﻿ The Energy Density of Fields
San José State University

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The Energy Density of Fields

This material investigates the matter of the energy density of fields. The analysis for an electromagnetic field is well known. The question is whether or not it can be generalized.

## 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 (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)

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

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

## The Energy Density of a Gravitational Field

The preceding derivation may also be carried out for mass. The force between a mass m uniformly distributed over a spherical ball centered on the origin and an increment of mass dm at a distance r from the origin is

#### −γmdm/r²

where γ is the gravitational constant.

The negative sign indicates the force between m and dm is an attraction. The energy released upon bringing dm from infinity to surface of the mass m at R is

#### γmdm/R

The total energy SR released on raising the mass at R from 0 to M is

#### SR = ½γM²/R

Whereas the total energy of the field of the charged particle is positive, for a mass particle it is negative.

If one takes into account the radius R of the ball with a mass m is given by

#### (4/3)πR³ρ = m and hence R = [(3/4)(m/(πρ)]⅓

where ρ is the mass density.

The total energy V is then

## Maxwell Equations for a Gravitational Field?

The equations of a gravitational field which correspond to the Maxwell equations for an electromagnetic field are:

#### ∇·G = ρ/γ ∇×G = 0 0 = F + γ(∂G/∂t) or, equivalently (∂G/∂t) = −F/γ

where G is the gravitational field intensity, ρ is the mass density, F is the mass flow and γ is the gravitational constant.

It is clear what the fourth of the Maxwell equations means; i.e., an electrical current will generate a magnetic field around it and a rate of change in the electric field will also generate a magnetic field. The effects of these two sources may offset each other, even to the point that no magnetic field is produced. But there is no reason to believe one effect could drive the other to maintain a zero magnetic field. However this what is supposed in the set of equations for a gravitational field. The following could be considered an exercise to see what the consequences would be.

Let {mi and vi for i=1, 2, … N} be the masses and their velocities of the N particles.

The equation of motion for the i-th mass particle is

#### mi(dvi/dt) = miG and hence (dvi/dt) = G

As noted before the kinetic energy K is ½mv² and hence

#### (dK/dt) = mv·(dv/dt) = v·(m(dv/dt))

The mass flow F is given by

#### F = Σ mivi

Now (dot) multiply the equation (∂G/∂t) = −F/γ by G and integrate over all space. The result is

But

Therefore

#### ½(∂/∂t)∫R³G²dV = −(1/γ)∫R³(dK/dt)dV which is equivalent to (∂/∂t)∫R³[(γ/2)G²dV + K]dV = 0

Since the second term in the integral is the kinetic energy of the mass particles the first term (γ/2)G² must be the energy density of the gravitational field created by those mass particles.

## Generalization

See Field Energy for a General Field.