A stirring supernovae cannot pay for

Star-forming discs at redshift two turn more slowly than their masses require, and the reason is that part of their weight is carried by the random motions of their gas rather than by rotation. Their gas moves at fifty to a hundred kilometres a second in random directions — five to ten times the dispersion of the gas in a spiral today — and a disc that turbulent is half held up by its own pressure. Correcting for that pressure restores them to the relation between rotation and mass, and the essay that made the correction ended on the question it raised: turbulence decays, so what was keeping these discs stirred?

The question has two candidate answers that have been argued over for fifteen years. One is the energy released by young stars, mostly by supernovae, which galaxies forming stars at a hundred solar masses a year have in abundance. The other is gravity: a disc too dense to be stable fragments into clumps and spirals, and the gas flowing inward through it releases orbital energy as random motion. This essay works out what each can supply and where each runs out, and finds that the answer is not a matter of degree. Supernovae run out at a fixed dispersion, far below what is observed. Above that, only gravity can pay — and the way it pays changes what the disc becomes.

Turbulence has to be paid for continuously #

Supersonic turbulence in a gas layer loses its energy in shocks, and it loses it fast: in about the time a parcel of gas takes to cross the layer’s thickness at the turbulent speed. For the gas disc of a spiral like the Milky Way, a hundred parsecs thick with a dispersion of ten kilometres a second, that is ten million years. For a redshift-two disc, a kiloparsec thick at fifty kilometres a second, it is twenty million. Both are short compared with the time either disc takes to turn its gas into stars, so whatever dispersion is observed is being continuously resupplied, and the resupply rate is the dissipation rate.

That rate is easy to write down. A layer of surface density $\Sigma_g$ holds kinetic energy $\tfrac32 \Sigma_g\sigma^2$ per unit area, and loses it in a crossing time $H/\sigma$, so the power per unit area needed to hold the dispersion is

$$D = \frac{3}{2}\frac{\Sigma_g\sigma^3}{H}.$$

The thickness is itself set by the dispersion. A layer supported by random motions in its own gravity and that of the stars around it has $H = \sigma^2/(\pi G \Sigma_{\rm tot})$ — a hotter layer is thicker — so the cost of holding the dispersion is proportional to $\Sigma_g \Sigma_{\rm tot}\,\sigma$. Doubling the dispersion doubles its cost. That linearity, which comes from the layer’s thickness growing with its turbulence, is the fact the whole argument rests on.

What supernovae can supply #

A massive star ends as a supernova releasing about $10^{51}$ ergs, and a population of new stars makes about one such supernova for every hundred solar masses formed. Most of the energy is radiated away from the hot interior of the expanding remnant; what survives as bulk motion of the surrounding gas, in simulations of remnants expanding into a turbulent medium, is a few per cent to a tenth. The power supernovae supply as turbulence is that surviving fraction times the energy per unit mass of stars formed times the star-formation rate per unit area.

The star-formation rate depends on the dispersion too, and in the opposite direction. Stars form at a small fraction — about one per cent — of the gas per free-fall time, and the free-fall time is set by the midplane density, which is the surface density divided by the thickness. A hotter layer is thicker, less dense, slower to collapse, and forms fewer stars. So the power supernovae supply falls as the dispersion rises — inversely, in a disc where gas dominates the gravity.

The two lines cross once, and the crossing is the dispersion supernovae can hold: below it they supply more than the turbulence costs and the dispersion grows; above it they fall short and it decays. For the redshift-two disc drawn it is thirteen kilometres a second.

The striking thing is how little the crossing depends on how much gas the disc has. Both the cost and the supply scale with the square of the surface density in a gas-dominated disc — the cost through the density and the thickness, the supply through the star-formation rate — so the surface density drops out of the balance, and the dispersion supernovae can hold is set by the coupling efficiency and the star-formation efficiency alone. That is why, in the hero figure, the dashed lines are flat. A disc forming stars a hundred times faster, with a hundred times more supernovae, has the same supernova-driven dispersion as one forming stars slowly: it has more energy to give and exactly as much more to spend.

The Milky Way’s gas, as a check #

The local numbers make the balance concrete, and they are worth having in view because the Galaxy’s disc is the one place every term is measured. Near the Sun the gas has a surface density of about thirteen solar masses per square parsec, the stars about thirty-five, and the gas dispersion is seven to ten kilometres a second. A layer that hot in that much gravity is about a hundred and fifty parsecs thick, its midplane density is a few hundredths of a solar mass per cubic parsec, and its free-fall time is a few tens of millions of years. One per cent of the gas per free-fall time then gives a few thousandths of a solar mass per year per square kiloparsec — which is the star-formation rate measured near the Sun. Feeding a tenth of the resulting supernova energy into the gas and dissipating it in a crossing time balances near ten kilometres a second.

