Take idea of relativistic magnetism, for example. How can a magnetic field be generated due to the electrons undergoing length contraction relative to the positive charges if the electrons moving through the wire themselves have no length?


2 Answers 2


Their three-dimensional fields get squished.

  • $\begingroup$ For physics undergrads unfamiliar with quantum field theory but competent in basic quantum mechanics, is there a slightly more intuitively satisfying way to explain that concept? $\endgroup$
    – Dutonic
    Sep 2, 2022 at 2:24
  • $\begingroup$ @Dutonic This is a classic relativity problem, and quantum mechanics is irrelevant. The entire problem can be understood by transforming the linear motions of charges. I posted it on PSE, but don;t know how to link to it. There are reams of wrongheaded explanations on the internet, yet no one just does Lorentz transform. $\endgroup$
    – JEB
    Sep 2, 2022 at 3:38
  • 1
    $\begingroup$ @Dutonic niels means the (classical) electromagnetic field, not the quantum electron field. $\endgroup$
    – d_b
    Sep 2, 2022 at 3:40

I assume your talking about a positive ion lattice with free electrons moving at $\beta$. This is one of the most misunderstood SR problems on the internet.

In the wire lattice frame, the proton and electron densities are the same:


(by definition…it is stipulated in the problem that the wire is neutral in its rest frame). Hence, there is just a magnetic field $B$.

Now in a primed frame moving at $\beta$, so that the current electrons are at rest, the positive lattice is Lorentz contracted, and the density increases:

$$ \rho_+’ = \gamma\rho_+$$

Moreover, they are moving at $-\beta$, so they create a current:

$$ I_+=-\beta\rho_+’$$

Fine. It’s the electrons that confuse everyone.

They are not in a lattice. They are not one physical object. They are many independent objects that, when the current is turned on, accelerate with identical profiles in the initial rest frame of the wire. Again, this is stipulated b/c the wire remains neutral.

At this point, you need to master Bell’s spaceship paradox, where each electron is a spaceship accelerating uniformly in the wire frame. If you do not understand Bell’s spaceship paradox, you will not understand this problem. Period.

It means in any electron’s rest frame, the other electrons spread out and the charge density decreases to:

$$ \rho_-‘ = \rho_-/\gamma $$

With zero current from electrons:

$$ I’_- = 0 $$

Now use those charge and current densities to compute $B’$ and $E’$. You will find they are the exact transforms of $B$ and $E=0$, by $\beta$.

Everything works out just fine,as it always does.

In thinking about the problem: in the wire rest frame, the moving electrons pass the ions all simultaneously, so that in the electron restframe: they don’t. That might clear up apparent contradictions.


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