# EM Fields in a Rotating Frame of Reference

I'm struggling on my approach to the problem of figuring out E and B fields in a non-relativistic way for a rotating frame of reference in the x-y plane around the z-axis. I am attempting to do this without the use of Tensors:

Consider a particle at rest with respect to the rotating frame. Since there is no velocity there is no force felt from $\vec{B'}$. Do I have to include the pseudo forces in this equation? $$\vec{F} = m \vec{a} = q \vec{E'} = q(\vec E+\vec v \times \vec B) \\ \vec{E'} = \vec E+\vec v \times \vec B$$

Additionally, considering a stationary particle in the inertial frame of reference, with pseudo forces included: $$\vec{F} = m \vec a + m \vec \omega \times \vec v + m \vec \omega \times \vec v + m\vec \omega \times \vec \omega \times \vec x \\ = q(\vec{E'}+\vec v \times \vec{B'}) = q\vec E$$

However if I just plug in $\vec{E'}$ into the second equation, $\vec{B'}$ = $\vec{B}$ which seems incorrect. Looking at the non-relativistic portions of a Lorentz transformations has $\vec{B'}$ = $\vec{B} - \vec v \times \vec{E}$

• @ TTa: "I am attempting to do this without the use of Tensors" It's going to be hard that way. I believe that that your best chance is to try calculating the 4-potentials, transforming, and then re-deriving the expressions for $\vec E$ and $\vec B$ – Hydro Guy May 5 '13 at 19:27
• You definitely should not do this without tensors. It will be easier to learn tensors and then do this problem than to do this problem without them. And afterwards you will be able to apply your knowledge of tensors to other problems – Dale Oct 10 '18 at 2:40

## 3 Answers

If you want to persist in solving the problem without using relativity or Minkowski space, of course you can do that. I've been known to be stubborn myself in a similar way. But it can't hurt to know the answer first, so that working in your chosen method, you can eventually figure out if you're on the right track.

Let there be an electromagnetic bivector $f = \frac{E}{c} \wedge e^t + e^{xyz} B$. Let's check this against a current $j = \rho c e_t$:

$$F = f \cdot j = (\frac{E}{c} \rho c \wedge e^t) \cdot e_t = E \rho$$

so that checks, sign-wise.

We define the primed frame by the following transformations:

$$x'= x \cos \omega t - y \sin \omega t, \quad y' = y \cos \omega t + x \sin \omega t, \quad z' = z, \quad t' = t$$

Collectively, let $s = x e_x + y e_y + z e_z + ct e_t$, and let $s' = g(s)$. We find the Jacobian of the transformation:

\begin{align*}\underline g(e_t) &= \frac{1}{c} \partial_t s' = e_t + \frac{\omega}{c}(-y' e_x + x' e_y) \\ \underline g(e_x) &= \partial_x s' = e_x \cos \omega t + e_y \sin \omega t \\ \underline g(e_y) &= \partial_y s'= e_y \cos \omega t - e_x \sin \omega t \\ \underline g(e_z) &= \partial_z s' = e_z\end{align*}

Currents will transform according to the Jacobian, so you can easily choose a stationary particle in the stationary frame and map it to its four-current in the primed frame. To consider a stationary particle in the primed frame, however, you will need to compute the inverse. Yuck.

To compute the transformation of the Faraday field, we'll need the adjoint transformation--this is kinda like a transpose, but slightly different due to the metric. The adjoint is more fundamental, however--this is the same reason conjugate transposes are used in QM, instead of plain transposes.

\begin{align*} \overline g(e^t) &= e^t \\ \overline g(e^x) &= -y' \frac{\omega}{c} e^t + e^x \cos \omega t - e^y \sin \omega t \\ \overline g(e^y) &= x' \frac{\omega}{c} e^t + e^x \sin \omega t + e^y \cos \omega t \\ \overline g(e^z) &= e^z\end{align*}

Luckily, the determinant of this operator is $1$, so the inverse is equal to the adjoint. That actually simplifies things considerably! This is, as a result, an orthogonal transformation--it is a position dependent rotation. Not a surprise; we designed it that way. That means we can map from the original, unprimed space to the primed space using only $\underline g$ for all objects.

There are some immediate consequences of this. First, you should see that $B_z$ remains invariant--the xy plane is mapped to itself, and that is exactly the component that corresponds to $f_{xy}$. (Edit: $E_z$ however does not remain invariant, as $tz$ gets mixed with $xz$ and $yz$, so some electric effects appear magnetic in the rotating frame.)

One insight from relativity is that we should always think about the EM field as planes, and knowing which planes are left invariant by a transformation tells us immediately which components stay the same and which change.

I was actually doing something just like this earlier today, and yes, the metric tensor is the best way to go about this.

As for your approach, I am a little confused by your use of the Lorentz force law to define the electric and magnetic fields. It turns out, in a non-inertial frame, this isn't quite right. As was pointed out by Muphrid, fields get mixed and it's not necessarily clear what is $\vec{E}$ and what is $\vec{B}$ anymore. It may become clearer if you take a Lagrangian approach, so you can encode the rotating frame into your coordinates. But from there, you still have to go back to Maxwell's equations in covariant form.

May I ask why you want to avoid tensors?

It's ok to change $$E$$ and not $$B$$ to first order in $$v/c \ll 1$$. This is the Euclidean or galilean boost and group. The full relativistic stuff (Feynman lectures) transforms $$B$$ as well as $$E$$ and includes $$\gamma$$ and $$\beta$$ factors.