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Math Help - Multi variable integration?

  1. #1
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    Multi variable integration?

    I am trying to integrate

    \frac{d\tau}{dp} = \left(\frac{2\ r_g}{3(p-\tau)}\right)^{1/3}

    where r_g is a constant. I am trying to obtain an expression that can be plotted on a \tau versus p chart.

    Attempt at a solution:

    My naive attempt was to simply integrate wrt p and I obtained

    \tau = \int \left(\frac{2\ r_g}{3(p-\tau)}\right)^{1/3} dp \ = \ \left(\frac{3(p-\tau)}{2}\right)^{2/3} r_g^{1/3} +C

    which then solves to

     p = \tau \pm \frac{2 \ (\tau-C)^{3/2}}{3\sqrt{r_g}}

    Unfortunately when plotted it seems unphysical, and I think I may have done the integration wrong. I suspect the problem is that the two variables p and \tau are not independent and multivariable integration is required, but I have no experience of that and no idea where to start. Can anyone help?

    Background:

    References http://www.phys.au.dk/~fedorov/GTR/09/note11.pdf and Lemaitre metric - Wikipedia, the free encyclopedia

    I am using the simplified version of the Lemaitre metric

    ds^{2} = d\tau^{2} - \frac{r_{g}}{r} dp^{2}

    which ignores angular motion and I am trying to obtain an expression for light paths which always have ds=0, so that for light:

    dp/d\tau = \sqrt{ \frac{r}{r_g}}

    and from the references

    r = \left(\frac{3}{2}(p-\tau)\right)^{2/3}\ r_g^{1/3}

    which I substituted into the dp/d \tau equation to obtain the expression that I am trying to integrate. If it helps any, the value 2 can be substituted for r_g. Any help would be greatly appreciated.
    Last edited by kev p; December 22nd 2009 at 10:12 PM.
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  2. #2
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    There is a mistake in your solution

    \tau = \int \left(\frac{2\ r_g}{3(p-\tau)}\right)^{1/3} dp
    It is wrong because (p - \tau) you can find it from the integrand is not just a function of  p , it is a function of  p and  \tau


    To solve this differential equation ,

    <br />
\frac{d\tau}{dp} = \left(\frac{2\ r_g}{3(p-\tau)}\right)^{1/3}<br />

    , substitute  p - \tau = u

    differentiate both sides with respect to  p

     1 - \frac{d\tau}{dp} = \frac{du}{dp}

     \frac{d\tau}{dp} = 1 - \frac{du}{dp}

    therefore ,

    1 - \frac{du}{dp} =  \left(\frac{2\ r_g}{3(u)}\right)^{1/3}

     \left[ 1 - \left(\frac{2\ r_g}{3(u)}\right)^{1/3} \right ] = \frac{du}{dp}

    Then solve it by separating the variables ..
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  3. #3
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    Quote Originally Posted by simplependulum View Post
     \left[ 1 - \left(\frac{2\ r_g}{3(u)}\right)^{1/3} \right ] = \frac{du}{dp}

    Then solve it by separating the variables ..
    so

    du = \left[ 1 - \left(\frac{2\ r_g}{3(p-\tau)}\right)^{1/3} \right ] dp

    and then

    u = (p-\tau) = \int \left[ 1 - \left(\frac{2\ r_g}{3(p-\tau)}\right)^{1/3} \right ] dp

    or am I on the wrong track?
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  4. #4
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    Still lost..

    Quote Originally Posted by simplependulum View Post
    To solve this differential equation ,

    <br />
\frac{d\tau}{dp} = \left(\frac{2\ r_g}{3(p-\tau)}\right)^{1/3}<br />

    , substitute  p - \tau = u

    differentiate both sides with respect to  p

     1 - \frac{d\tau}{dp} = \frac{du}{dp}
    ..
    Could you flesh this part out a bit? I am still not getting it...
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  5. #5
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    Quote Originally Posted by kev p View Post
    Could you flesh this part out a bit? I am still not getting it...
    simplependulum said it clearly, differentiate both sides with respect to p
    you need to simplify \frac{d}{dp}\{p - \tau \} = \frac{du}{dp}
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  6. #6
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    Quote Originally Posted by dedust View Post
    ...
    you need to simplify \frac{d}{dp}\{p - \tau \} = \frac{du}{dp}
    Ah! I get it now. For some reason I read
    Quote Originally Posted by simplependulum View Post
    \frac{d\tau}{dp} = \left(\frac{2\ r_g}{3(p-\tau)}\right)^{1/3}

    , substitute  p - \tau = u

    differentiate both sides with respect to  p
    as substitute  p - \tau = u (into the equation above it, and then) differentiate both sides (of the equation above it) with respect to  p rather than just differentiating both sides of  p - \tau = u . Sorry, my mistake. I hope you will bear with me as it is a very long time since I have done any serious calculus.

    O.K. so now

    \frac{du}{dp} = \left[ 1 - \left(\frac{2\ r_g}{3(u)}\right)^{1/3} \right ]

    \frac{dp}{du} = \left[ 1 - \left(\frac{2\ r_g}{3(u)}\right)^{1/3} \right ] ^{-1}

    p = \int \left[ 1 - \left(\frac{2\ r_g}{3(u)}\right)^{1/3} \right ] ^{-1} du

    Is that the correct way to go?
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  7. #7
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    Quote Originally Posted by kev p View Post
    p = \int \left[ 1 - \left(\frac{2\ r_g}{3(u)}\right)^{1/3} \right ] ^{-1} du

    Is that the correct way to go?

    I think it is correct
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  8. #8
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    p = \int \left[ 1 - \left(\frac{2\ r_g}{3(u)}\right)^{1/3} \right ] ^{-1} du
    When I carry out the above indefinite integration, the result is quite a large equation in which the u variable appears numerous times. It seams to be impossible to solve the equation for p or \tau after substituting (p-\tau) back in. Well, Mathematica and Maple seem unable to solve it anyway.

    Would it be acceptable to perform the whole operation again, but this time obtain an expression in terms of \tau and u and then use u as a parametric parameter in order to plot p versus \tau?
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