Why is $p_\phi$ conserved in a Schwarzschild orbit?

Question

This arises from the question What is the relationship between $a$ and $m$, which I'm afraid I answered just by looking it up in Schutz's book. However Schutz (as he frequently does) glosses over details he thinks are irrelevant or too simple to be worth explaining, and I have realised I don't understand an assumption he makes.

Schutz states without proof that if we have an equatorial orbit in a Schwartzschild metric then:

Independence of the metric of the angle $\phi$ about the axis implies that $p_\phi$ is constant.

In the non-relativistic world I assume this corresponds to angular momentum being constant in a central potential. So far so good. But why is it the component of the dual vector $p_\phi$ that is constant rather than $p^\phi$? The component $p^\phi$ is presumably not constant since (in this case) $p^\phi = p_\phi/r^2$.

Bonus points for also explaining his similar claim that time independance means that $p_t$ is constant rather than $p^t$.

I fear that Schutz didn't explain because it's an insultingly simple question, but if someone can provide a nice intuitive explanation I would be very pleased to read it.

Answer 1

But why is it the component of the dual vector $p_\phi$ that is constant rather than $p^\phi$?

From the bottom of page 189:

The geodesic equation can thus, in complete generality, be written

$$m \frac{dp_\beta}{d\tau} = \frac{1}{2}g_{\nu \alpha,\beta}\;p^\nu p^\alpha$$

We therefore have the following important result: if all of the components $g_{\mu \nu}$ are independent of $x^\beta$ for some fixed index $\beta$, then $p_\beta$ is a constant along any particle's trajectory

Also, be aware that, in the relevant section on equatorial orbits in the Schwarzschild geometry, Schutz is working in a coordinate basis and not a unit basis.

In the case that $\theta = \frac{\pi}{2}$ (as in this example), we have

$$\vec e_\phi \cdot \vec e_\phi = r^2$$

which is why, I believe, $p^\phi$ is $r$ dependent.

Answer 2

Let $\xi^\alpha$ be a Killing vector of a metric $g_{\mu\nu}$, i.e. it satisfies $$ \nabla_\mu \xi_\nu + \nabla_\nu \xi_\mu = g_{\mu\alpha} \partial_\nu \xi^\alpha + g_{\nu\alpha} \partial_\mu \xi^\alpha + \xi^\alpha \partial_\alpha g_{\mu\nu} $$ Then the quantity $$ Q = \xi^\alpha u_\alpha $$ is conserved along any geodesic. To see this, we can compute $$ u^\alpha \nabla_\alpha Q = u^\alpha u^\beta \nabla_\alpha \xi_\beta + u^\alpha \nabla_\alpha u_\beta \xi^\beta $$ The first term above is zero because I can symmetrize $\nabla_\alpha \xi_\beta \to \frac{1}{2} \left( \nabla_\alpha \xi_\beta + \nabla_\beta \xi_\alpha \right)$ which is then zero since $\xi$ is a Killing vector. The second term is zero due to the geodesic equation. Thus $$ u^\alpha \nabla_\alpha Q = 0 $$ Finally, we note that if the metric $g_{\mu\nu}$ is independent of a particular coordinate $\phi$, then $K^\alpha = \delta^\alpha_\phi$ is a Killing vector. We can see this by simply plugging this into the Killing equation and we find $$ g_{\mu\alpha} \partial_\nu K^\alpha + g_{\nu\alpha} \partial_\mu K^\alpha + K^\alpha \partial_\alpha g_{\mu\nu} = \partial_\phi g_{\mu\nu} = 0 $$ The first two terms vanish since $K$ is a constant. The last term vanishes by assumption.

Therefore, if the metric is independent of $\phi$, then $K^\alpha = \delta^\alpha_\phi$ is a Killing vector and $$ Q = K^\alpha u_\alpha = u_\phi \propto p_\phi $$ is a conserved quantity.