| Topic: |
Science > Physics |
| User: |
"Richard" |
| Date: |
21 Sep 2003 09:33:38 AM |
| Object: |
Relative ticking rate nonsense |
_____K'____
1) | |
|___________|--> v
oo oo ____________
| |
2) |_____K______|
oo oo
Two railway carriages are in motion wrt each other along the x axis, as
depicted in the diagram above. According to K (the rest frame of
carriage 2), carriage 1 and carriage 2 are equal in length (this is
taking into account the relative contraction of carriage 1 wrt K).
As 1 passes by the left side of 2 the time interval between the
correspondence of the right and left sides of 1 with the left side of 2
wrt K, is
dt = L_o/v
The time interval of the same two events wrt K' (the rest frame of
carriage 1) is
dt' = L_o gamma/v
Thus K' measures a longer time interval than does K between these two
events, and it follows that the K' clock is ticking faster than the K
clock. Thus in this experiment it is the moving clock that ticks faster.
Moreover the ratio of intervals is provided by
L_o gamma/dt' = L_o/dt
dt' = dt L_o gamma/L_o
or
dt' = dt gamma
OTOH we have also in the same sequence the additional two events which
are the correspondence of the left and right sides of 2 with the right
side of 1. Wrt K this time interval is exactly the same as the first,
since according to him the carriages are equal in length. However,
assuming reciprocity of length contraction the time interval wrt K' is
dt' = L_o/gamma v
But L_o = dtv, just as before
Thus
dt' = dt/gamma
Which is opposite in sense to the ticking rate that was derived above.
Thus which clock ticks faster seems to depend entirely upon the motion
of some arbitrary extended object wrt each. If the object is at rest wrt
K then the moving clock is ticking slower, but if it is at rest wrt K'
then the moving clock is ticking slower. If there are two objects, one
at rest wrt each, then the clocks are simultaneously ticking slower and
faster than each other. Moreover, the ticking ratio only assumes the
simplified form above when the object(s) is/are moving at v wrt one of
the frames, i.e. given an object that is moving wrt both frames, the
ticking rate does not resolve to either of the above forms, but assumes
a different value for every alternate value of x/t, or x'/t' that is not
equal to v. IOW SR is gibberish.
Richard Perry
http://www.cswnet.com/~rper
.
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| User: "Paul B. Andersen" |
|
| Title: Re: Relative ticking rate nonsense |
21 Sep 2003 04:37:52 PM |
|
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"Richard" <no_mail_no_spam@yahoo.com> skrev i melding news:3F6DB6C2.988405A6@yahoo.com...
_____K'____
1) | |
|___________|--> v
oo oo ____________
| |
2) |_____K______|
oo oo
Two railway carriages are in motion wrt each other along the x axis, as
depicted in the diagram above. According to K (the rest frame of
carriage 2), carriage 1 and carriage 2 are equal in length (this is
taking into account the relative contraction of carriage 1 wrt K).
As 1 passes by the left side of 2 the time interval between the
correspondence of the right and left sides of 1 with the left side of 2
wrt K, is
dt = L_o/v
Right.
Notice that in K, the spatial interval between the events is zero,
that is, the time L_o/v is a proper time, which can be measured
with one clock (At the left end of 2.).
The space time interval is L_o*c/v
The time interval of the same two events wrt K' (the rest frame of
carriage 1) is
dt' = L_o gamma/v
Right.
The temporal interval between the event in K' is L_o*gamma/v.
Notice however that the spatial inerval between the events in K'
is gamma*L_o. That means that the temporal interval is not
a proper time, it is the difference between the reading of two
separate clocks. (One at each end of carriage 1.)
The space time interval is:
sqrt((L_o*gamma*c/v)^2 - (gamma*L_o)^2) = L_o*c/v
Thus K' measures a longer time interval than does K between these two
events,
Right.
and it follows that the K' clock is ticking faster than the K
clock.
Wrong!
There are TWO "moving clocks".
You cannot conclude anything about what the rate of a clock
is without having two different readings _of the same clock_!
You have only one reading of each clock.
Note that "the rate of a moving clock" is:
dt'/dt when dx' = 0
dt/dt' when dx = 0
or in other words:
to measure the rate of a moving clock you must
compare two readings on the moving clock to
two reading on two separate (coordinate) clocks!
You have only got one reading from each of the TWO
moving clocks. You can conclude nothing about their
rate observed in K from that.
Thus in this experiment it is the moving clock that ticks faster.
Moreover the ratio of intervals is provided by
L_o gamma/dt' = L_o/dt
dt' = dt L_o gamma/L_o
or
dt' = dt gamma
Exactly.
And since dt is read off one clock while dt' is read off
two separate clocks, we can conclude:
The rate of the moving clock in 2 is dt/dt' = 1/gamma
as observed in K'.
OTOH we have also in the same sequence the additional two events which
are the correspondence of the left and right sides of 2 with the right
side of 1. Wrt K this time interval is exactly the same as the first,
since according to him the carriages are equal in length. However,
assuming reciprocity of length contraction the time interval wrt K' is
dt' = L_o/gamma v
But L_o = dtv, just as before
Thus
dt' = dt/gamma
Exactly.
Since dt' in this case is read of one clock while dt is read of two
separate clocks, we can conclude:
The rate of the moving clock in 1 is dt'/dt = 1/gamma
as observed in K'.
Which is opposite in sense to the ticking rate that was derived above.
No.
The ticking rates of the single, moving clock are in both cases
the same, namely 1/gamma.
Thus which clock ticks faster seems to depend entirely upon the motion
of some arbitrary extended object wrt each. If the object is at rest wrt
K then the moving clock is ticking slower, but if it is at rest wrt K'
then the moving clock is ticking slower. If there are two objects, one
at rest wrt each, then the clocks are simultaneously ticking slower and
faster than each other. Moreover, the ticking ratio only assumes the
simplified form above when the object(s) is/are moving at v wrt one of
the frames, i.e. given an object that is moving wrt both frames, the
ticking rate does not resolve to either of the above forms, but assumes
a different value for every alternate value of x/t, or x'/t' that is not
equal to v. IOW SR is gibberish.
If you are a sensible guy, you will have realized wher you went wrong.
Are you?
Paul
.
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| User: "Paul B. Andersen" |
|
| Title: Re: Relative ticking rate nonsense |
22 Sep 2003 04:41:53 AM |
|
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"Paul B. Andersen" <paul.b.andersen@hia.no> skrev i melding news:bkl5oc$38o$1@dolly.uninett.no...
"Richard" <no_mail_no_spam@yahoo.com> skrev i melding news:3F6DB6C2.988405A6@yahoo.com...
Moreover the ratio of intervals is provided by
L_o gamma/dt' = L_o/dt
dt' = dt L_o gamma/L_o
or
dt' = dt gamma
Exactly.
And since dt is read off one clock while dt' is read off
two separate clocks, we can conclude:
The rate of the moving clock in 2 is dt/dt' = 1/gamma
as observed in K'.
OTOH we have also in the same sequence the additional two events which
are the correspondence of the left and right sides of 2 with the right
side of 1. Wrt K this time interval is exactly the same as the first,
since according to him the carriages are equal in length. However,
assuming reciprocity of length contraction the time interval wrt K' is
dt' = L_o/gamma v
But L_o = dtv, just as before
Thus
dt' = dt/gamma
Exactly.
Since dt' in this case is read of one clock while dt is read of two
separate clocks, we can conclude:
The rate of the moving clock in 1 is dt'/dt = 1/gamma
as observed in K'.
Oooops.
Should be:
The rate of the moving clock in 1 is dt'/dt = 1/gamma
as observed in K.
Paul
.
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| User: "Wilson" |
|
| Title: Re: Relative ticking rate nonsense |
22 Sep 2003 05:00:42 PM |
|
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On Mon, 22 Sep 2003 11:41:53 +0200, "Paul B. Andersen" <paul.b.andersen@hia.no>
wrote:
"Paul B. Andersen" <paul.b.andersen@hia.no> skrev i melding news:bkl5oc$38o$1@dolly.uninett.no...
"Richard" <no_mail_no_spam@yahoo.com> skrev i melding news:3F6DB6C2.988405A6@yahoo.com...
Moreover the ratio of intervals is provided by
L_o gamma/dt' = L_o/dt
dt' = dt L_o gamma/L_o
or
dt' = dt gamma
Exactly.
And since dt is read off one clock while dt' is read off
two separate clocks, we can conclude:
The rate of the moving clock in 2 is dt/dt' = 1/gamma
as observed in K'.
OTOH we have also in the same sequence the additional two events which
are the correspondence of the left and right sides of 2 with the right
side of 1. Wrt K this time interval is exactly the same as the first,
since according to him the carriages are equal in length. However,
assuming reciprocity of length contraction the time interval wrt K' is
dt' = L_o/gamma v
But L_o = dtv, just as before
Thus
dt' = dt/gamma
Exactly.
Since dt' in this case is read of one clock while dt is read of two
separate clocks, we can conclude:
The rate of the moving clock in 1 is dt'/dt = 1/gamma
as observed in K'.
