Simultaneity of Relativity
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From: mpc755 on 9 Oct 2009 15:29 On Oct 9, 3:25 pm, PD <thedraperfam...(a)gmail.com> wrote: > On Oct 9, 2:02 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > > On Oct 9, 2:29 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > On Oct 9, 12:26 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > > On Oct 9, 1:17 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > On Oct 9, 12:08 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > > > > On Oct 9, 12:30 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > On Oct 9, 10:06 am, mpc755 <mpc...(a)gmail.com> wrote: > > > > > > > > > On Oct 9, 10:52 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > > > On Oct 9, 8:53 am, mpc755 <mpc...(a)gmail.com> wrote: > > > > > > > > > > > On Oct 9, 8:57 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > > > > > On Oct 8, 8:34 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > > > > > > > > > > > > > > > > > > > Sorry, I've fouled this up. > > > > > > > > > Yes, the light from A and B reaches M' at different times, but the > > > > > > > > light from A' and B' reaches M' simultaneously. > > > > > > > > A and A' are a single strike. B and B' are a single strike. > > > > > > > Yes, A and A' are a single strike but A and A' are two different > > > > > > locations in three dimensional space. A in on the embankment and A' is > > > > > > on the train. > > > > > > No, they are not. You have not read the gedanken carefully. A lighting > > > > > strike hits in ONE place, not two. A and A' are two labels for the > > > > > same point. In the original gedanken, A and A' label the point where > > > > > the train meets the track at one end of the train. That is ONE POINT. > > > > > > Please reread it and pay more attention than what you have been doing > > > > > so far. > > > > > The lightning strike leaves marks at A, A', B, and B'. This is four > > > > different locations in three dimensional space. A and A' and B and B' > > > > were co-located at the time of the strikes, but they are four > > > > different locations. > > > > A and A' are colocated but they are two different locations? Listen to > > > yourself! LOL! > > > > Lightning strikes at ONE POINT at the end of the train, where one end > > > of the train meets the track. There is a scorch mark left on the track > > > (A) and a scorch mark left on the train (A') where this ONE spot > > > occurred. > > > > Another lightning bolt strikes at ONE point at the other end of the > > > train, where that end of the train meets the track. > > > > Here's another way you can think of it. > > > Set up a graph with an x and y axis. The x and y axis meet at ONE > > > POINT. That point is labeled (y=0) for the y-axis and is labeled (x=0) > > > for the x-axis. Does the fact that there are two labels mean that it > > > is ONE point or TWO points where the axes meet? > > > Yes, two points where the embankment and the train meet at A/A' and > > two points where the embankment and the train meet at B/B'. > > > There is a mark on the train and a mark on the embankment where the > > lightning strike occurs at A/A' and there is a mark on the train and > > on the embankment where the lightning strike occurs at B/B'. > > > Do you understand there is a mark at A on the embankment and A' on the > > train? Do you understand that one of these points in on the embankment > > and one of these points is on the train? > > Where the x and y axis meet on a graph, (y=0) marks that place on the > y-axis and (x=0) marks that place on the x-axis. Is that one location > or two? > > LOL!! > How many marks are there? Two on the train and two on the embankment. Two marks are on the train where the aether is stationary relative to the train frame of reference. Two marks are on the embankment where the aether is stationary relative to the embankment frame of reference. > > > > The point on the embankment is in the embankment reference frame and > > the aether is stationary on the embankment in the embankment reference > > frame. The point on the train is in the train reference frame and the > > aether is stationary on the train in the train reference frame. > > > Just as you can have a mark at A on the embankment and A' on the train > > you can have stationary aether on the embankment and stationary aether > > on the train. > > Lovely! You're so fun! > So we have ONE aether and it is stationary relative to the embankment > and stationary relative to the train at the same time! > Tell me, can you do that with one ball, too? > > LOL!! If the embankment is knee deep in water, the water will be stationary in the embankment frame of reference. If the train is knee deep in water, where the train is enclosed, the water will be stationary in the train frame of reference. Replace one medium with another. Replace water with aether.
