Question

Consider a two-dimensional space with coordinates x µ = (θ, φ), for which the only nonvanishing Christoffel symbols are

3. Consider a two-dimensional space with coordinates Ζμ-(9,0), for which the only non- vanishing Christoffel symbols are tan θ Write down the two components of the geodesic equation, and find a set a solutions.

Answer 1

The geodesic equation is given by

where, repeated indices are summed over.
And for a 2D space with coordinates $x^\mu =(\theta, \phi)$ , we write the geodesic equations explicitly as

$\frac{\mathrm{d}^2 \theta }{\mathrm{d} s^2}+\Gamma^{\theta}_{\phi\phi}\frac{\mathrm{d} \phi}{\mathrm{d} s}\frac{\mathrm{d} \phi}{\mathrm{d} s}=0$
   $\Rightarrow \frac{\mathrm{d}^2 \theta }{\mathrm{d} s^2}-\sin\theta\cos\theta\left (\frac{\mathrm{d} \phi}{\mathrm{d} s} \right )^2=0$
And similarly
   $\frac{\mathrm{d}^2 \phi }{\mathrm{d} s^2}+\Gamma^{\phi}_{\theta\phi}\frac{\mathrm{d} \theta}{\mathrm{d} s}\frac{\mathrm{d} \phi}{\mathrm{d} s}+\Gamma^{\phi}_{\phi\theta}\frac{\mathrm{d} \phi}{\mathrm{d} s}\frac{\mathrm{d} \theta}{\mathrm{d} s}=0$
   $\Rightarrow \frac{\mathrm{d}^2 \phi }{\mathrm{d} s^2}+\frac{2}{\tan\theta}\frac{\mathrm{d} \phi}{\mathrm{d} s}\frac{\mathrm{d} \theta}{\mathrm{d} s}=0$
   $\Rightarrow \sin\theta\frac{\mathrm{d}^2 \phi }{\mathrm{d} s^2}+2\cos\theta\frac{\mathrm{d} \phi}{\mathrm{d} s}\frac{\mathrm{d} \theta}{\mathrm{d} s}=0$
   $\Rightarrow \sin^2\theta\frac{\mathrm{d}^2 \phi }{\mathrm{d} s^2}+2\sin\theta\cos\theta\frac{\mathrm{d} \phi}{\mathrm{d} s}\frac{\mathrm{d} \theta}{\mathrm{d} s}=0$
$\Rightarrow \frac{\mathrm{d} }{\mathrm{d} s}\left (\sin^2\theta\frac{\mathrm{d} \phi }{\mathrm{d} s} \right )=0$

By integrating this equation, we get
    $\Rightarrow\sin^2\theta\frac{\mathrm{d} \phi }{\mathrm{d} s} =\text{const.}\equiv\frac{c}{R}$
where, the constant is this way for convenience.
So, we get
    $\Rightarrow \frac{\mathrm{d} \phi }{\mathrm{d} s} =\frac{c}{R\sin^2\theta}$
Now if we substitute this back in the equation for \theta, then, we get
$\Rightarrow \frac{\mathrm{d}^2 \theta }{\mathrm{d} s^2}-\sin\theta\cos\theta\left (\frac{\mathrm{d} \phi}{\mathrm{d} s} \right )^2=0$
   $\Rightarrow \frac{\mathrm{d}^2 \theta }{\mathrm{d} s^2}=\sin\theta\cos\theta\left (\frac{c}{R\sin^2\theta} \right )^2$
   $\Rightarrow \frac{\mathrm{d}^2 \theta }{\mathrm{d} s^2}=\frac{c^2}{R^2}\frac{\cos\theta}{\sin^3\theta}$
This equation is in general difficult to solve. However, we will use the definition of path length in the 2D space to solve this equation.
    $ds^2=g_{\mu\nu}dx^\mu dx^\nu =R^2d\theta^2+R^2\sin^2\theta ~d\phi^2$
    $\Rightarrow R^2\left (\frac{d\theta}{ds} \right )^2+R^2\sin^2\theta ~\left (\frac{d\phi}{ds} \right )^2=1$
Now substituting for the derivative of \phi, we get
  
