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In Heaviside-Lorentz units the Maxwell's equations are:

$$\nabla \cdot \vec{E} = \rho $$ $$ \nabla \times \vec{B} - \frac{\partial \vec{E}}{\partial t} = \vec{J}$$ $$ \nabla \times \vec{E} + \frac{\partial \vec{B}}{\partial t} = 0 $$ $$ \nabla \cdot \vec{B} = 0$$

From EM Lagrangian density: $$\mathcal{L} = \frac{-1}{4} F^{\mu \nu}F_{\mu\nu} - J^\mu A_\mu$$

I can derive the first two equations from the variation of the action integral: $S[A] = \int \mathcal{L} \, d^4x$. Is it possible to derive the last two equations from it?

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    $\begingroup$ @LarryHarson Since you seem pretty annoyed by the absence of clarity of the present answers, I would like to make a sum-up of them. So the question is is it possible to derive the Maxwell's equations without source from the action $S(A)$. The answer is YOU DON'T NEED, because they are contained in the writing $S(A)$: they are definition of the fields in terms of the potentials. That's what everybody below have tried to say: since $F=dA$, then $dF=0$, called the Bianchi equality. $\endgroup$
    – FraSchelle
    Jul 20, 2013 at 7:24
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    $\begingroup$ @LarryHarson Otherwise how could you start from an action $S(A)$ depending on the potential, and end with equations for fields, if you don't choose $B=\nabla\times A$ and $E=\partial A + \nabla\phi$ ? These two field <-> potential relations\definitions impose $\nabla\cdot B =0$ and $\partial B + \nabla\times E =0$. $\endgroup$
    – FraSchelle
    Jul 20, 2013 at 7:26

4 Answers 4

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Assume for simplicity that the speed of light $c=1$. The existence of the gauge $4$-potential $A^{\mu}=(\phi, \vec{A})$ alone implies that the source-free Maxwell equations $$\vec{\nabla} \cdot \vec{B} ~=~ 0 \qquad ``\text{no magnetic monopole"}$$

$$ \vec{\nabla} \times \vec{E} + \frac{\partial \vec{B}}{\partial t} ~=~ \vec{0}\qquad ``\text{Faraday's law"}$$

are already identically satisfied. To prove them, just use the definition of the electric field

$$\vec{E}~:=~-\vec{\nabla}\phi-\frac{\partial \vec{A}}{\partial t},$$

and the magnetic field

$$\vec{B}~:=~\vec{\nabla}\times\vec{A}$$

in terms of the gauge $4$-potential $A^{\mu}=(\phi, \vec{A})$.

The above is more naturally discussed in a manifestly Lorentz-covariant notation. OP might also find this Phys.SE post interesting.

Thus, to repeat, even before starting varying the Maxwell action $S[A]$, the fact that the action $S[A]$ is formulated in terms the gauge $4$-potential $A^{\mu}$ means that the source-free Maxwell equations are identically satisfied. Phrased differently, since the source-free Maxwell equations are manifestly implemented from the very beginning in this approach, varying the Maxwell action $S[A]$ will not affect the status of the source-free Maxwell equations whatsoever.

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Note that since $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$, we have the equation $$ \partial_\mu F_{\nu\alpha} + \partial_\alpha F_{\mu\nu} + \partial_\nu F_{\alpha\mu} = 0 $$ This equation is called the Bianchi Identity. This equation is separate from the equations of motion one obtains from varying the action. It can be shown that the Bianchi equation is equivalent to the last two equations you have mentioned in your question.

NOTE: To elaborate on @Qmechanic's answer, the very fact that $F_{\mu\nu}$ can be written as $\partial_\mu A_\nu - \partial_\nu A_\mu$ is itself a consequence of the last two equations.

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    $\begingroup$ -1 The question asked if it's possible to derive the last two equations using variation of the electromagnetic action. $\endgroup$ Jul 19, 2013 at 21:12
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    $\begingroup$ @LarryHarson you know that it is completely legitimate to write partial answers to a question? Downvoting all answers that do not completely answer the question is almost trolling ... $\endgroup$
    – Dilaton
    Jul 19, 2013 at 21:27
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    $\begingroup$ @LarryHarson: Since the OP has clarified what he wanted, your downvote was unjustified Keep checking this answer daily to see if it has been edited so that you can reverse your downvote. . EDIT: I see that you downvoted after the clarificaiton... How does this not answer the question ? . $\endgroup$ Jul 22, 2013 at 13:55
  • $\begingroup$ @Dimension10 and prahar: yeah, I'll upvote this when it gets edited. $\endgroup$ Jul 22, 2013 at 13:59
  • $\begingroup$ @LarryHarson: I've edited the <s>question</s> ANSWER now. Please reverse your downvote . $\endgroup$ Aug 14, 2013 at 12:15
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From a geometric perspective, the last two equations are a consequence of:

(1)F = dA (Faraday tensor is the exterior derivative of the four-potential)

(2) dd = 0 (the exterior derivative of the exterior derivative vanishes)

Thus

(3) dF = 0

which gives the last two equations in your question.

