Show charge conservation in a curved spacetime

[etotheipi]

Homework Statement

Given Maxwell's equations $\nabla_a F^{ab} = -4\pi j^b$ and $\nabla_{[a}F_{bc]} = 0$

Relevant Equations

N/A

For the flat spacetime we could just use that partial derivatives commute as well as the antisymmetry of $F^{ab}$, i.e. $\partial_b \partial_a F^{ab} = -\partial_b \partial_a F^{ba} = -\partial_a \partial_b F^{ba} = -\partial_b \partial_a F^{ab} \implies \partial_b \partial_a F^{ab} = - 4\pi \partial_b j^b = 0 \implies \partial_b j^b = 0$.

The covariant derivative doesn't necessarily commute, I can only get $\left[\nabla_b \nabla_a + \nabla_a \nabla_b\right] F^{ab} = 0$ from the antisymmetry of $F^{ab}$. But I tried to use the Riemann tensor to write down$$[\nabla_b \nabla_a - \nabla_a \nabla_b]F^{cd} = -{R_{bae}}^{c} F^{ed} - {R_{bae}}^d F^{ce}$$ $$\nabla_a \nabla_b F^{ab} = \nabla_b \nabla_a F^{ab} + {R_{bae}}^a F^{eb} + {R_{bae}}^b F^{ae}$$So we have$$0 = \left[\nabla_b \nabla_a + \nabla_a \nabla_b\right] F^{ab} = 2\nabla_b \nabla_a F^{ab} + {R_{bae}}^a F^{eb} + {R_{bae}}^b F^{ae} = 2\nabla_b \nabla_a F^{ab} + R_{be} F^{eb} - {R_{be}} F^{be}$$where in the last term I relabelled the dummy indices from $a$ to $b$. Why aren't the last two terms vanishing, instead they sum to $R_{be} F^{eb} - {R_{be}} F^{be} = 2R_{be} F^{eb}$? Thanks!

 

[TSny]

$$0 = \left[\nabla_b \nabla_a + \nabla_a \nabla_b\right] F^{ab} = 2\nabla_b \nabla_a F^{ab} + {R_{bae}}^a F^{eb} + {R_{bae}}^b F^{ae} = 2\nabla_b \nabla_a F^{ab} + R_{be} F^{eb} - {R_{be}} F^{be}$$where in the last term I relabelled the dummy indices from $a$ to $b$. Why aren't the last two terms vanishing?

Does $R_{a b}$ have any symmetry properties?

 

[etotheipi]

Ah, that's it! The properties of the Riemann tensor are, ${R_{abc}}^d = - {R_{bac}}^d$, and secondly ${R_{[abc]}}^d = 0$. Thirdly, if the covariant derivative is chosen to be the one that is metric compatible (i.e. $\nabla_a g_{bc} = 0$), then $R_{abcd} = -R_{abdc}$, and finally we have the Bianchi identity $\nabla_{[a} {R_{bc]d}}^e$.

Let's start then with the second property,$$\frac{1}{6} \left[ {R_{abc}}^d - {R_{acb}}^d + {R_{bca}}^d - {R_{bac}}^d + {R_{cab}}^d - {R_{cba}}^d \right] = 0$$Now we use the antisymmetry in the first two arguments (property 1), multiply out the irrelevant numerical factors,$$\begin{align*}{R_{abc}}^d + {R_{bca}}^d + {R_{cab}}^d = 0 \implies {R_{abc}}^d - {R_{cba}}^d + {R_{cab}}^d

\end{align*}$$After re-writing the second term, then letting $d=b$ and summing over $b$,$$R_{ac} - R_{ca} + {R_{cab}}^b = 0$$However, since$${R_{cab}}^b = g^{be}R_{cabe} = -g^{be}R_{caeb} = - {R_{cae}}^{e} = - {R_{cab}}^{b}$$we must have ${R_{cab}}^b=0$, and thus$$R_{ac} = R_{ca}$$Hence, going back to the question,$$\begin{align*}0 = 2\nabla_b \nabla_a F^{ab} + R_{be} F^{eb} - {R_{be}} F^{be} &= 2\nabla_b \nabla_a F^{ab} + R_{be} F^{eb} - {R_{eb}} F^{be} \\

&= 2\nabla_b \nabla_a F^{ab} + R_{be} F^{eb} - {R_{be}} F^{eb} = 2\nabla_b \nabla_a F^{ab}

\end{align*}$$And thus $\nabla_b j^b = 0$, as required. Thanks again, I hope the above is alright!

 

[TSny]

That looks good.

An alternate approach to showing that $\nabla_b J^b = 0$ follows from $\nabla_a F^{ab} = -4 \pi J^b$ is to use a couple of well-known identities regarding divergences. These identities are derived in the standard texts.

