Simultaneity of Relativity
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From: mpc755 on 5 Nov 2009 00:53 On Nov 5, 12:16 am, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > On Nov 4, 7:57 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > > On Nov 4, 7:30 pm, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > > > > On Nov 3, 11:38 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > This additional observer, which I will lable N, will se the flash when > > > M' sees it, but he will see it coming from a different direction. A' > > > is perpendicular to M' but A is not perpendicular to N, so he will > > > have to look back at an angle to see The flash which was perpendicular > > > to M in the track frame. > > > When the Observer at M sees the light from A and B his clock will read > > 12:00:01. When the Observer at M' sees the light from A' and B' his > > clock will read 12:00:01. Since the embankment frame of reference and > > the train frame of reference are equal in all respects, this means the > > Observer at M sees the light from A and B at the same time as the > > Observer at M' sees the light from A' and B'. How can the Observer at > > N be seeing the light from A and B at the same instant the Observer at > > M' is seeing the light from A' and B' if at this instant the Observer > > at M' clock reads 12:00:01 and the Observer at M clock reads 12:00:01 > > and the light from A and B has just reached the Observer at M? > > M and M' are at the same place at the same instant, so they see the > same light arive from A/A'. They see it coming from different > directions due to their different states of motion. They both > consider themselves to be at rest, but obviously they are moving > relative to each other so they will see things differently. When you > sit at rest in your moving car you see rain drops falling diagonally > while the person on the side of the road sees them falling > vertically. The same thing happens with sound waves. If there is a > loud bang of to the side as you ride by you will hear it comeing from > a different direction than a person standing on the side of the road. > > As for it being the same time at M and M', it's not really, any more > than x=1 and x'=1 are the same place. Relative simultanity makes time > position dependent. Look at the Lorentz transformation for converting > the time coordinate from one frame to another. It contains an x which > means the time depends on the position. A/A' is a single event. B/B' is a single event. In my train thought experiment the Observer at M' is not hastening towards B and away from A. The Observer at M' remains equidistant from A and B at all times. A and M, B and M, A' and M', and B' and M' are equi-distant from each other. In my train thought experiment, the light from a lightning strike at A/ A' must take the same amount of time to travel the same distance from A to M as it does from A' to M'. If the light does not take the same amount of time to travel from A to M as it does from A' to M', then the light has not traveled at 'c' in one or both frames. The same is true for the lightning strike at B/B' and the light that travels from B to M and the light that travels from B' to M'. It has to take the same amount of time to reach both Observers. Now we have an Observer at N who is at the exact same location as M' is when the light from A' and B' reaches M'. In order for the light from A and B to reach the Observer at N after the light from A and B reaches the Observer at M, the light from A and B must reach M prior to the light from A' and B' reaching M'. This means the lightning strike at both A and B in the embankment frame of reference occurred prior to the lightning strike at A' and B' in the train frame of reference. If you consider that to be possible, we can continue with the analogy. There is an Observer at N' who is at the exact same location as M is when the light from A and B reaches M. The Observer at N' is at rest relative to the train. In the above scenario, since the light from A and B reaches M prior to the light from A' and B' reaching M', this means the light from A' and B' reaches N' prior to the light from A' and B' reaching M'. This is impossible. You can try and coordinate the events anyway you want to, but with Observers at N and N', it is physically impossible. And you do not have to use clocks or time. You just have to set when the events occur in terms of which lightning strike occurred prior to which lightning strike relative to both frames but the fact remains this cannot be resolved. When the Observer at M' sees the light from A' and B', the Observer at N sees the light from A and B. This means the Observer at M had to have seen the light from A and B prior to the Observer at M' seeing the light from A' and B'. But this means the Observer at N' sees the light from A' and B' prior to the Observer at M' seeing the light from A' and B'.
