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Interior metrics and circular orbits

  1. Oct 9, 2012 #1
    Hello,

    I have a couple of questions regarding the calculation of circular orbits in the Schwarzschild exterior spacetime and then the extension of these arguments to other (interior) metrics.

    First of all, in a few different books/sets of notes there seems to be a bit of 'drift' in the nomenclature which I was hoping to clear up. Specifically, is the condition that r-double-dot (w.r.t tau) = 0 a condition for a circular orbit to exist, or a condition for a stable circular orbit. I suspect the former ( can have max or min, i.e. stable or unstable, part of the potential).

    Secondly, the standard way to construct the effective potential in the Schwarzschild case is to identify symmetries and thus constants of the motion, and then to solve for dr/dtau either using some conserved four-momentum or equivalently looking at timelike (for massive particles ofc) curves and setting the spacetime interval to something.

    This turns out quite nicely as the schwarzschild metric (ds^2 = -a(r) dt^2 +b(r)dr^2...) has the nice property that a(r)b(r) = 1. In the case of say the Schwarzschild interior metric, or it's neater companion the Florides metric, this is not true.

    Thus when you follow the usual procedure it is difficult to separate out a constant 'energy' part and effective potential to then look for circular orbits in.

    I was wondering if there is a standard trick to getting around this. I thought of using tortoise coordinates (or more generally, coordinates that put your metric in a conformally flat form) which works well in say the wave equation for a scalar particle, but here I don't see how it would help.

    Sorry for the tl;dr, and thanks in advance for help/advice!
    -FD
     
  2. jcsd
  3. Oct 9, 2012 #2

    Mentz114

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    The energy condition just means that the 4-velocity is normalised. If you start with an ansatz that the orbit is
    [tex]
    U=\dot{t}\partial_t+\dot{\phi}\partial_\phi
    [/tex]
    then normalisation gives [itex]\dot{t}[/itex] in terms of [itex]\dot{\phi}[/itex]. Putting this into the geodesic equations can give an easy ODE, if you're lucky. It is easy for the exterior Schwarzschild.

    [Edit]
    For the Florides interior I find a circular geodesic in the equatorial plane (θ = ∏/2)
    [tex]
    \dot{t}=\frac{{\left( {R}^{3}-2\,m\,{r}^{2}\right) }^{\frac{3}{4}}}{{\left( R-2\,m\right) }^{\frac{3}{4}}\,\sqrt{{R}^{3}-3\,m\,{r}^{2}}},\ \ \dot{\phi}=\frac{\sqrt{m}}{\sqrt{{R}^{3}-3\,m\,{r}^{2}}}
    [/tex]
     
    Last edited: Oct 9, 2012
  4. Oct 9, 2012 #3

    pervect

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    The condition for a circular orbit is [itex]dR/d\tau=0[/itex], because it implies that R is constant. If you say you have an effective potential E(r) though, as per for instance http://www.fourmilab.ch/gravitation/orbits/, you can ALSO say that [itex]dE/dr[/itex] is a condition for a circular orbit, i.e. that they occur at maximum and minimum of the effective potential. (I've played some capitalization games to make this easier for me to read - r is equal to R in this case).

    Thus [itex]d^2E/dr^2[/itex] gives you the stability of the orbit by its sign, rather than what you wrote, because the stable orbits are at a minimum of the effective potential.
    The symmetries still exist if you change coordinates, they're just slightly more difficult to find. They're usually known as Killing vectors.

    Killing vectors transform just like other vectors, so you can with just a bit of vector transformation find out what they are in any coordinates that you have transformation equations for.

    For instance, in the Schwarzschild metric, none of the metric coefficients is a function of the time coordinate, x^0. This implies that it dual, x_0, is a Killing one form. (Note that x^0 isn't a Killing vector, the dual x_0 is.)

