I tried 3 MARKS from The Asian Physics Olympiad in India

Added: 2026-05-12

[00:00:00]Let's solve a problem from one of the hardest physics Olympiads, the Asian Physics Olympiad. This problem is on one of my all-time favorite concepts, i.e. the Chandrasekhar limit. Okay, we are given quite a lot of information over here. His famous work was carried out in 1930. The Indian physicist Professor Subrahmanyan Chandrasekhar studied the stability of stars. The problem will help you construct a simplified version of his analysis. Now, the first part is a classic physical problem where we need to consider a spherical star of uniform density R and mass M. Derive an expression for its gravitational potential energy EG due to its own gravitational field self-energy. This is a problem that we've actually solved before as part of the International Physics Olympiad and it's a classic.

[00:00:47]Okay, now consider that this here is the star and typically the way we solve these problems is using calculus and building shells around the star. Because of the star is of uniform density, my first trick would be just to write down that the density is equal to the total mass divided by the total spherical volume, which is just 4/3 pi R cubed. Then we're going to take a um a sphere of radius R. I we're sort of building our um our star and we're going to say that the mass as a function of the radius to what? Let's just give it maybe a lowercase m. As a function of the radius will be given by the density multiplied by 4/3 pi R cubed. I'm also going to consider a small shell which is just around it and then we're going to consider the amount of work that the gravitational field will do to clump everything together.

[00:01:56]This uh little mass over here, dm, well, that's going to be equal to the density multiplied by its infinitesimal volume, which is just about 4 pi r squared dr. So, how much energy would it be to add one shell? Let's say that this little amount of energy is equal to de. Well, this here will be minus g and then one of the masses, which is just m of r, and then dm, and then we're going to be dividing this by r. This is just our general formula for gravitational potential energy.

[00:02:38]Now, this here is supposed to be a dm. It's really important those problems to um express everything in terms of distance.

[00:02:46]So, we can get rid of this dm and m is a function of r, and suddenly this entire expression will be in the r world, and then we'll be able to do one of my favorite things ever, and that is to integrate. Okay. So, uh de will be given by minus g. Rather than m of r, I'm going to write the density 4 uh over 3 pi r cubed. Oops, this here is a cube.

[00:03:17]And then dm is equal to the density 4 pi r squared dr. And then we're going to be dividing this by r. Okay, let's tidy up some of those constants. So, those r's here are going to go. So, what are we going to get?

[00:03:40]We're going to get the de will be minus g rho squared. The four and the four is going to give us a 16. The pi and the pi is going to give us a pi squared r cubed and r. This here is going to be r to the power of four. Then we're just left with this three over here.

[00:04:02]Oh, and this dr of course. And you can probably see what's coming next and that is to take this expression and integrate it from zero to r.

[00:04:18]So, we're summing up all of those energies to add a shell from zero to r.

[00:04:23]Physics is just beautiful, isn't it? So, what is that going to give me? It's minus g row squared 16 over three pi squared. And now we just have the integral from zero to r of r to the four dr, which is a nice and easy integral.

[00:04:41]So, this here is just minus g row squared 16 pi over three times five.

[00:04:50]The five is just coming from the integration because uh the integral of r to the four is just r to the five divided by five. But, this is evaluated between zero and r. So, I'm just going to write this as r to the fifth.

[00:05:07]Okay, we're almost there. The only the final thing that we need to do is just take this row and then substitute that back into here. So, this here will be minus g row squared. What is row squared? Okay, so we're going to have m squared at the top 16 pi.

[00:05:29]Ooh, I have a feeling this here should be a pi squared cuz that was a pi squared here.

[00:05:34]Like it's that and then r to the five.

[00:05:39]And then at the bottom we're going to have 15 times This line over here 4 pi r cubed is going to turn to 16 pi squared. So, those are just going to cancel. And then, we are going to have r cubed, which is squared, so that's going to be r to the power of 6.

[00:06:03]I think that's right. The only thing Actually, the only thing that's missing is this factor of uh r uh this factor of 3, which is going to uh appear at the top of the fraction.

[00:06:14]So, this here should be multiplied by 9.

[00:06:17]Okay, the 16 pi's are canceled out.

[00:06:20]This here cancels out. Almost everything uh here seems to cancel out. So, what are we left with? Minus uh the 9 and the 5 uh should give me just 3 over 5 after dividing by 3, g m squared divided by r. And this here is a classic result for the uh overall gravitational potential energy. The next part involves some really beautiful physics. Let's have a look. So, we assume that the star is made up of only hydrogen and that all the hydrogen is in ionized form. Okay, so we've got a whole bunch of protons and a whole bunch of electrons.