Oooops.
Should be:
The rate of the moving clock in 1 is dt'/dt = 1/gamma
as observed in K.
What a load of crap.
Under conditions of local source dependency, a clock moving away at v appears
to be running slower by (c-v)/c.
One approaching at v appears fast by (c+v)/c.
If you don't believe me, try using Jupiter's rotation as your clock. Or maybe
you don't believe that what one OBSERVES is also what one SEES.
Note: without source dependency, the above ratios are, c/(c+v) and c/(c-v)
respectively. At large distances, the whole situation is complicated by the
intricacies of my H-aether theory.
Paul
.
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| User: "Paul B. Andersen" |
|
| Title: Re: Relative ticking rate nonsense |
28 Sep 2003 07:46:04 AM |
|
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"Henry Wilson" <HW@..> skrev i melding news:s4rumv8vqrpgqpo7jodrloe3vchlhlr7kt@4ax.com...
On Mon, 22 Sep 2003 11:41:53 +0200, "Paul B. Andersen" <paul.b.andersen@hia.no>
wrote:
"Paul B. Andersen" <paul.b.andersen@hia.no> skrev i melding news:bkl5oc$38o$1@dolly.uninett.no...
"Richard" <no_mail_no_spam@yahoo.com> skrev i melding news:3F6DB6C2.988405A6@yahoo.com...
Moreover the ratio of intervals is provided by
L_o gamma/dt' = L_o/dt
dt' = dt L_o gamma/L_o
or
dt' = dt gamma
Exactly.
And since dt is read off one clock while dt' is read off
two separate clocks, we can conclude:
The rate of the moving clock in 2 is dt/dt' = 1/gamma
as observed in K'.
OTOH we have also in the same sequence the additional two events which
are the correspondence of the left and right sides of 2 with the right
side of 1. Wrt K this time interval is exactly the same as the first,
since according to him the carriages are equal in length. However,
assuming reciprocity of length contraction the time interval wrt K' is
dt' = L_o/gamma v
But L_o = dtv, just as before
Thus
dt' = dt/gamma
Exactly.
Since dt' in this case is read of one clock while dt is read of two
separate clocks, we can conclude:
The rate of the moving clock in 1 is dt'/dt = 1/gamma
as observed in K'.
Oooops.
Should be:
The rate of the moving clock in 1 is dt'/dt = 1/gamma
as observed in K.
What a load of crap.
Of course you would think so.
Nobody expects you to understand this, Henry.
We know you are utterly incapable of understanding
anything related to SR and the Lorentz transform.
Under conditions of local source dependency, a clock moving away at v appears
to be running slower by (c-v)/c.
One approaching at v appears fast by (c+v)/c.
If you don't believe me, try using Jupiter's rotation as your clock. Or maybe
you don't believe that what one OBSERVES is also what one SEES
... as you so vividly demonstate here. :-)
"when the issue is beyond you - talk about soemething else"
Note: without source dependency, the above ratios are, c/(c+v) and c/(c-v)
respectively. At large distances, the whole situation is complicated by the
intricacies of my H-aether theory.
Quite.
The whole situation is too complicated for your H-aether
theory to predict anything sensible.
I will take your word for that.
But since the predictions of SR/GR has proven correct
every time they have been tested, the failure of your H-aether
theory to predict anything at all doesn't bother me much.
Paul
.
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| User: "Dirk Van de moortel" |
|
| Title: Re: Relative ticking rate nonsense |
22 Sep 2003 03:56:24 AM |
|
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"Paul B. Andersen" <paul.b.andersen@hia.no> wrote in message news:bkl5oc$38o$1@dolly.uninett.no...
"Richard" <no_mail_no_spam@yahoo.com> skrev i melding news:3F6DB6C2.988405A6@yahoo.com...
[snip]
Thus which clock ticks faster seems to depend entirely upon the motion
of some arbitrary extended object wrt each. If the object is at rest wrt
K then the moving clock is ticking slower, but if it is at rest wrt K'
then the moving clock is ticking slower. If there are two objects, one
at rest wrt each, then the clocks are simultaneously ticking slower and
faster than each other. Moreover, the ticking ratio only assumes the
simplified form above when the object(s) is/are moving at v wrt one of
the frames, i.e. given an object that is moving wrt both frames, the
ticking rate does not resolve to either of the above forms, but assumes
a different value for every alternate value of x/t, or x'/t' that is not
equal to v. IOW SR is gibberish.
If you are a sensible guy, you will have realized wher you went wrong.
Are you?
Ha, Paul, glad to see that you already took care of it.
That saves me the trouble of trying to understand the
remainder of that message.
Apparently I went too fast on this one.
Dirk Vdm
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| User: "Richard" |
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| Title: Re: Relative ticking rate nonsense |
22 Sep 2003 01:07:05 PM |
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"Paul B. Andersen" wrote:
"Richard" <no_mail_no_spam@yahoo.com> skrev i melding news:3F6DB6C2.988405A6@yahoo.com...
_____K'____
1) | |
|___________|--> v
oo oo ____________
| |
2) |_____K______|
oo oo
Two railway carriages are in motion wrt each other along the x axis, as
depicted in the diagram above. According to K (the rest frame of
carriage 2), carriage 1 and carriage 2 are equal in length (this is
taking into account the relative contraction of carriage 1 wrt K).
As 1 passes by the left side of 2 the time interval between the
correspondence of the right and left sides of 1 with the left side of 2
wrt K, is
dt = L_o/v
Right.
Notice that in K, the spatial interval between the events is zero,
that is, the time L_o/v is a proper time, which can be measured
with one clock (At the left end of 2.).
The space time interval is L_o*c/v
The time interval of the same two events wrt K' (the rest frame of
carriage 1) is
dt' = L_o gamma/v
Right.
The temporal interval between the event in K' is L_o*gamma/v.
Notice however that the spatial inerval between the events in K'
is gamma*L_o. That means that the temporal interval is not
a proper time, it is the difference between the reading of two
separate clocks. (One at each end of carriage 1.)
The space time interval is:
sqrt((L_o*gamma*c/v)^2 - (gamma*L_o)^2) = L_o*c/v
Thus K' measures a longer time interval than does K between these two
events,
Right.
and it follows that the K' clock is ticking faster than the K
clock.
Wrong!
There are TWO "moving clocks".
You cannot conclude anything about what the rate of a clock
is without having two different readings _of the same clock_!
You have only one reading of each clock.
Note that "the rate of a moving clock" is:
dt'/dt when dx' = 0
dt/dt' when dx = 0
or in other words:
to measure the rate of a moving clock you must
compare two readings on the moving clock to
two reading on two separate (coordinate) clocks!
You have only got one reading from each of the TWO
moving clocks. You can conclude nothing about their
rate observed in K from that.
Thus in this experiment it is the moving clock that ticks faster.
Moreover the ratio of intervals is provided by
L_o gamma/dt' = L_o/dt
dt' = dt L_o gamma/L_o
or
dt' = dt gamma
Exactly.
And since dt is read off one clock while dt' is read off
two separate clocks, we can conclude:
The rate of the moving clock in 2 is dt/dt' = 1/gamma
as observed in K'.
OTOH we have also in the same sequence the additional two events which
are the correspondence of the left and right sides of 2 with the right
side of 1. Wrt K this time interval is exactly the same as the first,
since according to him the carriages are equal in length. However,
assuming reciprocity of length contraction the time interval wrt K' is
dt' = L_o/gamma v
But L_o = dtv, just as before
Thus
dt' = dt/gamma
Exactly.
Since dt' in this case is read of one clock while dt is read of two
separate clocks, we can conclude:
The rate of the moving clock in 1 is dt'/dt = 1/gamma
as observed in K'.
Which is opposite in sense to the ticking rate that was derived above.
No.
The ticking rates of the single, moving clock are in both cases
the same, namely 1/gamma.
Thus which clock ticks faster seems to depend entirely upon the motion
of some arbitrary extended object wrt each. If the object is at rest wrt
K then the moving clock is ticking slower, but if it is at rest wrt K'
then the moving clock is ticking slower. If there are two objects, one
at rest wrt each, then the clocks are simultaneously ticking slower and
faster than each other. Moreover, the ticking ratio only assumes the
simplified form above when the object(s) is/are moving at v wrt one of
the frames, i.e. given an object that is moving wrt both frames, the
ticking rate does not resolve to either of the above forms, but assumes
a different value for every alternate value of x/t, or x'/t' that is not
equal to v. IOW SR is gibberish.
If you are a sensible guy, you will have realized wher you went wrong.
Are you?
Paul
Oh, yeah, I see the error, I forgot to stick my head up my ***** like you
just did! Sorry.
The clocks are both set to zero upon the intersection of the left side
of 2 and the right side of 1, an objective event that occurs in both
frames. This is reading 1 that you requested.
