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Accelerations: Transformation of Newton potential

  1. Feb 16, 2016 #1

    haushofer

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    Dear all,

    I'm taking a second look at the Newtonian limit of GR and the covariance group. My main interest is to see how the Newton potential transforms under the covariance group of Newtonian gravity. I know that the gradient of the potential is given by the Christoffel symbol [tex]\Gamma^i_{00} = \partial^i \Phi [/tex]
    from which you can derive how this gradient transforms under linear accelerations [itex] x^i \rightarrow x^{'i} = x^i + a^i(t) [/itex] (x^0 is called t):

    [tex] \Gamma^{'i}_{00} = \Gamma^{i}_{00} + \frac{\partial^2 a^i}{\partial t^2} \ \ \ \ (1) [/tex]

    However, from the Poisson equation [tex] \partial_i \partial ^i \Phi = 4 \pi G \rho[/tex] I know that the gradient is not only allowed to transform under only time-dependent accelerations in order to keep the Poisson eqn. covariant, but under accelerations with vanishing divergence:
    [tex] \partial_i a^i (r,t) = 0 [/tex]
    Apart from the Galilei transformations these restricted accelerations also keep the Poisson eqn. covariant. This makes good sense, since a freely falling observer in a Newtonian gravitational field caused by a mass M undergoes an acceleration

    [tex]
    a(t,r) = \frac{1}{2} g(r) t^2, \ \ \ \ g(r) = - \frac{GM}{r} \ \ \ \ \ \ (2)
    [/tex]

    which indeed is divergenceless. My question is: how does Phi transform under these accelerations? The covariance group of Newtonian gravity would then be the Galilei group plus 'divergence-free' but otherwise arbitrary accelerations, where the coefficient of the acceleration is given by g(r) of (2). So I'd say we cannot just integrate eqn.(1) to find the transformation of \Phi. I also tried to find the transformation from [itex]g_{00} = [/itex] but I'm not getting any sensible answer. Apparently, all the literature I know about of the Newtonian limit doesn't mention how the general coordinate transformations are broken down to this "acceleration-extended Galilei symmetries". Does anybody have an idea?
     
    Last edited: Feb 17, 2016
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  3. Feb 16, 2016 #2

    haushofer

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    I made a silly mistake; the Laplace operator is not invariant under my r-dep. accelerations. But now i'm confused; shouldn't the equiv.principle allow us to set \Phi=0 by an acceleration for arbitrary radial distance r?
     
  4. Feb 16, 2016 #3

    haushofer

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    Let me rephrase the question. In GR we can construct a vielbein, which takes us to inertial coordinates, making the metric locally the Minkowski-metric. The Newtonian analog of that would be an acceleration which puts [itex] \Phi \equiv 0 [/itex]. I'd say that the corresponding coordinate transformation (the radial component, the angular ones are zero) is given by

    [tex]
    r \rightarrow r' = r - \frac{1}{2} g(r) t^2, \ \ \ \ g(r) = - \frac{GM}{r}
    [/tex]

    such that the EOM of a particle in this gravitational field transforms as

    [tex]
    \ddot{r} + g(r) = 0 \rightarrow \ddot{r}' = 0
    [/tex]

    The GR-analog of this would be that in the geodesic eqn. the connection components disappear in these inertial coordinates. However, this coordinate transformation doesn't seem to keep the Poisson equation
    [tex]
    \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 \frac{\partial\Phi}{\partial r}) = 4 \pi G \rho
    [/tex]
    covariant under such a transformation. What am I missing? It should be covariant, right, as in
    [tex]
    \frac{1}{r'^2} \frac{\partial}{\partial r'} (r'^2 \frac{\partial\Phi'(r')}{\partial r'}) = 4 \pi G \rho
    [/tex]
    ?I'm feeling a bit silly now :P
     
  5. Feb 16, 2016 #4

    stevendaryl

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    If [itex]\Phi[/itex] obeys the Poisson equation, then so does [itex]\Phi + \vec{a} \cdot \vec{r}[/itex], since the additional term disappears when you let [itex]\nabla^2[/itex] operate on it.
     
  6. Feb 16, 2016 #5

    haushofer

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    Thanks, I see! This only works if a ~ 1/r, and it does. But doesn't the corresponding coordinate transformation also transform the Laplacian transformation, such that the Poisson eqn. is not covariant under it?

    Something else which confuses me is the fact that the product ar is not r-dependent anymore but only depends on time, and as such cannot be used to put \Phi'(r') = 0.
     
  7. Feb 16, 2016 #6

    stevendaryl

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    Here's what Wikipedia says about Newton-Cartan (which is what you get if you try to make Newton's theory of gravity generally covariant):
    https://en.wikipedia.org/wiki/Newton–Cartan_theory
     
  8. Feb 17, 2016 #7

    haushofer

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    Yes, I am familiar with the Newton-Cartan formulation, but this is just old-fashioned Newtonian gravity, and I want to understand it in this framework which should be simpler :P

    Are you agreeing with the fact that one should always be able to transform to an accelerating observer such that [itex]\Phi'(r') \equiv 0 [/itex]? This is not possible with the transformation you gave me, because in your transformation which I write as

    [tex]
    \Phi(r) \rightarrow \Phi'(r') = \Phi(r) + g(r)\, r
    [/tex]

    the term [itex] g(r)\, r [/itex] does not have an r-dependency anymore, while [itex] \Phi(r) \sim \frac{1}{r}[/itex] . As such you can only put [itex]\Phi'(r') \equiv 0 [/itex] locally (i.e. for constant r!). To make a relativistic analogy: the vielbeins of the Schwarzschild solution also have an r-dependency, such that at every point r it gives the transformation to inertial coordinates. Why can't I do this here?

