Skip to main content
1 of 2
moonblink
  • 1.1k
  • 6
  • 17

According to Einstein [the circumference] should be greater than $2 \pi R$ for a co-rotating observer … On the other hand, there's a nice explanation here on SE … but it concludes the opposite: $L' = L/\gamma$.

The Lorentz factor starts from $1$ at rest and decreases towards zero the faster you go (reaching $0$ only at the speed of light, so never). Being at the denominator in the equation, the final circumference will be greater the smaller the Lorentz factor is, i.e. the faster the body spins.

\begin{equation}L^\prime = \frac{L}{\gamma} = \frac{2\pi R}{\sqrt{1-\frac{v^2}{c^2}}}\end{equation}

Or, alternatively,

\begin{equation}L^\prime = \frac{L}{\gamma} = \frac{2\pi R}{\sqrt{1-\frac{R^2\omega^2}{c^2}}}\end{equation}

PSR J1748−2446ad is the fastest neutron star known; if you walked around its equator you would walk three kilometres more than you had predicted by staring at it from Earth.

This implies the impossibility of rigid rotation. Neutron stars are not rigid, and so they are free to deform to fit for the larger circumference.

More on this topic:

moonblink
  • 1.1k
  • 6
  • 17