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marcus
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#72
Jan21-05, 02:01 PM
Astronomy
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some special numbers go with these units, most are pure math numbers and would be factors in the equation no matter what system of units, but here they sometimes jump out a little more clearly.

80/pi this tells the evap time of a BH. cube the mass and multiply by 80/pi

pi2/15 tells the per-volume radiant energy density at some temp. quart the temp (raise to fourth) and multiply by pi2/15

pi2/60 tells the brightness at some temp (power radiated per unit area). quart the temp (raise to fourth) and multiply by pi2/60

3zeta(4)/zeta(3) = 2.701 tells the average photon energy at some temp.
just multiply the temperature by 2.701. Since sun temp is 2E-28, the average sunlight photon has energy 5.402E-28---anyway that's the idea.

1 tells the bekenhawking temperature of a BH. just take 1 over the mass.

1/4pi tells the Schw. radius of a BH. just take that times the mass.

1/4pi tells the area of the BH. take that times the square of the mass.

3 tells the critical density of the universe. just multiply 3 by the square of the hubble parameter

6 tells the density of a round planet. divide 6 by the square of the radian time in low orbit.

9 or thereabouts is the heat capacity of a molecule of water

29 is the molecular weight of air. It is handy to know.
(atomic and molecular weights generally are)

Oh, they tell us that the density of the universe is at or very close to the critical value. So 3 also tells the actual density of the universe.

1/137 (more exactly 1/137.036...) is the coulomb constant. it tells the force between two charges separated by a distance. just multiply the charges by 1/137 and divide by the square of the distance.

1/137 also tells the force between parallel currents (measured on a test segment with length equal half the separation). just multiply the currents by 1/137

(1/137)2 tells the energy needed to ionize a hydrogen atom. multiply the rest energy of an electron (2.1E-22) by it and you get a quantity of energy called the Hartree----which is twice the ionization energy (so you still need to divide by two)

in each case i am assuming that the calculation is done in natural units terms, so that I don't have to specify the units each time I say something.

there's lots more but maybe this is enough for now