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

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

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

And the next statement reads

"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 give, 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. 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....

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    $\begingroup$ Please, please refrain from putting unnecessary dollar signs in your LaTeX equations. You only need one or two at the beginning and end. (See my edit.) $\endgroup$ – JamalS Jun 9 '14 at 16:29
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Try working backwards. Take the solutions given and take the derivatives of both sides with respect to $\sigma$. Then use the product rule on the left-hand side.

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  • $\begingroup$ I tried it already and got the expected result, but can you please help with the INTEGRATION??That is the real problem..thanx $\endgroup$ – GRrocks Jun 9 '14 at 16:28
  • $\begingroup$ I'm not sure what you mean, I think Spain is just referring to how you for example multiply equation 1 by $r^2$ then "integrate" $\frac{d}{d\sigma}(r^2\frac{d\psi}{d\sigma})=0$ with respect to $\sigma$. If you mean you want to "solve" the resultant 1st order ODEs, you will need to look at the geodesic equation for $r$ first. This will give you $r$ in terms of $\sigma$, and that will allow you to solve for $\psi$ and $t$. $\endgroup$ – user82235 Jun 9 '14 at 16:34
  • $\begingroup$ What about the other terms when you integrate(as you said in you comment)??Can you please show the integration, cuz all I m getting is logs and $d\psi/d\sigma$.. $\endgroup$ – GRrocks Jun 10 '14 at 14:21
  • $\begingroup$ Ok..I got it..we don't need do work back ..but thanx anyways @user82235 $\endgroup$ – GRrocks Jun 13 '14 at 13:26
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OK, here's my answer..

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

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

After a little work,

$\frac{d}{d\sigma}(log(\frac{d\psi}{d\sigma}) + \int\frac{2}{r}dr)$ = $0$

Finally, integrating wrt $d\sigma$ gives the required answer.

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