For an isothermal collisionless tracer without streaming or an escape cutoff, integrate its three-dimensional Maxwell-Boltzmann velocity distribution over the two transverse velocities. The remaining marginal is the displayed normal distribution. Its width and mean do not depend on position. Integrating along the line of sight therefore only multiplies the marginal by the projected surface density; the normalized observed velocity distribution remains the same Gaussian at every projected radius.
Use relative energy , with . If the galactic distribution function depends on alone, rotations and sign reversals in velocity space leave it unchanged. Thus all mean velocities vanish, and the three component second moments are equal. The anisotropy term in the Spherical Jeans equation cancels. With constant radial velocity dispersion , it reduces to
Integration gives the isothermal collisionless tracer density
The constant changes if a constant is added to the potential; this has no effect on the velocity distribution.
Apply the given untruncated Eddington inversion, retaining its lower limit . Define its integral before differentiation by . With , the Gaussian integral gives
Differentiating and multiplying by yields
Indeed, at a fixed position this is
whose integral over all velocities is and whose component second moments are . This checks both the normalization and the velocity dispersion.
Let be the line of sight component of velocity. Integrate this Maxwell-Boltzmann velocity distribution over the two transverse components:
At projected radius , integrating along the line of sight simply replaces by the projected surface density . Dividing by gives the spectroscopist's normalized Gaussian line-of-sight velocity distribution:
It is a zero-mean normal distribution with width , independent of projected radius. This is the untruncated model specified by the lower limit in the inversion formula: no escape-energy cutoff has been imposed. A population restricted to positive binding energy would have truncated velocity tails and would not obey this exact Gaussian density law in a finite escape potential.