I read in the book Introduction to Electrodynamics by D. J, Griffiths the process of solving the four Maxwell equations in the most general form:

Firstly, the four equations were simplified using potentials as

$$\nabla ^2 \vec A-\mu_0 \varepsilon_0 \frac{\partial^2 \vec A} {\partial t^2} - \nabla \left({\nabla \cdot \vec A + \mu_0 \varepsilon_0 \frac{\partial V}{\partial t} }\right)=- \mu_0\vec J_c $$

$$\nabla^2 V +\frac{\partial{\left(\nabla\cdot \vec A\right)} } {\partial t} = - \frac{\rho} {\varepsilon_0} $$

Then the Lorenz gauge condition

$$\nabla \cdot \vec A = - \mu_0 \varepsilon_0 \frac{\partial V}{\partial t}\tag1$$

was used saying that this does not change the solution.

I understand that the only condition on $\vec A$ is that $\vec \nabla \times\vec A$ should give $\vec B$ and hence, $\vec \nabla \cdot \vec A$ could be any function and (1) is a good candidate. Also I see how this transformation makes the two above equations symmetric and easier.


But it seems to me that (1) imposes a new constraint on the value of $V$ at any point. It seems that eq (1) partially determines $\vec A(x, y, z, t) $ at a point from $V(x, y, z, t)$ and vice versa $\left( \text{just like }\ \vec \nabla \times \vec E = - \frac{\partial B}{\partial t}\right) $.

Since all of Maxwell's equations are used up, this new relationship between $A$ and $V$ seems to be a new important relation. This new relation is imposing a new constraint on possible values of $A$ and $V$ apart from the two already present.


Is (1) a completely new constrain which $A$ and $V$ follows apart from the above two equations?

If yes, then why shouldn't we consider it as another Maxwell’s equation and provide a strong proof for this relation rather than calling it a transformation that gives same solutions?

If no, then it should not depend on whether we do Lorenz gauge or not - we should get the answer just from the other two equations. Is it possible?

Note: If the added function (to $A$) was just some $f(x, y, z, t)$ with proper gauge conditions, I would not have this doubt. But here the gauge condition is more of a relation between $A$ and $V$. I'm finding it hard to understand.

  • 3
    $\begingroup$ Also, I'm not sure I see why you think that this doesn't impose a constraint on $V$. Indeed, both $\mathbf{A}$ and $V$ are constrained by Equation (1). You could, for example, read the equation as saying this: Given a value of $\mathbf{A}$, you need to choose a $V$ that satisfies (1) to have the Lorenz gauge condition met. Or have I missed something in your question? $\endgroup$
    – Philip
    Oct 13, 2020 at 13:01
  • $\begingroup$ @Philip : thanks for the clarification. I've edited my answer. Your way of choosing (1) makes bit sense. I have mentioned the issue I'm facing at the end... please go through $\endgroup$ Oct 13, 2020 at 14:22
  • $\begingroup$ I think this question can be salvaged, but you need to edit it. In fact, I think the entire first part is a red herring. Fundamentally, as I understand it, you're asking whether the gauge condition is a further equation for electrodynamics in addition to Maxwell's Equations. The answer is no, it can be derived from those equations and as such isn't separate. $\endgroup$
    – Philip
    Oct 13, 2020 at 15:31

1 Answer 1


Yes, it is a new constraint, but it does not follow from Maxwell's equations. Like any gauge condition, it's an optional constraint, which you can choose to obey if it makes your life easier, and disobey if it doesn't.

In a way, it's a tradeoff. You can choose to take the Lorenz gauge; if you do that, the equations for the fields will become simpler, but you have to make sure that the fields satisfy the gauge condition. Or, you can choose not to bother with gauge conditions, but you'll have to use more complicated equations.

You seem concerned that the gauge condition should already appear in the solutions to the equations. But an important fact that is not often discussed is that a solution of the most general equations for the potentials will have arbitrary, undetermined functions. This is exactly analogous to the arbitrary constant we find in the electrostatic potential: to fix it, we have to specify the value of the potential at some point, which is our gauge condition. The constant is completely unphysical, so we can choose it at will. Or we can not choose it, which is fine too, we'll just have to drag it around everywhere.

In fact, the analogy is stronger: the fundamental, non-gauge-fixed equation in electrostatics is $\nabla^2 V = -\rho$. The general solution for this is

$$V(\mathbf{x}) = \int d^3\mathbf{x}' \frac{\rho(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|} + f(t),$$

where $f$ is an arbitrary function of time. You cannot derive $f$ from the equation, it's undetermined - and of course, this is not a problem, since any $f$ will give the same field. What you can do (which is what we do implicitly all the time) is use a gauge condition $\partial V/\partial t = 0$, which will force $f$ to be a constant. Again, this condition doesn't come from the equations: you choose it yourself, to remove some of the arbitrariness in the solution.

  • $\begingroup$ I understand that adding functions that doesn't change $E$ (or $V$) and $B$ (or $A$) are physically insignificant and undeterministic. But here the added function to $A$ at any point is dependent on the value of $V$ at that point. Its like in the example suggested, we replace $f(t)$ with $\nabla \cdot B$ saying that it does not change the output since $B$ would be zero in electrostatics. I really need time to understand that. But "you have to make sure Gauge condition is satisfied" made sense. thank you for the answer. $\endgroup$ Oct 14, 2020 at 3:36
  • $\begingroup$ I mean $\nabla \cdot B$ is always zero... Let's say adding $\left|\vec B\right|$ because $B$ is zero in electrostatics $\endgroup$ Oct 14, 2020 at 3:41
  • $\begingroup$ @RishabNavaneet The arbitrary function that can be added doesn't depend on the other potentials, it's arbitrary. If, like in your question, you choose to use a gauge, then yes, you impose conditions on the arbitrary functions. That's the point of the gauge condition. $\endgroup$
    – Javier
    Oct 15, 2020 at 14:18

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