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COMPARING DISTANCES USING FLUXES AND LUMINOSITIES
Every luminous object's flux and luminosity obeys the inverse square law.
So if we compare two objects - let's call them A and B - then
(L of object A)
(F of object A) = -------------------------- and
4 * pi * (r of object A)^2
(L of object B)
(F of object B) = -------------------------- are both true.
4 * pi * (r of object B)^2
Let's use F(A), L(A), r(A), and F(B), L(B), r(B) in our notation to
keep things shorter. Now, if F(B) = F(A), then the relations can be
combined to produce a new relation:
L(A) L(B) L(A) L(B)
--------------- = --------------- , or just ------ = ------ .
4 * pi * r(A)^2 4 * pi * r(B)^2 r(A)^2 r(B)^2
So if A is twice as far away as B [ that is, r(A) = 2 * r(B) ], which
object is more luminous, and by how much? Well,
L(A) L(B)
------------ = -----------; so L(A) = 4 * L(B).
(2 * r(B))^2 r(B)^2
On the other hand, if A is twice as luminous as B [L(A) = 2 * L(B)],
which object is more distant, and by how much? Well,
2 * L(B) L(B)
---------- = --------; so r(A)^2 = 2* r(B)^2, or r(A) = sqrt(2) * r(B).
r(A)^2 r(B)^2
Now we can get a bit more complicated. Let's say the flux you measure
from A is five times more than the flux you measure from B, but you
know their luminosities are the same. Let's further say that:
right next to A there's a star with a flux of 500 W/m^2, and
right next to B there's a star with a flux of 100 W/m^2.
Which star is more luminous? Well, F(A) = 5 * F(B), so
L(A) 5 * L(B)
------ = ---------- ; and since L(A) = L(B), we can deduce
r(A)^2 r(B)^2
5 * r(A)^2 = r(B)^2 so sqrt(5) * r(A) = r(B).
So the star near A has luminosity
L = 4 * pi * r(A)^2 * 500 W/m^2.
and the star near B has luminosity
L = 4 * pi * ( sqrt(5) * r(A) )^2 * 100 W/m^2.
Apparently, according to our calculations, the stars are equally luminous!
We come to this conclusion without knowing what r(A) and r(B) actually may
be; it only matters what they are relative to each other, if we carefully
use the inverse square relationship between flux and luminosity.
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