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

Useful relativistic gravity equation of motion

  1. Dec 24, 2014 #1

    Jonathan Scott

    User Avatar
    Gold Member

    In a recent thread, I referred to what I thought was a well-known very simple way of describing how objects move in a gravitational field, in terms of the rate of change of their coordinate momentum in isotropic coordinates. It seems that this result is not widely known (and I can't see a reference on the internet, although I don't have access to papers behind paywalls), so I'd like to show how to derive this result directly. I've discussed this result with physicists before and believe it to be correct, but if anyone sees any problem with it, please say so. I was originally told how to do this, mostly in tensor notation, by a colleague at work (with a PhD in Physics) in around 1988 when I wanted to understand how different theories of gravity gave different perihelion precession values for Mercury, but I later found that one could express the same mathematical result in a simpler form, much more directly comparable with the Newtonian form, by adopting somewhat unconventional notation where all symbols including the speed of light ##c## refer to coordinate values unless specifically marked otherwise.

    This method requires an isotropic coordinate system, in which local distances in space are simply some multiple ##\sqrt{g_{xx}} = \Phi_x## of the distances measured in the frame of the observer, regardless of the direction in space, and the local time is simply some multiple ##\sqrt{-g_{tt}} = \Phi_t## of the time measured by the observer. This applies for example to the Schwarzschild solution in isotropic coordinates, and also to solutions for multiple static sources in the linearized approximation.

    In isotropic cartesian coordinates, the metric can be written in time units as follows, where primed values are values as observed locally (so ##c\,^\prime## is the standard constant local value of the speed of light but ##c##, the coordinate speed of light, varies with potential).
    $$d\tau^2 = \Phi_t^2 dt^2 - \Phi_x^2 \frac{dx^2 + dy^2 + dz^2}{{c\,^\prime}^2}$$
    The equation of motion for a test mass ##m## with velocity ##\mathbf{v}## can then be calculated using the relevant Lagrangian, which for this purpose is the usual one for Special Relativity except that the coordinate value of ##c## must be used (which then makes the coordinate action exactly equal to the local action):
    $$L = mc^2 \sqrt{1 - v^2/c^2}$$
    The general Euler-Lagrange equations of motion are as follows:
    $$\frac{\partial L}{\partial x_i}
    - \frac{d}{dt} \left( \frac{\partial L}{\partial v_i} \right) = 0 \, \, \, \mbox{(i = 1, 2, 3)}$$
    or in a more convenient form:
    $$\frac{d}{dt} \left( -\frac{\partial L}{\partial v_i} \right) = -\nabla L$$
    In this case, the expression ##-\partial L/\partial v_i## is simply the relativistic momentum as usual for Special Relativity:
    $$ -\frac{\partial L}{\partial v_i} = \frac{m{\bf v}}{\sqrt{1-v^2/c^2}} = \frac{E{\bf v}}{c^2} = {\bf p}$$
    The rest mass ##m^\prime## is constant in the local frame, as is the rest energy ##m^\prime {c^\prime}^2##, which varies between frames like frequency. This means that in our coordinate system ##mc^2## varies as ##\Phi_t## and ##\nabla (mc^2) = (mc^2/\Phi_t) \, \nabla \Phi_t##. Note also that the value of ##c## in ##v^2/c^2## varies with potential.
    $$\begin{align}
    -\nabla L & = -\nabla (mc^2) \sqrt{1-v^2/c^2} - mc^2\, \nabla \left( \sqrt{1-v^2/c^2} \right) \\
    & = -mc^2 \, \frac{1}{\Phi_t}\,\nabla \Phi_t\, \sqrt{1-v^2/c^2}
    - mc^2 \, \frac{v^2}{c^2} \, \frac{1}{c} \, \nabla c \, \frac{1}{\sqrt{1-v^2/c^2}} \\
    & = -\frac{mc^2}{\sqrt{1-v^2/c^2}} \, \left [ \left ( 1 - \frac{v^2}{c^2} \right) \, \frac{1}{\Phi_t}\,\nabla \Phi_t\,
    + \frac{v^2}{c^2} \, \left ( \frac{1}{\Phi_t} \, \nabla \Phi_t
    - \frac{1}{\Phi_x} \, \nabla{\Phi_x} \right ) \right ] \\
    & = -E \left( \frac{1}{\Phi_t} \, \nabla \Phi_t
    - \frac{v^2}{c^2} \frac{1}{\Phi_x} \, \nabla \Phi_x \right)
    \mbox{ where the energy } E = \frac{mc^2}{\sqrt{1-v^2/c^2}}
    \end{align}$$
    This gives the following equation of motion:
    $$ \frac{d \mathbf{p}}{dt} = -E \left( \frac{1}{\Phi_t} \nabla \Phi_t
    - \frac{v^2}{c^2} \frac{1}{\Phi_x} \nabla \Phi_x \right)$$
    So far, this applies not only to GR but also to any other metric theory of gravity which can be represented in isotropic coordinates, and it applies even for strong fields.

    To match Newtonian theory, the value of ##\Phi_t## at distance ##r## from a central mass ##M## must be approximately equal to ##(1-GM/rc^2)##, and to match GR and the well-known amount by which light is bent by the sun, the value of ##\Phi_x## must be approximately equal to ##1/\Phi_t##, at least to the first order in ##Gm/rc^2##. Different theories of gravity give slightly different expressions for ##\Phi_x## but this approximation is sufficiently accurate to determine the effect on perihelion precession of variations in the PPN ##\beta## parameter.