    One of the properties of Killing vectors is that the dot product of a tangent vector of a geodesic and a Killing vector is a constant along the geodesic. Knowing this, you can transform the Killing vector to different coordinates, and use the above fact to find the conserved quantities.
     
  5. Oct 9, 2012 #4
    Hi guys,

    Thanks very much for your posts.

    Mentz114: I was wondering if you could please elucidate your calculation a bit more? Specifically the normalization of the four velocity ( [itex] U^{\mu} U_{\mu} =-1= -\dot{t}^2+\dot{\phi}^2 [/itex] or something?) Sorry if i'm missing something simple...
     
    Last edited: Oct 9, 2012
  6. Oct 9, 2012 #5

    Mentz114

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    Yes, that's what I meant.
    I was a bit skimpy with the detail. By 'normalisation' I mean ensuring that the magnitude of the 4-velocity of the worldline is always -1, viz gαβUαUβ = -1. This means that the total energy of the particle on the WL is conserved. See eqn (1)-(4) of the attached ( you're most likely familiar with this ... )

    This enables expressing the 4-velocity entirely in terms of [itex]\dot{\phi}[/itex] which we assume depends on r only. The next step is to calculate the proper acceleration aμ of this WL from

    aμ = ∇αUμUα

    where ∇ is the covariant derivative. This leaves one or more equations in [itex]\dot{\phi}[/itex] and (maybe) its derivatives to be solved to set aμ to zero.

    If you can find a potential, though, it's probably easier to use the dynamic methods.
     

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  7. Oct 10, 2012 #6
    Hey mentz114, thanks!

    Is it also valid to simply set [itex] \dot{r} = 0, \ddot{r}=0 [\itex] and simply solve that set of simultaneous equations for conserved energy and r?

    Thanks
    -FD
     
  8. Oct 10, 2012 #7

    Bill_K

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    I think you meant to write aμ = UααUμ instead, and this is the form in which it is often quoted, but is conceptually incorrect.

    The covariant derivative is a four-dimensional gradient operation and as such can only be applied to quantities that are functions of all four dimensions, such as f(x,y,z,t). This formula for aμ only makes sense if you are talking about the motion of a continuous fluid, in which the velocity is a field, Uμ(x,y,z,t), and then ∇αUμ is defined and meaningful, and denotes a rank two tensor field.

    But we are talking here about the motion of a single point particle, in which Uμ(τ) is a function only of the proper time along the world line, and is undefined elsewhere. By analogy, in the Newtonian mechanics of projectile motion, it would make no sense at all to talk about the "gradient" of the projectile's velocity vector.

    The correct concept is not the covariant derivative but the absolute derivative, D/Dτ, and this is what must be used to differentiate world-line functions. In a manner similar to the covariant derivative, it involves Christoffel symbols, one for each tensor index. E.g. for any vector Vμ(τ) defined along a world line with tangent vector Uμ, the absolute derivative of Vμ is

    DVμ/Dτ ≡ dVμ/dτ + Γμνσ UνUσ
     
  9. Oct 10, 2012 #8

    Mentz114

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    Bill_K, thanks for your remarks. I've got questions about this which I'll put in a new topic.

    I'm not sure which set of equations you mean. But it sounds OK. Weinstein's equation for E2 (eqn (5) ), with [itex] \dot{r} = 0[/itex] (modified for the interrior metric) seems a good starting point. If E is constant then maybe you can differentiate this and set it to zero.

    The geodesic worldline I gave in post#2 actually satisfies both UααUμ=0 and [itex]{\Gamma^\mu}_{\alpha\beta}U^\alpha U^\beta=0[/itex].
     
    Last edited: Oct 10, 2012
  10. Oct 10, 2012 #9

    pervect

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    Covariant derivatives

    I had a question / comment about Bill's response that I'm moving to a new thread...
     
  11. Oct 15, 2012 #10
    Hi guys,

    Sorry to dig this up again, but I recently came across a paper called "On the Energy Extraction from the Interior Schwarzschild Metric" which considers the case of an interior metric and orbits therein.