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  • $\begingroup$ @LarryHarson: How does this not address the question? . $\endgroup$ Jul 22, 2013 at 13:59
  • $\begingroup$ I've removed my downvote. $\endgroup$ Jul 22, 2013 at 14:04
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    $\begingroup$ A very much belated comment: This notation is so succinct that it requires a manual. $\endgroup$
    – my2cts
    Apr 5, 2021 at 11:17
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The first scalar and vector equation come from the action principle, the last scalar and vector equation do not - and cannot.

In fact, a simple counting argument should already make that clear. How many things do you have available to vary, in the action principle? Answer: the four components of $A$. So, how many Euler-Lagrange equations will come out from the action principle? Answer: four. That includes the first scalar equation and the three components of the first vector equation, and you're done.

The last vector and scalar equation are a direct consequence of the definition of $F$ in terms of $A$ and are already there before you set up any action principle.

It helps to clarify, by distinguishing between the different objects that appear in this setting. The fields that go into the action principle may be said to comprise the "kinematics" for the field theory - by analogy with mechanics. So, the components of the potential $A$: $(Ο†,𝐀)$ and of $F$: $(𝐁,𝐄)$ are the "kinematic" fields. The condition that relates $F$ to $A$ may, then, be considered as a "kinematic constraint", $dA = F$, or in component form $𝐁 = βˆ‡Γ—π€$ and $𝐄 = -βˆ‡Ο† - βˆ‚π€/βˆ‚t$. And, since $dF = 0$ is a consequence of $dA = F$, then $dF = 0$ may be considered as a "secondary" kinematic constraint. In component form that's $βˆ‡Β·π = 0$ and $βˆ‡Γ—π„ + βˆ‚π/βˆ‚t = 𝟎$.

All of that is there first, before the action principle is employed. Moreover, the constraint is factored into the variational, itself: namely, that $Ξ”F$ and $Ξ”A$ should satisfy the same equation as do $F$ and $A$, since $Ξ”F = Ξ”(dA) = d(Ξ”A)$. So the variations of $𝐁$ and $𝐄$ are reduced to those of the potentials $𝐀$ and $Ο†$ under the constraint as $Δ𝐁 = βˆ‡Γ—(Δ𝐀)$ and $Δ𝐄 = -βˆ‡(Δφ) - βˆ‚(Δ𝐀)/βˆ‚t$ ... so that you can eventually do an integration by parts. That also means that $βˆ‡Β·(Δ𝐁)$ and $βˆ‡Γ—(Δ𝐄) + βˆ‚(Δ𝐁)/βˆ‚t = 𝟎$ are both already true - as part of the variation, arising from the secondary kinematic constraint.

The other fields are derived as derivatives with respect to the Lagrangian density: $$ρ = -\frac{βˆ‚π”}{βˆ‚Ο†}, \hspace 1em 𝐉 = \frac{βˆ‚π”}{βˆ‚π€}, \hspace 1em 𝐇 = -\frac{βˆ‚π”}{βˆ‚π}, \hspace 1em 𝐃 = \frac{βˆ‚π”}{βˆ‚π„}.$$ These are the "dynamic" fields. They play a role analogous to the role played by mass, energy and momentum in mechanics.

Only the dynamic fields have Euler-Lagrange equations! The Lagrangian density plays two roles here: (1) to establish the framework for those equations $$βˆ‡Β·πƒ = ρ, \hspace 1em βˆ‡Γ—π‡ - \frac{βˆ‚πƒ}{βˆ‚t} = 𝐉,$$ a form that it will have irrespective of the Lagrangian and (2) to establish a relation between the dynamic fields and kinematic fields (i.e. the "constitutive relations") that gives the Euler-Lagrange equations teeth. The constitutive relations are the part that's Lagrangian-dependent.

With the Lagrangian density you're using, which works out in component form as $$𝔏 = Β½ Ξ΅_0 \left(|𝐄|^2 - |𝐁|^2 c^2\right) + \bar{𝐉}·𝐀 - \bar{ρ}Ο†,$$ after re-inserting the multiplier $Ξ΅_0$, where your $J$ refers to an already-determined external field put into the Lagrangian by hand, whose components I will label $\bar{𝐉}$ and $\bar{ρ}$, to make that distinction clear, the constitutive relations work out to be: $$𝐃 = Ξ΅_0 𝐄, \hspace 1em 𝐇 = Ξ΅_0 c^2 𝐁, \hspace 1em 𝐉 = \bar{𝐉}, \hspace 1em ρ = \bar{ρ},$$ so that $𝐃$ and $𝐇$ are linked to the kinematic fields, while $𝐉$ and $ρ$ are linked to the external fields (and not to any kinematic fields at all).

In fact, the external fields are really nothing more than a place-holder that accounts for and collates the effects that the other fields that go into the Lagrangian density, but are left unmentioned, may have. The sources for $𝐉$ and $ρ$ come from them, and the constitutive laws tell you how they are defined in terms of the other fields. So, they too are actually linked to "kinematic" fields from elsewhere.

Somewhere down the line, time permitting, I may need to write a monograph on all this. These are fundamentals that are not really spelled out clearly anywhere, even though they are tacit in the whole enterprise.

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