 |
|
| Science Forum Index » Physics - Relativity Forum » Simultaneity of Relativity... |
|
Page 28 of 32 Goto page Previous 1, 2, 3 ... 27, 28, 29, 30, 31, 32 Next |
|
| Author |
Message |
| Bruce Richmond... |
 Posted: Thu Nov 05, 2009 3:37 pm |
|
|
|
Guest
|
On Nov 5, 7:36 pm, kenseto <kens... at (no spam) erinet.com> wrote:
[quote]On Nov 3, 7:29 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
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.
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]
Because from the perspective of M he is. That doesn't prevent M' from
constructing his own coordinate system and measuring the speed of
light to be c with respect to himself.
[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 -- Hide quoted text -
- Show quoted text -- Hide quoted text -
- Show quoted text -[/quote] |
|
|
| Back to top |
|
|
|
| Bruce Richmond... |
| Posted: Thu Nov 05, 2009 3:42 pm |
|
|
|
Guest
|
On Nov 5, 7:43 pm, kenseto <kens... at (no spam) erinet.com> wrote:
[quote]On Nov 3, 7:39 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
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.
Hey idiot....mpc755 asked me why the moving clock is running slow in
my theory.
Ken Seto
[/quote]
You jumped into a discussion that was not about your theory so he had
no way of knowing that you were discussing your theory.
[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 -- Hide quoted text -
- Show quoted text -- Hide quoted text -
- Show quoted text -[/quote] |
|
|
| Back to top |
|
|
|
| mpc755... |
| Posted: Thu Nov 05, 2009 3:46 pm |
|
|
|
Guest
|
On Nov 5, 8:30 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
[quote]On Nov 5, 11:27 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
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 -
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.
[/quote]
I get around the c+v problem by realizing tying the emission point of
the light wave to a point in three dimensional space is incorrect. A
pebble is dropped into the pool on the train. If an Observer on the
embankment was unable to detect the moving water and was only able to
detect the wave in the water, he would conclude the wave originated
from where the center of the pool is when he detected the wave. Any
Observer in any frame of reference who detects the wave will all
conclude the wave originated from where the center of the pool is when
they detect the wave. And all Observers will conclude the wave
traveled at the same speed from the center of the pool, thus 'c' is
maintained for light.
[quote]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.[/quote] |
|
|
| Back to top |
|
|
|
| Bruce Richmond... |
| Posted: Thu Nov 05, 2009 3:59 pm |
|
|
|
Guest
|
On Nov 5, 8:46 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
[quote]On Nov 5, 8:30 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 5, 11:27 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
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 -
Not even if the train frame considers the clocks at M and N to be out
of sync?
I think you will agree that using just one coordinate system it is
physically impossible for light to be measured as traveling at c with
respect to both M and M'. That is why you wanted to add a second
aether, or a pond, to provide a second reference point to use when
measuring the speed of light with respect to M'. But if you use time
and space the way it was used prior to SR you end up getting that
light in the train frame can travel at c+v as measured in the track
frame. And that does not agree with our previous claim that light
always travels at c with respect to the track frame.
I get around the c+v problem by realizing tying the emission point of
the light wave to a point in three dimensional space is incorrect. A
pebble is dropped into the pool on the train. If an Observer on the
embankment was unable to detect the moving water and was only able to
detect the wave in the water, he would conclude the wave originated
from where the center of the pool is when he detected the wave. Any
Observer in any frame of reference who detects the wave will all
conclude the wave originated from where the center of the pool is when
they detect the wave. And all Observers will conclude the wave
traveled at the same speed from the center of the pool, thus 'c' is
maintained for light.
