How Do You Calculate the D'Alembert Operator in a Given Metric for GR?

[ChrisJ]

Was not sure weather to post, this here or in differential geometry, but is related to a GR course, so...

I am having some trouble reproducing a result, I think it is mainly down to being very new to tensor notation and operations.

But, given the metric $ds^2 = -dudv + \frac{(v-u)^2}{4} \left( d\theta^2 + \sin^2 \theta d\phi^2 \right)$

<br /> g_{\alpha\beta} =<br /> <br /> \begin{pmatrix}<br /> 0 &amp; -\frac{1}{2} &amp; 0 &amp; 0 \\<br /> -\frac{1}{2} &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; \frac{(v-u)^2}{4} &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; \frac{(v-u)^2}{4} \sin^2 \theta<br /> \end{pmatrix}<br />

and given this definition of the d'Alambert operator $\Box := g^{\alpha\beta}\partial_{\alpha}\partial_{\beta}$ , reproduce the following given the d'Alambert acting on a function $f(u,v)$

<br /> <br /> \Box f(u,v) = 4 \left( -\frac{1}{v-u} \frac{\partial f}{\partial u} + \frac{\partial f}{\partial v} - \frac{\partial^2 f}{\partial u \partial v} \right)<br />

And when I try to to reproduce it, I can see from the definition that the only non-zero parts are where the inverse metric components are $g^{uv} = -2$ and $g^{vu} = -2$ . The $g^{\theta \theta}$ and $g^{\phi \phi}$ bits would be zero since the function is just of $u$ and $v$.

So what I get is this...
<br /> \Box := g^{\alpha\beta}\partial_{\alpha}\partial_{\beta}\\<br /> \Box f(u,v) = g^{uv}\partial_{u}\partial_{v} f+ g^{vu}\partial_{v}\partial_{u}f = -4 \frac{\partial^2 f}{\partial u \partial v}<br />

And I can't seem to see what I am missing here? Any help is really appreciated. Thanks.

 

[Orodruin]

The d'Alembertian is not $\Box = g^{\alpha\beta} \partial_\alpha \partial_\beta$, it is $\Box = \nabla^a \nabla_a$, which for scalar fields reduces to
$$
\Box f = g^{ab} \nabla_b \nabla_a = g^{ab} \nabla_b \partial_a f = g^{ab}(\partial_b\partial_a f - \Gamma_{ba}^c \partial_c f).
$$

 

[ChrisJ]

The d'Alembertian is not $\Box = g^{\alpha\beta} \partial_\alpha \partial_\beta$, it is $\Box = \nabla^a \nabla_a$, which for scalar fields reduces to
$$
\Box f = g^{ab} \nabla_b \nabla_a = g^{ab} \nabla_b \partial_a f = g^{ab}(\partial_b\partial_a f - \Gamma_{ba}^c \partial_c f).
$$

Oh. Ok.. thanks will give that a go!

 

[ChrisJ]

Oh. Ok.. thanks will give that a go!

Previous to trying this I found all of the christoffel symbols for the metric define in the OP, so now trying with this new definition, it still simplifies to $g^{ab}\partial_a \partial_b$ because $a$ and $b$ can only take on $u$ or $v$ since there are not partials wrt to the others, and there are no non-zero Christoffel symbols for $\Gamma^c_{uv}$ , so I must still be missing something

 

[Orodruin]

Do you have any Christoffel symbols of the form $\Gamma^u_{ab}$?

 

[ChrisJ]

Do you have any Christoffel symbols of the form $\Gamma^u_{ab}$?

Yes, but only for $\Gamma^u_{\theta\theta}$ and $\Gamma^u_{\phi\phi}$ (and the same, but with with $v$ as $c$). But that would lead to partials wrt to $\theta$ and $\phi$, which don't appear in what I am trying to reproduce.

 

[Orodruin]

But that would lead to partials wrt to $theta$ and $\phi$

No, this is wrong.

Edit: To be more specific, for example $\Gamma^u_{\theta\theta}$ leads to the term
$$
-\Gamma^{u}_{\theta\theta} g^{\theta\theta} \partial_u f,
$$
which includes a derivative wrt $u$.

 

[ChrisJ]

No, this is wrong.

Sorry! Yes, I just saw it, the $\partial_c$ bit,

No, this is wrong.

Edit: To be more specific, for example $\Gamma^u_{\theta\theta}$ leads to the term
$$
-\Gamma^{u}_{\theta\theta} g^{\theta\theta} \partial_u f,
$$
which includes a derivative wrt $u$.

Thank you! I just saw it, ok, so the terms with $\Gamma^u_{\theta\theta}$ and $\Gamma^u_{\phi\phi}$ lead to $\frac{-4}{v-u}\frac{\partial f}{\partial u}$ (and the same for the ones with $\Gamma^v_{\theta\theta}$ and $\Gamma^v_{\phi\phi}$ lead to $\frac{4}{v-u}\frac{\partial f}{\partial u}$. And then the one that I already found in the OP with the zero christoffel symbol, leads to $-4\frac{\partial^2 f}{\partial u \partial v}$

 

[Orodruin]

So everything works out now?

 

[ChrisJ]

So everything works out now?

Yes, thank you!