# How much does rotation and heat contribute to an overall gravitational

1. Sep 17, 2014

### povillsss

So let's say we have a neutron star. A neutron star is very massive and very dense therefore it has a very strong gravitational field. But also it is rotating very fast and it really hot. This means it has more energy than if it was stationary and cold. Of course if this neutron star has more energy its gravitational field strength must also be stronger. I'm wondering by how much? If you could give me some formulae that would be great :D

P.S. first time writing.
Sorry for my bad English I'm not a native speaker.

2. Sep 17, 2014

### Staff: Mentor

Welcome to the forums!

Rotation is a small contribution. The fastest neutron stars rotate at about 500 revolutions per second. With a radius of roughly 10 kilometers, this gives them an outer speed if 30000km/s or .1c. You can calculate the Lorentz-factor for this speed, it is very close to 1 (where 1 is "just mass contributes").

Heat is problematic to define - the inside does not follow the classical "this is cold matter, we can heat it up to get warm matter" pattern.

3. Sep 17, 2014

### PAllen

Well, for heat, one possible definition is the the decrease in gravitational mass measured by orbital testers between a 'new' neutron star, and the same star if if were isolated and allowed to radiative for e.g. 10 billion years (or reach equlibrium with CMB - its radiation being essentially indistinguishable from CMB).

I think the answer is that the contribution of head defined this way, to the mass of a new neutron star, is small. I don't have any numbers though.

4. Sep 17, 2014

### Staff: Mentor

That's probably a good approach. But then you might get contraction (and therefore the release of gravitational energy) as another contribution.

I agree that the heat contribution by this definition is probably small.

5. Sep 17, 2014

### ChrisVer

Well the energy density and stuff enter the Energy momentum tensor in Einstein's equation.
The metric is the Kerr-Metric +a term which is due to the existence of matter, and so I think you have to solve for $G_{\mu \nu} = a T_{\mu \nu}$ to find this extra term...although that's somewhat perturbative.

6. Sep 17, 2014

### PAllen

I'd be perfectly willing to accept 'gravitational heat'. Contraction does lead to heating in normal bodies, which then may be radiated. If in a more extreme regime, you get some direct GW emission, so what? I could also argue that what initially keeps a body un-contracted is some form of 'heat', so it is being converted to GW.

I don't mean to push this too far, just argue that if you accept by heat 'any energy (all forms) that radiates away until equilibrium with a very cold reservoir', then you can discuss heat for content for exotic bodies, and the corresponding mass equivalent (measured via influence on test bodies at great distance - to abstract from frame dragging effects of high rotation).

Last edited: Sep 18, 2014
7. Sep 19, 2014

### stevebd1

When considering black hole thermodynamics, spin and charge can contribute the BHs gravity-

$$dM=\frac{\kappa}{8\pi}\,dA\,+\,\Omega\,dJ\,+\,\Phi\,dQ$$

where the first term is related to irreducible mass- $M_{ir}=\sqrt(A/16\pi)$, the second to rotation- $J$ and the third to charge- $Q$. The irreducible mass would be mass left if both the properties of spin and charge were extracted from the black hole so it's not too much of a stretch to say that spin and heat contribute to the gravity of a neutron star. Pressure would also make a contribution to the gravity field. A basic algebraic interpretation of Einsteins law of gravity is $g=\rho c^2+3P$ where $\rho$ is density and $P$ is pressure, also pressure can be described as being synonymous with energy density.

Source-
Black Hole Thermodynamics by Narit Pidokrajt
http://www.physto.se/~narit/bh.pdf [Broken] pages 9-12

Last edited by a moderator: May 6, 2017