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Eq 1: $ da_d = G \rho dz / h^2 dz = G \rho / ( r^2 + z^2 ) dz $ in the diagonal direction Eq 1: $ da_d = G \rho dz / h^2 dz ~ = ~ G \rho / ( r^2 + z^2 ) dz ~ ~ ~ ~ ~ $ in the diagonal direction
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Eq 3: $ da_d = cos() G \rho r / ( r^2 + z^2 )^{3/2} dz Eq 3: $ da_d = cos() G \rho r / ( r^2 + z^2 )^{3/2} dz $
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Eq 4: $ a_d = \int_{-inf}^{+inf} da_d = \int_{- \inf}^{+ \inf} G \rho r / ( r^2 + z^2 )^{3/2} dz Eq 4: $ a_d = \int_{-inf}^{+inf} da_d = \int_{- \inf}^{+ \inf} G \rho r / ( r^2 + z^2 )^{3/2} dz $
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Eq 9: a_d = 2 G \rho / r Eq 9: $ a_d = 2 G \rho / r $
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Units check: $ ( m s^{-2} ) = ( m^3 kg^{-1} s{-2} ) ( kg m^{-1} ) ( m^{-1} ) = ( m s^{-2} ) $ Units check: $ ( m s^{-2} ) = ( m^3 kg^{-1} s{-2} ) ( kg m^{-1} ) ( m^{-1} ) = ( m s^{-2} ) ~ ~ ~ $  .. they check!

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{ r2 + z2 } 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 -\inf and +\inf

Eq 4: a_d = \int_{-inf}^{+inf} da_d = \int_{- \inf}^{+ \inf} 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 4: a_d = \int_{-inf}^{+inf} da_d = \int_{- \inf}^{+ \inf} 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 5: a_d = G \rho \int_{- \inf}^{+ \inf} r / ( r^2 ( 1 + {z'}^2 ))^{3/2} r dz'

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

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

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

I cheated and looked the integral up on Wolfram alpha:

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

so:

Eq 9: 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.

FilamentGravity (last edited 2019-03-21 05:30:39 by KeithLofstrom)