For a single charge, the rate of doing work by external fields
**B**
and **E** is q**v∙E**, in
which **v** is the velocity of the charge.

(a) Find the corresponding expression for a continuous distribution of
charge and current **and** interpret it physically.

(b) Use Maxwell's equations to express the result from part (a) in terms of the
fields alone.

(c) From your result in part b, verify Poynting's theorem (∂u/∂t) +
**∇∙S**
= -**E∙j**.

Find expressions for the terms u and **S** on the left-hand side, interpret those terms and
ultimately the physical significance of the theorem that
(∂u/∂t) + **∇∙S**
= -**E∙j**
represents.

Solution:

- Concepts:

Maxwell's equations, energy density and energy flux in the electromagnetic field - Reasoning:

We are explicitly instructed to use Maxwell's equations to convert the expression obtained in part (a) into one that contains the fields alone. When work is done on a charge by the field, field energy is converted into other forms of energy. Energy conservation applies. - Details of the calculation:

(a) Let ρdV = dq be the amount of charge in a volume element dV.

dW/dt = dq**v∙E**= ρ**v∙E**dV =**j∙E**dV = rate at which work is done by the field on the charges in dV.

(b)**E∙j**=**E∙**(1/μ_{0})(**∇**×**B**) - ε_{0}(∂**E**/∂t)**∙E**

= -(1/μ_{0})**∇∙**(**E**×**B**) +**B∙**(1/μ_{0})(**∇**×**E**) - ε_{0}(∂**E**/∂t)**∙E**= -(1/μ

_{0})**∇∙**(**E**×**B**) - (1/μ_{0})**B∙**(1/μ_{0})(∂**B**/∂t) - ε_{0}(∂**E**/∂t)**∙E**= -(1/μ

_{0})**∇∙**(**E**×**B**) - (∂/∂t)(1/(2μ_{0}))B^{2}+ (ε_{0}/2)E^{2})

Therefore

(∂/∂t)(1/(2μ_{0}))B^{2}+ (ε_{0}/2)E^{2}) + (1/μ_{0})**∇∙**(**E**×**B**) = -**E∙j.**

(c) Interpretation:

Let u = (1/(2μ_{0}))B^{2}+ (ε_{0}/2)E^{2}be the energy density in the electromagnetic field.

Let**S**= (1/μ_{0})(**E**×**B**) be the energy flux.

Then (∂u/∂t) +**∇∙S**= -**E∙j**.

∫_{V}**∇∙S**dV = ∮**S**·**n**dA, -∫_{V}∂u/∂t dV = ∮**S**·**n**dA + ∫_{V}**E∙j**dV

This is a statement of energy conservation. The rate at which the field energy in a volume decreases equals the rate at which it leaves across the boundaries plus the rate at which it gets converted into other forms.

Beginning with Newton's
second law and the force law for a charged particle in an
electromagnetic field, show that the time rate of change for the total
momentum of the field plus the charge distribution in a volume V is given by

d/dt(**P**_{field} +
**P**_{charge})_{i}
= ∮_{A}
∑_{j}T_{ij }**n**_{j}
dA,

where T is Maxwell's stress tensor given by

T_{ij} = ε_{0}E_{i}E_{j} + (1/μ_{0})B_{i}B_{j}
- ½(ε_{0}E^{2} + (1/μ_{0})B^{2})δ_{ij}.

Give the physical significance of the integral on the
right-hand side.

Solution:

- Concepts:

Newton's 2nd law, the Lorentz force, Maxwell's equations - Reasoning:

If we denote the momentum of all the particles in a volume V by**P**_{charge}then

d**P**_{charge}/dt = ∑_{q}q(**E**+**v**×**B**) = ∫_{V}d^{3}r(ρ**E**+**j**×**B**), with**j**= ρ**v**. We can now use Maxwell's equations to replace ρ and**j**by expressions in terms of the fields. - Details of the calculation:

ρ = ε_{0}**∇∙E**,**j**= (1/μ_{0})(**∇**×**B**- (1/c^{2})∂**E**/∂t)

ρ**E**+**j**×**B**= ε_{0}**E**(**∇∙E**) + (1/μ_{0})(**∇**×**B**- (1/c^{2})∂**E**/∂t)×**B**= ε

