Question 8.18: Work out the formulas for u, S, g, and ↔T in the presence of...

Work out the formulas for u, S, g, and \overleftrightarrow{ T } in the presence of magnetic charge. [Hint: Start with the generalized Maxwell equations (7.44) and Lorentz force law (Eq. 8.44), and follow the derivations in Sections 8.1.2, 8.2.2, and 8.2.3.]

\left. \begin{matrix} \text { (i) } \nabla \cdot E =\frac{1}{\epsilon_{0}} \rho_{e} \text {, } & \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} . \end{matrix} \right\}                    (7.44)

F =q_{e}( E + v \times B )+q_{m}\left( B -\epsilon_{0} \mu_{0} v \times E \right)                           (8.44)

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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 ) .

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