# Relatvistic Electrodynamics

1. Mar 14, 2005

### touqra

I hope I am in the right forum.

In relativistic electrodynamics, if the electric, E and magnetic, B fields are perpendicular to each other in reference frame S, can there be a reference frame, S' where E' = 0? If so, what will be the relative velocity, v?

2. Mar 14, 2005

### pmb_phy

Yes. A sheet of current might just be an example of such a case. I havent' aanalyzed it so you might want to give it a whirl and see for yourself.

Pete

3. Mar 14, 2005

### Andrew Mason

The short answer is No. An em wave will always appear as an em wave travelling at c regardless of the speed of the observer. The energy or wavelength of the light will be affected by the speed of S' relative to S. But it will always have E and B fields at right angles to each other propagating at the speed of light.

AM

4. Mar 14, 2005

### pmb_phy

The short answer is yes.

In matter the permittivity of free space is $\epsilon_0$ and the permittivity of free space is $\mu_0$. The speed of light in matter is a function of both and has the value c = $1/\sqrt{\epsilon_0\mu_0}}$. These for zero for water.

Pete

5. Mar 14, 2005

### Andrew Mason

I am not sure what you mean by your last sentence. Besides, the question asked about a frame of reference change. You are adding a change of medium as well.

Furthermore, the speed of a photon is always c, even in water.

AM

6. Mar 14, 2005

### krab

Yes. The velocity of the frame is v=E/B.

edit: Frankly, I don't know what the other guys are talking about...

7. Mar 14, 2005

### SpaceTiger

Staff Emeritus
My impression is that he's talking about a given instant in time, not a time-averaged wave. If you freeze an EM wave and consider the directions of the E and B vectors, it seems reasonable to think that there's a frame in which the E vector is 0, at least for that instant.

8. Mar 14, 2005

### dextercioby

Search for the LT applied to the $$F_{\mu\nu }$$ tensor and written in component form.I'm sure the electric field can't be zero in a certain reference frame.But one of his components can.

Daniel.

Last edited: Mar 14, 2005
9. Mar 14, 2005

### Andrew Mason

In other words, No. There is no inertial frame in which v = E/B. E/B = c.

AM

10. Mar 14, 2005

### dextercioby

I documented my answer and agree with Andrew.I advise everyone to do the same.A look in Jackson's book will be more than enough.

Daniel.

11. Mar 14, 2005

### SpaceTiger

Staff Emeritus
That's only a special case of light in a vacuum. More generally,

$$\bold{\overline{E}}=\bold{v} \times \bold{\overline{B}}$$

When E=0 in another frame, that's the electric field in the non-zero frame. It's equation 12.110 of Griffiths.

12. Mar 14, 2005

### krab

So what do I do with the devices in my lab called Wien filters? Pretend they don't exist? Sorry I'm still not getting why you don't get it. And who said v=c?

Here's the problem as I understand it from post #1: You have a constant electric field E and a constant magnetic field B. They are at right angles to each other.

Here's the question as I understand it: Can you travel at a speed where there is no electric field?

Answer: The force on a moving charged particle is qE from electric and qvB from magnetic (q is the charge, v is the velocity). They are anti-parallel by choice of the sign of v. So the total force when traveling at velocity v is zero. Now let's go into the frame that has velocity v. Our particle now has v=0 so magnetic force is zero. Also, it continues on undeflected, as we know from the lab frame. Therefore, E in the primed frame must be zero.

The only connection I see with what AM and Daniel are saying is that if E/B is too large, specifically, larger than c, then there no longer exists a frame where E'=0.

13. Mar 14, 2005

### Tom Mattson

Staff Emeritus
I think that Andrew is thinking about electromagnetic waves, in which E/B=c. But of course that's not true for an arbitrary E and B field.

14. Mar 14, 2005

### SpaceTiger

Staff Emeritus
...in a vacuum.

15. Mar 15, 2005

### touqra

I think I agree with krab, SpaceTiger and Tom...

16. Mar 15, 2005

### marlon

Yes, if v = E/B

Then you have two forces coming from the electric and magnetic field E and B : qE and vqB.

So chose v so that the total force on the particle is zero : hence v=E/B.

There is one problem though : you need to be sure that qE and qvB have opposite sign. This is easy to accomplish if you know that vB is a vector perpendicular to the plane defined by v and B. So if E is along X and B is along Y then v is negative with respect to the Z-axis.

In the labframe, the particle's trajectory is undeflected.

