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Derivation

Flat Rotation Curves and Dark Matter Halos

D-349 Home PU-308 Threads force · matter Depends on newtonian-gravitation, Virial Theorem for Bound Systems
Statement

For a test star on a circular orbit in a spherically symmetric mass distribution, Newtonian gravity fixes the circular speed through \( v_c^2(r) = G\,M(r)/r \). If the observed rotation curve is flat, \( v_c(r) = v_0 = \text{const} \) over some range of radii, then the enclosed mass must grow linearly with radius, \( M(r) = v_0^2 r / G \), and the local density must fall as \( \rho(r) = v_0^2 / (4\pi G r^2) \propto r^{-2} \) — the density profile of a singular isothermal sphere.

Why it matters

Luminous matter in a spiral galaxy — stars and gas — is concentrated toward the centre, so beyond the visible disc one expects \( M(r) \) to saturate and \( v_c \propto r^{-1/2} \), the Keplerian falloff seen in the Solar System. Instead, 21 cm and optical rotation curves stay flat far past the edge of the light. The derivation below turns that single empirical fact into a sharp statement about the invisible mass: it is distributed as \( r^{-2} \), extends much farther than the light, and dominates the outer galaxy.

This is the cleanest dynamical argument for dark matter at galactic scales, and the \( \rho \propto r^{-2} \) isothermal halo it produces became the reference model against which cosmological simulations (NFW, Einasto) are still compared today.

Assumptions
Spherical symmetry of the mass distribution.If dropped, the potential is not radial, \( v_c^2 = G M(r)/r \) no longer holds exactly, and one must use the actual disc-plus-halo potential; the flat-curve conclusion survives qualitatively but the mapping \( v_c \to M(r) \) picks up geometric factors of order unity.
Circular, closed orbits — negligible radial velocity and pressure support.If the tracers have significant random motions (as in a hot gas or a dispersion-supported system), the centrifugal balance is replaced by hydrostatic or Jeans equilibrium and \( v_c \) must be corrected by a pressure/dispersion term.
Newtonian gravity in the weak-field, low-velocity regime.If dropped for a modified-gravity theory (MOND) or strong fields, the acceleration law changes and a flat curve no longer implies any dark mass at all — the same data admit a different reading.
Time-independent, dynamically relaxed potential.If the halo is still assembling or strongly perturbed by a recent merger, \( M(r) \) inferred from an instantaneous \( v_c \) is not the equilibrium mass and the isothermal identification is only approximate.
Test-particle tracer — the orbiting star or gas contributes negligibly to the local field.If the tracer mass is not negligible its self-gravity must be added, but for individual stars and diffuse HI against a galactic halo this is entirely safe.
Derivation
1
\[ \frac{v_c^2}{r} = \frac{G\,M(r)}{r^2} \]
Centripetal acceleration of a circular orbit equals the inward gravitational acceleration; for spherical symmetry only the enclosed mass \( M(r) \) contributes (Newton's shell theorem, from newtonian-gravitation). A
2
\[ v_c^2(r) = \frac{G\,M(r)}{r} \]
Multiply both sides by \( r \). This is the exact circular-speed relation for any spherical mass profile — the master equation of the argument. A
3
\[ v_c(r) = v_0 = \text{const} \quad\Longrightarrow\quad M(r) = \frac{v_0^2\, r}{G} \]
Impose the observed flat rotation curve and solve step 2 for \( M(r) \). Enclosed mass rises linearly with radius. A
4
\[ M(r) = \int_0^r 4\pi r'^2\, \rho(r')\, dr' \quad\Longrightarrow\quad \frac{dM}{dr} = 4\pi r^2 \rho(r) \]
Definition of enclosed mass for a spherical density field, differentiated by the fundamental theorem of calculus (Leibniz rule on the radial integral). B
5
\[ \frac{dM}{dr} = \frac{d}{dr}\!\left( \frac{v_0^2 r}{G} \right) = \frac{v_0^2}{G} \]
Differentiate the linear mass law from step 3; the enclosed-mass gradient is a constant set by the flat speed. A
6
\[ 4\pi r^2 \rho(r) = \frac{v_0^2}{G} \quad\Longrightarrow\quad \rho(r) = \frac{v_0^2}{4\pi G\, r^2} \]
Equate the two expressions for \( dM/dr \) (steps 4 and 5) and solve for the local density. The \( r^{-2} \) profile is forced by the flat curve alone, with no further assumption about the halo. B
7
\[ \sigma^2 = \frac{v_0^2}{2}, \qquad \rho(r) = \frac{\sigma^2}{2\pi G\, r^2} \]
Identify \( v_0 \) with the velocity dispersion of a self-gravitating isothermal gas: solving the isothermal Lane–Emden / Jeans equation \( \frac{1}{r^2}\frac{d}{dr}\!\left(r^2 \frac{d\ln\rho}{dr}\right) = -\frac{4\pi G}{\sigma^2}\rho \) yields exactly \( \rho \propto r^{-2} \) with \( v_0^2 = 2\sigma^2 \). The empirical halo is the singular isothermal sphere; consistency with the virial theorem for a self-gravitating system fixes the numerical relation between orbital speed and dispersion. C
Result
\[ M(r) = \frac{v_0^2\, r}{G}, \qquad \rho(r) = \frac{v_0^2}{4\pi G\, r^2} \propto r^{-2} \]