What SR/LET do to get around the problem is to give each frame their
own time. The seconds in two different frames are different from each
other in about the same way that 1 mile north is different from 1 mile
east, but the Lorentz Transformations can be used to convert
coordinates from one frame into those of another. When everything is
done correctly it works.- Hide quoted text -
- Show quoted text -- Hide quoted text -
- Show quoted text -
[/quote]
Which puts you right back to a two aether theory which we have already
seen is impossible when one frame is passing right through the middle
of another. Sorry but I have no interest in your theory. Bye. |
|
|
| Back to top |
|
|
|
| Nick |
| Posted: Thu Nov 05, 2009 4:01 pm |
|
|
|
Joined: 17 Apr 2005
Posts: 3267
|
On Nov 5, 5:47 pm, "Inertial" <relativ... at (no spam) rest.com> wrote:
[quote]"kenseto" <kens... at (no spam) erinet.com> wrote in message
news:d7367991-b54a-4bac-a3c5-6da83e00444c at (no spam) c3g2000yqd.googlegroups.com...
On Nov 3, 7:39 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
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.
Hey idiot....mpc755 asked me why the moving clock is running slow in
my theory.
Liar .. he did not- Hide quoted text -
- Show quoted text -
[/quote]
If you move ahead of a light beam you leave it behind and there is
more space for it to travel before reaching you. Move toward light and
you and the light come together faster. This is vice versa. This is
the basis of relativity of simultaneity. Connectedness is at the speed
of light.
Mitch Raemsch |
|
|
| Back to top |
|
|
|
| mpc755... |
| Posted: Thu Nov 05, 2009 4:07 pm |
|
|
|
Guest
|
On Nov 5, 8:59 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
[quote]On Nov 5, 8:46 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 5, 8:30 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 5, 11:27 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
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 -
Not even if the train frame considers the clocks at M and N to be out
of sync?
I think you will agree that using just one coordinate system it is
physically impossible for light to be measured as traveling at c with
respect to both M and M'. That is why you wanted to add a second
aether, or a pond, to provide a second reference point to use when
measuring the speed of light with respect to M'. But if you use time
and space the way it was used prior to SR you end up getting that
light in the train frame can travel at c+v as measured in the track
frame. And that does not agree with our previous claim that light
always travels at c with respect to the track frame.
I get around the c+v problem by realizing tying the emission point of
the light wave to a point in three dimensional space is incorrect. A
pebble is dropped into the pool on the train. If an Observer on the
embankment was unable to detect the moving water and was only able to
detect the wave in the water, he would conclude the wave originated
from where the center of the pool is when he detected the wave. Any
Observer in any frame of reference who detects the wave will all
conclude the wave originated from where the center of the pool is when
they detect the wave. And all Observers will conclude the wave
traveled at the same speed from the center of the pool, thus 'c' is
maintained for light.
What SR/LET do to get around the problem is to give each frame their
own time. The seconds in two different frames are different from each
other in about the same way that 1 mile north is different from 1 mile
east, but the Lorentz Transformations can be used to convert
coordinates from one frame into those of another. When everything is
done correctly it works.- Hide quoted text -
- Show quoted text -- Hide quoted text -
- Show quoted text -
Which puts you right back to a two aether theory which we have already
seen is impossible when one frame is passing right through the middle
of another. Sorry but I have no interest in your theory. Bye.
[/quote]
There is only one aether. The aether is at rest relative to the train.
When a pebble is dropped into the pool, the center of the pool is at A/
A'. When the wave reaches the Observer at M, the Observer at M
correctly measures the distance the wave traveled as the distance M
was from A' when the wave was detected. Bye. |
|
|
| Back to top |
|
|
|
| Nick |
| Posted: Thu Nov 05, 2009 4:22 pm |
|
|
|
Joined: 17 Apr 2005
Posts: 3267
|
On Nov 5, 5:46 pm, "Inertial" <relativ... at (no spam) rest.com> wrote:
[quote]"kenseto" <kens... at (no spam) erinet.com> wrote in message
news:787ac87c-9042-4072-9ab4-e1687e06d4da at (no spam) m26g2000yqb.googlegroups.com...
On Nov 3, 7:12 pm, "Inertial" <relativ... at (no spam) rest.com> wrote:
"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.
That's why IRT has two equations for the rate of an observed clock.
And why it is useless
One for the observed clock to run slow and the other for the obsrved
clock to run faster than the observer's clock.
And in this case, you don't know which will happen.