    I have the feeling I'm making a stupid calculational mistake....

    -edit ... and I did. I was overlooking one factor of r,

    [tex]
    g(r) = \partial_r \Phi (r) = \frac{GM}{r^2}
    [/tex]

    such that indeed

    [tex]
    \Phi' (r') = \Phi (r) - r\, g(r) \equiv 0
    [/tex]
    for these freely-falling observer. And of course, the Poisson is not covariant (not a tensor eqn.) under the transformation

    [tex]
    r \rightarrow r'= r - \frac{1}{2} g(r) t^2 \ \ ;
    [/tex]

    otherwise you wouldn't be able to choose coordinates to put the eqn. to zero in the first place. I guess I was confused by the fact that the Einstein eqns. are tensorial under gct's.
     
    Last edited: Feb 17, 2016
  9. Feb 17, 2016 #8

    stevendaryl

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    Newton-Cartan is (as far as I know) ordinary Newtonian mechanics plus gravity. It's just a different way of presenting it. The only thing that makes it slightly different from Newton's approach is that the equivalence principle (all things fall at the same rate under gravity) is moved from being an empirical fact to a necessity.

    No, I'm saying that the proper way to deal with accelerated observers is not to use potentials, but to use connections in the equations of motion (as Newton-Cartan does).
     
  10. Feb 17, 2016 #9

    haushofer

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    So then one should always be able to transform to an observer for which \Gamma'^i_{00} =0, where the connection components are proportional to 1/r^2. In my approach I derived the transformation of the potential from the Newtonian limit of GR, so I'm using the transformation properties of \Gamma^i_{00}.

    I don't blame you if you think I'm being silly here, I'm just a bit confused :P
     
    Last edited: Feb 17, 2016
  11. Feb 17, 2016 #10

    stevendaryl

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    No, the connection coefficients don't have to be proportional to [itex]1/r^2[/itex]. Newtonian gravity is only proportional to [itex]\frac{1}{r^2}[/itex] if you are in vacuum outside of a spherically symmetric mass.

    No, you're not being silly.
     
  12. Feb 17, 2016 #11

    haushofer

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    Thanks, that's comforting to hear :P Yes, you are right, I'm being too restrictive. Anyway, I calculated how the Christoffel connection of interest transforms under

    [tex]
    t \rightarrow t' = t \\
    r \rightarrow r'= r - \frac{1}{2} g(r) t^2 \\
    \Omega \rightarrow \Omega ' = \Omega \,,
    [/tex]

    and it becomes

    [tex]
    \Gamma'^{r}_{00} = \frac{\partial r'}{\partial r} \Gamma^r_{tt} - \frac{\partial^2 r' }{\partial t \partial t } = [1 - \frac{1}{2}\partial_r g(r) t^2]\Gamma^r_{tt} + g(r)
    [/tex]

    (The second term [itex]- \frac{1}{2}\partial_r g(r) t^2 \Gamma^r_{tt} [/itex] is of second order in the gravitational field; the other two terms are just the terms I expected to arise) What confuses me also is that if I look for references on symmetries of Newtonian gravity, only purely time-dependent accelerations are considered, not the accelerations I use here. I get the feeling I'm confusing global and local statements.
     
  13. Feb 17, 2016 #12

    stevendaryl

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    You can use whatever coordinate transformation you like, but what is the purpose in switching to the coordinate [itex]r' = r - \frac{1}{2} g(r) t^2[/itex]? In the case where [itex]g[/itex] is a constant, then the nice thing about the new coordinates is that an object in freefall under gravity looks like an object at rest in gravity-free space in the new coordinate system. If [itex]g[/itex] is not constant, then that transformation doesn't result in an object at rest in the new coordinate system, so it's not clear what's the point of that transformation. It's only when you ignore the spatial dependency of [itex]g[/itex] that such a transformation has nice properties.
     
  14. Feb 17, 2016 #13

    haushofer

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    Yes, I think I'm confusing the fact that one can only locally 'switch off gravity' with wanting to do so globally. I have to think about it a bit more, but thanks a lot for your insights!
     
  15. Feb 17, 2016 #14

    haushofer

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    Having digested all of this, I keep wondering: in GR we have vielbeins, which transform us to flat spacetime locally. These vielbeins are functions of the coordinates; e.g., for the Schwarzschild eqn they depend on the radial distance r, telling us for every r (outside the horizon) how to transform to flat coordinates.

    I always regarded these accelerations as the Newtonian analog of the Vielbein for the Newton Potential. Is this correct, and why can it then not depend on r anymore?
     
