# Anisotropic instead of isotropic metric deriving acceleration

1. May 9, 2013

### Agerhell

In this documentation from Nasa a procedure to get to what I guess is the gravitational acceleration according to the post-Newtonian expansion at the 1PN-level for the spherically symmetric case is found:

http://descanso.jpl.nasa.gov/Monograph/series2/Descanso2_all.pdf [Broken]

The procedure is based on using the metric shown in expression (4.60) on page (4.42). The metric is a low order expansion of the isotropic Schwarzschild metric. The procecure to get to the expression for the acceleration is a bit beyond me.

My question is:

What expression for the gravitational gravitation do you get if you use the anisotropic Schwarzschild metric and apply the same procedure?

I would also like to know why the isotropic Schwarzschild metric is used to derive an expression for the acceleration and not the anisotropic metric, which is more common in basic textbooks...

Last edited by a moderator: May 6, 2017
2. May 9, 2013

### Bill_K

Why don't you try it and let us know what you get?

Probably because the reference they cribbed it from ("HRTW") used isotropic coordinates.

3. May 9, 2013

### Agerhell

Hmm... I will start. Maybe someone can continue or ells I will continue later.

According to the book you assume:

$$L^2=\frac{ds^2}{dt^2}$$

Then you get the equations of motion from the expression:

$$\frac{d}{dt}(L\frac{\partial L}{\partial \dot{x_i}})-(\frac{\dot{L}}{L})(L\frac{\partial L}{\partial \dot{x_i}})- L\frac{\partial L}{\partial x_i}=0, x_i = x,y,z$$

There is some approximation involved:

$$\frac{\dot{L}}{L}\approx \frac{L\dot{L}}{c^2}$$

The expression on the right side just above replaces the expression to the left. The book also states that "$L\dot{L}$ is obtained by differentiating a simplified expression for $L^2$ containing terms to order $1/c^0$ only."

Assuming:

$$ds^2=(1-\frac{2GM}{rc^2}+2\frac{(GM)^2}{r^2c^4})c^2dt^2-(1-\frac{2GM}{rc^2})(dx^2+dy^2+dz^2)$$

according to the book this will result in:

$${\bf \ddot{r}}=\frac{GM}{r^3c^2}((\frac{4GM}{r}-v^2){\bf r} +4({\bf r\cdot\dot{r}}){\bf \dot{r}})$$

I might have done some mistake here, interpreting what I read, but this is sort of standard procedure for how you get from a metric via some Lagrangian method to equations of motions? If anyone have a link to page describing this procedure in general it would be helpful.

I vaguely remember dealing with Lagrangians back at university. If I am trying to use polar coordinates (anisotropic Schwarzschild) I am sure there will be some extra complications...