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mysearch

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## Main Question or Discussion Point

I am trying to draw one of those nice plots of effective potential [Vr] against radius [r], which suggest the position of stable and quasi-stable orbits. Have no trouble getting the classical curve showing a minimum coinciding with a radius corresponding to a given angular momentum [L], but having trouble getting the equivalent min/max curve associated with general relativity. By way of reference, I was initially looking at the following Wikipedia page:

http://en.wikipedia.org/wiki/Kepler_problem_in_general_relativity

This page has two sub-titles that are relevant:

Under the first sub-title, the following equation for the effective potential is listed as:

[tex]Vr = -\frac{GMm}{r} + \frac{L^2}{2mr^2} - \frac{GML^2}{c^2mr^3} [/tex]

The first term relates to gravitational potential energy, the second relates to kinetic energy of rotation, while the third is described as

Gravitational Constant [G] = 6.67E-11

Black Hole Mass [M] = 7.92E+30 (4 solar masses)

Light [c] = 2.99E8

Schwarzschild Radius [Rs] = 1.18E+04 [tex]2GM/c^2[/tex]

Angular Momentum [L] = 4.08E+12 (r=2Rs??)

In classical physics, the angular momentum of a circular orbit L = mvr and, as such, [L] must change for each fixed circular orbit [r]. Only in a closed system is angular momentum [L] constant and therefore any decrease in radius [r] means an increase in velocity [v]. Normally, the

If the assumption L=mvr for a circular orbit is valid, then the equation for the effective potential above could be simplify to the form:

[tex]Vr = -\frac{GMm}{r} + 1/2 mv^2} - \frac{GMm}{r}(\frac{v^2}{c^2}) [/tex]

Where [v] is the orbital velocity, while the radial velocity [dr/dt] is set to zero for a circular orbit. If so, can this equation also be written in the form?

[tex]Vr = 1/2mv^2} - \frac{GMm}{r}(1 + \frac{v^2}{c^2}) [/tex]

This equation appears analogous to the classical form, but with an additional relativistic component. However, can this equation also be converted into an equivalent expression showing the balance between the `

[tex] \frac{mv^2}{r} = \frac{GMm}{r^2}(1 + \frac{v^2}{c^2}) [/tex]

If the assumptions made are correct, it would suggest that a larger gravitational force/curvature is required to counter the

[tex] v^2 = \frac{Rs*c^2}{(2r-Rs)} [/tex]

Therefore, I not sure any of the assumptions made are correct as I have not seen this approach presented in any standard text. Therefore, I would appreciate any help that could be given to clarify the situation.

http://en.wikipedia.org/wiki/Kepler_problem_in_general_relativity

This page has two sub-titles that are relevant:

*`Relation to classical mechanics and precession of elliptical orbits`*

`Circular orbits and their stability``Circular orbits and their stability`

Under the first sub-title, the following equation for the effective potential is listed as:

[tex]Vr = -\frac{GMm}{r} + \frac{L^2}{2mr^2} - \frac{GML^2}{c^2mr^3} [/tex]

The first term relates to gravitational potential energy, the second relates to kinetic energy of rotation, while the third is described as

*"an attractive energy unique to general relativity"*. It is stated that the 3 terms, containing different powers of radius [r], combine to give a curve with a minimum corresponding to a stable orbit and a maximum that relates to a quasi-knife-edge orbit. However, when I put the figures into a spreadsheet I don’t get this min/max curve.__Key values used:__Gravitational Constant [G] = 6.67E-11

Black Hole Mass [M] = 7.92E+30 (4 solar masses)

Light [c] = 2.99E8

Schwarzschild Radius [Rs] = 1.18E+04 [tex]2GM/c^2[/tex]

Angular Momentum [L] = 4.08E+12 (r=2Rs??)

__Assumptions about [L] for verification:__In classical physics, the angular momentum of a circular orbit L = mvr and, as such, [L] must change for each fixed circular orbit [r]. Only in a closed system is angular momentum [L] constant and therefore any decrease in radius [r] means an increase in velocity [v]. Normally, the

*'natural*` orbit of a satellite can be determined by balancing the outward `*centrifugal*` force [[tex]mv^2/r [/tex]] with the inward pull of gravity [GMm/r]. The effective potential (V) curve can show this balance in terms of energy plotted for fixed values of angular momentum [L]. In the classical case, the value of [L] inserted is only correct at one value of [r] corresponding to an energy rate of change [F=dE/dr=0]. With the addition of the relativistic component, the curve is said to produce a max/min curve, which I am not getting, but I can’t see my mistake.__Some other points for verification:__If the assumption L=mvr for a circular orbit is valid, then the equation for the effective potential above could be simplify to the form:

[tex]Vr = -\frac{GMm}{r} + 1/2 mv^2} - \frac{GMm}{r}(\frac{v^2}{c^2}) [/tex]

Where [v] is the orbital velocity, while the radial velocity [dr/dt] is set to zero for a circular orbit. If so, can this equation also be written in the form?

[tex]Vr = 1/2mv^2} - \frac{GMm}{r}(1 + \frac{v^2}{c^2}) [/tex]

This equation appears analogous to the classical form, but with an additional relativistic component. However, can this equation also be converted into an equivalent expression showing the balance between the `

*centrifugal*` force and gravitational pull associated with a circular orbit?[tex] \frac{mv^2}{r} = \frac{GMm}{r^2}(1 + \frac{v^2}{c^2}) [/tex]

If the assumptions made are correct, it would suggest that a larger gravitational force/curvature is required to counter the

*`centrifugal*` force. This factor would appear to only range between [1:2]. Equally, this expression could be solved without the use of the quadratic approach normally employed, e.g.[tex] v^2 = \frac{Rs*c^2}{(2r-Rs)} [/tex]

Therefore, I not sure any of the assumptions made are correct as I have not seen this approach presented in any standard text. Therefore, I would appreciate any help that could be given to clarify the situation.