For any vector $J^a$, $$\nabla_a J^a = \frac{1}{\sqrt{|g|}}\partial_a \left(\sqrt{|g|} \, J^a \right) \qquad \qquad [1]$$

For any antisymmetric tensor $F^{ab}$, $$\nabla_a F^{ab} = \frac{1}{\sqrt{|g|}}\partial_a \left(\sqrt{|g|} \, F^{ab} \right) \qquad \quad[2]$$

You don't need to invoke the curvature tensor if you use these.

 

[JD_PM]

Quick question: was this exercise aimed at showing that $\nabla_{\mu} j^{\mu} = 0$ by means of Maxwell's equations (i.e. $\nabla_{\mu} F^{\mu \nu} = -4\pi j^{\nu}$?

 

[etotheipi]

That looks good.

An alternate approach to showing that $\nabla_b J^b = 0$ follows from $\nabla_a F^{ab} = -4 \pi J^b$ is to use a couple of well-known identities regarding divergences. These identities are derived in the standard texts.

For any vector $J^a$, $$\nabla_a J^a = \frac{1}{\sqrt{|g|}}\partial_a \left(\sqrt{|g|} \, J^a \right) \qquad \qquad [1]$$

For any antisymmetric tensor $F^{ab}$, $$\nabla_a F^{ab} = \frac{1}{\sqrt{|g|}}\partial_a \left(\sqrt{|g|} \, F^{ab} \right) \qquad \quad[2]$$

You don't need to invoke the curvature tensor if you use these.

Would that be,$$\nabla_b J^b = \frac{1}{\sqrt{|g|}} \partial_b \left( -\frac{\sqrt{|g|}}{4\pi} \cdot \frac{1}{\sqrt{|g|}} \partial_a \left( \sqrt{|g|} F^{ab}\right) \right) = -\frac{1}{4\pi} \partial_b \partial_a F^{ab} = 0$$where I needed to use that $\partial_a \sqrt{|g|} \equiv \sqrt{|g|} \partial_a$, because the determinant of the metric doesn't depend on the coordinates.

That's pretty cool!

Quick question: was this exercise aimed at showing that $\nabla_{\mu} j^{\mu} = 0$ by means of Maxwell's equations (i.e. $\nabla_{\mu} F^{\mu \nu} = -4\pi j^{\nu}$?

Yeah, it's just that. The more advanced exercise might be to do it from Noether's theorem, but I don't know how do that yet!

 

[JD_PM]

If you wish to prove TSny $[1]$ formula, you may need to show first that the Christoffel symbol takes the form

$$\Gamma_{\mu \nu}^{\mu} =\frac{1}{\sqrt{|g|}}\partial_{\nu} \left(\sqrt{|g|}\right)$$

 

[etotheipi]

Okay, I'm a bit stuck. I did the easy part,$$\begin{align*}

\Gamma^{\mu}_{\mu \nu} &= \frac{1}{2} g^{\mu \rho}(\partial_{\mu} g_{\nu \rho} + \partial_{\nu} g_{\mu \rho} - \partial_{\rho} g_{\mu \nu}) \\

&= \left[ \frac{1}{2} g^{\mu \rho}\partial_{\mu} g_{\nu \rho} - \frac{1}{2} g^{\rho \mu} \partial_{\mu} g_{\rho \nu} \right] + \frac{1}{2} g^{\mu \rho} \partial_{\nu} g_{\mu \rho} = \frac{1}{2} g^{\mu \rho} \partial_{\nu} g_{\mu \rho}

\end{align*}$$but how do I get that into the form of the determinant? Maybe I need to go see if Tong mentioned it anywhere

 

[etotheipi]

Oh, Tong did mention it on page 111 of his third set of notes, the identity$$\text{tr}(\text{log}(A)) = \text{log}(\text{det}(A))$$for any diagonalisable matrix $A$

 

[TSny]

Would that be,$$\nabla_b J^b = \frac{1}{\sqrt{|g|}} \partial_b \left( -\frac{\sqrt{|g|}}{4\pi} \cdot \frac{1}{\sqrt{|g|}} \partial_a \left( \sqrt{|g|} F^{ab}\right) \right) = -\frac{1}{4\pi} \partial_b \partial_a F^{ab} = 0$$where I needed to use that $\partial_a \sqrt{|g|} \equiv \sqrt{|g|} \partial_a$, because the determinant of the metric doesn't depend on the coordinates.

That's almost it. But, in general, $g$ depends on the coordinates. For example, using spherical coordinates in Minkowski spacetime, $g = -r^4 \sin^2 \theta$.

So, you cannot "factor out" $\sqrt{|g|}$ in the expression $\partial_a \left( \sqrt{|g|} F^{ab}\right)$. But you should still be able to get the result.

