# Maxwell's Equations and the Variation of Metric Determinant

• etotheipi
Ah, the old "let's try to derive the equations of motion from the action" trick. I always forget about that one. Thanks!In summary, the source-free Maxwell equations can be derived from the action $$S = - \frac{1}{2} \int F \wedge \star F$$ by varying the action with respect to the dynamical variables and the background metric. Using the wedge product and the Hodge star, we can write the action as $$S[A] = - \int_{\Omega} \left( \frac{1}{2} F \wedge (\ast F) + A \wedge (\ast j)\right)$$ where ##\ast j## is the 3-form associated
etotheipi
Homework Statement
Derive the source-free Maxwell equations from the action$$S = - \frac{1}{2} \int F \wedge \star F$$
Relevant Equations
Some wedgey things
First, I let ##\omega = \sqrt{-g} dx^0 \wedge \dots \wedge dx^3## be the top-form on ##M##, and making use of the inner-product on the space of forms I can write\begin{align*} F \wedge \star F = g(F,F) \omega &= \frac{1}{4} F_{\mu \nu} F_{\rho \sigma} \begin{vmatrix} g(dx^{\mu}, dx^{\rho}) & g(dx^{\mu}, dx^{\sigma})\\ g(dx^{\nu}, dx^{\rho}) & g(dx^{\nu}, dx^{\sigma}) \end{vmatrix} \omega \\ &= \frac{1}{4} F_{\mu \nu} F_{\rho \sigma} (\delta^{\mu \rho} \delta^{\nu \sigma} - \delta^{\nu \rho} \delta^{\mu \sigma}) \omega\\ &= \frac{1}{4}\left( F^{\rho \sigma} F_{\rho \sigma} - F^{\sigma \rho} F_{\rho \sigma} \right) \omega = \frac{1}{2} F^{\sigma \rho} F_{\rho \sigma} \omega \end{align*}due to the anti-symmetry of ##F^{\rho \sigma}##. Replacing ##\omega## gives ##F \wedge \star F = \frac{1}{2} F^{\rho \sigma} F_{\rho \sigma} \sqrt{-g} d^4 x##, where I abbreviated ##d^4 x \equiv dx^0 \wedge \dots \wedge dx^3##. Now we can vary the action$$\delta S = - \frac{1}{4} \int d^4 x \left(\delta(\sqrt{-g}) F^{\rho \sigma} F_{\rho \sigma} + \sqrt{-g} F^{\rho \sigma} \delta(F_{\rho \sigma}) + \sqrt{-g} \delta(F^{\rho \sigma}) F_{\rho \sigma}\right)$$Because of following property of diagonal matrices ##A##,$$\frac{1}{\mathrm{det} A} \delta(\mathrm{det} A) = \mathrm{tr}(A^{-1} \delta A)$$we can write down the variation of the metric determinant$$\delta(\sqrt{-g}) = - \frac{1}{2} \sqrt{-g} g_{\mu \nu} \delta(g^{\mu \nu})$$But I don't know how to re-write the variations of the ##F_{\rho \sigma}## and ##F^{\rho \sigma}##. Wondered if someone could give a few pointers? Thanks

(Also, whilst I'm here, is there a faster/better way to work out ##F \wedge \star F##?)

vanhees71, JD_PM and Delta2
atyy said:

Ooh, those look quite nice! Give me like, half an hour or so to go through them. The problem is that right now I don't really have an intuitive feel for exterior calculus (I can't feel it in my bones yet..., you know ), so I'm really uncertain about many of these manipulations.

Delta2
etotheipi said:
Homework Statement:: Derive the source-free Maxwell equations from the action$$S = - \frac{1}{2} \int F \wedge \star F$$
Relevant Equations:: Some wedgey things
$$F \wedge \star F = \frac{1}{2} F^{\sigma \rho} F_{\sigma \rho} \omega .$$
Now we can vary the action
$$\delta S = - \frac{1}{4} \int d^4 x \left(\delta(\sqrt{-g}) F^{\rho \sigma} F_{\rho \sigma} + \sqrt{-g} F^{\rho \sigma} \delta(F_{\rho \sigma}) + \sqrt{-g} \delta(F^{\rho \sigma}) F_{\rho \sigma}\right)$$
But I don't know how to re-write the variations of the ##F_{\rho \sigma}## and ##F^{\rho \sigma}##. Wondered if someone could give a few pointers? Thanks
(Also, whilst I'm here, is there a faster/better way to work out ##F \wedge \star F##?)
1) To obtain the equations of motion, you need to vary the action with respect to the dynamical variables.