The K' clock is situated at the right side of 1, and the K clock is
situated at the right side of 2. When these clocks' positions coincide,
both sets of events dealt with above will have simultaneously concluded
wrt K, and at that point the clocks will be adjacent, and thus can be
compared. This is the second reading that you requested. Why you ever
though there was only one reading I'll never know, I can only assume
that you felt the need to pull at least something out of your *****, right
or wrong. Is that it? LOL.
The predicted difference in the clock readings upon coincidence of the
clocks, according to the lorentz transform, depends entirely upon which
of the pair of events is considered, but these pairs of events are
concurrent wrt K, and exactly equal in duration, and thus the observer
in K seeks an objective difference in his clock reading wrt the other,
i.e. that he was promised by you. SR cannot decide what to tell him
though, i.e. whether the other clock has ticked more or whether it has
ticked less than his. QED.
You guys are morons, do you understand what a contradiction is? I'll
give you a clue, it looks something like
a =/= 0
b =/= 0
c =/= 1
a = bc
a = b/c
This contradiction 'should' look familiar, it was just derived before
your eyes from the lorentz fucking transform.
Richard Perry
http://www.cswnet.com/~rper
.
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| User: "Paul B. Andersen" |
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| Title: Re: Relative ticking rate nonsense |
23 Sep 2003 09:25:36 AM |
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"Richard" <no_mail_no_spam@yahoo.com> skrev i melding news:3F6F3A49.7549BA13@yahoo.com...
"Paul B. Andersen" wrote:
If you are a sensible guy, you will have realized where you went wrong.
Are you?
I forgot to stick my head up my ***** like you
just did!
[..]
I can only assume
that you felt the need to pull at least something out of your *****,
[..]
You guys are morons,
[..]
the lorentz fucking transform.
I guess this answers my question.
Paul
.
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| User: "Dirk Van de moortel" |
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| Title: Re: Relative ticking rate nonsense |
23 Sep 2003 11:17:21 AM |
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"Paul B. Andersen" <paul.b.andersen@hia.no> wrote in message news:bkpl5t$994$1@dolly.uninett.no...
"Richard" <no_mail_no_spam@yahoo.com> skrev i melding news:3F6F3A49.7549BA13@yahoo.com...
"Paul B. Andersen" wrote:
If you are a sensible guy, you will have realized where you went wrong.
Are you?
I forgot to stick my head up my ***** like you
just did!
[..]
I can only assume
that you felt the need to pull at least something out of your *****,
[..]
You guys are morons,
[..]
the lorentz fucking transform.
I guess this answers my question.
Paul
hm... I was referring to the other details for you to take
care off, but never mind, I'll wait for an answer to my
question to him...
Dirk Vdm
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| User: "Richard" |
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| Title: Re: Relative ticking rate nonsense |
23 Sep 2003 01:14:05 PM |
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Dirk Van de moortel wrote:
"Paul B. Andersen" <paul.b.andersen@hia.no> wrote in message news:bkpl5t$994$1@dolly.uninett.no...
"Richard" <no_mail_no_spam@yahoo.com> skrev i melding news:3F6F3A49.7549BA13@yahoo.com...
"Paul B. Andersen" wrote:
If you are a sensible guy, you will have realized where you went wrong.
Are you?
I forgot to stick my head up my ***** like you
just did!
[..]
I can only assume
that you felt the need to pull at least something out of your *****,
[..]
You guys are morons,
[..]
the lorentz fucking transform.
I guess this answers my question.
Paul
hm... I was referring to the other details for you to take
care off, but never mind, I'll wait for an answer to my
question to him...
Dirk Vdm
I'll get back to you on that when I get time Dirk. BTW, I meant the term
'morons' to be taken in a friendly sort of way, forgot the :),
In short, you aren't following the argument Dirk, more later:)
Richard Perry
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| User: "Paul B. Andersen" |
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| Title: Re: Relative ticking rate nonsense |
23 Sep 2003 03:21:30 PM |
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"Richard" <no_mail_no_spam@yahoo.com> skrev i melding news:3F708D6D.3930E4BB@yahoo.com...
Dirk Van de moortel wrote:
"Paul B. Andersen" <paul.b.andersen@hia.no> wrote in message news:bkpl5t$994$1@dolly.uninett.no...
"Richard" <no_mail_no_spam@yahoo.com> skrev i melding news:3F6F3A49.7549BA13@yahoo.com...
"Paul B. Andersen" wrote:
If you are a sensible guy, you will have realized where you went wrong.
Are you?
I forgot to stick my head up my ***** like you
just did!
[..]
I can only assume
that you felt the need to pull at least something out of your *****,
[..]
You guys are morons,
[..]
the lorentz fucking transform.
I guess this answers my question.
Paul
hm... I was referring to the other details for you to take
care off, but never mind, I'll wait for an answer to my
question to him...
Dirk Vdm
I'll get back to you on that when I get time Dirk. BTW, I meant the term
'morons' to be taken in a friendly sort of way, forgot the :),
In short, you aren't following the argument Dirk, more later:)
You seem to have forgotten a lot of smileys.
But an ***** with a smiley attached is still an *****.
I have addressed the issues raised by you in a posting a couple
of years ago. The scenario is for all practical purposes identical
to yours. Notice that I have specifically addressed the issues
you find contradictory.
Here it is again: (best read with a fixed width font)
The "mutual time dilation" is no problem if you
realize what you are comparing.
Let's have two synchronized clocks in each of two frames
of reference, let the clocks be a proper distance d from each other
in their respective frames, let the frames move with a relative
speed v.
-d 0 x'
K' frame: B'------A'--> -> v
K frame: A ------B--> 0 d x
There are three events of interest:
E1: A and A' adjacent
E2: A and B' adjacent
E3: B and A' adjacent
Let's calculate what the clocks will show at these events:
E1: A = t1 = 0, A' = t1' = 0 (by fiat, we set the clocks thus)
E2:
In the K' frame, A will be at the position -d at t2' = d/v
LT: t2 = (d/v + (-d)*v/c^2)/sqrt(1 - v^2/c^2)
t2 = (d/v)*sqrt(1 - v^2/c^2)
E3:
In the K frame, A' will be at the position d at t3 = d/v
LT: t3' = (d/v - d*v/c^2)/sqrt(1 - v^2/c^2)
t3' = (d/v)*sqrt(1 - v^2/c^2)
Summing up, the readings of the clocks will be:
E1: A = t1 = 0, A'= t1'= 0
E2: A = t2 =(d/v)*sqrt(1-v^2/c^2),B'= t2'= d/v
E3: B = t3 = d/v, A'= t3'=(d/v)*sqrt(1-v^2/c^2)
The symmetry is obvious.
So which clock is running slow or fast relative to which?
The answer depend on how we compare the clocks!
===============================================
In the K-frame, we can measure the rate dt'/dt of
the moving A' clock by comparing the reading of A'
with the _two_ clocks A and B as it passes them:
dt'/dt = (t3' - t1')/(t3 - t1) = sqrt(1 - v^2/c^2)
Conclusion #1:
A' runs slow as measured in the K frame.
=======================================
In the K'-frame, we can measure the rate dt/dt' of
the moving A clock by comparing the reading of A
with the _two_ clocks A' and B' as it passes them:
dt/dt' = (t2 - t1)/(t2' - t1') = sqrt(1 - v^2/c^2)
Conclusion #2;
A runs slow as measured in the K' frame.
=======================================
Conclusion #1 does not contradict conclusion #2.
They state in fact two different things.
But we can draw more conclusions:
We can measure the rate R' at which an observer in K'
will see the co-ordinate time of K run by reading the
clocks A and B as they passes the A' clock:
R' = (t3 - t1)/(t3' - t1') = 1/sqrt(1 - v^2/c^2)
Conclusion #3:
The co-ordinate time of K runs _fast_ as measured in the K'-frame
=================================================================
We can measure the rate R at which an observer in K
will see the co-ordinate time of K' run by reading the
clocks A' and B' as they passes the A clock:
R = (t2' - t1')/(t2 - t1) = 1/sqrt(1 - v^2/c^2)
Conclusion #4:
The co-ordinate time of K' runs _fast_ as measured in the K-frame
=================================================================
There is nothing contradictory between conclusion #3 and #4 either,
as they too state different things.
It is in fact conclusions #1 and #3 and conclusions #2 and #4
respectively that state the same facts.
Paul
.
|
|
|
| User: "Richard" |
|
| Title: Re: Relative ticking rate nonsense |
24 Sep 2003 07:50:30 AM |
|
|
"Paul B. Andersen" wrote:
"Richard" <no_mail_no_spam@yahoo.com> skrev i melding news:3F708D6D.3930E4BB@yahoo.com...
Dirk Van de moortel wrote:
"Paul B. Andersen" <paul.b.andersen@hia.no> wrote in message news:bkpl5t$994$1@dolly.uninett.no...
"Richard" <no_mail_no_spam@yahoo.com> skrev i melding news:3F6F3A49.7549BA13@yahoo.com...
"Paul B. Andersen" wrote:
If you are a sensible guy, you will have realized where you went wrong.
Are you?
I forgot to stick my head up my ***** like you
just did!
