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 object 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:

Eq 10: a_d = 2 G { \large \rho } / r ~ ~ ~ the acceleration from an infinite line mass is proportional to 1 / r

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 ionized and vaguely detectable in bulk), 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.

Are the filaments perpendicular to rotation? If not, they may tilt themselves and the galaxy until they are. Keep in mind that we have NOT detected filaments associated with our own galaxy, we just detect them statistically between millions of pairs of other distant galaxies. We only see the radial rotation of other galaxies, we are imbedded in ours, and can only infer whole-Milky-Way behavior from the observed behavior of others.

FilamentGravity (last edited 2017-11-01 19:36:34 by KeithLofstrom)