Lagrangian of EM field: Why the $B$-field term has a minus sign in front of it in the Lagrangian?

I know that $$L = T - U$$ and that, in the non-relativistic case

$$L= \frac{1}2mv^2 - q\phi(r,t) + q\vec{v}\cdot\vec{A}(r,t).\tag{1}$$

My lecturer used the following form of the Lagrangian density to derive Maxwell's equations:

$$L = \vec{j}(r,t)\vec{A}(r,t) - \vec{\rho}(r,t)\vec{\phi}(r,t) + \frac{\epsilon}2 \vec{E}^2(r,t)-\frac{1}{2\mu}\vec{B}^2(r,t). \tag{2}$$

Comparing the two equations for $$L$$, I can see that the KE term in the first equation is substituted for the energy density of the EM field. What I do not understand is why the $$B$$-field term has a minus sign in front of it in the Lagrangian (2)?

Can someone please shed some light on this for me please?

P.S - I have checked the related posts and none of them address my issue.

• If it had a plus sign, minimizing the action would mean that both E and B are equal to zero. – Javier Nov 18 '18 at 14:12
• If $L$ had a plus sign there, then one of Maxwell's equations would turn out with the wrong sign at $\mu$: $-\frac{1}{\mu} \vec{\nabla} \times \vec{B} = \vec{j} + \epsilon \frac{\partial}{\partial t} \vec{E}$ – Thomas Fritsch Nov 18 '18 at 15:12
• @Javier That statement ignores that the variational requires boundary conditions as additional input (the same logic would seem to indicate that the classical solutions of electromagnetism has infinite magnetic field and vanishing electric field). In fact, if the minus sign is gone, the action remaining is Euclidean electromagnetism, which is a perfectly well defined field theory. – Bob Knighton Apr 26 at 9:06

In the gauge $$\phi=0$$, the $$E$$ term is $$\frac12\dot A^2$$, which is kinetic energy, and the $$B$$ term is $$(\nabla\times A)^2$$, which is potential energy and therefore gets a minus sign.

• why did they choose this form of lagrangian? is there no deeper meaning? most literature I come across just states it without giving explanation, the literature which derives the form of the lagrangian for a particle in an EM field derives the first equation I listed above. My question is still unanswered...... – Ralph Ramlal Nov 18 '18 at 11:32
• If you want your Lagrangian to be Lorentz-invariant, I think the only option is $\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$, where $F_{\mu\nu}=\partial_\mu A_\nu - \partial_\nu A_\mu$, and $A_\mu = (\phi, \vec{A})$. If you expand that out, you get the Lagrangian above. For a non-relativistic argument, just check that it correctly gives you Maxwell's equations as the equations of motion. Any other Lagrangian would not. – Eric David Kramer Nov 18 '18 at 13:09

The only reason why Lagrangians are what they are is because they give the correct equations of motion.

In addition you may want to require that certain symmetries are preserved, but there is not much more to that. Actually, one can prove that for many systems there are infinitely many (different) equivalent Lagrangians giving raise to the same equations of motion (so you could pick any of them).

First, to clarify, the first Lagrangian you gave is the Lagrangian describing a non-relativistic point particle moving in a fixed background potential $$(\phi,\textbf{A})$$. The second Lagrangian describes the dynamics of the electromagnetic field in the presence of a fixed background source $$(\rho,\textbf{j})$$.

Now, the short answer to why the minus sign is there is that, without it, the equations obtained aren't Maxwell's equations.

A slightly longer answer is that the minus sign exists to ensure relativistic invariance. Under a Lorentz transformation, the electromagnetic fields transform as

$$\textbf{E}_{\perp}\to\gamma\left(\textbf{E}_{\perp}+\textbf{v}\times\textbf{B}\right),\hspace{0.5cm}\textbf{B}_{\perp}\to\gamma\left(\textbf{B}_{\perp}-\frac{1}{c^2}\textbf{v}\times\textbf{E}\right),$$

where $$\perp$$ denotes the component of the field perpendicular to boost velocity $$\textbf{v}$$. Under this transformation, we have

$$\frac{\epsilon_0}{2}\textbf{E}^2-\frac{1}{2\mu_0}\textbf{B}^2=\left(\frac{\epsilon_0}{2}E_{\parallel}^2-\frac{1}{2\mu_0}B_{\parallel}^2\right)+\frac{\gamma^2\epsilon_0}{2}\left(\textbf{E}_{\perp}+\textbf{v}\times\textbf{B}\right)^2-\frac{\gamma}{2\mu_0}\left(\textbf{B}_{\perp}-\frac{1}{c^2}\textbf{v}\times\textbf{E}\right)^2.$$

Being careful with our cross product identities, we can show

$$(\textbf{E}_{\perp}+\textbf{v}\times\textbf{B})^2=\textbf{E}_{\perp}^2+\textbf{v}^2\textbf{B}^2-\left(\textbf{v}\cdot\textbf{B}\right)^2+2\textbf{E}\cdot\left(\textbf{v}\times\textbf{B}\right)$$ $$\left(\textbf{B}_{\perp}-\frac{1}{c^2}\textbf{v}\times\textbf{E}\right)=\textbf{B}_{\perp}^2+\frac{1}{c^4}\textbf{v}^2\textbf{E}^2-\frac{1}{c^4}\left(\textbf{v}\cdot\textbf{E}\right)^2-\frac{2}{c^2}\textbf{B}\cdot\left(\textbf{v}\times\textbf{E}\right).$$

Plugging this directly into the transformation of the Lagrangian shows that the $$\gamma$$ factors cancel, and the Lagrangian is relativistically invariant.

The longest answer requires the language of differential forms to truly appreciate. If $$A$$ is the electromagnetic 1-form, and $$F=\mathrm{d}A$$ is its corresponding field strength, the canonical action for this $$U(1)$$ gauge theory is given by

$$S=-\frac{1}{2}\int_{\mathcal{M}} F\wedge\star F.$$

Now, to express this in terms of the electric and magnetic field, we note that if we have an orthonormal basis $$\{\mathrm{d}x^\mu\}$$ of $$\Omega^{1}(\mathcal{M})$$ (the space of one-forms on $$\mathcal{M}$$), we can define an electric field one form

$$E=E_{1}\mathrm{d}x^1+E_{2}\mathrm{d}x^2+E_{3}\mathrm{d}x^3.$$

Next, we can define a magnetic field 2-form

$$B=B_1\mathrm{d}x^2\wedge\mathrm{d}x^{3}+B_2\mathrm{d}x^3\wedge\mathrm{d}x^1+B_3\mathrm{d}x^1\wedge\mathrm{d}x^2.$$

In terms of these variables, the field strength is decomposed as

$$F=B+E\wedge\mathrm{d}x^0.$$

Now, when evaluating $$F\wedge\star F$$, the hodge star of $$E\wedge\mathrm{d}x^0$$ will have an extra minus sign relative to the $$B$$ term, coming from the minus sign in the metric tensor $$g=\text{diag}(-1,1,1,1)$$. This is the mathematical origin of the minus sign.

In summary, there are three answers to this question, with three different levels of intuitive satisfaction:

• The minus sign is there to ensure that the equations of motion obtained are Maxwell's equations (not that satisfying).
• The minus sign is there to ensure relativistic invariance of the action (somewhat satisfying).
• The minus sign comes from the minus sign in the Minkowski space metric tensor (very satisfying).