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


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?

  • $\begingroup$ What if you insert $\vec \beta \propto \vec E \times \vec B$ and use the triple product formula? en.wikipedia.org/wiki/Triple_product#Vector_triple_product $\endgroup$ Jul 3, 2014 at 1:59
  • $\begingroup$ Are you sure you should drop the $\bar{\vec{E}}_{\perp}\times\bar{\vec{B}}_{\perp}$ term? After all, if motion is in the $x$ direction, there are two independent perpendicular directions, $y$ and $z$, whose cross product is nonzero. $\endgroup$
    – G. Paily
    Jul 3, 2014 at 8:24
  • $\begingroup$ I think I should drop it as it contains cross product of two field in the direction of rocket motion. $\endgroup$
    – user45765
    Jul 3, 2014 at 14:16

1 Answer 1


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: $$ \bar{\mathbf{E}} \times \bar{\mathbf{B}} = \frac{1}{1 - \beta^2} \left( \frac{\boldsymbol{\beta}\left(\mathbf{E}^2 + \mathbf{B}^2\right)} {1 + \beta^2} + \boldsymbol{\beta}\left( \frac{\beta^2\left(\mathbf{E}^2 + \mathbf{B}^2\right)}{1 + \beta^2}\right) -\boldsymbol{\beta} \left( \mathbf{E}^2 + \mathbf{B}^2 \right) \right) = 0 $$


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