The virial theorem relates the kinetic energy of a system to the total potential energy of the system: $ \Delta K = -\frac{1}{2}\Delta V $

so it has lots of uses in mechanics, thermodynamics and astrophysics.

My main question is what is the meaning of the word virial in this context?

Related; Quiescent galaxies in a virialized cluster at redshift 2: Evidence for accelerated size-grow refers to a 'virialized cluster' of galaxies. What does it mean here?


We present an analysis of the galaxy population in XLSSC 122, an X-ray selected, virialized cluster at redshift z = 1.98. We utilize HST WFC3 photometry to characterize the activity and morphology of spectroscopically confirmed cluster members. The quiescent fraction is found to be 88+4−20per cent within 0.5r500, significantly enhanced over the field value of 20+2−2 per cent at z ∼ 2. We find an excess of “bulge-like” quiescent cluster members with Sersic index n > 2 relative to the field. These galaxies are found to be larger than their field counterparts at 99.6 per cent confidence, being on average 63+31−24 per cent larger at a fixed mass of M⋆ = 5 × 1010 M⊙. This suggests that these cluster member galaxies have experienced an accelerated size evolution relative to the field at z > 2. We discuss minor mergers as a possible mechanism underlying this disproportionate size growth.

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    $\begingroup$ Interesting question! I've extended the title to cover the scope of your question, and added more information from your link. It's always better to include minimal material within the question post itself, rather than force each and every reader to go offsite. Please feel free to edit further. Thanks! $\endgroup$
    – uhoh
    Commented Sep 10, 2021 at 1:05
  • $\begingroup$ Hopefully your next questions will be "How does one measure the degree of virialization of an astronomical object?" and/or "Can entire galaxies be virialized? Can solar systems?" $\endgroup$
    – uhoh
    Commented Sep 10, 2021 at 1:08
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    $\begingroup$ Hmm, interesting comments. It's left me wondering: How does one measure the degree of virialization of an astronomical object? Can entire galaxies be virialized? Can solar systems? $\endgroup$
    – Jim421616
    Commented Sep 10, 2021 at 3:18

3 Answers 3


A slightly modified version of the virial theorem that you cite states that for a system of N bodyes (galaxies in a cluster) autogravotating

$${1 \over 2} \ddot I = 2K + V$$

Where $K$ is the total kinetic energy, $V$ is the total potential energy and $I$ is the "scalar moment of inertia" of the system, defined as

$$I = \sum_i^N m_i r_i^2.$$

When $\ddot I > 0$ the system tends to expand, when $\ddot I < 0$ it tends to contract. In these cases we say that the system is not virialized (yet).

If the system is left to evolve without external disturbances, the simulations show that $\ddot I$ tends to zero in a time the order of the crossing time of the system, following a curve similar to the one of a damped oscillator. (The crossing time is the characteristic time a typical galaxy takes to cross the system. It can be defined as the size if the cluster divided by the velocity dispersion)

When the system has settled on $\ddot I =0$, and therefore $2K = - V$, the system is said to be virialized. The system as reached a particular kind of statistical equilibrium.

In astrophysics is important to know whether a system is virialized, because we usually want to apply the virial theorem and find the potential energy from the kinetic one, and this can be done only if the system is old enough to be virialized.

  • $\begingroup$ Is that the second time derivative of the moment of inertia you compare to 0 ? $\endgroup$
    – Stian
    Commented Sep 10, 2021 at 9:44
  • $\begingroup$ @StianYttervik yes, that's correct. If you need it I can add the proof of that statement $\endgroup$
    – Prallax
    Commented Sep 10, 2021 at 18:04
  • $\begingroup$ No, I am just unused to the dot notation. Thanks. $\endgroup$
    – Stian
    Commented Sep 10, 2021 at 18:10

According to the Wikipedia article on the Virial Theorem:

The word virial for the right-hand side of the equation derives from vis, the Latin word for "force" or "energy", and was given its technical definition by Rudolf Clausius in 1870.

Investigating this further, one can see from the original publication of Rudolf Clausius that virial and vis viva are two older names referring to the potential energy and the kinetic energy, respectively. My source on that is a quote from Clausius' paper where he states, "The mean vis viva of the system is equal to its virial" as the 'virial theorem'. Since vis viva is an older notion part of the formation of the idea of kinetic energy (see the Wikipedia article on Vis Viva), this leaves the potential energy side to be the virial.

In all fairness, this is a bit of an understatement for the implications Clausius suggested in his time regarding a virial, but these notions of virial and vis viva were stepping stones and evolved into potential and kinetic energy as we know it today. Modern usage of the word probably has different implications, referring less to this historical idea of a 'virial' and more to systems where the virial theorem is relevant. As often is done in astronomy, it seems perhaps tradition has kept the word in association with the theorem around, while the idea of a virial has been less relevant in comparison to the idea of potential energy.

As far as virialized clusters, perhaps the following Wikipedia excerpt from the article on Virial Mass might be enlightening:

A spherical "top hat" density perturbation destined to become a galaxy begins to expand, but the expansion is halted and reversed due to the mass collapsing under gravity until the sphere reaches equilibrium – it is said to be virialized.

In other words, it looks like a system that obeys the virial theorem that’s in equilibrium. For your specific example, the galaxy cluster seems to be satisfying the theorem, and thus be gravitationally bound to one another.

Rudolf Clausius' paper introducing the virial theorem for the first time in this paper:

R. Clausius (1870) XVI. On a mechanical theorem applicable to heat, The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science, 40:265, 122-127, DOI: 10.1080/14786447008640370

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    $\begingroup$ A conclusion from your answer is still lacking. Is it supposed to mean 'in force equilibrium' from the base latin word then (which it obivously isnt)? $\endgroup$ Commented Sep 9, 2021 at 23:09
  • $\begingroup$ After some digging, I have found the original paper that coined the word and will edit with some clarification on the word specifically. $\endgroup$
    – Justin T
    Commented Sep 10, 2021 at 6:45

If you were free to set the galaxies in this cluster at any desired positions and in any desired states of motion, then you could easily violate the virial theorem. If the cluster was then left alone for a long enough time, then it would converge to a state in which it was bound and obeyed the virial theorem, and it would stay in that state for a very long (but not infinite) time.

The answers by Prallax and Justin Tackett state incorrectly that this is a matter of thermal equilibrium. Actually, such a system's thermal equilibrium state is one in which the cluster has evaporated. As time goes on, each galaxy randomly exchanges energy with other galaxies, and over a long enough time, every galaxy will experience a fluctuation that is extreme enough so that it escapes from the cluster. This is also easy to verify based on the second law of thermodynamics, since the phase space of the external universe is infinite.

The time scale for virialization is many orders of magnitude shorter than the time scale for evaporation, however.

Some galaxy clusters may also not be sufficiently strongly bound to avoid breaking up due to cosmological expansion, although I believe this is more common for so-called superclusters, which are mostly not really bound objects.

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    $\begingroup$ Hi, I would like to point out that I have never stated that virialization is a matter of thermal equilibrium, and I apologize if this is the message I have conveyed to you. It doesn't seem that Justin Tackett's answer contains this assertion either. $\endgroup$
    – Prallax
    Commented Sep 12, 2021 at 16:07

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