# EM determined by gauge freedom?

1. Nov 16, 2015

### maline

I've heard the claim that the gauge freedom of the general Lagrangian can be used to derive the Lorentz force on a charged particle. I understand that Langrangian gauge freedom allows A⋅v -∇φ to be unaffected by the gauge freedom in defining the EM potentials, but this seems like a convenient result, not something that can be used as a derivation. Can someone clarify?

2. Nov 16, 2015

### fzero

I don't think that the Lagrangian for a charged particle is completely determined by invariance under gauge and Lorentz transformations. I think there is some discussion in Landau and Lifschitz about an additional term that would be consistent with the symmetries but appears to be absent experimentally.

You are working in a non-Lorentz covariant form, so I will too. Suppose we have a charged particle in the fields described by the potentials $\mathbf{A}$ and $\phi$. The gauge transformations are $\mathbf{A}\rightarrow \mathbf{A} + \nabla f$ and $\phi\rightarrow \phi - \dot{f}$.

The Lagrangian can be discussed term by term. The kinetic term is (I take $c=1$)
$$- m \sqrt{ 1- |\dot{\mathbf{r}}|^2 } \approx \frac{m |\dot{\mathbf{r}}|^2 }{2} -m,$$
where the RHS is the nonrelativistic approximation (from which we typically drop the $-m$, since it doesn't affect the equations of motion).
Next, we can assume that we already know from electrostatics that there is a $- q \phi(\mathbf{r})$ term. The gauge transformation of this term is $q \dot{f}$, which is a total time derivative. This is almost as good as being gauge invariant, since a total time derivative will not contribute to the equations of motion.

Finally, we must introduce the vector potential. We will make the assumption that the term must be linear in the potential, since Maxwell's equations are linear. So we add a term $q \mathbf{A} \cdot \mathbf{g}$, for some vector-valued function $\mathbf{g} =\mathbf{g}(\mathbf{r},\dot{\mathbf{r}})$. We've made the further assumption that $\mathbf{g}$ is not explicitly time-dependent, but only depends on the coordinates and velocity of the particle, which seems reasonable. The gauge variation of this term is $q \nabla f \cdot \mathbf{g}$. We can't set this to zero for any $f$ without having $\mathbf{g}=0$, but if we note that for $f=f(\mathbf{r})$, we have $\dot{f} = \nabla f \cdot \dot{\mathbf{r}}$, then the variation is a total derivative if $\mathbf{g} = k \dot{\mathbf{r}}$, for some constant $k$.

Now, we can fix $k=1$ by demanding Lorentz invariance. I won't try to work through that, but we can note that the equations of motion following from the above Lagrangian can be written in the form
$$m \ddot{\mathbf{r}} = q \left( - \nabla_{\mathbf{r}} + \frac{d}{dt} \nabla_{\dot{\mathbf{r}}} \right) ( \phi - k \dot{\mathbf{r}} \cdot \mathbf{A} ).$$
The gauge variation of $\phi - k \dot{\mathbf{r}} \cdot \mathbf{A}$ is $(1-k) \dot{f}$, so gauge invariance also requires $k=1$.

So in a sense, we are led to the precise terms by gauge invariance. However, we had to start with the electrostatic potential term and we have not shown that there are no other invariant terms that we could have considered. So this is not a complete proof.

3. Nov 17, 2015

### maline

Thanks! Would you mind elaborating on this point:
How do we know that L should depend on A as opposed to E & B? Also, why does linearity of L, which "represents" forces on the particle, follow from linearity in the interactions between the fields?
Is this necessary or just for ease of presentation? What are the actual minimal assumptions here?

4. Nov 17, 2015

### fzero

I don't know if you have seen the Lagrangian description of the electromagnetic field itself. The Lagrangian (really Lagrangian density, but I won't make this distinction below) is
$$L_\text{Max.} = - \frac{1}{4} F_{\mu\nu}F^{\mu\nu} = \frac{1}{2}( \mathbf{E}^2 - \mathbf{B}^2 ),$$
where in the middle is the Lorentz covariant expression, where $F_{\mu\nu} = \partial_{\mu} A_\nu - \partial_nu A_\mu$ is the field strength and the 4-vector potential has components $A_0 = \phi$, $A_i = (\mathbf{A})_i$. The last expression is what we get in terms of the electromagnetic fields.

Now, even though the Lagrangian is a simple expression of the fields, there's no variation that we can do with respect to the fields that would lead to the Maxwell equations. It is quadratic, without derivatives of the fields, so we can't get expressions like divergence or curl out. However, the variation with respect to the potential $A_\mu$ leads quite naturally to Maxwell's equations. We could say that the components of the 4-potential are the natural Lagrangian degrees of freedom, instead of the fields.

It also turns out that we can get the Maxwell equations with a current $j^\mu$ out of the Lagrangian if we just add the term $j^\mu A_\mu$. This would give us another way to derive the Lagrangian for the charged point particle, presuming that we use physical arguments to express the current in terms of the charge and velocity.

As I just said in the last paragraph above, if you could argue on physical grounds that the component of current $j^0 = -q$, then you could use a Lorentz transformation to work out the other components in a moving frame. This would also be a derivation of the coupling terms, but from a different assumption. I personally think this more physical approach is more illuminating, but it still assumes the way that the current appears in Maxwell's equations to motivate the expression for the Lagrangian.

