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Far Gravitational field of a moving mass

  1. Nov 9, 2004 #1

    pervect

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    I've been trying to figure this one out for a while, I could use a bit of help. I think I'm getting close.

    If we consider the metric

    (1+2*m/R)*(dx^2+dy^2+dz^2) + (-1+2*m/R) dt^2, we have the metric of the nearly-Newtonian far-field of a body with mass m.

    If the mass isn't moving, we can write R = x^2 + y^2 + z^2

    We can "boost" the solution by substituting x = gamma(x'+vt'), t = gamma(t'+vx'). (At least I think we can do this safely, being interested only in the far field.)

    This makes R, which was originally a function of x,y,z, now a function of x',y,z,t'

    Finding the resulting Christoffel symbols of this metric is not too bad (with the computer to help).

    Now it comes time to interpret them.

    [rewrite]
    I think [tex]\Gamma_{x'tt}[/tex] should do the job?
     
    Last edited: Nov 9, 2004
  2. jcsd
  3. Nov 10, 2004 #2
    Yes. You can do this. Its a simple background Lorentz transformation. To reassure yourself that this is legitimate take a look at Schutz's text.
    Makes sense since the field is no longer static as observed from the new frame of reference.
    Why do you need a computer to find them? Are the too difficult to calculate by hand?
    [/quote]
    Do the job? What job? The quantity you speak of, i.e. [tex]\Gamma_{x'tt}[/tex], represents the field strength at the particular location, i.e. the gravitational acceleration/force. But I would think that there are other terms, i.e. other Christoffel symbols. That this should be so can be seen from the fact that there is a flow of mass which can be thought of as a mass current. As is the case in electromagnetism this current represents a magnetic field, or in your case a gravitomagnetic field. This means that falling particles are swept to to the side as well as being accelerated towards the body.

    There should be a xt cross term in the resulting metric after the background Lorentz transformation. Such a cross term is an indication of frame dragging effects, i.e. a gravitomagnetic effect.

    It'd be worth your while to consider simpler cases which are more readily analyzable such as a long rod or a uniform gravitational field produced by a sheet of mass. I've done this in several cases in the past and it proved to be most enlightening.

    Pete
     
  4. Nov 10, 2004 #3

    pervect

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    Yes, I start out with ds^2 = (1+2m/R)(dxx^2+dyy^2+dzz^2) + (-1+2m/R)dtt^2, and after I susbstitue

    dxx := (dx+v*dt)/sqrt(1-v^2)
    dyy := dy
    dzz := dz
    dttt := (dt+v*dx)/sqrt(1-v^2)

    I wind up with

    [tex]
    \left( 1+2\,{\frac { \left( 1+{v}^{2} \right) m}{ \left( 1-{v}^{2} \right) R}} \right) dx^2+8\,{
    \frac {vm dx dt}{ \left( 1-{v}^{2} \right) R}}+ \left( 1+2\,
    {\frac {m}{R}} \right) dy^2+
    \left( 1+2\,{\frac {m}{R}} \right) dz^2+ \left( -1+2\,{\frac { \left( 1+{v}^{2} \right) m}{
    \left( 1-{v}^{2} \right) R}} \right) dt^2
    [/tex]

    There are a whole raft of Christoffel symbols, but the ones for xtt, ytt, and ztt are:

    [tex]
    \Gamma_{xtt} =
    {\frac { \left( 1+{v}^{2} \right) m{\frac {\partial }{\partial x}}R
    \left( x,y,z,t \right) }{ \left( -1+{v}^{2} \right) \left( R \left(
    x,y,z,t \right) \right) ^{2}}}
    [/tex]

    [tex]
    \Gamma_{ytt} =
    {\frac { \left( 1+{v}^{2} \right) m{\frac {\partial }{\partial y}}R
    \left( x,y,z,t \right) }{ \left( -1+{v}^{2} \right) \left( R \left(
    x,y,z,t \right) \right) ^{2}}}
    [/tex]

