In hindsight, I really regret looking directly at the sun

For realises though, I'm gonna spend some time actually studying the star, because I'm here on this very disappointingly empty planet with nothing and noone on it and I might as well.

Let's start out with an HR-diagram. There were these two guys, Hertzsprung and Russel who decided to put different stars on a diagram showing the colour on the x-axis and the luminosity (how bright they were) on they-axis, and what they found were some very striking trends:

You can see the diagonal line going from top left to bottom right, right? That's the main sequence, where most stars we know about are. There are also the giants up top right, that are really bloody big stars that are also surpsingly cold for their size. Finally we have the dwarfs down bottom left. These are very small stars that are usually the remnants of supernova explosions or just dead stars and usually follow after a star has been a giant (in fact, most stars become a giant after being in the main sequence).

Anyways, I wanna point to where my star is on that diagram:

How do I know where it is placed in the diagram? Well, I already know the temperature from my databases as 4098 kelvin, and you can use the temperature to calculate the luminosity thusly: (actual definition of luminosity is amount of energy radiated per second, but I just say that it's how bright it is, because it really is) (also wow thusly is a fancy word)

At any one point of the stars surface the amount of energy radiated, or pointflux if you will, is proportional to the temperature raised to the power of four: \(\phi = \sigma T^4\), where \(\sigma\) is the universal Boltzmann constant and \(\phi\) is the pointflux.

if we assume that the temperature across the surface is constant, which is not crazy (you're crazy shut up), we can get the luminosity for the entire star by multiplying by the pointflux with the surface area of the star:

\(L = 4\pi R^2\sigma T^4 = 0.0844\), where the number is in relation to the actual suns luminosity

When you have both the temperature and the luminosity, you can just plug the point right into the diagram and presto!

Through some fancy mathemagics (that I will not include, the math is magical and also a bitch) you can express luminosity as aproximately proportional to mass and temperature separately: \(L \propto M^4, L \propto T^2\). Using some star we already know a lot about (the sun, duh), we can find the proportionality constant and use that to find an approximation for the luminosity where we don't have to know silly things like the radius and.... well, we don't have to know the radius at least.

If we test this with the sun we get that 

\(\frac{L}{L_\odot} \propto M^4 = 0.0415\)

\(\frac{L}{L_\odot} \propto T^2 = 0.5029\)

which, when dealing with logarithmic diagrams, isn't too far off. I mean, technically M should be raised to some constant \(\beta\) that ranges from 3 to 4 and the stars in the universe just seem to have a preference for 4, but if we choose \(\beta\) to be closer to 3 we get a much better answer so hey the star's a bit abnormal don't judge him for it, just leave him alone.
The temperature proportion, meanwhile, is just relatively fucked.

When stars get born (when a daddy star and a mommy star love eachother very much they blow eachother up and create new stars from eachothers carcasses), it usually begins as a cloud of dust particles just hanging round in space, not being up to much but probably being able to fit something into their very loose schedule sometime in the future. If this cloud is too thin, the particles may never get together, so there is probably some limit to how thin or thicc it can be before it actually can collapse. I'm also kinda curious about how large that cloud could have been if it eventually formed my star, so we might as well give that a shot.

We'll start off with an assumption, because the math is annoying, tedious and the voices in my head say I don't have to do it;

\(K = -\frac{1}{2}U\) (this is called the virial theorem btw, look it up if you want to)

This means that the total amount of kinetic energy is equal to minus a half of the potential energy (which is negative, and I can never wrap my head around why the fuck we do it that way, having a separate form of energy just seems so much easier but that's just me)

and assumption for this virial theorem is that the cloud is stable. This means that if it had some more kinetic energy it would fly apart (you can reason your way to that one) and if it had less it would collapse! If we know that you can find the total kinetic energy in a gas from the temperature (\(K = \frac{3MkT}{2\mu}\), where M is the total mass of the cloud, k is the Boltzmann constant, T is the temperature and \(\mu\) is the average mass of all the particles) and the total potential energy from the mass and radius (\(U = \frac{3GM^2}{5R}\), where G is the gravitational constant, M is the mass of the cloud again and R is the radius of the cloud because of course this cloud is spherical and has evenly distributed mass. This is physics, do you think we do things complex here?) (I will show you how to work out these expressions later, btw) 

Given this we can say that

\(\frac{3MkT}{\mu} < \frac{3GM^2}{5R}\)

where we can isolate the M or R or whatever to find that as long as we have everything else! If we assume that the cloud that created my star didn't lose too much mass, that it had a pretty average composition of 75% hydrogen and 25% helium and that it had a fairly low temperature like 10 kelvin, we can isolate the R so that \(R < \frac{MG\mu}{5kT} = 806258.579767\), which is cool I guess.

