tl;dr

Here is a haiku:

Charge remains the same

with respect to speed, but the

EM waves change form.

Could your electronics tell if you were trapped on a relativistic spaceship?

I had the thought the other day - if you were in a closed room in a relativistic space ship and all you had were basic home appliances and electronics, could you tell that you were traveling at relativistic speeds? I thought I would try to answer this in the style of xkcd’s Randall Munroe.

We know that the Doppler effect applies as a simple application of special relativity with respect to Doppler shifts of emitted radiation of stars that are moving towards/away from us. This is a simple consequence of time dilation from special relativity, where the dilated time $t’$ is: $$ t’ = t \; \sqrt{1-v^2/c^2} $$ where $t$ is time in a stationary frame, $v$ is the velocity of the relativistic frame and $c$ is the speed of light.

You learn in E&M class that Maxwell’s equations play nicely with special relativity (gravity and relativity make up the more difficult field of general relativity). Applying relativity to Maxwell’s equations is a well known problem using the electromagnetic tensor - a 4 dimensional tensor of $(x,y,z,t)$ first introduced by Minkowski. The EM 4-tensor $F$ can be written as:

$$ F = \begin{bmatrix} 0 & -E_x/c & -E_y/c & -E_z/c & \\ E_x/c & 0 & -B_z & B_y \\ E_y/c & B_z & 0 & -B_x \\ E_z/c & -B_y & B_x & 0 \end{bmatrix} $$

The EM tensor $F$ is defined as the gradient of the four-potential $A$ (in the same way for the non-relativistic case where the standard electric field $E = -\nabla V$ is the gradient of the electromagnetic potential). From the EM tensor, we can get out the $E,B$ fields as: $$ E_i = c F_{0,i} $$ $$ B_i = -1/2 \epsilon_{ijk} F_{j,k} $$ where $\epsilon_{ijk}$ is the Levi-Civita tensor. It is a good reminder to remember that tensors as essentially generalized matrices - however, some physics tensors occupy a large $n$-space that are represented in $d>2$ dimensions.

When dealing with relativistic problems, it is important to remember that charge and mass are Lorentz invariant - i.e. mass and charge are the same in all inertial reference frames. This means that observers in different reference frames will observe the same mass/charge for a relativistic particle. However, our EM tensor will tell us that the observed EM field from a relativistic particle will be different for a stationary observer versus a moving observer at relativistic speeds.

Back to our problem - let’s start with the case of an electron at rest with respect to our relativistic space ship traveling at velocity $v$. In the rest frame of the space ship, the electron is stationary and the magnitude electric field is $$ E = \frac{1}{4\pi\epsilon_0} \frac{e}{r^2} $$ where $r$ is the distance from the electron (yes, I know that the electric field is actually a vector, but I am only going to consider the magnitude). However, in a stationary rest frame, we know that the magnetic field is the cross product $$ B = (1/c^2) \, v \times E $$ where the direction is given by the right-hand rule.

Conclusion

What this means is that for our observer on the relativistic spaceship, everything will appear to be behaving the same as if they were at rest. However, all of the emitted EM radiation from the electronics on the spaceship to an outside observer will be redshifted/blueshifted in frequency and also the relative magnitude of the $E,B$ fields. Something like a cell phone would still work - battery, PCBs, etc. all would still behave the same to the observer within the space ship. Good luck get any bars though…