Why is a pre-main sequence star brighter than it will be when it reaches the main sequence?
[Physics] Why are pre-main sequence stars brighter than they will be on the main sequence
astronomystarsstellar-evolution
Related Solutions
Well, you would need an accurate stellar evolution model and you would need the age, chemical composition and probably the rotation rate too.
The mass-radius relationship is an area of intense research. Masses and radii are only determined for stars in eclipsing binary systems. These could define an empirical mass-radius relationship, but stellar evolution models tell us that the radius also depends on age. eg pre main sequence stars are larger, and even whilst still on the main sequence, stars gradually get larger by tens of percent over the course of their main sequence lifetime. Practically, this means that knowing the age would be important for stars of mass of about 80% of the Sun or larger. Lower mass main sequence stars would not have departed significantly from their initial main sequence radii even if they were as old as the Galaxy.
The radius will also depend on the chemical composition - both the overall metallicity and the fractional helium abundance are important, through their effects on atmospheric opacities and mean atomic masses. This is an effect at the level of 10-20%.
Lastly, it is becoming apparent that rotation probably plays an important role. In stars less massive than the Sun it appears that rotation-induced magnetic activity inflates stars (for poorly understood reasons). In higher mass stars rotation induces interior mixing that alters internal composition gradients and changes the evolution (as well as centrifugal distortion). Again, these are phenomena that affect radii at the level of 10% or so.
If you want something rough and ready, which at least allows you to account for the effects of age and metallicity, you could interpolate the results tables of radius versus mass that are easily obtained from this website, that reports calculations from the models of Siess, Dufour & Forestini (2000).
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, $\mu$ 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 $$ L \propto \frac{\mu^4}{\kappa}M^3,\ \ \ \ \ \ \ \ \ \ (2)$$ 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 more difficult and ultimately does depend on the details of the nuclear fusion reactions. Hydrostatic equilibrium and the virial theorem tell 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 because a very small increase in $T_c$ can provide the increased luminosity. Hence if $RT_c$ increases in proportion to $\mu$ then $R$ 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 contrast, for stars like the Sun, where H-burning via the pp-chain has $\alpha \sim 4$, the central temperature increases much more as $\mu$ and $\rho$ increase, and so the radius goes up but not by very much.
Best Answer
It is not generally true that a Pre-Main Sequence (PMS) star is brighter than the corresponding Zero-Age Main Sequence (ZAMS) star - whether this is the case depends on the mass. The main source of energy in this phase is the gravitational potential energy of the gas cloud being converted into random kinetic, i.e. thermal, energy. The main question is how fast this energy is transported out of the protostar, compared to how fast the star is contracting and the temperature rising.
The figure in the question shows, correctly, an evolutionary track in the H-R diagram of a PMS object. However, this track is only valid for a certain mass.
The figure above, from the Wikipedia Hayashi Track entry, shows representative PMS evolutionary tracks for different masses. The blue lines are tracks through the H-R diagram for PMS objects of different masses; they start on the upper diagonal, called the birth line (the time when the surrounding clouds get cleared away and the system becomes visible), and end on the lower black diagonal, the Zero Age Main Sequence (the time when Hydrogen fusion sets in). The blue numbers below the ZAMS show the final mass of the star in Solar masses. The red lines show isochrones; where the different evolutionary tracks crosses the same isochrone, the corresponding objects have the same age. You can see that a $6.0 M_{\odot}$ star reaches the ZAMS after only a hundred thousand years, a time when a $0.1 M_{\odot}$ star has not left the birthline yet. It takes 10 million years for a $2.0 M_{\odot}$ star and 100 million years for stars of $M \lesssim 1.5 M_{\odot}$ to reach the ZAMS.
The almost-vertical parts of the tracks are called Hayashi tracks. In this phase, the contraction happens more or less isothermally. The objects on these stages are convective, such that the heat generated by contraction is transported from the core to the outer layers efficiently enough that the object stays roughly at the same temperature. Therefore, its surface brightness also stays unchanged, and the luminosity simply scales with the surface area.
For higher mass systems, however, the contraction happens too fast, the temperature in the core becomes so hot that a radiative zone develops, and convection becomes inefficient and surplus energy is trapped in the system. As it contracts, the temperature therefore rises steeply, and the surface brightness rises with it. This is partly or completely cancelled by the shrinking surface area, meaning that the total luminosity stays roughly unchanged or growing slightly, but the temperature rises steeply. This is seen as a more or less horizontal track going right-to-left in the diagram; these are called Heyney Tracks
As the figure shows, intermediate mass systems are relatively cool in the beginning, allowing for convection, so they start on the Hayashi Track, but as the temperature gets larger, convection is broken, and it turns onto a Heyney track as the heat cannot escape efficiently anymore.
EDIT: i found this figure from this publication which shows some more detail.
Here, the isochrones are gone, but the diagram is separated into regions corresponding to four different scenarios: Fully convective, fully radiative, Radiative core with convective outer layers, and convective core with radiative outer layers. All stars with $M \leq 3.5 M_{\odot}$ go through some phase with a convective core. All stars of $M \geq 1.5 M_{\odot}$ go through several phases, from possibly being fully convective, over having a radiative core and convective outer layers, over being fully radiative, to developing a convective core and radiative outer layers, before settling on the ZAMS.