Question 12.70: (a) Construct a tensor D^μν (analogous to F^μν) out of D and...

(a)  \left.\begin{array}{l} D =\epsilon_{0} E + P \text { suggests } E \rightarrow \frac{1}{\epsilon_{0}} D \\H =\frac{1}{u_{0}} B – M \text { suggests } B \rightarrow \mu_{0} H\end{array}\right\} but it’s a little cleaner if we divide by \mu_{0} while we’re at it, so that

E \rightarrow \frac{1}{\mu_{0} \epsilon_{0}} D =c^{2} D , B \rightarrow H . Then: D^{\mu \nu}=\left\{\begin{array}{cccc}0 & c D_{x} & c D_{y} & c D_{z} \\-c D_{x} & 0 & H_{z} & -H_{y} \\-c D_{y} & -H_{z} & 0 & H_{x} \\-c D_{z} & H_{y} & -H_{x} & 0\end{array}\right\}

Then (following the derivation in Sect. 12.3.4):

\frac{\partial}{\partial x^{\nu}} D^{0 \nu}=c \nabla \cdot D =c \rho_{f}=J_{f}^{0} ; \frac{\partial}{\partial x^{\nu}} D^{1 \nu}=\frac{1}{c} \frac{\partial}{\partial t}\left(-c D_{x}\right)+( \nabla \times H )_{x}=\left(J_{f}\right)_{x} ; so \frac{\partial D_{\mu \nu}}{\partial x^{\nu}}=J_{f}^{\mu},

where J_{f}^{\mu}=\left(c \rho_{f}, J _{f}\right) . Meanwhile, the homogeneous Maxwell equations \left( \nabla \cdot B =0, E =-\frac{\partial B }{\partial t}\right) are unchanged, and hence \frac{\partial G^{\mu \nu}}{\partial x^{\nu}}=0.

(b)    H^{\mu \nu}=\left\{\begin{array}{cccc}0 & H_{x} & H_{y} & H_{z} \\-H_{x} & 0 & -c D_{z} & c D_{y} \\-H_{y} & c D_{z} & 0 & -c D_{x} \\-H_{z} & -c D_{y} & c D_{x} & 0\end{array}\right\}

(c) If the material is at rest, \eta_{\nu}=(-c, 0,0,0), \text { and the sum over } \nu collapses to a single term:

D^{\mu 0} \eta_{0}=c^{2} \epsilon F^{\mu 0} \eta_{0} \Rightarrow D^{\mu 0}=c^{2} \epsilon F^{\mu 0} \Rightarrow-c D =-c^{2} \epsilon \frac{ E }{c} \Rightarrow D =\epsilon E (\text { Eq. } 4.32),

H^{\mu 0} \eta_{0}=\frac{1}{\mu} G^{\mu 0} \eta_{0} \Rightarrow H^{\mu 0}=\frac{1}{\mu} G^{\mu 0} \Rightarrow- H =-\frac{1}{\mu} B \Rightarrow H =\frac{1}{\mu} B (\text { Eq. } 6.31).

(d) In general, \eta_{\nu}=\gamma(-c, u ), \text { so, for } \mu=0:

D^{0 \nu} \eta_{\nu}=D^{01} \eta_{1}+D^{02} \eta_{2}+D^{03} \eta_{3}=c D_{x}\left(\gamma u_{x}\right)+c D_{y}\left(\gamma u_{y}\right)+c D_{z}\left(\gamma u_{z}\right)=\gamma c( D \cdot u ),

F^{0 \nu} \eta_{\nu}=F^{01} \eta_{1}+F^{02} \eta_{2}+F^{03} \eta_{3}=\frac{E_{x}}{c}\left(\gamma u_{x}\right)+\frac{E_{y}}{c}\left(\gamma u_{y}\right)+\frac{E_{z}}{c}\left(\gamma u_{z}\right)=\frac{\gamma}{c}( E \cdot u ) , so

D^{0 \nu} \eta_{\nu}=c^{2} \epsilon F^{0 \nu} \eta_{\nu} \Rightarrow \gamma c( D \cdot u )=c^{2} \epsilon\left(\frac{\gamma}{c}\right)( E \cdot u ) \Rightarrow D \cdot u =\epsilon( E \cdot u ).                         [1]