    [tex]
    \Gamma_{ztt} =
    {\frac { \left( 1+{v}^{2} \right) m{\frac {\partial }{\partial z}}R
    \left( x,y,z,t \right) }{ \left( -1+{v}^{2} \right) \left( R \left(
    x,y,z,t \right) \right) ^{2}}}
    [/tex]

    Because the metrix is nearly flat, the symbols for [tex]\Gamma^x{}_{tt}[/tex] are only slightly different, [tex]\Gamma_{xtt} [/tex] reduces to F=GM/R^2, the newtonian result when v=0, the other ones don't quite.

    Substituing
    [tex]
    R = \sqrt {{\frac { \left( x+vt \right) ^{2}}{1-{v}^{2}}}+{y}^{2}+{z}^{2}}
    [/tex]

    gives the final result
     
    Last edited: Nov 10, 2004
  5. Nov 10, 2004 #4

    pervect

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    A few things I wanted to add:

    With the metric

    ds^2 = (1+2m/R) (dx^2+dy^2+dz^2)+(-1+2m/R)dt^2

    [tex]
    \Gamma^x{}_{tt} = -\frac{m x}{R^2(R+2m)}
    [/tex]

    while

    [tex]
    \Gamma_{xtt} = -\frac{m x}{R^3}
    [/tex]

    That's why I used [tex]\Gamma_{xtt} [/tex], so that the result would be the Newtonian answer [tex]\frac{GM}{R^2} \hat r [/tex] for a stationary mass.

    At the moment I'm only (hopefully!) calculating the field for stationary objects, so I'm ignoring the gravitomagnetic components.

    The final results for the boosted metric (i.e. the moving mass) that I'm getting are:

    [tex]
    \Gamma_{xtt} =-\frac{ \left( 1+{v}^{2} \right) m \left( x+vt \right)}{ \left( {\frac { \left( x+vt \right) ^{2}}{1-{v}^{2}}}+{y}^{2}+{z}^{2} \right) ^{3/2}
    \left( 1-{v}^{2} \right)^{2} }
    [/tex]

    [tex]
    \Gamma_{ytt} = -\frac{ \left( 1+{v}^{2} \right) m y}{ \left( {\frac { \left( x+vt \right) ^{2}}{1-{v}^{2}}}+{y}^{2}+{z}^{2} \right) ^{3/2}
    \left( 1-{v}^{2} \right) }
    [/tex]

    [tex]
    \Gamma_{ztt} = -\frac{ \left( 1+{v}^{2} \right) m z}{ \left( {\frac { \left( x+vt \right) ^{2}}{1-{v}^{2}}}+{y}^{2}+{z}^{2} \right) ^{3/2}
    \left( 1-{v}^{2} \right) }
    [/tex]

    Note that the field doesn't always point towards the mass. This can be seen by considering what happens when

    x=y=a, z=t=0

    [tex]\Gamma_{xtt}[/tex] and [tex]\Gamma_{ytt}[/tex] would have to be the same if the field pointed towards the mass, but they are different by a factor of 1/(1-v^2)
     
  6. Nov 16, 2004 #5
    I've printed this out and will be taking it home to study.

    But in a quick glance - So far what you wrote seems somewhat reasonable except for a few points. One point is "Note that the field doesn't always point towards the mass."

    If I were you I'd look at this in terms of gravitomagnetism. Compare the velocity dependant acceleration to the acceleration when v = 0. The bodies should, in some respects, behave like charged particles. If the source of the field is moving then there should be a gravitomagnetic field. A test particle at rest should experience only a gravitoelectric field. If the test particle is moving then there will be both a gravitomagnetic and gravitoelectric force on it. These will change the direction of the force from what it was when the test particle was at rest. I.e. compare your expected predictions with the EM analogy. Think of charge as you would think of relativistic mass (this is where rel-mass comes into play in GR - See Rindler's new SR/GR text on this point). What directions do you expect the charged particle to accelerate in a direction which is independant of its velocity.

