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How the heck do I integrate this?

  1. Nov 9, 2008 #1

    I am trying to find a solution for this equation in terms of r or t:

    [tex] \frac{d^2r}{dt^2} = \frac{m}{r(r-2m)}\left( \left( \frac{dr}{dt}\right)^2-\left(1-\frac{2m}{R}\right)\right)[/tex]

    It seems I should integrate it twice with respect t, but I have no idea how to do that with a dr/dt term in the middle of the right hand side. I need to get rid of all dr and dt variables and obtain an answer that contains just r,t and R variables, if that is possible.

    One possible (given) solution that I am not convinced is correct is:


    but I have no idea how the solution was obtained.
  2. jcsd
  3. Nov 9, 2008 #2

    George Jones

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    Since the independent variable [itex]t[/itex] doesn't appear, try the substitutions [itex]dr/dt = p[/itex] and

    [tex]\frac{d^2 r}{dt^2} = \frac{dp}{dt} = \frac{dr}{dt} \frac{dp}{dr} = p \frac{dp}{dr}.[/tex]
  4. Nov 12, 2008 #3
    Hi George,

    I am still getting nowhere with this derivation and maybe you could provide some more hints?

    Here is some some more info:

    The equation that I am looking for the derivation of is actually given as a parametric equation

    [tex]r = \frac{R}{2}(1+cos(\alpha)) , \tau = \frac{R}{2}\sqrt{\frac{R}{2m}}(\alpha+sin(\alpha))[/tex]

    on this mathpages page http://www.mathpages.com/rr/s6-04/6-04.htm

    and the equation it is derived from appears to be equation (5) on that page.

    Any ideas?
    Last edited: Nov 12, 2008
  5. Nov 12, 2008 #4


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    Putting aside the complicated coefficients, letting p= dx/dt as George Jones suggested,
    the equation becomes pdp/dx= A(p2- B) which is equivalent to
    [tex]\frac{p dp}{Ap^2- B}= dx[/tex] which is easy to integrate.
    Once you have found p= dx/dt, integrate again.
    Last edited by a moderator: Nov 13, 2008
  6. Nov 12, 2008 #5
    Is this on the right track?

    [tex] p \frac{dp}{dx} = A(p^2-B) [/tex]

    [tex] p^2 = \int{A(p^2-B)}dx = C[x,p] [/tex]

    where C[x,p] is some function of x and p.

    Solve for p

    [tex] p = D[x] = \frac{dx}{dt}[/tex] (assume t is shorthand for proper time tau)

    [tex] x = \int{ D[x] } dt = E[x,t] [/tex]

    Solve for x or t

    [tex] x = F[t] [/tex] or [tex] t = G[x] [/tex]

    I also need to know if a indefinite integration is OK for the first integration or if a definite integration is more rigorous. When I try a definite integral

    [tex] \int_{0}^{R} {A(p^2-B)} dx [/tex] or [tex] \int_{2m}^{R} {A(p^2-B)} dx [/tex]

    the software complains about division by zero errors. Presumably that is singularity problems at x=0 and x=2m?

    Should I allow for the constant of integration at the first integration or just add one in at the final integration?

    I hope that makes some sort of sense. I am a bit out of my depth here and need all the help I can get from you guys ;)

    P.S. I am aware some of these problems can be made a lot simpler if a parametric form can be found before integration. Is the parameter [itex]\alpha[/itex] in the mathpages parametric equation the same as the p = dr/dt introduced by George?
    Last edited: Nov 12, 2008
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