From: mpc755 on 9 Oct 2009 15:33 On Oct 9, 3:26 pm, PD <thedraperfam...(a)gmail.com> wrote: > On Oct 9, 2:04 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > > On Oct 9, 2:38 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > On Oct 9, 12:26 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > > > > > That's just it. Whether they are simultaneous or not depends on the > > > > > > > signals they actually receive at their locations. That is, the > > > > > > > observers M and M' are not AT the locations A, A', B, B'. They are at > > > > > > > the places marked M and M'. What they know is what happens where they > > > > > > > are, and THAT tells them what happens at A, A', B, and B'. They have > > > > > > > no other way of knowing. > > > > > > > There are observers at A, A', B and B' and they all have clickers and > > > > > > they all click their clickers when the pebbles hit the water. It is > > > > > > determined all four observers hit their clickers at the same time. > > > > > > How is it determined that they hit their clickers at the same time? > > > > > What procedure would you need to ensure that? > > > > > I want you to think about this very carefully... > > > > > There are observer as close to each contact point between the pebble > > > > and the water as possible. The length of wire from the clicker to the > > > > clicker response unit is the same for all observers. > > > > And the speed of the signal is the same through the wire in both > > > cases. > > > > So if the speed in the wire is the same, and the length of the wire is > > > the same, then you know that the time propagation through the wires > > > would be the same, right? > > > > And if this is the case, then you know the following: > > > 1. If the clicker response unit records signals from the clickers at > > > the same time, THEN you know that the pebbles landed at the same time.. > > > 2. If the clicker response unit records signals from the clickers at > > > different times, THEN you know that the pebbles landed at different > > > times. > > > > Right? This is how you determine from the clicker and clicker response > > > system whether the pebbles really landed at the same time or not. > > > > OK, so here's the situation with the lightning strikes: > > > You've got ONE lightning strike at one end of the train, and ONE > > > lightning strike at the other end of the train. > > > The path length (just like the wire length) from one lightning strike > > > to the observer M is the same as the path length from the other > > > lightning strike to the observer M. Equal path lengths, just like > > > equal wire lengths. > > > The path length (just like the wire length) from one lightning strike > > > to the observer M' is the same as the path length from the other > > > lightning strike to the observer M'. Equal path lengths, just like > > > equal wire lengths. > > > And you know the speed of the signal from one lightning strike to M is > > > the same as the speed of the signal from the other lightning strike to > > > M. > > > And you know the speed of the signal from one lightning strike to M' > > > is the same as the speed of the signal from the other lightning strike > > > to M'. > > > > The problem is, in experiment, M says he received the signals at the > > > same time (clicker case (1)), and M' says he received the signals at > > > different times (clicker case (2)). > > > Just like there are four observers at A, A', B, and B' > > There aren't four observers at A, A', B and B'. > Read it again! > You asked me how the pebbles dropped at A, A', B, and B' could be determined to be simultaneous and I explained how with four observers, one at each pebble drop. > > the lightning > > strikes occur at 4 points. > > Wow! One lightning strike strikes in two places at once! > LOL!! > Yes. When it leaves a mark on the train and on the embankment. This is two different places. > > When the light from the lightning strikes > > reach M and M' are dependent on the medium the light travels through. > > Especially when it's ONE medium! > > You're a hoot, you are! > Water is one medium, but if it is on the train and stationary relative to the train and water is on the embankment stationary relative to the embankment, pebbles dropped at A, A', B, and B' will have their associated waves from A and B reach M and from A' and B' reach M' simultaneously. > > > > > > If the clicks are > > > > determined to be simultaneous at the clicker response unit, then the > > > > pebbles each hit the water simultaneously. > >
From: mpc755 on 9 Oct 2009 15:47 On Oct 9, 3:33 pm, mpc755 <mpc...(a)gmail.com> wrote: > On Oct 9, 3:26 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Oct 9, 2:04 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > On Oct 9, 2:38 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Oct 9, 12:26 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > > > > > > That's just it. Whether they are simultaneous or not depends on the > > > > > > > > signals they actually receive at their locations. That is, the > > > > > > > > observers M and M' are not AT the locations A, A', B, B'. They are at > > > > > > > > the places marked M and M'. What they know is what happens where they > > > > > > > > are, and THAT tells them what happens at A, A', B, and B'. They have > > > > > > > > no other way of knowing. > > > > > > > > There are observers at A, A', B and B' and they all have clickers and > > > > > > > they all click their clickers when the pebbles hit the water. It is > > > > > > > determined all four observers hit their clickers at the same time. > > > > > > > How is it determined that they hit their clickers at the same time? > > > > > > What procedure would you need to ensure that? > > > > > > I want