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Schwarzschild/post-Newtonian circular orbital velocity

  1. Apr 13, 2013 #1
    The post-Newtonian expansion at the first post-Newtonian 1PN level under Schwarzschild conditions, a test-body in a spherically symmetric gravitational field, gives the following expression for the gravitational acceleration of the test-body: [tex]\frac{{\rm d}\bar{v}}{{\rm d}t}=-\frac{GM}{r^2}\left(1-\frac{4GM}{rc^2}+\frac{v^2}{c^2}\right)\hat{r}+\frac{4GM}{r^2}(\hat{r}{\cdot \hat{v}})\frac{v^2}{c^2}\hat{v}.[/tex]

    For circular orbits this gives (setting the right hand terms equal to the centripetal acceleration ##v^2/r##) an orbital velocity of:

    [tex]v=\sqrt{\frac{GM}{r}\frac{[1-4GM/(rc^2)]}{[1-GM/(rc^2)]}}[/tex]

    Using the same procedure on the post-Newtonian expansion taken to the third post-Newtonian level, (see for instance http://arxiv.org/abs/gr-qc/0209089) one gets the orbital velocity of circular orbits:

    [tex]v=\sqrt{\frac{GM}{r}\left[1-\frac{4GM}{rc^2}+9\left(\frac{GM}{rc^2}\right)^2-16\left(\frac{GM}{rc^2}\right)^3\right]}\left(1-\frac{GM}{rc^2}\right)^{-1/2}.[/tex]

    Now in Schwarzschild coordinates the orbital velocity really should be ##v=\sqrt{GM/r}##, however the post-Newtonian expansion is derived using isotropic coordinates so maybe the orbital velocity for circular orbits should be different? Now my personal alternative to the post-Newtonian expansion expression for gravitational acceleration looks like:

    [tex] \frac{{\rm d}\bar{v}}{{\rm d}t}=-\frac{GM}{r^2}(\hat{r}\cdot\hat{v})\left(1-3\frac{v^2}{c^2[1-2GM/(rc^2)]}+\frac{v^4}{c^4[1-2GM/(rc^2)]^2}\right)\hat{v}\nonumber \\+\frac{GM}{r^2}(\hat{r}\times\hat{v})\times \hat{v}.\hspace{4mm}[/tex]
    (for details: http://vixra.org/abs/1303.0004)
    which gives the classical orbital velocity for circular orbits, ##v=\sqrt{GM/r}##.

    Now, which expression for gravitational acceleration results in the more correct expression for the orbital velocity of a body in circular orbit?

    I tried to compare my expression to the post-Newtonian when I sent my paper in for peer-review. However, the reviewers report stated:

    Now, are the reviewers arguments valid? Is the post-Newtonian expansion to be treated like some kind of golden standard that can not be put into question?
     
  2. jcsd
  3. Apr 13, 2013 #2

    pervect

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    Do you have a reference for this? Or is this something you calculated yourself?

    If there is a reference for this, it would be helpful to include it, I have (as usual) some doubts as to how you've defined things.
     
  4. Apr 13, 2013 #3
    That would be equation 4-61 on page 4-42 in the official documentation of the Jet Propulsion Laboratory:

    http://descanso.jpl.nasa.gov/Monograph/series2/Descanso2_all.pdf

    I included the classical Newtonian term.
     
    Last edited: Apr 13, 2013
  5. Apr 13, 2013 #4

    PeterDonis

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    No, it shouldn't. That's the large-r approximation. The correct exact expression in Schwarzschild coordinates is:

    [tex]v = \sqrt{\frac{GM}{r - \frac{2 G M}{c^2}}}[/tex]
     
  6. Apr 14, 2013 #5
    Darn. This paper (http://www.astro.umd.edu/~miller/teaching/astr498/lecture10.pdf) expression 6 and this paper (http://arxiv.org/pdf/1303.4068v1.pdf) expression 3.6 states that the correct exact expression for angular velocity of a body in circular orbit in Schwarzschild coordinates is:

    [tex]\Omega_s=\sqrt{\frac{GM}{r^3}}[/tex]

    This strongly suggests that ##v=\sqrt{GM/r}## even for small r, but maybe this is incorrect?

    You can supposedly get to the post-Newtonian expression for gravitational acceleration (the first expressios I wrote above) from the "Schwarzshild isotropic one-body point mass metric." What is the correct expression for the orbital velocity of a circular orbit in isotropic coordinates?

    Note that according to the post-Newtonian expression the orbital velocity is less than the classical value for small r but in Schwarzschild coordinates, according to the formula you gave me, the orbital velocity is larger than the classical value for small r.
     