 

[etotheipi]

Oh, Tong did mention it on page 111 of his third set of notes, the identity$$\text{tr}(\text{log}(A)) = \text{log}(\text{det}(A))$$for any diagonalisable matrix $A$

Okay, in that case,$$\begin{align*}

\nabla_a J^a = \partial_a J^a + J^c \Gamma^{a}_{ac} &= \partial_a J^a + \frac{J^a}{\sqrt{|g|}} \partial_a \left(\sqrt{|g|} \right) \\

&= \frac{1}{\sqrt{|g|}} \left[ \sqrt{|g|}\partial_a J^a + J^a \partial_a \left(\sqrt{|g|} \right)\right] \\

&= \frac{1}{\sqrt{|g|}}\partial_a \left(\sqrt{|g|} J^a \right)

\end{align*}$$

 

[etotheipi]

That's almost it. But, in general, $g$ depends on the coordinates. For example, using spherical coordinates in Minkowski spacetime, $g = -r^4 \sin^2 \theta$.

So, you cannot "factor out" $\sqrt{|g|}$ in the expression $\partial_a \left( \sqrt{|g|} F^{ab}\right)$. But you should still be able to get the result.

That makes sense! I tried$$\partial_b \partial_a \left( F^{ab} \sqrt{|g|} \right) = \partial_b \left( \sqrt{|g|} \right) \partial_a F^{ab} + \sqrt{|g|} \partial_b \partial_a F^{ab} + \partial_a \left( \sqrt{|g|} \right) \partial_b F^{ab} + F^{ab} \partial_b \partial_a \left( \sqrt{|g|} \right)$$The second term on the RHS is zero, whilst the first and third terms sum to $$\begin{align*}

\partial_b \left( \sqrt{|g|} \right) \partial_a F^{ab} + \partial_a \left( \sqrt{|g|} \right) \partial_b F^{ab} &=

\partial_b \left( \sqrt{|g|} \right) \partial_a F^{ab} + \partial_b \left( \sqrt{|g|} \right) \partial_a F^{ba} \\

&= \partial_b \left( \sqrt{|g|} \right) \partial_a F^{ab} - \partial_b \left( \sqrt{|g|} \right) \partial_a F^{ab} = 0
\end{align*}$$and thus$$\partial_b \partial_a \left( F^{ab} \sqrt{|g|} \right) = F^{ab} \partial_b \partial_a \left( \sqrt{|g|} \right)$$However, $$F^{ab} \partial_b \partial_a \left( \sqrt{|g|} \right) = - F^{ba} \partial_b \partial_a \left( \sqrt{|g|} \right) = - F^{ba} \partial_a \partial_b \left( \sqrt{|g|} \right) = - F^{ab} \partial_b \partial_a \left( \sqrt{|g|} \right)$$so ´$F^{ab} \partial_b \partial_a \left( \sqrt{|g|} \right) = 0$ and thus $\partial_b \partial_a \left( F^{ab} \sqrt{|g|} \right) = 0$.

 

[TSny]

$$\partial_b \partial_a \left( F^{ab} \sqrt{|g|} \right) $$

Note that the quantity inside the parentheses is antisymmetric in $a$ and $b$.

 

[etotheipi]

$$\partial_b \partial_a \left( F^{ab} \sqrt{|g|} \right) $$

Note that the quantity inside the parentheses is antisymmetric in $a$ and $b$.

Ah, whoops , well it was good practice anyway . Yeah, I probably should have seen that. Oh well. Thanks for explaining the second approach to the problem, the more methods the merrier!

 

[JD_PM]

Okay, I'm a bit stuck. I did the easy part,$$\begin{align*}

\Gamma^{\mu}_{\mu \nu} &= \frac{1}{2} g^{\mu \rho}(\partial_{\mu} g_{\nu \rho} + \partial_{\nu} g_{\mu \rho} - \partial_{\rho} g_{\mu \nu}) \\

&= \left[ \frac{1}{2} g^{\mu \rho}\partial_{\mu} g_{\nu \rho} - \frac{1}{2} g^{\rho \mu} \partial_{\mu} g_{\rho \nu} \right] + \frac{1}{2} g^{\mu \rho} \partial_{\nu} g_{\mu \rho} = \frac{1}{2} g^{\mu \rho} \partial_{\nu} g_{\mu \rho}

\end{align*}$$but how do I get that into the form of the determinant? Maybe I need to go see if Tong mentioned it anywhere

Good job, you are halfway!

Oh, Tong did mention it on page 111 of his third set of notes, the identity$$\text{tr}(\text{log}(A)) = \text{log}(\text{det}(A))$$for any diagonalisable matrix $A$

Indeed, we need to use Jacobi's identity to prove the formula at #7. I suggest you massage it a bit, so that it is easier to apply to our specific problem.

Spoiler: Check your second part of the proof

Let us use the chain rule on the RHS

\begin{equation*}
\frac{1}{\sqrt{|g|}}\partial_{\nu} \left(\sqrt{|g|}\right) = \frac{1}{\sqrt{|g|}}\left( \frac{1}{2 \sqrt{|g|}}\partial_{\nu} g \right) = \frac{1}{2g} \partial_{\nu} g
\end{equation*}

Now we make the following observations: the metric $g_{\mu \nu}$ is an invertible square matrix $\text{M}$ (i.e. $g^{\mu \rho}g_{\rho \nu}=\delta_{\nu}^{\mu}$), the derivative is a linear operator $\text{D}$ and the determinant of the metric is $\sqrt{|g|}=\det(g_{\mu \nu}):=g$.

