Form of the Classical EM Lagrangian

Question

So I know that for an electromagnetic field in a vacuum the Lagrangian is $$\mathcal L=-\frac 1 4 F^{\mu\nu} F_{\mu\nu},$$ the standard model tells me this. What I want to know is if there is an elementary argument (based on symmetry perhaps) as to why it has this form. I have done some searching/reading on this, but have only ever found authors jumping straight to the expression and sometimes saying something to the effect that it is the "simplest possible".

Answer 1

The Lagrangian for Electromagnetism follows uniquely from requiring renormalizability and gauge invariance (plus parity time reversal)

U(1) gauge Invariance

if you require your Lagrangian to be locally invariant under symmetry operations of the unitary group U(1) that is under

$$\phi\to e^{i\alpha(x)}\phi$$

all derivatives $\partial_\mu$ have to be replaced by the covariant derivative $D_\mu = \partial_\mu+ieA_\mu$, where, in order to save local invariance the gauge field is introduced. Loosely speaking this is necessary to make fields at different spacetiem points comparable Since two points may have an arbitrary phase difference, due to the fact that we can set $\alpha(x)$ as we wish, something has to compensate this difference, before we can compare fields, which is what differentiation basically does. This is similar to parallel transport in general relativity (the mathematical keyword is connection see wiki: Connection (wiki) The gauge field $A_\mu$ transforms as $A_\mu \to A_\mu - \frac{1}{e}\partial_\mu\alpha(x)$.

Now the question is what kind of Lagrangians we can build with this requirement. For matter (i.e. non-gauge) fields it's easy to construct gauge invariant quantities by just replacing the derivatives with the covariant derivatives, i.e.

$$\bar{\psi}\partial_\mu\gamma^\mu\psi\to \bar{\psi}D_\mu\gamma^\mu\psi$$,

this will yield kinetic terms for the field (the part with the normal derivative), and interactions terms between matter fields and the gauge field.

Gauge-Field only terms

the remaining question is how to construct terms involving only the gauge field and no matter fields (i.e. the 'source-free' terms your question is about). For this we must construct gauge-invariant germs of $A_\mu$.

Once $\alpha(x)$ is chosen we can imagine starting from a point and walking on a loop back to that same point (this is called a wilson loop (wiki)). This must necessarily be gauge invariant since any phase that we pick up on the way we must also loose on the way back. It turns out, that this is exactly the term $F_{\mu\nu}$, i.e. the field strength. (the calculation is a little longer, see Peskin & Schroeder page 484). Actually this is only true for abelian symmetries such as U(1), for non abelian ones such as SU(3) we will get some interaction terms between the gauge fields which is why light does not interact with itself but gluons do.

Bilinear mass terms such as $A_\mu A^\mu$ are not gauge invariant (in the end this is the need for the Higgs meachanism)

Renormalizability

If we wish that our theory is renormalizable, we can only include terms into the lagrangian up to mass dimension 4. Now listing all terms up to mass dimension 4 we arrive at

$$\mathcal{L} = \cdot\bar{\psi}D_\mu\psi - m\bar{\psi}\psi - b\cdot F_{\mu\nu}F^{\mu\nu} + d\cdot \epsilon^{\alpha\beta\gamma\delta}F_{\alpha\beta}F_{\gamma\delta}$$

the last term involves the anti-symmetric tensor $\epsilon^{\alpha\beta\gamma\delta}$ and is therefore not time and parity invariant.

Note that we have not included linear terms here since we will be expanding around a local minimum anyways, so that the linear term will vanish.

Conclusion

if we require U(1) gauge invariance and renormalizability (mass dimension up to 4) and time and parity invariance we only get

$$\mathcal{L} = \cdot\bar{\psi}D_\mu\psi - m\bar{\psi}\psi - b\cdot F_{\mu\nu}F^{\mu\nu}$$

In the source-free case this is

$$\mathcal{L} = - b\cdot F_{\mu\nu}F^{\mu\nu}$$

the overall factor $\frac{1}{4}$ is not important.