2) Varying the action with respect to the background metric gives you the energy-momentum tensor.

3) If $\alpha , \beta \in \Lambda^{p}(M^{n})$, one can show (with some work) that $$\alpha \wedge \ast \beta = \beta \wedge \ast \alpha = (\alpha , \beta ) \ \mu , \ \ \ \ \ (1)$$ where $$(\alpha , \beta)(x) = \frac{1}{p!} \alpha_{\mu_{1} \cdots \mu_{p}}(x) \ \beta^{\mu_{1} \cdots \mu_{p}}(x),$$ $$\mu = \frac{\sqrt{-g}}{n!} \ \epsilon_{\mu_{1} \cdots \mu_{n}} \ dx^{\mu_{1}} \wedge \cdots \wedge dx^{\mu_{n}} = \sqrt{-g} \ d^{n}x .$$ So for $F \in \Lambda^{2}(M^{1,3})$, the Hodge star of $F$ is given by $$\ast F = \frac{1}{2} \mathscr{F}_{\mu\nu} \ dx^{\mu} \wedge dx^{\nu} ,$$ where $$\mathscr{F}_{\mu\nu} = \frac{\sqrt{-g}}{2} \ \epsilon_{\mu\nu\rho\sigma} F^{\rho\sigma}.$$ Thus, from (1), you get $$F \wedge \ast F = (F , F) \ \mu = \frac{\sqrt{-g}}{2} F_{\mu\nu} \ F^{\mu\nu} \ d^{4}x . \ \ \ \ \ (2)$$

4) Without using (2), Maxwell’s equations can be derived directly from the action $$S[A] = - \int_{\Omega} \left( \frac{1}{2}F \wedge (\ast F) + A \wedge (\ast j)\right) ,$$ where $$\ast j = \frac{1}{3!} j^{\mu} \ \epsilon_{\mu\nu\rho\sigma} \ dx^{\nu} \wedge dx^{\rho} \wedge dx^{\sigma}$$ is the 3-form associated with the 4-vector current $j^{\mu}$. Under $A \to A + \epsilon \alpha$, with $\epsilon \ll 1$ and $\alpha|_{\partial \Omega} = 0$ but otherwise arbitrary 1-form, we have (using $F = dA$) $$S[A + \epsilon \alpha ] = S[A] - \epsilon \left( \int_{\Omega} \left( \frac{1}{2} F \wedge \ast d\alpha + \frac{1}{2} d\alpha \wedge \ast F + \alpha \wedge \ast j \right) \right) + \mathcal{O}(\epsilon^{2}) .$$ Use (1) to obtain $$\delta S = \lim_{\epsilon \to 0} \frac{S[A + \epsilon \alpha] - S[A]}{\epsilon} = - \int_{\Omega} \left( d\alpha \wedge \ast F + \alpha \wedge \ast j \right).$$ Now, integrate the first term by parts using $d\left( \alpha \wedge \ast F \right) = d\alpha \wedge \ast F - \alpha \wedge d \ast F$, then use Stokes’ theorem: $$\delta S[A] = - \int_{\partial \Omega} \alpha \wedge \ast F - \int_{\Omega} \alpha \wedge \left( d \ast F + \ast j \right) .$$ The first integral vanish because $\alpha |_{\partial \Omega} = 0$. Thus, the action is stationary if $$d \ast F + \ast j = 0. \ \ \ \ (3)$$ Since $j$ is 1-form on the 4-dimensional Minkowski space, then $\ast^{2}j = j$. Therefore (3) may also be written in the form $$\ast d \ast F = \delta F = - j , \ \ \ \ \ (4)$$ where $\delta : \Lambda^{p}(M^{n}) \to \Lambda^{p - 1}(M^{n})$ is the so-called codifferential operator.

Here is an exercise for you: Show that (3) [or (4)] reproduces the Maxwell equation $$\partial_{\mu}F^{\mu\nu} = j^{\nu} .$$

dextercioby, TSny, vanhees71 and 4 others
I start to love the good old Ricci calculus more and more ;-)). SCNR.