As you pointed out, at this stage we need Jacobi's identity. Let us though give it in a more convenient form

\begin{equation*}
\text{D} \det(\text{M}) = \det(\text{M}) tr(\text{M}^{-1} \text{D} \text{M})
\end{equation*}

Applying it to our particular problem we get

\begin{align*}
\partial_{\nu} g &= g tr(g^{\mu \rho} \partial_{\nu} g_{\rho \sigma}) \\
&= g g^{\mu \rho} \partial_{\nu} g_{\rho \mu}
\end{align*}

Thus we get

\begin{equation*}
\frac{1}{\sqrt{|g|}}\partial_{\nu} \left(\sqrt{|g|}\right) = \frac{1}{2g} g g^{\mu \rho} \partial_{\nu} g_{\rho \mu}=\frac{1}{2} g^{\mu \rho} \partial_{\nu} g_{\rho \mu}
\end{equation*}

Notice that what we have just obtained is precisely the same that you got at #8. Thus we only have to equate both equations i.e.

\begin{equation*}
\Gamma^{\mu}_{\mu \nu} = \frac{1}{2} g^{\mu \rho} \partial_{\nu} g_{\rho \mu} = \frac{1}{\sqrt{|g|}}\partial_{\nu} \left(\sqrt{|g|}\right)
\end{equation*}

QED.

 

[JD_PM]

Still hungry? Here's a quick snack!

Show that the Klein Gordon operator (i.e. $\Box := \nabla_{\mu}\nabla^{\mu}$), when applied to a scalar field $\phi(x)$, takes the following form

\begin{equation*}
\Box \phi = \frac{1}{\sqrt{|g|}} \partial_{\mu} \left(\sqrt{|g|} g^{\mu \nu} \partial_{\nu} \phi \right)
\end{equation*}

Spoiler: Hint

The approach is very similar to the one you used to show TSny $[1]$ formula. You will also need to use the Christoffel symbol formula we have just proven

 

[vanhees71]

Another hint (much less tedious). Use the action principle for the free KG field. In flat space the action reads
$$A=\int \mathrm{d}^4 x \frac{1}{2} (\partial_{\mu} \phi)(\partial^{\mu} \phi)-\frac{m}{2} \phi^2.$$
Just write it in a generally covariant form for arbitrary spacetime coordinates $q^{\mu}$ and then evaluate the Euler-Lagrange equation.

 

[etotheipi]

Show that the Klein Gordon operator (i.e. $\Box := \nabla_{\mu}\nabla^{\mu}$), when applied to a scalar field $\phi(x)$, takes the following form

\begin{equation*}
\Box \phi = \frac{1}{\sqrt{|g|}} \partial_{\mu} \left(\sqrt{|g|} g^{\mu \nu} \partial_{\nu} \phi \right)
\end{equation*}

I tried, given the covariant derivative of a scalar field is just a partial derivative, and using the result we already worked for @TSny's equation [1], $$\begin{align*}

\nabla_{\mu} \nabla^{\mu} \phi = \nabla_{\mu} \partial^{\mu} \phi &= \frac{1}{\sqrt{|g|}} \partial_{\mu} \left(

\sqrt{|g|} \partial^{\mu} \phi \right) \\

&= \frac{1}{\sqrt{|g|}} \partial_{\mu} \left(\sqrt{|g|} g^{\mu \nu} \partial_{\nu} \phi \right)\end{align*}$$
Also, one notational question, do we define $g \equiv \text{det}[(g_{\mu \nu})]$, and then I guess $|g|$ would just be the modulus, which we need to use when we have a Lorentzian metric?

Another hint (much less tedious). Use the action principle for the free KG field. In flat space the action reads$$A=\int \mathrm{d}^4 x \frac{1}{2} (\partial_{\mu} \phi)(\partial^{\mu} \phi)-\frac{m}{2} \phi^2.$$
Just write it in a generally covariant form for arbitrary spacetime coordinates $q^{\mu}$ and then evaluate the Euler-Lagrange equation.

