Circular orbit in schwarzschild solution

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SUMMARY

The discussion centers on the analysis of circular orbits in Schwarzschild spacetime, particularly in relation to the infrared flare detected from SgrA*. The observed quasiperiodic oscillations with a period of approximately 17 minutes suggest that SgrA* may not be a Schwarzschild black hole. The participant outlines a method to calculate the period of a particle in a circular orbit using the metric derived from Schwarzschild geometry, ultimately leading to the equation for the period: Δt = (18πME/√(3/2E² - 1))(G/c³). The challenge remains in determining the energy parameter E to match the observed period.

PREREQUISITES
  • Understanding of Schwarzschild geometry and black hole physics
  • Familiarity with general relativity concepts, particularly circular orbits
  • Knowledge of the metric tensor and its application in spacetime calculations
  • Basic proficiency in integrating functions and solving differential equations
NEXT STEPS
  • Explore the derivation of the Schwarzschild metric in detail
  • Study the implications of energy conservation in general relativity
  • Investigate methods for determining the energy parameter E in orbital mechanics
  • Learn about observational techniques for detecting quasiperiodic oscillations in astrophysical phenomena
USEFUL FOR

Astronomers, physicists, and students of general relativity interested in black hole dynamics and the behavior of matter in extreme gravitational fields will benefit from this discussion.

Pietjuh
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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? :)
 
Last edited:
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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:
 
Last edited:
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 ??
 

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