    Last edited: Apr 14, 2013
  7. Apr 14, 2013 #6

    Bill_K

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    I believe that for a circular orbit in Schwarzschild, (dφ/dt)2 = GM/r3. In these terms, Kepler's Law holds exactly.
     
  8. Apr 14, 2013 #7

    Mentz114

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    I think it is ##{\dot{\phi}}^2=\frac{m}{{r}^{2}\,\left( r-3\,m\right) }##. Is what you've stated a large ##r## value ? ( I'm writing ##m## for your ##GM##, obviously).

    [eidt]I just noticed you have ##d\phi/dt## and I've written ##\dot{\phi}=d\phi/d\tau##.
     
    Last edited: Apr 14, 2013
  9. Apr 14, 2013 #8

    PeterDonis

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    It is incorrect, because the angular velocity you wrote down is with respect to coordinate time, but the orbital velocity formula I wrote down is with respect to proper time.

    More precisely, the orbital velocity formula I wrote down is the relative velocity at which a "hovering" observer at radial coordinate r would measure the orbiting observer to be passing him.

    As Mentz114 noted, there is also an angular velocity formula with respect to proper time, which is the *angular* velocity at which a "hovering" observer at radial coordinate r would measure the orbiting observer to be passing him.
     
  10. Apr 14, 2013 #9

    Bill_K

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    This is a definition of the word "incorrect" with which I was previously unaware. :wink:

    When doing post-Newtonian approximation, which time you use to write the result is part of the choice of gauge. You can certainly use any one of the three times we've mentioned.
     
  11. Apr 14, 2013 #10

    PeterDonis

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    But if you use coordinate time, the "velocity" you get is not really a velocity, is it? By which I mean, it's not going to correspond to anybody's actual measurement. Also, it won't be limited to less than the speed of light, which makes its interpretation as a "velocity" problematic (and leads to lots of threads here on PF that go on and on because that point is not understood).

    The velocity with respect to proper time (more precisely, with respect to a "hovering" observer's proper time) does correspond to an actual measurement, and it is limited to less than c, as any directly measured relative velocity must be.

    Yes, the word "incorrect" may have been a little strong. The OP used "more correct", so perhaps I should have used "less correct" to compensate. :wink:
     
  12. Apr 14, 2013 #11
    In a simulation, coordinate time makes most sense, typically what you want to do is to is to calculate the positions and velocities of celestial objects at some point in the future, from your estimates of the current positions and velocities.

    Ideally, you could find an expression for the gravitational acceleration that prohibits objects to be accelerated to superluminal speeds gravitationally. Using the expression for gravitational acceleration in the spherically symmetric case from the post-Newtonian exapansion at the first post-Newtonian level, seen above, I think you could get superluminal speeds easily, but the post-Newtonian expansion is not supposed to work in really strong gravitational fields.

    Note that the speed of light in the gravitational field varies in a different manner in isotropic coordinates (the set of coordinates that the post-Newtonian expansion is said to be based on) compared to how it varies in Schwarzschild coordinates.

    Isotropic coordinates on Wikipedia

    Now, to get this thread back on track, what is the correct expression for the orbital velocity of a body in circular orbit if isotropic coordinates (and not Schwarzschild coordinates) are used?
     
  13. Apr 14, 2013 #12

    pervect

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    OK, this is PPN, which is similar to isotropic Schwarzschild. But not exactly the same.

    My concerns:

    The notation you use is confusing and not the same as the papers.
    [itex]r[/itex] in the paper is the PPN coordinates r = (t,x,y,z) with the metric given in eq (2-1) through (2-6)

    It really IS clearier if you write [itex]\dot{r}[/itex] rather than v, and [itex]\ddot{r}[/itex] than a, and explain (as the paper does) that differentiation is being done with respect to coordinate time.

    And it's rather helpful if you write out the r vector as the paper does in (2-7). (Giving the metric as the paper does helps enormously too).

    Similarly [itex]\dot{s_j}[/itex] is PROBABLY what you meant when you write v. See (2-9). Also see the notation for [itex]r_{12}[/itex] in 2-8.

    Perhaps I'm wrong about where you went wrong on the notation - in which case what you're saying doesn't make any sense at all to me :-(.

    That's my best (and possibly misguided) effort to interpret what you "really mean", which is needed due to the lack of details and definitions in your post.

    In comparing this to ISOTROPIC Schwarzschild (it's not the standard Schwarzschild!), I still have some concerns.

