# Is a Kerr ring singularity a closed string?

I’m obviously not a scientist, but has anyone considered that a Kerr ring singularity might basically be a closed string?

• The singularity spins in one direction only and is incredibly flat and thin (one dimensional).

• Its size is on the order of Planck’s length, like a string.

• Its spinning surface is wriggling quantum foam and a string’s surface vibrates.

• A closed string vibrates to represent nuclear particles and the higher the frequency the more mass the particle has. Since they are one dimensional, if all the closed strings in a star were crushed onto each other by gravity, then they would become essentially one closed string. However, the frequency of the vibration would multiply and become almost infinitely high, thus representing a particle of incredible mass, like a singularity.

• We already know that gravity can implode a star into neutrons (and perhaps quarks), so why not even the smallest basic structure?

• I think there is an interesting question within this seemingly naive question: Can a Kerr singularity have hair in the form of vibrational anular modes? – lurscher Mar 22 '18 at 18:20
• and with 'hair' I mean, can it have observational consequences for far-away observers? – lurscher Mar 22 '18 at 18:21

## 1 Answer

A Kerr singularity is a ring that in Cartesian coordinates is described as
$x^2 + y^2 = a^2$ and $z = 0$
where $a$ is the angular momentum per unit mass of the black hole.
It is flat, but the size is not of the order of Planck's length. Reason being that the parameter $a$ is not infinitesimal, thus defining the ring as a macroscopic figure.
As per above, I do not think that you can conceive a Kerr black hole as a string.