TSny
Thanks, that's great! Okay, for the final part, I found it a little bit tricky to keep track of all of the ##\sqrt{-g}##'s..., but I'll show what I tried in any case. Using equation (2.27) from Tong's GR notes, ##(d\omega)_{\mu_1 \dots \mu_{p+1}} = (p+1) \partial_{[ \mu_1} \omega_{\mu_2 \dots, \mu_{p+1}]}##, as well as the identity ##\epsilon_{abcd} \epsilon^{defg} = 6\delta_{[a}^e \delta_{b}^{f} \delta_{c]}^{g}##, we obtain\begin{align*} (\star j)_{\nu \rho \sigma} = \sqrt{-g} \epsilon_{\mu \nu \rho \sigma} j^{\mu} = - (d \star F)_{\nu \rho \sigma} = -3 \partial_{[\nu} (\star F)_{\rho \sigma]} &= - \frac{(-g)}{2} \epsilon_{\nu \rho \sigma \mu}\, \epsilon^{\mu \alpha \beta \gamma} \partial_{\alpha} (\star F)_{\beta \gamma} \\ &= (-g) \epsilon_{\nu \rho \sigma \mu} \partial_{\alpha} \left(- \frac{1}{2} \epsilon^{\mu \alpha \beta \gamma} (\star F)_{\beta \gamma}\right) \end{align*}Now since ##\star(\star F) = \pm F##, taking the Hodge dual of the definition of ##\star F##, and then using the metric to raise the free indices and also to raise/lower both pairs of dummy indices, gives$$(\star F)_{\beta \gamma} = \frac{\sqrt{-g}}{2} \epsilon_{\beta \gamma \delta \varepsilon} F^{\delta \varepsilon} \implies F^{\beta \gamma} = \frac{\sqrt{-g}}{2} \epsilon^{\beta \gamma \delta \varepsilon} (\star F)_{\delta \varepsilon}$$In other words, the bit inside the brackets can be re-written as$$-\frac{1}{2} \epsilon^{\mu \alpha \beta \gamma} (\star F)_{\beta \gamma} = \frac{-1}{\sqrt{-g}} F^{\mu \alpha} = \frac{1}{\sqrt{-g}} F^{\alpha \mu}$$So the first equation becomes, after cycling each of the indices on the Levi-Civita symbol on the RHS one place to the right, and cancelling the ##\sqrt{-g}## terms on either side,$$\epsilon_{\mu \nu \rho \sigma} j^{\mu} = \epsilon_{\mu \nu \rho \sigma} \partial_{\alpha} F^{\alpha \mu} \implies j^{\mu} = \partial_{\alpha} F^{\alpha \mu}$$How does that look?

Last edited by a moderator:
etotheipi said:
I found it a little bit tricky to keep track of all of the ##\sqrt{-g}##'s
The exercise asks you to reproduce Maxwell’s equation on the 4-dimensional Minkowski space. So, you can set $\sqrt{-g} = 1$. If you keep $\sqrt{-g(x)}$, then Eq(3) or (4) reproduce the corresponding Maxwell’s equation on the 4-dimensional pseudo-Riemannian space $\nabla_{\mu}F^{\mu\nu} = j^{\nu}$, as explained bellow.
How does that look?
I did not check it because it looks very messy.

I will choose to derive $\nabla_{\mu}F^{\mu\nu} = j^{\nu}$ from Eq(4) because it is lengthier and require repeated application of the $\ast$ operator. This will let you learn how to keep track of the many metric tensors and the $\sqrt{-g}$ factors. Let’s start from the basic facts. We have, for $F \in \Lambda^{2}(M^{(1,3)})$,

$$F = \frac{1}{2!} F_{\mu\nu} \ dx^{\mu} \wedge dx^{\nu} .$$ Thus,

$$\ast F = \frac{1}{2} F_{\mu\nu} \ \ast (dx^{\mu} \wedge dx^{\nu}) .$$

Since $\ast : \Lambda^{p}(M^{n}) \to \Lambda^{n - p}(M^{n})$, then the action of $\ast$ on the element $dx^{\mu} \wedge dx^{\nu}$ of the local basis of $\Lambda^{2}(M^{(1,3)})$ is given by

$$\ast (dx^{\mu} \wedge dx^{\nu}) = \frac{\sqrt{-g}}{(4 - 2)!} \ g^{\mu\rho}g^{\nu\sigma} \ \epsilon_{\rho\sigma\alpha\beta} \ dx^{\alpha} \wedge dx^{\beta} .$$