    However, I do not understand their derivation of equation (5), given by

    [itex]U_{\textrm{eff}}=\sqrt{e^{\nu} (m^2+\frac{L^2}{r^2})}[/itex]

    where their metric is defined as
    [itex]ds^2=e^\nu dt^2-e^\lambda dr^2-r^2 d\Omega^2[/itex],
    and was hoping for some help in that department. Specifically, if I try to derive the equation of motion for a general metric, as they've written in in (1), then I would get

    [itex]\frac{dr}{d\tau} = E^2 e^{-(\nu+\lambda)}-\frac{L^2 e^{\lambda}}{r^2}-e^{-\lambda}[/itex].

    For the Schwarzschild exterior [itex]e^{-(\nu+\lambda)}=1[/itex] and so you get a nice equation of motion with a clearly identifiable total energy (independent of radius) and potential. Again, in the Schwarzschild exterior this agrees with what is written in the paper listed above.

    However, I do not see how you can identify the potential, in the way they have, in general; are you allowed to simply take all terms that don't depend on your total energy (which is really just the constant of the motion you say is the energy) is your potential?

    Hope this made sense!
     
  12. Oct 15, 2012 #11

    pervect

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    Sorry, I'm not following you at all. You can write the energy momentum vector
    p^a = m u^a You can use the metric to lower the indicies, p_b = g_ab p^a

    If none of the metric coefficeints is a function of t, ie. x^0, then the conjugate momentum p_0 is a constant, which you can call E

    Similarly, if none of the metric coefficients is a function of x^4 ([itex\phi[/itex]) then p_4 is a constant, which you can call L.

    You can then use the equation p^a p_a = m^2 u^i u_i to find the equations of motion. Depending on your sign convention, u^i u_i will be + or - 1.

    See for instance MTW"s treatment , or your favorite textbook if you don't have MTW.
     
  13. Oct 15, 2012 #12
    The method you outline would be equivalent to solving the geodesic equation, correct?
     
  14. Oct 15, 2012 #13

    pervect

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    Yes - basically you wind up showing that many of the geodesic equations can be put in the form DE/dtau = 0. DE/dtau is just the total derivative along the curve with respect to proper time tau, and E is some function of the position coordinates as a function of tau, and the derivatives of the position coordinates with respect to tau (which are also functions of tau).

    The fact that the geodesic equations make dE/dtau zero show that it's constant along the curve.
     
  15. Oct 17, 2012 #14

    pervect

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    To expand on my previous point a bit by example, in the Schwarzschild metric

    [tex] E = \left[1-\frac{2m}{r(\tau)}\right] \frac{dt}{d\tau} [/tex] is a constant of motion.

    If you grok Killing vectors, it corresponds to setting the dot product of the Killing vector [itex]\xi^a[/itex] = [1,0,0,0] (which does not have unit length!) with the four velocity u = [itex] [dt/d\tau, dr/d\tau, d\theta/d\tau, d\phi/d\tau] [/itex] equal to a constant. The dot product of a Killing vector and a tangent vector is constant everywhere along a geodesic.

    If you don't grok Killing vectors, it doesn't really matter where you get E, as long as you can verify that it's constant by the procedure below,.


    Setting [itex]dE/d\tau = 0[/itex] and applying the chain rule recovers one of the geodesic equations (after multiplying through some constants).

    [tex]t'' + \frac{2\,m\,\dot{r}\dot{t}}{r^2\left(1-2m/r\right)} = 0[/tex]

    which is necessary and sufficient to show that E is a constant of motion.
     