I haven't learned anything about field theory so I'm probably going to get a lot of this wrong, but here's what I tried. I don't know what it means to write it in general covariant form, so instead I just tried to see what I could do in the special case that $\nabla \equiv \partial$...$$\mathcal{L} = \frac{1}{2} g^{\mu \nu} (\partial_{\mu} \phi) (\partial_{\nu} \phi) - \frac{m}{2} \phi^2$$Then we'd have something like$$\frac{\partial \mathcal{L}}{\partial \phi} = -m\phi$$and$$\begin{align*}

\partial_{\sigma} \frac{\partial \mathcal{L}}{\partial(\partial_{\sigma} \phi)} =

\partial_{\sigma} \left[ \frac{1}{2} g^{\mu \nu} \cdot 2 \frac{\partial(\partial_{\mu} \phi)}{\partial(\partial_{\sigma} \phi)} \partial_{\nu} \phi \right] &= \partial_{\sigma} \left[ g^{\mu \nu} \delta^{\sigma}_{\mu} \partial_{\nu} \phi \right] \\

&= \partial_{\mu}(g^{\mu \nu} \partial_{\nu} \phi) \\

&= \partial^{\nu} \partial_{\nu} \phi + \left(\partial_{\mu} g^{\mu \nu}\right) \left(\partial_{\nu} \phi \right)

\end{align*}$$So EL equations give$$-m\phi = \partial^{\nu} \partial_{\nu} \phi + \left(\partial_{\mu} g^{\mu \nu}\right) \left(\partial_{\nu} \phi \right)$$Is that completely wrong?

 

[TSny]

I tried, given the covariant derivative of a scalar field is just a partial derivative, and using the result we already worked for @TSny's equation [1], $$\begin{align*}

\nabla_{\mu} \nabla^{\mu} \phi = \nabla_{\mu} \partial^{\mu} \phi &= \frac{1}{\sqrt{|g|}} \partial_{\mu} \left(

\sqrt{|g|} \partial^{\mu} \phi \right) \\

&= \frac{1}{\sqrt{|g|}} \partial_{\mu} \left(\sqrt{|g|} g^{\mu \nu} \partial_{\nu} \phi \right)\end{align*}$$

Looks good to me.

Also, one notational question, do we define $g \equiv \text{det}[(g_{\mu \nu})]$, and then I guess $|g|$ would just be the modulus, which we need to use when we have a Lorentzian metric?

Yes

I haven't learned anything about field theory so I'm probably going to get a lot of this wrong, but here's what I tried. I don't know what it means to write it in general covariant form

The non-covariant form is
$$A=\int \mathrm{d}^4 x\left[ \frac{1}{2} (\partial_{\mu} \phi)(\partial^{\mu} \phi)-\frac{m}{2} \phi^2\right]$$
As you noted, the integrand is already covariant due to $\nabla_{\nu} \phi = \partial_{\nu} \phi$. However, the volume element $d^4x$ is not covariant. But this can be fixed by replacing $d^4x$ by $d^4x \sqrt{|g|}$, which is covariant. That is, in switching from coordinates $x^{\mu}$ to $\overline x^{\mu}$, you find $d^4x \sqrt{|g|} = d^4 \overline x \sqrt{|\overline g|}$. The covariant form of $A$ then turns out to be $$A=\int \mathrm{d}^4 x \sqrt{|g|} \left[\frac{1}{2} (\partial_{\mu} \phi)(\partial^{\mu} \phi)-\frac{m}{2} \phi^2 \right] = \int \mathrm{d}^4 x \left[\sqrt{|g|} \frac{1}{2} (\partial_{\mu} \phi)(\partial^{\mu} \phi)-\sqrt{|g|}\frac{m}{2} \phi^2 \right] $$ You can then apply the Euler-Lagrange equations to the quantity in brackets on the far right.

 

[etotheipi]

Thank you, that's pretty interesting! So covariance in this context means that the volume element is an invariant. I'll try to check the explicit transformation later. I'm not sure about what I did$$\frac{\partial \mathcal{L}}{\partial \phi} = \nabla_{\mu} \left(\frac{\partial \mathcal{L}}{\partial(\nabla_{\mu} \phi)} \right) \equiv \nabla_{\mu} \left(\frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)} \right)$$The LHS is fine,$$\frac{\partial \mathcal{L}}{\partial \phi} = - \sqrt{|g|} m\phi$$The RHS is$$\begin{align*}

\nabla_{\mu} \left( \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)} \right) &=
\nabla_{\mu} \left( \sqrt{|g|} g^{\mu \nu} \partial_{\nu} \phi \right) \\

&= \partial_{\mu} \left( \sqrt{|g|} g^{\mu \nu} \partial_{\nu} \phi \right) + \frac{\sqrt{|g|} g^{\mu \nu} \partial_{\nu} \phi}{\sqrt{|g|}} \partial_{\mu} \left(\sqrt{|g|} \right) \\

&= 2\partial_{\mu} \left( \sqrt{|g|} g^{\mu \nu} \partial_{\nu} \phi \right) - \sqrt{|g|} \partial^{\nu} \partial_{\nu} \phi \quad \overset{!}{=} - \sqrt{|g|} m\phi

\end{align*}$$So$$\partial^{\nu} \partial_{\nu} \phi \equiv \partial_{\nu} \partial^{\nu} \phi = m \phi + \frac{2}{\sqrt{|g|}}\partial_{\mu} \left( \sqrt{|g|} g^{\mu \nu} \partial_{\nu} \phi \right) $$Going to be honest, I don't know where I'm going with this, since from the sounds of it the derivation should be straightforward...