[..]
I can only assume
that you felt the need to pull at least something out of your *****,
[..]
You guys are morons,
[..]
the lorentz fucking transform.
I guess this answers my question.
Paul
hm... I was referring to the other details for you to take
care off, but never mind, I'll wait for an answer to my
question to him...
Dirk Vdm
I'll get back to you on that when I get time Dirk. BTW, I meant the term
'morons' to be taken in a friendly sort of way, forgot the :),
In short, you aren't following the argument Dirk, more later:)
You seem to have forgotten a lot of smileys.
But an ***** with a smiley attached is still an *****.
I have addressed the issues raised by you in a posting a couple
of years ago. The scenario is for all practical purposes identical
to yours. Notice that I have specifically addressed the issues
you find contradictory.
Here it is again: (best read with a fixed width font)
The "mutual time dilation" is no problem if you
realize what you are comparing.
Let's have two synchronized clocks in each of two frames
of reference, let the clocks be a proper distance d from each other
in their respective frames, let the frames move with a relative
speed v.
-d 0 x'
K' frame: B'------A'--> -> v
K frame: A ------B--> 0 d x
There are three events of interest:
E1: A and A' adjacent
E2: A and B' adjacent
E3: B and A' adjacent
Let's calculate what the clocks will show at these events:
E1: A = t1 = 0, A' = t1' = 0 (by fiat, we set the clocks thus)
E2:
In the K' frame, A will be at the position -d at t2' = d/v
LT: t2 = (d/v + (-d)*v/c^2)/sqrt(1 - v^2/c^2)
t2 = (d/v)*sqrt(1 - v^2/c^2)
E3:
In the K frame, A' will be at the position d at t3 = d/v
LT: t3' = (d/v - d*v/c^2)/sqrt(1 - v^2/c^2)
t3' = (d/v)*sqrt(1 - v^2/c^2)
Summing up, the readings of the clocks will be:
E1: A = t1 = 0, A'= t1'= 0
E2: A = t2 =(d/v)*sqrt(1-v^2/c^2),B'= t2'= d/v
E3: B = t3 = d/v, A'= t3'=(d/v)*sqrt(1-v^2/c^2)
The symmetry is obvious.
So which clock is running slow or fast relative to which?
The answer depend on how we compare the clocks!
===============================================
In the K-frame, we can measure the rate dt'/dt of
the moving A' clock by comparing the reading of A'
with the _two_ clocks A and B as it passes them:
dt'/dt = (t3' - t1')/(t3 - t1) = sqrt(1 - v^2/c^2)
Conclusion #1:
A' runs slow as measured in the K frame.
=======================================
In the K'-frame, we can measure the rate dt/dt' of
the moving A clock by comparing the reading of A
with the _two_ clocks A' and B' as it passes them:
dt/dt' = (t2 - t1)/(t2' - t1') = sqrt(1 - v^2/c^2)
Conclusion #2;
A runs slow as measured in the K' frame.
=======================================
Conclusion #1 does not contradict conclusion #2.
They state in fact two different things.
But we can draw more conclusions:
We can measure the rate R' at which an observer in K'
will see the co-ordinate time of K run by reading the
clocks A and B as they passes the A' clock:
R' = (t3 - t1)/(t3' - t1') = 1/sqrt(1 - v^2/c^2)
Conclusion #3:
The co-ordinate time of K runs _fast_ as measured in the K'-frame
=================================================================
We can measure the rate R at which an observer in K
will see the co-ordinate time of K' run by reading the
clocks A' and B' as they passes the A clock:
R = (t2' - t1')/(t2 - t1) = 1/sqrt(1 - v^2/c^2)
Conclusion #4:
The co-ordinate time of K' runs _fast_ as measured in the K-frame
=================================================================
There is nothing contradictory between conclusion #3 and #4 either,
as they too state different things.
It is in fact conclusions #1 and #3 and conclusions #2 and #4
respectively that state the same facts.
Paul
Wrong.
In order to eliminate the confusion between interval ratios, and ticking
ratios, we'll let the events be the very ticks of the clocks.
Two identical laser clocks are oriented so that the photons are
propagating perpendicular to the x axis, and thus along the y axis. Both
clocks have counters attached that literally ticks off numbers as the
photon reflects of off the bottom end of each clock.
One clock (K') is set into motion at v wrt the other (K), and towards
it.
The clock on the right, the rest clock (frame K) has a fixed limb
extending toward the left clock which is parallel to the x axis, and its
length is L_o wrt K, and thus L_o/gamma wrt K'.
The K' clock's counter is switched on as it passes the end of the
extended limb.
According to K' and to K, the number of ticks on the K' counter will be
the same wrt both frames, as you also derived above, however neither of
these observers will agree upon the number of ticks counted on the K
counter. K predicts that his counter will be the one with the higher
reading when the clocks intersect, but K' predicts that his counter
will have the higher reading.
Why? Well from the K frame his prediction is obvious, i.e. the K' clock
is ticking slower.
But from the K' frame it is the K clock that is ticking slower.
Noting that both frames predict the same number of K' ticks is just a
diversion away from the fact that both frames predict a different number
of K ticks.
Richard Perry
.
|
|
|
| User: "Paul B. Andersen" |
|
| Title: Re: Relative ticking rate nonsense |
24 Sep 2003 09:00:50 AM |
|
|
"Richard" <no_mail_no_spam@yahoo.com> skrev i melding news:3F719316.37F69FAB@yahoo.com...
"Paul B. Andersen" wrote:
I have addressed the issues raised by you in a posting a couple
of years ago. The scenario is for all practical purposes identical
to yours. Notice that I have specifically addressed the issues
you find contradictory.
Here it is again: (best read with a fixed width font)
The "mutual time dilation" is no problem if you
realize what you are comparing.
Let's have two synchronized clocks in each of two frames
of reference, let the clocks be a proper distance d from each other
in their respective frames, let the frames move with a relative
speed v.
-d 0 x'
K' frame: B'------A'--> -> v
K frame: A ------B--> 0 d x
There are three events of interest:
E1: A and A' adjacent
E2: A and B' adjacent
E3: B and A' adjacent
Let's calculate what the clocks will show at these events:
E1: A = t1 = 0, A' = t1' = 0 (by fiat, we set the clocks thus)
E2:
In the K' frame, A will be at the position -d at t2' = d/v
LT: t2 = (d/v + (-d)*v/c^2)/sqrt(1 - v^2/c^2)
t2 = (d/v)*sqrt(1 - v^2/c^2)
E3:
In the K frame, A' will be at the position d at t3 = d/v
LT: t3' = (d/v - d*v/c^2)/sqrt(1 - v^2/c^2)
t3' = (d/v)*sqrt(1 - v^2/c^2)
Summing up, the readings of the clocks will be:
E1: A = t1 = 0, A'= t1'= 0
E2: A = t2 =(d/v)*sqrt(1-v^2/c^2),B'= t2'= d/v
E3: B = t3 = d/v, A'= t3'=(d/v)*sqrt(1-v^2/c^2)
The symmetry is obvious.
So which clock is running slow or fast relative to which?
The answer depend on how we compare the clocks!
===============================================
In the K-frame, we can measure the rate dt'/dt of
the moving A' clock by comparing the reading of A'
with the _two_ clocks A and B as it passes them:
dt'/dt = (t3' - t1')/(t3 - t1) = sqrt(1 - v^2/c^2)
Conclusion #1:
A' runs slow as measured in the K frame.
=======================================
In the K'-frame, we can measure the rate dt/dt' of
the moving A clock by comparing the reading of A
with the _two_ clocks A' and B' as it passes them:
dt/dt' = (t2 - t1)/(t2' - t1') = sqrt(1 - v^2/c^2)
Conclusion #2;
A runs slow as measured in the K' frame.
=======================================
Conclusion #1 does not contradict conclusion #2.
They state in fact two different things.
But we can draw more conclusions:
We can measure the rate R' at which an observer in K'
will see the co-ordinate time of K run by reading the
clocks A and B as they passes the A' clock:
R' = (t3 - t1)/(t3' - t1') = 1/sqrt(1 - v^2/c^2)
Conclusion #3:
The co-ordinate time of K runs _fast_ as measured in the K'-frame
=================================================================
We can measure the rate R at which an observer in K
will see the co-ordinate time of K' run by reading the
clocks A' and B' as they passes the A clock:
R = (t2' - t1')/(t2 - t1) = 1/sqrt(1 - v^2/c^2)
Conclusion #4:
The co-ordinate time of K' runs _fast_ as measured in the K-frame
=================================================================
There is nothing contradictory between conclusion #3 and #4 either,
as they too state different things.
It is in fact conclusions #1 and #3 and conclusions #2 and #4
respectively that state the same facts.
Paul
Wrong.
But you are not able to point out what is wrong,
because you didn't actually read it.
If you had, you would have seen that it is identical to your
scenario, but described in an unambiguous manner,
and it is define how "tick rate" is measured.
