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← Revision 10 as of 20181210 07:58:31 ⇥
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That seems like a lot, but if the line mass was a cylinder 20 kiloparsecs ( 6.2e20 m) radius, that is 1e22 kg/m³. The proton mass is 1.67e27 kg, so this is 60,000 protons per cubic meter, or 0.06 protons per cubic centimeter. That may still be too much, compared to those papers ...  That seems like a lot, but if the line mass was a cylinder 20 kiloparsecs ( 6.2e20 m) radius, that is 1e22 kg/m³. The proton mass is 1.67e27 kg, so this is 60,000 protons per cubic meter, or 0.06 protons per cubic centimeter. The papers claim less. Hmm. 
Cold Dark Matter
New Scientist 2017 11 October
Hideki Tanimura at the Institute of Space Astrophysics in Orsay, France
A Search for Warm/Hot Gas Filaments Between Pairs of SDSS Luminous Red Galaxies
Detection of intercluster gas in superclusters using the thermal SunyaevZel'dovich effect
Anna de Graaff at the University of Edinburgh, UK.
There may be no exotic dark matter, just gas that is very difficult to detect without large, sensitive instruments in space. Planck can only image some of this "dim" matter by overlaying hundreds of images, which cannot detect other baryons (mostly protons) that are not clustered this way.
Rotation Curves around a Line Mass
"Heavy" line mass filaments perpendicular to a galaxy produce rotation curves with constant velocity at any radius.
\mu 
Standard Gravitational Constant 
\lambda 
line mass density, i.e., kg/m 
r 
radius of body from line mass 
y 
vertical offset from mass element 
l = \sqrt{ r^2 + y^2 } 
distance from mass element 
b 
integration substitution variable 
a 
acceleration towards line mass 
v 
orbital velocity around line mass 
da ~=~ gravity \times cos( angle )
da ~=~ \Large { { \mu \lambda dy } \over { l^2 } } { r \over l }
a ~=~ \mu \lambda r \Large \int_{\infty}^{+\infty} { dy \over l^3 }
substitute y = b r :
dy ~=~ r ~ db
l ~=~ \sqrt{ r^2 + y^2 } = \sqrt{ r^2 + b^2 r^2 } = r \sqrt{ 1 + b^2 }
a ~=~ \mu \lambda r \Large \int_{\infty}^{+\infty} { r db \over { r^3 ( 1 + b^2 )^{3/2} } }
a ~=~ \Large { { \mu \lambda } \over r } \int_{\infty}^{+\infty} { db \over { ( 1 + b^2 )^{3/2} } }
a ~=~ \Large { { 2 \mu \lambda } \over r }
For circular orbital motion, ~ a ~=~ v^2 / r ~ , so v^2 ~ = ~ 2 \mu \lambda
Therefore, for an infinite line mass, rotation velocity is constant, regardless of radius.
And this matches galactic rotation curves, for example:
How much line mass would explain that graph?
λ = v² / 2 μ ~ = ~ ( 1.2e5 m/s )² / ( 2 * 6.674e−11 m³/kg s² ) = 1.2e20 kg/m
That seems like a lot, but if the line mass was a cylinder 20 kiloparsecs ( 6.2e20 m) radius, that is 1e22 kg/m³. The proton mass is 1.67e27 kg, so this is 60,000 protons per cubic meter, or 0.06 protons per cubic centimeter. The papers claim less. Hmm.