I'll try again in the morning anyway, because I'm going to go to sleep early today

 

[vanhees71]

The point is that you have
$$\partial_{\mu} \frac{\partial \mathcal{L}}{\partial_{\partial_{\mu} \phi}}=\frac{\partial \mathcal{L}}{\partial \phi}$$
with the generally covariant action principle, where
$$\mathcal{L}=\sqrt{|g|} \left [\frac{1}{2} g^{\mu \nu} (\partial_{\mu} \phi) (\partial_{\nu}\phi)-\frac{m^2}{2} \phi^2 \right].$$
The variational principle doesn't care about covariant derivatives, i.e., the EL equations are with the partial derivatives. If the action is generally covariant you get of course generally covariant equations.

 

[etotheipi]

Thanks for the tips! Unfortunately, I'm still slightly too stupid to pick up on the hints you're dropping . I guessed the other thing I could try would be to think about if the metric depends explicitly on $\phi$ and $\partial_{\mu} \phi$, i.e. whether these terms have non-zero partial derivatives.$$\mathcal{L} = \frac{1}{2} \sqrt{|g|} g^{\mu \nu} (\partial_{\mu} \phi)(\partial_{\nu} \phi) - \frac{1}{2} m \sqrt{|g|} \phi^2$$With respect to $\phi$,$$\frac{\partial \mathcal{L}}{\partial \phi} = -m\sqrt{|g|} \phi - \frac{1}{2} m \phi^2 \partial_{\phi} \left( \sqrt{|g|} \right) + \frac{1}{2} (\partial_{\mu} \phi) (\partial_{\nu} \phi) \partial_{\phi} \left( \sqrt{|g|} g^{\mu \nu} \right)$$With respect to $\partial_{\sigma} \phi$,$$
\frac{\partial \mathcal{L}}{\partial(\partial_{\sigma} \phi)} = \sqrt{|g|} g^{\sigma \nu} (\partial_{\nu} \phi) + \frac{1}{2} (\partial_{\mu} \phi)(\partial_{\nu} \phi) \partial_{(\partial_{\sigma} \phi)} \left( \sqrt{|g|} g^{\mu \nu} \right) - \frac{1}{2}m \phi^2 \partial_{(\partial_{\sigma} \phi)} \left( \sqrt{|g|} \right)
$$With respect to $q^{\sigma}$,$$\begin{align*}

\partial_{\sigma} \left( \frac{\partial \mathcal{L}}{\partial(\partial_{\sigma} \phi)} \right)

= \partial_{\sigma} \left( \sqrt{|g|} g^{\sigma \nu} (\partial_{\nu} \phi) \right)

&+ \frac{1}{2}(\partial_{\mu} \phi)(\partial_{\nu} \phi) \partial_{\sigma} \partial_{(\partial_{\sigma} \phi)}\left( \sqrt{|g|} g^{\mu \nu} \right) \\

&- \frac{1}{2}m \phi^2 \partial_{\sigma} \partial_{(\partial_{\sigma} \phi)} \left( \sqrt{|g|} \right) \\

&- m\phi (\partial_{\sigma} \phi) \partial_{(\partial_{\sigma} \phi)}\left( \sqrt{|g|} \right)\end{align*}
$$When I look to try and insert these into the EL condition for the extremum, I got stuck again because I don't know how to deal with $\partial_{\sigma} \partial_{(\partial_{\sigma} \phi)}$ and $(\partial_{\sigma} \phi)\partial_{(\partial_{\sigma} \phi)}$.

Anyway sorry I couldn't get any further, if this is still too far off then it's probably better that I try and do some even more basic questions first, because I don't think there's much else you can say without giving away the answer

 

[TSny]

The metric doesn't depend on $\phi$. So, $\partial_{\phi} \left( \sqrt{|g|}\right) = 0$, etc.

There was a typo in the original expression for $\mathcal{L}$. The mass $m$ should be squared.

 

[etotheipi]

The metric doesn't depend on $\phi$. So, $\partial_{\phi} \left( \sqrt{|g|}\right) = 0$, etc.

There was a typo in the original expression for $\mathcal{L}$. The mass $m$ should be squared.

But doesn't that just give$$-m^2 \phi = \frac{1}{\sqrt{|g|}} \partial_{\mu} \left( \sqrt{|g|} g^{\mu \nu} \partial_{\nu} \phi \right)$$which is what we want on the RHS but definitely not what we want on the LHS?

 

[TSny]

But doesn't that just give$$-m^2 \phi = \frac{1}{\sqrt{|g|}} \partial_{\mu} \left( \sqrt{|g|} g^{\mu \nu} \partial_{\nu} \phi \right)$$which is what we want on the RHS but definitely not what we want on the LHS?

