- #1
cr7einstein
- 87
- 2
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 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 don't 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...
$$\frac{d^2\psi}{d\sigma^2} + \frac{2}{r}\frac{dr}{d\sigma}\frac{d\psi}{d\sigma} = 0,$$
$$\frac{d^2t}{d\sigma^2} + \frac{2m}{c^2r}\left(1-\frac{2m}{c^2r}\right)^{-1}\frac{dr}{d\sigma}\frac{dt}{d\sigma} = 0,$$
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 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 don't 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...
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