# I Why transformed E and B are asymmetric?

1. Nov 7, 2017

### Luis Babboni

Hi people,

Thanks to use your time in me!

I saw and understood the maths to arrive to this:

Where primed fields are the electromagnetics fields saw in a system S´in movement respect another system S with velocity in X axe thats coincides with X´s axe.

My question is about a thing that bother me that is the assimetry I saw in Zs and Ys components with a plus in one and a minus in the other instead of minus or plus in both.

I can´t understand the physical reason behind.

Thanks!

2. Nov 8, 2017

### A.T.

Consider two perpendicular unit vectors in the YZ plane, and how their components are related.

3. Nov 8, 2017

### vanhees71

The particular signs in the transformation come from the realization of the Lorentz group on the field components $\vec{E}$ and $\vec{B}$. You can understand this in two ways. The first is using the tensor representation of the electromagnetic field, i.e., the Faraday tensor in Minkowski space. In terms of the potentials its defined by
$$F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}.$$
In 3D vector calculus on the other hand you have (I'm using Heaviside Lorentz units)
$$\vec{E}=-\frac{1}{c} \partial_t \vec{A} - \vec{\nabla} A^0, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
Now you can relate the $F_{\mu \nu}$ with $\vec{E}$ and $\vec{B}$ by splitting the Faraday tensor in temporal and spatial components (everything is of course in an inertial reference frame):
$$F_{0j}=\frac{1}{c} \partial_t A_j - \partial_j A_0=-\frac{1}{c} \partial_t A^j - \partial_j A^0=E^j,\\ F_{jk}=\partial_j A_k - \partial_k A_j=-\partial_j A^k + \partial_k A^j=-\epsilon^{jkl} B^l.$$
If ${\Lambda^{\mu}}_{\nu}$ is the matrix of the Lorentz transformation $x'=\hat{\Lambda} x$, you can get the transformation of the $\vec{E}$ and $\vec{B}$ by applying the Lorentz transformation to the Faraday-tensor components and translate the result back into the 3D formalism:
$$F^{\prime \mu \nu}(x')={\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} F^{\rho \sigma}(x) = {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} F^{\rho \sigma}( \hat{\Lambda}^{-1}x').$$
Another, much more elegant way is to introduce the complex valued 3D Riemann-Silberstein vector
$$\vec{F}=\vec{E}+\mathrm{i} \vec{B}.$$
You can show by direct comparison with the above given covariant method that this vector transforms under Lorentz transformations with a $\mathrm{SO}(3,\mathbb{C})$ matrix $\hat{L}$. For real rotation angles you get of course a usual real orthogonal $\mathrm{SO}(3)$ matrix, and $\vec{E}$ and $\vec{B}$ of course transform as vectors under rotations. For purely imaginary rotation angles you get Lorentz boosts, and the imaginary rotation angle becomes $\mathrm{i} \eta$, where $\eta$ is the rapidity of the boost, related to the velocity of the boost by $\tanh \eta=v/c$, i.e.,
$$\vec{F}'(x')=\hat{L} \vec{F}(x)=\hat{L} \vec{F}(\Lambda^{-1} x').$$

4. Nov 8, 2017

### Luis Babboni

Sorry, not understand. :-(
But.... why perpendicular vectors?
As far as I understood, those E and B vectors could be even parallels. Are any possible general fields. No?

5. Nov 8, 2017

### Luis Babboni

But even not understand your explanation, may be 30 years ago I could did it but now after 30 years of not doing more than sums and substracts, I´m near to need to start learing math again. I feel that the "physics" behind those equatinos remains hide.
I mean, I do not see any asymmetry in the situation that justify this asymetri in the fields.... or there is not any asymmetry?

Imagine B=(1,1,1) and E=(1,1,1), then:

B´x´ = 1
By= gamma (1 + beta / c )
B´z´= gamma (1 - beta / c )

So B´y´> B´z´ ! :-/

Where is my mistake?

Thanks again and sorry for ask for what I bet is a stupid confusion.

6. Nov 8, 2017

### Luis Babboni

But even not understand your explanation, may be 30 years ago I could did it but now after 30 years of not doing more than sums and substracts, I´m near to need to start learing math again. I feel that the "physics" behind those equatinos remains hide.
I mean, I do not see any asymmetry in the situation that justify this asymetri in the fields.... or there is not any asymetry?

7. Nov 8, 2017

### Luis Babboni

I think I could explain my problem a little better.
See this:

You have for O Ex = Ey = Ez and By = Bz and Bx = 0.
For O´moving to positive Xs axes with velocity v, E´x´ = Ex and E´y´< Ey and E´z´ > Ez.. . being Ey = Ez.
So my problem is that E´z´> E´y´.... why?! I see a completely axial simetry around X axe that coincides with X´ axe.

8. Nov 8, 2017

### Luis Babboni

Mmmmm.... there is not completely simmetry!!

9. Nov 8, 2017

### Luis Babboni

Understood!! :-)

How to edit the tittle to say I understood it?