In Scott Dodleson's book, he considers that baryons are all tightly coupled. I know this can be proved by calculating the interaction time scales and comparing with Hubble's Expansion time scale.

However I don't understand how we can neglect repulsive interaction like e-e or p-p. Shouldn't they destablise the baryon fluid because e-e interaction will perhaps be the strongest and make electrons fly away.

Can someone explain?

  • $\begingroup$ Can you please mention the specific segment of Dodelson's book you are referring to? This will help to understand the particular context better in order to answer your question. And also, firstly, electrons are not baryons. Second, some electromagnetic interactions are later taken care of in the book; probably in Ch. 4 as far as I can remember. Third, which interactions will be predominant depends upon various factors such as the strength of the scattering etc. Fourth, baryons are coupled via elecromagnetic (photon exchange) and weak interactions (n-p conversion) etc. $\endgroup$ – VacuuM Aug 11 '18 at 12:01
  • $\begingroup$ @VacuuM I am referring to the 4th Chapter (4.6 to be precise), as you remember correctly. And also as far as I have read until now all kinds of matter excluding Dark Matter are equivalently called Baryons, at least in elementary levels. I mean they always take electrons and protons as a single baryonic fluid. I believe the answer to my question is that all electrons are considered as a single electronic fluid by considering only their group properties and not micro interactions. However I am not convinced. $\endgroup$ – Indigo1729 Aug 11 '18 at 13:46
  • $\begingroup$ Try looking up topics like Debye shielding as a starting point. $\endgroup$ – honeste_vivere Aug 11 '18 at 21:10

In the integrated Boltzmann equation for electrons, the corresponding collision term for electron-electron repulsion $e(\vec{p})+e(\vec{q})\leftrightarrow e(\vec{p}^\prime)+e(\vec{q}^\prime)$ would be, following the notation given in section 4.6 in Scott Dodelson's book, $$\langle c_{ee}\rangle_{pqp^\prime q^\prime}=\int\frac{d^3p}{(2\pi)^3}\int\frac{d^3q}{(2\pi)^3}\int\frac{d^3p^\prime}{(2\pi)^3}\int\frac{d^3q^\prime}{(2\pi)^3}\delta^{(4)}(p+q-p^\prime-q^\prime)|\mathcal M|^2\\\frac{1}{8E(p)E(q)E(p^\prime)E(q^\prime)} \left[f(p^\prime)f(q^\prime)-f(p)f(q)\right].$$ Notice that all the distribution functions ($f$) and the energy functions are exactly same as we are considering scattering within the same species. Now it is important to understand that the variables $\vec{p},\vec{q},\vec{p}^\prime$, and $\vec{p}^\prime$ are dummy variables and can be interchanged, as follows, in principle without any resulting change in the integral value, $p\leftrightarrow p^\prime,q\leftrightarrow q^\prime$. This alteration keeps the integration measure unchanged but the integrand, being antisymmetric in the exchange, gains a negative sign. Thus, the integral value is its own negative, i.e., the integral vanishes. It makes sense as a scattering process which which keeps the number of particles conserved shouldn't change the nuber density of the particles, i.e., $\frac{dn_e}{dt}$ should have no contribution from such a collision process.

Similarly, the argument also works for proton-proton scattering. And that's why one can be excused from not mentioning this detail.

Caveat: However, this is only the zeroth moment of the Boltzmann equation. There is no guarantee that the higher order moments of the equation will also vanish because in higher moments this nice antisymmetric property of the integrand may disappear. This should not be a huge problem in the approximations considered here: in this particular situation, the contribution from Compton scattering predominates over that from Coulomb's scattering. And also the velocity of Baryons is pretty small.


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