Simultaneity of Relativity
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From: Bruce Richmond on 5 Nov 2009 20:59 On Nov 5, 8:46 pm, mpc755 <mpc...(a)gmail.com> wrote: > On Nov 5, 8:30 pm, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > > > > > > > On Nov 5, 11:27 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. 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.- Hide quoted text - > > > > - Show quoted text - > > > Not even if the train frame considers the clocks at M and N to be out > > of sync? > > > I think you will agree that using just one coordinate system it is > > physically impossible for light to be measured as traveling at c with > > respect to both M and M'. That is why you wanted to add a second > > aether, or a pond, to provide a second reference point to use when > > measuring the speed of light with respect to M'. But if you use time > > and space the way it was used prior to SR you end up getting that > > light in the train frame can travel at c+v as measured in the track > > frame. And that does not agree with our previous claim that light > > always travels at c with respect to the track frame. > > I get around the c+v problem by realizing tying the emission point of > the light wave to a point in three dimensional space is incorrect. A > pebble is dropped into the pool on the train. If an Observer on the > embankment was unable to detect the moving water and was only able to > detect the wave in the water, he would conclude the wave originated > from where the center of the pool is when he detected the wave. Any > Observer in any frame of reference who detects the wave will all > conclude the wave originated from where the center of the pool is when > they detect the wave. And all Observers will conclude the wave > traveled at the same speed from the center of the pool, thus 'c' is > maintained for light. > > > > > What SR/LET do to get around the problem is to give each frame their > > own time. The seconds in two different frames are different from each > > other in about the same way that 1 mile north is different from 1 mile > > east, but the Lorentz Transformations can be used to convert > > coordinates from one frame into those of another. When everything is > > done correctly it works.- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text - Which puts you right back to a two aether theory which we have already seen is impossible when one frame is passing right through the middle of another. Sorry but I have no interest in your theory. Bye.
From: BURT on 5 Nov 2009 21:01 On Nov 5, 5:47 pm, "Inertial" <relativ...(a)rest.com> wrote: > "kenseto" <kens...(a)erinet.com> wrote in message > > news:d7367991-b54a-4bac-a3c5-6da83e00444c(a)c3g2000yqd.googlegroups.com... > > > > > > > On Nov 3, 7:39 pm, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > >> On Nov 3, 10:32 am, kenseto <kens...(a)erinet.com> wrote: > > >> > 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. > > >> Nobody said anything about the absolute state of motion of any clock. > >> Quit making things up. > > > Hey idiot....mpc755 asked me why the moving clock is running slow in > > my theory. > > Liar .. he did not- Hide quoted text - > > - Show quoted text - If you move ahead of a light beam you leave it behind and there is more space for it to travel before reaching you. Move toward light and you and the light come together faster. This is vice versa. This is the basis of relativity of simultaneity. Connectedness is at the speed of light. Mitch Raemsch
From: mpc755 on 5 Nov 2009 21:07 On Nov 5, 8:59 pm, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > On Nov 5, 8:46 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > > On Nov 5, 8:30 pm, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > > > > On Nov 5, 11:27 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. 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.- Hide quoted text - > > > > > - Show quoted text - > > > > Not even if the train frame considers the clocks at M and N to be out > > > of sync? > > > > I think you will agree that using just one coordinate system it is > > > physically impossible for light to be measured as traveling at c with > > > respect to both M and M'. That is why you wanted to add a second > > > aether, or a pond, to provide a second reference point to use when > > > measuring the speed of light with respect to M'. But if you use time > > > and space the way it was used prior to SR you end up getting that > > > light in the train frame can travel at c+v as measured in the track > > > frame. And that does not agree with our previous claim that light > > > always travels at c with respect to the track frame. > > > I get around the c+v problem by realizing tying the emission point of > > the light wave to a point in three dimensional space is incorrect. A > > pebble is dropped into the pool on the train. If an Observer on the > > embankment was unable to detect the moving water and was only able to > > detect the wave in the water, he would conclude the wave originated > > from where the center of the pool is when he detected the wave. Any > > Observer in any frame of reference who detects the wave will all > > conclude the wave originated from where the center of the pool is when > > they detect the wave. And all Observers will conclude the wave > > traveled at the same speed from the center of the pool, thus 'c' is > > maintained for light. > > > > What SR/LET do to get around the problem is to give each frame their > > > own time. The seconds in two different frames are different from each > > > other in about the same way that 1 mile north is different from 1 mile > > > east, but the Lorentz Transformations can be used to convert > > > coordinates from one frame into those of another. When everything is > > > done correctly it works.- Hide quoted text - > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text - > > Which puts you right back to a two aether theory which we have already > seen is impossible when one frame is passing right through the middle > of another. Sorry but I have no interest in your theory. Bye. There is only one aether. The aether is at rest relative to the train. When a pebble is dropped into the pool, the center of the pool is at A/ A'. When the wave reaches the Observer at M, the Observer at M correctly measures the distance the wave traveled as the distance M was from A' when the wave was detected. Bye.