    As ##\Phi_x \approx 1/\Phi_t## for small potential differences, ##1/\Phi_x \,\nabla \Phi_x \approx -1/\Phi_t \,\nabla \Phi_t## and the equation of motion may be simplified to the following in such cases:
    $$ \frac{d \mathbf{p}}{dt} = \frac{E}{c^2} \, {\bf g}
    \left( 1 + \frac{v^2}{c^2} \right)
    \mbox{ where } {\bf g} = -\frac{c^2}{\Phi_t} \nabla \Phi_t $$
    This equation works even for particles with no rest mass travelling at ##c##, showing that the assumption that ##\Phi_x \approx 1/\Phi_t## leads directly to the correct result that the deflection relative to the observer's flat coordinate system is twice that which would be expected from Newtonian gravitational theory.

    As for Newtonian gravity, this equation has the neat property that the direction of the rate of change of coordinate momentum is directly towards the source regardless of the direction of travel, which also means that coordinate angular momentum is preserved.

    Note that since the total energy is constant, one may also divide by the energy of the test object:
    $$ \frac{d}{dt} \left ( \frac{\bf v}{c^2} \right ) = \frac{\bf g}{c^2} \,
    \left( 1 + \frac{v^2}{c^2} \right) $$
    This shows that the motion is much simpler when expressed in terms of ##\mathbf{v}/c^2## rather than just ##\mathbf{v}##.
     
  2. jcsd
  3. Dec 24, 2014 #2
    Jonathan, I'm not familiar with PHI, which seems to be a four dimensional. Is there a correspondence principle to newtonian mechanics, especially the time component?
    However, there's a lot of resemblances to the "gravitational index of refraction" here;
    https://www.physicsforums.com/threads/gravitational-index-of-refraction.264511/
    Is PHI proportional to the index or similar? I'm thinking your assumption could be found in optics.
     
  4. Dec 24, 2014 #3

    Jonathan Scott

    User Avatar
    Gold Member

    I'm just using the notation ##\Phi_t## to represent the multiplicative potential, equivalent to the time dilation factor and to the Newtonian factor ##(1 + \phi/c^2) = (1 - \sum_i G m_i/r_i c^2)##. Similarly, I'm using ##\Phi_x## to represent the equivalent scale factor for space, which is approximately ##1/\Phi_t##. In this notation, the coordinate value of ##c## is ##\Phi_t/\Phi_x \approx (1 - 2 \sum_i G m_i/r_i c^2)## times the standard value.

    [Edited to add a couple of missing factors of ##c^2##]
     
  5. Dec 24, 2014 #4

    Dale

    Staff: Mentor

    Do you have a reference for this?
     
  6. Dec 24, 2014 #5

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    I would suggest the following approach instead - I'm not sure if it's equivalent, but I believe it's correct.

    We start with at the same place, with the isotropic metric written in cartesian variables rather than spherical variables..
    $$d\tau^2 = f^2 dt^2 - g^2 \left( dx^2 + dy^2 + dz^2 \right)$$

    Specifying the metric gives you the coordinate speed of light, which must satisfy the equation ##d\tau = 0##

    ##f^2 dt^2 = g^2 \left(dx^2 + dy^2 + dz^2\right)##

    which leads to
    $$ \left( \frac{f}{g} \right)^2 = \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2$$

    Given that you define "coordinate speeds" formally as
    ##\vec{v} = \left( dx/dt, dy/dt, dz/dt \right)## and
    ##v = \sqrt{
    \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2}##

    the right hand side of this is the square of the "coordinate speed of light", which we derive directly from the metric.

    As for the Lagrangian, we simply use Hamilton's principle, which states that ##L = - m c^2 \int d\tau## and use the metric to solve for ##d\tau## in terms of the coordinates. Basically, we are saying the equations of motion are the ones that extremize proper time.

    Since
    ##\left( \frac{d\tau}{dt} \right)^2 = f^2 - g^2 v^2##

    we can write

    ##L = -m\,c^2 \int \frac{d\tau}{dt} dt = -m\,c^2 \int \sqrt{f^2 - g^2 \, v^2} dt = - m\,c^2 \int f \sqrt{1-v^2 g^2 / f^2} dt## where we can interpret ##v^2 g^2 / f^2 ## as ##v^2 / c'^2 ##, c' being the variable "coordinate speed of light" (which we earlier derived as being equal to f/g).

    However we see that f, the time dilation factor, appears in our expression for the Lagrangian. It must, I think, if the Lagrangian is to extremize proper time, as it should.
     
  7. Dec 24, 2014 #6

    Jonathan Scott

    User Avatar
    Gold Member

    Thanks for checking this.

    My own notes include both the derivation of the coordinate speed of light from the metric exactly as you show (except that I'm using different symbols for your f and g) and the working from L in the form you gave to the form I actually used, which is very straightforward of course. I had hoped that both of these steps were sufficiently obvious that I could omit them, but I guess that when I'm using unconventional notation I should explain every step.

    And of course your expansion in terms of f and g is equivalent to my own expansion in terms of the coordinate values. I deliberately chose to use primes for the standard, local values, not for the coordinate values, in order to make the relationship to the Newtonian form obvious.
     
  8. Dec 24, 2014 #7

    Dale

    Staff: Mentor

    Since you are still posting stuff without references this thread is closed. Do not start another one on this topic without references.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Useful relativistic gravity equation of motion
  1. Relativistic motion (Replies: 5)

  2. Relativistic Gravity (Replies: 28)

Loading...