    If we imagine only one body, the sun, I read the paper as using a heliocentric coordinate time. This will have a different scale factor for the metric and all coordinate clocks than the usual isotropic scharzschild metric.

    If this is correct, I'd expect g_44 not to go to infinity as r-> infinity, I'm really ot sure if it does or not.

    In addition, [itex]r_{ij}[/itex] isn't the same as the isotropic r coordinate. It doesn't make any sense to use the metric in the dot product because the metric varies with position, so I'm pretty sure that [itex]r_{ij}[/itex] is just sqrt((x_j - x_i)^2 + (y_j - y_i)^2 + (z_j - z_i)^2).

    I don't know what the correct expression for circular orbits, is, but v^2/r
    (or as the paper would put it, assuming body 1 is the heavy body and body 2 is the moving test mass, [itex]\dot{s_{2}}^2 / r_{12}[/itex] is almost certainly wrong. It's probably good to Newtonian order (by design), but there's no way it's going to explain the observed "double deflection" of light, so additional correction factors are almost certainly needed.

    You'd need to go back to the original metric and get the geodesic equations and solve them to really answer this properly, I think.
     
    Last edited: Apr 14, 2013
  14. Apr 14, 2013 #13

    Mentz114

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    Using the first type of isotropic coords from the Wiki page, I get for the circular orbit with ##\theta=\pi/2##

    ##{\dot{\phi}}^2=\frac{2048\,{r}^{3}\,r_s}{{\left( r_s+4\,r\right) }^{4}\,\left( {r_s}^{2}-16\,r\,r_s+16\,{r}^{2}\right) },\ \ {\dot{t}}^2=\frac{{\left( r_s+4\,r\right) }^{2}}{{r_s}^{2}-16\,r\,r_s+16\,{r}^{2}}##

    I'm pretty sure this is correct, but it's late and I can't check it until tomorrow.
     
    Last edited: Apr 14, 2013
  15. Apr 14, 2013 #14

    pervect

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    If we have one massive body of mass M, at rest, then the metric from the paper should be:

    coordinates: (x,y,z,ct)

    [tex]
    g_{11} = g_{22} = g_{33} = -1 + \left( \frac{2 \gamma} {c^2} \right) \left( \frac{GM}{r_{12}} \right )
    [/tex]
    [tex]
    g_{44} = 1 - \left( \frac{2} {c^2} \right) \left( \frac{G M}{r_{12}}\right) + \left( \frac{2 \beta}{c^4} \right) \left( \frac{G M}{r_{12}}\right) ^2 - \left( \frac{1}{c^4} \right) \left( G M \frac{\partial^2 r_{12}}{\partial t^2} \right)
    [/tex]

    where [itex]r^2_{12} = x^2 + y^2 + z^2[/itex].

    The metric should be a function of x,y,z (and not t, because it's stationary). Because of radial symmetry, it should only be a function of r_12. In that context I can't make any sense out of the last term with the partial derivative.

    MTW's seems to indicate that this term is zero as well.

    I can also get this as a series approximation from the isotropic Schwarzschild metric assuming [itex]\beta = \gamma = 1[/itex]


    I'll try to work out cirucular orbits in this metric later, in the meantime we might see if there's general agreement that this is in fact the "right" metric , in particular the last term in the original paper cited isn't making any sense to me at the moment.
     
    Last edited: Apr 15, 2013
  16. Apr 15, 2013 #15

    pervect

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    I"m pretty sure i can guess the answer, but your "dot" represents diffentiation with respect to proper time, correct?
     
  17. Apr 15, 2013 #16

    Mentz114

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    Yes. I'm bothered by the fact that the angular velocity seems to increase with r. Could it be because of the weird r coordinate ?
     
  18. Apr 15, 2013 #17

    pervect

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    I'll try to post again when it doesn't eat my posts when I hit "Go Advanced". Though in not as much detail
     
  19. Apr 15, 2013 #18

    pervect

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    OK, basically I converted the cartesian like metric into a polar one, using the wiki suggested substitutions

    (1)
    [tex]
    x = r_1 \, \sin\theta \, \cos\phi \,, \quad y = r_1 \, \sin\theta \, \sin\phi \,, \quad z = r_1 \, \cos\theta
    [/tex]

    I then considered a symbolic metric for my convenience

    f(r) dt^2 - g(r) dr^2 - h(r) d phi^2

    This is the metric of the equatorial plane - theta=0. As we are just interested in circular orbits, this simplifies the calculations.

    Setting dr / dtau = 0 made it immediatly obvious that dt / dtau and dphi / dtau were constant.

    The ratio of (d phi)/ (d tau) / (dt / dtau) = d phi / dt came direclty out of the geodesic equation for d^2 r / d tau^2 = 0.

    [tex]
    \left( \frac{d\phi}{dt} \right)^2 = \left( \frac{\partial f}{\partial r} \right) / \left( \frac{\partial h}{\partial r}\right)
    [/tex]

    [add]This is because the geodesic equation for r for the above metric basically becomes

    [tex]
    \Gamma^r{}_{tt} \left( \frac{dt}{d\tau} \right)^2 + \Gamma^r{}_{\phi\phi} \left( \frac{d \phi}{d \tau} \right)^2 = 0
    [/tex]

    when [itex] dr / d \tau = 0 [/itex] which imples [itex]d^2 r / d \tau^2=0[/itex] as well.

    Using the polar form of the rectangular metric (which I don't feel like reposting at the moment but will if anyone asks) gave the following.

    [tex]
    r^2 \left( \frac{d\phi}{dt} \right)^2 = s^2 = \frac{GM}{r} \left( \frac{r-2GM/c^2}{r-GM/c^2} \right) \approx \frac{GM}{r} \left( 1 - \frac{GM}{c^2r} \right)
    [/tex]

    (1) justifies and formalizes the conversion back and forth from polar PPN to cartesian PPN coordinates.

    r^2 is just as simple as x^2 + y^2 + z^2, though (I may have made mistaken remarks about that in the past).

    The polar form of the metric is essentially a series expansion of the isotropic method given in the wiki

    http://en.wikipedia.org/w/index.php?title=Schwarzschild_metric&oldid=550530310

    if you substitue for M in the proper place.

    s^2 should be PPN's notion of "velocity" if your coordinate clock ticks proper time at infinity. You'll need more adjustments if your ephermis clock keeps TAI time or TCB time. I"m afraid I'm not sure what standard people actually use for space tracking and such. The paper referenced earlier in the thread talked a lot about different timescales, but I didn't get a sense of which one was actually used .

    Knowing g_44 of the PPN metric would settle the issue, but there's some terms in the original paper I don't understand, so I don't consider the issue settled.

    s^2 is "just" (the square of) a coordinate velocity, but it's a coordinate velocity in a well defined coordinate system - the PPN coordinate system. (Well, it's the PPN velocity modulo the question about where , in this coordinate system, do coordinate clocks keep proper time, which isn't settled to my satisfaction yet.)


    Also, I can't guarantee that any of this was error free, take it for what it's worth.
     
    Last edited: Apr 16, 2013
  20. Apr 16, 2013 #19
    Thank you. I am not qualified to check your calculations. Note that the expression for ##s^2## for a body in circular orbit given above by Pervect is:

    [tex]s^2= \frac{GM}{r}\frac{1-2GM/(rc^2)}{1-GM/(rc^2)}[/tex]

    which is similar to what you get for the squared velocity if you use the post-Newtonian expansion at the first post-Newtonian level, as I stated in the first post of the thread:

    [tex]v^2=\frac{GM}{r}\frac{1-4GM/(rc^2)}{1-GM/(rc^2)}[/tex]

    Now I am using "v" again and not "s" but this is the interpretation people actually using the post-Newtonian expansion seem to be doing on a regular basis.

    There is a difference in that there is a factor ##1-2GM/(rc^2)## in the first expression but a factor of ##1-4GM/(rc^2)## in the last expression. I did not quite follow. Is this difference because of the fact that deriving the post-Newtonian expansion some other form of isotropic coordinates are used than the isotropic coordinates found on Wikipedia?

    The JPL documentation says on page 4-42 that the post-Newtonian expression is found by using "the Schwarzschild isotropic one-body point-mass metric, which has been expanded, retaining all terms to order 1/c^2". I do not now how throwing away higer order terms already in the metric influences the final expression for gravitational acceleration and thus also the final expression for orbital velocity of a body in circular orbit.
     
  21. Apr 16, 2013 #20

    pervect

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    We're back to "what do you mean when you say v"? "The same thing as all those other people" will work, I suppose, if you can give some published examples of what "the other people" meant when they used it.

    The paper you cited doesn't seem to be one of those that use v. (Unless you count (2-21), but that just seems to be s^2 in the context of the velocity of an earth tracking station)

    On a closely related note, your source for your expression you quote

    gives an expression for [itex] \ddot{r}[/itex] as a function of r, , [itex]\dot{r}[/itex], r, and [itex]\dot{s}[/itex]. It's not the same as your expression at all, so the source for your expression remains a mystery.
     
    Last edited by a moderator: May 6, 2017
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