We can do a bit better if we were studying the point particle described by a field itself, as one does in quantum field theory. Then Lorentz invariance and gauge invariance do really constrain the possible terms of the Lagrangian. The lowest-order terms give the Maxwell equations where the current is now described in terms of the particle field. But the details are a bit much to try to get into here. However, it is truly a case where you can use mathematics to fix the Lagrangian and then interpret the results physically. However, for the point particle represented by position and velocity as we've been discussing, you really need the physical interpretation before you can specify the mathematics.

5. Nov 17, 2015

### maline

Hmm... well what are the minimal assumptions needed to derive this Lagrangian? Do we need the full Maxwell equations, or can we use "gauge invariance" for this as well?

Does this refer to the gauge invariance of the potential, or just that of Lagrangians in general? If the first, why is "gauge symmetry" a reasonable assumption before we know anything about the dynamics?
Sorry if I'm flooding you with questions! I hope you're also enjoying this

6. Nov 17, 2015

### fzero

From the classical perspective, we impose gauge invariance by hand, typically because we already know that it is a symmetry of the Maxwell equations. If we were doing quantum field theory, we would see that we really need gauge invariance, otherwise the quantum theory doesn't make sense.

So we want the Lagrangian to be invariant under $A_\mu \rightarrow A_\mu + \partial_\mu f$. The field strength $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ is the simplest gauge invariant object that we can form. Similarly $F_{\mu\nu}F^{\mu\nu}$ is the simplest Lorentz invariant that we can construct from the field strength. The coefficient $-1/4$ is chosen by convention, since it makes the formulas in the quantum theory look nice. Therefore, gauge and Lorentz invariance naturally lead to the Maxwell equations.

We need invariance under the gauge transformation of the potential $A_\mu$. As I mentioned above, classically we have to impose this because we want to describe electrodynamics, but quantum mechanically we would see that gauge invariance is the solution to a problem that arises. Once we couple the potential to another field, say one that would describe an electron, then we would see that gauge invariance of the full Lagrangian requires that the electron field must also undergo a simultaneous gauge transformation along with the potential.

7. Nov 17, 2015

### maline

But for a charged particle, the lagrangian is not invariant under that transformation- it changes by a total derivative. The same is true for the Galilean transformation in classical dynamics. Why do we demand actual invariance here?

Last edited: Nov 17, 2015
8. Nov 17, 2015

### fzero

Let's go through the Lorentz-invariant terms that we can have from scratch. At linear order we have $j^\mu A_\mu$, which is how we include the current. The gauge variation of this term is $j^\mu \partial_\mu f= \partial_\mu (j^\mu f) - f \partial_\mu j^\mu$. The first term is a total derivative, so that's ok. For the 2nd term to vanish, we require that the current is conserved, $\partial_\mu j^\mu$. So we get some additional physics out of requiring gauge invariance.

We could also have linear terms with some derivatives, like $\partial^\mu A_\mu$, but all of these terms are total derivatives themselves, so we don't have to bother with them.

At quadratic order, we could have $A^\mu A_\mu$, which has an interpretation in QFT as a mass term for the photon. The gauge variation is, keeping all terms,
$$2 A^\mu \partial_\mu f + (\partial_\mu f)^2.$$
We could integrate these terms by parts to look for conditions for which they vanish up to partial derivatives, but the 2nd term would require $\partial^\mu \partial^\mu f =0$ and we're not allowed to put any restrictions on the function $f$ if we want true gauge invariance of the theory.

Similarly, with two derivatives, we could consider
$$L_{a,b} = a (\partial^\mu A^\nu)(\partial_\mu A_\nu) +b (\partial^\mu A^\nu)(\partial_\nu A_\mu),$$
which has gauge variation
$$(a+b) [ 2 (\partial_\mu A_\nu) \partial^\mu\partial^\nu f + (\partial^\mu\partial^\nu f)^2 ].$$
Again, there is no way to make terms quadratic in $f$ vanish without imposing restrictions on $f$, except in the case where $b = -a$, which is the case we had before with $(F_{\mu\nu})^2$.

We could consider other terms, but it's clear that they will suffer from the same problems unless we can write them in terms of an actual gauge invariant like $F_{\mu\nu}$, or introduce some new fields that participate in the gauge transformation, as I mentioned earlier.

9. Nov 18, 2015

### maline

Great!
So to summarize: Given gauge freedom of A which we know from Maxwell's equations, plus the Lorentz transformation of physics in SR, and given that the Lagrangian should be written in terms of A and j (because F has more components than d.o.f's), we are very limited in our options. The simplest of these options is the one we choose, and sure enough, it spits out Maxwell's equations & the Lorentz force density, which reduces to the Lorentz force if j is a delta function.
In classical EM, starting with gauge symmetry as a given would not work if we didn't already know Maxwell's equations, but in QED the gauge symmetry of Lagrangian mechanics leads to a demand that any vector field coupled to matter should have gauge symmetry. This can then be used as a first-principles derivation of EM, accurate to first order.
Did I get it straight?

10. Nov 18, 2015

### fzero

Yes, except that I would rephrase

It's not quite that it would not work, it's that the classical theory of a vector field does not require gauge symmetry, But, of course, we need to include gauge invariance if we want the theory to describe classical electrodynamics.

Physical examples of vector fields without gauge invariance are massive vector fields like the W and Z bosons, or the photon in a superconducting material. The quantum theory in those cases makes sense because the gauge symmetry is restored at a high energy scale (the symmetry is said to be spontaneously broken at low scales)..