<p>Stars do not <em>get</em> hot because of nuclear fusion, they <em>become</em> hot enough to sustain nuclear fusion and this process <em>maintains</em> their temperatures. Nuclear fusion actually <em>stops</em> a star getting hotter.</p>

<p>Protostars (before nuclear fusion) <em>get</em> hot because of a well known statistical relationship between the gravitational potential energy of a gas and the internal kinetic energy of the particles that make up the gas. [In an ideal gas, the kinetic energy of the particles is directly proportional to the temperature of the gas.] This is known as the <a href="https://en.wikipedia.org/wiki/Virial_theorem#In_astrophysics">virial theorem</a>, which says that twice the summed kinetic energy of particles ($K$) plus the gravitational potential energy ($\Omega$, which is a negative quantity for a bound object) equals zero.
$$ 2K + \Omega = 0$$</p>

<p>Now you can write down the <em>total</em> energy of the system as
$$ E_{tot} = K + \Omega$$
and hence from the virial theorem that
$$E_{tot} = \frac{\Omega}{2},$$
which is also negative.</p>

<p>If we now <em>remove</em> energy from the system, for instance by allowing the gas to radiate away energy, such that $\Delta E_{tot}$ is <em>negative</em>, then we see that 
$$\Delta E_{tot} = \frac{1}{2} \Delta \Omega$$</p>

<p>So $\Omega$ becomes <em>more negative</em> - which is another way of saying that the protostar attains a <em>more collapsed</em> configuration.</p>

<p>Oddly, at the same time, we can use the virial theorem to see that
$$ \Delta K = -\frac{1}{2} \Delta \Omega = -\Delta E_{tot}$$
is <em>positive</em>. i.e. the kinetic energies of particles in the gas (<strong>and hence their temperatures</strong>) actually become hotter. In other words, the gas has a negative heat capacity. But a hotter temperature usually means more radiation is produced and if the energy losses continue, then so does the collapse.</p>

<p>This process is ultimately arrested in a star by the onset of nuclear fusion. This replaces the radiative losses with nuclear energy and the star attains a quasi-equilibrium that lasts as long as it has nuclear fuel to burn.</p>

CC BY-SA 3.0
2016-04-18T11:02:36.493Z
It's a comprehensive answer, but possibly too complicated
Tanenthor
CC BY-SA 3.0
2016-04-18T12:22:25.890Z
@Tanenthor "Astronomy Stack Exchange is a question and answer site for astronomers and astrophysicists. It's built and run by you as part of the Stack Exchange network of Q&A sites. With your help, we're working together to build a library of detailed answers to every question about astronomy." It is *lack* of detail that is notable about many answers on Astronomy SE.
ProfRob
CC BY-SA 3.0
2016-04-18T09:02:19.857Z
2016-04-18T09:02:19.857Z
ProfRob
<p>Before the nuclear fusion in the core starts, the heat of the star comes from the contraction of the original nebula. When the matter comes closer together, the potential energy of it decreases, just like when you drop a rock. Energy is however constant, so it has to go somewhere. That "somewhere" is the heat in the newborn star.</p>

CC BY-SA 3.0
2016-04-17T23:20:29.233Z
So you're implying that the birth of a star is somewhat violent and not gradual or am I just interpreting it wrong?
Badr B
CC BY-SA 3.0
2016-04-17T23:21:54.570Z
@ReadySetPawn Nope, I have not said anything about how long the contraction phase lasts.
SE - stop firing the good guys
CC BY-SA 3.0
2016-04-18T07:00:42.323Z
@ReadySetPawn  while it's a different question, yes the birth of stars can be very violent and temporarily a lot brighter than when the star settles into it's main sequence.   Jupiter, for example is 1/75th the mass of the lightest red-dwarf stars, but the heat from formation is sufficient that Jupiter still emits 4 times the energy that it receives from the sun.   The energy and heat created when enough matter to form a star coalesces under gravity is impressive.
userLTK
CC BY-SA 3.0
2016-04-17T22:25:27.797Z
2016-04-17T22:25:27.797Z
SE - stop firing the good guys
<p>Simple question, but can't seem to find the answer anywhere. It can't be nuclear fusion because nuclear fusion occurs as a result of the heat. And it can't be because of gravity because it's believed that the core of black holes are nearly absolute zero. So what makes stars hot? </p>

CC BY-SA 3.0
2016-04-18T02:44:27.813Z
Who believes that the core of black holes is nearly absolute zero?
Sir Cumference
CC BY-SA 3.0
2016-04-18T09:04:00.790Z
Turns out to be a conceptually very interesting question with a non-obvious answer.
ProfRob
<star><heat>
What Makes Stars Hot?
Badr B
# What Makes Stars Hot?