Reading. A flat rotation curve is not a statement that gravity is weak far out — it is a statement that mass keeps accumulating. Every time you double the radius you double the enclosed mass, so the invisible halo must extend well beyond the luminous disc and its density must decline as the inverse square of radius. The luminous matter alone would give \( M(r)\to\text{const} \) and \( v_c \propto r^{-1/2} \); the excess is the dark-matter halo.

Units check. \( [\rho] = \dfrac{(\mathrm{m\,s^{-1}})^2}{(\mathrm{m^3\,kg^{-1}\,s^{-2}})(\mathrm{m^2})} = \dfrac{\mathrm{m^2\,s^{-2}}}{\mathrm{m^5\,kg^{-1}\,s^{-2}}} = \mathrm{kg\,m^{-3}} \). For \( M \): \( \dfrac{(\mathrm{m\,s^{-1}})^2\,\mathrm{m}}{\mathrm{m^3\,kg^{-1}\,s^{-2}}} = \mathrm{kg} \). Both correct.

Limiting cases
  • Point mass / Keplerian limit. If \( M(r)\to M_\ast = \text{const} \) beyond the light, then \( v_c = \sqrt{GM_\ast/r} \propto r^{-1/2} \) — the falling curve of the Solar System. A flat curve is precisely the failure of this limit.
  • Uniform-density core, \( r \to 0 \). The singular \( \rho \propto r^{-2} \) diverges at the centre; realistic halos flatten to a constant-density core, giving \( M \propto r^3 \) and a rising \( v_c \propto r \) (solid-body rotation) in the inner galaxy.
  • Outer truncation, large \( r \). The isothermal profile has divergent total mass \( M(r)\propto r \); physically the halo must steepen (e.g. to \( \rho\propto r^{-3} \), as in NFW) so \( M \) converges and \( v_c \) eventually falls.
  • Zero dark matter. \( v_0 \to 0 \) recovers \( \rho \to 0 \) and a pure Keplerian disc — the halo term vanishes smoothly.
Breaks when
  • Inner galaxy (small \( r \)). The \( r^{-2} \) law predicts an unphysical central density spike and a linearly rising \( v_c \); real curves show cored profiles, and the "core–cusp problem" is exactly this breakdown of the singular isothermal fit.
  • Very large radius. Enclosed mass \( M \propto r \) diverges, so the pure isothermal sphere has infinite mass and energy. Beyond the flat region the curve turns over, requiring a steeper outer profile and a finite halo mass.
  • Dispersion-supported or non-circular systems. In ellipticals, bars, or warped/lopsided discs the assumption of cold circular orbits fails and \( v_c^2 = GM/r \) must be replaced by the Jeans equation with pressure and anisotropy terms.
  • Modified-gravity interpretation. If the acceleration law itself departs from Newton at low \( a \) (MOND regime, \( a \lesssim a_0 \sim 10^{-10}\,\mathrm{m\,s^{-2}} \)), the same flat curve implies no dark mass; the derivation's conclusion is only as safe as its Newtonian premise.
Failure modes
  • Using total mass instead of enclosed mass. Writing \( v_c^2 = GM_\text{tot}/r \) with the full galaxy mass ignores the shell theorem; only \( M(<r) \) exerts a net inward pull on a circular orbit.
  • Confusing "flat curve" with "constant mass". Flat \( v_c \) means \( M\propto r \), i.e. mass still increasing — students often read the plateau as "no more mass out there," which is the opposite.
  • Dividing \( M(r) \) by volume \( \tfrac{4}{3}\pi r^3 \) to get \( \rho \). That yields the mean density \( \bar\rho \propto r^{-2} \), not the local density; the correct local \( \rho \) comes from \( dM/dr = 4\pi r^2\rho \). The two happen to share the \( r^{-2} \) scaling here but differ by a factor of 3.
  • Forgetting the \( 4\pi \). Dropping the spherical solid-angle factor gives \( \rho \) too large by \( 4\pi \).
  • Applying the Solar-System intuition. Assuming \( v_c \propto r^{-1/2} \) must hold because "most planetary mass is central" — true for the Sun, false for a galaxy with an extended halo.
  • Mixing velocity dispersion and circular speed. Setting \( v_0 = \sigma \) rather than \( v_0^2 = 2\sigma^2 \) misreads the isothermal-sphere normalization by \( \sqrt2 \).
Discussion

The power of this derivation is that a single, robustly observed feature — the flatness of \( v_c(r) \) — is converted into a specific density law with no adjustable shape parameters. The mapping runs entirely through the exact spherical relation \( v_c^2 = GM(r)/r \): flat speed forces linear mass, linear mass forces \( \rho \propto r^{-2} \). The only freedom is the overall scale \( v_0 \), which sets both the halo's density normalization and, through \( v_0^2 = 2\sigma^2 \), the temperature of the equivalent isothermal gas.