_{0}**E**(**∇∙E**) - (1/μ_{0})**B**×**∇**×**B**+ (1/(c^{2}μ_{0}))**B**×(∂**E**/∂t)

= ε_{0}**E**(**∇∙E**) + (1/μ_{0})**B**(**∇∙B**) - (1/μ_{0})**B**×**∇**×**B**- ε_{0}∂**E**/∂t×**B**)

= ε_{0}**E**(**∇∙E**) + (1/μ_{0})**B**(**∇∙B**) - (1/μ_{0})**B**×**∇**×**B**- ε_{0}∂/∂t(**E**×**B**) + ε_{0}**E**×∂**B**/∂t

= -ε_{0}∂/∂t(**E**×**B**) - (1/μ_{0})**B**×**∇**×**B**- ε_{0}**E**×**∇**×**E**+ ε_{0}**E**(**∇∙E**) + (1/μ_{0})**B**(**∇∙B**).

Note: We can add the term (1/μ_{0})**B**(**∇∙B**) since**∇∙B**= 0.

Interpretation:

(1/μ_{0})(**E**×**B**) = energy flux, (1/(cμ_{0}))(**E**×**B**) = momentum flux,

(1/(c^{2}μ_{0}))(**E**×**B**) = ε_{0}(**E**×**B**) = momentum density.

We have

d**P**_{field}/dt = d/dt ∫_{V}d^{3}r(1/(c^{2}μ_{0}))(**E**×**B**) = ∫_{V}d^{3}r∂/∂t[(1/(c^{2}μ_{0}))(**E**×**B**)]

in some fixed volume V.

We can therefore write

d/dt(**P**_{field}+**P**_{charge}) = ∫_{V}d^{3}r[ε_{0}(**E**(**∇∙E**) -**E**×**∇**×**E**) + (1/μ_{0})(**B**(**∇∙B**) -**B**×**∇**×**B**)].

Write the right hand side in component form.

(**E**(**∇∙E**) -**E**×**∇**×**E**)_{i}= E_{i}(∂E_{i}/∂x_{i}+ ∂E_{j}/∂x_{j}+ ∂E_{k}/∂x_{k}) - E_{j}(**∇**×**E**)_{k}+ E_{k}(**∇**×**E**)_{j}

= E_{i}(∂E_{i}/∂x_{i}+ ∂E_{j}/∂x_{j}+ ∂E_{k}/∂x_{k}) - E_{j}(∂E_{j}/∂x_{i}- ∂E_{i}/∂x_{j}) - E_{k}(∂E_{k}/∂x_{i}- ∂E_{i}/∂x_{k})

= ½∂E_{i}^{2}/∂x_{i}+ ∂E_{i}E_{j}/∂x_{j}+ ∂E_{i}E_{k}/∂x_{k}- ½∂/∂x_{i}(E_{j}^{2 }+ E_{k}^{2})

= ∂E_{i}^{2}/∂x_{i}+ ∂E_{i}E_{j}/∂x_{j}+ ∂E_{i}E_{k}/∂x_{k}- ½∂/∂x_{i}(E_{i}^{2 }+ E_{j}^{2 }+ E_{k}^{2})

= ∑_{l}∂/∂x_{l}[E_{i}E_{l}) - ½**E∙E**δ_{il}]

Similarly

**B**(**∇∙B**) -**B**×**∇**×**B**)_{i}= ∑_{l}∂/∂x_{l}[B_{i}B_{l}) - ½**B∙B**δ_{il}]

Define:

T_{ij}= ε_{0}E_{i}E_{j}+ (1/μ_{0})B_{i}B_{j}- ½(ε_{0}E^{2}+ (1/μ_{0})B^{2})δ_{ij},

then d/dt(**P**_{field}+**P**_{charge})_{i}= ∑_{l }∫_{V}∂/∂x_{l}T_{il }dV.

The components T_{ii}, T_{ij}, T_{ik }form a vector.

∑_{l }∂/∂x_{l}T_{il}is the divergence of a vector field.

Use Stoke's theorem.

∑_{l }∫_{V}∂/∂x_{l}T_{il }dV =_{ }∮_{A}∑_{l}T_{il }n_{l}dA.

d/dt(**P**_{field}+**P**_{charge})_{i}= ∮_{A}∑_{l}T_{il }n_{l}dA.

The integral represents the flow of the i = (x, y, z) component of the momentum across the surface of the volume V.