(Asked by: username_0 on 2016-04-17)

Simple question, but can't seem to find the answer anywhere. It can't be nuclear fusion because nuclear fusion occurs as a result of the heat. And it can't be because of gravity because it's believed that the core of black holes are nearly absolute zero. So what makes stars hot?

**Tags:** star, heat
> **Comments:**
> **username_1**: Who believes that the core of black holes is nearly absolute zero?  
> **username_2**: Turns out to be a conceptually very interesting question with a non-obvious answer.  

---

## 2 answers

### Answer by username_2 on 2016-04-18. Score: 7

Stars do not *get* hot because of nuclear fusion, they *become* hot enough to sustain nuclear fusion and this process *maintains* their temperatures. Nuclear fusion actually *stops* a star getting hotter.

Protostars (before nuclear fusion) *get* hot because of a well known statistical relationship between the gravitational potential energy of a gas and the internal kinetic energy of the particles that make up the gas. [In an ideal gas, the kinetic energy of the particles is directly proportional to the temperature of the gas.] This is known as the [virial theorem](https://en.wikipedia.org/wiki/Virial_theorem#In_astrophysics), which says that twice the summed kinetic energy of particles ($K$) plus the gravitational potential energy ($\Omega$, which is a negative quantity for a bound object) equals zero.
$$ 2K + \Omega = 0$$

Now you can write down the *total* energy of the system as
$$ E\_{tot} = K + \Omega$$
and hence from the virial theorem that
$$E\_{tot} = \frac{\Omega}{2},$$
which is also negative.

If we now *remove* energy from the system, for instance by allowing the gas to radiate away energy, such that $\Delta E\_{tot}$ is *negative*, then we see that
$$\Delta E\_{tot} = \frac{1}{2} \Delta \Omega$$

So $\Omega$ becomes *more negative* - which is another way of saying that the protostar attains a *more collapsed* configuration.

Oddly, at the same time, we can use the virial theorem to see that
$$ \Delta K = -\frac{1}{2} \Delta \Omega = -\Delta E\_{tot}$$
is *positive*. i.e. the kinetic energies of particles in the gas (**and hence their temperatures**) actually become hotter. In other words, the gas has a negative heat capacity. But a hotter temperature usually means more radiation is produced and if the energy losses continue, then so does the collapse.

This process is ultimately arrested in a star by the onset of nuclear fusion. This replaces the radiative losses with nuclear energy and the star attains a quasi-equilibrium that lasts as long as it has nuclear fuel to burn.

> **Comments:**
> **username_3**: It's a comprehensive answer, but possibly too complicated  
> **username_2**: @Tanenthor "Astronomy Stack Exchange is a question and answer site for astronomers and astrophysicists. It's built and run by you as part of the Stack Exchange network of Q&A sites. With your help, we're working together to build a library of detailed answers to every question about astronomy." It is *lack* of detail that is notable about many answers on Astronomy SE.  

---

### Answer by username_4 on 2016-04-17. Score: 2

Before the nuclear fusion in the core starts, the heat of the star comes from the contraction of the original nebula. When the matter comes closer together, the potential energy of it decreases, just like when you drop a rock. Energy is however constant, so it has to go somewhere. That "somewhere" is the heat in the newborn star.

> **Comments:**
> **username_0**: So you're implying that the birth of a star is somewhat violent and not gradual or am I just interpreting it wrong?  
> **username_4**: @ReadySetPawn Nope, I have not said anything about how long the contraction phase lasts.  
> **username_5**: @ReadySetPawn  while it's a different question, yes the birth of stars can be very violent and temporarily a lot brighter than when the star settles into it's main sequence.   Jupiter, for example is 1/75th the mass of the lightest red-dwarf stars, but the heat from formation is sufficient that Jupiter still emits 4 times the energy that it receives from the sun.   The energy and heat created when enough matter to form a star coalesces under gravity is impressive.  

---

