How to solve the geodesic equations in 2D polar coordinates?

Question

Consider the XY-plane as a surface in 3D-space, parametrized with polar coordinates $u$ an $v$ as $\vec{r}=(v\cos(u),v\sin(u),0)$. For the geodesic equations I get

• Eq.1: $v~d^2u+2~du~dv=0$
• Eq.2: $d^2v-v(du)^2=0$.

[EDIT] Of course we all know that the general solution is the straight line $a=v\cos(u-\alpha)$ which passes through some point $(u,v)=(\alpha,a)$. But I would like to derive that solution, i.e. solve the system in general.

Moreover, after checking out quite a few texts (e.g. Dray) I could only find discussions where the straight line is parameterized by arc length, which I would like to avoid. Although it's obvious that using arc length simplifies such things tremendously, I like the idea of $v$ being a function of $u$, thus with $u$ as the main variable. So I would prefer to avoid using e.g. $ds$ in the derivation. (This also fits in my opinion that "the geodesic" is a geometric concept whose definition should be kept free from restrictions on the parameterization. But that's another discussion, and even my "heroes" Millman and Parker clutter their definition of a geodesic with an arc length parameterization.)

Any suggestions?

[EDIT] Nobody seems to care? It looks like everybody thinks I'm a student doing my homework? I know it might look like a simple question but I just can't get it right. And no, I'm not a student.

Consider a straight line in the plane, e.g. $a=v*cos(u-\alpha)$ which passes through some point $(u,v)=(\alpha,a)$. After differentiating twice and rearranging things a bit I get

• Eq.3: $sin(u-\alpha)*(v*d^2u+2*du*dv)+cos(u-\alpha)*(v*(du)^2-d^2v)=0$.

So, if the geodesic equations (1) and (2) hold, then (3) is fulfilled. But that's the wrong direction: in order to prove that the straight line is a geodesic we should be able to prove that (1) and (2) are fulfilled if we know that (3) holds, and not the opposite way around. But I can't really satisfactorily convince myself with my reasonings.

Answer 1

In the mean time I worked out the following. We have $a=v\ cos(u-\alpha)$ and need to prove

• Eq.1: $v\ d^2u+2\ du\ dv=0$
• Eq.2: $d^2v−v\ (du)^2=0$.

We differentiate $a=v\ cos(u-\alpha)$ and obtain $0=dv\ cos(u-\alpha) - v\ sin(u-\alpha)\ du$. We square both terms of the result and get Eq.4: $(dv)^2\ cos^2(u-\alpha)=v^2\ sin^2(u-\alpha)\ (du)^2$.

We have $(dv)^2 + (v\ du)^2 = (ds)^2$ from which follows $(dv)^2 = (ds)^2 - (v\ du)^2$ which we substitute in Eq.4 and get $((ds)^2 - (v\ du)^2)\ cos^2(u-\alpha)=v^2\ sin^2(u-\alpha)\ (du)^2$. Working this out and rearranging gives $(ds)^2\ cos^2(u-\alpha)=v^2\ (du)^2$. We eliminate the cosine from this with $a=v\ cos(u-\alpha)$ and obtain $(ds)^2\ a^2 = v^4\ (du)^2$, so $a\ ds=v^2\ du$. You can stick with arc length or set $s=b\ t+c$ so the curve is not unit speed anymore and get $a\ b\ dt = v^2\ du$ . Differentiating this gives $a\ b\ d^2t = 2\ v\ du\ dv + v^2\ (d^2u)$. We set $d^2t=0$, and get $0= 2\ v\ du\ dv + v^2\ (d^2u)$. We can divide by $v$ (because $v=0$ gives the origin; no solution) and $v\ d^2u + 2\ du\ dv=0$. So far for Eq.1.

Now we differentiate $(dv)^2 + (v\ du)^2 = (ds)^2$ thus $2\ dv\ d^2v + 2\ v\ du\ (du\ dv + v\ d^2u) = 2\ ds\ d^2s = 0$. Thus $dv\ d^2v + v\ (du)^2\ dv + v^2\ du\ d^2u = 0$. We use Eq.1 to change $v\ d^2u$ in $-2\ du\ dv$ and get $dv\ d^2v + v\ (du)^2\ dv + v\ du\ (-2\ du\ dv)=0$ which simplifies to $dv\ d^2v - v\ (du)^2\ dv=0$. We can divide by $dv$ (because $dv\neq0$ for our straight line) and we get Eq.2.

There are 2 free parameters $a$ and $\alpha$ in our proposed solution $a=v\ cos(u-\alpha)$ and so we found all solutions to the system Eq.1 & Eq.2 ([EDIT] maybe this is too big a shortcut).

Pfewww... I guess that's it! Any comments still welcome, though.