- #1

Jonathan Scott

Gold Member

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## Main Question or Discussion Point

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}##.

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}##.