So the simple balance is not a model fitted to the answer; it reproduces, with conventional efficiencies, the dispersion the one well-measured disc actually has. That matters for what follows, because the same balance, applied to a disc with ten or thirty times the gas, gives nearly the same dispersion, and the discs with that much gas have five times more.

The gas near the Sun is not a single phase, and the dispersion that matters is a mass-weighted average over gas at two very different temperatures — cold clouds moving through warm intercloud gas, each with its own dispersion. Supernovae drive the warm gas directly and the clouds through it. The balance treats the mixture as one layer, which is a simplification of the same kind as treating the star-formation efficiency as one number, and the agreement with the local measurement says both are adequate at this level.

How hard supernovae would have to push #

The value of thirteen kilometres a second depends on two efficiencies that are not well known, and the obvious response is to ask whether different values could rescue the supernovae.

They cannot. The balance gives the dispersion as the square root of the product of the two efficiencies, because the supply scales with their product and the cost with the dispersion squared once the star-formation rate’s own dependence is included. Fifty kilometres a second at a star-formation efficiency of one per cent would require more than all of the supernovae’s energy to remain in the gas as motion — not a tenth, not a half, but more than the total. Doubling the star-formation efficiency helps by a factor of 1.4. The square root makes the argument robust in the useful direction: a factor of ten of uncertainty in the coupling, which is roughly the real uncertainty, moves the answer by a factor of three, and three times thirteen is still below the dispersions observed.

The same arithmetic applied to a spiral today gives about the dispersion it has. Nearby spirals have gas dispersions of eight to twelve kilometres a second over most of their discs, and the dispersion is nearly the same across a wide range of star-formation rates — the flat lower branch of the hero figure, where the nearby spirals sit. Supernovae can hold the turbulence of a disc that does not need much, and the balance between the thermal, turbulent, magnetic and cosmic-ray support of the Galaxy’s gas layer is consistent with it. What they cannot do is make a disc five times more turbulent than that.

Gravity stirs a disc it cannot hold flat #

The second candidate starts from a different fact. A rotating disc of gas is stable against collapsing into clumps only while its dispersion is high enough: the Toomre condition, $Q = \kappa\sigma/(\pi G\Sigma_g) \gtrsim 1$, compares the stabilising effects of random motion and of rotation (through the epicyclic frequency $\kappa$) with the destabilising effect of the disc’s self-gravity. A disc whose dispersion falls below that threshold fragments: it forms clumps and spiral arms, which exert torques — the same torques by which a spiral arm undoes itself in the work it does — which move mass inward and angular momentum outward, which releases orbital energy — and that energy, stirred into the gas, raises the dispersion back to the threshold.

So a gravitationally unstable disc regulates itself to marginal stability, $Q \approx 1$, and its dispersion is set by its surface density and its rotation: $\sigma \approx \pi G\Sigma_g/\kappa$. Unlike the supernova-driven dispersion it rises in proportion to the surface density. The hero figure’s solid lines show what happens when both drivers are allowed. At low gas surface densities the supernovae hold the disc above marginal stability and the dispersion is flat; above a threshold — 72 solar masses per square parsec for the redshift-two disc drawn — the supernovae fall short of the dispersion stability requires, gravity takes over, and the dispersion climbs with the gas content and with the star-formation rate. The redshift-two disc reaches the observed range; the nearby spiral, with a fifth as much gas relative to its stars, mostly never leaves the flat branch.

The two drivers therefore predict different relations between dispersion and star formation, and that is how the observations separate them. Feedback alone predicts a dispersion independent of the star-formation rate. Gravity predicts one that rises with it, once the disc is gas-rich enough to be unstable. The measured dispersions of star-forming galaxies are nearly flat, at about ten kilometres a second, up to star-formation rates of a few solar masses a year, and rise above that — the combined shape.

The instability is not hypothetical in these discs; it is visible. Images of star-forming galaxies at redshift two show their light concentrated in a handful of giant clumps, each a few hundred million to a billion solar masses, where nearby spirals show arms of small clusters. The mass a marginally stable disc fragments into is set by the same quantities as its dispersion — roughly $\sigma^4/(G^2\Sigma_g)$, the gas inside one Toomre wavelength — and for fifty kilometres a second and three hundred solar masses per square parsec that is about a billion solar masses. The clumps are the size the instability predicts. The same gas held at ten kilometres a second would fragment into objects six hundred times smaller; the gas near the Sun, at that dispersion and a twentieth of the surface density, into complexes of a few tens of millions of solar masses, the largest a nearby spiral forms.

The price is inflow #

Gravity’s energy is not free either. It comes from gas moving inward through the disc’s potential, and in a flat rotation curve gas that crosses the disc from its edge to its centre releases about $V^2$ per unit mass. Holding the dispersion therefore requires a definite rate of inward flow.

