Why is the gauge field Lagrangian proportional to the square of the strength tensor?

Question

I am studying quantum field theory and was wondering about a mathematically rigorous explanation of why $$\mathcal{L}=-\frac{1}{4}F^{\mu\nu}_\alpha F^\alpha_{\mu\nu}$$ for gauge boson fields. I have experience in differential geometry and am studying Hamilton's Mathematical Gauge Theory. I understand the field strength tensor describes the curvature of the PFB $\pi:P\to M$ upon which the Lagrangian is invariant under transformation involving $\text{Aut}(P)$. However, I do not understand how the equations of motion and/or the Lagrangian for the local connection 1-forms are derived. Is there an explanation for the form of this Lagrangian other than "it works," and if so, how is it derived?

Answer 1

The simplest, most direct, answer is that if the Lagrangian density is Lorentz-invariant, then it should be a function only of the Lorentz invariants that can be formed of the fields you plug into the Lagrangian. The only Lorentz invariants formed purely from the gauge field strength $F^a_{μν}$ are quadratic: $g^{μρ}g^{νσ}F^a_{μν}F^b_{ρσ}$ and $ε^{μνρσ}F^a_{μν}F^b_{ρσ}$, which are most conveniently written as $$ℑ^{ab} = -\frac14 \sqrt{|g|} g^{μρ}g^{νσ}F^a_{μν}F^b_{ρσ},\quad 𝔍^{ab} = -\frac18 ε^{μνρσ}F^a_{μν}F^b_{ρσ}.$$

If it's linear in these invariants, then the terms linear in $𝔍^{ab}$ are total divergences and you have just a linear combination of the $ℑ^{ab}$ terms. If it's non-linear, then the linear terms are the low-order terms, and you have the gauge-analogue of non-linear electrodynamics; e.g. gauge analogues of the Born-Infeld Model (expanded on in Introductory Notes on Non-linear Electrodynamics and its Applications).

The derivatives of the Lagrangian density $𝔏$, with respect to the invariants: $$ k_{ab} = \frac{∂𝔏}{∂ℑ^{ab}},\quad θ_{ab} = \frac{∂𝔏}{∂𝔍^{ab}} $$ would be constant for the linear cases and variable for the non-linear cases.

The only place I've seen non-linear models mentioned is "Generalizations of Yang–Mills theory with nonlinear constitutive equations", Goldin and Shtelen ( J. Phys. A: Math. Gen. 37 10711), (PDF, author's copy). They were directing it to the question of finding non-relativistic versions of the gauge theory constitutive law, but only ended up arriving at the gauge-theoretic version of what would be called zero-permittivity media (example references: Zero permittivity materials: Band gaps at the visible, The rise of near-zero-index technologies, Nonlinear optical effects in epsilon-near-zero media).

The coefficients $k_{ab}$ are analogous to the permittivity, via the correspondence $k ↔ εc$, while $θ_{ab}$ play a role somewhat analogous to axions. For Yang-Mills gauge theories, $θ_{ab} = 0$ and $k_{ab}$ is a metric for the underlying Lie algebra that is assumed to be adjoint-invariant, i.e. where $k_{ad} f^d_{bc} = -k_{bd} f^d_{ac}$, where $f^a_{bc}$ are the structure coefficients for the underlying Lie algebra, given in terms of the Lie basis $Y_a$ as $[Y_a, Y_b] = f^c_{ab} Y_c$ (summation convention used). This is the metric used to raise and lower the Lie indices $a, b, c, ⋯$; so that $F^{μν}_a = k_{ab} g^{μρ}g^{νσ} F^b_{ρσ}$. Thus, for Yang-Mills, the Lagrangian density has the form $𝔏 = k_{ab} ℑ^{ab} = -¼\sqrt{|g|} F^{μν}_a F^a_{μν}$.

Answer 2

The Lagrangian for a gauge theory is constructed by adding the kinetic term for the gauge fields to the Lagrangian for the matter fields (such as fermions or scalars) that interact with them. The kinetic term for the gauge fields is given by the square of the field strength tensor, which is a measure of the curvature of the gauge connection on the principal fiber bundle. The field strength tensor is defined as

$$F^a_{\mu\nu} = \partial_\mu A^a_\nu - \partial_\nu A^a_\mu + g f^{abc} A^b_\mu A^c_\nu$$

where $A^a_\mu$ are the components of the gauge potential (the connection one-form), $g$ is the coupling constant, and $f^{abc}$ are the structure constants of the Lie algebra of the symmetry group. The index $a$ runs over the generators of the Lie algebra, and $\mu,\nu$ run over the spacetime coordinates.

The Lagrangian density for a gauge theory can then be written as

$$\mathcal{L} = \mathcal{L}_\text{matter} - \frac{1}{4} F^a_{\mu\nu} F^{a\mu\nu}$$

where $\mathcal{L}_\text{matter}$ is the Lagrangian density for the matter fields. The factor of $-\frac{1}{4}$ is chosen for convenience and convention. This Lagrangian density is invariant under local gauge transformations of the form

$$A^a_\mu \to A^a_\mu + \frac{1}{g} \partial_\mu \alpha^a + f^{abc} \alpha^b A^c_\mu$$

where $\alpha^a$ are arbitrary functions of spacetime that parametrize the symmetry group.

The equations of motion for the gauge fields can be derived from this Lagrangian density by applying the Euler-Lagrange equations, which state that

$$\frac{\partial \mathcal{L}}{\partial A^a_\mu} - \partial_\nu \frac{\partial \mathcal{L}}{\partial (\partial_\nu A^a_\mu)} = 0$$

Using this formula, we obtain

$$\partial_\nu F^{a\nu\mu} + g f^{abc} A^b_\nu F^{c\nu\mu} = J^{a\mu}$$

where $J^{a\mu}$ is the current density for the matter fields, defined as

$$J^{a\mu} = \frac{\delta \mathcal{L}_\text{matter}}{\delta A^a_\mu}$$

These are called the Yang-Mills equations, and they describe how the gauge fields evolve in response to their own curvature and the interaction with matter.

I hope this answer helps you understand why the Lagrangian for gauge boson fields has that form. It is not derived from any deeper principle, but rather it is postulated to be consistent with gauge invariance, unitarity and causality. It also agrees with experimental observations of elementary particles.