## Question

Using the tensor transformation law applied to ##F_{\mu\nu}##, show how the electric magnetic field 3-vectors ##E## and ##B## transform under

a) a rotation about the ##y##-axis,

b) a boost along the ##z##-axis.

## Answer

Electric and magnetic fields increase and contra-rotate perpendicular to direction of relative motion. Boost is relative speed in ##z##-direction as a fraction of ##c## speed of light. |

F_{\mu\nu}=\left[\begin{matrix}0&-E_1&-E_2&-E_3\\E_1&0&B_3&-B_2\\E_2&-B_3&0&B_1\\E_3&B_2&-B_1&0\\\end{matrix}\right]=-F_{\nu\mu}

$$and the Lorentz rotation transformation matrix $$

\Lambda_{\ \ \nu}^{\mu^\prime}=\left[\begin{matrix}1&0&0&0\\0&\cos{\theta}&0&\sin{\theta}\\0&0&1&0\\0&-\sin{\theta}&0&\cos{\theta}\\\end{matrix}\right]

$$and the Lorentz transformation under a boost ##v## along the ##z##-axis $$

\Lambda_{\ \ \mu}^{\mu^\prime}=\left[\begin{matrix}\cosh{\phi}&0&0&-\sinh{\phi}\\0&1&0&0\\0&0&1&0\\-\sinh{\phi}&0&0&\cosh{\phi}\\\end{matrix}\right]

$$Where the boost parameter ##\phi=\tanh^{-1}{v}##.

There are many ways to solve this problem and I learnt many lessons.

### Method 1

The Cartesian rotation matrix by geometry. |

\left(\begin{matrix}{X^\prime}_1\\{X^\prime}_2\\{X^\prime}_3\\\end{matrix}\right)=\left(\begin{matrix}X_1\cos{\theta}+X_3\sin{\theta}\\X_2\\-X_1\sin{\theta}+X_3\cos{\theta}\\\end{matrix}\right)

$$Put ##E## then ##B## into that and back into the equation for ##F_{\mu\nu}## and you get the answer. But that would be cheating and will not work for (b). On the other hand, it is useful to check the answer.

### Method 2

I found a crib-sheet that helped me here: One really should apply ##\Lambda## to ##F## but as it stands ##\Lambda_{\ \ \nu}^{\mu^\prime}## will only work on ##F^{\mu\nu}## not on ##F_{\mu\nu}##. So one needs to get ##\Lambda_{\mu^\prime}^{\ \ \ \ \nu}## to operate on##\ F_{\mu\nu}## or ##F^{\mu\nu}## to be operated on by ##\Lambda_{\ \ \nu}^{\mu^\prime}##. We have$$\ F^{\mu\nu}=\eta^{\nu\sigma}\eta^{\mu\rho}F_{\rho\sigma}

$$then we would get$$

F^{\mu^\prime\nu^\prime}=\Lambda_{\ \ \mu}^{\mu^\prime}\Lambda_{\ \ \nu}^{\nu^\prime}F^{\mu\nu}

$$and one can find the transformed ##E,B## by comparing ##F^{\mu^\prime\nu^\prime}## and ##F^{\mu\nu}##.

The crib-sheet recommended to just do the sums that are implicit in the repeated ##\mu,\nu## indices in second equation.

**It is definitely easier to do two separate matrix multiplications!**One has to be careful with where the indices are. I learnt about that as well.

### Method 3

Just before I finished I proved that the transformation ##\ \Lambda_{\mu^\prime}^{\ \ \ \ \nu}## to operate on the covariant ##F_{\mu\nu}## is simply the inverse of ##\Lambda_{\ \ \nu}^{\mu^\prime}## the original, useless, transformation. The inverse of ##\Lambda_{\ \ \nu}^{\mu^\prime}## can be found by making ##\theta\rightarrow-\theta## or ##\phi\rightarrow-\phi## - reversing the angle or boost. So the answer can be obtained with many fewer calculations.**Duh!**

Full answer at Ex 1.10 Transformations of Electro Magnetic Field Tensor.pdf (11 pages!)

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