Answer 2

The Maxwell term

$$\tag{1} {\cal L}_{\rm Maxwell}~=~-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$

emerges naturally for many reasons.

1) Pure EM without matter. There is a very short list of special relativistic renormalizable terms, that one can put in a local Lagrangian density with no higher-order time derivatives, and that is gauge-invariant (up to boundary terms). Among the shortlist of these candidates, the Lagrangian density (1) is the only one (modulo an overall normalization factor and modulo boundary terms) that leads to Maxwell equations (without source terms). The Born-infeld Lagrangian density is an example of a non-local candidate.

2) EM coupled to matter. Additional conditions arise when one tries to couple EM to point charges. One may argue that the relativistic Lagrangian for $n$ point charges $q_1, \ldots, q_n$, at positions ${\bf r}_1, \ldots, {\bf r}_n$, in an EM background is given as

$$\tag{2} L ~=~ -\sum_{i=1}^n \left(\frac{m_{0i}c^2}{\gamma({\bf v}_i)} +q_i\{\phi({\bf r}_i) - {\bf v}_i\cdot {\bf A}({\bf r}_i)\} \right).$$

See also this Phys.SE answer. This Lagrangian (2) e.g. reproduces correctly the Lorentz force. From eq. (2)it is only a small step to conclude that the interaction term ${\cal L}_{\rm int}$ between EM and matter [in the $(-,+,+,+)$ sign convention] is of the form

$$\tag{3} {\cal L}_{\rm int}~=~J^{\mu}A_{\mu}.$$

Also recall that Maxwell's equations with sources are

$$\tag{4} d_{\nu} F^{\nu\mu}~=~-J^{\mu}. $$

If the action

$$\tag{5} S[A]~=~\int\! d^4x~ {\cal L} $$

is supposed to be varied wrt. the $4$-gauge potential $A_{\mu}$, i.e. the $4$-gauge potential $A_{\mu}$ are the fundamental variables of the theory, and if moreover the corresponding Euler-Lagrange equations

$$\tag{6} 0~=~\frac{\delta S}{\delta A_{\mu}} ~\stackrel{?}{=}~d_{\nu} F^{\nu\mu}+J^{\mu}$$

are supposed to reproduce Maxwell's equations (4), it quickly becomes clear that

$$\tag{7} {\cal L}~=~{\cal L}_{\rm Maxwell}+{\cal L}_{\rm int} $$

is the appropriate Lagrangian density for EM with background sources $J^{\mu}$.

3) For a discussion of formulating EM without the $4$-gauge potential $A_{\mu}$, see also this Phys.SE post.

Answer 3

The most economical derivation of the Maxwell equations (i.e., depending on the least number of postulates) is known nowadays by the name of "Feynman's proof of the Maxwell equations". This "proof" was discovered by Feynman in 1948, but was not published because Feynman did not think that it leads to new physics. It was only in 1989 after Feynman's death when his proof was published by Dyson, please see the Dyson's article.

After being published by Dyson, the Feynman's proof was found to contain very deep ideas in Poisson geometry. As well as its generalization to the non-Abelian case leads to the Wong equations describing the motion of a point particle in an external Yang-Mills field, (please see the following exposition by: Montesinos and Abdel Perez-Lorenzana) which in turn is related to the Kaluza-Klein theory. It explains various mechanical ideas, for example the phenomenon of the falling cat. The whole subject is known today by the name of sub-Riemannian geometry.

The postulates of the Feynman's proof are the following:

1. The position and velocity of the particle satisfy the canonical Poisson bracket relation with the position.
2. The acceleration of a charged particle in an electromagnetic field is a function of the position and velocity only.

Under these assumptions only, Feynman proved (please see the Dyson's article for details) that the electromagnetic force must satisfy the Lorentz law and the electromagnetic field must satisfy the homogeneous Maxwell equations.

Answer 4

Yes, symmetry argument is correct - It has to be invariant under local gauge transformations.