And, of course, it can be any of an infinite number of values because not
only do we not know which has the greater absolute motion, we do not know
the direction or value of absolute motion for M and M' either.
Your theory basically says we cannot possibly know the relative rates at
which any pair of clocks tick.
Its useless.- Hide quoted text -
- Show quoted text -
[/quote]
When time slows down for accelerating energy it is absolute.
When time slows down for falling matter it is absolute.
These are the two times. both absolute GR and SR times.
Mitch Raemsch |
|
|
| Back to top |
|
|
|
| Inertial... |
| Posted: Thu Nov 05, 2009 8:43 pm |
|
|
|
Guest
|
"kenseto" <kenseto at (no spam) erinet.com> wrote in message
news:e8111b69-7e24-4c8e-bedd-18289d31c8af at (no spam) a21g2000yqc.googlegroups.com...
[quote]On Nov 3, 7:29 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
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.
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)???
[/quote]
Your ignorance is showing .. it is closing velocity (as observed by the
observer relative to whom M' is moving at v). Please try to keep up ..
these ideas have been around for over a hundred years, you'd think you'd
have heard of this before. |
|
|
| Back to top |
|
|
|
| Inertial... |
| Posted: Thu Nov 05, 2009 8:46 pm |
|
|
|
Guest
|
"kenseto" <kenseto at (no spam) erinet.com> wrote in message
news:787ac87c-9042-4072-9ab4-e1687e06d4da at (no spam) m26g2000yqb.googlegroups.com...
[quote]On Nov 3, 7:12 pm, "Inertial" <relativ... at (no spam) rest.com> wrote:
"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.
That's why IRT has two equations for the rate of an observed clock.
[/quote]
And why it is useless
[quote]One for the observed clock to run slow and the other for the obsrved
clock to run faster than the observer's clock.
[/quote]
And in this case, you don't know which will happen.
And, of course, it can be any of an infinite number of values because not
only do we not know which has the greater absolute motion, we do not know
the direction or value of absolute motion for M and M' either.
Your theory basically says we cannot possibly know the relative rates at
which any pair of clocks tick.
Its useless. |
|
|
| Back to top |
|
|
|
| Inertial... |
| Posted: Thu Nov 05, 2009 8:47 pm |
|
|
|
Guest
|
"kenseto" <kenseto at (no spam) erinet.com> wrote in message
news:d7367991-b54a-4bac-a3c5-6da83e00444c at (no spam) c3g2000yqd.googlegroups.com...
[quote]On Nov 3, 7:39 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
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.
Hey idiot....mpc755 asked me why the moving clock is running slow in
my theory.
[/quote]
Liar .. he did not |
|
|
| Back to top |
|
|
|
| mpc755... |
| Posted: Fri Nov 06, 2009 3:32 am |
|
|
|
Guest
|
On Nov 5, 9:07 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
[quote]On Nov 5, 8:59 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 5, 8:46 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 5, 8:30 pm, Bruce Richmond <bsr3... at (no spam) my-deja.com> wrote:
On Nov 5, 11:27 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
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 -
Not even if the train frame considers the clocks at M and N to be out
of sync?
I think you will agree that using just one coordinate system it is
physically impossible for light to be measured as traveling at c with
respect to both M and M'. That is why you wanted to add a second
aether, or a pond, to provide a second reference point to use when
measuring the speed of light with respect to M'. But if you use time
and space the way it was used prior to SR you end up getting that
light in the train frame can travel at c+v as measured in the track
frame. And that does not agree with our previous claim that light
always travels at c with respect to the track frame.
I get around the c+v problem by realizing tying the emission point of
the light wave to a point in three dimensional space is incorrect. A
pebble is dropped into the pool on the train. If an Observer on the
embankment was unable to detect the moving water and was only able to
detect the wave in the water, he would conclude the wave originated
from where the center of the pool is when he detected the wave. Any
Observer in any frame of reference who detects the wave will all
conclude the wave originated from where the center of the pool is when
they detect the wave. And all Observers will conclude the wave
traveled at the same speed from the center of the pool, thus 'c' is
maintained for light.
