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I am currently reading the 2nd volume of Feynman Lectures and I am stuck in the part where he solves the Maxwell's equation to find the potentials and wave equations. I can't understand the thing he calls choosing a gauge.

He changed both of the potentials in a way that despite the change the electric and magnetic field stays the same. The tranformation for the vector potential is :

$\vec A' = \vec A+ \nabla \psi$ and then the book says

We can restrict $\vec A$ by choosing arbitrarily what the divergence of $\vec A$ must be. We can always do that without changing $\vec B$. In fact, $\nabla. \vec A' = \nabla. \vec A + \nabla ^2 \psi$ and by a suitable choice of $\psi$ we can make $\nabla. \vec A'$ anything.

But I am having trouble finding how this is possible. Can anyone show how this is true? Also the books mentions two kinds of gauge fixing (coulomb and lorenz). How do I know when to use either of these gauges? It's really confusing.

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  • $\begingroup$ What about this is confusing? What you call $\phi$ (your gauge field) is arbitrary, and -- if you know $\vec{A}$ -- you can solve $\nabla^2 \phi = -\vec{\nabla} \cdot \vec{A}$ so that $\vec{\nabla}\cdot\vec{A}^\prime = 0$, for example. $\endgroup$
    – Philip
    Jun 11 '20 at 6:02
  • $\begingroup$ @Philip But won't making $\phi$ anything change the $\vec E$ as $ \vec E = -\nabla \phi - \frac{\partial \vec A}{\partial t}$ $\endgroup$ Jun 11 '20 at 6:34
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    $\begingroup$ That's a different function $\phi$, it represents the scalar electric potential, you've got your symbols a little mixed up. The gauge field is usually denoted by a different symbol ($\lambda$ or $\chi$), and it's not the same thing as the scalar potential $\phi$. $\endgroup$
    – Philip
    Jun 11 '20 at 6:55
  • $\begingroup$ @Philip Thanks for pointing out. It is $\psi$ according to the book. But still doesn't it change the fields? $\endgroup$ Jun 11 '20 at 7:00
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    $\begingroup$ You're right, changing $A$ alone does seem to change $\vec{E}$, but if you also change $\phi \to \phi - \frac{\partial \psi}{\partial t}$, it won't change. In other words, you must change both $\vec{A}$ and $\phi$. $\endgroup$
    – Philip
    Jun 11 '20 at 7:02
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As you point out, only shifting $\vec{A}$ is not sufficient to keep the $\vec{E}$ and $\vec{B}$ fields the same. Indeed, if you shift $\vec{A}$ by the gradient of a scalar $\psi$,

$$\vec{B^\prime} = \vec{\nabla}\times\vec{A^\prime} = \vec{\nabla}\times\left(\vec{A} + \vec{\nabla}\psi\right) = \vec{\nabla}\times\vec{A} = \vec{B},$$

but

$$\vec{E^\prime} = -\vec{\nabla}\phi -\frac{\partial \vec{A^\prime}}{\partial t} = -\vec{\nabla}\phi -\frac{\partial \vec{A}}{\partial t} + \vec{\nabla}\frac{\partial \psi}{\partial t} \neq \vec{E}.$$

Of course, if you also changed $$\phi \to \phi^\prime = \phi - \frac{\partial \psi}{\partial t},$$

then it's easy to show that $\vec{E^\prime} = \vec{E}$. In other words, the actual gauge transformation is simultaneously changing:

$$A\to A^\prime = A + \vec{\nabla}\psi\\ \phi \to \phi^\prime = \phi - \frac{\partial \psi}{\partial t}.$$

This is made a little clearer if you know that $\phi$ and $\vec{A}$ actually form a four-vector in Special Relativity, $A^\mu \equiv \begin{pmatrix}\phi/c & \vec{A}\end{pmatrix}^T,$ and so the gauge transformations are actually a transformation of the four-vector $A^\mu$:

$${A^\prime}^\mu = A^\mu + \partial^\mu \psi.$$

Coulomb and Lorenz Gauges

Coulomb Gauge Which gauge you use will depend quite strongly on the type of problem you're solving, and the methods available to you. For example, in electrostatics, you must know that you can solve for the electric potential given a charge distribution using the Poisson Equation: $$\nabla^2 \phi = -\frac{\rho}{\epsilon_0},$$

and the solution is well known. It is just:

$$\phi(\vec{r},t) = \frac{1}{4\pi\epsilon_0}\int \frac{\rho(\vec{r}^\prime)}{|\vec{r} - \vec{r^\prime}|} \text{d}^3 \vec{r},$$

and we can solve it if we can perform that integral. However, when there is a magnetic field, this is no longer true in general, since

$$ \vec{\nabla}\cdot \vec{E} = - \nabla^2\phi - \frac{\partial}{\partial t}\vec{\nabla}\cdot\vec{A} = \frac{\rho}{\epsilon_0}.$$

