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| mpc755... |
 Posted: Tue Nov 03, 2009 6:38 pm |
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Guest
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On Nov 3, 11:00 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
[quote]On Nov 3, 8:17 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 3, 7:29 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 3, 1:16 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 3, 12:33 am, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 2, 9:36 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 2, 12:16 am, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 1, 11:20 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 1, 10:57 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 1, 10:13 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 1, 7:32 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Oct 8, 11:49 am, mpc755 <mpc... at (no spam) 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.
[/quote]
The light travels from where A was but where A' is? Bull shit.
There is another Observer at rest relative to the embankment. This
Observer is standing exactly where the Observer at M' is when the
light from A' and B' reach M'. Does this additional Observer see the
light from A' and B' or is this Observer still waiting for the light
from A and B to arrive? What light waves from which lightning strikes
reach this Observer and is it 12:00:01 when the light does arrive?
[quote]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.
[/quote]
If the train frame of reference and the embankment frame of reference
occupy different areas of three dimensional space, the aether can be
entrained in both frames of reference, similar to the pool of water on
the train.
http://www.youtube.com/watch?v=jyWTaXMElUk
[quote]
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'.[/quote] |
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| Inertial... |
| Posted: Tue Nov 03, 2009 7:12 pm |
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Guest
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"kenseto" <kenseto at (no spam) erinet.com> wrote in message
news:58559b17-56a0-454c-a202-8884aa94d656 at (no spam) a31g2000yqn.googlegroups.com...
[quote]On Nov 3, 9:13 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 3, 8:37 am, kenseto <kens... at (no spam) erinet.com> wrote:
On Nov 3, 1:16 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 3, 12:33 am, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 2, 9:36 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 2, 12:16 am, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 1, 11:20 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 1, 10:57 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com
wrote:
On Nov 1, 10:13 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 1, 7:32 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Oct 8, 11:49 am, mpc755 <mpc... at (no spam) 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.
[/quote]
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. |
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| Bruce Richmond... |
| Posted: Wed Nov 04, 2009 2:30 pm |
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Guest
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On Nov 3, 11:38 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
[quote]On Nov 3, 11:00 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 3, 8:17 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 3, 7:29 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 3, 1:16 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 3, 12:33 am, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 2, 9:36 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 2, 12:16 am, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 1, 11:20 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 1, 10:57 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 1, 10:13 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 1, 7:32 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Oct 8, 11:49 am, mpc755 <mpc... at (no spam) 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 light travels from where A was but where A' is? Bull shit.
[/quote]
That's not what I wrote so it's your own bull shit. M' sees the light
from where A was and from where A' was when he was with A. A' hasn't
moved in the frame of M', so his space coordinates are still the same,
but his time coordinate has changed.
[quote]There is another Observer at rest relative to the embankment. This
Observer is standing exactly where the Observer at M' is when the
light from A' and B' reach M'. Does this additional Observer see the
light from A' and B' or is this Observer still waiting for the light
from A and B to arrive?
[/quote]
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.
[quote]What light waves from which lightning strikes
reach this Observer
[/quote]
There is only one flash coming from A/A' so when he sees the flash
from A he is also seeing the flash from A'.
[quote]and is it 12:00:01 when the light does arrive?
[/quote]
Thank you for bringing this up since it demonstrates relative
simultanity in your new set-up. The clock at M was reading 12:00:01
when the flash from A/A' arrived there. N is further from A than M is
so the light from A/A' will take longer to reach him than M and will
read something later than 12:00:01. The clock at M' will read
12:00:01 so they do not agree. The clock at N did not speed up, nor
did the one at M' slow down. They read different because they were
synchronized in different frames.
[quote]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.
If the train frame of reference and the embankment frame of reference
occupy different areas of three dimensional space, the aether can be
entrained in both frames of reference, similar to the pool of water on
the train.
[/quote]
In our most recent version of the experiment, which you proposed, M'
passes right between A and B. There is no way you can come up with
seperate aethers for the two frames in such a set-up. So your theory
is impossible to implement, but LET works just fine with one aether.
