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GR equations of motion

  1. Feb 5, 2008 #1


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    These are the EOMs for a 4D space-time that has a plane singularity at x=0. They are simpler than most, but as I've confessed before, I need help with even the simplest DE's. This is not a home-work question.

    [tex]x\frac{d^2\tau}{dt^2} - \frac{d\tau}{dt}\frac{dx}{dt} = 0[/tex]

    [tex]x^3\frac{d^2x}{dt^2} + \frac{m^2}{2}(\frac{d\tau}{dt})^2 - \frac{x^2}{2}(\frac{dx}{dt})^2 = 0[/tex]

    Any solutions gratefully received and will be acknowledged.

  2. jcsd
  3. Feb 6, 2008 #2
    If [itex]x(t)=c_1,\,\tau(t)=c_2[/itex] then the equations are satisfied.
    If [itex]x(t)\neq 0,\,\tau(t) \neq c_2[/itex] then from the 1st equation we have

    [tex]\frac{\tau''}{\tau'}=\frac{x'}{x}\Rightarrow x=C\,\tau'[/tex]

    Plugging this result to the 2nd equation we have

    [tex]\tau'\,(2\,C^4\,\tau'\,\tau'''-C^4\,(\tau'')^2+m^2)=0\Rightarrow \tau'\,\tau'''-\frac{1}{2}\,(\tau'')^2+\frac{m^2}{2\,C^4}=0 \quad \text{since} \quad \tau'\neq 0[/tex]

    Now set [itex]\tau'=\frac{m}{C^2}\,y[/itex], thus

    [tex]y\,y''-\frac{1}{2}\,y'^2+\frac{1}{2}=0 \quad (\ast)[/tex]

    With the substitution
    [tex]y(t) \rightarrow \theta,\,t\rightarrow u(\theta),\,y'(t) \rightarrow \frac{1}{u'(\theta)},\, y''(t) \rightarrow -\frac{u''(\theta)}{u'(\theta)^3}[/tex]
    makes the last equation


    which is a 1st order separable ODE.
    Thus from the last one we can find [itex]u(\theta)[/itex] and from [itex](\ast), y(t)[/itex] and finally [itex]\tau(t)[/itex]
    Last edited: Feb 7, 2008
  4. Feb 6, 2008 #3


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    Thanks again, RainbowChild. Unfortunately what you've left looks just as difficult as the original to me and I wouldn't know where to begin to solve it.

    I'll go and look up 'separable'.

    So, I need to write the last equation the form

    [tex]\frac{dy}{dx} = h(x)g(y)[/tex]

    but I don't understand your last equation, are the primes still differentiation wrt t ?

    the nearest I can get is this

    [tex]\frac{dU}{dt} = \frac{1}{\theta}U(U^2-1)[/tex]

    I'm sorry, I'm at sea here.
    Last edited: Feb 6, 2008
  5. Feb 7, 2008 #4
    Ok, let me throw you a life jacket! :smile:



    let [itex]u'(\theta)=z(\theta)[/itex] thus

    [tex]\frac{z'(\theta)}{z(\theta)\,(z(\theta)^2-1)}=\frac{1}{2\,\theta}\Rightarrow \int \frac{d\,z}{z\,(z^2-1)}=\int \frac{d\,\theta}{2\,\theta} \Rightarrow \ln\frac{z^2-1}{z^2}=\ln\theta+C_1 \Rightarrow[/tex]

    [tex]z(\theta)=\pm \frac{1}{\sqrt{1-C_1\,\theta}}\Rightarrow u'(\theta)=\pm \frac{1}{\sqrt{1-C_1\,\theta}}\Rightarrow u(\theta)=\pm \frac{2}{C_1}\,\sqrt{1-C_1\,\theta}-C_2[/tex]

    which yields to

    [tex]t=\pm \frac{2}{C_1}\,\sqrt{1-C_1\,y(t)}-C_2\Rightarrow y(t)=-\frac{1}{4\,C_1}\,\left((t+C_2)^2\,C_1^2-4\right)[/tex]

    Now use [itex]\tau'=\frac{m}{C^2}\,y[/itex], in order to calculate [itex]\tau(t)[/itex]

    I hope that clears up things! :smile:
  6. Feb 7, 2008 #5


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    That helps a lot, thanks. My head is above water.

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