# The definition of the Lorenz gauge condition

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.

• Hint: Distinguish between upper and lower Lorentz indices. – Qmechanic Jun 25 '17 at 12:51

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 $\eta = \mathrm{diag}(\mp1,\pm1,\pm1,\pm1).$

• $$\frac{\partial A^0}{\partial t} + \nabla \cdot \vec A = 0$$ should be – Stephen Wynn Jun 30 '17 at 17:41
• "The Lorenz gauge condition is written as, $\partial_\mu A^\mu = 0$ " is wrong. – Stephen Wynn Jun 30 '17 at 19:22
• @StephenWynn See damtp.cam.ac.uk/user/tong/qft/qft.pdf, the section on electrodynamics. Moreover, the fact you don't clearly understand the difference between upper and lower indices which is where the confusion stems from, indicates you don't have the prerequisite knowledge to even identify whether that gauge condition is wrong. – JamalS Jun 30 '17 at 19:31

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.

• $\frac{\partial\phi}{\partial t}+\boldsymbol{\nabla}\cdot\textbf{A}$ – Stephen Wynn Jun 28 '17 at 18:24
• I'm not sure I understand your comment. – Bob Knighton Jun 28 '17 at 18:29
• $$\frac{\partial\phi}{\partial t}+\boldsymbol{\nabla}\cdot\textbf{A}$$ is the divergence for the orthogonal group o(4) . For the Lorentz group we need $$- \frac{\partial\phi}{\partial t}+\boldsymbol{\nabla}\cdot\textbf{A}=0.$$ – Stephen Wynn Jun 28 '17 at 18:36
• The divergence in any signature is the same, assuming it's the divergence of a contravarient object. – Bob Knighton Jun 28 '17 at 19:58
• That is true if I am taking the divergence of ta covariant object, but when I take the divergence of a contravariant object that's not the case. Did you read my (or any other) answer? – Bob Knighton Jun 29 '17 at 14:17

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?

• A co-vector is a linear map from a vector space to its field of scalars,, which defines the product of a co-vector and a vector. – Stephen Wynn Jun 29 '17 at 10:07

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.

## protected by Qmechanic♦Jun 30 '17 at 17:51

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