Merlyn wrote:
No, The principle that ring-laser gyro's work under is
that the signals are no longer in ophase when they
arrive. This can be accomplished either by allowing
them to arrive at different times, or by allowing them
to experience a different passage of time during their
travel.
If two wave fronts aren't in phase when the arrive back at the starting
point, then the crests arrived at different times. Draw a picture
showing the crests in the waves, if that helps; if they're in phase the
crests arrive at the same time. If the two fronts take the same time to
get there then they are in phase.
Wave crests don't "age" so the amount of time they experience doesn't
affect the result.
For problems of this sort a light ray can be viewed as a series of
crests, and the events are the arrival of the crests. You can model it
as a string of beads.
For Hafele's clocks to show a difference in time
passing, after they are brought back together, it does
not actually matter whether the planes land at
different times or even traveled at different speeds,
because neither of which would affect the clocks.
The solution probably has to do with the difference in
gravity experienced by the planes, as the one
traveling with the earth's rotation experiences a
decrease in gravity (due to an increased centrepetal
force) while the other plane experiences the opposite.
Acceleration doesn't affect clocks. That's been verified (can't cite
references, sorry). A clock in a centrifuge slows only as a result of
the speed at which it's traveling, not as a result of the centripetal force.
Clock rates in a gravitational field are affected by the gravitational
potential, not the local gravitational field strength.
Horace Heffner wrote:
>
> I have to wonder what the data actually looks like.
Me, too. In the absence of seeing it, I'm going to go out on a limb
and make a rough calculation of what the data _should_ look like,
according to SR. This is also a demonstration of how simple the
problem really must be in SR, assuming, of course, that SR is correct
and consistent. (If it's consistent then we can pick any point of
view we choose, in order to simplify the problem... if it's not then
we can't do that!)
The Earth is about 25000 miles in circumference, and the equator is
moving at about 1000 miles per hour. I'll take that as a starting
point; this is a _rough_ estimate I'm working out here. (The
assumptions are valid -- it's the arithmetic where I'm cutting
corners.)
Peel off a strip of Earth at the equator and straighten it out, but
don't stop it from moving -- let it go shooting off into space. Now
we have a straight rod, 25,000 miles long, traveling at 1,000 miles
per hour. From the POV of a "stationary" observer, let's work out the
"sync error" between clocks at the two ends.
First, we need to fix the units. Convert everything to light-seconds,
seconds, and fractions of C:
v = 1000 mph = 0.3 miles/sec = 1.6 * 10^-6 C
x = 25,000 miles / 186,000 miles/sec = 1.34 * 10^-1 light-seconds
gamma = 1/sqrt(1-v^2) ~ 1 + (1/2)v^2 = 1 + 1.3*10^-12
or, for our purposes, gamma == 1 -- our accuracy is a lot less than
one part in 10^12 here.
Finally, at time 0 in the "stationary" frame, if time is 0 at one end
of the moving "rod", then time at the other end must be
t_e = gamma*(t - v*x) = -v * x
= 1.6 * 10^-6 * 1.34 * 10^-1
= 2.1 * 10^-7
That's the clock synchronization error going around the world _one_
_way_. If two planes circle the Earth in opposite directions they'll
_each_ encounter that error, in opposite directions. So the total
error will be double that, and we have
Sagnac gap time = 2 * t_e = 0.4 microseconds
The "gap time" is independent of the speed of the aircraft.
****************************************************************
Now if we want to know how far off _ONE_ airplane's clock will be
relative to a ground-based observer upon return from a trip around the
world, we _do_ need to worry about how fast the plane is traveling.
There are a couple of simple ways to do it. Looking at it from a
stationary observation point at Earth's center is straightforward and
provides a check on the computation above.
There's one slightly tricky point, which is we need to worry about the
velocity of the traveler as measured by the central observer, and the
velocity relative to the Earth's surface:
V_earth = 1000 mph = 1.6 * 10^-6 C
V_nominal = velocity of traveler relative to surface
V_actual = V_nominal + V_earth
Time lost by a traveler going around the Earth, from the POV of the
"man in the middle", is just that traveler's gamma, minus 1, times the
time the trip takes:
offset = (gamma[plane] - 1) * (x/V_nominal)
For low speeds (all we care about here!) we have
gamma ~ 1 + (1/2)V_actual^2
so that reduces to
offset = (1/2)V_actual^2 * x / (V_actual - V_earth)
For 1000 MPH flights, this works out as:
1000 MPH westbound => V_actual=0
=> offset = 0
1000 MPH eastbound => V_actual=2000 MPH
=> offset = 0.429 uSec
Earth surface, during 1000 MPH flight:
V_actual=1000 mph, x/v is as above, and offset = 0.107 uSec
Eastbound plane clock is off by 0.429 - 0.107 = 0.322 uSec
Westbond plane clock is off by 0 - 0.107 = -0.107 uSec
Difference in planes' clocks is 0.429 - 0 = 0.429 uSec
=> gap time = 0.429 uSec
For 500 MPH flights, this works out as:
500 MPH westbound
=> offset = 0.0536 uSec
500 MPH eastbound
=> offset = 0.482 uSec
Stationary on ground => V_actual = 1000 mph, V_nom = 500 mph
=> offset = 0.214 uSec
And so we have:
Eastbound plane clock is off by 0.482 - 0.214 = 0.268 uSec
Westbound plane clock is off by 0.0536 - 0.214 = -0.160 uSec
Difference in plane's clocks is 0.482 - 0.0536 = 0.428 uSec
=> gap time = 0.428 uSec
