First off, I realize that there is already a number of questions relating to gravitational binding in cosmology:

However, none of these go in quite the same direction that I’m wondering about here (despite some very general titles), and neither do the corresponding answers.

This was prompted when I saw these lines in the GADGET-4 code that calculates the gravitational binding energy for a set of $N$-body particles. Translating to equations, it looks like it’s using the following definitions: $$ \begin{align} E &= T + V \\ T &= \frac{1}{2} m (\vec{v}_{\text{pec}} - \vec{v}_{\text{pec},\text{com}} + H(\vec{r} - \vec{r}_{\text{com}}))^2 \\ V &\le 0 \\ \text{“gravitationally bound”} &\Leftrightarrow E < 0 \end{align} $$ I’m trying to understand how this criterion can be derived. I think I have two main concerns that I’m not sure about:

  1. Since energy is not conserved in an expanding spacetime, does it still make sense to compare energies like this? In other words, could masses become unbound (according to this definition) due to the decay of peculiar velocities or an increase in $H$?

  2. Should a Hubble flow velocity $H(\vec{r} - \vec{r}_{\text{com}})$ really be included here? This answer seems to suggest that, when self-consistently taking into account the effect of the (non-homogenous, clumped) matter on the metric, there shouldn’t be any Hubble expansion on small scales (how small?). Further complicating the issue are statements like in this Wikipedia article commenting on “Effects of expansion on small scales”

    Once objects are formed and bound by gravity, they “drop out” of the expansion and do not subsequently expand under the influence of the cosmological metric, there being no force compelling them to do so.

    and elsewhere, suggesting that gravitationally bound objects do not expand – see also these questions:

I suppose in some sense both of these aspects go back to the issue to which extent the idealized FLRW relations apply locally in a collapsed, overdense object. Slightly complicating things is that in the framework of the the code, a Newtonian approximation with a static FLRW background is used – perhaps this subtly changes the answer?

What I’d really like to see is a mathematical derivation that gives the equations above, but I suspect this may not be straightforward if this definition is not rigorous in the first place.


2 Answers 2


The $E<0$ criterion for boundness is approximate, but not because the universe is expanding.

We should first note that all dynamics in Gadget-4 are Newtonian, so I will frame this discussion in terms of Newtonian gravity. General relativity is an extremely minor correction in the context of galactic dynamics and large-scale structure. It is only relevant to cosmological calculations at scales comparable to $1/H$, and even then, it is possible to pick coordinates ("N-body gauge") in which the naive Newtonian evolution is accurate (at least as long as the perturbations at the relevant scales are small).

Also, expansion of space is not a physical phenomenon. It's just a choice of coordinates. The decay of peculiar momenta (leading to decay of peculiar velocities and the cosmological redshift) just arises from the fictitious force associated with picking an expanding coordinate system. If you pick a static coordinate system, there is no longer any such effect. Incidentally, that's the reason it's appropriate to consider physical velocities (which include the Hubble flow) instead of peculiar velocities.

The main reason to worry about boundness in a cosmological context is not cosmic expansion but rather the fact that there is no reference potential at infinite distance. At large enough scales, the potential always scales as $$\Phi\propto H^2 r^2\tag{1}$$ at distance $r$ due to the homogeneity of the universe. This diverges as $r\to\infty$. The coefficient of this proportionality could be positive or negative, depending on the universe's content (it's positive for matter domination and negative for dark energy domination). Shouldn't this mean that either every particle in the universe is bound or that every particle is unbound? After all, the idea of setting $E<0$ as a criterion for boundness was that it means the particle does not have enough energy to escape to infinity.

However, that is irrelevant to what Gadget-4 is doing. Gadget-4 is trying to identify which particles are bound to a "subhalo" of dark matter. Gadget-4 only searches for subhalos inside "groups", which are identified by essentially a density criterion (the "friends-of-friends linking length") that guarantees that a group is orders of magnitude denser than the cosmological average. The cosmological background has no relevance because

  1. due to the Newtonian shell theorem, it has no gravitational influence within the group or subhalo;
  2. while its presence could be relevant to particles that escape the group, such particles would have already escaped the subhalo.

