As the title of the question suggest; how you could determine if a gauge fixing is a condition or a requirement. Let me explain.

Imagine you are working with Maxwell's Equations. By the definition of the fields: $$ \mathbf E = -\nabla\phi - \frac{\partial \mathbf A}{\partial t} $$

$$ \mathbf B = \nabla \times \mathbf A $$ and by using the Maxwell's Equations, you arrive to the following expressions: $$ \Box \mathbf A -\nabla \left ( \nabla \cdot \mathbf A + \frac{1}{c^2} \frac{\partial \phi}{\partial t} \right ) = -\mu_0\mathbf J $$

$$ \frac{\rho}{\varepsilon_0} = -\nabla^2\phi - \frac{\partial}{\partial t} (\nabla \cdot\mathbf A). $$ By imposing the Lorenz gauge fixing condition: $$ \nabla\cdot\mathbf A = - \frac{1}{c^2}\frac{\partial \phi}{\partial t} $$

We get: $$ \Box \mathbf A = -\mu_0 \mathbf J \\ \; \\ \Box \phi = -\frac{\rho}{\varepsilon_0}. $$ My question is, once our gauge is fix, and we get the simplified equations, does the gauge condition satisfies itself automatically, or after solving the resultant equations, you will have to go back to the gauge fixing equations and require them (by hand) to satisfy as well? How would you prove this?

  • $\begingroup$ The gauge is entirely unphysical, it is a purely a choice of 'representation', if you will. When you solve the equation, to get a unique solution, the gauge condition is PART of the equations you solve. By definition then, the gauge condition will hold for all times Nothing stops you from changing gauge condition at any arbitrary time. But the point is that which gauge is satisfied, it is purely your choice. In your example, you should solve the simplified equations AND the gauge conditions. It is true that in some cases, this might be automatic, but generally it shouldn't be the case. $\endgroup$
    – Frotaur
    Sep 29, 2023 at 13:34

3 Answers 3


This was stated in a comment, but I wish to make it really clear.

It is not sufficient to look for potentials merely satisfying $$ \Box \mathbf A = -\mu_0 \mathbf J \\ \; \\ \Box \phi = -\frac{\rho}{\varepsilon_0}. $$ You have to find potentials satisfying those equations and the gauge condition $$ \nabla\cdot\mathbf A = - \frac{1}{c^2}\frac{\partial \phi}{\partial t}. $$ Thus you have three equations to solve, not two.

Of course you can if you like replace the potentials with some other set related to the first by a gauge transformation. In that case your new potentials don't necessarily satisfy any of the above! An exception is when you replace a set of potentials by another (related by a gauge transformation) but which still satisfy the Lorenz gauge condition. In that case the new potentials will satisfy all the above, but I think they won't necessarily join smoothly to your first set. (I'm not sure of the correctness of that last clause but it seems reasonable).

Added comment

After pondering this a little it occurred to me that I can offer an easy example. In empty space the first two equations become wave equations and then there are plane wave solutions (among other solutions). If you take a plane wave solution for $\phi$ (at some non-zero frequency) and the solution ${\bf A}=0$ for $\bf A$ then the first two equations are satisfied but this is is not a valid solution overall because it does not satisfy the gauge condition.

  • $\begingroup$ How can you be sure that solutions of equations AND gauge conditions simultaneously exist? $\endgroup$ Sep 30, 2023 at 13:37
  • $\begingroup$ @ValterMoretti There's a proof that if you have some potentials not in Lorenz gauge, then there always exists a function which does the transformation to Lorenz gauge. One shows that the gauge function has to satisfy a wave equation with a source, and there are solutions for any source. $\endgroup$ Sep 30, 2023 at 15:01
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    $\begingroup$ Indeed I posted my own proof above...But it contains some subtle points as you can read. and not all initial conditions are permitted. $\endgroup$ Sep 30, 2023 at 15:09

This is a very important issue that is usually overlooked in almost all books even if it being of fundamental relevance in my view. Also with a great impact on the quantization procedure. (Also for other fields than Maxwell one