    I'll have more to say on this point when I actually examine all of this at home when I can look into the derivations and details.

    Pete

    ps - I'll be back online soon since I can sit for short periods now. I'd like to discuss this example much more then. Good example by the way!
     
  7. Nov 17, 2004 #6
  8. Nov 17, 2004 #7

    pervect

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    Assuming that the gamma in your eq 9b is [tex]\gamma_v[/tex] - it's written as [tex]\gamma_s[/tex] and also assuming that in 9c [tex] \Phi = \gamma_v \Phi ' [/tex] not [tex]\gamma_v \Phi_0[/tex] I get:

    [tex]\Phi= \frac{G M_0 (1+\beta^2)}{(1-\beta^2) R}[/tex]

    [tex]R = \sqrt{\frac{(x-vt)^2}{1-\beta^2}+y^2+z^2}[/tex]

    Given this assumption, I find my results agree with yours by taking the force components as [tex]\frac{\partial \Phi}{\partial x}, \frac{\partial \Phi}{\partial y}, \frac{\partial \Phi}{\partial z}[/tex] , using geometric units, and reversing the sign of v. (Because I use geometric units, my -v is your [tex]\beta[/tex]).


    Unfortunately this still means that if one evaluates the force components at x = y = a, z=t=0, that the x force component doesn't equal the y force component.
     
  9. Nov 18, 2004 #8

    sal

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    Pete asked me to suggest that you look in Jackson. He says the transformation of the EM field yields the same equations as the transformation of the G-force.

    (Don't shoot me, I'm only the messenger -- haven't even been following this thread and I don't personally own a copy of Jackson. Talked to Pete on the phone today, tho -- he should have his internet connection back in a day or two.)
     
  10. Nov 18, 2004 #9
    Thanks sal. I happened to be downtown today unexpectedly.
    I don't see why this is a problem. I wouldn't expect the x and y components to be the same. If they were then the field would be spherically symmetrical and it isn't.

    See Jackson's text on the field of a moving charge. You'll see similar equations. I'll try to up date my page to demonstrate this more clearly.

    Pete
     
  11. Nov 18, 2004 #10

    pervect

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    One big giant "Huh"?????

    If a force points towards the center, then the force is some scalar times the radius vector.

    [tex] f = k \hat r [/tex]

    In cartesian coordinates, the x component of the force is kx, and the y component of the force is ky. When x=y, kx=ky. This is true via the defintion of what it means to be a central force.

    I'm also fairly sure that Jackson makes a point of saying that the gravitational force is NOT the same as the electromagnetic force. As I recall his argument was very general, though, he just pointed out that the source of gravity, the stress energy tensor, had a different dimensionality than the current density J.

    The interesting thing to me is that the force in this case is the gradient of a potential. That's something I wouldn't have noticed without your approach. I'm still trying to justify, though, in my own mind, exactly why this is true (but it's obvious from differentiating your expresion that it is true).
     
  12. Nov 18, 2004 #11

    pervect

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    Upon review, it turns out the Christoffel symbols I've been automatically computing are backwards from the standard convetion. (Dont ask me why the program was written that way, I think it's terribly confusing). Anyway I think we should have



    [tex]\Gamma_{xtt} = \frac{1}{2} (-\frac{\partial g_{tt}}{\partial x} + 2\frac{\partial g_{xt}}{\partial t} )[/tex]

    which IS NOT just the gradient of gtt, so the force isn't just the gradient of a potential because of the dx*dt terms in the metric.

    (See the other thread about Christoffel symbols).

    Anyway, more when I think I may have a better answer.
     
  13. Nov 19, 2004 #12

    pervect

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    Taking a totally different approach, I think I can conclude that there is something definitley wrong with the results to date, though I don't see yet what is going wrong.

