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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. |