H^{0 \nu} \eta_{\nu}=H^{01} \eta_{1}+H^{02} \eta_{2}+H^{03} \eta_{3}=H_{x}\left(\gamma u_{x}\right)+H_{y}\left(\gamma u_{y}\right)+H_{z}\left(\gamma u_{z}\right)=\gamma( H \cdot u ),

G^{0 \nu} \eta_{\nu}=G^{01} \eta_{1}+G^{02} \eta_{2}+G^{03} \eta_{3}=B_{x}\left(\gamma u_{x}\right)+B_{y}\left(\gamma u_{y}\right)+B_{z}\left(\gamma u_{z}\right)=\gamma( B \cdot u ) , so

H^{0 \nu} \eta_{\nu}=\frac{1}{\mu} G^{0 \nu} \eta_{\nu} \Rightarrow \gamma( H \cdot u )=\frac{1}{\mu} \gamma( B \cdot u ) \Rightarrow H \cdot u =\frac{1}{\mu}( B \cdot u ).                         [2]

Similarly, for µ = 1:

=\gamma\left[c^{2} D +( u \times H )\right]_{x},

\gamma\left[c^{2} D +( u \times H )\right]_{x}=c^{2} \epsilon \gamma[ E +( u \times B )]_{x} \Rightarrow D +\frac{1}{c^{2}}( u \times H )=\epsilon[ E +( u \times B )].                                        [3]

\begin{aligned}H^{1 \nu} \eta_{\nu} &=H^{10} \eta_{0}+H^{12} \eta_{2}+H^{13} \eta_{3}=\left(-H_{x}\right)(-\gamma c)+\left(-c D_{z}\right)\left(\gamma u_{y}\right)+\left(c D_{y}\right)\left(\gamma u_{z}\right) \\&=\gamma c\left(H_{x}-u_{y} D_{z}+u_{z} D_{y}\right)=\gamma c[ H -( u \times D )]_{x}\end{aligned},

\gamma c[ H -( u \times D )]_{x}=\frac{1}{\mu} \frac{\gamma}{c}\left[c^{2} B -( u \times E )\right]_{x} \Rightarrow H -( u \times D )=\frac{1}{\mu}\left[ B -\frac{1}{c^{2}}( u \times E )\right].                                  [4]

Use Eq. [4] as an expression for H, plug this into Eq. [3], and solve for D:

D +\frac{1}{c^{2}} u \times\left\{( u \times D )+\frac{1}{\mu}\left[ B -\frac{1}{c^{2}}( u \times E )\right]\right\}=\epsilon[ E +( u \times B )];

D +\frac{1}{c^{2}}\left[( u \cdot D ) u -u^{2} D \right]=\epsilon[ E +( u \times B )]-\frac{1}{\mu c^{2}}( u \times B )+\frac{1}{\mu c^{4}}[ u \times( u \times E )].

Using Eq. [1] to rewrite (u · D):

Let \gamma \equiv \frac{1}{\sqrt{1-u^{2} / c^{2}}}, \quad v \equiv \frac{1}{\sqrt{\epsilon \mu}} . Then

D =\gamma^{2} \epsilon\left\{\left(1-\frac{u^{2} v^{2}}{c^{4}}\right) E +\left(1-\frac{v^{2}}{c^{2}}\right)\left[( u \times B )-\frac{1}{c^{2}}( E \cdot u ) u \right]\right\}.

Now use Eq. [3] as an expression for D, plug this into Eq. [4], and solve for H:

H – u \times\left\{-\frac{1}{c^{2}}( u \times H )+\epsilon[ E +( u \times B )]\right\}=\frac{1}{\mu}\left[ B -\frac{1}{c^{2}}( u \times E )\right];

H +\frac{1}{c^{2}}\left[( u \cdot H ) u -u^{2} H \right]=\frac{1}{\mu}\left[ B -\frac{1}{c^{2}}( u \times E )\right]+\epsilon( u \times E )+\epsilon[ u \times( u \times B )].

Using Eq. [2] to rewrite (u · H):

H =\frac{\gamma^{2}}{\mu}\left\{\left(1-\frac{u^{2}}{v^{2}}\right) B +\left(\frac{1}{v^{2}}-\frac{1}{c^{2}}\right)[( u \times E )+( B \cdot u ) u ]\right\}.