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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? :)
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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