In order to eliminate the confusion between interval ratios, and ticking
ratios,
... you are introducing yet another scenario where you demonstrate
your confusions about the very issues my posting was meant to clear up,
a scenario which I therefore
[snip]
I suggest you go back to my version of your original scenario,
and point out exactly what you think is wrong.
But you won't do that of course.
Because you won't find any errors to point out.
Paul
.
|
|
|
| User: "Richard" |
|
| Title: Re: Relative ticking rate nonsense |
24 Sep 2003 12:16:28 PM |
|
|
"Paul B. Andersen" wrote:
"Richard" <no_mail_no_spam@yahoo.com> skrev i melding news:3F719316.37F69FAB@yahoo.com...
"Paul B. Andersen" wrote:
I have addressed the issues raised by you in a posting a couple
of years ago. The scenario is for all practical purposes identical
to yours. Notice that I have specifically addressed the issues
you find contradictory.
Here it is again: (best read with a fixed width font)
The "mutual time dilation" is no problem if you
realize what you are comparing.
Let's have two synchronized clocks in each of two frames
of reference, let the clocks be a proper distance d from each other
in their respective frames, let the frames move with a relative
speed v.
-d 0 x'
K' frame: B'------A'--> -> v
K frame: A ------B--> 0 d x
There are three events of interest:
E1: A and A' adjacent
E2: A and B' adjacent
E3: B and A' adjacent
Let's calculate what the clocks will show at these events:
E1: A = t1 = 0, A' = t1' = 0 (by fiat, we set the clocks thus)
E2:
In the K' frame, A will be at the position -d at t2' = d/v
LT: t2 = (d/v + (-d)*v/c^2)/sqrt(1 - v^2/c^2)
t2 = (d/v)*sqrt(1 - v^2/c^2)
E3:
In the K frame, A' will be at the position d at t3 = d/v
LT: t3' = (d/v - d*v/c^2)/sqrt(1 - v^2/c^2)
t3' = (d/v)*sqrt(1 - v^2/c^2)
Summing up, the readings of the clocks will be:
E1: A = t1 = 0, A'= t1'= 0
E2: A = t2 =(d/v)*sqrt(1-v^2/c^2),B'= t2'= d/v
E3: B = t3 = d/v, A'= t3'=(d/v)*sqrt(1-v^2/c^2)
The symmetry is obvious.
So which clock is running slow or fast relative to which?
The answer depend on how we compare the clocks!
===============================================
In the K-frame, we can measure the rate dt'/dt of
the moving A' clock by comparing the reading of A'
with the _two_ clocks A and B as it passes them:
dt'/dt = (t3' - t1')/(t3 - t1) = sqrt(1 - v^2/c^2)
Conclusion #1:
A' runs slow as measured in the K frame.
=======================================
In the K'-frame, we can measure the rate dt/dt' of
the moving A clock by comparing the reading of A
with the _two_ clocks A' and B' as it passes them:
dt/dt' = (t2 - t1)/(t2' - t1') = sqrt(1 - v^2/c^2)
Conclusion #2;
A runs slow as measured in the K' frame.
=======================================
Conclusion #1 does not contradict conclusion #2.
They state in fact two different things.
But we can draw more conclusions:
We can measure the rate R' at which an observer in K'
will see the co-ordinate time of K run by reading the
clocks A and B as they passes the A' clock:
R' = (t3 - t1)/(t3' - t1') = 1/sqrt(1 - v^2/c^2)
Conclusion #3:
The co-ordinate time of K runs _fast_ as measured in the K'-frame
=================================================================
We can measure the rate R at which an observer in K
will see the co-ordinate time of K' run by reading the
clocks A' and B' as they passes the A clock:
R = (t2' - t1')/(t2 - t1) = 1/sqrt(1 - v^2/c^2)
Conclusion #4:
The co-ordinate time of K' runs _fast_ as measured in the K-frame
=================================================================
There is nothing contradictory between conclusion #3 and #4 either,
as they too state different things.
It is in fact conclusions #1 and #3 and conclusions #2 and #4
respectively that state the same facts.
Paul
Wrong.
But you are not able to point out what is wrong,
because you didn't actually read it.
Oh yes, I read it, you plainly produced the same contradiction, in that
each frame will expect the other's clock to have ticked less upon the
intersection of the clocks.
If you had, you would have seen that it is identical to your
scenario, but described in an unambiguous manner,
and it is define how "tick rate" is measured.
No it isn't identical, but it works every bit as well to provide the
contradiction.
In order to eliminate the confusion between interval ratios, and ticking
ratios,
.. you are introducing yet another scenario where you demonstrate
your confusions about the very issues my posting was meant to clear up,
a scenario which I therefore
[snip]
I suggest you go back to my version of your original scenario,
and point out exactly what you think is wrong.
What is wrong, is :
"The co-ordinate time of K runs _fast_ as measured in the K'-frame"
"The co-ordinate time of K' runs _fast_ as measured in the K-frame"
Reciprocity, and it is simply not achievable in a causal universe.
But you won't do that of course.
Because you won't find any errors to point out.
Paul
Your math was sound, it soundly led you directly into positing
contradictory statements.
Richard Perry
.
|
|
|
|
| User: "Richard" |
|
| Title: Re: Relative ticking rate nonsense |
24 Sep 2003 12:17:35 PM |
|
|
"Paul B. Andersen" wrote:
"Richard" <no_mail_no_spam@yahoo.com> skrev i melding news:3F719316.37F69FAB@yahoo.com...
"Paul B. Andersen" wrote:
I have addressed the issues raised by you in a posting a couple
of years ago. The scenario is for all practical purposes identical
to yours. Notice that I have specifically addressed the issues
you find contradictory.
Here it is again: (best read with a fixed width font)
The "mutual time dilation" is no problem if you
realize what you are comparing.
Let's have two synchronized clocks in each of two frames
of reference, let the clocks be a proper distance d from each other
in their respective frames, let the frames move with a relative
speed v.
-d 0 x'
K' frame: B'------A'--> -> v
K frame: A ------B--> 0 d x
There are three events of interest:
E1: A and A' adjacent
E2: A and B' adjacent
E3: B and A' adjacent
Let's calculate what the clocks will show at these events:
E1: A = t1 = 0, A' = t1' = 0 (by fiat, we set the clocks thus)
E2:
In the K' frame, A will be at the position -d at t2' = d/v
LT: t2 = (d/v + (-d)*v/c^2)/sqrt(1 - v^2/c^2)
t2 = (d/v)*sqrt(1 - v^2/c^2)
E3:
In the K frame, A' will be at the position d at t3 = d/v
LT: t3' = (d/v - d*v/c^2)/sqrt(1 - v^2/c^2)
t3' = (d/v)*sqrt(1 - v^2/c^2)
Summing up, the readings of the clocks will be:
E1: A = t1 = 0, A'= t1'= 0
E2: A = t2 =(d/v)*sqrt(1-v^2/c^2),B'= t2'= d/v
E3: B = t3 = d/v, A'= t3'=(d/v)*sqrt(1-v^2/c^2)
The symmetry is obvious.
So which clock is running slow or fast relative to which?
The answer depend on how we compare the clocks!
===============================================
In the K-frame, we can measure the rate dt'/dt of
the moving A' clock by comparing the reading of A'
with the _two_ clocks A and B as it passes them:
dt'/dt = (t3' - t1')/(t3 - t1) = sqrt(1 - v^2/c^2)
Conclusion #1:
A' runs slow as measured in the K frame.
=======================================
In the K'-frame, we can measure the rate dt/dt' of
the moving A clock by comparing the reading of A
with the _two_ clocks A' and B' as it passes them:
dt/dt' = (t2 - t1)/(t2' - t1') = sqrt(1 - v^2/c^2)
Conclusion #2;
A runs slow as measured in the K' frame.
=======================================
Conclusion #1 does not contradict conclusion #2.
They state in fact two different things.
But we can draw more conclusions:
We can measure the rate R' at which an observer in K'
will see the co-ordinate time of K run by reading the
clocks A and B as they passes the A' clock:
R' = (t3 - t1)/(t3' - t1') = 1/sqrt(1 - v^2/c^2)
Conclusion #3:
The co-ordinate time of K runs _fast_ as measured in the K'-frame
=================================================================
We can measure the rate R at which an observer in K
will see the co-ordinate time of K' run by reading the
clocks A' and B' as they passes the A clock:
R = (t2' - t1')/(t2 - t1) = 1/sqrt(1 - v^2/c^2)
Conclusion #4:
The co-ordinate time of K' runs _fast_ as measured in the K-frame
=================================================================
There is nothing contradictory between conclusion #3 and #4 either,
as they too state different things.
It is in fact conclusions #1 and #3 and conclusions #2 and #4
respectively that state the same facts.
Paul
Wrong.
But you are not able to point out what is wrong,
because you didn't actually read it.
Oh yes, I read it, you plainly produced the same contradiction, in that
each frame will expect the other's clock to have ticked less upon the
intersection of the clocks.
If you had, you would have seen that it is identical to your
scenario, but described in an unambiguous manner,
and it is define how "tick rate" is measured.