Answer 5

Yes, it is based on symmetry, with the primary player being the Utiyama Theorem. You're actually not pulling the issue back far enough, because you're already assuming the $F$ fields. Even that has to be included in your question. Nor does it include consideration of other fields that may be present. The Utiyama Theorem has a material bearing on them, too.

I addressed the matter here, in greater depth, in my reply to the following similar query, where I included an extra field with the electromagnetic field just for illustration's sake:

Deriving Lagrangian density for electromagnetic field
Using Utiyama's Theorem

So, I will only lay out the basics in cursory form here, with additional remarks.

The question on hand is: assume that the dynamics are governed by a Lagrangian density $𝔏(A,∂A,q,∂q)$ that is a function of the electromagnetic potentials $A$, their gradients $∂A$ and other fields, denoted here as $q$ and their gradients $∂q$. What governs the form of the Lagrangian density, and what forms may it have?

The potentials $A$ possess gauge symmetries that the laws of electrodynamics are invariant with respect to. So, we first require that the Lagrangian density be gauge invariant. As a minor footnote, this is actually slightly stronger of a condition than what is required, since it is actually the action integral $S = \int 𝔏 d^4x$ that one wants invariance for. But I'll ignore the subtlety and keep to the stronger condition.

The Utiyama theorem mandates that the most general gauge invariant Lagrangian must be one that (1) has no explicit dependency on $A$, (2) whose dependency on $∂A$ may only occur through the anti-symmetric combinations of the gradients $F_{μν} = ∂_μA_ν - ∂_νA_μ$, (3) whose dependency on the other gradients $∂q$ of the other fields is restricted to "gauge covariant" derivatives $a = ∇_Aq$, and (4) through those means, a tacit dependence on $A$ may occur. Thus, $𝔏$ reduces to a function of the form $𝔏\left(F,q,a\right)$. The field $F$, itself, may be regarded as the gauge-covariant derivative of $A$.

This, and this alone, is already enough to determine all of the equations of the field theory. The only Lagrangian-dependency that exists is - ultimately - what's locked up in a set of coefficients, which may be deemed the "constitutive coefficients". Apart from that, all of the equations have the same form.

Label the fields $A$, $F$, $q$ and $a$ the "kinematic" fields. The derivatives of the Lagrangian density with respect to them may then be defined as the "sources" (at least in the case of $A$) and "response fields" (at least for $G$). Together they comprise the "dynamic fields". $$J = \frac{∂𝔏}{∂A}, \hspace 1em G = -\frac{∂𝔏}{∂F}, \hspace 1em p = \frac{∂𝔏}{∂a}, \hspace 1em f = \frac{∂𝔏}{∂q}.$$ The Euler-Lagrange equations for the Lagrangian will then yield a set of equations $$J = ∇_AG, \hspace 1em f = ∇_Ap,$$ that are independent of $𝔏$.

The whole point of $𝔏$, in reality, is to serve as a means to generate the relation between the dynamic and kinematic fields.

For the electromagnetic field, the relevant relations are: $$𝐃 = \frac{∂𝔏}{∂𝐄}, \hspace 1em 𝐇 = -\frac{∂𝔏}{∂𝐁}, \hspace 1em ρ = -\frac{∂𝔏}{∂φ}, \hspace 1em 𝐉 = \frac{∂𝔏}{∂𝐀},$$ where the electric potential $φ$ and magnetic potential $𝐀$ together make up $A$, while the magnetic induction $𝐁$ and the electric field $𝐄$ together make up $F$. The response fields are the displacement field $𝐃$, and magnetic field $𝐇$, which make up $G$ and the charge density $ρ$ and current density $𝐉$, which make up the source $J$.

The Maxwell equations involving those four fields arise solely from this definition, as the Euler-Lagrange equations of the Lagrangian, and their form is independent of any consideration of what the Lagrangian density is.

There may be other gauge symmetries, to consider, besides these. And some could also involve the electromagnetic field. So, this is not the end-all of this process, but just the first step.

