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?
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Qmechanic
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moboDawn_φ
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1Well, it is a straight-forward generalization of Maxwell's theory to non-abelian gauge groups. The Lagrangian of Maxwell's theory has the same form, where $F$ is the curvature of an $U(1)$ bundle. Alternatively, it can also be viewed as the simplest possible gauge-invariant Lagrangian quadratic in the fields (together with the topological Yang Mills term, which is usually called theta-term and ignored for physical reasons, c.f. "strong CP problem") – G. Blaickner Jul 21 '23 at 22:18
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2A Lagrangian is not really derived mathematically from scratch. It's actually just a good guess, which is later corroborated by several consistency and experimental checks. – Avantgarde Jul 21 '23 at 22:31
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Doesn’t your book write $F^{\mu\nu}a$ rather than $F^{\mu\nu}\alpha$? The usual convention is that spacetime indices are Greek and gauge indices are Roman. – Ghoster Jul 21 '23 at 22:57
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On non effective field theories you normally include all renormalizable terms that are invariant under the summery group, here F^2 is the only such term normally e.g. a linear is not invariant under gauge Trafo. – Thomas Tappeiner Jul 21 '23 at 23:04
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Related question for abelian gauge theory: https://physics.stackexchange.com/q/55291/2451 and links therein. – Qmechanic Jul 22 '23 at 01:34
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For one you need renormalizability, so given a curvature, which you can think of in terms of the covariant derivatives, there is only so much that will give you renormalizability. As mentioned you also want gauge invariance, which often is accommodated by having such a curvature dependence. – MrDBrane Nov 22 '23 at 07:14
1 Answers
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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.
