Astronomical magnitude formula convertion

Question

It is a question asked out of pure frustration, but I really want to know the steps leading from the expression

$$m_1-m_2=2.5\log\bigg(\frac{F_2}{F_1}\bigg)$$

to this one

$$\frac{F_1}{F_2}=2.5^{m_2-m_1}$$

Do you mean $\log_{2.5}\big(\frac{F_2}{F_1}\big)$? – Pavel Feb 14 '16 at 01:38 — Pavel, Feb 14 '16 at 01:38
No, I mean what I explicitly wrote. – almighty_dollar Feb 14 '16 at 01:41 — almighty_dollar, Feb 14 '16 at 01:41

score 3 · Accepted Answer · answered Feb 14 '16 at 01:42

This is an approximate formula, and formulas like these are common in the practical sciences (like Astronomy and Applied Physics).

Those should be base-$10$ logs.

$\displaystyle m_1-m_2=2.5\log(\frac{F_2}{F_1})$

Rearrange,

$\displaystyle \log(\frac{F_2}{F_1}) = \frac{m_1 - m_2}{2.5}$

Take antilogs of both sides:

$\displaystyle \frac{F_2}{F_1} = 10^{\frac{m_1 - m_2}{2.5}}$

Take reciprocals of both sides and use $\displaystyle \frac{1}{x^y} = x^{-y}$:

$\displaystyle \frac{F_1}{F_2} = 10^{\frac{m_2 - m_1}{2.5}}$

(Here, $\displaystyle m_2 - m_1 = -(m_1 - m_2)$)

Now use $\displaystyle x^{yz} = {(x^y)}^z$:

$\displaystyle \frac{F_1}{F_2} = (10^{\frac{1}{2.5}})^{m_2 - m_1}$

And note that $\displaystyle (10^{\frac{1}{2.5}}) \approx 2.5$ to give:

$\displaystyle \frac{F_1}{F_2} \approx 2.5^{m_2 - m_1}$

1 Answers1