But here is the most important question of all, the question with so much impact, the question you've all been waiting for!

where would a cloud of that maximum size fit into the HR-diagram!??!?!??!?!!?!!?!?!?!?!?!?!?!!

Truly, we struggle with the great and important questions here today. Please refrain from crying at the greatness of it all.

Anyways, we can find how much the gas cloud radiates in the same way as for a star; we find how much it radiates in a point, and then take the sum of all the points on the surface. Which means assuming the thing is spherical and that the temperature is constant all along the surface, and just dragging out the same formula we used last time: \(L = 4\pi R^2\sigma T^4 \)

\(L = 4\pi (\frac{MG\mu}{5kT})^2\sigma T^4 \)

\(\frac{L}{L_\odot} = \frac {4\pi (\frac{MG\mu}{5kT})^2\sigma T^4 } {L_\odot} = 0.0585\)

So plugging in numbers gives us that the cloud would have 0.0585 of the suns luminosity, which is less than it currently has which I guess happens. Then again, we used a fuckton of approximations

Now, onto something completely related; can man be hot? Or can sun be hot? How hot is the centre of this star? We can find out, assuming some very gross assumptions;

• The density is the same throughout the entire thing
• The only pressure is gas pressure
• The only thing inside the star are protons

Now that we have a very accurate representation of the star, we can get to work. First order of business is find an expression for the total mass inside a radius r:
Since the density is the same throughout, we can just take the density multiplied by the volume of the sphere given by the radius r:

\(M = \frac{4}{3}\pi r^3 \rho_0\)

Since we assumed the only pressure is the gas pressure from an ideal gas, we can use ideal gas laws concering hydrostatic equilibrium, meaning that the forces of gravytea are held back be the asocial-ness of particles.

Ideal gas law for the pressure: \(P = \frac{\rho k T}{\mu} <=> \frac{dP}{dr} = \frac{\rho k}{\mu}\frac{dT}{dr}\)

Hydrostatic equilibrium for pressure: \(\frac{dP}{dr} = -\rho(r)g(r)\), where \(g(r) = G\frac{M(r)}{r^2} = G\frac{4}{3}\pi r\rho_0\)

if we combine these equations, we get that:

\(\frac{dT}{dr} = -\frac{4\mu}{3k}G\pi r\rho(r)\) where \(\rho(r) = \rho_0\)

If we then integrate this from the core (\(r = 0\)) to the surface (\(r = R\)) we get that 

\(T(R) - T(0) = -\frac{4\mu}{3k}G\pi \rho_0\frac{R^2}{2}\)

\(T(0) = T_{core} = T(R)+\frac{2\mu}{3k}G\pi\rho_0R^2\)

Since we have assumed that the star just exists of a bunch of protons, we have everything we need! We use the protons to find the mean molecular mass (\(\mu\), the proper name for it), check the surface temperature for T(R), use the mass and the radius to find \(\rho \) and baddabim baddabom, we get that the core temperature is just above 9 million kelvin.

We wanna find the luminosity due to the fusion at the core of the star. Since the core is below 90 million kelving (by a long shot) the prevailing reactions are the pp-chain (I know what you're thinking, and I completely agree with you. It's fucking hilarious) and the CNO-cycle. We also add the assumption that the core radius is 0.2 times the radius of the star, because we don't wanna get too accurate here, right?

These reactions give out an energy given as (just take these formulas, they're strange):

\(\epsilon_{pp} \approx \epsilon_{0,pp} X_H^2\rho T_6^4\)

\(\epsilon_{CNO} = \epsilon_{0,CNO}X_HX_{CNO}\rho T_6^{20}\)

where both \(\epsilon_0\)'s are constants:

 \(\epsilon_{0,pp} = 1.08e-12 W m^3/kg^2\)

\( \epsilon_{0,CNO} = 8.24e-31 W m^3/kg^2\)

The X's showing to how much a particletype makes up of the total mass and the T refers to the temperature. The 6 next to the T means that we measure it in millions of kelvin, which means that we put in \(9^{20}\) there, since we have 9 000 000 kelvin in the core.