The identification with the singular isothermal sphere is not a coincidence. Independently solving hydrostatic equilibrium for a self-gravitating isothermal gas — pressure gradient balancing gravity with \( P = \rho\sigma^2 \) — returns exactly \( \rho \propto r^{-2} \). So the halo dynamically inferred from rotation curves is the same object statistical mechanics predicts for a relaxed, constant-temperature self-gravitating system, tying this result to the virial theorem for self-gravitating matter: \( 2K + U = 0 \) with \( K \sim \tfrac12 M\sigma^2 \) fixes the depth of the potential well the flat curve is probing.

Physically, the flat curve tells us the halo is far more extended than the light. Integrating \( M(r) = v_0^2 r/G \) out to the last measured HI point typically gives a dynamical mass several times the luminous mass, and the "edge" of the halo is set not by where the mass runs out but by where the profile finally steepens. Cosmological \( N \)-body simulations replace the singular isothermal sphere with the NFW profile \( \rho \propto r^{-1}(1+r/r_s)^{-2} \), which behaves like \( r^{-2} \) near \( r_s \) — reproducing the flat curve over the observed range — but has a shallower cusp inside and a steeper \( r^{-3} \) tail outside, curing both divergences of the pure isothermal model.

The deepest tension the result exposes is the core–cusp problem. The clean \( r^{-2} \) (and the NFW \( r^{-1} \)) both predict a diverging central density, whereas high-resolution rotation curves of low-surface-brightness and dwarf galaxies favour constant-density cores. Whether this reflects baryonic feedback flattening an initially cuspy halo, the self-interaction cross-section of the dark matter, or a genuine failure of the collisionless cold-dark-matter picture remains open — and the argument here is precisely where the two pictures are forced to disagree, because the inner rotation curve is where the isothermal derivation breaks. Common misconceptions: a flat curve does not mean "gravity is constant" or "mass stops" — it means mass keeps growing linearly; and \( \rho\propto r^{-2} \) is the local density, distinct from the mean density inside \( r \), even though both scale as \( r^{-2} \) for this special profile.

Worked examples
1
A spiral galaxy has a flat rotation curve at \( v_0 = 220~\mathrm{km\,s^{-1}} \) (a Milky-Way-like value). Find the dark-halo mass enclosed within \( r = 30~\mathrm{kpc} \), and the local dark-matter density there. A
\[ M(r) = \frac{v_0^2 r}{G}, \qquad \rho(r) = \frac{v_0^2}{4\pi G r^2} \]
Symbols first (step 3 and step 6 results); now insert numbers with \( v_0 = 2.2\times10^5~\mathrm{m\,s^{-1}} \), \( r = 30~\mathrm{kpc} = 30\times3.086\times10^{19}~\mathrm{m} = 9.26\times10^{20}~\mathrm{m} \), \( G = 6.674\times10^{-11}~\mathrm{m^3\,kg^{-1}\,s^{-2}} \).
\[ M = \frac{(2.2\times10^5)^2 (9.26\times10^{20})}{6.674\times10^{-11}} = \frac{(4.84\times10^{10})(9.26\times10^{20})}{6.674\times10^{-11}} \approx 6.7\times10^{41}~\mathrm{kg} \]
Convert to solar masses, \( M_\odot = 1.989\times10^{30}~\mathrm{kg} \): \( M \approx 3.4\times10^{11}~M_\odot \).
\[ \rho = \frac{4.84\times10^{10}}{4\pi (6.674\times10^{-11})(9.26\times10^{20})^2} \approx 6.7\times10^{-22}~\mathrm{kg\,m^{-3}} \]
\[ M(<30~\mathrm{kpc}) \approx 3.4\times10^{11}~M_\odot, \qquad \rho \approx 6.7\times10^{-22}~\mathrm{kg\,m^{-3}} \]

Reading. Several times the Milky Way's stellar mass sits within 30 kpc, most of it dark. The density there, \( \sim 10^{-2}~M_\odot\,\mathrm{pc^{-3}} \), is far below the mean stellar density of the disc — the halo is diffuse but voluminous.

2
The same halo is observed to keep its flat curve out to \( r_1 = 10~\mathrm{kpc} \) and \( r_2 = 50~\mathrm{kpc} \). By what factor does the enclosed mass grow, and what is the mean density ratio? A
\[ \frac{M(r_2)}{M(r_1)} = \frac{v_0^2 r_2/G}{v_0^2 r_1/G} = \frac{r_2}{r_1} \]
Symbolic first: because \( M\propto r \), the mass ratio is just the radius ratio, independent of \( v_0 \) and \( G \).
\[ \frac{M(r_2)}{M(r_1)} = \frac{50}{10} = 5 \]
Mean density inside \( r \) is \( \bar\rho = M/(\tfrac{4}{3}\pi r^3) \propto r/r^3 = r^{-2} \), so the ratio is \( (r_1/r_2)^2 \).
\[ \frac{\bar\rho(r_2)}{\bar\rho(r_1)} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{10}{50}\right)^2 = \frac{1}{25} \]
\[ \frac{M(r_2)}{M(r_1)} = 5, \qquad \frac{\bar\rho(r_2)}{\bar\rho(r_1)} = \frac{1}{25} \]

Reading. Extending the flat curve fivefold in radius adds five times the mass while the mean density drops by \( 25\times \) — mass is dominated by the outermost shells even as the halo thins out. This is why the total mass is so sensitive to where the curve is last measured.

Problems
  1. A dwarf galaxy has a flat rotation curve at \( v_0 = 60~\mathrm{km\,s^{-1}} \). Find the enclosed mass within \( 5~\mathrm{kpc} \) in solar masses.
    Solution \( M = v_0^2 r/G \). With \( v_0 = 6.0\times10^4~\mathrm{m\,s^{-1}} \), \( r = 5\times3.086\times10^{19} = 1.543\times10^{20}~\mathrm{m} \): \( M = (3.6\times10^9)(1.543\times10^{20})/(6.674\times10^{-11}) = 8.3\times10^{39}~\mathrm{kg} = 4.2\times10^{9}~M_\odot \).
  2. Show that for the singular isothermal sphere the mean density inside \( r \) equals three times the local density at \( r \).
    Solution Local: \( \rho(r) = v_0^2/(4\pi G r^2) \). Mean: \( \bar\rho = M/(\tfrac{4}{3}\pi r^3) = (v_0^2 r/G)/(\tfrac{4}{3}\pi r^3) = 3v_0^2/(4\pi G r^2) = 3\rho(r) \). Hence \( \bar\rho = 3\rho \), the factor students miss when they compute local density by dividing mass by volume.
  3. A galaxy's curve is flat at \( v_0 = 200~\mathrm{km\,s^{-1}} \) out to \( 20~\mathrm{kpc} \), then falls Keplerian (\( M \) constant) beyond. What is \( v_c \) at \( 40~\mathrm{kpc} \)?
    Solution Beyond the edge \( M = M(20~\mathrm{kpc}) = \text{const} \), so \( v_c = \sqrt{GM/r} \propto r^{-1/2} \). Thus \( v_c(40) = v_0\sqrt{20/40} = 200/\sqrt2 = 141~\mathrm{km\,s^{-1}} \).
  4. Compute the velocity dispersion \( \sigma \) of the equivalent isothermal gas for a halo with \( v_0 = 220~\mathrm{km\,s^{-1}} \), and comment on its magnitude relative to \( v_0 \).
    Solution From step 7, \( v_0^2 = 2\sigma^2 \Rightarrow \sigma = v_0/\sqrt2 = 220/1.414 = 156~\mathrm{km\,s^{-1}} \). The dispersion is smaller than the circular speed by \( \sqrt2 \): the ordered rotation of the tracers exceeds the random motion of the equivalent thermal support, as expected for a rotation-supported disc embedded in a pressure-supported halo.
  5. The isothermal profile is truncated at \( r_\text{max} = 100~\mathrm{kpc} \). Using \( v_0 = 180~\mathrm{km\,s^{-1}} \), find the total halo mass and estimate the fractional error if the truncation is ignored versus placed at \( 200~\mathrm{kpc} \).
    Solution \( M(r_\text{max}) = v_0^2 r_\text{max}/G \). With \( v_0 = 1.8\times10^5 \), \( r_\text{max} = 100\times3.086\times10^{19} = 3.086\times10^{21}~\mathrm{m} \): \( M = (3.24\times10^{10})(3.086\times10^{21})/(6.674\times10^{-11}) = 1.50\times10^{42}~\mathrm{kg} = 7.5\times10^{11}~M_\odot \). Since \( M\propto r_\text{max} \), doubling the truncation radius to 200 kpc doubles the mass to \( 1.5\times10^{12}~M_\odot \) — a 100% change. This linear sensitivity is exactly why the pure isothermal sphere has divergent total mass and why the outer truncation (or a steeper NFW tail) is physically essential.