$\Rightarrow R^2\left (\frac{d\theta}{ds} \right )^2+R^2\sin^2\theta ~\left (\frac{c}{R\sin^2\theta} \right )^2=1$
      $\Rightarrow R^2\left (\frac{d\theta}{ds} \right )^2+\frac{c^2}{\sin^2\theta}=1$
      $\Rightarrow R^2\left (\frac{d\theta}{ds} \right )^2=1-\frac{c^2}{\sin^2\theta}$
   $\Rightarrow \frac{d\theta}{ds} =\pm \frac{1}{R\sin\theta}\sqrt{\sin^2\theta-c^2}$
   $\Rightarrow ds=\pm \frac{R\sin\theta d\theta}{\sqrt{\sin^2\theta-c^2}}$
   $\Rightarrow ds=\pm \frac{R\sin\theta d\theta}{\sqrt{1-\cos^2\theta-c^2}}$
   $\Rightarrow ds=\pm \frac{R\sin\theta d\theta}{\sqrt{a^2-\cos^2\theta}}$
where, a^2 = 1 - c^2. So, now
   we substitute
   $u=\cos\theta \Rightarrow -\sin\theta d\theta =du$
So, in terms of u, we get
   $\Rightarrow ds=\pm \frac{Rdu}{\sqrt{a^2-u^2}}$
   $\Rightarrow \int ds=\pm \int \frac{Rdu}{\sqrt{a^2-u^2}}$
    $\Rightarrow s=\pm R\sin^{-1}\left ( \frac{u}{a} \right )+k$
where, k is the integration constant. So, we get by putting u
    $\Rightarrow s=\pm R\sin^{-1}\left ( \frac{\cos\theta}{a} \right )+k$
Now we choose the constant k such that when
      $\cos\theta = 0$ , s = 0. And that implies k = 0. So, we get
$\Rightarrow s=\pm R\sin^{-1}\left ( \frac{\cos\theta}{a} \right )$
   $\Rightarrow \cos\theta =\pm a\sin\left ( \frac{s}{R} \right )$
Now from the \phi equation of motion, we get
   $\frac{\mathrm{d} \phi }{\mathrm{d} s} =\frac{c}{R\sin^2\theta}=\frac{c}{R(1-\cos^2\theta)}$
   $\Rightarrow \frac{\mathrm{d} \phi }{\mathrm{d} s} =\frac{c}{R(1-a^2\sin^2\left ( \frac{s}{R} \right ))}$
Now if we substitute back the definition of a^2 = 1- c^2, then, we get
   $\Rightarrow \frac{\mathrm{d} \phi }{\mathrm{d} s} =\frac{c}{R(1-(1-c^2)\sin^2\left ( \frac{s}{R} \right ))}$
   $\Rightarrow \frac{\mathrm{d} \phi }{\mathrm{d} s} =\frac{c}{R(\cos^2\left ( \frac{s}{R} \right )-c^2\sin^2\left ( \frac{s}{R} \right ))}$
   $\Rightarrow \frac{\mathrm{d} \phi }{\mathrm{d} s} =\frac{1}{\cos^2\left ( \frac{s}{R} \right )}\frac{c}{R(1-c^2\tan^2\left ( \frac{s}{R} \right ))}$
   $\Rightarrow \frac{\mathrm{d} \phi }{\mathrm{d} s} =\frac{c\sec^2\left ( \frac{s}{R} \right )}{R(1-c^2\tan^2\left ( \frac{s}{R} \right ))}$
   $\Rightarrow d\phi =\frac{c\sec^2\left ( \frac{s}{R} \right )}{R(1-c^2\tan^2\left ( \frac{s}{R} \right ))}ds$
   $\Rightarrow \int d\phi =c\int \frac{\sec^2\left ( \frac{s}{R} \right )}{(1-c^2\tan^2\left ( \frac{s}{R} \right ))}d\left ( \frac{s}{R} \right )$
   $\Rightarrow \phi-\phi_0 =\tan^{-1}\left (c\tan\left ( \frac{s}{R} \right )\right)$
   $\Rightarrow \tan(\phi-\phi_0 )=c\tan\left ( \frac{s}{R} \right )$
   

The geodesic equation is given by d^2 x^mu/ds^2 + gamma_alpha beta^mu dx^alpha/ds dx^beta/ds = 0 where, repeated indices are summed over And for a 2D space with coordinated x^mu = (theta, phi), we write the geodesic equations explicity as d^2 theta/ds^2 + gamma_phi phi^theta d phi d phi/ds ds = 0 d^2 theta/ds^2 - sin theta cos theta (d phi/ds)^2 = 0 And similarly d^2 phi/ds^2 + gamma_theta phi^phi d theta/ds d phi/ds + gamma_phi theta^phi d phi/ds d theta/ds = 0 d^2 phi/ds^2 + 2/tan theta d phi/ds d theta/ds = 0 sin theta d^2 phi/ds^2 + 2 cos theta d phi/ds d theta/ds = 0 sin^2 theta d^2 phi/ds^2 + 2 sin theta cos theta d phi/ds d theta/ds = 0 d/ds(sin^2 theta d phi/ds) = 0 By integrating this equation, we get sin^2 theta d phi/ds = const . identicalto c/R where, the constant is this way for convenience So, we get d phi/ds = c/R sin^