From: mpc755 on 5 Nov 2009 11:27 On Nov 5, 12:53 am, mpc755 <mpc...(a)gmail.com> wrote: > On Nov 5, 12:16 am, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > > > > > On Nov 4, 7:57 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > On Nov 4, 7:30 pm, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > > > > > On Nov 3, 11:38 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > > This additional observer, which I will lable N, will se the flash when > > > > M' sees it, but he will see it coming from a different direction. A' > > > > is perpendicular to M' but A is not perpendicular to N, so he will > > > > have to look back at an angle to see The flash which was perpendicular > > > > to M in the track frame. > > > > When the Observer at M sees the light from A and B his clock will read > > > 12:00:01. When the Observer at M' sees the light from A' and B' his > > > clock will read 12:00:01. Since the embankment frame of reference and > > > the train frame of reference are equal in all respects, this means the > > > Observer at M sees the light from A and B at the same time as the > > > Observer at M' sees the light from A' and B'. How can the Observer at > > > N be seeing the light from A and B at the same instant the Observer at > > > M' is seeing the light from A' and B' if at this instant the Observer > > > at M' clock reads 12:00:01 and the Observer at M clock reads 12:00:01 > > > and the light from A and B has just reached the Observer at M? > > > M and M' are at the same place at the same instant, so they see the > > same light arive from A/A'. They see it coming from different > > directions due to their different states of motion. They both > > consider themselves to be at rest, but obviously they are moving > > relative to each other so they will see things differently. When you > > sit at rest in your moving car you see rain drops falling diagonally > > while the person on the side of the road sees them falling > > vertically. The same thing happens with sound waves. If there is a > > loud bang of to the side as you ride by you will hear it comeing from > > a different direction than a person standing on the side of the road. > > > As for it being the same time at M and M', it's not really, any more > > than x=1 and x'=1 are the same place. Relative simultanity makes time > > position dependent. Look at the Lorentz transformation for converting > > the time coordinate from one frame to another. It contains an x which > > means the time depends on the position. > > A/A' is a single event. B/B' is a single event. > > In my train thought experiment the Observer at M' is not hastening > towards B and away from A. The Observer at M' remains equidistant from > A and B at all times. > > A and M, B and M, A' and M', and B' and M' are equi-distant from each > other. > > In my train thought experiment, the light from a lightning strike at A/ > A' must take the same amount of time to travel the same distance from > A to M as it does from A' to M'. If the light does not take the same > amount of time to travel from A to M as it does from A' to M', then > the light has not traveled at 'c' in one or both frames. The same is > true for the lightning strike at B/B' and the light that travels from > B to M and the light that travels from B' to M'. It has to take the > same amount of time to reach both Observers. > > Now we have an Observer at N who is at the exact same location as M' > is when the light from A' and B' reaches M'. > > In order for the light from A and B to reach the Observer at N after > the light from A and B reaches the Observer at M, the light from A and > B must reach M prior to the light from A' and B' reaching M'. This > means the lightning strike at both A and B in the embankment frame of > reference occurred prior to the lightning strike at A' and B' in the > train frame of reference. > > If you consider that to be possible, we can continue with the > analogy. > > There is an Observer at N' who is at the exact same location as M is > when the light from A and B reaches M. The Observer at N' is at rest > relative to the train. > > In the above scenario, since the light from A and B reaches M prior to > the light from A' and B' reaching M', this means the light from A' and > B' reaches N' prior to the light from A' and B' reaching M'. This is > impossible. > > You can try and coordinate the events anyway you want to, but with > Observers at N and N', it is physically impossible. > > And you do not have to use clocks or time. You just have to set when > the events occur in terms of which lightning strike occurred prior to > which lightning strike relative to both frames but the fact remains > this cannot be resolved. > > When the Observer at M' sees the light from A' and B', the Observer at > N sees the light from A and B. This means the Observer at M had to > have seen the light from A and B prior to the Observer at M' seeing > the light from A' and B'. But this means the Observer at N' sees the > light from A' and B' prior to the Observer at M' seeing the light from > A' and B'. How this works in SR is the following. Since The Observer at M and the Observer at N are in the same frame of reference, their clocks maintain the same time. When the light from A and B reaches the Observer at M, his clock reads 12:00:01:00. When the light from A and B reaches the Observer at N, his clock reads 12:00:01:03. From the perspective of the train frame of reference, the Observer at M is hastening away from the lightning strikes which occurred at A' and B'. From the perspective of the train frame of reference, the Observer at M' and the Observer at N see the light from the lightning strike at A' and B' prior to the Observer at M seeing the light from the lightning strike. From the perspective of the train frame of reference, when the Observer at N sees the light, he looks at his watch and it reads 12:00:01:03. From the perspective of the train frame of reference, later on, when the Observer at M sees the light from the lightning strikes he looks down at his watch and it reads 12:00:01:00. I say this is physically impossible.
