I'm having a bit of trouble understanding how the gauge condition is found.

Consider the potentials $V$ and $\vec A$ and $V'$ and $\vec A'$ so that
$$\vec E = -\vec\nabla V -\frac{\partial\vec A}{\partial t} = -\vec\nabla V' -\frac{\partial\vec A'}{\partial t}$$ $$\vec B = \vec\nabla \times \vec A=\vec\nabla \times \vec A'$$

The 2 sets of potentials are then related like this:

$$\vec A' = \vec A+\vec\nabla\theta$$ $$V' = V-\frac{\partial\theta}{\partial t}$$ For a function $\theta$.

The equations for the potentials are the following:

$$\vec\nabla^2\vec A-\varepsilon\mu\frac{\partial^2 \vec A}{\partial t^2}=-\mu\vec J+\vec \nabla(\vec \nabla.\vec A+\varepsilon\mu\frac{\partial V}{\partial t})$$

$$\vec\nabla^2V -\varepsilon\mu\frac{\partial^2 V}{\partial t^2}=-\frac{\rho}{\varepsilon} - \frac{\partial}{\partial t}(\vec\nabla . \vec A+\varepsilon\mu\frac{\partial V}{\partial t})$$

They can be decoupled by choosing the right potentials so that $(\vec\nabla . \vec A+\varepsilon\mu\frac{\partial V}{\partial t})=0$

The above part is the theory in general.

Now to find the correct potentials where this is the case, let's say that this last part is true for potentials $\vec A'$ and $V'$, i.e.

$$\vec\nabla . \vec A' + \varepsilon\mu\frac{\partial V'}{\partial t}=0$$

Filling in the relations between the 2 sets of potentials from earlier gives:

$$\vec\nabla . \vec A + \nabla^2\theta + \varepsilon\mu\frac{\partial V}{\partial t} -\varepsilon\mu\frac{\partial^2\theta}{\partial t^2}=0$$

Yet my book tells me that this means that

$$\nabla^2\theta - \varepsilon\mu\frac{\partial^2\theta}{\partial t^2}=0$$

I don't see why this is the case.

  • 1
    $\begingroup$ Because the Lorenz gauge tells you that the 1st and 3rd terms sum to zero, therefore the 2nd and 4th terms must too? $\endgroup$
    – ProfRob
    Jun 17, 2015 at 23:06
  • $\begingroup$ It doesn't, that's why I explicitly stated that the "above part is the theory in general". In the part where I search for the Lorenz Gauge it's not the equation with $\vec A$ and $V$ that is 0 but the one with $\vec A'$ and $V'$ $\endgroup$
    – Joshua
    Jun 18, 2015 at 8:48
  • $\begingroup$ Not with you. You can hypothesise any scalar relationship between A and V, which must also be true for A' and V'. This works for the Lorenz gauge if $\theta$ obeys the relationship at the end. $\endgroup$
    – ProfRob
    Jun 18, 2015 at 8:56

1 Answer 1


You've just run into the fact that the Lorenz gauge is only a partial gauge choice; it does not uniquely specify the vector and scalar potentials. The condition given on $\theta$ ensures that, if your original $A, V$ obeyed the Lorenz condition, then the new potentials $A', V'$ will also obey the Lorenz condition.

Unique solutions for $A, V$ arise only with additional conditions imposed (for instance, vanishing sufficiently fast at infinity).

  • $\begingroup$ So basically $(\vec\nabla . \vec A+\varepsilon\mu\frac{\partial V}{\partial t})$ also equals $0$? $\endgroup$
    – Joshua
    Jun 18, 2015 at 10:52

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