Thus $$\ast F = \frac{\sqrt{-g}}{4} F^{\rho\sigma} \ \epsilon_{\rho\sigma\alpha\beta} \ dx^{\alpha} \wedge dx^{\beta}.$$ Taking the exterior derivative $d : \Lambda^{p}(M^{n}) \to \Lambda^{p + 1}(M^{n})$, we get

$$d \ast F = \frac{1}{4} \partial_{\tau}(\sqrt{-g}F^{\rho\sigma}) \ \epsilon_{\rho\sigma\alpha\beta} \ dx^{\tau} \wedge dx^{\alpha} \wedge dx^{\beta} .$$ Operating with $\ast$: $$\ast d \ast F = \frac{1}{4} \partial_{\tau}(\sqrt{-g}F^{\mu\nu}) \ \epsilon_{\mu\nu\alpha\beta} \ \ast (dx^{\tau} \wedge dx^{\alpha} \wedge dx^{\beta}) . \ \ \ (1)$$ Now

$$\ast (dx^{\tau} \wedge dx^{\alpha} \wedge dx^{\beta}) = \frac{\sqrt{-g}}{(4 - 3)!} \ g^{\tau \mu} g^{\alpha \nu} g^{\beta \gamma} \ \epsilon_{\mu \nu \gamma \eta} \ dx^{\eta} . \ \ \ (2)$$ From the definition of the determinate of the inverse metric: $$\epsilon_{\mu \nu \gamma \eta} \ g^{\mu \tau} g^{\nu \alpha} g^{\gamma \beta} g^{\eta \delta} = \frac{1}{g} \ \epsilon^{\delta \tau \alpha \beta} ,$$ it follows that $$\epsilon_{\mu \nu \gamma \eta} \ g^{\mu\tau} g^{\nu \alpha} g^{\gamma \beta} = \frac{1}{g} \ \epsilon^{\rho \tau \alpha \beta} \ g_{\rho \eta} .$$ Substituting in (2), we get $$\ast (dx^{\tau} \wedge dx^{\alpha} \wedge dx^{\beta}) = - \frac{1}{\sqrt{-g}} \ \epsilon^{\rho \tau \alpha \beta} \ g_{\rho \eta} \ dx^{\eta} .$$ Substituting in (1) leads to $$\ast d \ast F = \frac{-1}{4 \sqrt{-g}} \partial_{\tau}(\sqrt{-g}F^{\mu\nu}) \ \epsilon_{\mu \nu \alpha \beta} \ \epsilon^{\rho \tau \alpha \beta} \ g_{\rho \eta} \ dx^{\eta} .$$ Finally, we make use of the identity $$\epsilon_{\mu \nu \alpha \beta} \ \epsilon^{\rho \tau \alpha \beta} \ g_{\rho \eta} = (-2!) \left( \delta^{\tau}_{\nu} \ g_{\mu\eta} - \delta^{\tau}_{\mu} \ g_{\nu\eta} \right),$$ to obtain $$\ast d \ast F = - \left[ \frac{1}{\sqrt{-g}} \ \partial_{\nu} \left(\sqrt{-g} F^{\nu\mu}\right)\right] \ g_{\mu\eta} \ dx^{\eta} .$$ Since the tensor $F^{\mu\nu}$ is antisymmetric, the expression in the big brackets is nothing but the (spacetime) covariant expression $\nabla_{\nu}F^{\nu\mu}$. So, we are almost done: $$\ast d \ast F = - \nabla_{\nu}F^{\nu\mu} \ g_{\mu\eta} \ dx^{\eta} . \ \ \ \ \ \ \ (3)$$ Now, the 1-form associated with the current vector is given by $$j = j^{\mu} \ g_{\mu \eta} \ dx^{\eta} . \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$$ So, from $\ast d \ast F + j = 0$, it follows that $$\nabla_{\nu}F^{\nu\mu} = j^{\mu} .$$

Exercise: Start with the 3-form $$\ast j = \frac{\sqrt{-g}}{3!} \ j^{\mu} \ \epsilon_{\mu\nu\rho\sigma} \ dx^{\nu} \wedge dx^{\rho} \wedge dx^{\sigma} .$$ Show that operating with $\ast$ gives you $$j = j^{\mu} \ g_{\mu \eta} \ dx^{\eta} .$$

JD_PM, etotheipi, dextercioby and 1 other person
samalkhaiat said:
Exercise: Start with the 3-form $$\ast j = \frac{\sqrt{-g}}{3!} \ j^{\mu} \ \epsilon_{\mu\nu\rho\sigma} \ dx^{\nu} \wedge dx^{\rho} \wedge dx^{\sigma} .$$ Show that operating with $\ast$ gives you $$j = j^{\mu} \ g_{\mu \eta} \ dx^{\eta} .$$

