Natural Science Forum / Physics / Research / November 2007

Thus spake Andrew Harland <leohar@rogers.com>

The fundamental postulate is the general principle of relativity.
According to this the local laws of physics are the same for all
observers. So one clock, as seen by an observer moving with it, must
behave in exactly the same way as a second, identical, clock as seen by
an observer moving with the second clock.

However, when these observers look at each others clocks, they find time
dilation effects, both the special relativistic effect from motion and
the gravitational effect. It is not right to say that motion has
effected the clocks, because motion of matter only makes sense relative
to other matter. It would be better to say that motion affects the way
in which a clock is seen by a moving observer, and that position affects
the way in which it is seen by a distant observer.

Regards

In a broad sense it depends on the clocks, eg, pendulum clocks behave
quite differently to atomic clocks under motion (try dropping a
pendulum clock - it won't work at all in freefall).  So, let's stick
to atomic clocks.

An atomic clock in space follows a freely falling trajectory, while an
atomic clock on Earth does not, and hence the difference in the two
elapsed times is nontrivial to take into account.

However, this trajectory dependence does not necessarily affect the
_local_ operation of such a clock.

For freely-falling clocks, one can always choose coordinates in a
small neighbourhood of a given spacetime position such that the clock
is motionless to first-order, and the metric tensor is just the
Minkonski tensor to the same order:
    g_uv =  diag[1, -1, -1, -1] .
Hence, the local physics of the operation of such a clock can be
described as if the clock was motionless in a flat spacetime - and so
doesn't depend on any details of its motion - one only needs special
relativity.  In this sense the Taylor-Wheeler statement is correct
(but would fail if the clock's operation depended on large scale
features outside this neighbourhood, eg, reflection of light pulses
from a mirror on the moon).

However, if the clock is NOT falling freely, such "locally-flat
comoving coordinates cannot be found in general - the clock will act
as if subjected to an inertial force. This underlies the central
concept of general relativity, that freely-falling particles are not
being acted on by a gravitational "force", they are merely moving
freely in a curved spacetime.

Note that even for freely falling clocks, one must keep varying the
above "small neighbourhood" to track the clock as it moves, and hence
keep varying the corresponding coordinate system as the clock moves.
Hence the total elapsed time is trajectory-dependent (the exception
being in a flat spacetime, where the coordinate system can be extended
over the whole spactime).  The time elapsed on any clock, as it
follows some trajectory, is given by
    elapsed time = integral dt sqrt{ g_uv dx^u/dt dx^v/t } ,
where x^u is the spacetime coordinate (t, x,y,z) relative to some
coordinate system (c=1), and g_uv is the metric tensor of the
spacetime for that coordinate system.

The figures you are quoting would apply if one would compare the
satellite clock directly with the receiver clock. However, this is not
how GPS works in practice. The receiver clock is in fact very much
irrelevant, as the position is obtained by observing the difference
between the time signals obtained from a number of different
satellites.

Consider for simplicity a one dimensional problem where the receiver
is located somewhere on the line connecting the two transmitters. In
this case the signal from transmitter 1 reaches the receiver at time

(1)     t1 = t0+ x1/c

and the signal from transmitter 2 reaches the receiver at time

(2)     t2 = t0+ x2/c ,

where t0 is the time the signal is being sent out (assuming both
transmitter clocks are synchronized), x1 is the distance of the
receiver from transmitter 1, x2 the distance of the receiver from
transmitte 2, and c the speed of light.
Now if one subtracts Eqs.(1) and (2) one gets

(3)     x1-x2 = c*(t1-t2) .

One knows therefore the position of the receiver just by comparing the
time signals from the two transmitters (if t1=t2, the receiver would
be exactly in the middle between the two transmitters). The receiver
clock is thus completely irrelevant for determining the position.

If one assumes now that the transmitter clocks are running fast or
slow by a relative factor (1+r), one has instead:

(4)     x1-x2 = c*[(1+r)*t1 -(1+r)*t2] = c*(1+r)*(t1-t2)

which means that the position will simply be wrong by a relative
factor r,  but there is obviously no accumulation as the transmitter
clocks run at the same rate relatively to each other.

Now the usually quoted relativistic overall correction of 38
microseconds/day corresponds to r=4.4*10^-10. As the satellites are at
a distance of around 20000 km (=2*10^9 cm), the positional error due
to relativity should actually only be 4.4*10^-10 *2*10^9 cm = 0.8 cm.
This is  much less than the presently claimed accuracy of the GPS of a
few meters, so relativistic effects should actually not be relevant at
all (which would be anyway largely due to General Relativity, i.e. not
be associated with the motion).

Thomas

I don't have a copy of the Taylor/Wheeler book to check the context of
this quote, but it is clear from everything I've read that relative
motion between the observer and a clock, whatever type or mechanism,
certainly does affect the observed rate.  Could the quote merely mean
that the operation of a clock in the same frame as the observer (i.e.
not moving wrt the observer)?  Are they perhaps making the point that
even though a moving clock is observed to be going slowly, that to an
observer moving with the clock it would appear to be unaffected?  Or
perhaps they are making the point that there is no preferred rest
frame in the universe and that the physics (and hence the operation of
clocks) is the same even for a clock moving at very high speeds?

Rich L.

The essence of Taylor and Wheeler's argument, as explained in the
two preceding paragraps, is that the hardware (and thus the structure
and operation) of a clock in motion cannot be affected by that motion
because different observers measure different effects. When you
change your state of motion with respect to the clock, you get a
different result. What is really -operationaly- affected, is time interval
measured with your clock between two ticks of the other clock.
Nothing more, nothing less.

Dirk Vdm

<http://relativity.livingreviews.org/open?pubNo=lrr-2003-1&page=node5.html>
Relativistic effects on orbital clocks

The clock that traverses the most space records the least time
elapsed.

A synchronous motor clock with a very long cord is not
affected by motion wrt the generating plant that drives it.

Sue...