Simultaneity of Relativity
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From: mpc755 on 3 Nov 2009 20:17 On Nov 3, 7:29 pm, Bruce Richmond <bsr3...(a)my-deja.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'. > > Agreed. > > > When the Observers get back together they each conclude the light > > reached them at 12:00:01. But the Observer on the embankment insists > > the light from the lightning strikes must have reached the Observer at > > M' after 12:00:01 because the light had further to travel from A and B > > to the Observer at M' than it did to the Observer at M and the > > Observer on the train insists the light from the lightning strikes had > > to have reached the Observer at M after 12:00:01 because the light had > > further to travel from A' and B' to the Observer at M than it did to > > the Observer at M'. > > The observers can insist all they want, it doesn't change the fact > that both frames see the strikes at 12:00:01 as measured in their own > coordinate systems. > And that is what I am saying is impossible. I'm saying it is impossible for the light to travel from A and B to M and also for the light to travel from A' and B' to M' at the same time the light from A and B does not travel to M' and the light from A' and B' does not travel to M. I am saying the light wave will travel at 'c' from the point where the lightning strike occurred in three dimensional space relative to the aether. > Since A and A' are on lines perpendicular to the direction of motion > there will be no length contraction to consider. When A and A' meet > so do M/M' and B/B', all at the same instant. All agree that the > distance from A/A' to M/M' is the same as the distance from A'/A to M'/ > M. The same reasoning applies to B/B'. > > M can see A' moving toward A. When they meet there is the flash of > the strike, and A' continues to move on. A marks the spot in the > track frame where the strike happen. From the track frame M' sees A > moving toward A'. When they meet there is a flash and A continues to > move on. A' marks the spot in the train frame where the strike > happen. > > > The Observer on the embankment and the Observer on the train both > > correctly conclude the light could not have reached both Observers > > simultaneously because it did not travel the same distance to both > > Observers, but their clocks say otherwise. > > > How do you resolve this in SR and Relativity of Simultaneity? > > Try using the correct coordinates. The train measures things in train > coordinates while the tracks measure things in track coordinates. The > train frame does not agree with the track frame on the distance the > light has to travel to reach M'.
From: Bruce Richmond on 3 Nov 2009 23:00 On Nov 3, 8:17 pm, mpc755 <mpc...(a)gmail.com> wrote: > On Nov 3, 7:29 pm, Bruce Richmond <bsr3...(a)my-deja.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'. > > > Agreed. > > > > When the Observers get back together they each conclude the light > > > reached them at 12:00:01. But the Observer on the embankment insists > > > the light from the lightning strikes must have reached the Observer at > > > M' after 12:00:01 because the light had further to travel from A and B > > > to the Observer at M' than it did to the Observer at M and the > > > Observer on the train insists the light from the lightning strikes had > > > to have reached the Observer at M after 12:00:01 because the light had > > > further to travel from A' and B' to the Observer at M than it did to > > > the Observer at M'. > > > The observers can insist all they want, it doesn't change the fact > > that both frames see the strikes at 12:00:01 as measured in their own > > coordinate systems. > > And that is what I am saying is impossible. I'm saying it is > impossible for the light to travel from A and B to M and also for the > light to travel from A' and B' to M' at the same time the light from A > and B does not travel to M' and the light from A' and B' does not > travel to M. When M' sees the flash from A/A' he is seeing the light from A and he is seeing the light from A' both at the same time. There is nothing magical about that, A and A' were together when the light was emitted. A has moved on. The light from A traveled from where A was when he was together with A', not where he was at some later time. The two frames disagree on when and where things happen because they disagree on which frame the speed of light is c in. There is no reason to expect them to agree since they are using two different coordinate systems. > I am saying the light wave will travel at 'c' from the point where the > lightning strike occurred in three dimensional space relative to the > aether. > And where is that? How do you know where a point is in the aether? Claiming that each frame has its own aether is bullshit. > > > Since A and A' are on lines perpendicular to the direction of motion > > there will be no length contraction to consider. When A and A' meet > > so do M/M' and B/B', all at the same instant. All agree that the > > distance from A/A' to M/M' is the same as the distance from A'/A to M'/ > > M. The same reasoning applies to B/B'. > > > M can see A' moving toward A. When they meet there is the flash of > > the strike, and A' continues to move on. A marks the spot in the > > track frame where the strike happen. From the track frame M' sees A > > moving toward A'. When they meet there is a flash and A continues to > > move on. A' marks the spot in the train frame where the strike > > happen. > > > > The Observer on the embankment and the Observer on the train both > > > correctly conclude the light could not have reached both Observers > > > simultaneously because it did not travel the same distance to both > > > Observers, but their clocks say otherwise. > > > > How do you resolve this in SR and Relativity of Simultaneity? > > > Try using the correct coordinates. The train measures things in train > > coordinates while the tracks measure things in track coordinates. The > > train frame does not agree with the track frame on the distance the > > light has to travel to reach M'.- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: mpc755 on 4 Nov 2009 19:57 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?
From: mpc755 on 4 Nov 2009 23:15 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? How does the Observer at M see the light from A and B, the Observer at M' see the light from A' and B', and the Observer at N see the light from A and B all at the same instant.
From: Bruce Richmond on 5 Nov 2009 00:16
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. |