From: PD on 5 Nov 2009 11:37 On Nov 5, 10:27 am, mpc755 <mpc...(a)gmail.com> wrote: > On Nov 5, 12:53 am, mpc755 <mpc...(a)gmail.com> wrote: > > > > > On Nov 5, 12:16 am, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > > > > On Nov 4, 7:57 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > > On Nov 4, 7:30 pm, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > > > > > > On Nov 3, 11:38 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > > > This additional observer, which I will lable N, will se the flash when > > > > > M' sees it, but he will see it coming from a different direction. A' > > > > > is perpendicular to M' but A is not perpendicular to N, so he will > > > > > have to look back at an angle to see The flash which was perpendicular > > > > > to M in the track frame. > > > > > When the Observer at M sees the light from A and B his clock will read > > > > 12:00:01. When the Observer at M' sees the light from A' and B' his > > > > clock will read 12:00:01. Since the embankment frame of reference and > > > > the train frame of reference are equal in all respects, this means the > > > > Observer at M sees the light from A and B at the same time as the > > > > Observer at M' sees the light from A' and B'. How can the Observer at > > > > N be seeing the light from A and B at the same instant the Observer at > > > > M' is seeing the light from A' and B' if at this instant the Observer > > > > at M' clock reads 12:00:01 and the Observer at M clock reads 12:00:01 > > > > and the light from A and B has just reached the Observer at M? > > > > M and M' are at the same place at the same instant, so they see the > > > same light arive from A/A'. They see it coming from different > > > directions due to their different states of motion. They both > > > consider themselves to be at rest, but obviously they are moving > > > relative to each other so they will see things differently. When you > > > sit at rest in your moving car you see rain drops falling diagonally > > > while the person on the side of the road sees them falling > > > vertically. The same thing happens with sound waves. If there is a > > > loud bang of to the side as you ride by you will hear it comeing from > > > a different direction than a person standing on the side of the road. > > > > As for it being the same time at M and M', it's not really, any more > > > than x=1 and x'=1 are the same place. Relative simultanity makes time > > > position dependent. Look at the Lorentz transformation for converting > > > the time coordinate from one frame to another. It contains an x which > > > means the time depends on the position. > > > A/A' is a single event. B/B' is a single event. > > > In my train thought experiment the Observer at M' is not hastening > > towards B and away from A. The Observer at M' remains equidistant from > > A and B at all times. > > > A and M, B and M, A' and M', and B' and M' are equi-distant from each > > other. > > > In my train thought experiment, the light from a lightning strike at A/ > > A' must take the same amount of time to travel the same distance from > > A to M as it does from A' to M'. If the light does not take the same > > amount of time to travel from A to M as it does from A' to M', then > > the light has not traveled at 'c' in one or both frames. The same is > > true for the lightning strike at B/B' and the light that travels from > > B to M and the light that travels from B' to M'. It has to take the > > same amount of time to reach both Observers. > > > Now we have an Observer at N who is at the exact same location as M' > > is when the light from A' and B' reaches M'. > > > In order for the light from A and B to reach the Observer at N after > > the light from A and B reaches the Observer at M, the light from A and > > B must reach M prior to the light from A' and B' reaching M'. This > > means the lightning strike at both A and B in the embankment frame of > > reference occurred prior to the lightning strike at A' and B' in the > > train frame of reference. > > > If you consider that to be possible, we can continue with the > > analogy. > > > There is an Observer at N' who is at the exact same location as M is > > when the light from A and B reaches M. The Observer at N' is at rest > > relative to the train. > > > In the above scenario, since the light from A and B reaches M prior to > > the light from A' and B' reaching M', this means the light from A' and > > B' reaches N' prior to the light from A' and B' reaching M'. This is > > impossible. > > > You can try and coordinate the events anyway you want to, but with > > Observers at N and N', it is physically impossible. > > > And you do not have to use clocks or time. You just have to set when > > the events occur in terms of which lightning strike occurred prior to > > which lightning strike relative to both frames but the fact remains > > this cannot be resolved. > > > When the Observer at M' sees the light from A' and B', the Observer at > > N sees the light from A and B. This means the Observer at M had to > > have seen the light from A and B prior to the Observer at M' seeing > > the light from A' and B'. But this means the Observer at N' sees the > > light from A' and B' prior to the Observer at M' seeing the light from > > A' and B'. > > How this works in SR is the following. No, that's not at all how it works in SR. That's how it works in the empty pecan shell you call a cranium. > Since The Observer at M and the > Observer at N are in the same frame of reference, their clocks > maintain the same time. When the light from A and B reaches the > Observer at M, his clock reads 12:00:01:00. When the light from A and > B reaches the Observer at N, his clock reads 12:00:01:03. > > From the perspective of the train frame of reference, the Observer at > M is hastening away from the lightning strikes which occurred at A' > and B'. From the perspective of the train frame of reference, the > Observer at M' and the Observer at N see the light from the lightning > strike at A' and B' prior to the Observer at M seeing the light from > the lightning strike. From the perspective of the train frame of > reference, when the Observer at N sees the light, he looks at his > watch and it reads 12:00:01:03. From the perspective of the train > frame of reference, later on, when the Observer at M sees the light > from the lightning strikes he looks down at his watch and it reads > 12:00:01:00. I say this is physically impossible.
