# Most proper time in Schwarzchild metric

1. Jan 28, 2014

### stallm

Hi
In the Schwarzchild metric, the proper time is given by
$c^{2}dτ^{2} = (1- \frac{2\Phi}{c^2})c^2 dt^2 - r^2 dθ^2$
with where $\Phi$ is the gravitational potential. I have left out the $d\phi$ and $dr$ terms.
If there is a particle moving in a circle of radius $R$ at constant angular velocity ω, the proper time experienced by the particle completing one whole orbit will be
$τ = (1- \frac{2\Phi(R)}{c^2} - R^2 \frac{ω^2}{c^2}) t$
where t is the period of the orbit. If the particle were to remain stationary for time t instead it would experience proper time
$τ = (1- \frac{2\Phi(R)}{c^2}) t$
Which is more than the value for the circular orbit. Since the circular motion is inertial and the stationary motion isn't, doesn't this violate the principle of most proper time?
Thanks

2. Jan 28, 2014

### WannabeNewton

No because all the variational principle says is if there are two fixed points in space-time then there exists an open subset of space-time containing these two points such that amongst all curves contained in this open subset, a geodesic will be the curve of longest length between these two points. In other words the variational principle says geodesics maximizes length amongst all curves in space-time nearby the geodesic. The world line of the static observer in your example is certainly not nearby the world line of the observer in circular orbit between the two fixed events for which you measure the elapsed proper times so there's no problem.

3. Jan 28, 2014

### stallm

Thank you, I understand now

4. Jan 28, 2014

### PAllen

If there is a curve that is a global maximum of proper time between two events (there need not be; you can have a situation where there is a least upper bound, but no curve delivers that least upper bound), then that curve will be geodesic. Note that there can be many geodesics between two events in GR. This says that IF there is a global maximum, it is one of the geodesics. [edit: actually, there could be multiple geodesics that deliver the global maximum; and also geodesics that don't. Any curve that delivers the global maximum will be a geodesic.]

In the Schwarzschild metric, I think there are some cases of no global maximum. However, for two events on a static world line, there is always a global maximum geodesic. It is simply the 'ballistic' trajectory: shoot radially outward and fall back, all under free fall.

Last edited: Jan 28, 2014
5. Feb 1, 2014

### waitedavid137

You missed your sign on the $\Phi(R)$
It should be
$c^{2}dτ^{2} = (1+ \frac{2\Phi}{c^2})c^2 dt^2 - r^2 dθ^2$
where you have
$\Phi(R)=-\frac{GM}{R}$
Yielding
$τ = (1- \frac{2GM}{c^2 R} - R^2 \frac{ω^2}{c^2}) t$

6. Feb 1, 2014

### pervect

Staff Emeritus
The geodesic between points (t,r,theta, phi) and (t+delta,r,theta, phi) in the Schwarzschild geoemtry that exists for all values of delta is easy to describe. It's the geodesic that travels radially outwards, reaches some maximum height, and falls back in to the starting height, reaching it's starting height again after the specified coordinate time interval delta.

This radial geodesic is the path that actually maximizes proper time between the two endpoints. You can see that for a small enough delta, the orbital geodesic won't exist (the points are too close in time to be connected by an orbit).

When delta becomes large enough so that orbital geodesics also occur (in addition to the radial geodesic) one has multiple (two in this case) geodesics between the same pair of points - the proper time-maximizing radial geodesic, and the proper time extremizing (but not maximizing) orbital geodesic. This marks the end of the "normal convex neighborhood" in which there is a unique geodesic between any two points, because now there are two geodesics connecting the same two points.

7. Feb 1, 2014

### PAllen

Please note what I said: "However, for two events on a static world line, there is always a global maximum geodesic. It is simply the 'ballistic' trajectory: shoot radially outward and fall back, all under free fall. "

That is what you are describing. What I was thinking about (in reference to no global maximum), but didn't compute carefully enough to resolve, was events on opposite sides of the horizon, close enough together in SC time that there are no free fall trajectories at all connecting them. Give enough initial velocity to have chance, and the trajectory is not bent enough by the central field. The trajectory that would work for Newtonian physics goes inside the horizon instead.
Did you read anything I wrote carefully? I describe exactly such cases. However, it is still true that for events on a static world line, the ballistic trajectory is a global maximum. It is also true, that if there is any curve delivering a global maximum, then every such curve is a geodesic (out of possibly more than one delivering said maximum [I don't think this case occurs in SC metric, but does in GR in general], and there may also be multiple geodesics not delivering the maximum).

Last edited: Feb 1, 2014
8. Feb 1, 2014

### pervect

Staff Emeritus
Apparently not - sorry. A bit distracted today.

9. Feb 1, 2014

Sorry, is there power $\frac{1}{2}$ in the coefficients of the above equations?

10. Feb 1, 2014

### waitedavid137

Oh and a square root on that last one as well.

11. Feb 1, 2014

### waitedavid137

Yep, a sign missed and the power.