What Will an Observer Measure When Riding on a Moving Electron in a Wire?

[marcusl]

I am puzzled by all the confusion in some replies in this thread. The answer above is the only answer consistent with SR. Is that people don't believe SR or that they don't understand SR?

Thank you Meir. I also cannot understand what the issue is in this thread...

 

[masudr]

Meir Achuz, marcusl:

So B' = \gamma B

What happens in the non-relativistic limit, as \gamma \rightarrow 1?

B' = B.

That's part of what is being discussed here. No one is denying the transformation properties of electromagnetic fields.

 

[marcusl]

E and B mix under boost (they are components of electromagnetic tensor), as pervect noted earlier in this thread. If \gamma = 1, the test charge is at rest and cannot detect any magnetic force; SR applies for every other value. There is no non-relativistic limit for magnetism! Classical magnetism comes from Coulomb's Law and SR, and is entirely a relativistic effect even at low speeds. That's the end of the story.

 

[masudr]

If the test charge is moving at v = 100 m/s, \gamma = 1

 

[marcusl]

I assume you know that's wrong. Gamma is actually about 1+5e-13. The small size of (v_drift/c)^2 explains why the magnetic force between two current carrying wires is so much smaller than electric force for comparable amounts of charges (moving charges in one case, stationary charge in the other). When \gamma = 1 the magnetic force is identically zero.
Edit: sorry, meant 1 + 5e-14.

 

[pervect]

Another way to simply work out the charge density and show that it varies when you do a boost (and work out quantitatively HOW it varies)

The 4-current vector - in MKS units (c rho, Jx, Jy, Jz) transforms as a 4-vector

Here J is the current density (amps/m^2), rho is the charge density (columbs/meters^3).The invariant of this 4-vector is [correction] (c * rho)^2 - (Jx^2 + Jy^2 + Jz^2). The factors of c are annoying, but probably less annoying than dealing with non-standard units.

See for instance Griffiths , "Introduction to Electrodynamics", or the wikipedia http://en.wikipedia.org/wiki/4-current

 

[masudr]

So, the bottom line is, if you have a wire with current, you cannot transform, via a boost, into any inertial frame whereby B stays the same, even for boosts corresponding to non-relativistic velocities.

I shall write out my calculations:

We have current flowing down a wire along the x-axis. This gives

j^a = [c\rho, J_x,0,0]

and a boost along the x-axis of v=\beta c gives

j'^a = [c\gamma\rho-\beta\gamma{J_x}, -c\beta\gamma\rho + \gamma{J_x},0,0]

We try and choose a \beta such that J'_x=0:

<br /> \begin{array}{rl}<br /> -c\beta\gamma\rho + \gamma J_x &amp;= 0 \\<br /> c\beta\gamma\rho &amp;= \gamma J_x \\<br /> \beta &amp;= J_x/c\rho \\<br /> v &amp;= J_x/\rho<br /> \end{array}<br />

This means the charge density in the boosted frame, c\rho&#039; is

<br /> \begin{array}{rl}<br /> c\rho&#039; &amp;= c\gamma\rho-\beta\gamma{J_x} \\<br /> &amp;= c\gamma\rho-J_x/c\rho\cdot\gamma J_x \\<br /> &amp;= \gamma(c\rho-J_x^2/c\rho) \\<br /> \rho&#039; &amp;= \gamma(\rho-J_x^2/\rho c^2)<br /> \end{array}<br />

So, my conclusion is that in a frame where we can make the current vanish, there is still some effect left over from the charge density contraction, which means there is no frame where the B-field remains constant and the E-field remains zero.

Thanks to all (notably pervect) for their patience.

It'd be nice if I could finally show the E-field emerging from this contracted charge density.

 

[bernhard.rothenstein]

relativistic electrodynamics

So, the bottom line is, if you have a wire with current, you cannot transform, via a boost, into any inertial frame whereby B stays the same, even for boosts corresponding to non-relativistic velocities.

I shall write out my calculations:

We have current flowing down a wire along the x-axis. This gives

j^a = [c\rho, J_x,0,0]

and a boost along the x-axis of v=\beta c gives

j&#039;^a = [c\gamma\rho-\beta\gamma{J_x}, -c\beta\gamma\rho + \gamma{J_x},0,0]

We try and choose a \beta such that J&#039;_x=0:

<br /> \begin{array}{rl}<br /> -c\beta\gamma\rho + \gamma J_x &amp;= 0 \\<br /> c\beta\gamma\rho &amp;= \gamma J_x \\<br /> \beta &amp;= J_x/c\rho \\<br /> v &amp;= J_x/\rho<br /> \end{array}<br />

This means the charge density in the boosted frame, c\rho&#039; is

<br /> \begin{array}{rl}<br /> c\rho&#039; &amp;= c\gamma\rho-\beta\gamma{J_x} \\<br /> &amp;= c\gamma\rho-J_x/c\rho\cdot\gamma J_x \\<br /> &amp;= \gamma(c\rho-J_x^2/c\rho) \\<br /> \rho&#039; &amp;= \gamma(\rho-J_x^2/\rho c^2)<br /> \end{array}<br />

So, my conclusion is that in a frame where we can make the current vanish, there is still some effect left over from the charge density contraction, which means there is no frame where the B-field remains constant and the E-field remains zero.

Thanks to all (notably pervect) for their patience.

It'd be nice if I could finally show the E-field emerging from this contracted charge density.

You can find the derivations presented above in W.G.V. Rosser "Classical Electromagnetism via Relativity" London Butterworth 1968 p.165(old man has old editions). He starts with two relativistic identities obtained by expressing
1/(1-uu/cc)^1/2 and u/(1-uu/cc)^1/2 as a function of u' via the addition law of relativistic velocities. Multiplying both sides of the mentioned identities with the proper value of the charge density he obtains your equations. I have tried to extend Rosser's method multiplying both sides of the identities with the proper values of the OY component of the electric field, obtaining directly the transformation equations for E and B. Did you see such a derivation in the literature of the subject? If you are interested in my approach I could give you a link to it. If not elegant it is time saving and transparent showing how relativity relates different physical quantities introduced in order to characterize electric charges and the fields produced by them when they move.
sine ira et studio

 

[masudr]

Did you see such a derivation in the literature of the subject? If you are interested in my approach I could give you a link to it.

I have not seen such a derivation; I would very much appreciate such a link.

 

[bernhard.rothenstein]

relativistic electrodynamics

I have not seen such a derivation; I would very much appreciate such a link.

Please have a critical look at

arXiv.org > physics > physics/0505130
Physics, abstract
physics/0505130


Relativistic velocity transformation as a genitor of transformation equations (electrodynamics)
Authors: Bernhard Rothenstein, Aldo De Sabata
Comments: 6 pages
Subj-class: Physics Education

We show that invariance of the electric charge and relativistic kinematics lead to the transformation equations for electric field intensity and the magnetic induction.
My invitations are allways for a critical look.

 

[Meir Achuz]

I guess my comment meant that, in SR, you can usually do something very simply or you can try to do it in a complicated "interesting" (critical?) way, leading to this inexhaustible thread. That is like Iraq. The simple solution is to get out today. The Baker commision has 79 more interesting options.