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 \rho , and perpendicular axis z . Calculate the radial gravitational acceleration a .

Assume a mass element dm = \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 \rho dz / h^2 dz ~ = ~ G \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_d = cos() G \rho r / ( r^2 + z^2 )^{3/2} dz

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

Eq 4: a_d = \int_{-\infty}^{+\infty} da_d = \int_{-\infty}^{+\infty} G \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 = \int_{-\infty}^{+\infty} da_d = \int_{-\infty}^{+\infty} G \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 \rho \int_{-\infty}^{+\infty} r / ( r^2 ( 1 + {z'}^2 ))^{3/2} r dz'

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

Eq 8: a_d = G \rho ( r^2 / r^3 ) \int_{-\infty}^{+\infty} 1 / ( 1 + {z'}^2)^{3/2} dz'

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

I cheated and looked the integral up on Wolfram alpha:

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

so:

Eq 10: a_d = 2 G \rho / r

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

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

As to what "holds the filaments up" over gigayears, I have no clue. They are hot and charged and detectable, so nature is smarter than I am.

And I have no idea whether they are 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.