The left side looks good. Are you worried about the negative sign on the left? That could be due to the choice of sign convention for the metric used by @vanhees71 when he wrote the Lagrangian. (+, -. -, -) vs. (-, +, +, +).

 

[etotheipi]

I'm just a bit confuzzled, because I thought the left hand side is supposed to be $\nabla_{\mu} \nabla^{\mu} \phi$. I'm not sure how I can extract the required identity from this last equation, I guess that's why I tried that strange (incorrect) form of the EL equations with the covariant derivative a few posts back, in order to get it to show up.

 

[etotheipi]

Oh, okay, I went to look at David Tong's QFT notes and on the first few pages he introduced the Klein-Gordon equation$$\square \phi = -m^2 \phi$$which is clearly what we want. Thanks for the help everyone! It's always fun to learn a little bit more than one expected to . Now if you'll excuse me, I'm just going to go cry for a little bit

 

[TSny]

I'm just a bit confuzzled, because I thought the left hand side is supposed to be $\nabla_{\mu} \nabla^{\mu} \phi$. I'm not sure how I can extract the required identity from this last equation, I guess that's why I tried that strange (incorrect) form of the EL equations with the covariant derivative a few posts back, in order to get it to show up.

Ok, I think I see what's bothering you. You were hoping that the EL equation for the Lagrangian would be a way to prove the identity $$\nabla_{\mu} \nabla^{\mu} \phi = \frac{1}{\sqrt{|g|}} \partial_{\mu} \left( \sqrt{|g|} g^{\mu \nu} \partial_{\nu} \phi \right)$$.

But, instead you find that EL yields $$-m^2 \phi = \frac{1}{\sqrt{|g|}} \partial_{\mu} \left( \sqrt{|g|} g^{\mu \nu} \partial_{\nu} \phi \right)$$

You already proved the first identity at the beginning of post #18. But you were expecting the Lagrangian approach to be another way to derive it.

All right, fair enough. You don't get the desired identity directly from EL. But if we assume that the generally covariant form of @vanhees71 's Lagrangian is correct for the Klein-Gordon field in curved spacetime, then the equation of motion must be the generally covariant form of the Klein-Gordon equation. In Minkowski spacetime, the Klein-Gordon equation is $\partial_{\mu} \partial^{\mu} \phi = -m^2 \phi$. The generally covariant form of this in curved spacetime is $\nabla_{\mu} \nabla^{\mu} \phi = -m^2 \phi$. Comparing this to what we got from EL, we can make the identification that $\nabla_{\mu} \nabla^{\mu} \phi = \frac{1}{\sqrt{|g|}} \partial_{\mu} \left( \sqrt{|g|} g^{\mu \nu} \partial_{\nu} \phi \right)$.

I'm not sure if this helps.

 

[etotheipi]

Yeah that helps, thanks . I have come across some variant of that equation before (when I asked a question here: https://www.physicsforums.com/threads/energy-current-for-field-satisfying-kg-equation.997421/) but it was written like$$\phi_{tt} - c^2 \nabla^2 \phi + m^2 c^4 \phi = 0$$So I guess we're using the special case where $\phi_{tt} = 0$. Anyway, guess I better study more. Thanks everyone!

 

[etotheipi]

Oh hang on, I just realized that the $\nabla^2$ here only acts on spatial components, and$$\phi_{tt} - c^2 \nabla^2 \phi + m^2 c^4 \phi = 0$$is equivalent to$$\square \phi = -m^2 \phi$$so we are certainly not using a special case where $\phi_{tt} = 0$. Sorry!

 

[TSny]

Anyway, guess I better study more.

Don't push yourself too much. You're doing amazingly well!

 

[vanhees71]

Ok, I think I see what's bothering you. You were hoping that the EL equation for the Lagrangian would be a way to prove the identity $$\nabla_{\mu} \nabla^{\mu} \phi = \frac{1}{\sqrt{|g|}} \partial_{\mu} \left( \sqrt{|g|} g^{\mu \nu} \partial_{\nu} \phi \right)$$.

If you want to do this an alternative way is to couple the field to an external scalar current $J$:
$$\mathcal{L}=\frac{1}{2} \sqrt{|g|} g^{\mu \nu} (\partial_{\mu} \phi)(\partial_{\nu} \phi)-\sqrt{|g|} J \phi.$$
Then you get
$$\Box \phi=J$$
as an equation of motion, and the variational principle tells you
$$\partial_{\mu} \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)}=\frac{\partial \mathcal{L}}{\partial \phi}.$$
Now
$$ \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)}=\sqrt{|g|} g^{\mu \nu} \partial_{\nu} \phi$$
and
$$\frac{\partial \mathcal{L}}{\partial \phi}=\sqrt{|g|} J,$$
from which
$$\partial_{\mu} (\sqrt{|g|} g^{\mu \nu} \partial_{\nu} \phi)=\sqrt{|g|} J \; \Rightarrow \; J=\Box \phi =\frac{1}{\sqrt{|g|}} \partial_{\mu} (\sqrt{|g|} g^{\mu \nu} \partial_{\nu} \phi).$$
QED

