A General wave equation

Your question is unclear. What do you want to derive the wave equation for? The wave equation itself is just the divergence of the covariant derivative. It is not really much to derive.

 

I watched most of the video, but I don't understand what part you are having trouble with. He's trying to figure out two things about a wave field:

1. How does gravity (metric + Christoffel symbols) affect a wave.
2. How does a wave affect gravity?

The second part is about how to write the stress-energy tensor [itex]T_{\mu \nu}[/itex] for a scalar field [itex]\phi[/itex]. It doesn't have much to do with the covariant derivative.

The first part, though, is about the covariant derivative. The wave equation in terms of the covariant derivative is just:

[itex]g^{\mu \nu} \nabla_\mu \nabla_\nu \phi = 0[/itex]

where [itex]\nabla_\mu[/itex] is the covariant derivative.

So what is your question about this?

 

So we come back to the original problem. That it is unclear what you mean. The wave equation itself is not something you derive, it is just a name you give to a certain equation. The wave equation in curved space-time is its natural generalisation.

Whether or not a physical quantity obeys the wave equation is a diffrent question and the derivation might in general depend on the quantity in question.

 

I'm not sure what you mean by deriving it. Deriving it starting from what?

As Susskind says in the video, the wave equation looks the same, whether you have curved spacetime or curvilinear coordinates in flat spacetime. So if you start with the flat spacetime, Cartesian coordinates form for the wave equation, you can derive the form in arbitrary curvilinear coordinates by just doing some calculus:

[itex]\eta^{a b} \partial_a \partial_b \phi = 0[/itex] in Cartesian coordinates. Now let's switch to curvilinear coordinates. I'll use Greek letters for the curvilinear indices, and Roman letters for the Caresian indices.

The relationship between the two kinds of derivatives is: [itex]\partial_b \phi = L^\mu_b \partial_\mu \phi[/itex] where [itex]L^\mu_b = \frac{\partial x^\mu}{\partial x^b}[/itex]. When you take a second derivative, you end up derivatives of [itex]L^\mu_b[/itex] in addition to derivatives of [itex]\phi[/itex]. The derivatives of [itex]L^\mu_b[/itex] become part of the Christoffel symbols in the curvilinear coordinates.

 

Okay. In Cartesian coordinates, there is no difference between partial derivatives and covariant derivatives. So if [itex]V^b[/itex] is a vector field, then
[itex]\nabla_a V^b = \partial_a V^b[/itex]

Now, switch to a curvilinear coordinate system. Let [itex]L^\mu_a = \partial_a x^\mu[/itex] and letting [itex]\tilde{L}^b_\nu = \partial_\nu x^b[/itex]. [itex]L[/itex] and [itex]\tilde{L}[/itex] are inverses, in the sense that:

[itex]L^\mu_a \tilde{L}^b_\mu = \delta^b_a[/itex]
[itex]L^\mu_a \tilde{L}^a_\nu = \delta^\mu_\nu[/itex]

Then since [itex]\nabla_a V^b[/itex] is a tensor, we can rewrite it in terms of the curvilinear coordinates:

[itex]\nabla_a V^b = L^\mu_a \tilde{L}^b_\nu \nabla_\mu V^\nu[/itex]

We can also write:

[itex]\partial_a V^b = L^\mu_a \partial_\mu (\tilde{L}^b_\nu V^\nu)[/itex]
[itex] = L^\mu_a (\partial_\mu \tilde{L}^b_\nu) V^\nu + L^\mu_a \tilde{L}^b_\nu \partial_\mu V^\nu[/itex]

Here's where I think we have to do some unmotivated manipulation. We want to pull out a factor of [itex]\tilde{L}^b_\nu[/itex]. So we can rewrite:

[itex](\partial_\mu \tilde{L}^b_\nu) V^\nu [/itex]
[itex]= (\partial_\mu \tilde{L}^b_\lambda) V^\lambda [/itex]
[itex]= \delta^b_c (\partial_\mu \tilde{L}^c_\lambda) V^\lambda[/itex]
[itex]= \tilde{L}^b_\nu L^\nu_c (\partial_\mu \tilde{L}^c_\lambda) V^\lambda[/itex]

So we can write:
[itex]\partial_a V^b = L^\mu_a \tilde{L}^b_\nu [ L^\nu_c (\partial_\mu \tilde{L}^c_\lambda) V^\lambda + \partial_\mu V^\nu][/itex]

Putting the facts together, we have:
[itex]L^\mu_a \tilde{L}^b_\nu \nabla_\mu V^\nu = L^\mu_a \tilde{L}^b_\nu [ L^\nu_c (\partial_\mu \tilde{L}^c_\lambda) V^\lambda + \partial_\mu V^\nu][/itex]

which implies that
[itex]\nabla_\mu V^\nu = L^\nu_c (\partial_\mu \tilde{L}^c_\lambda) V^\lambda + \partial_\mu V^\nu[/itex]

Comparing that with the usual expression for [itex]\nabla_\mu V^\nu[/itex] tells us that:

[itex]\Gamma^\nu_{\mu \lambda} = L^\nu_c (\partial_\mu \tilde{L}^c_\lambda)[/itex]

 

It's just [itex]g^{\mu \nu} \nabla_\mu \nabla_\nu \phi = 0[/itex]. Expanding in terms of Christoffel symbols, it's:

[itex]g^{\mu \nu} (\partial_\mu \partial_\nu \phi + \Gamma^\lambda_{\mu \nu} \partial_\lambda \phi) = 0[/itex]

[edit] I got a sign wrong. The covariant derivative of a covector has a minus sign:

[itex]g^{\mu \nu} (\partial_\mu \partial_\nu \phi - \Gamma^\lambda_{\mu \nu} \partial_\lambda \phi) = 0[/itex]