Orbital velocities in the Schwartzschild geometryby espen180 Tags: geometry, orbital, schwartzschild, velocities 

#19
Jun1210, 11:34 AM

P: 836





#20
Jun1210, 11:40 AM

P: 1,568





#21
Jun1210, 11:43 AM

P: 1,568





#22
Jun1210, 12:24 PM

P: 836

As for your earlier equation: [tex]v=\frac{r\omega}{\sqrt{1r_s/r}}[/tex] If I use this definition with my calculations I get [tex]v=c\sqrt{\frac{\frac{r_s}{2}}{\left(r\frac{r_s}{2}\right)\left(1\frac{r_s}{r}\right)}}=\sqrt{\frac{GM}{\left(r\frac{GM}{c^2}\right)\left(1\frac{2GM}{rc^2}\right)}}[/tex] If I let [tex]r=\frac{3GM}{c^2}[/tex] I get [tex]v=\sqrt{\frac{3}{2}}c[/tex] where it is expected to be v=c. So my calculations must be wrong since it produces that factor of [tex]\sqrt{\frac{3}{2}}[/tex]. The equations I based my calculations off of are identical to the ones given in George Jones' reference in post #2, so I don't get what's wrong here. I also don't understand your argument that the equations don't exist. 



#23
Jun1310, 03:48 PM

P: 3,967

You state "From the Shwarschild metric we can , by imposing dr/ds=0 and theta=pi/2 obtain the relation": [tex] ds^2 = dt^2  r^2 d\phi^2 \;\;\; (15)[/tex] where I am using units of G=c=1 and ds to mean proper time of the test particle. That should read: [tex] ds^2 = (12M/r) dt^2  r^2 d\phi^2 \;\;\; (15)[/tex] which gives: [tex] \left(\frac{dt}{ds}\right)^2 = \frac{1}{(12M/r)}\left(1 + r^2 \left(\frac{d\phi}{ds}\right)^2\right) \;\;\; (16)[/tex] I am not quite sure how you got equation (13) from (10) but there seems to be a problem there somewhere when cos(pi/2)=0 and dr/ds=0. 



#24
Jun1310, 04:14 PM

P: 3,967

[tex]1 = \alpha \dot{t}^2  \alpha^{1}\dot{r}^2  r^2 \dot{\phi}^2[/tex] We can of course multiply both sides by some multiple and obtain a different constant on the left. Eg if I use a multiple of 1/2 then: [tex]1/2 = (\alpha \dot{t}^2  \alpha^{1}\dot{r}^2  r^2 \dot{\phi}^2)/2[/tex] I can now declare L=1/2 and proceed from there if I wish. The important thing is that I arrive at an equation in a form with a constant on one side. For a massless particle such as a photon, ds=0 and I can write the metric as: [tex]0 = \alpha  \alpha^{1}\dot{r}^2  r^2 \dot{\phi}^2[/tex] and in this case I can use L=0. I can also obtain nonzero values of L here by adding a constant to both sides. Note that for a masslesss particle, the overdot means the derivative with respect to coordinate time (t) rather than proper time (s). Obtain a new constant of motion for the angular velocity of a photon in the Schwarzschild metric by taking the partial differential of the right hand side with respect to [tex]\dot{\phi}[/tex]. Using the constant, solve for dr^2/dt^2 and differentiate both sides with respect to r and divide both sides by 2 to obtain the radial acceleration of a light particle. For a circular orbit d^2(r)/d(t)^2 = 0 and by setting the acceleration to zero and solving for r, the result that the photon radius is r=3M comes out very simply and clearly. 



#25
Jun1310, 04:28 PM

P: 836

Thank you very much for pointing that error out for me.
As for (13), it was derived from (9), not (10). (10) reduces to 0=0 when the restrictions are imposed. Updated article. I corrected the error you pointed out, and another regarding a constant factor in one of the Christoffel symbol's entries. The new (19), though, still does not do what it's supposed to, giving infinite velocity at the photosphere radius. Regarding Starthaus' last equation in post #10, doesn't the two angular terms indicate that v=ωr , being the same as for flat spherical coordinates? 



#26
Jun1310, 04:40 PM

P: 3,967

You have to be careful when using ds rather than dt. In the example below for something moving at the speed of light, you should expect to obtain an infinite velocity r*d(phi)/0 rather than c. I am a bit short of time and that is why post #25 is a bit cryptic, but it might be interesting to carry out the calculations and obtain coordinate velocity = c for a photon at r=3m using the method described. P.S. There still seems to be a problem with (13) derived from (9) but I do not have time to check that at the moment. 



