Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homogeneous gravitational field and the geodesic deviation

  1. Jan 24, 2013 #1
    In General Relativity (GR), we have the _geodesic deviation equation_ (GDE)

    $$\tag{1}\frac{D^2\xi^{\alpha}}{d\tau^2}=R^{\alpha}_{\beta\gamma\delta}\frac{dx^{\beta}}{d\tau}\xi^{\gamma}\frac{dx^{\delta}}{d\tau}, $$

    see e.g. [Wikipedia](http://en.wikipedia.org/wiki/Geodesic_deviation_equation) or [MTW](http://en.wikipedia.org/wiki/Gravitation_(book)).

    Visually, this deviation can be imagined, to observe the motion of two test particles in the presence of a spherically symmetric mass.In the case of a homogeneous gravitational field, such as a geodesic deviation should not be, because acceleration due to gravity equal at every point.

    Clearly, such a field can be realized in a uniformly accelerated frame of reference. In this case, all components of the curvature tensor will be zero, and the equation (1) correctly states that the deviation will not be surviving.

    But if a homogeneous field will be created by infinite homogeneous flat layer, in this case, the components of the curvature tensor are non-zero, then by (1) will be a deviation. It turns out that such fields, even though they are homogeneous, can be discerned.
    I think this situation is paradoxical. Is there an explanation?
     
  2. jcsd
  3. Jan 24, 2013 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi sergiokapone! :smile:

    yes, the geodesic deviation in a uniform field should be zero
    are they?
     
  4. Jan 24, 2013 #3
    Equation of motion in the metric of this type:
    $$ds^2=(1-8gz)^{-1/4}dt^2-(1-8gz)^{1/2}(dx^2+dy^2)-(1-8gz)^{-5/4}dz^2$$
    in the Newtonian limit becomes as equation of motion in homogeneous field. However, this metric has a non-zero curvature. The metric of this type was obtained in 1971 by ukrainian astronomer Bogorodsky, in the case of an infinite plane. The article, unfortunately, in Russian.
    In English, a derivation Bogorodsky's metrics can be found in this article http://arxiv.org/pdf/gr-qc/0202058.pdf
    In his article, Bogorodsky gets two solutions, one of them has no curvature, and enters the corresponding transformation of the Minkowski metric, and another - a curvature. A very interesting fact.
     
  5. Jan 24, 2013 #4

    Bill_K

    User Avatar
    Science Advisor

    If Bogorodsky had checked the literature (always a good idea!) he would have found that such very simple solutions depending on only one variable z had been enumerated by Kasner back in 1925.
     
  6. Jan 24, 2013 #5
    But it does not matter now. I wonder why, in a uniform field of this type is the deviation of geodesics.
     
  7. Jan 24, 2013 #6

    Bill_K

    User Avatar
    Science Advisor

    Isn't it just because the Riemann tensor is nonzero? Do you want us to come up with an intuitive reason why it's nonzero?

    Well, by reflection symmetry there are geodesics in which a particle "falls" in the z direction, keeping the coordinate values x = const and y = const. Then from the (1−8gz)1/2(dx2+dy2) term in the metric, the distance between two such neighboring particles changes as they fall. (It would be much harder to explain if it did not!)
     
  8. Jan 24, 2013 #7
    Ok, can we call such a field as homogeneous in the GR-sense?
     
  9. Jan 24, 2013 #8

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You may find this relevant: http://www.lightandmatter.com/html_books/genrel/ch07/ch07.html#Section7.4 [Broken]

    You say "homogeneous," but do you really want something isotropic as well as homogeneous?

    The Ionescu paper's description of Bogorodskii's work is completely coordinate-based. To some extent this is inevitable in discussions of the GR equivalent of a Newtonian uniform field, since the Newtonian notion of a field is coordinate-dependent. However, one really wants to classify especially symmetric spacetimes according to their *intrinsic* symmetry, not their symmetry when expressed in some coordinates.

    A general Kasner metric lacks the high degree of symmetry we'd like for a truly uniform field. By fiddling with metrics of the Kasner form, you can get the Petrov metric described in the link above. The Petrov metric is basically the vacuum spacetime with the highest intrinsic symmetry you can get without having it be Minkowski space.

    The Newtonian result is that an infinite, flat sheet of mass has a uniform field on both sides. But I don't see any reason to think that anything similar holds in GR. It's not even obvious how to state such a notion in GR. For example, what does it mean for the sheet to be "flat?" A cylinder of dust is intrinsically flat, and the Petrov metric can be interpreted as the field of a certain rotating cylinder of dust. "Uniform field" would have to be translated into some appropriately coordinate-independent langage in GR, i.e., it would have to become something like a statement about the number of Killing vectors. This is the kind of criterion on which the Petrov metric becomes the best candidate for a uniform field in GR.
     
    Last edited by a moderator: May 6, 2017
  10. Jan 24, 2013 #9

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    I throw this metric into GRTensor and find all the components of R_{ab} are zero. (Except perhaps at the origin, where I suspect R is techinically undefined). R_{abcd) isn't zero, though.

