# Forms of Maxwell's equations

In my physics class, I was taught two forms of one of Maxwell's equations: Ampere's law

$$\vec{\nabla} \times \vec{B} = \mu J$$

and Maxwell-Ampere's law

$$c^2\vec{\nabla} \times \vec{B} = \dfrac{\vec{j}}{\epsilon_{0}} + \dfrac{\partial \vec{E}}{\partial t}.$$

What is the difference between these two equations? When can we use the simplified version (the first equation I posted), and when do we have to use the more complicated equation?

Can someone please give me an example in which the $\frac{\partial \vec{E}}{\partial t}$ term is needed?

• Ask yourself what the conditions are whereunder the second equation is equivalent to the first. They do overlap in applicability. – WetSavannaAnimal Nov 23 '17 at 0:53

When $\frac{\partial E}{\partial t}=0$, then the 1st equation is valid. That is, it is for magnetostatics, where currents (and fields) are not time-varying.

• Thanks. Could you give me a specific example as to when the second equation must be used? – joesx Nov 23 '17 at 0:38
• @Hat What's the magnetic field inside a charging capacitor?--it's not 0 even though there is no current in there. – JEB Nov 23 '17 at 12:09
• In particular, its curl is not 0 as well. – Photon Nov 25 '17 at 11:26

We know from the continuity equation that:

$$\tag{1}\nabla\cdot\vec{J}+\frac{\partial\rho}{\partial t}=0\label{1}$$

So if we apply the divergence operator to equation \ref{2}:

$$\tag{2}\vec{\nabla}\times\vec{B}=\mu_{0}\vec{J}\label{2}$$

We can see that this equation is valid only if $\frac{\partial\rho}{\partial t}=0.$

$$\tag{3}\nabla\cdot\left(\vec{\nabla}\times\vec{B}\right)=0\label{3}$$ $$\tag{4}\nabla\cdot\left(\mu_{0}\vec{J}\right)=-\mu_{0}\frac{\partial\rho}{\partial t}\label{4}$$

In general in order to make equation \ref{2} consistent, we have to make the following transformation:

$$\tag{5}\vec{J}\rightarrow \vec{J}+\epsilon_{0}\frac{\partial\vec{E}}{\partial t}\label{5}$$

So that equation \ref{2} becomes:

$$\tag{6}\vec{\nabla}\times\vec{B}=\mu_{0}\left(\vec{J}+\epsilon_{0}\frac{\partial\vec{E}}{\partial t}\right)\label{6}$$

Using Gauss' law and applying the divergence operator to equation \ref{6}, we get zero on both sides of the equation.

$$\tag{7}\nabla\left(\vec{J}+\epsilon_{0}\frac{\partial\vec{E}}{\partial t}\right)\label{7}=\nabla\cdot\vec{J}+\epsilon_{0}\frac{\partial}{\partial t}\nabla\cdot\vec{E}=\nabla\cdot\vec{J}+\frac{\partial\rho}{\partial t}=0$$

Gauss' law: $$\tag{8}\nabla\cdot\vec{E}=\frac{\rho}{\epsilon_{0}}\label{8}$$

$\newcommand{\vect}[1]{{\bf #1}}$

The first thing you should notice is that $c^2 = 1 / \mu_0\epsilon_0$, so that in the vacuum your second equation becomes

$$\nabla \times\vect{B} = \frac{1}{\epsilon_0 c^2}\vect{J} + \frac{1}{c^2}\frac{\partial \vect{E}}{\partial t} = \mu_0\vect{J} + \mu_0\epsilon_0 \frac{\partial \vect{E}}{\partial t} \tag{1}$$

Technically the equations you show are the same, it is just that in the first on the term associated with time-changes in the electric field is not present.

All this equation is saying is that there are two sources of magnetic fields: currents and time-varying electric fields. I will give you an example, perhaps it will help clarify when you can use these expressions

Currents (left case)

Consider a straight wire carrying a steady (time-independent) current $I$, in this example the term $\partial_t {\bf E} = 0$ and so we end up with the equation

$$\nabla \times \vect{B} = \mu_0 \vect{J}$$

By integrating on both sides over the area enclosed by the path $C$ we have

$$\int {\rm d}\vect{A}~\cdot \nabla \times \vect{B}= \oint_C\vect{\rm d}\vect{l}\cdot \vect{B} = \mu_0\int{\rm d}\vect{A}\cdot \vect{J} = \mu_0 I$$

Because of the cylindrical symmetry you can argue that $\vect{B}$ is constant along the loop $C$, moreover $\vect{B}$ and ${\rm d}\vect{l}$ are parallel

$$B\int_C {\rm d}l = \mu_0 I ~~~~\Rightarrow~~~~ B = \frac{\mu_0 I}{2\pi r} \tag{2}$$

Time-varying electric fields (right case)

In the example of the sketch you can see that the capacitor will charge, opposite charges will appear on the plates of the capacitor, and the electric field will change as a consequence. Despite the fact that no-actual charges flow through the left fact of the blue surface a magnetic field will be present.

I will let you do the math, now you have to take into account the term $\partial_t\vect{E}$ in Eq. (1), but the result is that this magnetic field is generated by the change in $\vect{E}$. You can actually see that the result is that the resulting $\vect{B}$ has a similar form to that in Eq. (2)

$$B = \frac{\mu_0 I_D}{2\pi r}$$

where

$$I_D = \frac{\epsilon_0 \dot{E}}{2\pi r}$$

is called the displacement current

The simplified equation you mention in your question is part of the quasistatic approximation of the Maxwell equations. Let me explain. In vacuum the full Maxwell equations are $$\nabla \epsilon_0 \bf E=\rho$$ $$\nabla \wedge \bf E = -\frac {\partial \mu_0 H}{\partial t}$$$$\nabla \mu_0 \bf H=0$$$$\nabla \wedge \bf H = J+ \frac {\partial E}{\partial t}.$$ As you can see, these equations imply that varying electrical fields do generate magnetic fields and the other way around. This mechanism is responsible for the existence of electromagnetic waves. In the quasistatic approximation we assume that the time derivative of $\bf E$ is small wherever $\bf H$ is significant and that the time derivative of $\bf H$ is small wherever $\bf E$ is significant, so that the coupling between $\bf E$ and $\bf H$ can be neglected. In that case, the equations for the electrical field are$$\nabla \epsilon_0 \bf E=\rho$$ $$\nabla \wedge \bf E = 0$$ $$\nabla \bf J + \frac{\partial \rho}{\partial t}=0$$where the last equation is obtained by taking the divergence of the fourth above. The equations for the magnetic field are $$\nabla \wedge \bf E = -\frac {\partial \mu_0 H}{\partial t}$$$$\nabla \mu_0 \bf H=0$$$$\nabla \wedge \bf H = J.$$ As you can see, the coupling that causes varying electrical field to generate varying magnetic fields and so on has been removed. The removal of the coupling significantly reduces the complexity of the set of equations.

Your question is now, when can we use this quasistatic approximation. One can show (reference) that this approximation can be used when circuit dimensions are small compared to the wavelength of the electromagnetic waves involved. Let us look at the power grid as an example. All fields are varying with a frequency of 50Hz. The associated wavelength is $3.10^8/50 m$ or $6000\; km$. So as long as we are studying phenomena in the power grid over distances far below $6000\; km$, we can use this approximation. Practically speaking, the full Maxwell equations are only required at radiofrequencies or higher (e.g. optics).

One example is a plane electromagnetic wave in vacuum (see, e.g., http://physics.oregonstate.edu/~minote/COURSES/ph632/lib/exe/fetch.php?media=emch7.pdf)

What is the difference between these two equations?

The first form (simplified version) is only rigorously valid when the current density do not change with time, i.e., de current density is constant in time. It was first discovered by Ampere and remained in this form until Maxwell spot some faults on it:

Why there are 2 different equations

First, the first equation is incompatible with the continuity equation (see Andrei Geanta’s answer). In order to make it compatible with this fundamental principle of nature the current density should be redefined to include a term of $\frac{\partial\vec{E}}{\partial t}$.

Second when Maxwell collected and summarized those equations scattered on piles of electromagnetic books, which he considered to be the most fundamental ones, he got the following:

$$\nabla\cdot\vec{E}=\frac{\rho}{\epsilon_{0}}\tag{1}$$ $$\nabla\cdot\vec{B}=0\tag{2}$$ $$\nabla\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}\tag{3}$$ $$\nabla\times\vec{B}=\mu_{0}\vec{J}\tag{4}$$

In vacuum where there is no charges nor currents ($\rho=0$ and $\vec{J}=0$) these equations looks like this $$\nabla\cdot\vec{E}=0\tag{5}$$ $$\nabla\cdot\vec{B}=0\tag{6}$$ $$\nabla\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}\tag{7}$$ $$\nabla\times\vec{B}=0\tag{8}$$

Comparing (7) and (8) a striking difference appears. There is a time dependent term in the curl of an electric field, but no equivalent term in the curl of a magnetic field. This is a break of symmetry in an otherwise “beautiful” set of equations.

By similitude, if the rate of change of a magnetic field creates an electric field, why the rate of change of an electric field not causing any magnetic field. It seems that, to make these equations fully symmetric a term of $\frac{\partial\vec{E}}{\partial t}$ should be added to the right side of equation (8). However, this was exactly what the continuity (conservation) of electric charges demanded. Following this sort of reasoning’s, Maxwell decided to correct Ampere’s law to the second form you presented. It solves both problems that we were analyzing. So, equation (4) becomes

$$\vec{\nabla}\times\vec{B}=\mu_{0}\left(\vec{J}+\epsilon_{0}\frac{\partial\vec{E}}{\partial t}\right)\tag{9}$$

Maxwell called it’s addition to Ampere’s law, displacement current.

It is cool to notice, that by the time Maxwell did it, there was no experimental backup whatsoever to accept this. It was a pure theoretical development by Maxwell, based on the symmetry of the first set of equations

When can we use the simplified version?

The simplified version, should be used when the electric current do not change considerably with time (mathematically $\frac{\partial\vec{E}}{\partial t}→0$), like in DC electric circuits.

The second form is the general one and should be used in all situations where the displacement current cannot be ignored (mathematically $\frac{\partial\vec{E}}{\partial t}≠0$), like in AC electric circuits.

Another important application of equation (4) is in electromagnetic radiation which makes the connection between light (optics) and electromagnetic phenomena. The thing is, that with the addition of displacement current to Ampere’s law Maxwell equations can be summarized into 2 wave equations for the electric and magnetic field, respectively, travelling with the speed of light.