# Planetary motion equations integral

1. Jun 12, 2014

### cr7einstein

I was reading Planetary Motion (page 117) in Barry Spain's *Tensor calculus*, and stupidly enough, I didn't understand this. The equations are:

$$\frac{d^2\psi}{d\sigma^2} + \frac{2}{r}\frac{dr}{d\sigma}\frac{d\psi}{d\sigma} = 0,$$

$$\frac{d^2t}{d\sigma^2} + \frac{2m}{c^2r}\left(1-\frac{2m}{c^2r}\right)^{-1}\frac{dr}{d\sigma}\frac{dt}{d\sigma} = 0,$$

>we integrate the above to get
>
>$$r^2\frac{d\psi}{d\sigma} = h, \quad \left(1-\frac{2m}{c^2r}\right) \frac{dt}{d\sigma}= k$$
>
>respectively, for constants $$k$$ and $$h$$.

I believe it is a simple question, as no steps are given, but I am unable to get it. I tried everything I could, substitutions et al, but to no avail. So please entertain this silly question. Thank you very much...

EDIT
-
Please show the integration, I dont want to confirm the validity of the answers by working backwards, but I want to establish them.

EDIT (2)

I want to see the actual integration. If I am right, then the second term($2r\frac{dr}{d\sigma}\frac{d\psi}{d\sigma}$) of the first equation yields the desired result which is $r^2\frac{d\psi}{d\sigma}$ when integrated wrt $d\sigma$. It means that the integral of $$r^2\frac{d^2\psi}{d\sigma^2}$$ is $$0$$...I think the integral simplifies to $$\frac{d}{d\sigma}\int{r^2d\psi}$$, and to get the required answer, it should be zero or a constant.....But HOW??? PLease help....

Last edited: Jun 12, 2014
2. Jun 12, 2014

### pasmith

The equations are each of the form $$\frac{d^2 u}{d\sigma^2} + f(v)\frac{dv}{d\sigma}\frac{du}{d\sigma} = 0.$$ This suggests that one should first divide by $\frac{du}{d\sigma}$ to obtain $$\left(\frac{du}{d\sigma}\right)^{-1} \frac{d}{d\sigma}\left(\frac{du}{d\sigma}\right) + f(v)\frac{dv}{d\sigma} = \frac{d}{d\sigma}\left(\log \left(\frac{du}{d\sigma}\right) + \int f(v)\,dv \right) = 0.$$

EDIT: Working back from the given answer leads me to conclude that the second equation should be $$\frac{d^2t}{d\sigma^2} - \frac{2m}{c^2r^2}\left(1-\frac{2m}{c^2r}\right)^{-1}\frac{dr}{d\sigma}\frac{dt}{d\sigma} = 0.$$

Last edited: Jun 12, 2014
3. Jun 12, 2014

### cr7einstein

Ok, thanks pasmith, I got it....

Last edited: Jun 13, 2014