# Calculating Core Temperature of a Brown Dwarf

1. Nov 29, 2015

### rpthomps

To begin with, I am an enthusiastic but very amateur physicist. I am working on my knowledge by taking online open courseware, buying textbooks and doing all the questions. I recently bought Astrophysics of Planet Formation which is no doubt out of my league but I like to pick at it from time to time. In the first page it talks about the maximum mass of a planet being at the deuterium burning threshold. I decided I wanted to do a simple calculation to see if I could get the core temperature of a “star” like this. I set the hydrostatic equilibrium equation to the ideal gas law and solved for the temperature. I was looking for a temperature of 10^6 K.

Is this approximation too simplistic? I am within an order of magnitude in my answer if I approximate the radius to be the size of the sun.

2. Nov 29, 2015

### phyzguy

I'm not quite clear what you did. When you say, "I set the hydrostatic equilibrium equation to the ideal gas law," can you be more specific. What equations did you use and how did you solve for T?

3. Nov 30, 2015

### Ken G

You will need to specify not only a mass, but also a radius. You can find the minimum radius by saying the gas is fully degenerate, but of course that's not the ideal gas law. The radius will need to be a lot larger than that to use the ideal gas law. If you make the radius too large, the temperature will be lower than its peak value. So the best you can do is benchmark estimates, but the exercise should be useful. As an even simpler approach, you can notice that gas giants have densities of around 1 g/cc, and just use that to get the radius. If you also assume it is an ideal gas, it's not going to be accurate, but you can get a ballpark maximum temperature.

4. Nov 30, 2015

### SteamKing

Staff Emeritus
All brown dwarf stars are similar in size to the planet Jupiter, perhaps a little larger, but nowhere near the radius of the sun. Brown dwarves more massive than about 13 Jovian masses are thought to be capable of fusing deuterium:

https://en.wikipedia.org/wiki/Brown_dwarf

5. Nov 30, 2015

### rpthomps

This is what I did:

Hydrostatic Equilibrium expression:

$P=R\rho g$

gravitation field intensity

$g=\frac{GM}{R^2}$

Ideal gas law

$P=\rho Tk$

Combine them all

$T=\frac{GM}{Rk}$

I get 3 million K which is 3 times the deuterium burning temperature.

6. Nov 30, 2015

### Ken G

There's a missing average mass per particle in your expressing, stemming from it being missing from the ideal gas law, so I'm not sure how your answer works out. Perhaps you are using an unusual version of the Boltzmann constant k. Anyway, if you stick in the average mass per particle, your analysis is fine as a good estimate-- as long as you can input a radius R. A logical way to do that is to use the mass-radius relationship for degenerate electrons. Even though you are in the ideal gas limit, that will give you the boundary of the crossover to degeneracy, which gives a benchmark of where the temperature will reach its maximum prior to the onset of degeneracy.

7. Nov 30, 2015

### phyzguy

The problem is that P, M, ρ, and T are all functions of R. You cannot equate them algebraically in this way, you have to write a set of differential equations which you can solve simultaneously. For example, it is not true that P = R ρ g, instead, the condition of hydrostatic equilibrium states that:
$$\frac{dP(r)}{dr}= − \frac{GM(r)\rho(r)}{r^2}$$, where:
$$\frac{d M(r)}{dr} = 4 \pi r^2\rho(r)$$
In the simplest case, solving these equations leads to the Lane-Emden equation. Here is a good reference for deriving and solving this equation.

8. Nov 30, 2015

### Ken G

You can equate them if you are just doing estimates. It would not be reasonable to ask for a full solution to the differential equations, because that is a research-level calculation. Even polytrope approximations are just that-- approximations. So if one chooses to approximate, one can select what level of approximation is being used, and it sounds like rpthomps is seeking a very rough level of approximation. That's perfectly valid, just less accurate than a polytrope, which is less accurate than a research-level solution.