Most proper time in Schwarzchild metric

In summary, the Schwarzschild metric provides a way to calculate proper time using the gravitational potential, and it has been shown that the world line of a stationary observer experiences more proper time than that of a particle in circular motion. However, this does not violate the principle of most proper time, as the variational principle states that geodesics, not necessarily the longest curves, maximize proper time between two fixed points in space-time. In the Schwarzschild metric, there may be cases where there is no global maximum for proper time, but for two events on a static world line, there is always a global maximum geodesic. This is known as the "ballistic" trajectory, where the particle shoots radially outward and falls back
  • #1
stallm
7
0
Hi
In the Schwarzschild metric, the proper time is given by
[itex]c^{2}dτ^{2} = (1- \frac{2\Phi}{c^2})c^2 dt^2 - r^2 dθ^2[/itex]
with where [itex]\Phi[/itex] is the gravitational potential. I have left out the [itex]d\phi[/itex] and [itex]dr[/itex] terms.
If there is a particle moving in a circle of radius [itex] R [/itex] at constant angular velocity ω, the proper time experienced by the particle completing one whole orbit will be
[itex] τ = (1- \frac{2\Phi(R)}{c^2} - R^2 \frac{ω^2}{c^2}) t [/itex]
where t is the period of the orbit. If the particle were to remain stationary for time t instead it would experience proper time
[itex] τ = (1- \frac{2\Phi(R)}{c^2}) t [/itex]
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
 
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  • #2
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
Thank you, I understand now
 
  • #4
One additional point:

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:
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  • #5
stallm said:
Hi
In the Schwarzschild metric, the proper time is given by
[itex]c^{2}dτ^{2} = (1- \frac{2\Phi}{c^2})c^2 dt^2 - r^2 dθ^2[/itex]
with where [itex]\Phi[/itex] is the gravitational potential. I have left out the [itex]d\phi[/itex] and [itex]dr[/itex] terms.
If there is a particle moving in a circle of radius [itex] R [/itex] at constant angular velocity ω, the proper time experienced by the particle completing one whole orbit will be
[itex] τ = (1- \frac{2\Phi(R)}{c^2} - R^2 \frac{ω^2}{c^2}) t [/itex]
where t is the period of the orbit. If the particle were to remain stationary for time t instead it would experience proper time
[itex] τ = (1- \frac{2\Phi(R)}{c^2}) t [/itex]
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

You missed your sign on the [itex]\Phi(R)[/itex]
It should be
[itex]c^{2}dτ^{2} = (1+ \frac{2\Phi}{c^2})c^2 dt^2 - r^2 dθ^2[/itex]
where you have
[itex]\Phi(R)=-\frac{GM}{R}[/itex]
Yielding
[itex] τ = (1- \frac{2GM}{c^2 R} - R^2 \frac{ω^2}{c^2}) t [/itex]
 
  • #6
PAllen said:
In the Schwarzschild metric, I think there are some cases of no global maximum.

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 into 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
pervect said:
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 into the starting height, reaching it's starting height again after the specified coordinate time interval delta.
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.
pervect said:
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.

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).
 
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  • #8
PAllen said:
Did you read anything I wrote carefully? I describe exactly such cases.

Apparently not - sorry. A bit distracted today.
 
  • #9
stallm said:
...
[itex] τ = (1- \frac{2\Phi(R)}{c^2} - R^2 \frac{ω^2}{c^2}) t [/itex]
where t is the period of the orbit. If the particle were to remain stationary for time t instead it would experience proper time
[itex] τ = (1- \frac{2\Phi(R)}{c^2}) t [/itex]
...
Thanks

Sorry, is there power [itex] \frac{1}{2}[/itex] in the coefficients of the above equations?
 
  • #10
waitedavid137 said:
You missed your sign on the [itex]\Phi(R)[/itex]
It should be
[itex]c^{2}dτ^{2} = (1+ \frac{2\Phi}{c^2})c^2 dt^2 - r^2 dθ^2[/itex]
where you have
[itex]\Phi(R)=-\frac{GM}{R}[/itex]
Yielding
[itex] τ = (1- \frac{2GM}{c^2 R} - R^2 \frac{ω^2}{c^2}) t [/itex]

Oh and a square root on that last one as well.
 
  • #11
jadyliber said:
Sorry, is there power [itex] \frac{1}{2}[/itex] in the coefficients of the above equations?

Yep, a sign missed and the power.
 

Related to Most proper time in Schwarzchild metric

1. What is the Most Proper Time in Schwarzchild Metric?

The Most Proper Time in Schwarzchild Metric is a concept in general relativity that refers to the longest time interval observed by an observer at a fixed distance from a massive object, such as a black hole. It is the time measured by a clock that is at rest relative to the massive object.

2. How is the Most Proper Time Calculated in Schwarzchild Metric?

The Most Proper Time in Schwarzchild Metric is calculated using the formula: Δt = Δτ / √(1 - (2GM/rc²)), where Δt is the coordinate time, Δτ is the proper time, G is the gravitational constant, M is the mass of the massive object, r is the distance from the object, and c is the speed of light.

3. Why is the Most Proper Time Important in Schwarzchild Metric?

The Most Proper Time in Schwarzchild Metric is important because it is the time interval that is measured by a clock that is at rest relative to the massive object. It is the longest time interval that can be measured by an observer at a fixed distance from the object, and it is a crucial concept in understanding the effects of gravity on time and space.

4. How does the Most Proper Time Differ from Coordinate Time in Schwarzchild Metric?

The Most Proper Time and Coordinate Time in Schwarzchild Metric are different because the Most Proper Time takes into account the effects of gravity on time, while Coordinate Time does not. The Most Proper Time is measured by a clock that is at rest relative to the massive object, while Coordinate Time is measured by a clock that is in motion relative to the object.

5. Can the Most Proper Time be Observed by an Observer Outside the Event Horizon of a Black Hole?

Yes, the Most Proper Time can be observed by an observer outside the event horizon of a black hole. However, as the observer gets closer to the event horizon, the Most Proper Time will appear to slow down due to the strong gravitational pull of the black hole. Eventually, at the event horizon, the Most Proper Time will appear to stop completely.

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