Orbital period around Schwarzschild radius

In summary, the problem is asking for the proper time for a moving observer at a radius of 3GM, with a constant angular velocity in the phi direction, to complete one orbit and return to their starting point. The observer's motion is governed by the Schwarzschild line element and the observer has a non-zero proper acceleration.
  • #1
Kyrios
28
0

Homework Statement


An observer is orbiting at a radius r = 3GM, [itex] \theta = \frac{\pi}{2} [/itex] and [itex] \phi = \omega t [/itex] where w is constant.
The observer sends a photon around the circular orbit in the positive [itex] \phi [/itex] direction. What is the proper time [itex] \Delta \tau [/itex] for the photon to complete one orbit and return to the observer?

Homework Equations


Schwarzschild line element where dr =0 and [itex]d\theta[/itex] =0.

The Attempt at a Solution



From the line element we have
[tex] \left(\frac{d\tau}{dt}\right)^2 = 1 - \frac{2GM}{r} - r^2 \left(\frac{d\phi}{dt}\right)^2 [/tex]
[tex] \left(\frac{d\tau}{dt}\right)^2 = 1 - \frac{2GM}{r} - r^2 \omega^2 [/tex]

I was trying to use [itex] \omega^2 = \frac{GM}{r^3} [/itex] but that just gives

[tex] \left(\frac{d\tau}{dt}\right)^2 = 1 - \frac{2GM}{r} - r^2 \frac{GM}{r^3} [/tex]
[tex] \left(\frac{d\tau}{dt}\right)^2 = 1 - \frac{2GM}{r} - \frac{GM}{r} [/tex]
[tex] \left(\frac{d\tau}{dt}\right)^2 = 1 - \frac{3GM}{3GM} [/tex]
as r = 3GM. This gives zero and I'm not really sure what to do with it. Have I gone wrong somewhere? What should I do with the w?
 
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  • #2
Light follows light-like (naturally) geodesics. These have the defining property of ds=0 so it shoukd come as no surprise that it has length zero.

This makes me suspect that the problem wants you to compute the proper time for an observer at fixed spatial coordinates before the light signal comes back from the opposite direction.
 
  • #3
I have already worked out the proper time for a stationary observer, [itex] \Delta \tau = 6 \pi GM[/itex]. The question specifically states that the observer is not at fixed coordinates, but moving with [itex] \phi = \omega t[/itex] and emitting a photon while orbiting.
 
  • #4
I see, so note that the observer according to the problem statement is not necessarily moving on a geodesic. When you put ##\omega=0## you should regain your result for the stationary observer.
 
  • #5
How can I simply assume that [itex] \omega = 0[/itex]? It says that w is a constant. Since the previous question was for the case of the stationary observer, I'm quite sure that this one isn't after the same answer. Although I cannot see any way around it
 
  • #6
You cannot. What I said was that, when you have solved your problem, letting ##\omega = 0## should give you the result you had before. Your answer to the current problem should depend on ##\omega##, which you can treat as an arbitrary constant (but with ##\omega^2 < 3GM/r^3## or you will have an observer following a non-time-like curve - as you found out for the limiting case already).

Edit: It is not necessary to solve the problem, but note that, regardless of the value of ##\omega##, the observer will have a non-zero proper acceleration.
 

Related to Orbital period around Schwarzschild radius

1. What is the orbital period around the Schwarzschild radius?

The orbital period around the Schwarzschild radius refers to the time it takes for an object to complete one full orbit around a non-rotating black hole. This is determined by the mass of the black hole and the distance of the object from the event horizon, or the Schwarzschild radius.

2. How is the orbital period around the Schwarzschild radius calculated?

The orbital period around the Schwarzschild radius can be calculated using Kepler's third law, which states that the square of the orbital period is proportional to the cube of the semi-major axis of the orbit. In the case of a black hole, the semi-major axis is equal to the distance from the event horizon to the object's orbit.

3. Can the orbital period around the Schwarzschild radius change?

Yes, the orbital period around the Schwarzschild radius can change if there is a change in the mass of the black hole or the distance of the object from the event horizon. However, for a stable orbit, the changes in these parameters must be very small in relation to the original values.

4. What is the significance of the orbital period around the Schwarzschild radius?

The orbital period around the Schwarzschild radius is significant because it provides insight into the gravitational pull of a black hole. The shorter the orbital period, the stronger the gravitational pull of the black hole. This can also be used to calculate the mass of the black hole.

5. Can objects have the same orbital period around the Schwarzschild radius?

Yes, objects can have the same orbital period around the Schwarzschild radius if they are at the same distance from the event horizon and have the same mass. This is known as a synchronous orbit and is similar to how the Moon orbits around the Earth at a constant rate.

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