- #1

jbergman

- 345

- 143

- Homework Statement
- Not homework but looking for some help on a problem.

- Relevant Equations
- $$ \Phi = \frac{-GM}{|\textbf{r}|} $$

$$R^j_{0k0}=-R^j_{00k} = \frac{\partial^2\Phi}{\partial x^j \partial x^k}$$

Hello, I am attempting to work through problem 12.6 in MTW which involves formulating Newtonian Gravity using Curvature as opposed to the standard formulation. This is a precursor before standard GR. In it he states that the curvature tensor in this formulation is as follows.

$$R^j_{0k0}=-R^j_{00k} = \frac{\partial^2\Phi}{\partial x^j \partial x^k}$$

For problem 12.6 I am trying to calculate this for the earth with the potential,

$$ \Phi = \frac{-GM}{|\textbf{r}|} $$

And for example, for one of the components I get.

$$R^j_{0k0}=-R^j_{00k} = \frac{\partial^2\Phi}{\partial x^j \partial x^k}$$

$$\frac{\partial^2}{\partial x^j\partial x^k}(-GM/|\textbf{r}|)=

\frac{\partial}{\partial x^k}\left( \frac{GM x^j}{|\textbf{r}|^3}\right) =

\frac{\partial}{\partial x^k}(x^j) \frac{GM}{|\textbf{r}|^3} -

\frac{3GM x^j x^k}{|\textbf{r}|^5}$$

$$R^x_{0x0} = \frac{GM}{|\textbf{r}|^3} -

\frac{3GM x^2}{|\textbf{r}|^5}$$

However, I am not convinced this is correct based on p.37 where it looks like,

$$R^x_{0x0} = \frac{GM}{|\textbf{r}|^3}$$

However, MTW states that z is in the radial direction which would suggest that ##x=0## and ##y=0##. So, it looks like MTW is using a specific position to simplify the problem. Is my understanding correct or did I botch the calculation.

$$R^j_{0k0}=-R^j_{00k} = \frac{\partial^2\Phi}{\partial x^j \partial x^k}$$

For problem 12.6 I am trying to calculate this for the earth with the potential,

$$ \Phi = \frac{-GM}{|\textbf{r}|} $$

And for example, for one of the components I get.

$$R^j_{0k0}=-R^j_{00k} = \frac{\partial^2\Phi}{\partial x^j \partial x^k}$$

$$\frac{\partial^2}{\partial x^j\partial x^k}(-GM/|\textbf{r}|)=

\frac{\partial}{\partial x^k}\left( \frac{GM x^j}{|\textbf{r}|^3}\right) =

\frac{\partial}{\partial x^k}(x^j) \frac{GM}{|\textbf{r}|^3} -

\frac{3GM x^j x^k}{|\textbf{r}|^5}$$

$$R^x_{0x0} = \frac{GM}{|\textbf{r}|^3} -

\frac{3GM x^2}{|\textbf{r}|^5}$$

However, I am not convinced this is correct based on p.37 where it looks like,

$$R^x_{0x0} = \frac{GM}{|\textbf{r}|^3}$$

However, MTW states that z is in the radial direction which would suggest that ##x=0## and ##y=0##. So, it looks like MTW is using a specific position to simplify the problem. Is my understanding correct or did I botch the calculation.

Last edited: