﻿ The Energy Density of Fields in General
San José State University

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Thayer Watkins
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The Energy Density of Fields in General

This material investigates the matter of the energy density of fields. The analysis for an electromagnetic field is well known. Special Relativity requires the existence of a magnetic field when an electrostatic charge is in motion. The question is whether or not it can be generalized. The same argument applies for the motion of any two-valued charge.

## The Energy Density of an Electromagnetic Field

Maxwell's equations in free space for an electromagnetic field of electric field intensity E and magnetic field intensity H are:

#### ∇·E = ρ/ε0 ∇·H = 0 ∇×E = −(1/c)(∂H/∂t) ∇×H = (1/c)(∂E/∂t) + (4π/c)j

where c is the speed of light in a vacuum, ρ is charge density and j is current density.

Now let the above equations apply for any two-valued charged particles. The parameters may have radically different values from thos for electromagnetism.

Now (dot) multiply both sides of the third equation by H, both sides of the fourth equation by E and subtract the two equations to obtain an equation that reduces to

#### E·∇×H − H·∇×E = (1/2c)(∂H²/∂t) + (1/2c)(∂E²/∂t) + (4π/c)j·E

By an identity in vector analysis the LHS of the above equation is equal to ∇·(E×H). The above equation can thus be rearranged to the form

#### (1/2c)(∂(E²+H²)/∂t) = −(4π/c)j·E − ∇·(E×H) and by multiplying through by (c/4π) to (∂((E²+H²)/(8π))/∂t) = −j·E −∇·((c/4π)(E×H)

In electromagnetism the vector (c/4π)(E×H) is known as the Poynting vector. Let it be denoted as S. The above equation is then

#### (∂((E²+H²)/(8π))/∂t) = −j·E −∇·S

Now integrate over some volume V. The LHS of the integrated equation would be, because of the interchangeability of the operation of differentiation with respect to time and integration over space,

The RHS is

#### −∫Vj·EdV − ∫V∇·SdV

By Gauss' Theorem the integral in the second term can be changed to

#### ∫∂VS·ndA

where ∂V is the boundary surface of V and n is a unit normal to the surface element dA.

Now let V go to the whole space R³. The integral of S goes to zero because E and H, and hence S, are zero at infinity.

Thus

#### (∂/(∂t)∫R³ ((E²+H²)/(8π))dV = −∫R³j·EdV

Clearly the field energy density is (E²+H²)/(8π).

## The Motion of Charged Particles

Now consider the motion and kinetic energies of the charges which determine the current. Let {qi, vi, mi for i=1, 2, … N} be the charges, their velocities and masses of the N particles.

The equation of motion for the i-th charge is

#### mi(dvi/dt) = qiE + (qi/c)(vi×H)

Note that the kinetic energy K is ½mv²=½mv·v and hence

#### (dK/dt) = mv·(dv/dt)

Thus for the i-th charge

#### (dKi/dt) = mivi·(dvi/dt) = vi·[midvi/dt)] and therefore (dKi/dt) = vi·[qiE + (qi/c)(vi×H] but vi·[vi×H] = 0 therefore (dKi/dt) = vi·[qiE]

But the current due to the i-th charge is qivi and the sum Σqivi is equal to the current j. Therefore the time rate of change of the kinetic energy of all the charges is

Therefore

#### (∂/(∂t)∫R³ ((E²+H²)/(8π))dV + ∫R³(dK/dt)dV = 0 or, equivalently (∂/(∂t)∫R³ [(E²+H²)/(8π)) + K]dV = 0

Thus the quantity in brackets is conserved. The second term is the kinetic energy of the charges. The first term must be the energy of the field and hence (E²+H²)/(8π)) is the energy density of the electromagnetic field.

Consider a stationary particle with charge Q. The electrical field intensity E at a distance r from the center of that particle is given by

#### E(r) = (1/(4πε0))Q/r²

where ε0 is a constant called the permittivity of space.

For a particle in empty space the energy density U is

#### U = ½ε0E² = (1/(32π²ε0))Q²/r4

The energy in a spherical shell of radius r and thickness dr is equal to 4πr²(Q²/(32π²r4) which reduces to Q²/(8πε0r²). The integration of these terms from R to ∞ gives a total energy TR

#### TR = ∫R∞U(4πr²)dr = Q²/(8πε0R)

If R→0, then TR → ∞. Thus a charged point particle, if such existed, would have infinite energy.

An alternate derivation of the formula for the total energy involves computing the work done in bringing the charge bit-by-bit from infinity to a sphere of radius R. Let q be the charge on the sphere. It is uniformly distributed over the sphere so its effect is the same as if it were concentrated at the center of the sphere at the origin. The force on an increment of charge dq at a distance r is

#### (1/(4πε0))qdq/r²

The sign of the force being positive is crucial. It indicates that the force between q and dq is repulsion and work has to be done to bring them together. The amount of energy required to bring the charge dq from +∞ to R is

#### ∫R∞dr/(4πε0))qdq/r² = (1/(4πε0))qdq/R

The energy required to raise the charge at R from 0 to Q is