frames because then math of indication of clock will be wrong. One event, two observers. The observation times don't agree, but this is not an issue of simultaneity as it's defined in relativity. That would require two events and two observers, as with the train example gifs Janus supplied.
DimaMazin Posted November 14, 2013 Posted November 14, 2013 (edited) One event, two observers. The observation times don't agree, but this is not an issue of simultaneity as it's defined in relativity. And so you approve that the flash is simultaneous in two frames and every observation of every of the observers is simultaneous in two frames.Let's check your math. to =0 - time of start of flash in two frames t1 - time of collision of O2 with light in O1 frame t2 - time of observation of O2 S - distance between observers and flash in O1 frame S' - distance between observers and flash in O2 frame t1=S/(c+v) t2=S/gamma(c+v) t2=S'/c t2=S/(gamma*c) S/gamma(c+v) doesn't equal S/(gamma*c) Edited November 14, 2013 by DimaMazin
yknot Posted November 14, 2013 Author Posted November 14, 2013 Again, because O2 is closer to the light source than O1. This has got to be simple trolling. This is too simple to misunderstand as repeatedly as you appear to. Yes, of course it's because O2 is closer to the source, but how did he get closer? My answer is simply that he moved relative to the source or, more specifically, rel. to light's emission point, or even more specifically, he moved closer or toward the tip of the approaching ray. In other words, it's observer motion relative to the ray's tip that matters here, and not observer motions relative to each other. We can see this clearly by using a single observer and a light source. The closer this obs. is the the source, the quicker he will see the light that is emitted (toward him). Yes, this is too simple, but it is also critical to understanding E's train ex. which purportedly showed rel. simult. whereas all it really showed was the very simple fact that observer motion rel. to light makes diff. obs's see the light differently. We can say more: It is wrong to use light rays from events + differently-moving obs's to judge the events' occurrence times if you refuse to take into account the fact that the obs's move diff'ly wrt to the sources. (The best way to correctly time events is by placing correctly synch'd clocks at the events.)
Strange Posted November 14, 2013 Posted November 14, 2013 (edited) You don't seem to understand that the relativity of simultaneity is caused by the effects of motion that you are talking about (combined with the constant speed of light for both observers). When the two observers take into account all the measurements they can make (distance from the point where the flashes occurred, speed of light, etc. they will come to different conclusions about the relative timing of the two flashes. (See that neat animation earlier.) It doesn't matter how often you deny that, it is still true. Yes, this is too simple, but it is also critical to understanding E's train ex. which purportedly showed rel. simult. whereas all it really showed was the very simple fact that observer motion rel. to light makes diff. obs's see the light differently. What do you mean by observers "see the light differently"? We can say more: It is wrong to use light rays from events + differently-moving obs's to judge the events' occurrence times if you refuse to take into account the fact that the obs's move diff'ly wrt to the sources. No one is refusing to take the movement of the observers into account. That relative movement is the cause of relativity of simultaneity (and time dilation, length contraction, etc) Edited November 14, 2013 by Strange
swansont Posted November 14, 2013 Posted November 14, 2013 And so you approve that the flash is simultaneous in two frames and every observation of every of the observers is simultaneous in two frames.Let's check your math. to =0 - time of start of flash in two frames t1 - time of collision of O2 with light in O1 frame t2 - time of observation of O2 S - distance between observers and flash in O1 frame S' - distance between observers and flash in O2 frame t1=S/(c+v) t2=S/gamma(c+v) t2=S'/c t2=S/(gamma*c) S/gamma(c+v) doesn't equal S/(gamma*c) I said the observation times DON'T agree, and there is no math to check. There is no issue of simultaneity, since there is only one event. Simultaneity requires two events. An observation doesn't get called an event in these problems. An event is the source of a signal. (The best way to correctly time events is by placing correctly synch'd clocks at the events.) No, this won't work. One of the ramifications of relativity is that it's not possible to synchronize clocks in different reference frames.
DimaMazin Posted November 14, 2013 Posted November 14, 2013 I said the observation times DON'T agree, and there is no math to check. There is no issue of simultaneity, since there is only one event. Simultaneity requires two events. An observation doesn't get called an event in these problems. An event is the source of a signal. Relativity of one event to different times of different frames is issue of simultaneity. Don't make axiom without math.
Strange Posted November 14, 2013 Posted November 14, 2013 Relativity of one event to different times of different frames is issue of simultaneity. But you can only measure the times that the event is observed in different frame with respect to some other event. Then it becomes a matter of relativity of simultaneity (between the two events).
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