<p>I know that with increasing age and depletion of hydrogen in the core the radiation inside a main sequence (MS) star increases pushing the envelope/surface of the star outwards. What I don't understand is why the temperature of the star shrinks due to this? I would expect that the temperature of the star to be constant in the process - the core of the star becomes more luminous, but the envelope is pushed further out. I know that this reasoning is a little bit flawed, because the hydrostatic equilibrium is not governed by radiation and temperature balancing each other, but by radiation and gravitation. But then on the other side, in the early phase of the star evolution there is also a change of the star size due to radiation and during this process of change the temperature remains constant (at least for low mass stars, see <a href="https://en.wikipedia.org/wiki/Hayashi_track" rel="nofollow noreferrer">Hayashi track</a>).</p>
<p>So I would like to have a quantitative answer, with equations, why the surface temperature of the star falls during the red giant phase. Tnx!</p>

CC BY-SA 4.0
2021-06-01T12:14:25.747Z
Are you interested in low mass stars (less than solar mass) or high mass stars (more than a solar mass)?
Daddy Kropotkin
CC BY-SA 4.0
2021-06-01T15:34:02.373Z
@DaddyKropotkin Any of the 2 options :)
NeStack
CC BY-SA 4.0
2021-06-01T19:11:42.417Z
Not a complete answer, but it comes down to energy density. How does the size increase with a specific energy increase and thus how does the energy density change? If more energy means lower energy density (because the size has gotten so massive), then the temperature goes down with more energy.
zephyr
<star><temperature><red-giant><hydrostatic-equilibrium><star-evolution>
Why red giants have cooler surface than MS stars?
NeStack
# Why red giants have cooler surface than MS stars?

(Asked by: username_0 on 2021-06-01)

I know that with increasing age and depletion of hydrogen in the core the radiation inside a main sequence (MS) star increases pushing the envelope/surface of the star outwards. What I don't understand is why the temperature of the star shrinks due to this? I would expect that the temperature of the star to be constant in the process - the core of the star becomes more luminous, but the envelope is pushed further out. I know that this reasoning is a little bit flawed, because the hydrostatic equilibrium is not governed by radiation and temperature balancing each other, but by radiation and gravitation. But then on the other side, in the early phase of the star evolution there is also a change of the star size due to radiation and during this process of change the temperature remains constant (at least for low mass stars, see [Hayashi track](https://en.wikipedia.org/wiki/Hayashi_track)).

So I would like to have a quantitative answer, with equations, why the surface temperature of the star falls during the red giant phase. Tnx!

**Tags:** star, temperature, red-giant, hydrostatic-equilibrium, star-evolution
> **Comments:**
> **username_1**: Are you interested in low mass stars (less than solar mass) or high mass stars (more than a solar mass)?  
> **username_0**: @DaddyKropotkin Any of the 2 options :)  
> **username_2**: Not a complete answer, but it comes down to energy density. How does the size increase with a specific energy increase and thus how does the energy density change? If more energy means lower energy density (because the size has gotten so massive), then the temperature goes down with more energy.  

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## 0 answers

