# When an electron is moving at relativistic speeds, is the electric field length-contracted as well?

Suppose I have an electron that is moving at relativistic speeds. SR says some degree of length contraction should take place.

Does this mean that the electric field around the electron should also be length contracted, e.g. instead of being spherically symmetric about the electron, it's now got an elliptical kind of shape? And if so, given that would mean things are "rolling off" faster in the direction of motion (to create such an elliptical shape), does that mean the length-contracted field is attenuated in some sense - since it rolls off faster - or does it somehow maintain the same net intensity?

The reason I ask is to try to make sense of questions like this about electrons in moving wires. Typically we talk about, e.g. protons or electrons being length-contracted from some other reference frame, causing some kind of net electric charge. But I don't quite get if we are viewing it as though the electric field around each particle is length contracted as well. (Otherwise, how could one length-contract an individual point?)

• Commented Jun 10 at 21:06

Yes, the electric field is length-contracted because it lives in physical space just like the charge itself. In other words, each point in space-time in which the electric field is present is a space-time event just like the point where the charge is located, and a Lorentz transformation affects all space-time events equally regardless of what is actually happening there.

To find the electric field of a charge moving at constant velocity, we consider an observer for whom the charge is constant; the field in that case is given by Coulomb's law. Then, for the observer who sees the charge moving at constant velocity, what they observe must be the Lorentz-boosted version of what the other observer observes. Because the boost will contract all lengths in the direction of the charge's velocity (since that's the direction of the boost) the electric field is affected as well.

Note, however, that not just the space in which the electric field lives is affected, but also the value of the field at each point, since the 3 components of the electric field are components of the order 2 electromagnetic tensor which also contains the magnetic field. Though it's not obvious from a purely mathematical basis what the result of the Lorentz transformation is on the value of the field (as most of us cannot visualize a Lorentz transformation of a tensor of order 2) it is somewhat more intuitive on physical grounds that the field must point directly toward/away from the charge just as it does when the charge is stationary. This is because two stationary charges acting on each other through their electrostatic fields don't experience any torque around a point that lies on the line between their positions (because the force is parallel to that line) and it would be absurd if there were such a torque in the boosted frame.

In effect, the electric field surrounding the moving charge looks as if you took the electric field lines of the stationary configuration and just squished them by a factor of $$\gamma$$.

• Thanks, that mostly answers it. The only thing I don't get is: does this mean the field strength is attenuated? For instance, if I get your explanation correctly, as $\gamma \to \infty$ the electric field is squished so much that it basically is nonzero only in the subspace orthogonal to the direction of motion. This would mean there is basically no electric charge in front of or behind the electron in its path of travel, only to the sides. Does the strength of the electric field somehow increase to preserve some idea of "net charge," or is the field just weaker for moving electrons? Commented Jun 11 at 6:01
• @MikeBattaglia Yes actually, I forgot to mention that. The orthogonal components of the field must also increase when boosted so that Gauss's law will still be satisfied. Commented Jun 11 at 14:01
• the electric field isn't squished. The longitudinal component is not attenuated; rather, the transverse component increases by $\gamma$. BTW, if this question was inspired by a recent Dialect yt video, you can just forget what he said in that. The idea of the field "pointing" to the charge being a-casual is negated by using 4-vector separations, where the time component automatically handles the relativity of simultaneity. Trying to "intuit it" in 3 + 1 is exactly how we get all our Special relativity Paradoxes, so being puzzled is a natural outcome.
– JEB
Commented Jun 13 at 16:28

For an electron that is moving at relativistic speeds, $${\bf E}=\frac{q\gamma{\bf r}}{({\bf r_\perp}^2 +\gamma^2{\bf r_\parallel}^2)^{\frac{3}{2}}} = \frac{q{\bf r}} {\gamma^2[r^2-({\bf v\times r})^2]^{\frac{3}{2}}}$$.

Although the direction of E is radial, It's strength is not spherically symmetric.

I don't think this is a fruitful way to look at it. If you want to look at fields, just transform charge distributions, say a line density of charge moving along its axis. All will be explained.

People always what to talk about electrons and drift velocities--none of that matters. Relativity does not depend on the microscopic quantum structure of matter. That only causes confusion.

Then when it comes to electrons: they're point particles: contraction is meaningless. You can't contract something with zero dimensions.

So you ask if "fields" are contracted. The Faraday tensor is an antisymmetric rank-2 4-tensors comprising $$\vec E$$ and $$\vec B$$, what does it even mean to contact that?

Rather, deal with things you can understand: charges and currents.

A stationary charge:

$$j^{\mu}(t) = (q, q\delta(x), q\delta(y), q\delta(z))$$

sources $$A^{\mu}$$, and that makes fields you can observe, Boost that.

But like I said in the beginning, starting with a linear charge density:

$$j^{\mu}(t) = (\lambda, \lambda\delta(x), \lambda\delta(y), \lambda)$$

and boosting along:

$$\vec v = (0, 0, \beta c)$$ is easiest.