What SR/LET do to get around the problem is to give each frame their
own time. The seconds in two different frames are different from each
other in about the same way that 1 mile north is different from 1 mile
east, but the Lorentz Transformations can be used to convert
coordinates from one frame into those of another. When everything is
done correctly it works.- Hide quoted text -
- Show quoted text -- Hide quoted text -
- Show quoted text -
Which puts you right back to a two aether theory which we have already
seen is impossible when one frame is passing right through the middle
of another. Sorry but I have no interest in your theory. Bye.
There is only one aether. The aether is at rest relative to the train.
When a pebble is dropped into the pool, the center of the pool is at A/
A'. When the wave reaches the Observer at M, the Observer at M
correctly measures the distance the wave traveled as the distance M
was from A' when the wave was detected. Bye.
[/quote]
There is only one aether. If the aether is at rest relative to the
embankment and a lightning strike occurs at A/A' the light wave
propagates outward at 'c' from A. When an Observer, regardless of
frame of reference, sees the light it has traveled from where A *is*.
If Observers on the train or the embankment do not know their state of
motion relative to the aether, they may approximate where the light
originated from by measuring to marks left on the embankment or left
on the train. If the Observer on the embankment concludes the
lightning strikes occurred simultaneously and the Observer on the
train concludes the lightning strike at B' occurred prior to the
lightning strike at A', then one or both of the Observers is incorrect. |
|
|
| Back to top |
|
|
|
| mpc755... |
| Posted: Fri Nov 06, 2009 3:36 am |
|
|
|
Guest
|
On Nov 5, 9:07 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
[quote]
There is only one aether. The aether is at rest relative to the train.
When a pebble is dropped into the pool, the center of the pool is at A/
A'. When the wave reaches the Observer at M, the Observer at M
correctly measures the distance the wave traveled as the distance M
was from A' when the wave was detected.
[/quote]
There is only one aether. If the aether is at rest relative to the
embankment and a lightning strike occurs at A/A' the light wave
propagates outward at 'c' from A. When an Observer, regardless of
frame of reference, sees the light it has traveled from where A *is*.
If Observers on the train or the embankment do not know their state of
motion relative to the aether, they may approximate where the light
originated from by measuring to marks left on the embankment or left
on the train. If the Observer on the embankment concludes the
lightning strikes occurred simultaneously and the Observer on the
train concludes the lightning strike at B' occurred prior to the
lightning strike at A', then one or both of the Observers is
incorrect.
Light does not travel at 'c' relative to frames of reference. Light
travels at 'c' relative to the aether. When an Observer sees the
light, the light wave will have traveled at 'c' to the Observer
relative to the aether. |
|
|
| Back to top |
|
|
|
| mpc755... |
| Posted: Fri Nov 06, 2009 4:31 am |
|
|
|
Guest
|
On Nov 6, 8:36 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
[quote]On Nov 5, 9:07 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
There is only one aether. The aether is at rest relative to the train.
When a pebble is dropped into the pool, the center of the pool is at A/
A'. When the wave reaches the Observer at M, the Observer at M
correctly measures the distance the wave traveled as the distance M
was from A' when the wave was detected.
There is only one aether. If the aether is at rest relative to the
embankment and a lightning strike occurs at A/A' the light wave
propagates outward at 'c' from A. When an Observer, regardless of
frame of reference, sees the light it has traveled from where A *is*.
If Observers on the train or the embankment do not know their state of
motion relative to the aether, they may approximate where the light
originated from by measuring to marks left on the embankment or left
on the train. If the Observer on the embankment concludes the
lightning strikes occurred simultaneously and the Observer on the
train concludes the lightning strike at B' occurred prior to the
lightning strike at A', then one or both of the Observers is
incorrect.
Light does not travel at 'c' relative to frames of reference. Light
travels at 'c' relative to the aether. When an Observer sees the
light, the light wave will have traveled at 'c' to the Observer
relative to the aether.
