Adding stellar, gas and halo gravity

Several gravitating components act on one tracer

Let $g_g,g_d,g_h$ be the signed inward accelerations from gas, a stellar disk and a spherical halo at radius $R$. Newtonian gravity superposes these accelerations. Multiplying each by the same radius gives contributions with units of speed squared.

Component tables often express those forces as equivalent circular speeds. They describe the speed supported by a component’s gravity, not three independent streams whose velocities should be added. A native stellar column is calculated first at unit stellar mass-to-light ratio and then scaled by the chosen ratio.

Write the numerical disk mass-to-light ratio in solar units as $\Upsilon_d$, the signed gas equivalent speed as $V_g$, and the unit-ratio stellar column as $V_d$. The fixed object used here has no bulge. Its native gas term already includes the helium factor $1.33$, so we do not apply that factor again.

Reading: Lelli, McGaugh & Schombert (2016), SPARC, AJ152,157 · Independent saved-observation rotation-curve calculation

An outward gas force has a negative contribution

Gas concentrated outside a test radius can pull a disk-plane particle outwards. The native table marks an outward contribution with a negative equivalent speed. Squaring every column unconditionally would reverse that force.

For a worked example, take $V_g=-20\,\mathrm{km\,s^{-1}}$, $V_d=80\,\mathrm{km\,s^{-1}}$, $\Upsilon_d=0.5$ and $v_h=100\,\mathrm{km\,s^{-1}}$. Their signed contributions are $-400$, $3200$ and $10000$, all in $(\mathrm{km\,s^{-1}})^2$. The sum is $12800$, giving $v_c=113.137\,\mathrm{km\,s^{-1}}$.

Turning the gas term positive would instead give $116.619\,\mathrm{km\,s^{-1}}$. If the signed sum is zero or negative, the prescribed field does not support this kind of circular solution at that radius. Taking an absolute value of the total would invent a different force.

For the stellar component alone, increasing $\Upsilon_d$ from $0.25$ to $1$ quadruples the force and doubles its equivalent speed. The total-speed multiplier still follows the combined gas, stellar and halo terms.

Reading: Lelli, McGaugh & Schombert (2016), SPARC, AJ152,157 · Independent saved-observation rotation-curve calculation

Integrate the NFW cusp before interpreting its speed

The NFW family has a density scale $\rho_s$ and radius scale $r_s$. Define $x=R/r_s$. Its inner density rises approximately in inverse proportion to radius; this central rise is called a cusp. The scale density is not a finite central density.

Changing to dimensionless radius in the shell integral leaves $\int_0^x t/(1+t)^2\,dt$. Split the numerator as $t=(1+t)-1$, integrate the logarithmic and inverse terms, then evaluate the endpoints. This produces the mass below.

Reading: Navarro, Frenk & White (1997), fixed astro-ph/9611107v1 · Independent saved-observation rotation-curve calculation

Cusp and core approach the centre differently

Near the centre, $\rho(R)\simeq\rho_s r_s/R$. A thin shell contributes $4\pi R^2\rho(R)\,dR\propto R\,dR$, so integrating outward gives mass proportional to $R^2$. The exact NFW mass bracket correspondingly approaches $x^2/2$. Enclosed mass therefore grows as $R^2$, and dividing by radius to obtain speed squared gives $v_h\propto R^{1/2}$. A density cusp does not require a divergent circular speed. The smaller central sphere contains less mass.

A second family has finite central density $\rho_0$. Set $x=R/r_c$. Its shell integral contains $\int_0^x t^2/(1+t^2)\,dt$. Writing the integrand as $1-1/(1+t^2)$ gives $x-\arctan x$. Here $\arctan x$ is the angle in radians whose tangent is $x$; its derivative is $1/(1+x^2)$.

Near the centre this bracket approaches $x^3/3$, so enclosed mass grows as $R^3$ and speed as $R$. These distinct slopes offer a possible observational difference. Resolving it also requires the stellar contribution, angular resolution and noncircular motions to be understood.

Reading: Navarro, Frenk & White (1997), fixed astro-ph/9611107v1 · Independent saved-observation rotation-curve calculation

The exponential disk keeps its disk geometry

The ideal thin disk has $\Sigma(R)=M_d e^{-R/R_d}/(2\pi R_d^2)$. Integrating its axisymmetric ring potential gives a convenient special-function expression. $I_n$ and $K_n$ are modified Bessel functions, standard functions that arise in cylindrical problems. The scientific code evaluates them with its numerical library.

An independent check takes a different route: integrate the Newtonian potential of actual circular rings, then differentiate with two radial steps. It compares the results over declared finite radii instead of evaluating the Bessel formula twice and calling the second evaluation a reference.

Native SPARC stellar columns preserve their original finite vertical thickness and radial structure. Reading and scaling those columns adopts that native component model. Using a thin exponential disk explains and tests a separate ideal geometry.

Reading: Independent saved-observation rotation-curve calculation · Lelli, McGaugh & Schombert (2016), SPARC, AJ152,157

Subtracting nearby floating-point numbers can lose the result

For very small $x$, direct evaluation of $x-\arctan x$ subtracts two nearly equal floating-point numbers. Significant digits can disappear. The code instead uses the equivalent convergent series $x^3/3-x^5/5+x^7/7-\cdots$, then checks it against positive-density integration.

This is a numerical-precision issue within the same physical core model. The failed direct-subtraction examples remain in the scientific output, with the original independent tolerance. Stable arithmetic is a prerequisite for asking what the observations say about the density profile.

Reading: Independent saved-observation rotation-curve calculation