In sci.physics.relativity, Jeckyl
wrote
on Thu, 31 Jan 2008 09:28:55 +1100
:
"The Ghost In The Machine" wrote in message
...
In sci.physics.relativity, Jeckyl
wrote
on Wed, 30 Jan 2008 21:28:28 +1100
:
"Ockham" wrote in message
k...
"snapdragon31" wrote in message
...
On Jan 29, 8:54 pm, Randy Poe wrote:
On Jan 29, 8:14 pm, HW@....(Dr. Henri Wilson) wrote:
According to relativists, GPS clocks GAIN 38us per day on the ground
clock.
That is due to two components, 45us for gravity and -7us for relative
speed.
Accordingly, an observer (OO) in GPS orbit would see the GC LOSING
52us
per
day.
After one year, the OO would calculate that the OC was about 19ms
ahead
of the
GC.
However, the GO would calculate that his GC was only 13ms behind.
What happens when the clocks are reunited?
Who is right?
Two people drive different routes from city A to
city B. When they are reunited, one odometer reads
220 km and the other reads 230 km. Which one is
right?
- Randy
| According to relativity, both odometer readings are wrong. They do
| not represent the true distance of the routes travelled because of the
| length contraction effect.
| According to Newton's law, both odometer readings are right.
| The GPS clock paradox is a variation of the twin paradox, so no valid
| solution.
The paradox resides in the third postulate.
Androcles .. we've told you .. there is no third postulate
Yes there is; it's not usually expressed as a postulate, but
it is a simple one:
- If a TWLS be conducted between a source and a moving mirror,
then the time taken (as observed by the source) of the
light beam from source to mirror and back to source is
exactly twice that of the time taken from source to
mirror. In other words, t_AB = t_BA.
That follows from the second postulate, that says the speed of light is c ..
as the distances are the same ,the time must be the same. Why do you need
another postulate forthat?
An interesting subpoint, that; I am not certain now. However, it's
clear that three events in A's space occur, and A can only observe
two of them.
These are, of course;
(0,0)_A
(L,L/c)_A
(0,2L/c)_A
for some value L. A cannot observe the second one directly.
Indeed .. SR and Einstein agrees with that. Time from A to B for light
is
only the same as the time from B to A when A and B both at rest in some
frame of reference (ie they are not moving relative to each other)
B does not have to be at rest.
If B is not at rest, it does raise some issue in determining the distance AB
without some definition of simultaneity.
It's not *that* difficult, though you're right; one can
get tied into semantical knots if one's not careful.
But given that A is motionless and B moves v away from A,
one can simply calculate that L, the distance the beam
travels from A to B, is such that
(0,t0)_A = (L,t0+L/c)_A = (gL-gv(t0+L/c), gL0+gL/c-gvL/c^2)_B
using the Lorentz.
Since B's origin reflects the beam, L can be calculated as follows:
gL-gv(t0+L/c) = 0
L-vt0-vL/c = 0
L(1-v/c) = vt0
L = vt0/(1-v/c)
according to SR. The total time is of course simply 2L/c, as L
was defined in A's reference frame.
But, yes, if A is at rest (ie the
light is returning to the same point in space and so travlling the same
distance) then that is sufficient for saying the time of each leg of the
trip is the same. In Einstiens paper, A and B were described as fixed
points in space.
Of course the actual time
at which the ray of light impacts B (and the position
of B at the point of impact) might be a little hard to
specify unless one has an alternate "infinite speed"
particle, which is currently (and probably forever will
be) impossible.
--
#191,
Useless C++ Programming Idea #23291:
void f(item *p) { if(p != 0) delete p; }
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