you to think about this very carefully... > > > > > > There are observer as close to each contact point between the pebble > > > > > and the water as possible. The length of wire from the clicker to the > > > > > clicker response unit is the same for all observers. > > > > > And the speed of the signal is the same through the wire in both > > > > cases. > > > > > So if the speed in the wire is the same, and the length of the wire is > > > > the same, then you know that the time propagation through the wires > > > > would be the same, right? > > > > > And if this is the case, then you know the following: > > > > 1. If the clicker response unit records signals from the clickers at > > > > the same time, THEN you know that the pebbles landed at the same time. > > > > 2. If the clicker response unit records signals from the clickers at > > > > different times, THEN you know that the pebbles landed at different > > > > times. > > > > > Right? This is how you determine from the clicker and clicker response > > > > system whether the pebbles really landed at the same time or not. > > > > > OK, so here's the situation with the lightning strikes: > > > > You've got ONE lightning strike at one end of the train, and ONE > > > > lightning strike at the other end of the train. > > > > The path length (just like the wire length) from one lightning strike > > > > to the observer M is the same as the path length from the other > > > > lightning strike to the observer M. Equal path lengths, just like > > > > equal wire lengths. > > > > The path length (just like the wire length) from one lightning strike > > > > to the observer M' is the same as the path length from the other > > > > lightning strike to the observer M'. Equal path lengths, just like > > > > equal wire lengths. > > > > And you know the speed of the signal from one lightning strike to M is > > > > the same as the speed of the signal from the other lightning strike to > > > > M. > > > > And you know the speed of the signal from one lightning strike to M' > > > > is the same as the speed of the signal from the other lightning strike > > > > to M'. > > > > > The problem is, in experiment, M says he received the signals at the > > > > same time (clicker case (1)), and M' says he received the signals at > > > > different times (clicker case (2)). > > > > Just like there are four observers at A, A', B, and B' > > > There aren't four observers at A, A', B and B'. > > Read it again! > > You asked me how the pebbles dropped at A, A', B, and B' could be > determined to be simultaneous and I explained how with four observers, > one at each pebble drop. > > > > the lightning > > > strikes occur at 4 points. > > > Wow! One lightning strike strikes in two places at once! > > LOL!! > > Yes. When it leaves a mark on the train and on the embankment. This is > two different places. > > > > When the light from the lightning strikes > > > reach M and M' are dependent on the medium the light travels through. > > > Especially when it's ONE medium! > > > You're a hoot, you are! > > Water is one medium, but if it is on the train and stationary relative > to the train and water is on the embankment stationary relative to the > embankment, pebbles dropped at A, A', B, and B' will have their > associated waves from A and B reach M and from A' and B' reach M' > simultaneously. > > Let's see if we can find any common ground. Forget Einstein's Train Thought Experiment for a second. We have an embankment that is knee deep in water. The water on the embankment is stationary relative to the embankment. We have an enclosed train that is need deep in water. The water on the train is stationary relative to the train. Four pebbles are dropped at A and B on the embankment and A' and B' on the train. A and B are equi-distant from M and A' and B' are equi- distant from M'. The distance from A and B to M is the same distance as A' and B' are to M'. The pebbles are dropped simultaneously at A and A'. The pebbles are dropped simultaneously at B and B'. If the waves from A and B reach M simultaneously, then the waves from A' and B' reach M' simultaneously.
From: paparios on 9 Oct 2009 15:47 On 9 oct, 14:58, mpc755 <mpc...(a)gmail.com> wrote: > On Oct 9, 2:23 pm, "papar...(a)gmail.com" <papar...(a)gmail.com> wrote: > > > > > > > On 9 oct, 13:26, mpc755 <mpc...(a)gmail.com> wrote: > > > > On Oct 9, 1:17 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > There is no enclosure implied in the gedanken. Please reread it > > > > carefully. > > > > There is no enclosure implied in Einstein's train thought experiment > > > because he did not realize the mistake he was making. Einstein states > > > the speed of light is 'c' regardless of the speed of the source. What > > > Einstein failed to realize is the speed of light is 'c' relative to > > > the medium it is propagating through: > > > That is, of course, totally wrong. Einstein clearly wrote in his 1905 > > paper: > > > "...In agreement with experience we further assume the quantity 2AB/ > > (t'_A-t_A)=c to be a universal constantthe velocity of light in empty > > space." > > > So the train gedanken, which was a pedagogical example for people like > > you, is clearly setup in deep space and far away from any > > gravitational mass. > > > The two events could be two nuclear blasts, observer M just floating > > in space on a small ship and observer M' moving at a speed v in a > > small ship. > > > So all your Aether nonsense is totally useless in this example. > > > Einstein himself also clearly wrote about it: > > > "The introduction of a luminiferous ether will prove to be > > superfluous inasmuch as the view here to be developed will not require > > an absolutely stationary space provided with special properties, nor > > assign a velocity-vector to a point of the empty space in which > > electromagnetic processes take place." > > > Miguel Rios > > Yes, and Einstein was mistaken. His whole concept of curved space-time > is simply displaced aether. Well obviously the whole scientific community considers that Einstein was right and that you are....... not even wrong!! Well I let you talking to yourself again.... Miguel Rios
