Circular orbit in schwarzschild solution

In summary, some astronomers have found an infrared flare from SgrA*m and have interpreted it as a shining plasma blob on a circular orbit around the black hole. If this interpretation is correct, SgrA* is not a schwarzschild black hole. I plan to use Newton's law to calculate the angular momentum of the plasma blob and then use classical arguments to find the approximate value of E.
  • #1
Pietjuh
76
0
Hello everybody, I've been working on a problem about circular orbits in schwarschild spacetime. Recently a infrared flare has been detected from SgrA*m and the lightcurve during the flare has shown some quasiperiodic oscillations with a period of about 17 minutes. Some astronomers interpreted this effect as being due to a shining blob of plasma which is on a circular orbit around SgrA*. Show that if this interpretation is correct, SgrA* is not a schwarzschild black hole.

My thought on how to solve this, is to calculate [itex]d\phi / dt[/itex], and then integrate this expression to phi from 0 to 2pi to find the period of a particle in a circular orbit. From the lecture notes I've got that the radius of such an orbit equals to 3r_g, with r_g the schwarzschild radius.

In order to obtain this, I used the symmetries of the problem:
1) [itex]u_0 = const = E[/itex]
2) [itex]u_{\phi} = const[/itex]
3) [itex]d\theta = 0[/itex] because [itex]\theta = \pi / 2[/itex]
4) [itex]dr = 0[/itex] because we are on a circular orbit

This gives the following metric:

[tex]ds^2 = (1 - \frac{r_g}{r}) dt^2 - r^2 d\phi^2[/tex]

Using the relation that the magnitude of the 4-velocity equals to one, I find that:
[tex](1 - \frac{r_g}{r}) (u^0 )^2 - r^2 \left(\frac{d\phi}{d\tau}\right)^2 = 1[/tex]
[tex]\frac{E^2}{1 - r_g / r} - r^2(u^{\phi})^2 = 1[/tex]
[tex]u^{\phi} = \frac{1}{r} \sqrt{\frac{E^2}{1 - r_g / r} - 1}[/tex]

Now it's easy to find [tex]d\phi / dt[/tex]:
[tex]\frac{d\phi}{dt} = \frac{u^{\phi}}{u^0} = \frac{1}{r} \sqrt{\frac{E^2}{1 - r_g / r} - 1} \frac{1 - r_g / r}{E}[/tex]

If we now integrate this to phi and plug in the correct factors of G and c and the value [itex]r = 3 r_g[/itex] we find the total period of the orbit in seconds:

[tex]\Delta t = \frac{18 \pi M E}{\sqrt{\frac{3}{2}E^2 - 1}} \frac{G}{c^3}[/tex]

Now the problem is this factor of E. I need to have this value to obtain a solution for the period. Is there any way to obtain this E, which is equal to the first covariant component of the 4-velocity vector?

For example if I take E = 1, the equation gives me a period of 26 minutes, but I think it also may be possible to choose E such that the period would be 17 minutes.

Can anybody help me on this? :)
 
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  • #2
I've thought of a way to obtain E using classical arguments. But I'm not sure if it's justifiable to do it this way.

Because the blob of plasma is on a circular orbit around the black hole, I intend te use just Newton's law to determine the approximately angular momentum of the blob. From equating the centripetal force with the gravitational force we find that the velocity equals to [itex]v = \sqrt{M / r_c}[/itex]. This gives that [itex]l^2 = r_c M[/itex].
If we know use the fact that the phi component of the 4-velocity equals the angelur momentum, we see that [itex]u^{\phi} = -\sqrt{r_c M}[/itex], which gives when doing some simplification that E = 7/9. This gives me a period of 35 minutes, which is reasonable :rolleyes:
 
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  • #3
Eghm ... does it have sense to solve the formula for E, imposing the period to be 17 minutes, and then trying to do some considerations about the values of E found ??
 

1. What is a circular orbit in the Schwarzschild solution?

A circular orbit in the Schwarzschild solution refers to the path of a particle or object around a spherically symmetric mass, such as a black hole, where the gravitational force is balanced by the centrifugal force, resulting in a circular path.

2. How is the circular orbit in the Schwarzschild solution different from a circular orbit in Newtonian mechanics?

The circular orbit in the Schwarzschild solution is different from a circular orbit in Newtonian mechanics because it takes into account the curvature of spacetime caused by the mass of the object, whereas Newtonian mechanics assumes a flat spacetime.

3. What is the significance of the radius of a circular orbit in the Schwarzschild solution?

The radius of a circular orbit in the Schwarzschild solution, also known as the photon sphere, is the distance from the center of the mass at which the gravitational pull is so strong that even light cannot escape. This is a key feature of black holes.

4. How does the mass of the object affect the circular orbit in the Schwarzschild solution?

The mass of the object directly affects the radius of the circular orbit in the Schwarzschild solution. The more massive the object, the larger the radius of the orbit will be.

5. Can there be multiple circular orbits in the Schwarzschild solution?

Yes, there can be multiple circular orbits in the Schwarzschild solution, depending on the mass and radius of the object. However, the innermost stable circular orbit is the closest orbit that can exist without the object collapsing into a singularity.

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