0
$\begingroup$

Consider the following line element:

$ds^2=dr^2+r^2d\theta^2 \tag1 $

From which I obtained the following non-zero Christoffel symbols:

$\Gamma^r_{\theta\theta}=-r\ \mathrm{and}\ \Gamma^\theta_{r\theta}=1/r\tag2$

This is my problem: I am required to show that $r=\sec(\theta)$, where is a constant, is a geodesic on the above metric.

The method that I tried to solve it is to work out the derivative:

$\dfrac{dr}{ds}\ ,\ \dfrac{d\theta}{ds}\ ,\ \dfrac{d^2r}{ds^2}\ ,\ \dfrac{d^2\theta}{ds^2}\ ,\ \dfrac{drd\theta}{ds^2} $

and substitute them in the equation of the geodesic and get 0. Is this correct, and if not what should I do differently please?

Thanks in advance

$\endgroup$
  • $\begingroup$ You could cheat, and see that this is the polar coordinates on the Euclidean plane, and transform the line element and curve to Cartesian coordinates. $\endgroup$ – Arthur Apr 21 '17 at 6:44
  • $\begingroup$ How would this help me though? Is the method which I was going to use correct? $\endgroup$ – Jeremy Curmi Apr 21 '17 at 6:57
  • $\begingroup$ You should make explicit which parameter is constant. $\endgroup$ – Thomas Apr 21 '17 at 7:18
1
$\begingroup$

It is convenient to stick to the independent variable given in first fundamental form metric 1) itself.

From curvature formula for geodesic straight line in the plane with $(r,\theta) $ polar coordinates, primed wrt $\theta$ we have

$$ \kappa_g= \dfrac{r^2+2r^{\prime^2}-rr^{\prime \prime}}{(r^2+r^{\prime^2})^{\frac32} }=0 $$

you need to only verify (plugging it in) that

$$ r^{\prime \prime}= \dfrac{2 r^{\prime}}{r}+r $$

for the general geodesic/straight line obtained by direct integration

$$ r = a \sec (\theta + \alpha) $$

where $ a,\alpha $ are two arbitrary constants. (They are pedal distance and initial rotation wrt pole respectively).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.