From: BURT on 5 Nov 2009 21:22 On Nov 5, 5:46 pm, "Inertial" <relativ...(a)rest.com> wrote: > "kenseto" <kens...(a)erinet.com> wrote in message > > news:787ac87c-9042-4072-9ab4-e1687e06d4da(a)m26g2000yqb.googlegroups.com... > > > > > > > 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. > > And why it is useless > > > One for the observed clock to run slow and the other for the obsrved > > clock to run faster than the observer's clock. > > And in this case, you don't know which will happen. > > And, of course, it can be any of an infinite number of values because not > only do we not know which has the greater absolute motion, we do not know > the direction or value of absolute motion for M and M' either. > > Your theory basically says we cannot possibly know the relative rates at > which any pair of clocks tick. > > Its useless.- Hide quoted text - > > - Show quoted text - When time slows down for accelerating energy it is absolute. When time slows down for falling matter it is absolute. These are the two times. both absolute GR and SR times. Mitch Raemsch
From: mpc755 on 6 Nov 2009 08:32
On Nov 5, 9:07 pm, mpc755 <mpc...(a)gmail.com> wrote: > On Nov 5, 8:59 pm, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > > > > > On Nov 5, 8:46 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > On Nov 5, 8:30 pm, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > > > > > On Nov 5, 11:27 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. 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.- Hide quoted text - > > > > > > - Show quoted text - > > > > > Not even if the train frame considers the clocks at M and N to be out > > > > of sync? > > > > > I think you will agree that using just one coordinate system it is > > > > physically impossible for light to be measured as traveling at c with > > > > respect to both M and M'. That is why you wanted to add a second > > > > aether, or a pond, to provide a second reference point to use when > > > > measuring the speed of light with respect to M'. But if you use time > > > > and space the way it was used prior to SR you end up getting that > > > > light in the train frame can travel at c+v as measured in the track > > > > frame. And that does not agree with our previous claim that light > > > > always travels at c with respect to the track frame. > > > > I get around the c+v problem by realizing tying the emission point of > > > the light wave to a point in three dimensional space is incorrect. A > > > pebble is dropped into the pool on the train. If an Observer on the > > > embankment was unable to detect the moving water and was only able to > > > detect the wave in the water, he would conclude the wave originated > > > from where the center of the pool is when he detected the wave. Any > > > Observer in any frame of reference who detects the wave will all > > > conclude the wave originated from where the center of the pool is when > > > they detect the wave. And all Observers will conclude the wave > > > traveled at the same speed from the center of the pool, thus 'c' is > > > maintained for light. > > > > > What SR/LET do to get around the problem is to give each frame their > > > > own time. The seconds in two different frames are different from each > > > > other in about the same way that 1 mile north is different from 1 mile > > > > east, but the Lorentz Transformations can be used to convert > > > > coordinates from one frame into those of another. When everything is > > > > done correctly it works.- Hide quoted text - > > > > - Show quoted text -- Hide quoted text - > > > > - Show quoted text - > > > Which puts you right back to a two aether theory which we have already > > seen is impossible when one frame is passing right through the middle > > of another. Sorry but I have no interest in your theory. Bye. > > There is only one aether. The aether is at rest relative to the train. > When a pebble is dropped into the pool, the center of the pool is at A/ > A'. When the wave reaches the Observer at M, the Observer at M > correctly measures the distance the wave traveled as the distance M > was from A' when the wave was detected. Bye. There is only one aether. If the aether is at rest relative to the embankment and a lightning strike occurs at A/A' the light wave propagates outward at 'c' from A. When an Observer, regardless of frame of reference, sees the light it has traveled from where A *is*. If Observers on the train or the embankment do not know their state of motion relative to the aether, they may approximate where the light originated from by measuring to marks left on the embankment or left on the train. If the Observer on the embankment concludes the lightning strikes occurred simultaneously and the Observer on the train concludes the lightning strike at B' occurred prior to the lightning strike at A', then one or both of the Observers is incorrect. |