Page 351 - Intro to Tensor Calculus

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Now integrate equation (2.6.83) over a volume and apply Gauss’s divergence theorem to obtain

ZZ ZZZ ZZZ
∂D i ∂B i
ijk E j H k n i dσ + J i E i dτ = − ( E i + H i ) dτ. (2.6.84)
S V V ∂t ∂t
The first term in equation (2.6.84) represents the outward flow of energy across the surface enclosing the
volume. The second term in equation (2.6.84) represents the loss by Joule heating and the right-hand side
is the rate of decrease of stored electric and magnetic energy. The equation (2.6.84) is known as Poynting’s
theorem and can be written in the vector form

ZZ ZZZ ∂D ∂B
~
~
~
~
~
~
~
~
(E × H) · ˆ n dσ = (−E · − H · − E · J) dτ. (2.6.85)
S V ∂t ∂t
For later use we define the quantity
2
~
~
~
S i = ijk E j H k or S = E × H [Watts/m ] (2.6.86)
as Poynting’s energy flux vector and note that S i is perpendicular to both E i and H i and represents units
of energy density per unit time which crosses a unit surface area within the electromagnetic field.
Electromagnetic Stress Tensor
Instead of calculating energy flow per unit volume, let us calculate force per unit volume. Consider a
region containing charges and currents but is free from dielectrics and magnetic materials. To obtain terms
with units of force per unit volume we take the cross product of equation (2.6.79) with D i and the cross
product of equation (2.6.80) with B i and subtract to obtain

∂D i ∂B s
− irs ijk (E k,j D s + H k,j B s )= ris J i B s + ris B s + D i
∂t ∂t
which simplifies using the e − δ identity to
∂
−(δ rj δ sk − δ rk δ sj )(E k,j D s + H k,j B s )= ris J i B s + ris (D i B s )
∂t
which further simplifies to
∂
−E s,r D s + E r,s D s − H s,r B s + H r,s B s = ris J i B s + ( ris D i B s ). (2.6.87)
∂t
Observe that the first two terms in the equation (2.6.87) can be written

E r,s D s − E s,r D s =E r,s D s − 0 E s,r E s
1
=(E r D s ) ,s − E r D s,s − 0 ( E s E s ) ,r
2
1
=(E r D s ) ,s − ρE r − (E j D j δ sr ) ,s
2
1
=(E r D s − E j D j δ rs ) ,s − ρE r
2
which can be expressed in the form

E
E r,s D s − E s,r D s = T rs,s − ρE r