The definition for the X's is \(X_A = \frac{n_Am_A}{nm} = \frac{\text{total mass in type A nuclei}}{\text{total mass}}\)

If we assume the core to be made up of 74.5% hydrogen, 25.3% helium and 0.2% carbon, oxygen and nitrogen collectively, then:

\(X_H = \frac{0.745m_H}{m_H + m_{He} + m_{CNO}} = \frac{0.745m_H}{\Sigma m}\)

\(X_{CNO} = \frac{0.002m_{CNO}}{m_H + m_{He} + m_{CNO}} = \frac{0.002m_{CNO}}{\Sigma m} \)

Since we already assumed the density to be constant, we already have the \(\rho\), and the \(T_6 = 9\), so we just plug in the numbers and everything should be fine! We get that 

\(\epsilon_{pp} \approx 1.07721192682\times 10^{-08}\)

\(\epsilon_{CNO} = 6.36250735969\times10^{-10}\)

which is the energy radiated per kg, so we still have to multiply by the total mass of the core: \(M_{core} = \rho\frac{4}{3}\pi R_{core}^3\)

\(L = M_{core} (\epsilon_{pp} + \epsilon_{CNO}) = 8.38423 \times 10^{19}\)

Which is wayyy less than the sun. Why is it less than the sun, you ask? welp, I have no fucking clue. it could be that my star is half as warm as the sun at the core (9 v 15 million kelvin), and fusion reactions are very heat sensitive, or maybe our assumptions fucked us up, or maybe I just made some calculation errors, who knows. My calculator shows this number and therefore it is gospel truth and nonbelievers shall be stoned to death.

An interesting question is if that is in fact the luminosity, how hot would the star be at the surface? If we just rework the formula used to find the luminosity earlier to find the temperature instead, we can find that out!

\(L = 4\pi R^2\sigma T^4 <=> T = \sqrt[\leftroot{-2}\uproot{2}4]{\frac{L}{4\pi R^2\sigma}} = 164k\)

...

so 164 kelvin seems reasonable, right?

that's how hot a star usually is, right?

I for one, think none of the error is my fault. I think that it's all because of those assumptions, especially that the only pressure acting outwards is gass pressure. I think that fucked up bigtime to be honest. Welp, done is done and spice is spice, the star is cold now, this is law.

So how old is our star, actually? You should never ask a lady her age, but the star is genderless, so in your face, star! 
Are you ready? Aight, I'mma explain the basic principles behind the mass of a star being proportional to its age. We start off with luminosity being proportional to mass, which we have already established. Then we assume (people say don't assume because you make an ass out of u and me, but I say butts are nice. Make more assumptions) that luminosity throughout the stars life is constant and that there is some fraction of the mass that is radiated away at any point in time, and that this fraction is also constant. We can use this fraction to calculate how much energy the star has/will radiated/radiate away over the course of its entire lifespan, and connect that to me luminosity (basically connecting  the energy radiated over a span of time to the energy radiated at one point in time, it's not that hard). And since the luminosity is already proportional to the mass, we can play connect the dots and find out that the lifespan of a star is proportional to its mass! Well, technically inversely proportional to the mass squared but who gives a fuck about that part. I mean, I'd show you the math on this, but the voices in my head are very insistent on only doing this part briefly, without calculations and with your own words. bah, humbug.

anyways, when I actually do the math, and compare it to the sun to get the proportionality constant, I get that my star will live 10 times as long as the sun. Cool. I can say in bed longer. (No, I'm not running out of zingy one-liners, it's just really really late and I'm hella tired so shuddup

I also wanna know how how my star will move through the HR-diagram as it grows older, but how do we do this?

 Well, the star started in the main sequence, where I pointed to earlier, and will eventually move its way towards the giants as it burns up its hydrogen. On its path towards becoming a giant, it hits a subgiant phase where the core is not hot enough to fuse helium, but a shell around the core is hot enough to fuse hydrogen. As the subgiant expands due to the hydrogen burning in a shell, it moves towards being an actual giant instead of a lowly subgiant. As the star expands, the temperature  on the surface will drop due to the huge increase in surface area. Eventually it reaches a large enough size and becomes a red giant! At this point, there are huge currents moving through the star, moving a fuckton of material around and increasing luminosity by a lot because all the energy gets moved outwards.

This part is common for all main sequence stars, but the following part depends on how massive the star is. If the star is, like my star, less massive than like two solar masses, the core becomes so dense that quantum effects set in; The core becomes electron deficient, which means that the electrons literaly can't get closer to eachother because quantum physics say so. This creates an outwards force that keeps the star stable that is indepedent of temperature, meaning that the star won't increase in size while this is happening, or while the temperature increases. This will continue until the sun is in fact warm enough to fuse helium, at which point the entire degenerate shell will blow up all at once LIKE A GIANT LAYER OF C4  OR ME AFTER A TRIP TO TACO BELL!!! This is called a helium flash. It's kinda cool.

