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Garth said:That is, of course, exactly what normally I understand!
However, in this case, where A is the COM observer and B the 'bobbing' observer and using a notation in which the time elapse between encounters is written, we have:
\Delta \tau_{A,A} being A's time elapse as measured by A,
\Delta \tau_{A,B} being A's time elapse as computed by B,
\Delta \tau_{B,B} being B's time elapse as measured by B,
\Delta \tau_{B,A} being B's time elapse as computed by A,
so \Delta \tau_{A,A} & \Delta \tau_{B,B} are measured by the identical clocks carried by A and B in their inertial frames of reference respectively.
We have said that:
\Delta \tau_{A,A} = \Delta \tau_{A,B} and
\Delta \tau_{B,B} = \Delta \tau_{B,A} as they are geometric objects independent of the coordinate systems in which they may be calculated.
Now either:
\Delta \tau_{A,A} > \Delta \tau_{B,B} or
\Delta \tau_{A,A} < \Delta \tau_{B,B} or
\Delta \tau_{A,A} = \Delta \tau_{B,B} and intuition tells you it is the first of these options that would be proved correct in an actual experiment.
Intuition isn't terribly reliable - I intuitively guessed the third option a while back, so my intuition was different than yours. And it was just as wrong, because after doing the calculation I'm convinced any of the above cases can be true depending on the density profile of the planet.
However that would also mean \Delta \tau_{A,B} > \Delta \tau_{B,B} as \Delta \tau_{A,A} = \Delta \tau_{A,B} would it not?
In which case in B's inertial frame of reference, A's space-time interval is computed by B to be greater than B's space-time interval, even though it is A that is moving and B that is stationary in that inertial frame of reference, against all space-time intuition about relative moving clock time dilation.The moving clock is 'ticking' faster than the stationary one, in other words, the moving observer A is aging more quickly than the stationary B!
This is perfectly possible. I'm having a bit of a problem untangling your notation or why you think there is a difficulty.
For any observer, using a +--- sign convention, we can write
d\tau^2 = g_{00} dt^2 + 2 g_{01} dt dx + g_{11} dx^2
which can be re-written as
d\tau = \sqrt{g_{00} + 2 g_{01} \frac{dx}{dt} + g_{11} (\frac{dx}{dt})^2 } dtThe numerical values of the g_ij will depend on the coordinate system used.
Let's look at it from the viewpoint of the stationary observer at the center of the planet, which we will call obsever A.
You are assuming, I believe, that we adopt coordinates so that g_00=1, g_01=0, and g_11 = 1 for the stationary observer. (We don't have to adopt such coordinates, but that's what I'm getting from what you wrote). Then the above intergal reduces to \tau_{A,A} = t.
Now let's consider the calculation of \tau_{B,A}
The moving obserer will have g_00, g_01, and g_11 as functions of time. We do not expect the metric coefficeints for observer B to represent a Minkowskian metric, because observer B is far away from A in a gravity field.
g_00 may be greater than 1 for the moving obsever. In fact, we expect g_00 to be greater than 1. Well, I expect it to be greater than 1.
Therfore it is possible in principle for dtau > dt.
The moving observer will have g_11 approximately equal to -1. Thus it is possible in principle for dtau < dt.
To decide which is the case, we need more information.
In english:
The moving observer is higher in a gravity well. Thus it's clock is ticking faster because of it's height. The moving obsever is also moving. Thus it's clock is ticking slower due to relativistic time dilation. Which effect dominates is not clear - it could be either one.