A more theoretical point -- relevant to the question of boundness in an expanding universe, but not relevant to what Gadget-4 is doing -- is that due to the shell theorem, the divergence of the potential at large distances is not even relevant to the dynamics of the cosmological background. Different concentric spherical shells that follow the Hubble flow do not cross, so to understand whether they are bound (i.e. whether the universe will collapse), it is only necessary to ask whether $E<0$ based on the mass that they enclose.

Why the $E<0$ criterion for boundness is approximate

As I mentioned above, subhalos exist inside groups. There are tidal forces exerted by the group and by other objects within the group. These forces reduce the energy threshold needed to escape. One nice way to visualize this is to think about the boosted potential: boosted potential Gadget-4 is considering only the self-potential (left). $E>0$ is the criterion to escape the self-potential, which is calculated based on the subhalo's mass alone. The boosted potential is calculated based on both the subhalo and its environment, but it's transformed ("boosted") into the subhalo's accelerated reference frame. Here it can be seen that a particle only needs to exceed the energy associated with the purple contour to escape toward the center of the group or the yellow contour to escape away from it.

Also, the subhalo changes over time. If it is the central subhalo of the group, it likely accretes material, deepening its potential over time. As a result, particles that were deemed to be unbound at some point might become bound later. If it is a satellite subhalo, it loses material over time due to tidal forces. Particles that were deemed to be bound at some point would become unbound later. In principle, none of a satellite's material is bound over sufficiently long time scales.

  • $\begingroup$ Thanks, that helps. Though I’m not sure in what way dark energy would be accounted for here. At least in the code, it doesn’t seem like it enters in an obvious way, since GADGET-4 just calculates the Newtonian expression for an N-body potential. Also, when you say that the change to the universe on average doesn’t matter, how can that be when $H$ appears in the definition of $T$? Wouldn’t a change in the universe’s average, i. e. $H$, lead to a different value of $T$? Or would this be compensated by a corresponding change in $V$ (again for the code, I don’t see that it would)? $\endgroup$
    – Socob
    Commented Oct 25, 2023 at 15:25
  • 1
    $\begingroup$ @Socob I restructured the answer to more directly address what's relevant to Gadget-4, and that removed the mention of dark energy. $H$ only appears in the calculation of the kinetic energy as part of the transformation from comoving to physical coordinates. Conceptually, it's subtracting out the influence of the expanding coordinate system. If you didn't include $H$, your result would improperly be affected by the expanding coordinate system. $\endgroup$
    – Sten
    Commented Oct 25, 2023 at 18:46
  • $\begingroup$ Ah, I think I got confused by the way the equation was written. Is the coordinate transformation just a Galilean transformation $\vec{r}' = \vec{r} - \vec{r}_{\text{com}} - (\vec{v}_{\text{pec},\text{com}} + H\vec{r}_{\text{com}}) t$, such that $\vec{v}' = (\vec{v}_{\text{pec}} + H\vec{r}) - (\vec{v}_{\text{pec},\text{com}} + H\vec{r}_{\text{com}})$? $\endgroup$
    – Socob
    Commented Oct 26, 2023 at 16:40

I think point 1.) in my question was addressed in Sten’s answer.

Concerning point 2.), I think I just got confused by the way the equation is written in the code, and it’s actually much less profound than I thought. It looks like what’s really going on is just that the energy is calculated within the center-of-mass frame of the halo. Switching to this frame is done using a Galilean transformation $$ \vec{r}' = \vec{r} - \vec{r}_{\text{com}} - (\vec{v}_{\text{pec},\text{com}} + H\vec{r}_{\text{com}}) t, $$ which implies for the velocities $$ \vec{v}' = (\vec{v}_{\text{pec}} + H\vec{r}) - (\vec{v}_{\text{pec},\text{com}} + H\vec{r}_{\text{com}}). $$ With this, the kinetic energy in the halo’s center-of-mass frame becomes $$ T = \frac{1}{2} m \vec{v}'^2 = \frac{1}{2} m \big((\vec{v}_{\text{pec}} + H\vec{r}) - (\vec{v}_{\text{pec},\text{com}} + H\vec{r}_{\text{com}})\big)^2 $$ which is exactly what is being calculated in the code!


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