No goto the frame with velocity v=E/B. The particle has v = 0, so qvB is zero. Only qE remains but the trajectory is undeflected so E needs to be 0

marlon

17. Mar 15, 2005

### Andrew Mason

You are right on both counts. I was pretty sure that was what the questioner had in mind, but I see from his subsequent posts that he had in mind static fields.

AM

18. Mar 15, 2005

### reilly

I've not thought this through fully, but consider, for simplicity, two || long wires, one with + charges with v, and - charges with -v, and close together. Both have + current, hence there Bs will reinforce. But their E fields will substantially cancel. Build a plane surface, a la pmb's suggestion, with these dipole wires and one should clearly get a non-zero magnetic field, and a very small E field.

Regards,
Reilly Atkinson

19. Mar 15, 2005

### Andrew Mason

Another way to look at it is that the moving particle sees each of the two fields as part magnetic and part electric. This determined by the Lorentz transformation applied to the rest frame of the particle. It is just that if v = E/B in the lab frame, the particle measures the two electric fields as equal and opposite so they cancel each other. (One can work this out by applying the Lorentz transformation ot the particle frame). Since the particle does not see itself as having a magnetic field, it experiences no interaction with either of the magnetic fields.

So long as v<c, this is possible. Otherwise, as krab has pointed out, there is no inertial frame in which the electric fields result in a net E of 0.

AM

20. Mar 17, 2005

### Crosson

Think of a long straight wire with a current through it. If I am at rest with respect to the wire, the is no electric field in my frame (wire is neutral).

Now suppose I start moving in the same direction as the current. As I speed up to relativistic speeds, I will observe the distance between positive charges as less contracted then distance between negative charges. The negatives are going in the opposite direction, so they have more length contraction!

This puts a net charge on the wire and I observe an electric field in this frame. All the E and B fields in this example were perpendicular, does this answer the original question?

21. Mar 17, 2005

### pervect

Staff Emeritus
I'm with all the people who answered "yes".

If you take the Faraday tensor

\left[ \begin {array}{cccc} 0&-{\it Ex}&-{\it Ey}&-{\it Ez}\\\noalign{\medskip}{\it Ex}&0&{\it Bz}&-{\it By}\\\noalign{\medskip}{ \it Ey}&-{\it Bz}&0&{\it Bx}\\\noalign{\medskip}{\it Ez}&{\it By}&-{ \it Bx}&0\end {array} \right]

and do a boost in the z direction (using a system of units where c=1) you get

\left[ \begin {array}{cccc} 0&{\frac {-{\it Ex}+{\it By}\,v}{\sqrt {1-{v}^{2}}}}&-{\frac {{\it Ey}+{\it Bx}\,v}{\sqrt {1-{v}^{2}}}}&-{\it Ez}\\\noalign{\medskip}-{\frac {-{\it Ex}+{\it By}\,v}{\sqrt {1-{v}^{2 }}}}&0&{\it Bz}&{\frac {{\it Ex}\,v-{\it By}}{\sqrt {1-{v}^{2}}}} \\\noalign{\medskip}{\frac {{\it Ey}+{\it Bx}\,v}{\sqrt {1-{v}^{2}}}}& -{\it Bz}&0&{\frac {{\it Ey}\,v+{\it Bx}}{\sqrt {1-{v}^{2}}}} \\\noalign{\medskip}{\it Ez}&-{\frac {{\it Ex}\,v-{\it By}}{\sqrt {1-{ v}^{2}}}}&-{\frac {{\it Ey}\,v+{\it Bx}}{\sqrt {1-{v}^{2}}}}&0 \end {array} \right]

BTW - I had this in a maple worksheet, so I didn't really have to do much work to give such a detailed answer - maple also took care of the latex formatting. In case the formatting isn't too legible, the transformed components of the electric field are

Ex' = (-Ex + By * v)/sqrt(1-v^2)
Ey' = -(Ey + Bx*v)/sqrt(1-v^2)
Ez' = Ez

If the original frame has only components Ex and By, it satisfies the condition that E and B are perpendicular.

If Ex = By*v, the transformed frame has no electric field.

Thus it's possible to have E and B perpendicular in a one frame, and to have E=0 in another by direct construction.

Obviously, v has to be less than 1 (less than c) for this to be possible.

Note: the webpage

http://scienceworld.wolfram.com/physics/ElectromagneticFieldTensor.html

also has the same information.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook