Why is the displacement current term needed in the Maxwell's equations?

Why did Maxwell believe that a displacement current term needed to be added to Ampere's circuital law?

I have found loads of answers online about the plates acting as capacitors but i don't understand it fully. Could anyone explain this to me in pure simple english?

Does it have something to do with the fact that Ampere's law is time-independent? Adding the differential does it gives time-dependency?

• – glS Jan 7 '15 at 12:40

Trying to explain in almost pure simple english, we have: $$\nabla\times\mathbf B = \mu_0\mathbf J$$

This means the current is the source of the magnetic field. $\nabla\times$ is an operator which says how the magnetic field will behave when there is a current. Basically, this operator is known as curl operator. It basically acts on "rotating" stuffs. So, if it acts on a non-rotating stuff, it must be zero. So, the curl of a irrotational field is zero.

An easy way to visualize the operation of the curl operator when current is impressed, is to analyse the magnetic field generated by a wire. The wire is straigh up (irrotational), and so as the current. Hence because of the curl operator acting on the magnetic field, the magnetic field generated is completely rotational.

The divergent operator $\nabla\cdot$ acts on non-rotating stuffs. This means, the divergence of a field which only has rotational contributions, is zero. Hence, when we apply the divergent on both sides:

$$\nabla\cdot\nabla\times\mathbf B = \mu_0\nabla\cdot\mathbf J = 0 \quad\Longrightarrow\quad \nabla\cdot\mathbf J = 0$$

This means the current must be divergence-free. Which in other words, "rotating" or "stopped". Of course, both divergent and curl operators are related to spatial variation rates. So, if the current is uniform, there will be no divergence and no curl. So, now we have the following conclusion: The current must be "rotating", or must be uniform.

However, from charge conservation we have: $$\nabla\cdot\mathbf J = -\frac{\partial\rho}{\partial t}$$

This means, the field won't be uniform if there is variation of charge density (which makes intuitive sense). Also, all "rotating" contributions will go to zero, because it is no longer divergence-free. Now, all we have to think, is a situation where the magnetic field is created and we have variation of charge.

This is quite simple: Imagine a capacitor being charged. Since the charge of the plates is varying with time, then its density is also varying. Thus, the current won't be uniform. Hence the current is not divergence-free. But we saw that, if this $\nabla\times\mathbf B = \mu_0\mathbf J$ is true, then the current must be divergence-free. Contradiction. Hence, $\nabla\times\mathbf B = \mu_0\mathbf J$ is not true.

To fix this, it is necessary to add a displacement current. Maxwell saw it, and fixed it.

Why is the displacement current term needed in the Maxwell's equations?

Without the displacement current term, we have

$$\nabla \times \vec H = \vec J$$

Taking the divergence of both sides yields

$$0 = \nabla \cdot \vec J$$

But, by the continuity equation (conservation of electric charge), we have

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

Thus, without the displacement current term, we have the result

$$\frac{\partial \rho}{\partial t} = 0$$

which is clearly false.

• if you were trying to explain it to a 6 year old what would you say? – liamsaaaa Jan 7 '15 at 17:47
• @liamsaaaa, I wouldn't try to explain it to a 6 year old. – Alfred Centauri Jan 7 '15 at 23:15
• if you were trying to explain it to a very clever 6 year old with a great interest in electromagnetics what would you say? – liamsaaaa Jan 8 '15 at 13:39

A clever 6 year old would have an open mind and so would understand the following; An electric pulse travels down a transmission line, (coaxial cable or parallel plate Capacitor) at the speed of light for the medium. Conventional theory states that current is electron flow, and current creates a magnetic field. This is physically impossible! The billiard ball model is flawed. What actually occurs, is that a TEM step or pulse travels in the space between the wires. Oliver Heaviside calls this Energy Current. it is known as the Poynting Vector ExH.

Because the conductors are not perfect guides, some of the energy leaks into the wire and energizes the atoms, hence the electrons jump from atom to atom latching on to the outer valence band in Copper which contains one electron. (Electrons have mass, and cannot travel at the speed of light from one atom to the next) This electron flow is an effect not a cause of the more fundamental flow of energy. Which flows at a speed defined by the medium. Any other explanation is false. The observed displacement current is merely the difference between the energy moving from left to right compared to that moving right to left. An AC impressed signal will always have a magnetic field observed in a capacitor because the energy moving through the open circuit transmission line is unequal (ie from right to left and left to right) Displacement Current needs to be assigned to the dustbin of history.

protected by Qmechanic♦Aug 8 '17 at 5:21

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