See also this Astraveo video:
A concept I've had to be thinking about a lot lately is something really cool called radiative levitation. This is when you have a light that's so bright that it can push atoms around and if its strong enough it can even counteract the force of gravity. You'll see this kind of concept come up not just in astrophysics but in other cool ideas like solar sails, that rely on the momentum from light to provide propulsion.
The reason I've been thinking about this is because in my research, I've been working with a system where a disk of matter is suspended on radiation coming from a compact object. I thought it would be fun to go through the math of that, and one of the easiest places to start is the Eddington luminosity, which is the maximum brightness a star can output before it starts to blow out the atoms its made of, producing something called a stellar wind. It's a really neat and straightforward derivation written down by Sir Arthur Eddington in the 1910s.
The basic idea is this. We're going to consider the simplest scenario, an electron floating around in a photon gas, bound by gravity to the star below it. This is called the classical Eddington limit. How bright do the photons have to be so that when they bump into the electron, they can transmit enough force to overcome the force of gravity? That's the fundamental question we're asking here.
Recall that force is simply a change in momentum with unit time, so we need to calculate how much momentum the photon gives to the electron. We'll assume that the post-interaction re-emitted photons don't have a preferred direction, so the average momentum of the emitted photons averages out to zero, and therefore we can assume the full incident momentum gets transferred to the electrons on average. To calculate the momentum $p$, we're going to use Einstein's equation from special relativity. You're probably familiar with the famous equation,
$$E = mc^2$$
but this is only part of the full equation, the part corresponding to the energy of a particle at rest. The full equation, where the total energy of a particle $E$ is its rest energy combined in quadrature with its energy from its momentum, is,
$$E^2 = (pc)^2 + (mc^2)^2,$$
Since a photon has no mass, $m=0$, and then $E^2 = (pc)^2 \implies E = pc$ and thus $p=E/c$. The energy of a photon comes from quantum mechanics and is simply $E_\gamma = h\nu$, where $\nu$ is the frequency of the photon. (So a higher frequency photon has more energy.) Thus our change in momentum $\delta p$ imparted to the electron from the collision is
$$\delta p \sim \frac{E_\gamma}{c}.$$
That gives us the change in momentum, but we need a timescale over which the change in momentum occurs to calculate a force, since $F = dp/dt$. When you have a gas of colliding particles—like our electron floating around in a sea of photons—one way you can think about the timescale for collisions is in terms of a mean free path $\lambda$, which is the average distance a particle will travel in a volume before it runs into another one. We can imagine that the electron will collide with a photon roughly once every mean free path, and since the photons are traveling at $c$, that collision timescale is,
$$t_{\rm coll} \sim \frac{\lambda}{c}.$$
The actual length of the path depends on two numbers, the density of the particles in the fluid and how much those particles interact. You can imagine that as the density of the particles in a volume goes up, the harder it is for particles to get through, and so the shorter the mean free path $\lambda$ is. If the number of photons in the gas per unit volume is $n_\gamma$, and the cross-sectional area that will likely produce an interaction is $\sigma_{\rm th}$ (I'll explain why its called this in a moment), then,
$$\lambda \sim \frac{1}{n_\gamma \sigma_{\rm th}}.$$
The reason we call this cross-sectional area $\sigma_{\rm th}$ is because it changes depending on the properties of the particles you're simulating, and this one in particular in the Thomson cross section, which is the cross-section related to electrons scattering with low-energy photons. That's why we're using it here. If the particles that are interacting have a stronger interaction than electrons and photons, for example, this area might be bigger. So now we have our collision timescale,
$$t_{\rm coll} \sim \frac{1}{n_\gamma \sigma_{\rm th} c}.$$
This lets us easily produce a force,
$$F \sim \frac{\delta p}{\delta t} \sim \frac{h\nu}{c} n_\gamma \sigma_{\rm th} c \sim h\nu\, n_\gamma \sigma_{\rm th}.$$
But remember, we're looking for a luminosity, or how bright the star has to be. So what luminosity will produce such a force? As it turns out, there's a nice way to relate luminosity and force, and it comes down to this thing called "flux". The flux $F_*$ can be thought of as measuring the amount of water flowing through a pipe with time. It's a measure of how many things pass through an area in a particular interval of time. You can think of this as sitting outside a door and counting how many people pass through in one hour. Or, another way to think about it is imagining a whole bus of people that pass through the door, bus after bus after bus. Now you don't need to count the people individually. You just need to know the density of the people on the bus, and multiply it by the velocity of the bus, which gives you the flux of people through the doorway with time. If instead of people we're talking about energetic particles in a fluid, we can use this analogy to relate the flux of energy to the energy density by
$$F_{*} = u v,$$
where $u$ is the energy density and $v$ is the velocity of the fluid. In physics terms, we would say this for a direct free streaming beam—its a little bit different if the particles are isotropically bouncing around.
So we can actually calculate this flux with all the ingredients we have above. We know the density of photons in the gas, which is $n_\gamma$. Since each photon has energy $E = h\nu$, the energy density is just $u = E n_\gamma$. And all photons move at the speed of light, so $v=c$. So we actually have this really nice relationship between flux and energy density,
$$F_* = h \nu\, n_\gamma c.$$
What can we do with this? Well, remember our force equation? If you look closely, you can see there's actually an $h\nu n_\gamma$ already in there. It turns out we were working with a flux (divided by $c$) the whole time and didn't realize it! So if substitute our new $F_*$ in, we find
$$F \sim h\nu\, n_\gamma \sigma_{\rm th} \sim \frac{F_* \sigma_{\rm th}}{c}.$$
Now, we can relate this directly to luminosity. A flux is the energy per unit area per unit time; a luminosity is just energy per unit time, so its just the flux added up over the whole area you're measuring (in our previous analogy, the door). In this case, its the area of the star at the radius where the electron is hovering, which we'll call $R$. So flux and luminosity are related by,
$$F_* = \frac{L}{4\pi R^2},$$
which gives us an easy way to put the luminosity in that force equation,
$$F \sim \frac{F_* \sigma_{\rm th}}{c} \sim \frac{L \sigma_{\rm th}}{4\pi R^2 c}.$$
So we've achieved the first part of our goal, which is answering the question: how much force does a luminosity generate on an electron? To make the electron levitate, that force needs to be greater than the gravitational force of the star pulling on the electron, which we'll remember from introductory physics is just
$$F_g = \frac{GMm_e}{R^2}.$$
So, $F > F_g$ for the electron to start levitating. If we find where that limit is, we see that
$$F = F_g \implies \frac{L \sigma_{\rm th}}{4\pi R^2 c} = \frac{GMm_e}{R^2} \implies L_{\rm edd} = \frac{4\pi G M m_e c}{\sigma_{\rm th}}.$$
And there you have it! The classical Eddington limit. Interestingly, the radius cancelled out—so it doesn't matter how big the star is, only how massive it is. If a star is brighter than this luminosity, it will be bright enough to overcome the force of gravity itself and shoot electrons off its surface! Now the final thing I want to note is a minor subtlety, which is that the Eddington limit isn't usually concerned with the electrons, but rather the protons they are attached to. But since the photons only really interact with the electrons, and the protons carry all the mass, we can just tweak this final formula to use the proton plus electron mass, and all the same logic holds.
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