1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding Geodesics/Solving Differential Equations

  1. Jun 16, 2014 #1
    I want to find geodesics on 2-dimensional Anti de-Sitter space. According to http://www.aei.mpg.de/~gielen/ads.pdf I can write the metric of this as

    [tex] ds^2 = - \cosh^2{r} dt^2 + dr^2 [/tex]

    Now, we know the geodesic equation is

    [tex] \frac{d^2 x^\mu}{d \tau^2} + \Gamma^\mu{}_{\nu \rho} \frac{d x^\nu}{d \tau} \frac{d x^\rho}{d \tau} = 0, \quad \Gamma^\mu{}_{\nu \rho} = \frac{1}{2} g^{\mu \sigma} \left( g_{\nu \sigma, \rho} + g_{\rho\ sigma, \nu} - g_{\nu \rho, \sigma} \right) [/tex]

    and where [itex] \tau [/itex] is an affine parameter.

    Let's consider the case [itex] \mu = r[/itex], then to work out the Christoffel symbols, I need to know that [itex] g^{rr} = 1[/itex] (using the fact that the metric is diagonal).

    Then I compute
    [tex] \Gamma^{r}_{tt} = \frac{1}{2} \left( g_{tr,t} + g_{tr,t} - g_{tt, r} \right) = \frac{1}{2} 2 \cosh{r} \sinh{r} = \cosh{r} \sinh{r} [/tex]
    [tex] \Gamma^{r}_{tr} = \frac{1}{2} \left( g_{tr,r} + g_{rr,t} -g_{tr,r} \right) = 0 [/tex]
    [tex] \Gamma^r{}_{rr} = \frac{1}{2} \left( g_{rr,r} + g_{rr,r} - g_{rr,r} \right) = 0 [/tex]

    This means that the Geodesic equation takes the form

    [tex] \ddot{r} + \left( \cosh{r} \sinh{r} \right) \dot{t}^2 = 0 [/tex]

    Where the dot denotes differentiation with respect to the affine parameter [itex]\tau[/itex].

    I don't know how to solve this equation. I can multiply the cosh and sinh to get it to the form [itex] \ddot{r} + \frac{1}{2} \left( \sinh{2r} \right) \dot{t}^2 = 0 [/itex] but that doesn't seem to help. Does anyone have any suggestions? Am I using a poor choice of coordinates - hopefully not as I'd ideally like this in terms of sinh and cosh if possible.

    Thanks.
     
  2. jcsd
  3. Jun 17, 2014 #2

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    You should have 2 geodesic equations though right? Where's your other geodesic equation? You have 2 variables, but only 1 equation.
     
  4. Jun 17, 2014 #3
    So I find [itex] g^{tt} = - \frac{1}{\cosh^2{r}}[/itex] and so

    [tex] \Gamma^t_{tt} = \frac{1}{2} g^{tt} g_{tt,t} = - \frac{1}{2 \cosh^2{r}} \left( -2 \cosh{r} \sinh{r} \frac{dr}{dt} \right) = \tanh{r} \frac{\dot{r}}{\dot{t}}[/tex]
    [tex] \Gamma^{t}_{tr} = \frac{1}{2} g^{tt} \left( g_{tt,r} + g_{tr,t} -g_{tr,t} \right) = - \frac{1}{2 \cosh^2{r}} \left( -2 \cosh{r} \sinh{r} \right) = \tanh{r} [/tex]
    [tex] \Gamma^{t}_{rr} = \frac{1}{2} g^{tt} \left( g_{tr,t} + g_{rt,t} - g_{rr,t} \right) =0 [/tex]

    This means the differential equations are
    1, [tex] \ddot{t} + \left(2 \tanh{r} \right) \dot{r} \dot{t} =0 [/tex]
    2, [tex] \ddot{r} + \cosh{r} \sinh{r} \dot{t}^2 =0 [/tex]

    Obviously these are coupled and I cannot see how to solve them - perhaps a suitable change of coordinates will improve things but I really cannot see it. Any ideas?

    Thanks.
     
  5. Jun 17, 2014 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since the first equation involves only derivatives of t, not t, itself, let [tex]u= \dot{t}[/tex] so that the equation is [tex]\dot{u}+ \left(2\tanh(r)\right)\dot{r}u= 0[/tex] or
    [tex]\dot{u}= \left(2\tanh(r)\right)\dot{r}u[/tex]
    [tex]\frac{du}{u}= 2\tanh(r) dr[/tex]

    That can be integrated for u in terms of r.
     
