A question on an exercise from Gravitation by Misner, Thorne and Wheeler My question is on problem 4.1 of Gravitation. In a generic case of electric field and magnetic field(i.e not $E=0$ or $B=0$ or $E$ and $B$ perpendicular), define the direction $\hat{n}$ unit vector ,
$$\hat{n}\tanh (2\alpha)=\frac{2\vec{E}\times\vec{B}}{\vec{E}^{2}+\vec{B}^2}$$
and $\vec{\beta}=\tanh(\alpha)\hat{n}$ 
is the velocity vector. 
Show in the frame of the rocket with velocity $\vec{\beta}$, the Poynting vector vanishes. 
I tried the following but I am stuck at the cancellation.
Let the $\bar{\vec{E}}$ and $\bar{\vec{B}}$ be the field in the rocket frame and the field without bars be the field of the rest frame. direction parallel along the velocity of rocket is denoted as subscript $\parallel$ and direction perpendicular to velocity of rocket direction is denoted by $\perp$ as the subscript. 
By lorentz transformation. 
\begin{align}
\bar{\vec{E_{\parallel}}}&=\vec{E_{\parallel}}, \\
\bar{\vec{E_{\perp}}}&=\frac{\vec{E_{\perp}}+\vec{\beta}\times\vec{B_\perp}}{\sqrt{1-\beta^{2}}},\\ 
\bar{\vec{B_{\parallel}}}&=\vec{B_{\parallel}}, \\ 
\bar{\vec{B_{\perp}}}&=\frac{\vec{B_{\perp}}-\vec{\beta}\times\vec{E_\perp}}{\sqrt{1-\beta^{2}}}.
\end{align}
Note $\beta\times X_{\perp}$=$\beta\times X$
In barred frame, 
\begin{align}
\bar{\vec{E}}\times\bar{\vec{B}}
&=(\bar{\vec{E_{\perp}}}+\bar{\vec{E_{\parallel}}})\times(\bar{\vec{B_{\perp}}}+\bar{\vec{B_{\parallel}}})\\
&=\bar{\vec{E_{\perp}}}\times\bar{\vec{B_{\parallel}}}+\bar{\vec{E_{\parallel}}}\times\bar{\vec{B_{\perp}}}.
\end{align}
Plug in the lorentz transformation. One gets
$$\frac{\vec{E}\times\vec{B}-\vec{E}_{\parallel}\times(\vec{\beta}\times\vec{E})-\vec{B}_{\parallel}\times(\vec{\beta}\times\vec{B})}{\sqrt{1-\beta^{2}}}.$$
Now the last two terms looks generically like $\vec{X_{\parallel}}\times(\vec{\beta}\times\vec{X}$) simplifies to $\vec{X_{\perp}}(\vec{X}\cdot\vec{\beta})$. 
However, I could not see the cancellation at this stage. Did I do something wrong? 
 A: First, note that $\mathbf{n}$ points in the direction $\mathbf{E} \times \mathbf{B}$,
so it's orthogonal to both $\mathbf{E}$ and $\mathbf{B}$.
This simplifies your decompositions, since we just have:
$$
\mathbf{E} = \mathbf{E}_{||} + \mathbf{E}_{\perp} = \mathbf{E}_{\perp} \\
\mathbf{B} = \mathbf{B}_{||} + \mathbf{B}_{\perp} = \mathbf{B}_{\perp}
$$
Since the Lorentz transform is a boost in the $\mathbf{n}$ direction, we
also have $\bar{\mathbf{E}}_{||} = \mathbf{E}_{||} = 0$ and
$\bar{\mathbf{B}}_{||} = \mathbf{B}_{||} = 0$. That means again,
$\bar{\mathbf{E}} = \bar{\mathbf{E}}_{\perp}$ and 
$\bar{\mathbf{B}} = \bar{\mathbf{B}}_{\perp}$
Evaluate:
$$
\bar{\mathbf{E}} \times \bar{\mathbf{B}} = 
\bar{\mathbf{E}}_{\perp} \times \bar{\mathbf{B}}_{\perp}
= \frac{1}{1 - \beta^2}\left( \mathbf{E}_\perp + \boldsymbol{\beta} \times \mathbf{B}_{\perp} \right)
\times \left( \mathbf{B}_\perp - \boldsymbol{\beta} \times \mathbf{E}_{\perp} \right) \\