  16. Feb 20, 2016 #15

    haushofer

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    Maybe a physical picture helps. Imagine I´m standing at the top of a very high tower. The height of the tower plus the distance to the middle of the Earth i´ll call R. Someone else, let´s call him Ed, stands at the bottom of the tower on the ground. I feel gravity pulling on me, with

    [tex]
    g(R) = \frac{GM}{R^2} \ m/s^2
    [/tex]

    Ed measures g = 9,81 m/s^2 at the ground. Now I fall of the tower towards Ed. I consider my fall as the coordinate transformation

    [tex]
    r \rightarrow r' = r - \frac{1}{2}g(r) t^2
    [/tex]
    where my trajectory is described by r(t), and r'(t=0) = R. Now, during my whole fall (!), so for r' ranging from R to the ground, I don't feel gravity. But according to the calculation given earlier,
    [tex]
    \Gamma^{'r}_{tt} = \partial^{'r} \Phi' (r') \neq 0
    [/tex]
    suggesting I do experience gravity! Apparently, only for g=constant I can locally, around some fixed radial distance, put [itex]\Gamma^{'r}_{tt} = [/itex] for a very short time. Also, the Poisson eqn. is not covariant with respect to this transformation, while transforming to inertial coordinates in GR keeps the Einstein eqns. covariant. What does this mean?

    Does it have to do with the fact that in Newton-Cartan theory the connection doesn't depend only on the metrics, but also on an extra ambiguity K, which is a closed two-form? So that choosing 'inertial coordinates' in NC doesn't allow one to make the connection to vanish anymore?
     
  17. Feb 20, 2016 #16

    stevendaryl

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    If [itex]g[/itex] varies with [itex]r[/itex], then the path of a falling object is not [itex]r = -\frac{1}{2} g t^2[/itex]. I don't see that the transformation [itex]r \rightarrow r - \frac{1}{2} g t^2[/itex] is of much use except in the special case [itex]g = [/itex] constant. (Or if you are considering [itex]t[/itex] to be so small that [itex]g[/itex] doesn't vary much while falling for a time [itex]t[/itex])
     
  18. Feb 20, 2016 #17

    haushofer

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    Indeed, I see, the path changes because I don't treat g as a constant anymore, but I'm still using the path r = R - 1/2gt^2.

    I guess your last statement is then a local (locally in t, and thus also in r) statement.
     
  19. Feb 20, 2016 #18

    stevendaryl

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    Maybe it's helpful to look at Newton's law of gravity in inertial coordinates, and see how it transforms in noninertial coordinates.

    In inertial coordinates, we have:

    [itex]\frac{d^2 x^j}{dt^2} + g^j = 0[/itex]

    where [itex]g^j[/itex] is the local acceleration due to gravity. If we transform to new coordinates [itex]y^u[/itex], let

    [itex]L^j_v = \frac{\partial x^j}{\partial y^v}[/itex]
    [itex]L^j_0 = \frac{\partial x^j}{\partial t}[/itex]

    Then in terms of [itex]y^u[/itex] this changes to:

    [itex]\frac{d^2 y^u}{dt^2} + \Gamma^u_{00} + 2 \Gamma^u_{0 v} \frac{dy^v}{dt} + \Gamma^u_{vw} \frac{dy^v}{dt} \frac{dy^w}{dt}[/itex]

    where [itex]\Gamma^u_{00} = L^u_i g^i + L^u_j \frac{\partial}{\partial t} L^j_0[/itex]
    [itex]\Gamma^u_{0v} = L^u_j \frac{\partial}{\partial y^v} L^j_0[/itex]
    [itex]\Gamma^u_{vw} = L^u_j \frac{\partial}{\partial y^v} L^j_w[/itex]

    Then Poisson's equation [itex]\dfrac{\partial g^i}{\partial x^i}= 4\pi \rho[/itex] becomes (I think!):

    [itex]\frac{\partial}{\partial y^u} \Gamma^u_{00} - \frac{\partial}{\partial t} \Gamma^u_{0u} +\Gamma^\beta_{\beta u} \Gamma^u_{00} - \Gamma^\beta_{0 \alpha} \Gamma^\alpha_{0 \beta} = 4 \pi \rho[/itex]
     
  20. Feb 20, 2016 #19

    haushofer

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    For a nice treatment on NC in non-inertial coordinates, see

    http://arxiv.org/abs/1412.8655

    The purely spatial connection coefficients are zero for inertial (=constant velocity) observers, and stay zero when you invoke time-dependent accelerations and rotations. Your equation is then just R_{00} in non-inertial coordinates, (where the purely spatial connection components can now be non-zero) so I'd say that is right.

    But I do see my mistake; I took the path r(t) for g=cst to construct my coordinate transfo and to draw conclusions, but for more general g(r) this path is only true locally in time (because then g(r) is approx. constant). This explains also why my transformation only seems to work locally. Thank you very much for your insights, it helped me a lot!
     
  21. Feb 20, 2016 #20

    stevendaryl

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    I'm not sure I understand the paper saying that Ricci mass density is different from Gauss mass density in noninertial frames.
     
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