Second, there is the requirement that the Lagrangian be invariant with respect to a certain set of geometric transforms. In general, this will imply that the Lagrangian reduces to a function of all of the independent geometric invariants that can be formed from the fields. Call them $ℑ^1$, $ℑ^2$, ⋯.

Thus, the Lagrangian density will have the form $𝔏\left(ℑ^1, ℑ^2, ⋯\right)$; and that's it.

For instance, for the field $F$, the constitutive law for $G$ will have a specific form involving the constitutive coefficients: $$ε_1 = \frac{∂𝔏}{∂ℑ^1}, \hspace 1em ε_2 = \frac{∂𝔏}{∂ℑ^2}, \hspace 1em ⋯,$$ with the relation given by $$G = ε_1 G^1 + ε_2 G^2 + ⋯,$$ where $G^1$, $G^2$, ⋯ are specific combinations of the other fields of the respective forms: $$G^1 = -\frac{∂ℑ^1}{∂F}, \hspace 1em G^2 = -\frac{∂ℑ^2}{∂F}, ⋯.$$

For the electromagnetic field, the two independent Lorentz invariants are $ℑ = ½(E^2 - B^2 c^2)$ and $𝔍 = 𝐄·𝐁$. The invariant $ℑ$ is the combination you cited, up to constant factors. The corresponding coefficients are $$ε = \frac{∂𝔏}{∂ℑ}, \hspace 1em θ = \frac{∂𝔏}{∂𝔍}.$$ In general they will each be functions of the invariants $ℑ$, $𝔍$ and whatever other invariants there are, and will satisfy the relation: $$\frac{∂ε}{∂𝔍} = \frac{∂θ}{∂ℑ},$$ as well as others involving the other constitutive coefficients. It just so happens and if $θ$ is constant, it can be set to $0$ without loss of generality.

The Lagrangian density has the general form $𝔏(ℑ,𝔍,⋯)$, where $(⋯)$ are the other geometric invariants formed from the fields $F$, $q$ and $a$.

The respective constitutive laws for the response fields are $$𝐃 = ε 𝐄 + θ 𝐁 + ⋯, \hspace 1em 𝐇 = ε c^2 𝐁 - θ 𝐄 + ⋯,$$ regardless of what the Lagrangian density is. The extra terms (if any) are determined by whatever other invariants are involved that contained $𝐁$ and $𝐄$ in them, mixed with other fields. It won't depend on the Lagrangian density, itself.

For the source fields $ρ$ and $𝐉$, the only contribution to them will arise from invariants that involve the gauge-covariant derivatives $a = ∇_Aq$, through the tacit dependency on $A$ in them.

As a footnote: the Proca Lagrangian density (i.e. for the version of the electromagnetic field whose "photons" have positive rest mass) has a dependency on the Lorentz invariant $|𝐀|^2 - φ^2/c^2$, but this breaks gauge invariance. It selects out the Lorenz gauge.

However, it is possible to get the effect of Proca by having this combination appear in the invariants involving $∇_Aq$. That's related to the issue of "symmetry breaking".

Answer 6

Varying it with respect to the four-potential yields Maxwell's equations. That's really the only answer to why any classical Lagrangian has the form that it does - because it yields the correct field equations.

Answer 7

The standard Lagrangian is the only form that

a) gives the correct equations of motion

b) is a Lorentz scalar density

c) is gauge invariant.

Requirement a) is obvious. b) is required for the action integral to be Lorentz invariant. In my paper published as Eur. Phys. J. D, vol. 8, p 9-12 (2000) I show that c) is not required.

Note that quantization of the theory requires the Gupta-Bleuler formalism, which involves infinitesimal gauge breaking terms. This solves the problem that the Lorenz, or any other, gauge cannot be imposed in operator form. This conflicts with c) although the end result does not depend on precisely how gauge invariance is broken. Taming the infrared divergencies requires infinitesimal photon mass, which also breaks gauge invariance condition c) (Itzykson&Zuber, p. 172).