10
$\begingroup$

I know black holes can be described by only a few parameters. Given the extreme conditions within a neutron star, how much hair does it have, ie. how much of the original star parameters and composition (e.g. metallicity) affects the features of a neutron star?

$\endgroup$
1

1 Answer 1

10
$\begingroup$

No complete answer from me, but some obvious examples of "hair" on a neutron star are:

The temperature - both internal and surface temperature can be measured (in principle). These are mainly a function of the age of the neutron star.

The magnetic field - young neutron stars have very strong magnetic fields that are probably an inherited property from the composition and rotation of the progenitor. The field decays with time.

The surface composition may vary from neutron star to neutron star. In principle this can be measured with X-ray spectroscopy. The composition may be connected with the progenitor, but could also reflect accreted material.

Neutron stars have a radius! The radius may or may not be a unique function of mass. It could depend on the rotation, composition and age.

I think though, the gist of your question is could one determine the progenitor of a neutron star from observations of the neutron star? At present the answer to that is no. Even the relationship between neutron star mass and progenitor mass is not understood; it is also unclear why some neutron stars have very high magnetic fields and so on. But ultimately that would be an aim.

$\endgroup$
2
  • $\begingroup$ Thank you for your answer. It sounds to me that we just don't know much yet to say much more than this about the subject. Beyond magnetic fields and thin atmosphere, would I be totally wrong if I'd say most of the properties are based on mass, rotation and age (~ temperature), but those fixed two neutron stars would look and feel pretty much the same? $\endgroup$
    – tuomas
    Commented Oct 25, 2016 at 9:38
  • $\begingroup$ @tuomas I asked something similar on Physics SE and got no conclusive answer! physics.stackexchange.com/questions/133324/… $\endgroup$
    – ProfRob
    Commented Oct 25, 2016 at 10:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .