Length Contraction in a Particle Accelerator

Consider $N$ particles equally spaced on a circle which are uniformly accelerated to $99\%$ the speed of light.

In Newtonian mechanics, the distance between the particles would be $2\pi r/N$ (for large $N$). If we add up all the distances between the particles we get the circumference of the particle accelerator. In special relativity however, distance is Lorentz contracted by $\sqrt{1-v^2/c^2}$. But from the frame of reference of the accelerator, the circumference must stay the same. How can all the distances between particles be contracted without $N$ or the circumference of the accelerator changing?

From the particle's frame of reference, time passes by normally, but the distance between each particle will be Lorentz contracted. Imagine a particle directly in front of it. Since they're traveling in "almost" the same direction, there won't be a whole lot of length contraction since they're pretty close together. But now imagine a particle at 0 degrees and one at 90 degrees (a quarter way around). One is traveling in the "x" direction and the other in the "y" direction. Since the first particle traveling in the x-direction has no velocity in the y-direction, there will be significant length contraction, and actually the exact amount contracted would be the Lorentz factor $\gamma$. Now two particles spaced on opposite sides are traveling towards each other, so length contraction is further exacerbated. Since all particles are evenly spaced, on average there is a factor of $\gamma$ of length contraction, so the circumference in the particle's frame appears to be $C/\gamma$.
• Any object with a "length" will contract. Depending on what kind of particles you are talking about (for example, an atom usually is on the order of an angstrom wide), they will contract by the same Lorentz factor, albeit one angstrom contracting will be very difficult to detect. Also, I think you got the gamma factor mixed up, it's actually $1/\sqrt{1-\frac{v^2}{c^2}}$, so it would actually get smaller since that number will be less than one – Spaderdabomb Aug 1 '14 at 5:10