  6. Jun 17, 2014 #5
    So [tex] \ln{u} = -\ln{\cosh{r}} + c[/tex]

    Exponentiating we get [tex] u = -A \cosh{r} [/tex]

    This becomes [tex] dt = -A \cosh{(r(\tau))} d \tau[/tex]

    So we get [tex] t(\tau) = -A \sinh{(r(\tau))} + C[/tex]

    Is that correct? I seem unable to use this to satisfy the other geodesic equation!
     
  7. Jun 18, 2014 #6
    Actually I realise I made a mistake in my above answer. I now find [itex]\ln{u} = -2 \ln{\cosh{r}} + C \Rightarrow u = A ( \cosh{r})^{-2}[/itex]

    Substituting into the other geodesic equation gives

    [tex] \ddot{r} + A^2 \frac{\sinh{r}}{\cosh^3{r}} =0[/tex]

    I don't know how to solve this and am now thinking that maybe I made a mistake earlier on - perhaps with the factor of 2 in the t geodesic equation as if this were a 1 then I would get a tanh in the above equation which would be easier to deal with. However, since we sum over the tr and rt contributions in the geodesic equation, I think it's correct that it should double up.

    What do you reckon?

    Thanks.
     
  8. Jun 18, 2014 #7
    ##g_{tt,t}## =0? Then no factor 2.
    ##\int cosh(r(\tau))\,d\tau \neq sinh(r(\tau))## since it's a composite function.
     
    Last edited: Jun 18, 2014
  9. Jun 18, 2014 #8
    I disagree. I find

    [tex] \Gamma^t_{tt} = \frac{1}{2} g^{tt} g_{tt,t} = - \frac{1}{2 \cosh^2{(r(\tau))}} \left( -2 \cosh{r} \sinh{r} \frac{dr(\tau)}{dt} \right) = \tanh{r} \frac{dr}{d \tau} \frac{d \tau}{dt} = \tanh{r} \frac{\dot{r}}{\dot{t}}[/tex]

    Does that make sense?
     
  10. Jun 18, 2014 #9
    It's confusing but "##,_t##" means partial derivative along the coordinate t and Christoffel's symbols are defined independently of geodesics.
    Maybe not very helpful..
    I get ##u = \frac{A}{cosh(r)}## and ##\ddot{r}+B\, tanh(r)=0##
     
    Last edited: Jun 18, 2014
  11. Jun 18, 2014 #10
    Your eqns are lacking any first order derivatives. How did you get them? Even accepting your dispute with my [itex]\ddot{t}[/itex] equation, I don't see how you get a [itex]\tanh{r}[/itex] in the [itex]\ddot{r}[/itex] equation?
     
  12. Jun 19, 2014 #11
    Trying to help without self confidence: not a good idea...
    I find ##\Gamma^t_{tt} = 0 = \Gamma^t_{rr}## and##\Gamma^t_{rt} = tanh(r) = \Gamma^t_{tr}##.
    Then ##\ddot{t} + tanh(r) \dot{t}\dot{r}+tanh(r)\dot{r}\dot{t} =0## where I missed the sum over the two gammas. Then as Hallsoflvy ##u=\dot{t}## that gives ##\frac{du}{u}=-2\,tanh(r)\,dr## and like you did ##ln(u)=-2\,ln(cosh(r))+A## then ##u=\frac{A}{cosh^2 (r)}## wich I put in the second equation(##u^2=\dot{t}^2##): ##\ddot{r}+tanh(r)\frac{B}{cosh^2(r)}=0##. That is ##\ddot{r}=C\, tanh(r)\,tanh'(r)## and multiplying both sides with ##\dot{r}## I get ##\ddot{r}\dot{r}=D\frac{d\,tanh(r)}{d\tau}\,tanh(r)##. This is ok with ##\dot{r}=thanh(r)##. Then ##\frac{dr}{tanh(r)}=cst\,d\tau## and ##r=arcsinh(exp(cst\,\tau))## and ##\dot{t}=\frac{A}{1+sinh^2(arcsinh(exp(\alpha\tau)))}## and so on
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Finding Geodesics/Solving Differential Equations
Loading...