No it isn't identical, but it works every bit as well to provide the
contradiction.
In order to eliminate the confusion between interval ratios, and ticking
ratios,
.. you are introducing yet another scenario where you demonstrate
your confusions about the very issues my posting was meant to clear up,
a scenario which I therefore
[snip]
I suggest you go back to my version of your original scenario,
and point out exactly what you think is wrong.
What is wrong, is :
"The co-ordinate time of K runs _fast_ as measured in the K'-frame"
"The co-ordinate time of K' runs _fast_ as measured in the K-frame"
Reciprocity, and it is simply not achievable in a causal universe.
But you won't do that of course.
Because you won't find any errors to point out.
Paul
Your math was sound, it soundly led you directly into positing
contradictory statements.
Richard Perry
.
|
|
|
| User: "Paul B. Andersen" |
|
| Title: Re: Relative ticking rate nonsense |
24 Sep 2003 04:37:05 PM |
|
|
"Richard" <no_mail_no_spam@yahoo.com> skrev i melding news:3F71D1AF.6994F368@yahoo.com...
"Paul B. Andersen" wrote:
"Richard" <no_mail_no_spam@yahoo.com> skrev i melding news:3F719316.37F69FAB@yahoo.com...
"Paul B. Andersen" wrote:
I have addressed the issues raised by you in a posting a couple
of years ago. The scenario is for all practical purposes identical
to yours. Notice that I have specifically addressed the issues
you find contradictory.
Here it is again: (best read with a fixed width font)
The "mutual time dilation" is no problem if you
realize what you are comparing.
Let's have two synchronized clocks in each of two frames
of reference, let the clocks be a proper distance d from each other
in their respective frames, let the frames move with a relative
speed v.
-d 0 x'
K' frame: B'------A'--> -> v
K frame: A ------B--> 0 d x
There are three events of interest:
E1: A and A' adjacent
E2: A and B' adjacent
E3: B and A' adjacent
Let's calculate what the clocks will show at these events:
E1: A = t1 = 0, A' = t1' = 0 (by fiat, we set the clocks thus)
E2:
In the K' frame, A will be at the position -d at t2' = d/v
LT: t2 = (d/v + (-d)*v/c^2)/sqrt(1 - v^2/c^2)
t2 = (d/v)*sqrt(1 - v^2/c^2)
E3:
In the K frame, A' will be at the position d at t3 = d/v
LT: t3' = (d/v - d*v/c^2)/sqrt(1 - v^2/c^2)
t3' = (d/v)*sqrt(1 - v^2/c^2)
Summing up, the readings of the clocks will be:
E1: A = t1 = 0, A'= t1'= 0
E2: A = t2 =(d/v)*sqrt(1-v^2/c^2),B'= t2'= d/v
E3: B = t3 = d/v, A'= t3'=(d/v)*sqrt(1-v^2/c^2)
The symmetry is obvious.
So which clock is running slow or fast relative to which?
The answer depend on how we compare the clocks!
===============================================
In the K-frame, we can measure the rate dt'/dt of
the moving A' clock by comparing the reading of A'
with the _two_ clocks A and B as it passes them:
dt'/dt = (t3' - t1')/(t3 - t1) = sqrt(1 - v^2/c^2)
Conclusion #1:
A' runs slow as measured in the K frame.
=======================================
In the K'-frame, we can measure the rate dt/dt' of
the moving A clock by comparing the reading of A
with the _two_ clocks A' and B' as it passes them:
dt/dt' = (t2 - t1)/(t2' - t1') = sqrt(1 - v^2/c^2)
Conclusion #2;
A runs slow as measured in the K' frame.
=======================================
Conclusion #1 does not contradict conclusion #2.
They state in fact two different things.
But we can draw more conclusions:
We can measure the rate R' at which an observer in K'
will see the co-ordinate time of K run by reading the
clocks A and B as they passes the A' clock:
R' = (t3 - t1)/(t3' - t1') = 1/sqrt(1 - v^2/c^2)
Conclusion #3:
The co-ordinate time of K runs _fast_ as measured in the K'-frame
=================================================================
We can measure the rate R at which an observer in K
will see the co-ordinate time of K' run by reading the
clocks A' and B' as they passes the A clock:
R = (t2' - t1')/(t2 - t1) = 1/sqrt(1 - v^2/c^2)
Conclusion #4:
The co-ordinate time of K' runs _fast_ as measured in the K-frame
=================================================================
There is nothing contradictory between conclusion #3 and #4 either,
as they too state different things.
It is in fact conclusions #1 and #3 and conclusions #2 and #4
respectively that state the same facts.
Paul
[..]
I suggest you go back to my version of your original scenario,
and point out exactly what you think is wrong.
I see you have accepted that the clocks will show what
I said they will show at the different events.
What is wrong, is :
"The co-ordinate time of K runs _fast_ as measured in the K'-frame"
This is a simple observation of the clocks.
Stationary, synchronized clocks ARE co-ordinate clocks.
They show the co-ordinate time at their location in the frame
in which they are stationary, in this case in the K frame.
When an observer stationary in the K' frame observe the readings
of the co-ordinate clocks as they pass by him, he WILL observe
that the co-ordinate time in the passing K frame are running faster
than his own clock.
"The co-ordinate time of K' runs _fast_ as measured in the K-frame"
Same as above.
If you accept that the clocks are reading what I said,
then this conclusion follows from a simple observation
of said clocks.
Reciprocity, and it is simply not achievable in a causal universe.
Notice that the two conclusions above are based on two
different set of clock readings, that is two different set of events.
They are NOT stating two reciprocal relationships between
the same set of events.
That's why they are NOT contrary to each other.
I specifically commented this fact in my posting.
But you won't do that of course.
Because you won't find any errors to point out.
Paul
Your math was sound, it soundly led you directly into positing
contradictory statements.
If you accept that there is nothing contradictory about that
the clocks shows what I said they show, then my four conclusions
are inescapable and cannot be contradictory as they are
simple observations of what the clock are showing.
If you want to prove the scenario and my conclusions
contradictory, you have to prove that the readings
of the clocks at the different events cannot be as I said
they was.
Paul
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| User: "Dirk Van de moortel" |
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| Title: Re: Relative ticking rate nonsense |
24 Sep 2003 12:40:11 PM |
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"Richard" <no_mail_no_spam@yahoo.com> wrote in message news:3F71D1AF.6994F368@yahoo.com...
What is wrong, is :
"The co-ordinate time of K runs _fast_ as measured in the K'-frame"
"The co-ordinate time of K' runs _fast_ as measured in the K-frame"
Reciprocity, and it is simply not achievable in a causal universe.
That's a problem many people don't understand.
And it's so simple: as soon as the clocks get and
stick together again, or even stop moving wrt each
other, they can be more meaningfully compared.
It's all caused by how you define measurements.
When the clocks get together again the measurements
become standard again, so everyone understands.
It can be compared with the situation where we (you
and me) are looking at each other from a distance
between our fingers, and we both see the other one
as being shorter than ourself... and calling this activity
a measurement. When we get together, we can do it
the proper way and get a more meaningful result.
Nothing mysterious about it.
But again, many people seem to have a problem with
it. Nothing to be ashamed about really :-)
Dirk Vdm
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| User: "Richard" |
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| Title: Re: Relative ticking rate nonsense |
25 Sep 2003 07:52:45 AM |
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Dirk Van de moortel wrote:
"Richard" <no_mail_no_spam@yahoo.com> wrote in message news:3F71D1AF.6994F368@yahoo.com...
What is wrong, is :
"The co-ordinate time of K runs _fast_ as measured in the K'-frame"
"The co-ordinate time of K' runs _fast_ as measured in the K-frame"
Reciprocity, and it is simply not achievable in a causal universe.
That's a problem many people don't understand.
And it's so simple: as soon as the clocks get and
stick together again, or even stop moving wrt each
other, they can be more meaningfully compared.
It's all caused by how you define measurements.
When the clocks get together again the measurements
become standard again, so everyone understands.
It can be compared with the situation where we (you
and me) are looking at each other from a distance
between our fingers, and we both see the other one
as being shorter than ourself... and calling this activity
a measurement. When we get together, we can do it
the proper way and get a more meaningful result.
Nothing mysterious about it.
But again, many people seem to have a problem with
it. Nothing to be ashamed about really :-)
Dirk Vdm
So you're saying that SR is the result of measurement error?
And you still haven't provided for the Hafele-Keating results, since
theoretically wrt some frame in motion wrt Earth that a jet spent the
majority of its time in, the Earth is the accelerated frame. Without the
acceleration histories of the frames you cannot dispute this. OTOH, if
you are correct then the Earth has never accelerated, hmmm. OTOH you are
not right, you just saw
dt = dt'/gamma
dt = dt' gamma
derived. If neither of these can be validly used to compute the relative
'ticking rate' of two clocks, then it must follow that the ticking rate
cannot be computed by any means whatsoever. What do these equations mean
to you?
