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Planetary motion equations integral

  1. Jun 12, 2014 #1
    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...

    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. jcsd
  3. Jun 12, 2014 #2


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    Homework Helper

    The equations are each of the form [tex]
    \frac{d^2 u}{d\sigma^2} + f(v)\frac{dv}{d\sigma}\frac{du}{d\sigma} = 0.[/tex] This suggests that one should first divide by [itex]\frac{du}{d\sigma}[/itex] to obtain [tex]
    \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.[/tex]

    EDIT: Working back from the given answer leads me to conclude that the second equation should be [tex]
    \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.[/tex]
    Last edited: Jun 12, 2014
  4. Jun 12, 2014 #3
    Ok, thanks pasmith, I got it....
    Last edited: Jun 13, 2014
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