[00:07:02]Electrons obey the Pauli exclusion principle, and the total energy can be can be can be computed using quantum statistics. I remember this expression from university, and you may take the total electronic energy, ignoring the protonic energy, to be given this scary-looking expression.

[00:07:19]Where n is the total number of electrons, and h bar is the reduced Planck's constant. Obtain the equilibrium condition of the star relating its radius, r white dwarf, to its mass. And this radius is called the white dwarf radius. The first thing to discuss here is our plan of action. So, if we have some sort of a total energy, let's call it E, the total energy is going to be the sum of the gravitational potential energy, let's call it E subscript G, plus the elec- the energy stemming from Pauli's exclusion principle, um, stemming from what is known as electron degeneracy pressure, the energy of those electrons. So, my plan is to use my result from the previous part, add this to our new expression given, and then find the derivative with respect to the radius. I'm going to set that derivative to be equal to zero.

[00:08:27]This here should give us the minimum energy condition, i.e. the equilibrium condition. Let's give it a shot. Okay, the first thing to write down was that EG was minus 3/5 GM^2 / R. Let's just double check that. Minus 3/5 GM^2 / R. E E on the other hand, that's going to be given by this massive expression h bar pi cubed over 10 m e 4 to the 2/3. These are all constants, so I'm just going to call that C in a minute.

[00:09:08]7/3 n e to the 5/3.

[00:09:15]Now, if we look at this, how many of those things do you think will depend on the radius of the star?

[00:09:22]All of this stuff here is a constant. h pi m e 3/5 pi 7/3 etc. So, I'm going to say that this here is equal to some constant. Well, let's just call it C, but the number of electrons should depend on the radius of the star because if the radius of the star is higher, then we're going to have a higher mass, i.e. we're going to have a higher number of protons in theory.

[00:09:50]I need to make sure to draw this clearly as a subscript by the way because this here is just simply the number of electrons, not the number multiplied by the charge. Okay, so how do we find this number?

[00:10:02]Well, because each of those electrons came from an ionized hydrogen, uh we can say that NE So, let's just write that down over here is going to be have to equal to the mass of the star M divided by the mass of an individual proton. Why am I dividing by the mass of a proton rather than an electron?

[00:10:29]Because most of the star is actually made out of Most of the mass from the star will come from protons rather than electrons. Okay, if we plug that into here, we should get that EE is going to be given by C N E Well, rather than NE, I'm just going to write M to the 5 over 3 divide that by MP to the power of 5 over 3 uh squared.

[00:11:01]Okay, now I think I'm ready to add those two up and then differentiate.

[00:11:08]Okay, we can find our dE by dr which we're going to be setting to zero. So, what is this here going to be? Minus 3 over 5 Gm squared and then the integral of 1 over r is going to be minus So, that's going to turn that into a plus r to the minus 2, which is just going to give us a factor of r squared. Plus Well, is it a plus actually? Because now we have this constant and then when differentiating r to the minus two, which is going to give us r to the minus three, and then we are also um going to uh get a factor of minus two. And then c m to the five over three and then we have m p to the power of five uh over three. Okay, now let's set the derivative, basically this thing here to be equal to zero. So, we can set those two expressions to be equal to one another. So, we can go three over five g m squared over r squared has to be equal to two c m to the five over three over m p raised to the five over three r cubed. Okay, let's bring this over on this side. So, what are we going to get?

[00:12:33]We're going to get r cubed over r squared, which is just going to be r and that's three over five g and that here will be equal to two c over m p five over three. This m squared I'm going to bring down over here. So, what I'm going to be left with? m to the five over three minus two. So, that's going to be uh six over three, which is going to give me minus one over three, I think.

[00:13:07]So, we're left with this expression that r is equal to two c m p to the five over three multiplied by five, divide that by three g m to the minus one over three. Really interesting relationship actually, but R seems to be proportional to M to the minus one over three. So, there's this highly interesting effect that when the radius increases, or when the mass increases, the radius must actually decrease. Who knew this? Now, this here with just a single Olympia problem. To appreciate some more absolutely beautiful physics, you need to have a look at this video in which we solve an entire IPhO problem on relativity. Have a look over here, enjoy the physics.