Thanks for that bcrowell, that was a useful thread.
This is what I came up with:
I bent a uniform linear charge density in a circle such that each charge is experiencing uniform circular motion. Two, actually, a positive one going clockwise, and a negative one going counter clockwise. Thus, the total current is 2 λ v in the CW direction.
Then, I wanted to represent this relativistically, so I built a four vector for both charge densities. Normally, they would be represented by J
α, but since these are linear charge densities, I use I
α:
$$I_{+}^{\alpha}=\lambda^{+}\left(c,-v\sin\phi,v\cos\phi,0\right)$$ and $$I_{-}^{\alpha}=-\lambda^{+}\left(c,v\sin\phi,-v\cos\phi,0\right)$$ which can be combined to find $$I_{tot}^{\alpha}=I_{+}^{\alpha}+I_{-}^{\alpha}= 2\lambda^+ v\left(0,-\sin\phi, \cos \phi,0\right)$$.
Near the center of the loop, close to the axis, the magnetic field is given by ##B=\frac{\mu_0 I}{2r}## with ##I=2\lambda^{+}v##. The zero in the first term of the four-vector shows that there is no net charge density, and thus no electric field anywhere.
What does this loop look like when boosted in the +x-direction (to the right in the figure) at speed ##u##? Simply perform the Lorentz transform ##L^{\alpha}_{\beta}I^{\beta}_{tot}## to obtain
$$I_{tot}^{' \alpha}=2\lambda^{+}v(\beta_{u}\gamma_{u}\sin\phi,\gamma_{u}\sin\phi, v\cos\phi,0)$$.
Looking at this carefully, we identify ##\lambda^{'}=\frac{I^{'0}_{tot}}{c}=\frac{2\lambda u v \gamma_u}{c^2}\sin\phi##. The ##\sin\phi## makes it so there is a varying amount of charge density that increases to a maximum at the bottom and top of the loop.
Any comments would be appreciated, but as this was ages ago I expect most people have moved on.