[/quote]
The problem with Einstein's train thought experiment is in order for
the Observer on the train to measure to A' and for the Observer on the
embankment to measure to A and for both measurements to be accurate,
the aether must be at rest relative to the train and at rest relative
to the embankment which is physically impossible if both frames
intersect and occupy the same three dimensional space. |
|
|
| Back to top |
|
|
|
| kenseto... |
| Posted: Fri Nov 06, 2009 4:34 am |
|
|
|
Guest
|
On Nov 5, 8:46 pm, "Inertial" <relativ... at (no spam) rest.com> wrote:
[quote]"kenseto" <kens... at (no spam) erinet.com> wrote in message
news:787ac87c-9042-4072-9ab4-e1687e06d4da at (no spam) m26g2000yqb.googlegroups.com...
On Nov 3, 7:12 pm, "Inertial" <relativ... at (no spam) rest.com> wrote:
"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.
That's why IRT has two equations for the rate of an observed clock.
And why it is useless
[/quote]
Fucking idiot....when you compare two clocks the following will
happen:
1. they are running at the same rate.
2. A is running at a faster rate than B.
3. B is running slower than A.
There is no way that A predicts that B is running slow and B predict
that A is running slow.
[quote]
One for the observed clock to run slow and the other for the obsrved
clock to run faster than the observer's clock.
[/quote]
Yes that's right.
[quote]
And in this case, you don't know which will happen.
[/quote]
Yes you don't know which will happen and that's why you have two
equations....and two soultions for the observed clock. You will have
to do both calculation to see which prediction is correct.
[quote]
And, of course, it can be any of an infinite number of values because not
only do we not know which has the greater absolute motion,
[/quote]
Fucking idiot there is only one Fab fr the observed clock. If the
observed clock is running slow then it is running slow by a factor of
Fab/Faa.
If the observed clock is running fast then it is running fast by a
factor of Faa/Fab.
You are so fucking stuid.
Ken Seto
[quote]we do not know
the direction or value of absolute motion for M and M' either.
Your theory basically says we cannot possibly know the relative rates at
which any pair of clocks tick.
Its useless.- Hide quoted text -
- Show quoted text -[/quote] |
|
|
| Back to top |
|
|
|
| mpc755... |
| Posted: Fri Nov 06, 2009 5:45 am |
|
|
|
Guest
|
On Nov 6, 9:31 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
[quote]On Nov 6, 8:36 am, mpc755 <mpc... at (no spam) gmail.com> wrote:
On Nov 5, 9:07 pm, mpc755 <mpc... at (no spam) gmail.com> wrote:
There is only one aether. The aether is at rest relative to the train..
When a pebble is dropped into the pool, the center of the pool is at A/
A'. When the wave reaches the Observer at M, the Observer at M
correctly measures the distance the wave traveled as the distance M
was from A' when the wave was detected.
There is only one aether. If the aether is at rest relative to the
embankment and a lightning strike occurs at A/A' the light wave
propagates outward at 'c' from A. When an Observer, regardless of
frame of reference, sees the light it has traveled from where A *is*.
If Observers on the train or the embankment do not know their state of
motion relative to the aether, they may approximate where the light
originated from by measuring to marks left on the embankment or left
on the train. If the Observer on the embankment concludes the
lightning strikes occurred simultaneously and the Observer on the
train concludes the lightning strike at B' occurred prior to the
lightning strike at A', then one or both of the Observers is
incorrect.
Light does not travel at 'c' relative to frames of reference. Light
travels at 'c' relative to the aether. When an Observer sees the
light, the light wave will have traveled at 'c' to the Observer
relative to the aether.
The problem with Einstein's train thought experiment is in order for
the Observer on the train to measure to A' and for the Observer on the
embankment to measure to A and for both measurements to be accurate,
the aether must be at rest relative to the train and at rest relative
to the embankment which is physically impossible if both frames
intersect and occupy the same three dimensional space.
[/quote]
The aether is the preferred frame.
The idea of motion may be applied to the aether. |
|
|
| Back to top |
|
|
|
|
|
All times are GMT - 5 Hours
The time now is Sat Nov 28, 2009 10:05 pm
|
|