Partindo do tensor de energia do campo eletromagnético
M^{\mu \nu}=\frac{1}{4 \pi}\left[F_{\alpha}^{\mu} F^{\alpha \nu}+\frac{1}{4} \eta^{\mu \nu} F_{\alpha \beta} F^{\alpha \beta}\right]
obtenha o vetor de Poynting e o tensor de stress, respectivamente:
\begin{aligned} s_{i} & =c M^{0 i}=\frac{c}{4 \pi}(\vec{E} \times \vec{B})_{i} \\\ p_{i j} & =M^{i j}=-\frac{1}{4 \pi}\left[E_{i} E_{j}+B_{i} B_{j}+\frac{1}{2} \eta^{i j}\left(E^{2}+B^{2}\right)\right] \end{aligned}

Passo 1

O enunciado nos dá o tensor de energia-momento do campo eletromagnético
M^{\mu\nu}=\frac{1}{4\pi}\Bigl[F^{\mu}{}_{\alpha}F^{\alpha\nu}+\frac14\,\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}\Bigr]
e pede para obter o vetor de Poynting
s_i=c\,M^{0i}
e o tensor de tensões
p_{ij}=M^{ij}.
Vamos usar as componentes do tensor de campo
\displaystyle F^{0i}=E^i,
\displaystyle F^{ij}=-\varepsilon^{ijk}B_k,
a métrica \eta^{\mu\nu}=\mathrm{diag}(+,-,-,-) e a identidade de Levi-Civita
\displaystyle \varepsilon^{ikl}\varepsilon^{kjm}=\delta^{ij}\delta^{lm}-\delta^{im}\delta^{lj}.

Passo 2

Cálculo de M^{0i}. Temos
M^{0i} =\frac{1}{4\pi}\Bigl[F^{0}{}_{\alpha}F^{\alpha i}+\frac14\,\eta^{0i}F_{\alpha\beta}F^{\alpha\beta}\Bigr].
Como \eta^{0i}=0, só resta
F^{0}{}_{j}F^{j i} =E_{j}\bigl(-\varepsilon^{j i k}B_{k}\bigr) =(\vec E\times\vec B)^{i}.
Logo
M^{0i}=\frac{1}{4\pi}(\vec E\times\vec B)^{i} \quad\Longrightarrow\quad s_{i}=c\,M^{0i}=\frac{c}{4\pi}(\vec E\times\vec B)_{i}.

Passo 3

Cálculo de M^{ij}. Escrevemos
M^{ij} =\frac{1}{4\pi}\Bigl[F^{i}{}_{\alpha}F^{\alpha j}+\frac14\,\eta^{ij}F_{\alpha\beta}F^{\alpha\beta}\Bigr].
  1. F^{i}{}_{0}F^{0j}=-E_{i}E^{j}.
  2. F^{i}{}_{k}F^{k j} =(-\varepsilon^{i k \ell}B_{\ell})(-\varepsilon^{k j m}B_{m}) =\bigl(B^{2}\delta^{ij}-B_{i}B_{j}\bigr).
  3. F_{\alpha\beta}F^{\alpha\beta}=2\,(B^{2}-E^{2}).
  4. Como \eta^{ij}=-\delta^{ij}, tem-se
\frac14\,\eta^{ij}F_{\alpha\beta}F^{\alpha\beta} =\frac12\,\eta^{ij}(B^{2}-E^{2}) =\frac12\,(E^{2}-B^{2})\,\delta^{ij}.
Somando os termos dentro dos colchetes:
-\,E_{i}E_{j} +\bigl(B^{2}\delta^{ij}-B_{i}B_{j}\bigr) +\tfrac12\,(E^{2}-B^{2})\,\delta^{ij} =-E_{i}E_{j}-B_{i}B_{j}+\tfrac12\,(E^{2}+B^{2})\,\delta^{ij}.
Portanto
M^{ij} =\frac{1}{4\pi}\Bigl[-E_{i}E_{j}-B_{i}B_{j}+\tfrac12\,(E^{2}+B^{2})\,\delta^{ij}\Bigr] =-\frac{1}{4\pi}\Bigl[E_{i}E_{j}+B_{i}B_{j}+\tfrac12\,\eta^{ij}(E^{2}+B^{2})\Bigr].

Resposta

s_{i}=\frac{c}{4\pi}(\vec E\times\vec B)_{i}, \qquad p_{ij}=-\frac{1}{4\pi}\Bigl[E_{i}E_{j}+B_{i}B_{j}+\tfrac12\,\eta^{ij}(E^{2}+B^{2})\Bigr].

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