Dot product of electric and magnetic field conserved in special relativity?

[ralph961]

An inertial reference frame 2 is moving along the x-axis with constant velocity v with respect to inertial reference frame 1.
......->-> ->->
How can i prove the E.H = E'.H' ?? (dot product)
using the 4 dimensional (Ax,Ay,Az,phi)
where E = -1/c dA/dt - gradiant(phi)
and H = curl(A)
Where E is the electric field and H the magnetic field.

 

[Andrew Mason]

An inertial reference frame 2 is moving along the x-axis with constant velocity v with respect to inertial reference frame 1.
......->-> ->->
How can i prove the E.H = E'.H' ?? (dot product)
using the 4 dimensional (Ax,Ay,Az,phi)
where E = -1/c dA/dt - gradiant(phi)
and H = curl(A)
Where E is the electric field and H the magnetic field.

If they weren't the same, inertial frames would not be equivalent.

The principle of relativity states that the laws of physics are the same in all frames of reference. The Lorentz transformations for Maxwell's equations were developed by applying that principle. So by applying the Lorentz transformations you don't prove the principle of relativity, you just confirm that the transformations were derived correctly.

The answer to your question is contained in http://www.fourmilab.ch/etexts/einstein/specrel/www/#SECTION20"

AM

 

[ralph961]

Thank you for replying,
It is noted by my professor that:
Note that this result is highly non-trivial since usually the 3-dimensional scalar
product is NOT preserved under the 4-dimensional Lorenz transformations, so the case of the scalar product (E.H) is very special - it is the same in all the reference frames. This important scalar product is known as one of the invariants of the electromagnetic field.

I need to prove that E.H= E'.H', not by theory but i have to derive that they are equal using lorentz transformation.
Is that possible and how?

 

[StatMechGuy]

Are you familiar with the electromagnetic tensors F^{\alpha \beta}? If you are, this is a result of that tensor and the tensor \mathfrak{F}^{\alpha \beta} being multiplied together, and taking the trace, and such.

 

[ralph961]

No I'm not familiar with that.
Sorry..
So the dot product E.B is always 0 in all reference frames?

 

[ralph961]

Special relativity help

A body travels at a speed c/10 from point A to point B distant 3 light years.
Is the time of the event "arrival of the body at B" with respect to the inertial reference frame at A, 30 years or 30.15 years?

 

[nrqed]

Thank you for replying,
It is noted by my professor that:
Note that this result is highly non-trivial since usually the 3-dimensional scalar
product is NOT preserved under the 4-dimensional Lorenz transformations, so the case of the scalar product (E.H) is very special - it is the same in all the reference frames. This important scalar product is known as one of the invariants of the electromagnetic field.

I need to prove that E.H= E'.H', not by theory but i have to derive that they are equal using lorentz transformation.
Is that possible and how?

Yes. You only need to use the transformations laws of H and E. Do you have those equations? It's just plugging in.

 

[ralph961]

I have the transformation laws of (Ax,Ay,Az,phi) which are like the (x,y,z,ct)
A=(Ax,Ay,Az) and phi are defined by
E = -1/c dA/dt - gradiant(phi)
and H = curl(A)

 

[Flux = Rad]

The only way I've seen it proved is with 4-tensors. However, maybe you have the results of the 4-tensor analysis without the analysis itself. Have you seen:
<br /> \vec{E}^\prime_\perp = \gamma_0 (\vec{E}_\perp + \vec{\beta}_0 \times c \vec{B}_\perp)<br />
<br /> c \vec{B}^\prime_\perp = \gamma_0 (c \vec{B}_\perp - \vec{\beta}_0 \times \vec{E}_\perp)<br />
<br /> E^\prime_\parallel = E_\parallel<br />
<br /> B^\prime_\parallel = B_\parallel<br />

These are all derived from knowing how the 4 vector you've mentioned (A) transforms, and how it relates to EM fields. If you've seen these, you can definitely use them to show that E.B is invariant.

 

[Mentz114]

Fixed your tex -

c \vec{B}^\prime_\perp = \gamma_0 (c \vec{B}_\perp - \vec{\beta}_0 \times \vec{E}_\perp)