Weber Electrodynamics

Correction To Gauss's Law

We wish to verify Gauss's Law, one of Maxwell's four differential equations. (We will work in the SI system of units.) But we find that a correction is required. We consider several possible corrections. In the end, we choose the simplest, least restrictive, approach: A simple notational change without imposing restrictions on other equations.

Setting Up The Mathematics

First, we will distinguish between the coordinates of the sources by using a subscript "S", and the coordinates of the detector (observation, or test) position by using a subscript "T". Then, for example, the \( \nabla \) vector operator, when referring to the detector (test) position parameters, will be written as, in a Cartesian coordinate system $$ \nabla_T \equiv \dfrac{\partial}{\partial x_T}\hat{x} + \dfrac{\partial}{\partial y_T}\hat{y} + \dfrac{\partial}{\partial z_T}\hat{z}. $$

Gauss's Law, one of the four Maxwell equations, can be written as $$ \nabla_T \bullet \vec{E} = \dfrac{1}{\epsilon_0} \rho_f, $$ where \( \rho_f \) is the free charge density function.

We use the scalar $$ \Phi \equiv \dfrac{1}{4 \pi \epsilon_0} \iiint \rho \dfrac{1}{r_{TS}} d\tau_S $$ and the vector $$ \vec{A} \equiv \dfrac{1}{4 \pi \epsilon_0 c^2} \iiint \vec{J} \dfrac{1}{r_{TS}} d\tau_S $$ potential function definitions, where \( r_{TS} \) is the distance from the source differential volume element \( d\tau_S \) to the detector (observation, or test) position, \( \rho \) is the charge density function, and \( \vec{J} \) is the current density function defined by $$ \vec{J} \equiv \rho \vec{v}_S $$ and \( \vec{v}_S \) is the "average drift velocity" of the source charges passing through the differential volume element \( d\tau_S .\)

The eletric field can be decomposed into the gradient of the scalar potential function and the time derivative of the vector potential function as $$ \vec{E} = -\nabla_T \Phi - \dfrac{\partial}{\partial t} \vec{A}. $$

Attempting Verification Of Gauss's Law

With these definitions, the left hand side of Gauss's law can be written as $$ \nabla_T \bullet \vec{E} = -\nabla_T \bullet \nabla_T \Phi - \nabla_T \bullet \dfrac{\partial}{\partial t} \vec{A}. $$ This can be written as $$ \nabla_T \bullet \vec{E} = -\nabla_T^2 \Phi - \nabla_T \bullet \dfrac{\partial}{\partial t} \vec{A}. $$

Using the definition of the scalar potential function, the first term can then be written as $$ -\nabla_T^2 \Phi = -\nabla_T^2 \dfrac{1}{4 \pi \epsilon_0} \iiint \rho \dfrac{1}{r_{TS}} d\tau_S. $$

Bringing the Laplacian operator, \( \nabla_T^2 \), under the integration, and noting that the charge density function is not a function of the detector (observation or test) position, we have $$ -\nabla_T^2 \Phi = - \dfrac{1}{4 \pi \epsilon_0} \iiint \rho \nabla_T^2 \dfrac{1}{r_{TS}} d\tau_S. $$

Next, using the relation $$ \nabla_T^2 \dfrac{1}{r_{TS}} = -4 \pi \delta(\vec{r}_T - \vec{r}_S) $$ we have $$ -\nabla_T^2 \Phi = - \dfrac{1}{4 \pi \epsilon_0} \iiint \rho \left(-4 \pi \delta(\vec{r}_T - \vec{r}_S) \right) d\tau_S. $$

Performing the integration gives $$ -\nabla_T^2 \Phi = \dfrac{1}{\epsilon_0} \rho(\vec{r}_T, t) $$ where the charge density function is now the "free" charge density function $$ \rho_f \equiv \rho(\vec{r}_T, t). $$

Using this result, we now have $$ \nabla_T \bullet \vec{E} = \dfrac{1}{\epsilon_0} \rho_f - \nabla_T \bullet \dfrac{\partial}{\partial t} \vec{A}. $$

We see that in order for this to match Gauss's Law given above, some special condition (also called a restriction) must be imposed on the second term involving the vector potential. For example, we can impose the restriction \( \vec{A} = \vec{0},\) or the restriction \( \partial \vec{A} / \partial t = \vec{0} \), or the restriction \( \nabla_T \bullet \vec{A} = 0 \) or combinations.