From: kenseto on 5 Nov 2009 19:36 On Nov 3, 7:29 pm, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > On Nov 3, 12:34 am, mpc755 <mpc...(a)gmail.com> wrote: > > > > > > > On Nov 3, 12:19 am, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > > > > On Nov 2, 9:16 am, mpc755 <mpc...(a)gmail.com> wrote: > > > > > On Nov 2, 12:15 am, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > > > > > > On Nov 1, 10:57 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > > > > On Nov 1, 10:47 pm, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > > > > > > > > > Since Einstein required the aether for the propagation of light, what > > > > > > > > you are referring to is an error of omission. > > > > > > > > Einstein did not require an aether for propagation of light. > > > > > > > What part of the next sentence don't you understand? > > > > > > Those words had not been written when he wrote the train experiment. > > > > > Also GR and SR are not the same thing. > > > > > In SR and the train experiment Einstein does not require an aether for > > > > propagation of light, but in GR space without aether is unthinkable > > > > for there would be no propagation of light? Am I understanding you > > > > correctly? > > > > That is pretty much what I wrote but I don't think you are > > > understanding it. > > > > When Einstein wrote SR there was still much dissagreement about how > > > light was transmitted. Experimental evidence had established that > > > however it got from place to place it always traveled at c, > > > reguardless of the state of motion of those making the measurement. > > > SR explained how that could happen based on c being a universal > > > constant. It didn't matter how light got from place to place, only > > > that it always traveled at the same speed. And not for just one > > > frame. Two frames moving relative to each other could both measure > > > the same beam to be traveling at c. > > > > Einstein's later quote does not support your theory. Many say he > > > didn't mean aether as proposed in any past or present aether theory. > > > Even if he did we know that it would have to agree with SR since he > > > never said that SR was wrong. So that would limit you to an aether > > > similar to LET, not a dragged aether theory like yours. > > > But Einstein believe there is an aether or there is no propagation of > > light, which means there is an aether in SR > > He could have believed that light was transported in buckets by little > blue fairies, it wouldn't have made any difference in the train > experiment. What matters is that both frames measure *the same light* > to travel at c in their own frame. So how come Einstein said that M' is rush toward the light from the front (c+v) AND RECEDING AWAY FROM THE LIGHT FROM THE REAR (C-V)??? Ken Seto > > > and if the idea of motion > > cannot be applied to the aether and the train frame of reference and > > the embankment frame of reference both occupy the same three > > dimensional space then this implies the aether is at rest in both > > frames which is impossible. > > The idea that motion cannot be applied to the aether is another way of > saying that no frame can exclude other frames from considering > themselves to be at rest. > > The track frame sees the train moving relative to the wave fronts and > sees that the M' does not see the flashes from A' and B' at the same > instant. There is no getting around that fact. There is only one > wave front moving out from the strike at A/A' and one from B/B'. > Those wave fronts meet at only one point on the tracks, and that is at > M. M' sees one flash before M and the other after M. > > Since A' and B' are equal distances from M', and light travels at c in > the train frame, the only explaination is that in the train frame the > strikes happen at different times. > > BTW, the tracks and the train do not occupy the same three dimensional > space. Each frame is using its own set of dimensions. In track > coordinates the x coordinate of M' is constantly changing since he is > moving at v. In the train frame M' is at rest so his coordinates > don't change. > > > I have tried to explain to you how light travels at 'c' relative to > > all Observer's but you are not understanding it. > > You have the waves traveling at c relative to the pond which is at > rest in the train frame but moving in the track frame. That results > in the leading edge of the waves traveling at c+v relative to the > tracks. The track frame makes measurements relative to the tracks, > not the train or the pond. > > > You are tying the > > emission point to a particular point in three dimensional space which > > is inaccurate. > > In the track frame I am tying the emission to the track coordinates > where the emission took place. In the train frame it is tied to the > train coordinates where it took place. They are both correct for > their respective frames. > > > Resolve the mpc755 train thought experiment in terms of SR and > > Relativity of Simultaneity. If you can't, then SR doesn't hold. > > > Since light travels at 'c' relative to the aether, the mpc755 train > > thought experiment is physically impossible for a single lightning > > strike at A/A' and a single lightning strike at B/B'. > > LET says otherwise.- Hide quoted text - > > - Show quoted text -
From: kenseto on 5 Nov 2009 19:40
On Nov 3, 7:12 pm, "Inertial" <relativ...(a)rest.com> wrote: > "kenseto" <kens...(a)erinet.com> wrote in message > > news:58559b17-56a0-454c-a202-8884aa94d656(a)a31g2000yqn.googlegroups.com... > > > > > > > On Nov 3, 9:13 am, mpc755 <mpc...(a)gmail.com> wrote: > >> On Nov 3, 8:37 am, kenseto <kens...(a)erinet.com> wrote: > > >> > On Nov 3, 1:16 am, mpc755 <mpc...(a)gmail.com> wrote: > > >> > > On Nov 3, 12:33 am, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > > >> > > > On Nov 2, 9:36 am, mpc755 <mpc...(a)gmail.com> wrote: > > >> > > > > On Nov 2, 12:16 am, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > > >> > > > > > On Nov 1, 11:20 pm, mpc755 <mpc...(a)gmail.com> wrote: > > >> > > > > > > On Nov 1, 10:57 pm, Bruce Richmond <bsr3...(a)my-deja.com> > >> > > > > > > wrote: > > >> > > > > > > > On Nov 1, 10:13 pm, mpc755 <mpc...(a)gmail.com> wrote: > > >> > > > > > > > > On Nov 1, 7:32 pm, mpc755 <mpc...(a)gmail.com> wrote: > > >> > > > > > > > > > On Oct 8, 11:49 am, mpc755 <mpc...(a)gmail.com> wrote: > > >> > > > > > > > > > > If the aether is stationary relative to the > >> > > > > > > > > > > embankment and stationary > >> > > > > > > > > > > relative to the train, this is what will occur in > >> > > > > > > > > > > Einstein's train > >> > > > > > > > > > > thought experiment: > > >> > > > > > > > > > >http://www.youtube.com/watch?v=jyWTaXMElUk > > >> > > > > > > > > > Einstein says in order for the propagation of light to > >> > > > > > > > > > exist there > >> > > > > > > > > > must be aether. Einstein also says the idea of motion > >> > > > > > > > > > may not be > >> > > > > > > > > > applied to aether. > > >> > > > > > > > > > I conclude this means aether must be at rest relative > >> > > > > > > > > > to the > >> > > > > > > > > > embankment and at rest relative to the train which is > >> > > > > > > > > > physically > >> > > > > > > > > > impossible if the embankment frame of reference and the > >> > > > > > > > > > train frame of > >> > > > > > > > > > reference occupy the same three dimensional space. > > >> > > > > > > > > mpc755 train thought experiment. > > >> > > > > > > > > The train is moving perpendicular to the line A and B > >> > > > > > > > > exist on. > >> > > > > > > > > The train is wide enough that A' and B' exist on opposite > >> > > > > > > > > sides of the > >> > > > > > > > > aisle. > > >> > > > > > > > > Here is an image of the train and the embankment and the > >> > > > > > > > > corresponding > >> > > > > > > > > locations prior to the lightning strikes. The arrows > >> > > > > > > > > represent the > >> > > > > > > > > train moving towards the embankment as viewed from the > >> > > > > > > > > embankment > >> > > > > > > > > frame of reference: > > >> > > > > > > > > A-----M-----B > >> > > > > > > > > ^ ^ ^ > >> > > > > > > > > | | | > >> > > > > > > > > | | | > >> > > > > > > > > A'----M'----B' > > >> > > > > > > > > When the lightning strike occurs at A/A', A and A' exist > >> > > > > > > > > at the same > >> > > > > > > > > point in three dimensional space. When the lightning > >> > > > > > > > > strike occurs at > >> > > > > > > > > B/B', B and B' exist at the same point in three > >> > > > > > > > > dimensional space. > > >> > > > > > > > > The train continues to move perpendicular to the line A > >> > > > > > > > > and B exist on > >> > > > > > > > > after the lightning strikes. > > >> > > > > > > > > This is what the embankment and train look like after the > >> > > > > > > > > lightning > >> > > > > > > > > strikes. The arrows indicate the train moving away from > >> > > > > > > > > the embankment > >> > > > > > > > > as viewed from the embankment frame of reference: > > >> > > > > > > > > A'----M'----B' > >> > > > > > > > > ^ ^ ^ > >> > > > > > > > > | | | > >> > > > > > > > > | | | > >> > > > > > > > > A-----M-----B > > >> > > > > > > > > If the light from A and B reaches M simultaneously, the > >> > > > > > > > > light from A' > >> > > > > > > > > and B' reaches M' simultaneously because A/A' was a > >> > > > > > > > > single lightning > >> > > > > > > > > strike and B/B' was a single lightning strike and A and > >> > > > > > > > > M, B and M, A' > >> > > > > > > > > and M', and B' and M' are equi-distant. But this requires > >> > > > > > > > > the light to > >> > > > > > > > > travel from four locations to each Observer. It is either > >> > > > > > > > > that or the > >> > > > > > > > > light travels from A and B to M and M', making the > >> > > > > > > > > embankment the > >> > > > > > > > > preferred frame or the light travels from A' and B' to M > >> > > > > > > > > and M', > >> > > > > > > > > making the train the preferred frame. > > >> > > > > > > > > I don't think this can be resolved in Relativity of > >> > > > > > > > > Simultaneity. > > >> > > > > > > > This has nothing to do with Einstein's train experiment or > >> > > > > > > > relative > >> > > > > > > > simultaneity. > > >> > > > > > > It has everything to do with Relativity of Simultaneity. > > >> > > > > > Nope, wrong set-up. > > >> > > > > Observers must be traveling along the line which intersects the > >> > > > > two > >> > > > > lightning strikes in order for Relativity of Simultaneity to be > >> > > > > correct?- Hide quoted text - > > >> > > > > - Show quoted text - > > >> > > > Nope. Relativity of Simultaneity would still exist, but your > >> > > > choice > >> > > > of event locations would not allow it to be observed. Your set-up > >> > > > is > >> > > > the special case where the distances from M' to A and B stay equal > >> > > > as > >> > > > M' passes between them. > > >> > > When the Observer at M on the embankment and the Observer at M' on > >> > > the > >> > > train pass one another at the instant of the lightning strikes at > >> > > A/A' > >> > > and B/B' the Observers synchronize their watches at 12:00:00. It > >> > > takes > >> > > one second for the light from A and B to reach M and one second for > >> > > the light from A' and B' to reach M'. > > >> > No M' clock is running slower than M's clock....that means that it > >> > takes (1/Gamma seconds on the train clock) for the light fronts to > >> > reach. > > >> > Ken Seto > > >> Why is M' clock running slower than Ms clock? Both frames of reference > >> are moving relative to one another. > > > Because M' is in a higher state of absolute motion than M. > > According to your theory one cannot tell from the scenario whether it is M > or M' that is in a higher state of absolute motion. So one cannot tell how > the clock rate relate at all. That's why IRT has two equations for the rate of an observed clock. One for the observed clock to run slow and the other for the obsrved clock to run faster than the observer's clock. Ken Seto - Hide quoted text - > > - Show quoted text - |