According to the original issue, we end up with this set of differential equations (I henceforth assume $c=1$) $$ \nabla\cdot\mathbf A = - \frac{\partial \phi}{\partial t} \tag{1} $$ $$ \Box \mathbf A = -\mu_0 \mathbf J \tag{2}$$ $$ \Box \phi = -\frac{\rho}{\varepsilon_0}. \tag{3} $$ They have to be solved simultaneously. As already noticed in @Sanchar's answer, the second pair of equations implies $$ \square \left( \boldsymbol{\nabla}\cdot\boldsymbol{A} +\frac{\partial \phi}{\partial t} \right) = -\mu_0 \boldsymbol{\nabla}\cdot\boldsymbol{J} - \frac{\partial_t\rho}{\varepsilon_0} = 0, \tag{4}$$ when one assumes that the continuity equation of the electric charge is valid, and this can be done with a suitable choice of the given sources $\rho$ and $J$.

It is not possible at the level of equations to get anything stronger than (4), in place of the wanted (1).

In summary, we cannot assume from scratch that the gauge condition (1) is satisfied by solutions of (2) and (3) with given $\rho$, $J$. Even if they satisfy the continuity law of the electric charge!

The only place where we can impose some further constraint in order to get (1) satisfied is while giving the initial conditions for the fields $\phi$ and ${\bf A}$.

To see how this machinery works, let me remind a fundamental theorem of existence and uniqueness for 2nd order hyperbolic differential equations specialized to the n+1 dimensional Minkowski spacetime and for the D'Alemebert equation (it can be formulated into a generalized version in any globally hyperbolic spacetime for normally hyperbolic PDEs).

THEOREM 1 Consider the system of PDEs for the vector valued function $\Phi \in C^\infty(\mathbb{R}\times \mathbb{R}^n; \mathbb{R}^k)$ $$-\frac{\partial^2 \Phi}{\partial t^2} + \Delta_{\vec{x}}\Phi = F(t,\vec{x}) \tag{E}$$ where $F \in C^\infty(\mathbb{R}\times \mathbb{R}^n; \mathbb{R}^k)$ is known and has compact support on every surface $t=t_0$. If $$\Phi(0, \vec{x}) = \Phi_0(\vec{x})\:, \quad \forall \vec{x} \in \mathbb{R}^n\tag{BC1}$$ and $$(\partial_t\Phi)(0, \vec{x}) = \Pi_0(\vec{x})\:, \quad \forall \vec{x} \in \mathbb{R}^n\tag{BC2},$$ where $\Phi_0,\Pi_0 \in C^\infty(\mathbb{R}^n; \mathbb{R}^k)$ are given compactly supported functions, then there exists a unique solution of the Cauchy problem (E)-(BC1)-(BC2).

The condition on the supports of the source $F$ and the initial conditions can be relaxed, but I will deal with this basic case.

Now consider the system (1)-(3). If we define $\Phi = (\phi, {\bf A})$ we can exploit Theorem 1 to prove that a solution of the system (2)-(3) exists and is unique for every choice of the initial conditions $$\phi(0, \cdot)=: \phi_0(\cdot)$$ $$\partial_t \phi(0, \cdot)=: \pi_0(\cdot) $$ $$A(0, \cdot)=: A_0(\cdot) $$ $$\partial_t A(0, \cdot)=: \Pi_0(\cdot)$$

If we wish that also (1) be valid, we can exploit Theorem 1 again for the scalar $G:= \boldsymbol{\nabla}\cdot\boldsymbol{A} + \frac{\partial \phi}{\partial t}$.

This should amount to impose some constraints on the initial conditions written above!

However, apparently, a problem pops out here. Because, to have that (1) is valid for all times as a consequence of THeorem 1 (applied to $\Phi=G$), we must impose two initial conditions. Evidently one is $$G(0, \vec{x})=0 \quad \mbox{that is} \quad \boldsymbol{\nabla}\cdot\boldsymbol{A} + \frac{\partial \phi}{\partial t}=0 \quad \mbox{for $t=0$}$$ The other should be the first time derivative of it $$(\partial_t G)(0, \vec{x})=0 \quad \mbox{that is} \quad \boldsymbol{\nabla}\cdot\boldsymbol{\partial_t {\bf A}} + \frac{\partial^2 \phi}{\partial t^2}=0 \quad \mbox{for $t=0$}\:.\tag{S}$$ The problem is that here the constraint is imposed on the second $t$-derivative of $\phi$ and we cannot arbitrarily fix it! We can only fix the fields at $t=0$ and their first $t$-derivatives at $t=0$.

