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## Main Question or Discussion Point

I've been working on a problem where I have to find the geodesics for a given Riemannian Manifold. To present my doubt, I tried to find a simpler example that would demonstrate my uncertainty but the one I found, and shall present bellow, has actually a simplification that my problem doesn't, so please ignore that simplification (I shall indicate it when it comes up).

Given a Riemannian manifold with metric

$$\tag{1} ds^{2}=\frac{dr^2}{a+r^2}+r^2\left(d \theta^2+\sin^2(\theta) \, d\phi^{2}\right),$$

consider a curve [itex]c(\lambda)=(r(\lambda),\phi(\lambda),\pi/2)[/itex] whose tangent vector is [itex]c'(\lambda)=(r'(\lambda),\phi'(\lambda),0)[/itex]. The geodesic equations (if I didn't mess up) are given by:

\begin{cases}

\tag{2} r''-\frac{r'^{2}}{2(a+r)}-r(a+r)\phi'^{2}=0,\\

\phi''+2\frac{r'}{r}\phi'=0.

\end{cases}

The second equation of Eq.[itex](2)[/itex] can be integrated, such that:

$$\tag{3} \phi'(\lambda)=\frac{C}{r^{2}(\lambda)},$$

where [itex]C[/itex] is a constant of integration.

Now I introduce a new equation:

$$\tag{4}r'^{2}\left(\frac{1}{a+r^{2}}\right)+\phi'^{2}r^{2}=1,$$

which is basically impose unit speed. Substituting Eq.[itex](3)[/itex] in Eq.[itex](4)[/itex] we have that:

$$\tag{5} r'=\sqrt{(a+r^{2})\left(1-\frac{C^{2}}{r^{2}}\right)}.$$

(Here is the difference from my case since Eq.[itex](5)[/itex] can be integrated analytically and in my case, the congener equation can't).

Dividing Eq.[itex](3)[/itex] by Eq.[itex](5)[/itex] we have that

$$\tag{6} \frac{d\phi}{dr}=\frac{C}{r^{2}(\lambda)}\frac{1}{\sqrt{(a+r^{2})\left(1-\frac{C^{2}}{r^{2}}\right)}}.$$

So my question is:Does Eq.[itex](6)[/itex] allow me to write the curve [itex]c[/itex] as [itex]c(r)=(r,\phi(r))[/itex], where [itex]\phi(r)[/itex] is given by [itex](6)[/itex] and [itex]c[/itex] is a geodesic?

Given a Riemannian manifold with metric

$$\tag{1} ds^{2}=\frac{dr^2}{a+r^2}+r^2\left(d \theta^2+\sin^2(\theta) \, d\phi^{2}\right),$$

consider a curve [itex]c(\lambda)=(r(\lambda),\phi(\lambda),\pi/2)[/itex] whose tangent vector is [itex]c'(\lambda)=(r'(\lambda),\phi'(\lambda),0)[/itex]. The geodesic equations (if I didn't mess up) are given by:

\begin{cases}

\tag{2} r''-\frac{r'^{2}}{2(a+r)}-r(a+r)\phi'^{2}=0,\\

\phi''+2\frac{r'}{r}\phi'=0.

\end{cases}

The second equation of Eq.[itex](2)[/itex] can be integrated, such that:

$$\tag{3} \phi'(\lambda)=\frac{C}{r^{2}(\lambda)},$$

where [itex]C[/itex] is a constant of integration.

Now I introduce a new equation:

$$\tag{4}r'^{2}\left(\frac{1}{a+r^{2}}\right)+\phi'^{2}r^{2}=1,$$

which is basically impose unit speed. Substituting Eq.[itex](3)[/itex] in Eq.[itex](4)[/itex] we have that:

$$\tag{5} r'=\sqrt{(a+r^{2})\left(1-\frac{C^{2}}{r^{2}}\right)}.$$

(Here is the difference from my case since Eq.[itex](5)[/itex] can be integrated analytically and in my case, the congener equation can't).

Dividing Eq.[itex](3)[/itex] by Eq.[itex](5)[/itex] we have that

$$\tag{6} \frac{d\phi}{dr}=\frac{C}{r^{2}(\lambda)}\frac{1}{\sqrt{(a+r^{2})\left(1-\frac{C^{2}}{r^{2}}\right)}}.$$

So my question is:Does Eq.[itex](6)[/itex] allow me to write the curve [itex]c[/itex] as [itex]c(r)=(r,\phi(r))[/itex], where [itex]\phi(r)[/itex] is given by [itex](6)[/itex] and [itex]c[/itex] is a geodesic?

**Note:**This treatment is based on the Clairaut parametrization but I'm not sure if I can do it in this kind of problem where the [itex]g_{rr}[/itex] component of the metric varies...
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