Note that in general (not applying any restriction) the second term is not zero. The only way we recover Gauss's Law is to apply a restriction. And once a restriction is applied, it must be carried over (applied to) all following equations during the development of an electrodynamics mathematical problem.

Whatever condition we wish to impose, it must be made explicit so that 1) it is not forgotten later in the development, and 2) other people will know what condition has been imposed.

First Approach

If we impose the restriction \( \vec{A} = \vec{0} \) then Gauss's Law needs to be written as $$ \nabla_T \bullet \vec{E} = \dfrac{1}{\epsilon_0} \rho_f \;\;\; restriction: \vec{A} = \vec{0}. $$

However, this approach also changes Faraday's Law $$ \nabla_T \times \vec{E} = -\nabla_T \times \nabla_T \Phi - \nabla_T \times \dfrac{\partial}{\partial t} \vec{A} $$ Since the curl of a gradient is zero, and for this case \( \vec{A} = \vec{0} \), we have $$ \nabla_T \times \vec{E} = \vec{0}. $$

This, then, is not a good approach because it eliminates Faraday's Law.

Second Approach

For the second approach, we consider \( \partial \vec{A} / \partial t = \vec{0}. \) We then write Gauss's Law as $$ \nabla_T \bullet \vec{E} = \dfrac{1}{\epsilon_0} \rho_f \;\;\; restriction: \partial \vec{A} / \partial t = \vec{0}. $$

However, this too changes Faraday's Law. Since $$ \vec{E} = -\nabla_T \Phi - \dfrac{\partial}{\partial t} \vec{A} $$ and imposing the restriction, we now have $$ \vec{E} = -\nabla_T \Phi. $$ Then Faraday's Law becomes $$ \nabla_T \times \vec{E} = -\nabla_T \times \nabla_T \Phi - \nabla_T \times \dfrac{\partial}{\partial t} \vec{A} $$ $$ \nabla_T \times \vec{E} = -\nabla_T \times \nabla_T \Phi, $$ $$ \nabla_T \times \vec{E} = \vec{0}. $$

As with the previous approach, this is not a good approach because it eliminates Faraday's Law.

Third Approach

For the third approach, we impose the restriction \( \nabla_T \bullet \vec{A} = 0. \)

Gauss's Law is then written as $$ \nabla_T \bullet \vec{E} = \dfrac{1}{\epsilon_0} \rho_f \;\;\; restriction: \nabla_T \bullet \vec{A} = 0. $$

As mentioned previously, this choice is called the Coulomb gauge. By making this choice, we have selected a specific gauge. This, then, removes the "gauge freedom" of Maxwell's equations because we have made a specific, fixed, choice for the gauge to make Gauss's Law valid.

We prefer to keep the gauge freedom of Maxwell's equations, if possible, and so this approach is not used.

Fourth Approach

It is for these reasons that we prefer to adopt the notation \( \vec{E}^{(C)} \) and define the symbols \( \vec{E}^{(C)} \) to mean the "Coulomb field" defined by $$ \vec{E}^{(C)} \equiv -\nabla_T \Phi. $$ This is simply a special case of the more general electrodynamic field given by $$ \vec{E} = -\nabla_T \Phi - \dfrac{\partial}{\partial t} \vec{A}, $$ that can also be written as $$ \vec{E} = \vec{E}^{(C)} - \dfrac{\partial}{\partial t} \vec{A}. $$

In this form, it is clear that, in general $$ \vec{E} \ne \vec{E}^{(C)}. $$

This is not indicating that there are different electric fields. It only indicates that the special case of the Coulomb field is not the same as the general case of the electrodynamic field. It is the same "field" under different imposed constraints.

Also, by using this \( \vec{E}^{(C)} \) definition, as being the Coulomb field, Gauss's Law, now written as $$ \nabla_T \bullet \vec{E}^{(C)} = \dfrac{1}{\epsilon_0} \rho_f, $$ remains valid in general and no restrictions need to be carried over to other equations. This appears to be the simplest, least impactful, approach.


In Maxwell's four differential equations, we find, as we have just demonstrated above, that we need to indicate when \( \vec{E} \) is referring to the special case \( \vec{E}^{(C)} \) and when it is refering to the general case \( \vec{E}. \) In the usual way of writing Maxwell's equations, there is no explicit indication when the special case (Coulomb field) is being refered to, nor when a special condition has been imposed. Thus some notational adjustment needs to be made to Maxwell's equations. Our preference, to use \( \vec{E}^{(C)} \) as the Coulomb field, is only one possibility, but appears to be the cleanest approach without imposing additional restrictions on any following equations.

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