Page 352 - Intro to Tensor Calculus

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where
1
E
T rs = E r D s − E j D j δ rs (2.6.88)
2
is called the electric stress tensor. In matrix form the stress tensor is written
 
1
E 1 D 1 − E j D j E 1 D 2 E 1 D 3
2
T E =  E 2 D 1 1 E 2 D 3  . (2.6.89)
rs E 2 D 2 − E j D j
2
1
E 3 D 1 E 3 D 2 E 3 D 3 − E j D j
2
By performing similar calculations we can transform the third and fourth terms in the equation (2.6.87) and
obtain
M
H r,s B s − H s,r B s = T rs,s (2.6.90)
where
1
M
T rs = H r B S − H j B j δ rs (2.6.91)
2
is the magnetic stress tensor. In matrix form the magnetic stress tensor is written
 
1
B 1 H 1 − B j H j B 1 H 2 B 1 H 3
2
T M =  B 2 H 1 1 B 2 H 3  . (2.6.92)
rs B 2 H 2 − B j H j
2
1
B 3 H 1 B 3 H 2 B 3 H 3 − B j H j
2
The total electromagnetic stress tensor is
T rs = T E + T M . (2.6.93)
rs rs
Then the equation (2.6.87) can be written in the form

∂
T rs,s − ρE r = ris J i B s + ( ris D i B s )
∂t

or
∂
ρE r + ris J i B S = T rs,s − ( ris D i B s ). (2.6.94)
∂t
For free space D i = 0 E i and B i = µ 0 H i so that the last term of equation (2.6.94) can be written in terms
of the Poynting vector as
∂
∂S r
µ 0 0 = ( ris D i B s ). (2.6.95)
∂t ∂t
Now integrate the equation (2.6.94) over the volume to obtain the total electromagnetic force
ZZZ ZZZ ZZZ ZZZ
∂S r
ρE r dτ + ris J i B s dτ = T rs,s dτ − µ 0 0 dτ.
V V V V ∂t
Applying the divergence theorem of Gauss gives

ZZZ ZZZ ZZ ZZZ
∂S r
ρE r dτ + ris J i B s dτ = T rsn s dσ − µ 0 0 dτ. (2.6.96)
V V S V ∂t
The left side of the equation (2.6.96) represents the forces acting on charges and currents contained within
the volume element. If the electric and magnetic fields do not vary with time, then the last term on the
right is zero. In this case the forces can be expressed as an integral of the electromagnetic stress tensor.