Rotating Nambu–Goto string

A straight rigidly rotating classical string has speed of light at a free endpoint. Its energy and angular momentum satisfy a Regge trajectory. A segment with one anchored and one free end has $J=E^2/(\pi T)$, twice the slope of the string with two free ends.

With string tension $T=1/(2\pi {\alpha }^{\prime })$, the free-ended open string $X^0=a\tau$, $X^1+i{X}^2=a\cos \sigma e^{i\tau }$, $0\le \sigma \le \pi$, is straight at every target time. Its Nambu–Goto equations of motion and Virasoro constraints hold, and its ends move at light speed. Integrating the target-space Noether charges of a string gives $M=\pi Ta$ and $J=\pi T{a}^2/2$. Thus $J={\alpha }^{\prime }{M}^2$. Extending the same map to $0\le \sigma \le 2\pi$ describes a double-covered folded closed string, with twice the mass and twice the angular momentum, hence $J={\alpha }^{\prime }{M}^2/2$. The different slopes record the different boundary topology.

If $A(t)$ and $B(t)$ are perpendicular four-dimensional spatial vectors of equal fixed magnitude, every constant-time slice is a planar circle. Opposite rotations in two orthogonal coordinate planes can produce a Nambu–Goto action solution with constant radius and a rotating plane. The induced worldsheet metric can be constant even though the embedding moves. Rotating Cartesian axes clarify its shape without removing its kinetic target-space energy.

A Nambu–Goto action solution with $X^0=t$ is $\mathbf{X}=L\sin (\sigma /L)(\cos (t/L),\sin (t/L))$, for $0\le \sigma \le \pi L/2$. It solves the wave equation, has a fixed endpoint at the origin, and obeys the free-end string boundary condition at the other end. Its proper length is $L$ and its free endpoint has angular speed $1/L$ and spatial speed one.