 

[JD_PM]

Hello! I was rereading this nice thread and wanted to discuss a bit further (if you are willing to do so as well of course! )

Mmm I am actually trying to show Maxwell's equations in vacuum i.e. $g^{\nu \mu}\nabla_{\mu} F_{\nu \sigma} = 0$ and $\nabla_{[\mu}F_{\nu \sigma]}=0$ using the same Riemann tensor property i.e. $\nabla_{\mu} \nabla_{\nu} A_{\sigma} = R_{\sigma \nu \mu \rho} A^{\rho}$, where $F_{\mu \nu} = \nabla_{\mu} A_{\nu} - \nabla_{\nu} A_{\mu}$.

Let's focus on $g^{\nu \mu}\nabla_{\mu} F_{\nu \sigma} = 0$ first.

We get

\begin{align*}
\nabla_{\mu} F_{\nu \sigma} &= \nabla_{\mu} \nabla_{\nu} A_{\sigma} - \nabla_{\mu} \nabla_{\sigma} A_{\nu} \\
&= R_{\sigma \nu \mu \rho} A^{\rho} - R_{\nu \sigma \mu \rho} A^{\rho} \\
&= 2R_{\sigma \nu \mu \rho} A^{\rho}
\end{align*}

Multiplying both sides by $g^{\nu \mu}$ yields

\begin{equation*}
\nabla^{\nu} F_{\nu \sigma} = 2 R_{\sigma \rho} A^{\rho} = 0
\end{equation*}

Where I used the fact that, in vacuum, $R_{\mu \nu} = 0$.

Does this proof look correct to you?

I am stuck on how to show that $\nabla_{[\mu}F_{\nu \sigma]}=0$ by means of $\nabla_{\mu} \nabla_{\nu} A_{\sigma} = R_{\sigma \nu \mu \rho} A^{\rho}$ though

I start expanding it out

\begin{align*}
\nabla_{[\mu } F_{\nu \rho]} &= \frac{1}{3!} \left( \nabla_{\mu} F_{\nu \rho} + \nabla_{\nu} F_{\rho \mu} + \nabla_{\rho} F_{\mu \nu} -\nabla_{\mu} F_{\rho \nu} -\nabla_{\nu} F_{\mu \rho} -\nabla_{\rho} F_{\nu \mu} \right) \\
&= \frac{1}{3} \left( R_{\rho \nu \mu \sigma} + R_{\mu \rho \nu \sigma} + R_{\nu \mu \rho \sigma} - R_{\nu \rho \mu \sigma} - R_{\rho \mu \nu \sigma} - R_{\mu \nu \rho \sigma}\right) A^{\sigma} \\
&= \frac{2}{3} \left( R_{\rho \nu \mu \sigma}+ R_{\mu \rho \nu \sigma} + R_{\nu \mu \rho \sigma}\right) A^{\sigma}
\end{align*}

But I do not see how to show that the last equation vanishes without having to explicitly write the Riemann tensor as a function of Christoffel symbols (which does not seem to be the best idea ).Regards.

JD.

 

[JD_PM]

\begin{align*}
\nabla_{[\mu } F_{\nu \rho]} &= \frac{1}{3!} \left( \nabla_{\mu} F_{\nu \rho} + \nabla_{\nu} F_{\rho \mu} + \nabla_{\rho} F_{\mu \nu} -\nabla_{\mu} F_{\rho \nu} -\nabla_{\nu} F_{\mu \rho} -\nabla_{\rho} F_{\nu \mu} \right) \\
&= \frac{1}{3} \left( R_{\rho \nu \mu \sigma} + R_{\mu \rho \nu \sigma} + R_{\nu \mu \rho \sigma} - R_{\nu \rho \mu \sigma} - R_{\rho \mu \nu \sigma} - R_{\mu \nu \rho \sigma}\right) A^{\sigma} \\
&= \frac{2}{3} \left( R_{\rho \nu \mu \sigma}+ R_{\mu \rho \nu \sigma} + R_{\nu \mu \rho \sigma}\right) A^{\sigma}
\end{align*}

But I do not see how to show that the last equation vanishes without having to explicitly write the Riemann tensor as a function of Christoffel symbols (which does not seem to be the best idea ).

Alright, I got it.

\begin{equation*}
R_{[\rho \nu \mu]\sigma} = 0
\end{equation*}

 

[etotheipi]

Just for fun:$$
\begin{align*}

\mathrm{d} \star F &= \star j \\

0 = \mathrm{d}\mathrm{d}\star F &= \mathrm{d} \star j \implies \int_{\Omega} \mathrm{d} \star j = \oint_{\partial \Omega} \star j = 0

\end{align*}$$