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The Transformation of Electromagnetic Fields

Now that we have this in hand, we can easily see how to transform the electric and magnetic fields when we boost a frame. Of course, that does not guarantee that the result will be simple.

To convert $F^{\alpha\beta}$ from $K$ to $K'$, we must contract its indices with the transformation tensors,

F'^{\alpha\beta} = \frac{\partial x^{\prime \alpha}}{\parti...
...tial x^{\prime \beta}}{\partial x^\delta} F^{\gamma \delta} .
\end{displaymath} (16.165)

Note that since $A$ is a linear transformation:
A^\alpha_{ \gamma} = \frac{\partial x^{\prime \alpha}}{\partial x^\gamma}
\end{displaymath} (16.166)

(where I have deliberately inserted a space to differentiate the first index from the second) we can write this in terms of the components of $A$ as:
$\displaystyle F'^{\alpha\beta}$ $\textstyle =$ $\displaystyle A^\alpha_{ \gamma} F^{\gamma \delta} A^{\
  $\textstyle =$ $\displaystyle A^\alpha_{ \gamma} F^{\gamma \delta}
\tilde{A}^\beta_{ \delta}$ (16.167)

or (in a compressed notation):
F' = AF\tilde{A}
\end{displaymath} (16.168)

This is just a specific case of the general rule that $A$ can be used in general to transform any nth rank tensor by contracting it appropriately with each index.

As we saw in our discussion of Thomas precession, we will have occasion to use this result for the particular case of a pure boost in an arbitrary direction that we can without loss of generality pick to be the 1 direction. Let's see how this goes. Recall that $A$ for a pure boost in the one direction is the matrix formed with a lower right quadrant identity and an upper left quadrant $2 \times 2$ with $\gamma$ on the diagonal and $-\gamma \beta$ on the corners). Thus: so:

$\displaystyle F'^{01}$ $\textstyle =$ $\displaystyle A^0_{ 0} F^{0 1} A^{ 1}_1 + A^0_{ 1}F^{1 0}A^{ 1}_1$  
$\displaystyle -\frac{E'_1}{c}$ $\textstyle =$ $\displaystyle - \gamma^2 \frac{E_1}{c} - \gamma^2\beta^2
$\displaystyle E'_1$ $\textstyle =$ $\displaystyle (\gamma^2 + \gamma^2\beta^2) E_1$  
$\displaystyle E'_1$ $\textstyle =$ $\displaystyle E_1$ (16.169)

Note that we have extracted the ordinary cartesian components of $\mbox{\boldmath$E$}$ and $\mbox{\boldmath$B$}$ from $F$ after transforming it. I leave the rest of them to work out yourself. You should be able to show that:
$\displaystyle E_1'$ $\textstyle =$ $\displaystyle E_1$ (16.170)
$\displaystyle E_2'$ $\textstyle =$ $\displaystyle \gamma(E_2 - \beta B_3)$ (16.171)
$\displaystyle E_3'$ $\textstyle =$ $\displaystyle \gamma(E_3 + \beta B_2)$ (16.172)
$\displaystyle B_1'$ $\textstyle =$ $\displaystyle B_1$ (16.173)
$\displaystyle B_2'$ $\textstyle =$ $\displaystyle \gamma(B_2 + \beta E_3)$ (16.174)
$\displaystyle B_3'$ $\textstyle =$ $\displaystyle \gamma(B_3 - \beta E_2)$ (16.175)

The component of the fields in the direction of the boost is unchanged, the perpendicular components of the field are mixed (almost as if they were space-time pieces) by the boost. If you use instead the general form of $A$ for a boost and express the components in terms of dot products, you should also show that the general transformation is given by:

$\displaystyle {\bf E}'$ $\textstyle =$ $\displaystyle \gamma({\bf E} + \mbox{\boldmath$\beta$} \times {\bf B}) -
...a^2}{\gamma + 1} \mbox{\boldmath$\beta$}(\mbox{\boldmath$\beta$} \cdot {\bf E})$ (16.176)
$\displaystyle {\bf B}'$ $\textstyle =$ $\displaystyle \gamma({\bf B} - \mbox{\boldmath$\beta$} \times {\bf E}) -
...2}{\gamma + 1} \mbox{\boldmath$\beta$}(\mbox{\boldmath$\beta$} \cdot {\bf B}) .$ (16.177)

A purely electric or magnetic field in one frame will thus be a mixture of electric and magnetic fields in another. We see that truly, there is little reason to distinguish them. We have to be a little careful, of course. If there is a monopolar (static) electric field in any frame, we cannot transform it completely into a magnetostatic field in another, for example. Why? Because the equations above will lead to some mixture for all $\beta < 1$, and $\beta < 1$ in nature as a constraint.

I encourage you to review the example given in Jackson and meditate upon the remarks therein. We will not spend valuable class time on this, however. Instead we will end this, after all, purely mathematical/geometrical kinematical interlude (no Hamiltonians or Lagrangians = no physics) and do some physics. Let us deduce the covariant dynamics of relativistic particles in (assumed fixed) electromagnetic fields.

next up previous contents
Next: Relativistic Dynamics Up: The Lorentz Group Previous: Covariant Formulation of Electrodynamics   Contents
Robert G. Brown 2007-12-28