There is no reference to lengths, x intervals, etc. in these equations,
since these have all canceled out. What is left is just the relative
ticking rates, period, and you see we have 'two'. This cannot possibly
correspond to objective reality. Sorry.
Richard Perry
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| User: "Dirk Van de moortel" |
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| Title: Re: Relative ticking rate nonsense |
25 Sep 2003 08:17:38 AM |
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"Richard" <no_mail_no_spam@yahoo.com> wrote in message news:3F72E51D.23A079A5@yahoo.com...
Dirk Van de moortel wrote:
"Richard" <no_mail_no_spam@yahoo.com> wrote in message news:3F71D1AF.6994F368@yahoo.com...
What is wrong, is :
"The co-ordinate time of K runs _fast_ as measured in the K'-frame"
"The co-ordinate time of K' runs _fast_ as measured in the K-frame"
Reciprocity, and it is simply not achievable in a causal universe.
That's a problem many people don't understand.
And it's so simple: as soon as the clocks get and
stick together again, or even stop moving wrt each
other, they can be more meaningfully compared.
It's all caused by how you define measurements.
When the clocks get together again the measurements
become standard again, so everyone understands.
It can be compared with the situation where we (you
and me) are looking at each other from a distance
between our fingers, and we both see the other one
as being shorter than ourself... and calling this activity
a measurement. When we get together, we can do it
the proper way and get a more meaningful result.
Nothing mysterious about it.
But again, many people seem to have a problem with
it. Nothing to be ashamed about really :-)
Dirk Vdm
So you're saying that SR is the result of measurement error?
no.
And you still haven't provided for the Hafele-Keating results, since
theoretically wrt some frame in motion wrt Earth that a jet spent the
majority of its time in, the Earth is the accelerated frame. Without the
acceleration histories of the frames you cannot dispute this. OTOH, if
you are correct then the Earth has never accelerated, hmmm. OTOH you are
not right, you just saw
dt = dt'/gamma
dt = dt' gamma
derived.
O Richard, how many times have we explained?
The first equation
dt = dt' / gamma
is only valid for two events that satisfy dx = 0.
Check it with the LT. Simple algebra.
The second equation
dt = dt' gamma
is only valid for two events that satisfy dx' = 0.
Check it with the LT. Simple algebra.
So, the equations are only valid *together* for
events that satisfy dx = dx' = 0, and thus also
dt = dt' = 0.
Check it with the LT. Simple algebra.
So unless gamma = 1 (trivial case with v=0), the two
equations
dt = dt'/gamma
dt = dt' gamma
when forced to be valid together, say nothing more than
0 = 0 / gamma
0 = 0 * gamma
It's not *that* difficult, is it?
Now go to:
http://groups.google.com/groups?&threadm=0fHbb.34177$0C.1626157@phobos.telenet-ops.be
and try to answer the questions.
Androcles is having the struggle of his life with this
(look all over the place). Surely *you* must be smarter
than Androcles, no?
Dirk Vdm
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| User: "Richard" |
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| Title: Re: Relative ticking rate nonsense |
25 Sep 2003 02:50:55 PM |
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Dirk Van de moortel wrote:
The first equation
dt = dt' / gamma
is only valid for two events that satisfy dx = 0.
Check it with the LT. Simple algebra.
The second equation
dt = dt' gamma
is only valid for two events that satisfy dx' = 0.
Check it with the LT. Simple algebra.
So, the equations are only valid *together* for
events that satisfy dx = dx' = 0, and thus also
dt = dt' = 0.
Check it with the LT. Simple algebra.
So unless gamma = 1 (trivial case with v=0), the two
equations
dt = dt'/gamma
dt = dt' gamma
when forced to be valid together, say nothing more than
0 = 0 / gamma
0 = 0 * gamma
It's not *that* difficult, is it?
Now go to:
http://groups.google.com/groups?&threadm=0fHbb.34177$0C.1626157@phobos.telenet-ops.be
and try to answer the questions.
Androcles is having the struggle of his life with this
(look all over the place). Surely *you* must be smarter
than Androcles, no?
Dirk Vdm
Yo DIRK!, Earth to Dirk, come in please.
Your conclusion is contrary to the premises, i.e. that dt and dt' are
'not' zero. You just don't get it do you? In 'none' of the experiments
did we begin with either dt = 0 or dt' = 0, and yet we derived the
simultaneous equations
dt'/dt = gamma
dt'/dt = 1/gamma
If you wish to begin with premises
dt' = 0
dt = 0
then I would agree that the conclusion agrees with the premises. OTOH,
we didn't begin with those premises, but rather have drawn them as
conclusions from the premises
dt=/=0
dt'=/=0
, conclusions that are contradictory to the premises. This is the
definition of a contradiction. Ieeee! Shoot me 'now', please, I can't
take it.;)
Richard Perry
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| User: "Dirk Van de moortel" |
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| Title: Re: Relative ticking rate nonsense |
25 Sep 2003 03:16:46 PM |
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"Richard" <no_mail_no_spam@yahoo.com> wrote in message news:3F73471F.591D3D30@yahoo.com...
Dirk Van de moortel wrote:
The first equation
dt = dt' / gamma
is only valid for two events that satisfy dx = 0.
Check it with the LT. Simple algebra.
The second equation
dt = dt' gamma
is only valid for two events that satisfy dx' = 0.
Check it with the LT. Simple algebra.
So, the equations are only valid *together* for
events that satisfy dx = dx' = 0, and thus also
dt = dt' = 0.
Check it with the LT. Simple algebra.
So unless gamma = 1 (trivial case with v=0), the two
equations
dt = dt'/gamma
dt = dt' gamma
when forced to be valid together, say nothing more than
0 = 0 / gamma
0 = 0 * gamma
It's not *that* difficult, is it?
Now go to:
http://groups.google.com/groups?&threadm=0fHbb.34177$0C.1626157@phobos.telenet-ops.be
and try to answer the questions.
Androcles is having the struggle of his life with this
(look all over the place). Surely *you* must be smarter
than Androcles, no?
Dirk Vdm
Yo DIRK!, Earth to Dirk, come in please.
Your conclusion is contrary to the premises, i.e. that dt and dt' are
'not' zero. You just don't get it do you? In 'none' of the experiments
did we begin with either dt = 0 or dt' = 0, and yet we derived the
simultaneous equations
dt'/dt = gamma
dt'/dt = 1/gamma
If you wish to begin with premises
dt' = 0
dt = 0
then I would agree that the conclusion agrees with the premises. OTOH,
we didn't begin with those premises, but rather have drawn them as
conclusions from the premises
dt=/=0
dt'=/=0
, conclusions that are contradictory to the premises. This is the
definition of a contradiction. Ieeee! Shoot me 'now', please, I can't
take it.;)
You can use the Lorentz transformation
dt' = g( dt - v/c^2 dx ) [1]
dx' = g( dx - v dt ) [2]
and the equivalent inverse:
dt = g( dt' + v/c^2 dx' ) [3]
dx = g( dx' + v dt' ) [4]
1) First give me any two events that satisfy
dt = dt' / g
without having dx = 0.
2) Then give me any two events that satisfy
dt = dt' * g
without having dx' = 0.
3) Finally give me any two events that satisfy together
dx = dx' = 0
without having
dt = dt = 0.
If you can do this, you have found a contradiction.
Dirk Vdm
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| User: "Jeff Relf" |
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| Title: It's the Physical Data That's Retarded . |
21 Sep 2003 05:20:41 PM |
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Hi Richard ,
Regarding a pair of non-accelerating particles
traveling near the speed of light , You say :
" If there are two objects
[ each at rest with respect to the other ]
then the clocks are simultaneously ticking slower
and faster than each other . "
The equations of relativity are all about the speed of light ,
and how long actual physical " Data " takes to propagate .
Because the speed of light is observed to be the
same in all inertial and free fall frames :
" Information " , i.e. a type of " Reality " if you will ,
is " Retarded " , held back by this limit .
Hypothetically ( In your mind ) you might know that
the two distant frames are " Simultaneous " ,
but the actual physical " Data "
always takes some Finite time to propagate .
Newtonian gravity's deficiency is that :
It assumes instantaneous transmission .
Check out Steve Carlip's Sci.Physics FAQ entry . ( below )
He says that the observed dampening of binary pulsar systems
shows that speed of gravity travels at the speed of light
... at least as far as the GR model is concerned .
( It's instant as far as the Newtonian model is concerned . )
And a GR-independent verification requires that
either LIGO et al. gain much more precision
or that a super nova happens real close .
A pulsar spins extremely fast ,
but the earth is spinning too .
Gravity Probe-B is designed to measure
how the earth's " wobbly spin " , i.e. it's precessions ,
effect the moon .
Correct me if I'm wrong ,
but I think Newtonian gravity ,
with it's instant speed of gravity ,
could not fully account for the moon's orbit .
....
From :
http://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html
" While current observations
do not yet provide a direct model-independent
measurement of the speed of gravity ,
a test within the framework of general relativity
can be made by observing the binary pulsar PSR 1913+16 .
