Gravity Field of a Filament Perpendicular to a galaxy

Integrate the gravitational acceleration from an object at radius r toward a line mass perpendicular to its orbit.

Assume gravitational constant G , radius r , line density { \large \rho } , and perpendicular axis z . Calculate the radial gravitational acceleration a .

Assume a mass element dm = { \large \rho } dz at position z . The distance between the particle and the mass element is the hypotenuse h = \sqrt{ r^{2 + z}2 } The diagonal gravitational acceleration towards dm is

Eq 1: da_d = G { \large \rho } dz / h^2 dz ~ = ~ G { \large \rho } / ( r^2 + z^2 ) dz ~ ~ ~ ~ ~ in the diagonal direction

This is force is diagonal to the center of the particle's orbit; however, we are only concerned with the force component towards the center in the radial direction, because there is equal mass in the plus and minus z direction. Hence, we only care about the "cosine" component in the radial direction,

Eq 2: cos( ) = r / \sqrt{ r^2 + z^2 }

So the force component in the r direction is:

Eq 3: da_r = cos() G { \large \rho } r / ( r^2 + z^2 ) ~ dz ~ = ~ G { \large \rho } r / ( r^2 + z^2 )^{3/2} ~ dz

The total acceleration is the integral of this between -\infty and +\infty :

Eq 4: a_r = {\Large \int_{-\infty}^{+\infty}} ~ da_d = {\Large \int_{-\infty}^{+\infty}} ~ G { \large \rho } r / ( r^2 + z^2 )^{3/2} ~ dz

I am a lazy fellow, so I will normalize z to units of r with z' = z / r or z = z' r :

Eq 5: a_d = {\Large \int_{-\infty}^{+\infty}} ~ G { \large \rho } r / ( r^2 + (z' r)^2 )^{3/2} ~ r ~ dz'

This allows us to pull everything out of the integral besides z' :

Eq 6: a_d = G { \large \rho } {\Large \int_{-\infty}^{+\infty}} ~ r / ( r^2 ( 1 + {z'}^2 ))^{3/2} ~ r ~ dz'

Eq 7: a_d = G { \large \rho } {\Large \int_{-\infty}^{+\infty}} ~ r / ( r^3 ( 1 + {z'}^2)^{3/2} ) ~ r ~ dz'

Eq 8: a_d = G { \large \rho } ( r^2 / r^3 ) {\Large \int_{-\infty}^{+\infty}}~ 1 / ( 1 + {z'}^2)^{3/2} ~ dz' ~ move the constant r out

Eq 9: a_d = ( G { \large \rho } / r ) {\Large \int_{-\infty}^{+\infty}} ~ 1 / ( 1 + {z'}^2)^{3/2} ~ dz'

I cheated and looked the integral up on Wolfram alpha:

{\Large \int_{-\infty}^{+\infty}} 1 / ( 1 + {z'}^2 )^{3/2} dz' = 2 ~ ~ ~ so:

Units check: ( m ~ s^{-2} ) = ( m^3 kg^{-1} s^{-2} ) ( kg ~ m^{-1} ) ( m^{-1} ) = ( m ~ s^{-2} ) ~ ~ ~ ... they match!

Am I supposed to put Q.E.D. here?

A reminder: for a general NON-Keplerian orbit, v^2 / r = a = 2 G { \large \rho } / r ~ so ~ v^2 = 2 G { \large \rho } = constant.

As to what "holds the filaments up" relative to a galaxy over gigayears, I have no clue. They are actually there (hot and charged and detectable), so nature is smarter than I am. We might expect the material close to the galaxy to fall into the central black hole, leaving a void; that will reduce the effect for the inner galactic material more than the outer material, so this may reduce inner rotation velocities relative to outer rotation velocities.

And I have no idea whether the filaments are actually massive enough to affect galactic rotation, but they seem to be massive enough to account for a large fraction of the "dark matter" the cosmologists are looking for. Perhaps all of it, if "dark energy" is an observational error because the SN1a "standard candle" claim is untrue.