    Last edited: Oct 17, 2012
  16. Oct 18, 2012 #15

    Mentz114

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    Pervect, that's a pretty good summary of stuff I've learnt (woefully) recently. If I can ask, given this vector (K is a constant )
    [tex]
    U^\mu = -\frac{\sqrt{r}\,\sqrt{2\,m\,\left( 1-r\,K\right) +r-2\,m}}{2\,m-r}\partial_t +\frac{\sqrt{2}\,\sqrt{m}\,\sqrt{1-r\,K}}{\sqrt{r}}\partial_r
    [/tex]
    This is a geodesic because its product with the Killing vector [itex]\xi^a=[1,0,0,0][/itex], [itex]\xi^aU_a=\xi_aU^a = -\sqrt{1-2\,K\,m}[/itex] is constant. The product with the other Killing vectors is trivially zero.

    How does it satisfiy the equation you've got for [itex]\ddot{t}[/itex] ? I'm missing something very obvious :redface:
     
    Last edited: Oct 18, 2012
  17. Oct 18, 2012 #16

    pervect

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    Basically, I think all that needs to be done is to translate from the shorthand notation I used for the geodesic equation (without really explaining it, which I probably should have done). And a whole lot of math, which I'll leave to you :-)

    We start with some curve t(tau), r(tau), theta(tau), phi(tau), or ust (t,r,theta,phi) if we omit the dependence on tau

    Given your expression for the 4-velocity, we can write the following:

    [tex]
    t' = \frac{dt}{d\tau} = -\frac{\sqrt{r}\,\sqrt{2\,m\,\left( 1-r\,K\right) +r-2\,m}}{2\,m-r}
    [/tex]
    [tex]
    r' = \frac{dr}{d\tau} = \frac{\sqrt{2}\,\sqrt{m}\,\sqrt{1-r\,K}}{\sqrt{r}}
    [/tex]

    This gives t' and r', now all we need to compute is t''. We should also check the normalization of the four-velocity while we're at it, i.e that [itex]u^a u_a = -1[/itex]. Or +1, depending on your sign conventions. I wasn't quite sure if you said this was satisfied, but if it's not then your expression isn't really the 4-velocity we need.

    To get t'', we just use the chain rule. The only thing in our expression for t' that is a function of tau is r - which is really r(tau).

    [tex]t'' = \frac{d}{d\tau} t' = \frac{\partial}{\partial r} \left( -\frac{\sqrt{r}\,\sqrt{2\,m\,\left( 1-r\,K\right) +r-2\,m}}{2\,m-r} \right) \frac{dr}{d\tau} [/tex]

    So now we have t'', t' and r', and we just need to show it satisfies the geodesic equation I gave earlier.

    Which I assume it should but I haven't gone through the math. If it doesn't, I'm missing something too. You might need to use [itex]u^a u_a = -1[/itex] somewhere along the line to show that it satisfies the geodesic equation as well.

    Another thing you could try is writing E = (1-2m/r) t' and demonstrating that E is constant. Which would require using the relationship about [itex]u^a u_a[/itex] being constant, I htink.
     
    Last edited: Oct 18, 2012
  18. Oct 19, 2012 #17

    Mentz114

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    Pervect, thanks a lot. I did those calculations earlier and botched them - hence my question. With the added motivation that I was doing the right thing I found the error and now it all works out exactly right.

    The radial equation has given me a small insight. The radial acceleration calculated via the geodesic equations (absolute derivative ?) is in two parts, [itex]d \dot{r}/d\tau[/itex] and [itex]\Gamma^r_{ab}\dot{u}^a\dot{u}^b[/itex] ( please excuse mixed notations ) which must cancel for a geodesic. In this case they both give [itex] m/r^2[/itex], but with opposite signs.

    Now I can check if the geodesic congruences found via the kinetic decomposition also solve the geodesic equations. I have yet to find a physical example where it doesn't, which means nothing without some generalization.

    I did this calculation some time ago for the circular orbital congruence and in that case all the geodesic equations come out as 0=0. I suspect this is because there's a Killing vector in the [itex]\phi[/itex] direction.
     
    Last edited: Oct 19, 2012
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