Calculating Core Temperature of a Brown Dwarf

In summary, you attempted to calculate the core temperature of a "star" by setting the hydrostatic equilibrium equation to the ideal gas law and solving for the temperature. You found that the temperature is within an order of magnitude of 10^6 K if you approximate the radius to be the size of the sun.
  • #1
rpthomps
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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.
 
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  • #2
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?
 
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  • #3
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.
 
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  • #4
rpthomps said:
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.
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
 
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  • #5
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
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.
 
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  • #7
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:
[tex] \frac{dP(r)}{dr}= − \frac{GM(r)\rho(r)}{r^2}[/tex], where:
[tex] \frac{d M(r)}{dr} = 4 \pi r^2\rho(r) [/tex]
In the simplest case, solving these equations leads to the Lane-Emden equation. http://www.astro.caltech.edu/~jspineda/ay123/notes/2011/ay123_polytropes_sep2010.pdf is a good reference for deriving and solving this equation.
 
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  • #8
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.
 
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1. How is the core temperature of a brown dwarf calculated?

The core temperature of a brown dwarf is typically calculated using theoretical models and equations that take into account factors such as the mass, radius, and composition of the brown dwarf. These models are based on our understanding of stellar evolution and the properties of matter under extreme conditions.

2. What is the typical range of core temperatures for brown dwarfs?

The core temperatures of brown dwarfs can vary greatly depending on their mass and age. However, the range is typically between 1,500 to 3,000 Kelvin (K), which is much lower than the core temperatures of stars like the Sun, which can reach up to 15 million K.

3. How does the core temperature of a brown dwarf affect its properties?

The core temperature of a brown dwarf is a key factor in determining its overall properties, such as its luminosity, radius, and atmospheric composition. A higher core temperature can lead to a higher luminosity and a larger radius, while a lower core temperature can result in a cooler and more compact brown dwarf.

4. Can the core temperature of a brown dwarf be directly measured?

No, the core temperature of a brown dwarf cannot be directly measured due to their extremely low luminosity and the fact that they do not undergo nuclear fusion like stars do. However, scientists can use indirect methods, such as spectroscopy, to infer the core temperature based on other observable properties.

5. How does the core temperature of a brown dwarf change over time?

The core temperature of a brown dwarf will gradually decrease over time as it loses energy through radiation. However, this process is much slower than the cooling of stars, so brown dwarfs can maintain their core temperature for billions of years. Additionally, as a brown dwarf ages, its core temperature may also decrease due to the settling of heavier elements towards the core.

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