    Let's consider a completely different problem, a particle falling directly towards a black hole in a schwarzschild metric with no angular momentum.

    We know that the equations of motion are (dr/dtau) = sqrt(E^2 - 1 + 2M/r)

    Let f = dr/dtau. Then df/dr = df/dtau * dtau/dr, or

    df/dtau = d^2r/dtau^2 = df/dr * f

    Working this out, one comes up with the result f * df/dr = -m/r^2
    This gives the result that dr/dtau^2 = -m/r^2


    To convert this into the acceleration as seen from a system co-moving with the particle, we only have to correct for things that affect 'r', since the result is already in terms of proper time. These effects are the length contraction which lowers the acceleration by a factor of sqrt(1-v^2), and effects due to g_rr component of the schwarzschild metric, which we can ignore far away from the mass.

    This should give us

    f = (1/gamma)*m/r^2 = gamma*m/r'^2, where r' = (1/gamma)*r, the length contracted distance to the mass.

    Even if I've managed to get the factor of gamma in the wrong place somehow, there's no way to allow for terms like (1+v^2) in the numerator

    [add]
    At least this term shouldn't exist in the radial component of the force - there's actually some reason to suspect that a term like this might exist in the non-radial components.
     
    Last edited: Nov 19, 2004
  14. Nov 19, 2004 #13

    pervect

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    Some more comments/musings....

    As far as whether or not the gravitational force looks like an electromagnetic force, the question boils down to whether a body attracted by another body by an electrostatic force moves in the same path as a body attracted to another body by a gravitational force.

    At low velocities this is defintely a good approximation, but at high velocities I believe the anology fails. Light deflects twice as much as it should according to a Newtonian calculation, for example. Exploring this question is part of the motivation that inspired the problem in the first place. I do not believe the electrostatic answer solves the problem - I don't think it gives the right geodesic paths.

    The electrostatic result is online

    here

    Correcting the expression for [tex]\Gamma_{xtt} [/tex] for the computer's re-shuffling of the order of the components did not, unfortunately, help the original calculation any - with the addition of the additonal term, the bizarre results had a (1-3v^2) term in them, which is total nonsense (it changes sign!).
     
  15. Nov 22, 2004 #14
    That is correct. And if you look at Jackson you'll see that this is the case and that the function in front of r is a function of the field point and depends only on the angle between r and v.

    I don't recall him saying that but in any case I was implying that there is an analogy between EM and GR in that the equations in the weak field limit are similar. What I said cannot be taken to mean that the grav-force is the same as the Lorentz force. If that was true then there wouldn't be two forces (GR and EM) there would be just one. Analogy does not mean exact identity. No analogy that exists is an identity. The EM analogy is good because it gives you a very strong feeling of quantitative behaviour and you can use it to verify your results using your EM intuition. It doesn't matter that the equations are not identical to the EM equations, i.e. if the field gets strong it simply means that the quantitative results are a bit wrong. But don't forget that the exact relationship between the gravitational force G_k and the Einstein potentials g_ab is

    G_k = (1/2)m g_ab,k V^a V^b

    where

    V = (c, v_x, v_y, v_z) and m = m0dt/dT

    That statement can be decieving. In SR inertial mass (i.e. "relativistic" mass) is a function of pressure. In GR active gravitational mass is also a function of pressure. Its for reasons like this that a second rank tensor is required. In the weak field limit the space-space component of Tab[/ab] is neglected. The (relativistic mass) density plays the role of "gravitational charge" and momentum plays the role similar to that played by current in EM.
    That's the nice thing about it. This is also true in the strong field of a uniform g-field.

    There is an article that you should read.

    Analogy Between General Relativity and Electromagnetism for Slowly Moving Particles in Weak Gravitational Fields, Edward G. Harris, American Journal of Physics, Vol. 59 No. 5, May 1991.

    Let me get back to you later on the rest. I should get my connection back in a few weeks and will have more time for this.

    Pete
     
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