    [add]
    So this is in fact a solution to EInstein's equation, and it does have curvature. But I have to agree that it's unclear if it represents an "infinite flat sheet". For instance, you see tidal forces in the x directions (R_txtx is nonzero). That's not something you'd expect in an infinite flat sheet.
     
    Last edited: Jan 24, 2013
  11. Jan 24, 2013 #10
    bcrowell, thank you! The question becomes clear to me.

    Why R_txtx? I thought that the tidal force along the x-direction must meet the components $$R^x_{yxy}$$ and $$R^x_{zxz}$$ which, according to my calculations is non-zero.
     
  12. Jan 24, 2013 #11

    Mentz114

    User Avatar
    Gold Member

    I agree with your tidal calculation. The values are ( for a hovering observer)

    [tex]T_{xx}=T_{yy}=\frac{2g^2}{\sqrt{1-8\,g\,z}}[/tex]

    But I also found 2 Killing vectors, one each in ∂x and ∂y directions. I guess that means the potential doesn't change in those directions ?
     
  13. Jan 24, 2013 #12

    Mentz114

    User Avatar
    Gold Member

    If we consider a stationary frame, then only the t-component of the 4-velocity U is non-zero and in the contraction Tab= RdacbUcUd we need to look only at R0x0y as Pervect has done ( some errors with the indexes there, but you know what I mean ...)
     
    Last edited: Jan 24, 2013
  14. Jan 24, 2013 #13
    Oh, I see.
     
  15. Jan 24, 2013 #14
    Yes, I think so. The px and py components of momentum are conserved, like in Newtonian homogeneous field.
     
  16. Jan 24, 2013 #15

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    I suppose that depends what you mean by potential. But you're right about it being a killing vector, thus [itex]g^{xx} dx / d\lambda[/itex] should be constant along a geodesic.

    This is easily confirmed, if we let our geodesic be [itex]t(\lambda)[/itex], [itex]x(\lambda)[/itex], [itex]y(\lambda)[/itex], [itex]z(\lambda)[/itex]

    then

    [tex]
    \frac{d}{d\lambda} \left[ \frac{\dot{x}}{\sqrt{1- 8 g z(\lambda)}} \right] = \frac{1}{\sqrt{1-8 g z(\lambda)}} \left[ \ddot{x} + \frac{4 \dot{x}\dot{z} } {\sqrt{1-8 g z(\lambda)} } \right]
    [/tex]

    where the rhs is zero because of the geodesic equation, thus insuring that the derivative is zero and
    [tex]\frac{\dot{x}}{\sqrt{1- 8 g z(\lambda)}}[/tex]

    is constant , and can be considered as the x-momentum.

    (I've worked this out more for the OP than Mentz, I hope it's somewhat clear. The "dots" represent derivatives with respect to [itex]\lambda[/itex].)

    Now I'm scratching my head about how there can be a tidal force, but I need to get back to the leaky faucet repair :-(.
     
    Last edited: Jan 24, 2013
  17. Jan 24, 2013 #16

    Mentz114

    User Avatar
    Gold Member

    Pervect, that's cool. The Killing vector

    [tex]K_\mu= C\sqrt{1-8\,g\,z}\ \partial_x[/tex] seems to tie in, with K.U = const ( U being the geodesic ). If we identify the constant C with the rest-mass m, we get the conserved momentum exactly.

    I solved the Killing equations [itex]K_{(a;b)}=0[/itex] to get K, and C is a constant of integration.
     
    Last edited: Jan 24, 2013
  18. Jan 24, 2013 #17

    Mentz114

    User Avatar
    Gold Member

    Right. Strange about the tidal forces. I must be misinterpreting something.
     
  19. Jan 24, 2013 #18
    Strange result. From geodesics eqn via mometums:
    $$m\frac{dp_{\beta}}{d\lambda}=\frac{1}{2}g_{\nu \alpha, \beta}p^{\nu}p^{\alpha}$$
    I see that there are conserve such values: $$p_t=g_{tt} \dot{t},$$ $$p_x=g_{xx} \dot{x}=\sqrt{1-8gz}\dot{x}$$ and $$p_y=g_{yy} \dot{y}=\sqrt{1-8gz}\dot{y}$$
     
  20. Jan 24, 2013 #19

    Mentz114

    User Avatar
    Gold Member

    Are you sure ?

    The conserved quantity found from the Killing vector is just [itex]m\dot{x}[/itex]. (I've edited my post above).
     
  21. Jan 24, 2013 #20

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    No luck with the faucet :-(. But basically what seems to be happening is that if [itex]\dot{x}[/itex] starts out as zero, it remains zero. Thus in this case, x remains constant along a geodesic. However, the separation between neighboring geodesics (both of which have constant x) changes with time for a free falling observer, due to the g_xx and g_yy metric coefficient dependence on z which changes in time. Hence, there really is a tidal force in the free-fall geodesic Fermi frame.

    What's really needed to give some intuitive significance to the metric is to calculate some Fermi-normal coordinates. However, this will probably wind up to be a real pain-in-the-rear to do.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Homogeneous gravitational field and the geodesic deviation
  1. Geodesic deviation (Replies: 2)

  2. Geodesic deviation (Replies: 1)

Loading...