# Newtonian Gravity from curved space?

Imagine you have the arc-length of a curve, in spherical, coordinates: $$s = \int_{\mathcal C}{d\tau \; \sqrt{f(r)^2 \left (\frac{dr}{d \tau} \right )^2 + r^2 \left (\frac{d \theta}{d \tau} \right )^2+ r^2 \sin^2({\theta}) \left (\frac{d \varphi}{d \tau} \right )^2}} \\ \frac{ds}{d\tau} = 1$$ Minimizing the Functional (will be done only with respect $$r$$ for simplicty): $$\frac{\partial F}{\partial r} -\frac{d}{d\tau}\frac{\partial F}{\partial \dot{r}} = 0 \\ \; \\ \ddot{r} f(r)^2 = r \left ( \dot{\theta}^2 + \sin^2{(\theta)} \dot{\varphi}^2 \right) - f(r)\frac{d f}{d\tau}\dot{r}$$ This reminds a lot of the two body problem of Newtonian Gravity.
$$\ddot{r} = r \left ( \dot{\theta}^2 + \sin^2{(\theta)} \dot{\varphi}^2 \right) - \frac{GM}{r^2}\\$$ But, $$f(r)$$, must satisfy the following: $$f(r)^2=1 \\ \; \\ f(r)\frac{df}{d\tau}\dot{r} = \frac{GM}{r^2}$$ It is possible to find this $$f(r)$$. Is this line of reasoning correct? (I know that the metric tensor exist, but I rather to this the "old-fashioned way").

• You are looking for f(r) not $f(\tau)$
– Eli
Commented Nov 19, 2022 at 9:30

## 1 Answer

taking $$~\theta=\frac\pi2~$$ you obtain those equations of motion (with $$~F=\frac{GM}{r^2}\,\vec e_r~)$$

$$f(r)^2\ddot{r}+f(r)\,\dot r^2\,\frac{d}{dr}\,f(r)-r\dot\varphi^2-\frac{G\,M}{r^2}=0\tag 1$$

and $$~r^2\,\dot\varphi=\text{constant}$$

equation (1) must be:

$$\ddot{r}-r\dot\varphi^2-\frac{G\,M}{r^2}=0$$

thus

$$f(r)=1\quad\Rightarrow\,\frac{d}{dr}f(r)=0$$