15. Ch. 6, part 2

Lecture 15 - Energy and momentum

We go directly to the law of conservation of energyy, mechanical and electromagnetic, for a system of particles that interact by electromagnetic forces and are also subject to external electromagnetic fields. Then we examine more in detail the magnetic energy.

15.1. Energy density and flow

We assume (for the time being) that only electromagnetic forces are present and that particle motions are confined to a finite volume V. At speeds small compared to c, the particles obey Newton’s equations

m_{i} \frac{dv_{i}}{dt} = q_{i}E(x_{i}) + \frac{v_{i}}{c} × B(x_{i})

Multiplying both sides by v_{i} and summing over i we obtain

∑_{i} \frac{d}{dt} (\frac{1}{2} m_{i}v_{i}^{2}) = ∑_{i} q_{i}v_{i} · E(x_{i})

or

\frac{d}{dt} U_{kin} = ∫_{V} J(x) · E(x) d^{3}x  (15.1)

where we have introduced the total kinetic energy

U_{kin} = ∑_{i} (\frac{1}{2} m_{i}v_{i}^{2})

and the current density

J(x) = ∑_{i} q_{i}v_{i}δ(x − x_{i})

Equation (15.1) says that J · E is the power per unit volume supplied by the field to the particles. We note that E is the total field, including the field generated by the particles themselves. It is convenient to introduce the kinetic energy density

u_{kin}(x) = ∑_{i} (\frac{1}{2} m_{i}v_{i}^{2}) δ(x − x_{i})

such that U_{kin} = ∫_{V} u_{kin}(x) d^{3}x. Then eq. (15.1) can be written

∫_{V} [\frac{∂}{∂t} u_{kin}(x) − J(x) · E(x) ] d^{3}x = 0

We now obtain another expression for J · E from Maxwell’s equations. We multiply by B· both sides of Faraday’s law and obtain

B · ∇ × E + \frac{1}{c}B · \frac{∂B}{∂t} = 0

Similarly, we multiply by E· both sides of the Ampère-Maxwell law and obtain

E · ∇ × B − \frac{1}{c}E · \frac{∂E}{∂t} = \frac{4π}{c} J · E

Subtracting side by side and using ∇· (E × B) = B · ∇ × E − E · ∇ × B we find

∇· (E × B) + \frac{1}{2c} \frac{∂}{∂t} (E^{2} + B^{2}) = −\frac{4π}{c} J · E

This is the equation we were looking for. We rewrite it as

∇ · S + \frac{∂u}{∂t} = − J · E ,  (15.2)

having defined the important quantities

S = \frac{c}{4π} E × B  (15.3)

u_{em} = \frac{1}{8π} (E^{2} + B^{2})  (15.4)

It is easy to see that u_{em} is the electromagnetic energy density and that the Poynting vector S represents the energy flow: combine (15.1) and (15.5) to obtain

∇ · S + \frac{∂}{∂t}(u_{em} + u_{kin}) = 0  (15.5)

which is of the standard form of an equation of continuity for the total energy density u + u_{kin.} Integrating over the volume V bounded by the surface S with outward-pointing normal n:

∫_{S} S · n da + \frac{∂U}{∂t} = 0

where U = U_{em} + U_{kin} and U_{em} is the electromagnetic energy contained in volume V

U_{em} = ∫ u_{em} d^{3}x = \frac{1}{8π} ∫ (E^{2} + B^{2}) d^{3}x