Why transformed E and B are asymmetric?

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  • Thread starter Luis Babboni
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Sorry, I am an AI and I do not have the ability to edit titles. You can simply mention in your response that you have understood the explanation. Is there anything else I can assist you with?
  • #1
Luis Babboni
Hi people,

Thanks to use your time in me!

I saw and understood the maths to arrive to this:
Transformaci%C3%B3n%20B%20y%20%20E%20asim%C3%A9trica.jpg

Where primed fields are the electromagnetics fields saw in a system S´in movement respect another system S with velocity in X axe that's 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!
 

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  • #2
Luis Babboni said:
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.
Consider two perpendicular unit vectors in the YZ plane, and how their components are related.
 
  • #3
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
A.T. said:
Consider two perpendicular unit vectors in the YZ plane, and how their components are related.

Thanks A.T. for your time!

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
Thanks for your time vanhees71!

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
B`y`= 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
Thanks for your time vanhees71!

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


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.
 

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  • #8
Mmmmm... there is not completely simmetry!
 
  • #9
Understood! :-)

How to edit the tittle to say I understood it?
 

1. Why are transformed E and B fields asymmetric?

The asymmetry of transformed electric and magnetic fields is due to the fact that they are not independent of each other. This is a fundamental characteristic of electromagnetic fields known as the Maxwell-Faraday equation, which states that a changing magnetic field induces an electric field, and vice versa. Therefore, the asymmetry arises from the interdependent nature of these fields.

2. How does the asymmetry of transformed E and B fields affect electromagnetic waves?

The asymmetry of transformed E and B fields is essential for the generation and propagation of electromagnetic waves. The changing electric field induces a magnetic field, which in turn induces an electric field, and this cycle continues to create a self-sustaining electromagnetic wave. Without this asymmetry, electromagnetic waves could not exist.

3. Are there any real-world applications that rely on the asymmetry of transformed E and B fields?

Yes, there are several real-world applications that rely on the asymmetry of transformed E and B fields. One example is in wireless communication, where electromagnetic waves are used to transmit information. Another example is in electromagnetic imaging techniques, such as magnetic resonance imaging (MRI), which utilize the asymmetry of transformed E and B fields to produce images of internal body structures.

4. How does the asymmetry of transformed E and B fields contribute to the behavior of light?

The asymmetry of transformed E and B fields plays a crucial role in the behavior of light. The varying electric and magnetic fields in an electromagnetic wave are perpendicular to each other and to the direction of propagation. This results in the wave's transverse nature, which allows light to be polarized and exhibit other unique properties.

5. Can the asymmetry of transformed E and B fields be altered or controlled?

Yes, the asymmetry of transformed E and B fields can be altered or controlled in various ways. For example, the use of optical materials with different refractive indices can change the speed of light and alter the asymmetry of the fields. Additionally, advanced technologies such as metamaterials can manipulate the asymmetry of transformed E and B fields to achieve desired effects, such as cloaking or super-lensing.

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