From: PD on 9 Oct 2009 16:06
On Oct 9, 2:29 pm, mpc755 <mpc...(a)gmail.com> wrote: > On Oct 9, 3:25 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Oct 9, 2:02 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > On Oct 9, 2:29 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Oct 9, 12:26 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > > > On Oct 9, 1:17 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > On Oct 9, 12:08 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > > > > > On Oct 9, 12:30 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > > On Oct 9, 10:06 am, mpc755 <mpc...(a)gmail.com> wrote: > > > > > > > > > > On Oct 9, 10:52 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > > > > On Oct 9, 8:53 am, mpc755 <mpc...(a)gmail.com> wrote: > > > > > > > > > > > > On Oct 9, 8:57 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > > > > > > On Oct 8, 8:34 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > > > > > > > > > > > > > > > > > > > > Sorry, I've fouled this up. > > > > > > > > > > Yes, the light from A and B reaches M' at different times, but the > > > > > > > > > light from A' and B' reaches M' simultaneously. > > > > > > > > > A and A' are a single strike. B and B' are a single strike. > > > > > > > > Yes, A and A' are a single strike but A and A' are two different > > > > > > > locations in three dimensional space. A in on the embankment and A' is > > > > > > > on the train. > > > > > > > No, they are not. You have not read the gedanken carefully. A lighting > > > > > > strike hits in ONE place, not two. A and A' are two labels for the > > > > > > same point. In the original gedanken, A and A' label the point where > > > > > > the train meets the track at one end of the train. That is ONE POINT. > > > > > > > Please reread it and pay more attention than what you have been doing > > > > > > so far. > > > > > > The lightning strike leaves marks at A, A', B, and B'. This is four > > > > > different locations in three dimensional space. A and A' and B and B' > > > > > were co-located at the time of the strikes, but they are four > > > > > different locations. > > > > > A and A' are colocated but they are two different locations? Listen to > > > > yourself! LOL! > > > > > Lightning strikes at ONE POINT at the end of the train, where one end > > > > of the train meets the track. There is a scorch mark left on the track > > > > (A) and a scorch mark left on the train (A') where this ONE spot > > > > occurred. > > > > > Another lightning bolt strikes at ONE point at the other end of the > > > > train, where that end of the train meets the track. > > > > > Here's another way you can think of it. > > > > Set up a graph with an x and y axis. The x and y axis meet at ONE > > > > POINT. That point is labeled (y=0) for the y-axis and is labeled (x=0) > > > > for the x-axis. Does the fact that there are two labels mean that it > > > > is ONE point or TWO points where the axes meet? > > > > Yes, two points where the embankment and the train meet at A/A' and > > > two points where the embankment and the train meet at B/B'. > > > > There is a mark on the train and a mark on the embankment where the > > > lightning strike occurs at A/A' and there is a mark on the train and > > > on the embankment where the lightning strike occurs at B/B'. > > > > Do you understand there is a mark at A on the embankment and A' on the > > > train? Do you understand that one of these points in on the embankment > > > and one of these points is on the train? > > > Where the x and y axis meet on a graph, (y=0) marks that place on the > > y-axis and (x=0) marks that place on the x-axis. Is that one location > > or two? Answer this question, MPC! The answer is obvious to a 4th grader! > > > LOL!! > > How many marks are there? > > Two on the train and two on the embankment. Right, and there are two marks for the place where the x-axis meets the y-axis. Are you an idiot? Why yes, I believe you are trying really hard to be one! > > Two marks are on the train where the aether is stationary relative to > the train frame of reference. Two marks are on the embankment where > the aether is stationary relative to the embankment frame of > reference. > > > > > > > > The point on the embankment is in the embankment reference frame and > > > the aether is stationary on the embankment in the embankment reference > > > frame. The point on the train is in the train reference frame and the > > > aether is stationary on the train in the train reference frame. > > > > Just as you can have a mark at A on the embankment and A' on the train > > > you can have stationary aether on the embankment and stationary aether > > > on the train. > > > Lovely! You're so fun! > > So we have ONE aether and it is stationary relative to the embankment > > and stationary relative to the train at the same time! > > Tell me, can you do that with one ball, too? > > > LOL!! > > If the embankment is knee deep in water, the water will be stationary > in the embankment frame of reference. Why? Why isn't it flowing with the train? > If the train is knee deep in > water, where the train is enclosed, Who said the train is enclosed? Are we repeating ourselves now? > the water will be stationary in > the train frame of reference. > > Replace one medium with another. Replace water with aether. |