Maxwell’s equations with magnetic charge (Eq. 7.44):
\text { (i) } \nabla \cdot E =\frac{1}{\epsilon_{0}} \rho_{e} , \text { (iii) } \nabla \times E =-\mu_{0} J _{m}-\frac{\partial B }{\partial t} ,
\text { (ii) } \nabla \cdot B =\mu_{0} \rho_{m} , \text { (iv) } \nabla \times B =\mu_{0} J _{e}+\mu_{0} \epsilon_{0} \frac{\partial E }{\partial t} .
The Lorentz force law becomes (Eq. 8.44)
F =q_{e}( E + v \times B )+q_{m}\left( B -\frac{1}{c^{2}} v \times E \right) .
Following the argument in Section 8.1.2:
F \cdot d l =\left[q_{e}( E + v \times B )+q_{m}\left( B -\frac{1}{c^{2}} v \times E \right)\right] \cdot v d t=\left(q_{e} E +q_{m} B \right) \cdot v d t ,
\frac{d W}{d t}=\int\left( E \cdot J _{e}+ B \cdot J _{m}\right) d \tau
(which generalizes Eq. 8.6). Use (iii) and (iv) to eliminate J _{e} \text { and } J _{m} :
\frac{d W}{d t}=\int_{ \nu }( E \cdot J ) d \tau (8.6)
\left( E \cdot J _{e}+ B \cdot J _{m}\right)=\frac{1}{\mu_{0}} E \cdot(\nabla \times B )-\epsilon_{0} E \cdot \frac{\partial E }{\partial t}-\frac{1}{\mu_{0}} B \cdot(\nabla \times E )-\frac{1}{\mu_{0}} B \cdot \frac{\partial B }{\partial t},
\text { but } \nabla \cdot( E \times B )= B \cdot(\nabla \times E )- E \cdot(\nabla \times B ) , so
\left( E \cdot J _{e}+ B \cdot J _{m}\right)=-\frac{1}{\mu_{0}} \nabla \cdot( E \times B )-\frac{1}{2} \frac{\partial}{\partial t}\left(\epsilon_{0} E^{2}+\frac{1}{\mu_{0}} B^{2}\right)
(generalizing Eq. 8.8). Thus
E \cdot J =-\frac{1}{2} \frac{\partial}{\partial t}\left(\epsilon_{0} E^{2}+\frac{1}{\mu_{0}} B^{2}\right)-\frac{1}{\mu_{0}} \nabla \cdot( E \times B ) (8.8)
\frac{d W}{d t}=-\frac{d}{d t} \int \frac{1}{2}\left(\epsilon_{0} E^{2}+\frac{1}{\mu_{0}} B^{2}\right) d \tau-\frac{1}{\mu_{0}} \oint( E \times B ) \cdot d a,
which is identical to Eq. 8.9. Evidently Poynting’s theorem is unchanged(!), and
\frac{d W}{d t}=-\frac{d}{d t} \int_{ \nu } \frac{1}{2}\left(\epsilon_{0} E^{2}+\frac{1}{\mu_{0}} B^{2}\right) d \tau-\frac{1}{\mu_{0}} \oint_{ S }( E \times B ) \cdot d a (8.9)
u=\frac{1}{2}\left(\epsilon_{0} E^{2}+\frac{1}{\mu_{0}} B^{2}\right), \quad S =\frac{1}{\mu_{0}}( E \times B ),
the same as before.
To construct the stress tensor we begin with the generalization of Eq. 8.13:
F =\int_{ \nu }( E + v \times B ) \rho d \tau=\int_{ \nu }(\rho E + J \times B ) d \tau (8.13)
F =\int\left[( E + v \times B ) \rho_{e}+\left( B -\frac{1}{c^{2}} v \times E \right) \rho_{m}\right] d \tau .
The force per unit volume (Eq. 8.14) becomes
f =\rho E + J \times B (8.14)
f =\left(\rho_{e} E + J _{e} \times B \right)+\left(\rho_{m} B -\frac{1}{c^{2}} J _{m} \times E \right)
=\epsilon_{0}( \nabla \cdot E ) E +\left(\frac{1}{\mu_{0}} \nabla \times B -\epsilon_{0} \frac{\partial E }{\partial t}\right) \times B +\frac{1}{\mu_{0}}( \nabla \cdot B ) B -\mu_{0} \epsilon_{0}\left(-\frac{1}{\mu_{0}} \nabla \times E -\frac{1}{\mu_{0}} \frac{\partial B }{\partial t}\right) \times E
=\epsilon_{0}[( \nabla \cdot E ) E – E \times( \nabla \times E )]+\frac{1}{\mu_{0}}[( \nabla \cdot B ) B – B \times( \nabla \times B )]-\epsilon_{0} \frac{\partial}{\partial t}( E \times B )
=\epsilon_{0}\left[( \nabla \cdot E ) E -\frac{1}{2} \nabla \left(E^{2}\right)+( E \cdot \nabla ) E \right]+\frac{1}{\mu_{0}}\left[( \nabla \cdot B ) B -\frac{1}{2} \nabla \left(B^{2}\right)+( B \cdot \nabla ) B \right]-\epsilon_{0} \frac{\partial}{\partial t}( E \times B ).
This is identical to Eq. 8.16, so the stress tensor is the same as before:
T_{i j} \equiv \epsilon_{0}\left(E_{i} E_{j}-\frac{1}{2} \delta_{i j} E^{2}\right)+\frac{1}{\mu_{0}}\left(B_{i} B_{j}-\frac{1}{2} \delta_{i j} B^{2}\right) .
Likewise, Eq. 8.20 is still valid. In fact, this argument is more straightforward when you include magnetic charge, since you don’t need artificially to insert the (∇ · B)B term (after Eq. 8.15).
F =\oint_{ S } \overleftrightarrow{ T } \cdot d a -\epsilon_{0} \mu_0\frac{d}{dt} \int_{\nu }{ S d \tau} (8.20)
f =\epsilon_{0}[(\nabla \cdot E ) E – E \times(\nabla \times E )]-\frac{1}{\mu_{0}}[ B \times(\nabla \times B )]-\epsilon_{0} \frac{\partial}{\partial t}( E \times B ) (8.15)
The electromagnetic momentum density (Eq. 8.29) also stays the same, since the argument in Section 8.2.3 is formulated entirely in terms of the fields:
g =\mu_{0} \epsilon_{0} S =\epsilon_{0}( E \times B ) (8.29)
g =\epsilon_{0}( E \times B ) .