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I have to calculate the geodesics of the metric:

$$\left(\matrix {1 &0\\0& x^2 }\right)$$

I've been able to derive its equations, which are:

$$\ddot x -x\dot y ^2=0$$ $$\ddot y+\frac{2}{x}\dot x\dot y=0$$

It's easy to check that lines of constant $y$ are solutions of that equation: all $(v_0t+x_0,y_0)$ satisfy the above, so geodesics between two points of the form $(x_1,y_1)$, $(x_2,y_1)$ are straight lines, but I can't get the general solutions.

Any help? Thanks in advance.

BTW, To avoid calculations on your side, the Riemann tensor, and therefore the Ricci tensor and scalar curvature, vanish., so $R^2$ with that metric is flat

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1 Answer 1

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Note that

$$ \ddot{y} + \frac{2}{x}\dot{x}\dot{y} = \frac{1}{x^2}\frac{d}{dt}\left(\dot{y}x^2\right) = 0$$

so we can now solve for $\dot{y}$:

$$\dot{y} = \frac{C}{x^2}$$ which gives $$\ddot{x} - x\dot{y}^2 = \ddot{x} - \frac{C^2}{x^3} = 0 $$

This is solved by

$$x(t) = \pm\frac{\sqrt{C^2 + a^2(t+b)^2}}{\sqrt{a}}$$

for integration constants $a,b$. The last equation I leave for you to solve

$$y(t) - y(0) = \int_0^t\frac{C a}{C^2 + a^2(t+b)^2}dt $$

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  • $\begingroup$ Thank you. How do you get from $\ddot x-C^2 /x^3=0$ to the solution $x(t)$? $\endgroup$ Jun 25, 2014 at 10:37
  • $\begingroup$ I just tried a test solution $\sqrt{f(t)} \to \frac{f''}{\sqrt{f}} - \frac{f'^2}{2\sqrt{f}f} = \frac{C^2}{\sqrt{f}f}$. This gives $f''f - f'^2/2 = C^2$ and from this one finds that $f$ is a quadratic polynomial in $t$ (just try $f = At^2 + Bt + D$) $\endgroup$
    – Winther
    Jun 25, 2014 at 10:41
  • $\begingroup$ btw when solving it you might want to get rid of $C$ by writing $C^2/a = x(-b)^2$ (insert it into $x(t)$ for $t=-b$ to see why this is true). And also maybe write $y(t) - y(-b) = \int_{-b}^t ...$ instead. This way you will get a 'nicer' answer in the end. Or by shifting the time-variable you are of course free to put $b=0$. $\endgroup$
    – Winther
    Jun 25, 2014 at 10:47
  • $\begingroup$ Ok, I see it. Thank you very much $\endgroup$ Jun 25, 2014 at 11:12

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