Thanks! I'll have a go at this exercise. Since ##*j \in \Lambda^3(M^{(3,1)})##, it satisfies ##*(*j) = j##. Let's also make use of the equation$$*(dx^{\nu} \wedge dx^{\rho} \wedge dx^{\sigma}) = \sqrt{-g} g^{\nu \alpha} g^{\rho \beta} g^{\sigma \gamma} \epsilon_{\alpha \beta \gamma \eta} dx^{\eta}$$in order to write down\begin{align*} j = *(*j) &= \frac{\sqrt{-g}}{3!} j^{\mu} \epsilon_{\mu \nu \rho \sigma} *(dx^{\nu} \wedge dx^{\rho} \wedge dx^{\sigma}) \\ \\ j &= \frac{(-g)}{3!} j^{\mu} \epsilon_{\mu \nu \rho \sigma} g^{\nu \alpha} g^{\rho \beta} g^{\sigma \gamma} \epsilon_{\alpha \beta \gamma \eta} dx^{\eta} \ \ \ (1) \end{align*}Now since$$g^{\nu \alpha} g^{\rho \beta} g^{\sigma \gamma} \epsilon_{\alpha \beta \gamma \eta} = \frac{1}{g} \epsilon^{\lambda \nu \rho \sigma} g_{\lambda \eta}$$we can re-write ##(1)## as$$j = \frac{(-g)}{3!} j^{\mu} \epsilon_{\mu \nu \rho \sigma} \left( \frac{1}{g} \epsilon^{\lambda \nu \rho \sigma} g_{\lambda \eta} \right) dx^{\eta} = \frac{(-1)}{3!} j^{\mu} \epsilon_{\mu \nu \rho \sigma} \epsilon^{\lambda \nu \rho \sigma} g_{\lambda \eta} dx^{\eta} \ \ \ (2)$$Then, finally we can use the identity$$\epsilon_{\mu \nu \rho \sigma} \epsilon^{\lambda \nu \rho \sigma} = (3!) \delta^{\lambda}_{\mu}$$to re-write (2) as$$j = \frac{(-1)}{3!} j^{\mu} (3!) \delta^{\lambda}_{\mu} g_{\lambda \eta} dx^{\eta} = - j^{\mu} g_{\mu \eta} dx^{\eta}$$I seem to have picked up an erroneous negative sign somewhere... ...

etotheipi said:
finally we can use the identity $\epsilon_{\mu \nu \rho \sigma} \epsilon^{\lambda \nu \rho \sigma} = (3!) \delta^{\lambda}_{\mu}$
I seem to have picked up an erroneous negative sign somewhere... ...
$\epsilon_{\mu \nu \rho \sigma} \ \epsilon^{\lambda \nu \rho \sigma} = (-3!) \ \delta^{\lambda}_{\mu}$
so that $\epsilon^{\mu\nu\rho\sigma} \ \epsilon_{\mu\nu\rho\sigma} = -4!$, with $\epsilon^{0123} = +1$.

vanhees71 and etotheipi
samalkhaiat said:
$\epsilon_{\mu \nu \rho \sigma} \ \epsilon^{\lambda \nu \rho \sigma} = (-3!) \ \delta^{\lambda}_{\mu}$
so that $\epsilon^{\mu\nu\rho\sigma} \ \epsilon_{\mu\nu\rho\sigma} = -4!$, with $\epsilon^{0123} = +1$.

Ah! Careless of me. Thank you

## 1. What are Maxwell's Equations?

Maxwell's Equations are a set of four fundamental equations in electromagnetism that describe the relationship between electric and magnetic fields, and how they are affected by electric charges and currents.

## 2. Who discovered Maxwell's Equations?

James Clerk Maxwell, a Scottish physicist, developed the equations in the 1860s based on the work of other scientists such as Michael Faraday and André-Marie Ampère.

## 3. What is the significance of the variation of metric determinant in Maxwell's Equations?

The variation of metric determinant, also known as the determinant of the metric tensor, is a mathematical concept that is crucial in understanding the behavior of electric and magnetic fields in different reference frames. It allows for the equations to be written in a covariant form, meaning they are valid in all reference frames.

## 4. How are Maxwell's Equations used in practical applications?

Maxwell's Equations have numerous practical applications, including the design of electronic devices, the development of wireless communication technologies, and the study of electromagnetic waves. They are also used in fields such as optics, plasma physics, and astrophysics.

## 5. Are Maxwell's Equations still relevant today?

Yes, Maxwell's Equations are still considered one of the cornerstones of modern physics and are used extensively in current research and technology. They have been tested and confirmed countless times and are considered to be one of the most accurate and reliable theories in science.

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