The orbit of this binary system is gradually decaying ,
and this behavior is attributed to the loss of energy
due to escaping gravitational radiation .
But in any field theory ,
radiation is intimately related to
the finite velocity of field propagation ,
and the orbital changes due to gravitational radiation
can equivalently be viewed as damping
caused by the finite propagation speed .
( In the discussion above ,
this damping represents a failure of the ' retardation '
and ' noncentral , velocity-dependent ' effects
to completely cancel . )
The rate of this damping can be computed ,
and one finds that it
depends sensitively on the speed of gravity .
The fact that gravitational damping is measured at all
is a strong indication that
the propagation speed of gravity is not infinite .
If the calculational framework
of general relativity is accepted ,
the damping can be used to calculate the speed ,
and the actual measurement confirms that the speed of gravity
is equal to the speed of light to within 1% .
( Measurements of at least one other binary pulsar system ,
PSR B1534+12 , confirm this result ,
although so far with less precision . ) "
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| User: "Richard" |
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| Title: Re: It's the Physical Data That's Retarded . |
21 Sep 2003 07:56:10 PM |
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Jeff Relf wrote:
Hi Richard ,
Regarding a pair of non-accelerating particles
traveling near the speed of light , You say :
" If there are two objects
[ each at rest with respect to the other ]
then the clocks are simultaneously ticking slower
and faster than each other . "
The equations of relativity are all about the speed of light ,
and how long actual physical " Data " takes to propagate .
Because the speed of light is observed to be the
same in all inertial and free fall frames :
" Information " , i.e. a type of " Reality " if you will ,
is " Retarded " , held back by this limit .
Hypothetically ( In your mind ) you might know that
the two distant frames are " Simultaneous " ,
but the actual physical " Data "
always takes some Finite time to propagate .
Newtonian gravity's deficiency is that :
It assumes instantaneous transmission .
Check out Steve Carlip's Sci.Physics FAQ entry . ( below )
He says that the observed dampening of binary pulsar systems
shows that speed of gravity travels at the speed of light
... at least as far as the GR model is concerned .
( It's instant as far as the Newtonian model is concerned . )
And a GR-independent verification requires that
either LIGO et al. gain much more precision
or that a super nova happens real close .
A pulsar spins extremely fast ,
but the earth is spinning too .
Gravity Probe-B is designed to measure
how the earth's " wobbly spin " , i.e. it's precessions ,
effect the moon .
Correct me if I'm wrong ,
but I think Newtonian gravity ,
with it's instant speed of gravity ,
could not fully account for the moon's orbit .
...
From :
http://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html
" While current observations
do not yet provide a direct model-independent
measurement of the speed of gravity ,
a test within the framework of general relativity
can be made by observing the binary pulsar PSR 1913+16 .
The orbit of this binary system is gradually decaying ,
and this behavior is attributed to the loss of energy
due to escaping gravitational radiation .
But in any field theory ,
radiation is intimately related to
the finite velocity of field propagation ,
and the orbital changes due to gravitational radiation
can equivalently be viewed as damping
caused by the finite propagation speed .
( In the discussion above ,
this damping represents a failure of the ' retardation '
and ' noncentral , velocity-dependent ' effects
to completely cancel . )
The rate of this damping can be computed ,
and one finds that it
depends sensitively on the speed of gravity .
The fact that gravitational damping is measured at all
is a strong indication that
the propagation speed of gravity is not infinite .
If the calculational framework
of general relativity is accepted ,
the damping can be used to calculate the speed ,
and the actual measurement confirms that the speed of gravity
is equal to the speed of light to within 1% .
( Measurements of at least one other binary pulsar system ,
PSR B1534+12 , confirm this result ,
although so far with less precision . ) "
The Newtonian formula is for the ideal case of 'point' masses 'at rest'.
It is derivable from first principles, i.e. if on atom experiences a
force of x, then an identical atom in the same position in the field
will experience the same force. Thus the force on two atoms, is
collectively twice the force on one atom. If mass is unaffected by
position wrt other atoms, then the mass of two identical atoms is also
twice that of one atom, thus the force is proportional to mass. The
inverse square field is 'defined' and it is arbitrarily defined to be
so. A 'flat' field will work just as well, we need only have the
'gravitational' mass change with vicinity to other masses. If for
instance the force is constant with r, and the gravitational mass
decreases as r, then the force still drops as r^2. The inverse square
field of Newton was 'assumed' but it is as good an assumption as any
other, all of them can be built upon when any one is taken to be a
definition. Empirical variations to the 'general' law, can always be
attributed to extraneous fields. Learn the difference between reality
and theory.
If you want to account for deviations in an orbit introduced by
differences in volumes, differences in density gradients and density
distributions within the masses, unaccounted for fields extending
through the solar system, zero point field variations, locally generated
electromagnetic fields, and temperature differences between the masses
as well as temperature gradients within the masses, particle exchange
with the surroundings, radiation pressure, chirality, and tidal
deformations, etc. then none of those are in themselves contradictory to
Newton, and none to the special case formula that neglects all of these,
among the many nonlisted extraneous other influences on the orbit.
Newton's system is true by definition, it is a logical system and simply
cannot be falsified. There are perhaps portions of it that should be
more clearly defined in order to rule out misinterpretations by others.
If he ill-defined a term or two, then the solution isn't to generate a
mutually exclusive alternate system, but simply to make the corrections
to the existing system, which I'll wager weren't even originally there,
but were introduced by some confused individuals that came after.
Richard Perry
http://www.cswnet.com/~rper
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| User: "Jeff Relf" |
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| Title: Instantaneous ? |
21 Sep 2003 11:49:18 PM |
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Hi Richard ,
You say : " Newton's system is true by definition "
You think the speed of gravity is instantaneous ?
You think gravity is " Spooky action at a distance " ,
rather than a local propagation ?
Even though we can imagine instant propagation ,
Nothing is known to propagate faster than
the speed of light .
While the mind is finite and fantastical ,
Nature's complexity and breadth are infinite .
So while nature likely has absolute material determinism ,
scientific determinism , i.e. the axioms of science ,
will always be finite and artificial .
And with absolute material determinism :
Natural time is spatial ... Static ... A timescape .
And there may even be cause to define
a fifth spatial dimension , a static heatscape ,
to describe the universe's " Transition "
from the effectively infinite heat of the big bang
to the effectively infinite cold of the big freeze .
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| User: "Androcles" |
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| Title: Re: Instantaneous ? |
22 Sep 2003 07:05:49 PM |
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"Jeff Relf" <__.Jeff-Relf@NCPlus.NET> wrote in message
news:1fl43ewhumoio$.dlg@__.Jeff.Relf...
Hi Richard ,
You say : " Newton's system is true by definition "
You think the speed of gravity is instantaneous ?
You think gravity is " Spooky action at a distance " ,
rather than a local propagation ?
Even though we can imagine instant propagation ,
Nothing is known to propagate faster than
the speed of light .
While the mind is finite and fantastical ,
Nature's complexity and breadth are infinite .
So while nature likely has absolute material determinism ,
scientific determinism , i.e. the axioms of science ,
will always be finite and artificial .
And with absolute material determinism :
Natural time is spatial ... Static ... A timescape .
And there may even be cause to define
a fifth spatial dimension , a static heatscape ,
to describe the universe's " Transition "
from the effectively infinite heat of the big bang
to the effectively infinite cold of the big freeze .
Hmm...
looks mildly interesting, mind if I poke in? Well, I will anyway.
Do we think the speed of gravity is instantaneous, huh?
Tough question, I don't know. Let's have a thought experiment.
Out there, some distance away, let's make it 4 light years, the distance to
the nearest star system, E = mc^2 suddenly happens. A star blows itself to
smithereens. Not quite sure what 'smithereens' are, but anyway, it goes
supernova and converts itself entirely to energy. No mass left. Just lots of
photons flying away. Having no mass, it now has no gravity. So its original
gravity, if it were detectable, has now vanished. A step function, from some
to none. But the star is (was) a long way off. How long (if we could detect
it anyway) would it take for us to notice that it's gravitational field was
missing?
Do you think the speed of gravity is instantaneous? if you answer 'no', why?
If you answer 'no', why not? On what basis (or prejudice) would you base
your assumption? If you answer 'Don't know', then I'll be in full agreement
with you.
Androcles
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| User: "\formerly" |
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| Title: Re: Instantaneous ? |
22 Sep 2003 07:13:26 PM |
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Dear Androcles:
"Androcles" <jp006f9750@blurbblueyonder.co.uk> wrote in message
news:Z7Mbb.764$yn2.64@news-binary.blueyonder.co.uk...
....
Do you think the speed of gravity is instantaneous? if you answer 'no',
why?
If you answer 'no', why not? On what basis (or prejudice) would you base
your assumption? If you answer 'Don't know', then I'll be in full
agreement
with you.
I will answer "doesn't matter". Since the space created by any collection
of mass/energy "orbits" with it (Mach, Einsten), and we cannot make the
stuff truly disappear, such a concept is a nonsequitur.
David A. Smith
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