The Mass of the Milky Way as a Function of Distance from Its Center

Dynamical Mass from Rotation

The observed rotation speed of the Milky Way tells us how much mass is required gravitationally.

For circular motion:

[latex]M_{\mathrm{dyn}}(

v_c(r) is the circular velocity at radius r.

G is the gravitational constant.

In practical units:

[latex]M_{\mathrm{dyn}}(If the rotation velocity is approximately flat:

[latex]v_c(r)\approx233\,\mathrm{km/s}\)

then:

\(M_{\mathrm{dyn}}(with r in kpc.

This means that if the rotation curve stays nearly flat, the dynamical mass grows almost linearly with radius.

A Wave-Based Extension of the Missing Mass

A disk model explains the visible mass. The missing mass is what remains after comparing this visible mass to the dynamical mass.

A wave-based model can describe the missing mass as an effective density generated by the visible disk.

The guiding idea is that each visible mass element generates an effective field decreasing with distance.

Let the distance between a source point r′ and an observation point r be:

[latex]D=|r-r’|\)

Then an elementary contribution can be written as:

\(d\rho_{\mathrm{wave}}(r)=\rho_{\mathrm{visible}}(r’)\,\lambda e^{-D/\ell}\,dV\)

• λ is a dimensionless coupling factor.
• ℓ is a coherence length.
• D is the distance between source and observation point.

This form means that the effective contribution decreases exponentially with distance:

\(e^{-D/\ell}\)

The parameter ℓ controls how far the effect extends.

Ring-by-Ring Interpretation

The disk can again be understood through rings.

A visible ring at radius R′ has mass:

\(dM_{\mathrm{visible}}=2\pi R’\Sigma(R’)\,dR’\)

In the wave-based extension, that ring contributes to the effective density around it.

The contribution is strongest near the ring and decreases with distance:

\(e^{-D/\ell}\)

So the effective density is not inserted by hand as a spherical halo. It is generated from the geometry of the disk itself.

At short distances, it follows the disk geometry. At larger distances, after integrating over many rings, the effective distribution can become smoother and more extended.

From Effective Density to Effective Mass

Once the effective density is known, the corresponding effective mass inside radius r can be written as:

\(M_{\mathrm{wave}}(In spherical coordinates:

[latex]M_{\mathrm{wave}}(This effective mass can then be compared with the observed missing mass:

[latex]M_{\mathrm{wave}}(That gives a testable condition.

Simple Numerical Example at the Sun’s Radius

The Sun is located at about:

[latex]R_0\simeq8.2\,\mathrm{kpc}\)

Using the stellar disk equation:

\(M_{\mathrm{disk,stars}}(<8.2)=3.52\times10^{10}\left[1-e^{-8.2/2.50}\left(1+\frac{8.2}{2.50}\right)\right]+1.05\times10^{10}\left[1-e^{-8.2/3.02}\left(1+\frac{8.2}{3.02}\right)\right][/latex]

This gives approximately:

[latex]M_{\mathrm{disk,stars}}(<8.2\,\mathrm{kpc})\approx3.7\times10^{10}M_\odot[/latex]

If the circular speed is:

[latex]v_c\simeq233\,\mathrm{km/s}\)

then the dynamical mass inside 8.2 kpc is:

\(M_{\mathrm{dyn}}(<8.2)=2.325\times10^5(233)^2(8.2)M_\odot[/latex] [latex]M_{\mathrm{dyn}}(<8.2)\approx1.03\times10^{11}M_\odot[/latex]

The difference shows why visible mass alone cannot explain the observed rotation.

Final Summary of the Main Equations

Visible stellar disk

\(M_{\mathrm{disk,stars}}(Full visible disk

[latex]M_{\mathrm{disk,visible}}(Dynamical mass

[latex]M_{\mathrm{dyn}}(Missing mass

[latex]M_{\mathrm{missing}}(Ring mass

[latex]dM=2\pi r\Sigma(r)\,dr\)

Exponential disk

\(\Sigma(r)=\Sigma_0e^{-r/R_d}\)

Wave-based effective density

\(\rho_{\mathrm{wave}}(R,z)=\frac{\lambda}{\ell}\int_0^\infty\int_0^{2\pi}\Sigma(R’)e^{-D/\ell}R’\,d\phi\,dR’\)

with:

\(D=\sqrt{R^2+R’^2-2RR’\cos\phi+z^2}\)