# Is Gauss' law valid for time-dependent electric fields?

The Maxwell's equation $\boldsymbol{\nabla}\cdot \textbf{E}(\textbf{r})=\frac{\rho(\textbf{r})}{\epsilon_0}$ is derived from the Gauss law in electrostatics (which is in turn derived from Coulomb's law). Therefore, $\textbf{E}$ must be an electrostatic field i.e., time-independent. Then how is this equation valid for the electric field $\textbf{E}(\textbf{r},t)$ which is time-dependent (for example, the electric field of an electromagnetic wave)? Can we prove that $\boldsymbol{\nabla}\cdot \textbf{E}(\textbf{r},t)=\frac{\rho(\textbf{r},t)}{\epsilon_0}$ ?

EDIT: I have changed $\boldsymbol{\nabla}\cdot \textbf{E}=0$ to $\boldsymbol{\nabla}\cdot \textbf{E}=\frac{\rho}{\epsilon_0}$ in the question.

• First off, $\vec{\nabla} \cdot \vec{E} = \rho$, which can be a function of time. However, this is not quite where electromagnetic radiation comes from... – dbq Jul 25 '15 at 5:06
• I have edited the post. Thanks for pointing out. Electromagnetic radiation was just to give an example of time-dependent electric field. I just wanted to know how do you apply Gauss' law to time-dependent fields? After all, it is derived from Coulomb's law. Right? – SRS Jul 25 '15 at 5:21
• Interesting stuff. Particularly if you check out atomic orbitals wherein electrons "exist as standing waves". Standing waves aren't really static. They look static, but if you open the cavity the photon is off at c from a standing start. Because it was always moving at c. IMHO you can say something similar for the electron, with its spin and magnetic moment. It's like the simplest most fundamental electrostatic field you can find isn't really static. It's time-dependent, with a dynamical heart, contrived to look static. – John Duffield Jul 25 '15 at 6:46
• Time dependent Gauss's law also derived from the experimental fact that Charge is invariant. that means the number of charges in an isolated system is independent of whether the charges are moving or not. – Shing Jul 25 '15 at 14:28

You need to watch what you mean by the ambiguous term "derive", which can mean either "was derived historically" (i.e. was motivated by or is a derivative of, in the non-mathematical sense) or "is derived logically/mathematically".

Historically, I think you are correct that $\boldsymbol{\nabla}\cdot \textbf{E}(\textbf{r},t)=\frac{\rho(\textbf{r},t)}{\epsilon_0}$ was "derived" by Maxwell from the electrostatic version $\boldsymbol{\nabla}\cdot \textbf{E}(\textbf{r})=\frac{\rho(\textbf{r})}{\epsilon_0}$, which in turn was "derived" from Coulomb's law.

Logically, it's the other way around. $\boldsymbol{\nabla}\cdot \textbf{E}(\textbf{r},t)=\frac{\rho(\textbf{r},t)}{\epsilon_0}$ is a fundamental law of the universe (at least it is in classical electromagnetism; in reality it is "derived" from quantum electrodynamics). The electrostatic version of it can be "derived" mathematically from this law as a special case. The same is true of Coulomb's law.

• Steve, can you give a reference for that QED derivation please? – John Duffield Jul 25 '15 at 12:37
• Any textbook about quantum field theory should have a proof that Maxwell's equations + Lorentz Force Law are the classical limit of QED. (Although sometimes it's left as a homework exercise.) – Steve Byrnes Jul 25 '15 at 14:10

No we cannot prove it; Maxwell postulated that it would hold dynamically because it made the most sense for it to do so as he pondered the displacement current problem. As you likely know, Maxwell pondered the inconsistency between Ampère's law for magnetostatics and the charge continuity equation. Ampère's law for magnetostatics reads $\nabla\times \vec{H}=\vec{J}$; when we take the divergence of both sides of this equation we get $0=\nabla\cdot\vec{J}$ for any magnetic field with continuous second derivatives. This violates the charge continuity equation; we need $0=\nabla\cdot\vec{J} + \partial_t\,\rho$. So we need to add a term to the right of Ampère's law in the dynamic case whose divergence is the charge density $\rho$. The easiest solution is to assume that Gauss's electrostatics law holds in the dynamical case: then we add the electric displacement to the RHS of Ampère's and it has the right divergence to make everything properly in keeping with the continuity equation. Note that we can also add an arbitrary vector of the form $\nabla\times \vec{N}$ to the electric displacement for this to work, but this degree of freedom doesn't affect Gauss's law.

Maxwell derived his equations from 1) charge conservation law; 2) Coulomb's law; 3) Bio--Savart--Laplace law; 4) Faraday's law of induction. The equation $\boldsymbol{\nabla}\cdot \textbf{E}(\textbf{r})=\frac{\rho(\textbf{r})}{\epsilon_0}$ was indeed derived from Coulomb's law and in its differential form is written using Gauss--Ostrogradskiy theorem. Maxwell made one step further and suggested (postulated) that the same law is true when $\mathbf{E}$ and $\mathbf{\rho}$ are functions of space AND time. It turned out to be a correct guess and it doesn't contradict the other equations. Indeed, from Bio--Savart--Laplace law Maxwell derived $\nabla\times \vec{H}=\vec{J}$. If you take the divergence of both sides of this equation it contradicts the charge conservation law, so an additional term must be added to this equation (the so-called displacement current). And then, the resultant equation will not contradict the equation we get after differentiating $\boldsymbol{\nabla}\cdot \textbf{E}(\textbf{r},t)=\frac{\rho(\textbf{r},t)}{\epsilon_0}$ with respect to time and, in fact, the comparison of these two equations enables us to find the displacement current.

Well I dont know if we can prove it but there is a much more elegant way of formulating EM which may be helpful here. As you may know there are two potentials on EM: the scalar potential $\phi$ and the vector potential $\vec{A}$, from which $\vec{E}(t,x)$ and $\vec{B}(t,x)$ are derived. From this two objects and following symmetry considerations you can construct a tensor $F$ called the field strength tensor (basically it is a 4x4 antisymmetric tensor which entries are the components of $\vec{E}$ and $\vec{B}$. The interesting thing is that this tensor satisfies a mathematical identity ($dF=0$) from which all the four Maxwell equations are obtained, and since we where considering time-dependent fields this may be kind of "proof" for your question.

• I don't see how this directly answers the question. – Kyle Kanos Jul 25 '15 at 12:42

The divergence of the electric field component of a TEM wave is zero (this field is only rotational and divergence free).

The Maxwell-Gauss law for dynamic fields should be valid. One can check this by deriving the Jefimenko electric field, see equation (10) and subsequently derive the divergence of the Jefimenko electric field, which should yield $\frac{\rho(\vec r,t)}{\epsilon_0}$. This is only a theory consistency check; the experimental proof for Gauss' law for dynamic fields is more important.

## protected by Qmechanic♦Jul 25 '15 at 12:00

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