[quote]http://www.youtube.com/watch?v=jyWTaXMElUk
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 -[/quote] |
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| mpc755... |
| Posted: Wed Nov 04, 2009 2:57 pm |
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Guest
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On Nov 4, 7:30 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
[quote]On Nov 3, 11:38 pm, mpc755 <mpc... at (no spam) 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.
[/quote]
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? |
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| mpc755... |
| Posted: Wed Nov 04, 2009 6:15 pm |
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Guest
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On Nov 4, 7:57 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
[quote]On Nov 4, 7:30 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 3, 11:38 pm, mpc755 <mpc... at (no spam) 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?
[/quote]
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. |
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| Bruce Richmond... |
| Posted: Wed Nov 04, 2009 7:16 pm |
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Guest
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On Nov 4, 7:57 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
[quote]On Nov 4, 7:30 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 3, 11:38 pm, mpc755 <mpc... at (no spam) 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?
[/quote]
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. |
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| mpc755... |
| Posted: Wed Nov 04, 2009 7:53 pm |
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Guest
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On Nov 5, 12:16 am, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
[quote]On Nov 4, 7:57 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 4, 7:30 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 3, 11:38 pm, mpc755 <mpc... at (no spam) 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.
[/quote]
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'. |
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| mpc755... |
| Posted: Thu Nov 05, 2009 6:27 am |
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Guest
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On Nov 5, 12:53 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
[quote]On Nov 5, 12:16 am, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 4, 7:57 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 4, 7:30 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 3, 11:38 pm, mpc755 <mpc... at (no spam) 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'.
[/quote]
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. |
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| PD... |
| Posted: Thu Nov 05, 2009 6:37 am |
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Guest
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On Nov 5, 10:27 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
[quote]On Nov 5, 12:53 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 5, 12:16 am, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 4, 7:57 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 4, 7:30 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 3, 11:38 pm, mpc755 <mpc... at (no spam) 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.
[/quote]
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.
[quote]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.[/quote] |
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| kenseto... |
| Posted: Thu Nov 05, 2009 2:36 pm |
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Guest
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On Nov 3, 7:29 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
[quote]On Nov 3, 12:34 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 3, 12:19 am, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 2, 9:16 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 2, 12:15 am, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 1, 10:57 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 1, 10:47 pm, Bruce Richmond <bsr3... at (no spam) 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.
[/quote]
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
[quote]
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 -[/quote] |
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| kenseto... |
| Posted: Thu Nov 05, 2009 2:40 pm |
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Guest
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On Nov 3, 7:12 pm, "Inertial" <relativ... at (no spam) rest.com> wrote:
[quote]"kenseto" <kens... at (no spam) erinet.com> wrote in message
news:58559b17-56a0-454c-a202-8884aa94d656 at (no spam) a31g2000yqn.googlegroups.com...
On Nov 3, 9:13 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 3, 8:37 am, kenseto <kens... at (no spam) erinet.com> wrote:
On Nov 3, 1:16 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 3, 12:33 am, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 2, 9:36 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 2, 12:16 am, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 1, 11:20 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 1, 10:57 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com
wrote:
On Nov 1, 10:13 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 1, 7:32 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Oct 8, 11:49 am, mpc755 <mpc... at (no spam) 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.
[/quote]
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 -
[quote]
- Show quoted text -[/quote] |
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| kenseto... |
| Posted: Thu Nov 05, 2009 2:43 pm |
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On Nov 3, 7:39 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
[quote]On Nov 3, 10:32 am, kenseto <kens... at (no spam) erinet.com> wrote:
On Nov 3, 9:13 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 3, 8:37 am, kenseto <kens... at (no spam) erinet.com> wrote:
On Nov 3, 1:16 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 3, 12:33 am, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 2, 9:36 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 2, 12:16 am, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 1, 11:20 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 1, 10:57 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 1, 10:13 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 1, 7:32 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Oct 8, 11:49 am, mpc755 <mpc... at (no spam) 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.
[/quote]
Hey idiot....mpc755 asked me why the moving clock is running slow in
my theory.
Ken Seto
[quote]
I think you are saying M' clock is
running slower from the perspective of the embankment frame of
reference, but the same must be true from the perspective of the train
frame of reference.
No that's not what I am saying. I said that the M' clock is really
running slower than the M clock. It is not a perstie effect. From M'
pint of view: the M clock is running fast compared to the M' clock.
Ken Seto
And you are wrong.
From the train frame of reference, the M clock
must be running slower than the M' clock. The frames are moving
relative to one another. Both frames, in order to be isotropic, must
be moving identically relative to one another.
To say the M' clock is running slower than M's clock is to have a
preferred frame of reference.- Hide quoted text -
- Show quoted text -- Hide quoted text -
- Show quoted text -- Hide quoted text -
- Show quoted text -- Hide quoted text -
- Show quoted text -- Hide quoted text -
- Show quoted text -[/quote] |
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| Bruce Richmond... |
| Posted: Thu Nov 05, 2009 2:49 pm |
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On Nov 5, 12:53 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
[quote]On Nov 5, 12:16 am, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 4, 7:57 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 4, 7:30 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 3, 11:38 pm, mpc755 <mpc... at (no spam) 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.
[/quote]
Agreed.
[quote]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.
[/quote]
But M' does hasten toward M and then away from him. They are not at
rest with respect to each other so they do not use the same coordinate
systems. By convention the direction of the relative motion is
defined as the x axis. By putting A/A' off to the side you have
shifted the event from being on the x axis to the y axis. The
position dependence of time that I mentioned above is along the x
axis, not the y axis.
In Einstein's experiment he put everything on the x axis so we only
have to deal with one dimension of space and one dimension of time for
each coordinate system. Moving the strikes off to the sides added a
second space dimension to each coordinate system and complicated
things a bit, but relative simultanity still shows up.
[quote]A and M, B and M, A' and M', and B' and M' are equi-distant from each
other.
[/quote]
Agreed.
[quote]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.
[/quote]
Nope, you are doing what is called frame jumping. That is incorrectly
applying measurements made in one frame to the other frame. For
starters t=1 is no more the same as t'=1 than 1 mile north is the same
as 1 mile east. You wrote above, correctly, that the strike at A/A'
was a single event. There is no way that a single event can happen
prior to itself.
I think I can explain SR/LET to you if you are interested in learning
it. If not I will go away.
[quote]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'.- Hide quoted text -
- Show quoted text -[/quote] |
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| mpc755... |
| Posted: Thu Nov 05, 2009 3:27 pm |
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Guest
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On Nov 5, 7:49 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
[quote]On Nov 5, 12:53 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 5, 12:16 am, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 4, 7:57 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 4, 7:30 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 3, 11:38 pm, mpc755 <mpc... at (no spam) 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.
Agreed.
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.
But M' does hasten toward M and then away from him. They are not at
rest with respect to each other so they do not use the same coordinate
systems. By convention the direction of the relative motion is
defined as the x axis. By putting A/A' off to the side you have
shifted the event from being on the x axis to the y axis. The
position dependence of time that I mentioned above is along the x
axis, not the y axis.
In Einstein's experiment he put everything on the x axis so we only
have to deal with one dimension of space and one dimension of time for
each coordinate system. Moving the strikes off to the sides added a
second space dimension to each coordinate system and complicated
things a bit, but relative simultanity still shows up.
A and M, B and M, A' and M', and B' and M' are equi-distant from each
other.
Agreed.
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.
Nope, you are doing what is called frame jumping. That is incorrectly
applying measurements made in one frame to the other frame. For
starters t=1 is no more the same as t'=1 than 1 mile north is the same
as 1 mile east. You wrote above, correctly, that the strike at A/A'
was a single event. There is no way that a single event can happen
prior to itself.
I think I can explain SR/LET to you if you are interested in learning
it. If not I will go away.
[/quote]
If you respond to my subsequent post, we can end the conversation
there, thanks.
[quote]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'.- Hide quoted text -
- Show quoted text -
[/quote] |
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| Bruce Richmond... |
| Posted: Thu Nov 05, 2009 3:30 pm |
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Guest
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On Nov 5, 11:27 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
[quote]On Nov 5, 12:53 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 5, 12:16 am, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 4, 7:57 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 4, 7:30 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 3, 11:38 pm, mpc755 <mpc... at (no spam) 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 -
[/quote]
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.
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. |
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