general relativity - Why is $p_phi$ conserved in a Schwarzschild orbit?

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

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.