At closed-string level $N=\widetilde N$, the <bosonic string mass spectrum> is
$$
M^2=\frac4{\alpha'}(N-1).
$$
The state on the leading <Regge trajectory> uses only level-one oscillators and has maximal spin $J=N+\widetilde N=2N$. Eliminating $N$ gives
$$
\boxed{J=2+\frac{\alpha'M^2}{2}}.
$$
Because $N$ is unbounded, the spectrum contains infinitely many particles of increasing mass and spin. The intercept $J=2$ at $M^2=0$ includes the graviton.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 306 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-306/3/b/solution