Christoffel symbols, Geodesic Equations, Tensors

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?

• 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. – Arthur Apr 21 '17 at 6:44
• How would this help me though? Is the method which I was going to use correct? – Jeremy Curmi Apr 21 '17 at 6:57
• You should make explicit which parameter is constant. – Thomas Apr 21 '17 at 7:18

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).