The inflow grows steeply. The power gravity has to supply is the dissipation, which in a marginally stable disc scales as the cube of the dispersion, and the energy released per unit mass is fixed by the rotation curve, so the inflow needed rises as $\sigma^3$. The star-formation rate, meanwhile, rises only about linearly. At fifty kilometres a second the inflow is several times the star-formation rate.

That inflow is not small even on cosmological scales. A galaxy of this mass at redshift two accretes gas from its halo at something like a hundred solar masses a year, and the disc would be moving gas inward at about twice that rate. It can do so only while it holds a reservoir, and it empties the reservoir in a few hundred million years — the same race between supply and consumption by which the gas runs out before the galaxy does — after which the dispersion falls unless accretion refills the disc. That ties the turbulence of a disc to its supply of gas from outside, which supernova-driven turbulence would not be.

Gas that reaches the centre before it forms stars #

The comparison that matters is between two times: how long the disc’s gas lasts at its current star-formation rate, and how long the inflow takes to carry it to the centre.

The answer depends on the star-formation law, and the figure draws two. The model’s law — one per cent of the gas per free-fall time of the whole layer — forms stars a few times more slowly than redshift-two discs are observed to, because their star formation is concentrated in massive clumps far denser than the layer’s mean. With the slow law the inflow wins at almost every dispersion gravity sets; with a law five times faster, above forty kilometres a second. Either way, the observed discs, at fifty to a hundred, are on the side where gas reaches the centre faster than it forms stars along the way.

This is the most consequential difference between the two drivers. Turbulence driven by supernovae holds the gas up where it is. Turbulence driven by gravity is paid for by moving the gas inward, so a disc at redshift two held at marginal stability is continuously building up its centre — the clumps migrating inward and merging, the gas flowing through to form a dense central component. That is one of the routes by which the slow gas a halo delivers too much of can end up in a bulge without ever forming one by collapse, and it predicts that the bulges of today’s massive discs were built during the epoch of their most turbulent gas.

What the driver means for the Tully–Fisher relation #

The question the slow rotation of these discs raised was practical: how much of a turbulent disc’s weight does its turbulence carry, and so how large a correction does its rotation need before it is placed on the Tully–Fisher relation?

If supernovae set the dispersion, the correction would never exceed a couple of per cent, the rotation of a turbulent disc would measure its circular speed directly, and the slow rotation of redshift-two discs would have to mean something else — less mass, or a different relation. If gravity sets it, the correction is tens of per cent in exactly the discs that are observed to rotate slowly, and it is large enough to account for most of their offset. The pressure correction that restored redshift-two discs to the relation is therefore not just an empirical fix: it is a statement that their dispersions are of the size only gravity can supply.

What the model leaves out #

The disc is described at one radius by mean quantities, with an exponential’s equivalent area standing in for the whole. Real discs have dispersions that vary with radius, and their marginal stability is a condition on a two-component system of gas and stars, which is more unstable than the gas alone; including the stars lowers the gas dispersion needed for $Q = 1$ by tens of per cent.

The supernova energy is coupled with a single efficiency. In reality the coupling depends on the density the supernovae explode into, on whether they are clustered — clustered supernovae build superbubbles that can break out of the disc and deposit their energy in the halo rather than the layer — and on the momentum rather than the energy, which some treatments argue is the conserved quantity that matters. The square root in the balance makes the conclusion insensitive to most of this.

The star-formation law is a fixed efficiency per free-fall time of the mean density. Redshift-two discs form stars in giant clumps, at densities ten or a hundred times the mean, and their measured star-formation rates are several times what the simple law gives. That changes where the inflow overtakes star formation, as the race figure shows, and it raises the supernova power by the same factor — which, by the square root, raises the supernova-driven dispersion by a factor of about two, still well short.

Other drivers are omitted: the kinetic energy of gas accreting onto the disc from the halo, which can contribute directly where the inflow lands; magnetorotational instability, which matters in the outer, gas-poor parts of discs; and the energy of the clumps’ own motions. Each has its advocates. None changes the conclusion that supernovae alone cannot hold the observed dispersions; each changes how the shortfall is shared.

Still open: one driver or a sequence #

The energy arithmetic rules out supernovae as the sole driver of the turbulence in redshift-two discs and allows them to drive it in spirals today. It does not say that gravity is the only other driver, and it does not say how the transition happens in a single galaxy as its gas runs down.

What would settle it is a measurement of the inflow itself. The gravity-driven picture requires gas moving inward through the disc at rates comparable to or larger than the star-formation rate, and that motion — a few kilometres a second of radial streaming on top of rotation and turbulence — is at the edge of what integral-field spectroscopy of redshift-two discs can resolve. A detection of radial inflow at the rates required, in discs whose dispersions are high, would make the disc’s own gravity the engine of both its turbulence and its bulge. A failure to find it would leave a gap in the energy budget that neither supernovae nor any known process fills. The dispersions themselves, measured from line widths that stand in for speeds, are no longer in doubt; what pays for them is.