Gadget by The Blog Doctor.

Thursday, August 23, 2007

Measuring distance to the stars: Distance Modulus

The apparent magnitude (brightness) of a star can be easily measured, as it is just the brightness of the star as seen from Earth. The absolute magnitude is the brightness of the star as seen from 10 parsecs away. It is possible to determine the absolute magnitude of a star by determining its spectral type and then reading off the luminosity in a HR diagram. The luminosity can then be easily converted to absolute magnitude.

Once these two measures of brightness have been determined the distance to the star can be calculated by the formula:
d = 10(m-M+5)/5
where d = distance (in parsecs), m = apparent magnitude and M = absolute magnitude.

Here are some sample calculations:

Proxima Centauri
m = 11.01, M = 15.53
d= 10(11.01 -15.53 + 5) / 5
= 1.247 parsecs
The distance determined from parallax is 1.3 parsecs so the Distance Modulus calculation is fairly accurate.
The source for the parallax distance is Research Consortium on Nearby Stars (RECONS).
Parallax = 0.76887 therefore distance = 1 / 0.76887 = 1.3 parsecs

m = -1.47, M = 1.48
d = 10(-1.47 - 1.48 + 5) /5
= 2.57
Paralax distance is 2.63 parsecs (1/0.38002) which is
also reasonably accurate.

This is all well and good, but how was the Distance Modulus formula determined?
The rest of the post attempts to answer this question.
Warning: the following is fairly mathematical!

Relating Flux to Distance
Flux is the energy (light) passing through an area in an amount of time
It is defined as:
F = L / (4πD2)
Note that 4πD2 is the surface area of a sphere.
The observed brightness of a light source is related to its distance by the inverse square law - a source twice as far away appears to be on quarter as bright. For a single object or two objects of the same luminosity:

The L terms cancel as do the 4π terms

Therefore F1 / F2 = (D2 / D1)2 - Equation 1

Converting Luminosity to Magnitude
It is important to realise that luminosity is linear while magnitude is logarithmic.
A difference of 5 magnitudes corresponds to a ratio of 100 in luminosity.
L1 / L2 = xΔM ΔM is a change in M
100 = x5 5 M difference = 100 L difference
2.5118875 = 100.0001132
It is conventional to abbreviate this to 2.55 which equals 97.66 .
L1 / L2 = 2.5ΔM - Equation 2

Relating Luminosity to Distance
We need to find the ratio of luminosity required to produce the same flux from different differences.
F = L / (4πD2)
L = F4πD2
L1 = F4πD12
L2 = F4πD22
Note: there is no F1 or F2 as we are looking for luminosity
required to produce the same flux from different differences, ie F = F
L1 / L2 = F4πD12 / F4πD22
= D12 / D22
= (D1 / D2) 2 - Equation 3
Note this is different to the flux / distance relationship.

Relating Magnitude to Distance
L1 / L2 = 2.5ΔM
L1 / L2 = (D1 / D2) 2
2.5ΔM = (D1 / D2) 2

ΔM = 2.5log10(D1 / D2) 2
= 5log10(D1 / D2)
Note that ΔM = M2 - M1 therefore
M2 - M1 = 5log10(D1 / D2)
Note: absolute magnitude is defined as the apparent magnitude of an object when seen at a distance of 10 parsecs so the magnitude equation can be written as:
m - M = 5log10(D/10)
= 5(log10D - log1010)
= 5(log10D - 1)
= 5log10D -5

Solving for Distance
if m - M = 5log10D -5 then
m - M + 5 = 5log10D
(m - M + 5)/5 = log10D
Therefore D = 10(m - M + 5)/5
which is the relationship that we set out to prove!

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