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How does fermion degeneracy affect elastic scattering in astrophysical compact objects?

+ 6 like - 0 dislike
94 views

As I currently understand it, absorption processes or inelastic scattering involving fermions in a degenerate gas are strongly inhibited, because to change the fermion energy (or create a new fermion) requires it to be placed in a vacant energy state, which in turn means the fermion must be close enough to the Fermi energy that it can attain such a state.

My question here is how are elastic scattering processes affected? And I want to consider two possible cases.

  1. The scattered particle (the one with a much lower mass) is a fermion in a degenerate gas. An example would be an electron in the interior of a white dwarf scattering off a carbon nucleus.

  2. The scattering (more massive) particle is a fermion in a degenerate gas. So examples here could include neutrino scattering from degenerate nucleons inside a neutron star or perhaps even photon scattering off degenerate nucleons.

Does the fact that the degenerate fermion must change its momentum, even though its energy stays (almost) the same, inhibit these processes?


This post imported from StackExchange Physics at 2017-05-08 20:26 (UTC), posted by SE-user Rob Jeffries

asked Apr 26 in Astronomy by Rob Jeffries (110 points) [ revision history ]
recategorized May 8 by Dilaton

1 Answer

+ 2 like - 0 dislike

Take any system which has states labelled by $i$ and occupation numbers labelled by $f_i$. I.e. $f_i$ is the expectation number of particles in the given state and is normalized not to one but as $\sum_i f_i = N$. Now let this system have a probability of transition per unit time from the state $i$ to another one $j$ denoted by $P_{ij}$. This means that $f_i P_{ij}$ particles will jump to $j$ from $f_i$ in unit time. On the other hand, $f_j P_{ji}$ particles will jump from $j$ to $i$ in the same time. Hence, we can characterize the time derivative of the occupation number as
$$\frac{d f_i}{d t} = \sum_j (P_{ji} f_j - P_{ij}f_i)$$
In many systems, particularly reversible ones we have $P_{ij}=P_{ji}$.

Imagine, however, that we are dealing with identical quantum particles, there we can derive from the permutation symmetries that the transition rates have to be modified as
$$\frac{d f_i}{d t} = \sum_j [P_{ji} f_j(1\pm f_i) - P_{ij}f_i(1 \pm f_j)]$$
where the plus is for Bose statistics and the minus for Fermi statistics. In particular, you can see that for fermions the probability of transition from $j$ to $i$ is zero if the state is already occupied.


The specific case of a gas in astrophysics can be modelled by the Boltzmann equation
$$\partial_t f + \partial_p H \partial_x f - \partial_x H \partial_p f = \delta f_{coll}$$
where $f(p,x)$ now stands for the occupation number in a phase-space cell of volume $\sim \hbar$ at $p,x$, and $H$ is some effective single-particle Hamiltonian which represents the free drifting of the microscopic particles in the macroscopic fields. The right-hand side is the collision term which (within a certain approximation) modulates the behaviour of the occupation number due to collision between particles.

Particles following Boltzmann statistics would have

$ \delta f_{coll}(p,x) = \int  Q(p,q \to p',q') [f(p,x) f(q,x)- f(p',x)f(q',x) ] dq' dq dp'$

where $Q(p,q \to p',q')$ is a scattering matrix computed for two particles scattering off each other while being alone in the universe. However, in analogy with the previous part of this answer, fermionic particles have

$\delta f_{coll}(p,x) = \int Q(p,q \to p',q')$

$\left[ f(p) f(q)(1-f(p'))(1-f(q')) - f(p')f(q')(1-f(p))(1-f(q)) \right] dq' dq dp'$

where I have written $f(q,x) \to f(q)$ for brevity.

I.e., the probability of scattering into a state occupied with density $f(p,x)$ will be modulated by a $1-f$ factor. If $f$ is close to one, this scattering will be essentially forbidden. In strongly degenerate gases this means that we can essentially neglect any scattering outcome in the Fermi phase-space surface, be it elastic or non-elastic.

answered May 10 by Void (1,505 points) [ revision history ]
edited May 12 by Void

If you post this on Physics SE you will likely receive a bounty. Your conclusion is clear enough, but I have just a couple of questions about the notation. Pardon my non-theoretical physicist's lack of knowledge here.

Looking at your initial equation I guess the RHS is actually some sort of term representing a small amount of scattering? Then I have to admit I don't understand what the second and third terms represent on the LHS. Could you fill me in?

Finally I guess Q is a transition probability assuming the initial state is full and the final state empty right?

@RobJeffries I expanded the post and reposted on SE, I think now it should be much clearer. Let me know if it makes more sense now.

Thanks, yes it is. I need to go and do some work to understand the origin of terms 2/3 in your Boltzmann equation. Anyway, I guess this reasonably explains why neutron stars are transparent to the neutrinos they produce, since $f \sim 1$ for most possible final momentum states of the neutron involved in any scattering.

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