The definition of the Lorenz gauge condition

Question

The inner product of two vectors in space-time is:

$$(x_1, y_1, z_1, t_1) \cdot (x_2, y_2, z_2, t_2) = x_1 x_2 + y_1 y_2 + z_1 z_2 - t_1 t_2$$

So

$$(\frac{\partial }{\partial x}, \frac{\partial }{\partial y}, \frac{\partial }{\partial z}, \frac 1c \frac{\partial }{\partial t}) \cdot (A_1, A_2, A_3, \phi) = \text{div}(\vec A) - \frac 1c \frac{\partial \phi}{\partial t}$$

is Lorentz invariant, where $\vec A=(A_1, A_2, A_3)$. But the [Lorenz gauge condition] (https://en.wikipedia.org/wiki/Lorenz_gauge_condition) is defined by $\text{div}(\vec A) + 1/c\ \partial_t \phi=0$. Why has the minus changed into plus? So there is apparently no longer invariance.

Answer 1

The Lorenz gauge condition is written as, $\partial_\mu A^\mu = 0$ which can be expanded as,

$$\partial_\mu A^\mu = \frac{\partial A^0}{\partial t} + \nabla \cdot \vec A = 0$$

in natural units, where we are simply doing what the Einstein summation convention instructed us to do, take a sum through the index, $\mu = 0, \dots, 3$. You can also write this as,

$$\partial_\mu A^\mu = \eta^{\mu\nu}\partial_\mu A_\nu$$

in which case you would get a minus sign, but notice the sum involves the co-vector $A_\nu$, a different quantity, related to $A^\mu$ by a change of sign, in flat spacetime since $\eta = \mathrm{diag}(\mp1,\pm1,\pm1,\pm1).$

Answer 2

The inner product of two spacetime vectors is given by

$$V^{\mu}W_{\mu}=V^{0}W_{0}+\textbf{V}\cdot\textbf{W}.$$

Note that there is no inherent minus sign in this definition. The minus signs only come in when your vector $W$ is naturally described with an upstairs (contravariant) index. In ths case, we would write

$$V^{\mu}W_{\mu}=-V^{0}W^{0}+\textbf{V}\cdot\textbf{W},$$

because our signature requires $W^{0}=-W_{0}$. Your example of $x\cdot y$ only has a minus sign because coordinate vectors are naturally contravariant.

Now on to your question: the Lorenz gauge is defined as

$$\partial_{\mu}A^{\mu}=\frac{\partial}{\partial t}A^{0}+\boldsymbol{\nabla}\cdot\textbf{A}=0.$$

Now, since the vector potential $A$ naturally has an upstairs index, we can write $A^0=\phi$. Thus, we have

$$\frac{\partial\phi}{\partial t}+\boldsymbol{\nabla}\cdot\textbf{A}=0.$$

No need for all of the finicky minus signs (they're annoying as hell)!

I hope this helped!

Note: I have used units where $c=1$ in this answer. Also note that I have a tendency to mix up the terms "covariant" and "contravariant." Please correct me if I've made a mistake.

Answer 3

In the dot product you described in the first line both objects are vectors i.e. they have upper indices, namely $x^\mu$.

In the dot product you describe in the second line one of the two objects is a co-vector, i.e. it has lower indices. Which one is it? How is the dot product between a vector and a co-vector defined?

Answer 4

This question was caused by an incorrect definition of the Lorenz gauge condition on Wikipedia. The minus should not have changed into plus.

Having a correct definition of the Lorenz gauge condition seems to have ramifications, which has resulted in a further question: The definition of E.