Now the paths for big and small stars realign themselves for a while. At this point the core starts expanding again, pushing hydrogen burning closer to the surface which makes the hydrogen burning slow down. Because Helium is not as good as hydrogen (get outta here, nerd) at burning, the luminosity will decrease, and it will start contracting again. This will increase temperatures and just generally move the star towards the left in the HR-diagram

Eventually it contracts enough to become electron degenerate again, and ensues a temperature-independent push outwards from the core. At this point the core is made up of mostly carbon and oxygen. There is also created another electron degenerate helium shell (aside from the electron degenerate oxygen-carbon core), that repeatedly blows up, like a very very very very slow machine gun that also blows a lot of mass away from the star. Eventually only the degenerate oxygen-carbon core remains. At this point the luminosity has decreased drastically and the star has moved towards its final phase at the bottom left of the HR-diagram as a white dwarf.

yay, that was a trip. Was it fun for you? was sure not fun or me. too much writing. Anyways, off to more math-y stuff! Like calculating how large a white dwarf would be!

Because lots of matter has been thrown off, we're gonna assume the white dwarf has the mass of like \(\frac{M}{8M_\odot}M_{Chandrasekhar}\), where M is the original mass, \(M_\odot\) is the solar mass, and \(M_{Chandrasekhar}\) is the largest mass a star can have while still becoming a white dwarf, called the Chandrasekhar limit. (Could you guess that from the subscript?)
To actually calculate the radius of a white dwarf, we have to know some stuff about electron degeneration. I think I might just spare you some of that (if by you I mean me, the formulas are a blight to the eye and a bitch to write down), and just explain the principles to ya.

\(R_{WD} = (\frac{3}{2\pi})^{\frac{4}{3}} \frac{h^2}{20m_eG} (\frac{Z}{Am_H})^{\frac{5}{3}} M^{-\frac{1}{3}}\)

This is the equation for the radius of a white dwarf. It has taken the formula for hydrostatic equilibrium and exchanged the factor for gas pressure with the formula for degeneration pressure, and it ends up with this thing. The final answer turns out to be something like \(R_{WD} \approx 3315 km\), which is a pretty ok-size white dwarf. When the sun turns into a white dwarf, it'll be something like 7000km, so I think our little star did pretty well for itself! I'm proud of it.

Fun notes:

• The gravitational acceleration on the surface would be just about 981 \(km/s^2\), which is a lot. Don't try to walk there
• one litre of white dwarf would weigh \(10^{12} kg\). Not only is the subject matter heavy to handle, but the subject matter is heavy to handle, hahahahahaha.

Tell you what, though, I promised some mathematics about the parts of the virial theorem, so here we go!

We start out with the potential, because it's the bichiest: If you have some infinitesimal piece of matter in a cloud, that piece, dm, would have a potential, du, equal to \(du = -G\frac{M(r)dm}{r}\). Since the density is equal in all parts of the cloud, as we assumed, M(r) becomes very manageable as \(M(r) = \frac{4}{3}\pi\rho r^3\). We can take this expression for a point and integrate it over an entire shell, and since the expression is only dependent on the radius, we might as well just multiply by the surface of that shell;

\(dU_{shell} = -G\frac{M(r)4\pi r^2 \rho(r)dr}{r}\)

If we then integrate that over the entire cloud, from r = 0 to r = R, we get the total potential!

\(U = -4\pi G \int_{0}^{R}M(r)\rho(r)rdr\)

Here we can remove \(\rho\) by expressing it in terms of M:

\(\rho = \frac{3M}{4\pi R^3}\)

\(U = -4\pi G (\frac{3M}{4\pi R^3})^2\frac{4}{3} \int_0^Rr^4dr = -\frac{3GM^2}{5R}\)

and then for the K:

From thermodynamics (this is as far as I'm taking ya, go look that up yourselves if you want), we can know that the total kinetic energy in a gas is given by \(K = \frac{3}{2}NkT\). You should also be able to see that \(N = \frac{M}{\mu}\), where N is the number of particles, M is the mass of the gas cloud and \(\mu \) is the average mass of a particle.

Therefore: 

\(2K = \frac{3MkT}{\mu}\)

hallelujah, it worked

anyways, I think that was all I felt like doing about the star. I'mma try giving some relativity studies a shot if I have the time. Maybe I won't, who knows.