= \frac{1}{1 - \beta^2} \Bigl(
\mathbf{E}_{\perp} \times \mathbf{B}_{\perp}
- \mathbf{E}_{\perp} \times \left(\boldsymbol{\beta} \times \mathbf{E}_{\perp} \right)
+ \left(\boldsymbol{\beta} \times \mathbf{B}_{\perp}\right) \times \mathbf{B}_{\perp} 
- \left(\boldsymbol{\beta} \times \mathbf{B}_{\perp}\right) \times 
\left(\boldsymbol{\beta} \times \mathbf{E}_{\perp} \right)
\Bigr)
$$
Using the BAC-CAB identity and the fact that $\boldsymbol{\beta}$ is orthogonal to $\mathbf{E}$ and
$\mathbf{B}$ in the second and third terms gives:
$$
- \mathbf{E}_{\perp} \times \left(\boldsymbol{\beta} \times \mathbf{E}_{\perp} \right)
 = -\boldsymbol{\beta} \mathbf{E}_{\perp}^2 \\
 \left(\boldsymbol{\beta} \times \mathbf{B}_{\perp}\right) \times \mathbf{B}_{\perp} =
- \boldsymbol{\beta} \mathbf{B}_{\perp}^2 \\
$$
On the fourth term, use BAC-CAB and some triple product identities (just the ones that come from a
triple product being the determinant of the three vectors), to get:
$$
- \left(\boldsymbol{\beta} \times \mathbf{B}_{\perp}\right) \times 
\left(\boldsymbol{\beta} \times \mathbf{E}_{\perp} \right)
= \mathbf{E}_{\perp} \left( \boldsymbol{\beta} \times \mathbf{B}_{\perp} \cdot \boldsymbol{\beta} 
\right)
- \boldsymbol{\beta}\left( \boldsymbol{\beta} \times \mathbf{B}_{\perp} \cdot 
\mathbf{E}_{\perp}\right) \\
= \mathbf{E}_{\perp} \left( \boldsymbol{\beta} \cdot \boldsymbol{\beta} \times \mathbf{B}_{\perp}
\right)
- \boldsymbol{\beta}\left( \mathbf{E}_{\perp} \cdot \boldsymbol{\beta} \times \mathbf{B}_{\perp}
\right) = \boldsymbol{\beta}\left( \boldsymbol{\beta} \cdot \mathbf{E}_{\perp} \times
 \mathbf{B}_{\perp}
\right)
$$
Substituting these into $\bar{\mathbf{E}} \times \bar{\mathbf{B}}$ and dropping the
unnecessary perps, you get:
$$
\bar{\mathbf{E}} \times \bar{\mathbf{B}} = 
\frac{1}{1 - \beta^2} \Bigl(
\mathbf{E} \times \mathbf{B}
+ \boldsymbol{\beta}\left( \boldsymbol{\beta} \cdot \mathbf{E} \times  \mathbf{B} \right)
-\boldsymbol{\beta} \left( \mathbf{E}^2 + \mathbf{B}^2 \right)
\Bigr)
$$
Recalling $\tanh \alpha = \beta$, the double-angle identity for tanh is:
$$
\tanh 2\alpha = \frac{2 \sinh \alpha \cosh \alpha}{2 \cosh^2 \alpha - 1}
= \frac{2\beta}{1 + \beta^2}
$$
Then the poynting ratio,
$$
\mathbf{n} \tanh 2\alpha = \frac{2 \mathbf{E} \times \mathbf{B}}{\mathbf{E}^2 + \mathbf{B}^2},
$$ gives
$$
\mathbf{E} \times \mathbf{B} = \frac{\boldsymbol{\beta}\left(\mathbf{E}^2 + \mathbf{B}^2\right)}
{1 + \beta^2} \\
\boldsymbol{\beta} \cdot \mathbf{E} \times \mathbf{B} = 
\frac{\beta^2\left(\mathbf{E}^2 + \mathbf{B}^2\right)}{1 + \beta^2}
$$
Substituting these into $\bar{\mathbf{E}} \times \bar{\mathbf{B}}$ gives:
$$
\bar{\mathbf{E}} \times \bar{\mathbf{B}} = 
\frac{1}{1 - \beta^2} \left(
\frac{\boldsymbol{\beta}\left(\mathbf{E}^2 + \mathbf{B}^2\right)}
{1 + \beta^2}
+ \boldsymbol{\beta}\left( \frac{\beta^2\left(\mathbf{E}^2 + \mathbf{B}^2\right)}{1 + \beta^2}\right)
-\boldsymbol{\beta} \left( \mathbf{E}^2 + \mathbf{B}^2 \right)
\right) = 0
$$