Which may be a more complicated equation to solve than the "simpler" Poisson Equation. Of course, if we have the liberty to choose an $\vec{A}$ whose divergence is zero, then we get back the old equation that we know how to solve! In other words, in the Coulomb Gauge, we can solve for the potential as we could in the electrostatic case, even when there are magnetic fields.

Another place that the Coulomb Gauge is useful is in Quantum Mechanics. The Hamiltonian of a particle in an electromagnetic field is given by $$\hat{H} = \frac{(\hat{p} - e\vec{A})^2}{2m} = -\frac{h^2}{2m}\left(\vec{\nabla} - i\frac{e}{h}\vec{A}\right)^2,$$

and since the operators $\hat{p}$ and $\hat{x}$ don't commute, there is no reason for this to look like the classical result. It turns out that in the Coulomb Gauge, the quantum mechanical Hamiltonian operator looks exactly like the classical Hamiltonian, and that makes this a little simpler to solve.

$$\hat{H} = \frac{\hat{p}^2 - 2 e \hat{A}\cdot\hat{p} + e^2 \hat{A}^2}{2m}$$

Lorenz Gauge There is a "problem", however, with the Coulomb Gauge as -- looking at Poisson's Equation -- as it looks like if the charge distribution is changed at some point in space, then the potential (and consequently $\vec{E}$) changes everywhere in space simultaneously. Of course, we know that this is not true since information can't travel faster than $c$! This would suggest that while the Coulomb Gauge may work well for static distributions, it is perhaps not the best gauge for distributions that change in time.

For this, it's best to use the Lorenz Gauge $$\partial_\mu A^\mu = 0, \quad \quad \text{i.e.}\quad\quad \vec{\nabla}\cdot \vec{A} = - \frac{1}{c^2} \frac{\partial \phi}{\partial t}.$$

In this case, it can be shown from Maxwell's Equations (it's a nice exercise) that the potentials satisfy

$$\nabla^2\phi - \frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2} = \frac{\rho}{\epsilon_0},\\ \nabla^2\vec{A} - \frac{1}{c^2}\frac{\partial^2 \vec{A}}{\partial t^2} = \mu_0 \vec{j},$$

or in other words the potentials satisfy wave equations that are "sourced" charge and current densities, and so information of changes in these distributions travel at the speed $c$.

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  • $\begingroup$ So we can always change $\vec A$ and compensate the change by taking a suitable $\psi$? But Suppose $\vec A$ is changed in a certain way and this change can't be compensated by any $\psi$. Can this happen? What would it mean if this happens? $\endgroup$ Jun 11 '20 at 15:26
  • $\begingroup$ How could that be possible? If you have a function $\psi(x,t)$ that is differentiable by both $x$ and $t$, then you could always shift $\vec{A}$ and $\phi$ using it to get some new potentials. And since the relations that we've used (the divergence of the curl of any vector field is zero, and the curl of the gradient of any scalar field is zero) are always true, there should be nothing stopping us from doing this. $\endgroup$
    – Philip
    Jun 11 '20 at 18:56
  • $\begingroup$ Let's consider an one dimensional case where $\vec A = x^x \hat i$ and now if I want to make $\vec A$ zero then I have to find a $\psi$ whose gradient $\frac{\partial \psi}{\partial x}\hat i$ will have to be $x^x \hat i$ which means $\psi$ will have to be the integal of $x^x$ with respect to $x$. But $x^x$ can't be integrated and so it will be impossible to find a suitable $\psi$. I maybe wrong and will be very relieved if you point it out. $\endgroup$ Jun 12 '20 at 7:13
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    $\begingroup$ There are many ideas to unpack here. Firstly, gauge invariance doesn't mean that we can always set the vector potential to 0! That wouldn't be possible if there actually was a magnetic field, since it would mean that people using different gauges would disagree on whether $\vec{B}$ existed! All it says is that we can set a constraint on one of its components (for example, by enforcing its divergence to be 0). $\endgroup$
    – Philip
    Jun 12 '20 at 7:24
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    $\begingroup$ Secondly, its best not to use a one dimensional case (EM becomes very different in different dimensions), so let's assume you're asking about a case where $\vec{A} = x^x \hat{x}$. Clearly, in this case $\vec{B}=\vec{\nabla}\times\vec{A}=0$, so it should be possible to set $\vec{A}=0$ using a gauge transformation. As you point out, we'd need $\psi = -\int x^x \text{d}x$, and while there is no closed form solution, this function can certainly be integrated! It's just that the solution can't be written in terms of nice well-known functions, but can be done numerically! $\endgroup$
    – Philip
    Jun 12 '20 at 7:27
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To turn @Philip's comments into a full answer:

The physical, measurable quantities in electromagnetism are the electric and magnetic fields, $\vec{E}$ and $\vec{B}$. As noted in the comments, they are defined in terms of the potentials as follows: $$\vec{B} = \nabla\times \vec{A}, \text{ and }\, \vec{E} =- \nabla\phi - \frac{\partial \vec{A}}{\partial t}.$$

The essence of a gauge transformation is that, when we simultaneously change $\vec{A}$ and $\phi$ in a particular way, the electric and magnetic fields remain the same. This particular way is: $$\vec{A} \rightarrow \vec{A} +\nabla \psi, \qquad \phi \rightarrow\phi -\frac{\partial \psi}{\partial t}.$$

Then one sees that $$\vec{B} \rightarrow \nabla \times \vec{A} + \nabla \times \nabla\psi = \nabla \times \vec{A}\\ \vec{E}\rightarrow - \nabla \phi - \frac{\partial \vec{A}}{\partial t} + \nabla \frac{\partial \psi}{\partial t} - \frac{\partial}{\partial t} \nabla \psi = - \nabla \phi - \frac{\partial \vec{A}}{\partial t},$$ where the last equality on each line follows from the fact that curl grad = 0 and that we can interchange the $\nabla$ with the $\frac{\partial}{\partial t}$, provided $\psi$ is at least twice differentiable.

This tells us that the above transformation does not change the physics and so we are free to pick the potential up to a gauge transformation.

Finally, Coloumb gauge and Lorenz gauges are choices of a function $\psi$ such that we end up having the following constraints on the gauge potential $$\nabla \cdot \vec{A}=0 \qquad \text{ Coulomb gauge},\\ \nabla \cdot \vec{A}=- \frac{1}{c^2}\frac{\partial \phi}{\partial t} \qquad \text{Lorenz gauge}.$$

The wiki links above should give you an idea of when each of those two gauges is useful.

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  • $\begingroup$ The gauge fixing seems to impose conditions only on divergence of $\vec A$, can't one impose a condition directly on $\vec A$ without chaning electric and magnetic field? $\endgroup$ Jun 11 '20 at 15:28
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    $\begingroup$ These examples don't. But one can do it in principle. Take for example symmetric gauge in the context of the Landau problem in two dimensional systems. One has a magnetic field perpendicular to the surface and the vector potential could point radially outwards in any direction and still produce the same $\vec{B}$ field. The symmetric gauge makes it so that the components of $\vec{A}$ are symmetric w.r.t. the x-y axes. Thus it imposes a condition on $\vec{A}$, as opposed to its divergence. $\endgroup$
    – Stratiev
    Jun 11 '20 at 17:18
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Fixing a gauge is essentially fully constraining the remaining free variables. Since $E$ and $B$ are defined through derivatives, we have some freedom in our choice of the scalar and vector potential that results in the same derivative. As an analogy, suppose we had the following problem:

$$\frac{\partial}{\partial x} f(x,y) = 2x$$

We could say that $f(x,y) = x^2+g(y)+c$, and our choice of $c$ and $g(y)$ would be the same as fixing a "gauge".

In electrodynamics, we have a similar result. We can choose some auxiliary field $\lambda$. Your mistake is that we need to transform both our scalar and vector potentials, as in the following

$$A \mapsto A + \nabla \lambda$$ $$V \mapsto V - \frac{\partial}{\partial t} \lambda$$

You could do a little work to convince yourself that this transformation results in the same results of $E$ and $B$. You can also show that there is some $\lambda$ that produces the Coulomb and the Lorentz gauge that you mention as well.

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