#27
Jun1410, 05:06 PM

P: 3,967

[tex]\frac{rd\phi}{ds} = c \sqrt{\frac{GM}{rc^2  3GM}[/tex] which gives the correct result that the velocity with respect to proper time (ds) of a particle orbiting at r=3GM/c^2 is infinite. If you convert the above equation to local velocity as measured by a stationary observer at r by using [tex]ds = dt'\sqrt{1(rd\phi)^2/(cdt')^2}[/tex] you get: [tex]\frac{rd\phi}{dt'} = c \sqrt{\frac{GM}{rc^2  2GM}[/tex] which gives the result that the local velocity of a particle orbiting at r=3GM/c^2 is c. If you convert the equation to coordinate velocity using [tex]dt = dt'\sqrt{12M/(rc^2)}[/itex] you get: [tex]\frac{rd\phi}{dt} = c \sqrt{\frac{GM}{r}[/tex] which is the same as the Newtonian result (if you use units of c=1). All the above are well known solutions, so it seems all is in order with your revised document (from (9) onwards anyway  I have not checked the preceding calculations). 



#28
Jun1410, 05:13 PM

P: 1,568

[tex]ds^2=(1r_s/R)dt^2R^2d\phi^2[/tex] where [tex]R[/tex] is a constant. This is important in the correct derivation of the Christoffel symbols, since you should get a lot more "zeroes" than you have in your writeup. So, your Christoffel symbols need to reflect that. They don't. You need to recalculate the Christoffel symbols based on the correct metric. 



#29
Jun1410, 05:56 PM

P: 3,967





#30
Jun1410, 06:30 PM

P: 1,568

Covariant derivatives and lagrangian methods, if done correctly, should produce the same results. 



#31
Jun1410, 08:52 PM

P: 836

@#27:
Thank you very much. I will study these in greater detail. 



#32
Jun1410, 09:23 PM

P: 1,568





#33
Jun1510, 04:59 PM

P: 3,967

Starting with Schwarzschild metric and assuming orbital motion in a plane about the equator such that [itex]\theta = \pi/2[/itex] and [itex]d\theta = 0[/itex] [tex]ds^2=\alpha dt^2\frac{1}{\alpha}dr^2r^2d\phi^2[/tex] [tex]\alpha=1\frac{2m}{r}[/tex] For a massless particle travelling at the speed of light, ds=0 so in this case: [tex]L = 0 =\alpha dt^2\frac{1}{\alpha}dr^2r^2d\phi^2[/tex] Divide both sides by dt^2: [tex]L = 0 =\alpha \frac{1}{\alpha}\frac{dr^2}{dt^2}r^2\frac{d\phi^2}{dt^2}[/tex] The metric is independent of [itex]\phi[/itex] and t, so the constant associated with the angular velocity of a photon is obtained by finding the partial derivative of L with respect to [itex]d\phi/dt[/itex] [tex]\frac{\delta{L}}{\delta(d\phi/dt)} = r^2 \frac{d\phi}{dt} = H_c [/tex] where [itex]H_c[/itex] is the specific form of the constant for angular velocity that applies to a massless particle. Substitute this constant into the metric for a massless particle above to obtain: [tex]0 =\alpha \frac{1}{\alpha}\frac{dr^2}{dt^2}\frac{H_c^2}{r^2}[/tex] and solve for (dr/dt)^2: [tex]({dr}/{dt})^2 = (12M/r)(12M/r H_c^2/r^2) [/tex] Differentiate the above with respect to r and divide by 2 (this is the same as differentiating dr/dt with respect to t, but is much quicker and simpler) to obtain the radial acceleration of a photon in the metric: [tex]\frac{d^2r}{dt^2}= \frac{H_c^2}{r^4}(r3M) [/tex] Setting d^2r/dt^2 = 0 (which is true for a circular orbit) and solving for r gives r=3M as the circular orbit radius of a particle travelling at the speed of light. 



#34
Jun1510, 05:25 PM

P: 1,568

How can this be since in the line above you declared [tex]L=0[/tex]? Is this again some sort of numerology that defies the rules of calculus? 



#35
Jun1510, 05:41 PM

P: 836

[tex]\frac{H_c^2}{4r^4}(r3M)=0[/tex] still gives [tex]r=3M[/tex] 



#36
Jun1510, 05:48 PM

P: 1,568




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