Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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...

    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. jcsd
  3. Jun 12, 2014 #2

    pasmith

    User Avatar
    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Planetary motion equations integral
  1. Integral Equation (Replies: 4)

  2. Motion equation (Replies: 7)

  3. Integral Equation (Replies: 6)

Loading...