However we have to keep in our mind that (S) does not come alone, but it is imposed together with (1),(2) and (3). In particular (3) yields (I assumed $c=1$) $$\frac{\partial^2\phi}{\partial t^2} = \Delta_{\vec{x}} \phi + \frac{\rho}{\epsilon_0}$$ Summing up, our initial Cauchy problem (1)-(2)-(3) yields the system $$ \Box\left(\nabla\cdot\mathbf A + \frac{1}{c^2}\frac{\partial \phi}{\partial t}\right)=0 \tag{G2} $$ $$ \Box \mathbf A = -\mu_0 \mathbf J $$ $$ \Box \phi = -\frac{\rho}{\varepsilon_0}. $$ with initial conditions $$\phi(0, \cdot)=: \phi_0(\cdot)$$ $$\partial_t \phi(0, \cdot)=: \pi_0(\cdot) $$ $$A(0, \cdot)=: A_0(\cdot) $$ $$\partial_t A(0, \cdot)=: \Pi_0(\cdot)$$ If these initial conditions satisfy the constraints $$\boldsymbol{\nabla}\cdot\boldsymbol{A}_0(\vec{x}) + \pi_0(\vec{x})=0 \qquad that \: is\: G(0, \vec{x})=0 \quad for \: t=0\tag{C1}$$ and $$\boldsymbol{\nabla}\cdot\boldsymbol{{\bf \Pi}_0}(\vec{x}) + \Delta_{\vec{x}} \phi_0(\vec{x}) + \frac{\rho(0,\vec{x})}{\epsilon_0}=0 \quad that\: \: is \:(\partial_t G)(0,\vec{x})=0 \: for \:t=0\:,\tag{C2}$$ Theorem 1 guarantees that there exists exactly one solution of the equation of motion and of the gauge constraint. If it happens, as the unique solution of (G2) with vanishing initial conditions guaranteed by the constraints (C1) and (C2) is $G\equiv 0$, we have found a (unique) solution of the initial gauged problem (1)-(2)-(3).

If the constraints (C1) and (C2) are not satisfied on the initial conditions of (1)-(3) -- as they arise from the request that $G=0$ at every time -- there is no solution of the system (1)-(2)-(3).

In summary, assuming the validity of the continuity of the electric charge, the answer to the initial question is the following one.

There is a (uniquely determined) solution of the Maxwell equations together with the validity of the (Lorenz) gauge condition if and only if the initial conditions of the fields $\phi$ and ${\bf A}$ satisfy some constraints. If these constraints are not fulfilled, there are no solutions of the overall problem.

  • $\begingroup$ +1, it is indeed important to specify and deal with things carefully at the level of initial data when dealing with evolution equations, rather than extraneously imposing extra conditions (also I fixed a few obvious typos regarding where $\partial_t$ should/should not be, notably equation 4- the continuity equation, and the constraint equation for $\partial_tG$, which shouldn’t have $\partial_t\rho$, only $\rho$). $\endgroup$
    – peek-a-boo
    Sep 30, 2023 at 20:42
  • $\begingroup$ Thanks, it s difficult to fix typos by phone. $\endgroup$ Sep 30, 2023 at 20:53
  • $\begingroup$ Also to see them… $\endgroup$ Sep 30, 2023 at 21:12

In the particular (Lorentz) gauge that you mentioned, indeed the gauge is satisfied at all times. This can be proven by first writing an equation of motion for the "gauge function", $$ \square \left( \boldsymbol{\nabla}\cdot\boldsymbol{A} + \frac{1}{c^2} \frac{\partial \phi}{\partial t} \right) = -\mu_0 \boldsymbol{\nabla}\cdot\boldsymbol{J} - \frac{\rho}{\varepsilon_0 c^2} = 0, $$ where the last equality follows from charge conservation and $c^2 \varepsilon_0 \mu_0 = 1$.

Let me call the "gauge function" as $G$. As an example, consider the boundary condition of $G=0$ at $t=0$ at all locations. Then, using the uniqueness theorem for differential equations, likely $G=0$ at all times. Although, I am struggling to find which precise uniqueness theorem to use here.

I am not sure if something like this holds for a general gauge, say $a^{\nu} \partial_{\nu} A^{\mu} = 0$ for some constants $a^{\nu}$.

Hope it helps!


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