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As stars age, the concentration of Hydrogen in the core decreases which lowers the power output, causing an imbalance between outward radiation pressure and inward gravitational pressure. This causes the core to collapse so that it becomes denser and hotter, which increases fusion rate until it generates enough pressure to counteract the enlarged gravitational pressure due to the core's smaller size and hence smaller surface area. What I don't understand is why would this increased power output be larger than before, making the star brighter and causing the outer layers to expand. Shouldn't the total force produced by the core add up to balance the gravity so that the star remains at the original size?

Edit: I am specifically asking about the hydrogen burning phase of main sequence stars, where this effect is still observed in.

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  • $\begingroup$ Are you just asking about the normal hydrogen fusion processes of a main sequence star? Some of the answers below are talking about red giants, and other fusion reactions like helium burning. $\endgroup$ – PM 2Ring Feb 26 at 10:09
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    $\begingroup$ I've deleted some comments that were offering partial or speculative answers to the question, as well as responses to those comments. Please keep in mind that answers should only be posted using the answer box; comments are meant for requesting clarification and suggesting improvements to the original post. $\endgroup$ – David Z Feb 27 at 10:31
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Why does the luminosity increase?

As core hydrogen burning proceeds, the number of mass units per particle in the core increases. i.e. 4 protons plus 4 electrons become 1 helium nucleus plus 2 electrons. But pressure depends on both temperature and the number density of particles. If the number of mass units per particle is $\mu$, then $$ P = \frac{\rho k_B T}{\mu m_u}, \ \ \ \ \ \ \ \ \ (1)$$ where $m_u$ is the atomic mass unit and $\rho$ is the mass density.

As hydrogen burning proceeds then $\mu$ increases from about 0.6 for the initial H/He mixture, towards 4/3 for a pure He core. Thus the pressure would fall unless $\rho T$ increases.

An increase in $\rho T$ naturally leads to an increase in the rate of nuclear fusion (which goes as something like $\rho^2 T^4$ in the Sun) and hence an increase in luminosity.

This is the crude argument used in most basic texts, but there is a better one.

The luminosity of a core burning star, whose energy output is transferred to the surface mainly via radiation (which is the case for the Sun, in which radiative transport dominates over the bulk of its mass) depends only on its mass and composition. It is easy to show, using the virial theorem for hydrostatic equilibrium and the relevant radiative transport equation (e.g. see p.105 of these lecture notes), that $$ L = \frac{\mu^4}{\kappa}M^3,\ \ \ \ \ \ \ \ \ \ (2)$$ where $\kappa$ is the average opacity in the star.

Thus the luminosity of a radiative star does not depend on the energy generation mechanism at all. As $\mu$ increases (and $\kappa$ decreases because of the removal of free electrons) the luminosity must increase.

Why does the radius increase?

Explaining this is a bit trickier and ultimately does depend on the details of the nuclear fusion reactions. Hydrostatic equilibrium and the virial theorem tells us that the central temperature depends on mass, radius and composition as $$T_c \propto \frac{\mu M}{R}$$ Thus for a fixed mass, as $\mu$ increases then the product $T_c R \propto \mu$ must also increase.

Using equation (2) we can see that if the nuclear generation rate and hence luminosity scales as $\rho^2 T_c^{\alpha}$ then if $\alpha$ is large, the central temperature can remain almost constant to provide the increased luminosity and hence $R$ must increase significantly. Thus massive main sequence stars, in which CNO cycle burning dominates and $\alpha>15$, experience a large change in radius during main sequence evolution. In stars like the Sun with $\alpha \sim 4$, the central temperature increases as $\mu$ and $\rho$ increases and the radius goes up, but not by very much.

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    $\begingroup$ @nevertookcoursesbutwhy 4H$^+$ + 4e = 0.5 mass units per particle. Changes into 1He$^{2+}$ + 2e = 1.33... mass units per particle $\endgroup$ – Rob Jeffries Feb 27 at 8:28
  • $\begingroup$ First sentence seems to state that it decreases. Third states that it increases. Perhaps I'm missing something? Do you mean the reciprocal? $\endgroup$ – Peter Bingham Feb 27 at 10:30
  • $\begingroup$ @PeterBingham Indeed, fixed. $\endgroup$ – Rob Jeffries Feb 27 at 11:16
  • $\begingroup$ Does this mean that stellar evolution must lead to this and that an energy output that can keep the star at constant size and brightness simply cannot exist? $\endgroup$ – never took courses but why Feb 27 at 11:50
  • $\begingroup$ The Russell-Vogt theorem says that a star's position on the HR diagram depends on its mass and composition. Since the composition changes with age, but the mass is (reasonably) constant on the main sequence, then the star must move on the HR diagram. @nevertookcoursesbutwhy $\endgroup$ – Rob Jeffries Feb 27 at 13:17
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Hydrogen fusion happens at relatively cool temperatures and low densities compared to the higher order fusion processes. To fuse helium and even heavier atoms, you need much, much more temperature.

Heat dissipation from the star's core to the shell is highly dependent on the temperature gradient. As such, a core at iron producing temperatures looses much more heat to the shell than a hydrogen fusing core does. And because the core looses heat much quicker, it also produces energy much quicker, it's a self regulating fusion reactor, after all.

However, since the shell is so much more effectively heated by the late stage core, and thus much hotter, it's also much less dense. So the total size of the star increases.


TL;DR:

high fusion temperature -> fast heat loss (-> fast energy production) -> high shell temperature -> low shell density -> large size

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    $\begingroup$ Why does the core losing more heat cause it to generate more heat? They seem like independent processes. I get that to maintain the star's temperature, this must be the case. However, it doesn't seem to be an explanation of why it happens. Also, I was asking about the hydrogen burning phase specifically. E.g. Our Sun is getting bigger and brighter continuously throughout its hydrogen burning phase. Does the same logic apply? $\endgroup$ – never took courses but why Feb 26 at 8:08
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    $\begingroup$ @nevertookcoursesbutwhy The point is, that stellar fusion is a self regulating process: If the core cools down, it shrinks, which ramps up the density, and thus the fusion, which increases the temperature, which provides the pressure that stops the shrinking. - As the sun ages, the density of hydrogen in its core decreases, so a higher temperature and/or density is required for the same fusion rate. And because of that higher temperature, we get higher heat losses, quicker energy production, and a larger star. Same mechanic, just much less extreme than with iron producing fusion. $\endgroup$ – cmaster - reinstate monica Feb 26 at 8:15
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    $\begingroup$ I don't quite understand. Ok so the core shrinks and heats up in order to increase its power output back up, but this doesn't necessarily mean its power output is increased beyond its initial power output. I would imagine it goes back to the initial power output (i.e. "same fusion rate" as you stated), which is clearly not the case, hence my confusion. Heat loss rate doesn't seem to affect this process, and a brighter and larger star is simply the consequences of an increased power output compared to initial power output. Can you elaborate on this power increase? $\endgroup$ – never took courses but why Feb 26 at 8:41
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    $\begingroup$ This does not address the question if why a main sequence star becomes more luminous as it consumes it's fuel. $\endgroup$ – Rob Jeffries Feb 26 at 8:42
  • $\begingroup$ @nevertookcoursesbutwhy As I said, the heat production follows the energy losses. And all the energy that the core loses, must in the end be radiated away by the surface. If the flux of heat from the core to the shell increases, so must the radiation escaping from the star's surface in the end. And as long as the star's surface is not large enough to radiate away all the energy, the shell will continue getting hotter and less dense, and the star will continue to grow. $\endgroup$ – cmaster - reinstate monica Feb 26 at 19:27
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The size of the star depends on the balance between the gravitational pressure that wants to make it smaller and radiation and thermal pressure from the nuclear reaction that want it to expand. What the nuclear reaction is affects the point at which the balance is achieved.

As stars form and their core gets denser and hotter, the temperature reaches the point when hydrogen is burned into helium. These are the stars of the main sequence.

However, eventually the amount of hydrogen in the core drops, and the energy production from burning hydrogen wanes. The gravity momentarily owerpowers and brings more hydrogen close to the core. This causes a situation when there is no longer a significant amount of hydrogen burnt in the core, but there are regions around the core where hydrogen is still being burnt. Overall, the amount of enrgy produced end up being greater than originally, which means that the star becomes brighter.

However the conservation of gravitational and thermal energy within the star (the radiated energy is much smaller than the total energy in the star, so even with the outflux of radiation the total energy of the star is mostly conserved) implies that when one part of the star contracts, a different part needs to expand. This is called the mirror principle and is predicted by star models. In this case, while the core contracts, the outer shell (the envelope) expands. The result is what is a typical red giant: a helium core, hydrogen burnt in a shell around the core, and a large envelope.

There are other types of red giants (some of them already start burning helium into carbon), but the principle is the same: radiation of the core stops being enough to balance the gravity, gravity collapses the core until new equilibrium is reached (with higher total brightness), energy is transfered to the outer shell, outer shell expands.

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    $\begingroup$ "Overall, the amount of enrgy produced end up being greater than originally" Why? $\endgroup$ – never took courses but why Feb 26 at 17:06
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    $\begingroup$ @nevertookcoursesbutwhy That's not so simple to explain. There are multiple factors that affect the amount of energy produced (the type of reaction, the temperature and preasure, the amount of fuel) and all of them change as the star evolves. One would just need to look at the star evolution models and see where the next equilibrium will be. In this case, since the core becomes heavier and denser, one can expect that a higher internal presure will be needed to keep it from further collapsing. This higher internal pressure requires higher overall energy production, but a proof is complicated. $\endgroup$ – Adam Latosiński Feb 27 at 6:59
  • $\begingroup$ So are you actually gonna answer my question? You aren't going to show proof for half the stuff you are claim just because it's "complicated"? $\endgroup$ – never took courses but why Feb 27 at 11:19
  • $\begingroup$ This explanation is the friendliest to my level of understanding. I didn't get how collapsing and becoming denser would make something bigger. But it's actually the core that collapses and the shell that gets bigger. Thanks! For the other part (why does the dense core get hotter), for me it's enough that something is burning and then starts burning hotter. Why would it always burn at the same rate as things change? $\endgroup$ – Daniel Darabos Feb 27 at 12:43
  • $\begingroup$ @nevertookcoursesbutwhy If you're looking for formulas, RobJeffries alredy put tsome in his answer, with links to more. I'm trying to keep my answer layman-friendly. And admittedly, I've seen this derivaqtion only once, and I'm not sure I could repeat it here. $\endgroup$ – Adam Latosiński Feb 28 at 0:08

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