In this article Neutrino Transport in core-collapse supernovae, in the description of core-collapse supernova mechanism, it is stated that

The velocity of infalling matter in the core increases as we move out from the core center.

I initially thought this velocity is the free-fall velocity. Since free-fall time varies as 1/sqrt(density) and density decreases as we move outwards from core centre, so free-fall time will be higher, and hence the free-fall velocity will be smaller.

But the infall velocity shows an opposite trend. What sets this 'infall velocity' and why it increases as we move out from the core centre?

  • 1
    $\begingroup$ Interesting question! Could it be that the infalling matter is fighting against radiation pressure from neutrinos, etc...? $\endgroup$ Jul 29 at 15:14
  • $\begingroup$ @ProfRob in some other text I found that this happens at the onset of core-collapse. $\endgroup$
    – Lekha
    Jul 30 at 7:20
  • $\begingroup$ I think you are right and I have been under a misapprehension. $\endgroup$
    – ProfRob
    Jul 30 at 11:53
  • $\begingroup$ so free-fall time will be higher, and hence the free-fall velocity will be smaller Why is that? Is the freefall velocity constant? $\endgroup$ Jul 30 at 15:49

You may want to look at "Stellar Structure and Evolution" by Kippenhahn, Weigert and Weiss. Paragraph 36.3.1 "Simple collapse solutions" contains just what you are looking for. In particular, figure 36.5 shows the velocity profile as a function of the radial mass coordinate. I have reproduced the figure here:

velocity profile as a function of radial mass coordinate

The reasoning is more or less as follows: you can model the collapsing core as a relativistic degenerate gas with equation of state

$$P = K \rho^{4/3}$$

You also need the equation of continuity of mass $$\frac{dm}{dr} = 4\pi r^2 \rho$$ and the Euler equation $${dv_r \over dt} + {1 \over \rho}{\partial P \over \partial r} + {\partial \Phi \over \partial r}= 0$$

If you define a scale factor $a$ and a radial coordinate $z$ such that $r=a(t)z$, you can see that the system of equations decouple into one equation for $a(t)$ that can be easily integrated, and one equation for $\rho(z)$:

$${1 \over z^2}{d \over dz}\left(z^2{dw \over dz}\right) + w^3 = \lambda$$

Which is very similar to Lane-Emden equation. Here $w(z)$ is defined such that $\rho(z) = \rho_c w^3(z)$, just like in Lane-Emden. The parameter $\lambda$ is a constant that measures in some way the deviation from hydrostatic equilibrium. Indeed, if $\lambda = 0$, you obtain the equilibrium Lane-Emden equation.

Just as with the Lane-Emden equation, this equation admits solutions only up to a total mass $M<M_{c}$, where $M_c$ is not much higher the the Chandrasekhar mass.

The result is that if the mass of the collapsing core is greater than $M_c$, the whole core cannot be described by the solution of the equation. The inner part of the core, up to $M_c$ will follow the outlined equation and will therefore fall homologously with $v_r = \frac{\dot{a}}{a}r$: the radial velocity is higher the further from the centre. The external part of the core, not described by the equation, will instead be nearly in free fall.

Furthermore, the Chandrasekhar mass is proportional to $\mu_e^{-2}$. But electrons are captured during the collapse, thus $\mu_e$ increases and $M_{c}$ decreases. This means that during the collapse the part of the core that collapses homologously will shrink.

This was just an outline of the process, if you want more details I vigorously encourage you to take a look at the Kippenhahn.

tl;dr: The inner part of the core collapses homologously with $v \propto r$, while the outer part is nearly in free fall with $v \propto r^{-1/2}$

Supersonic infall and some more references

Prompted by Lekha's comment, I have found more references on the subject. Kippenhahn(1) bases his argument on Goldreich(3), and Yahil(4)(5). The figure comes instead from Müller(2).

Following Yahil(4), since the velocity of the infall cannot exceed the free-fall velocity $v_{ff} \approx \sqrt{\frac{2Gm}{r}}$, the mass of the inner part of the core can be determined by finding where the velocity of homologous collapse $v \propto r$ equals $v_{ff}$. Numerical calculations show that in the outer part $\rho r^3 = const$. Reasoning roughly, this means that $m(r) \sim log(r)$ and $v_{ff} \sim \rho^{1/6}$. But also the sound speed goes as $c_s = {\partial P \over \partial \rho} \sim \rho^{1/6}$, therefore $\frac{v}{c_s} \approx const$. Indeed the results of the simulation show that the infall velocity of the outer core is constantly around Mach 2. Keep in mind, tough, that Yahil's model doesn't assume a politrope with $\gamma = \frac{4}{3}$ and is therefore more general than the one exposed here.


(1) Kippenhahn, R., Weigert, A., and Weiss, A., Stellar Structure and Evolution. 2012. pp. 462-465. doi:10.1007/978-3-642-30304-3.

(2) Müller, E. "Computational Methods for Astrophysical Fluid Flow, Saas Fee Advanced Course 27". 1997. ed. by LeVeque, R.J., Mihalas, D., Dorfi, E.A., Müller, E. (Springer, Berlin Heidelberg), p. 343

(3) Goldreich, P. and Weber, S. V., “Homologously collapsing stellar cores”, The Astrophysical Journal, vol. 238, pp. 991–997, 1980. doi:10.1086/158065.

(4) Yahil, A. and Lattimer, J. M., “Supernovae for pedestrians”, in Supernovae: A Survey of Current Research, 1982, vol. 90, pp. 53–70. (I couldn't find this paper, but it is referenced in the following one)

(5) Yahil, A., “Self-similar stellar collapse”, The Astrophysical Journal, vol. 265, pp. 1047–1055, 1983. doi:10.1086/160746.

  • $\begingroup$ So 'The velocity of infalling matter in the core increases as we move out from the core center.' must be true for the inner core, right? But the author writes 'Eventually, the infall velocity exceeds the local sound speed, i.e., the infall becomes supersonic.' So, how does the velocity of the outer part of the core, which is in free-fall, becomes supersonic? $\endgroup$
    – Lekha
    Jul 30 at 17:35
  • $\begingroup$ @Lekha yes, the outer part of the core is supersonic, because the density of the outer part is significantly less to the point that sound speed is lower than the free fall speed. I have looked up the topic and it seems that this results is mainly derived by numerical calulation. I will add a paragraph and some additional reference to the answer. $\endgroup$
    – Prallax
    Aug 1 at 15:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.