The brightness of a variable star

[Benzoate]

Homework Statement

A variable star changes in brightness by a factor of 4. What is the change in magnitude?

Homework Equations

m and n represent two stars of magnitude
m-n=2.5 log(f(n)/f(m))
log (f(n)/f(m))=.4*(m-n)

The Attempt at a Solution

I think m-n is suppose to represent a change in magnitude.perhaps a factor of 4 = log(f(n)/f(m) . Therefore , m-n=2.5*4=10 ?

 

[Andrew Mason]

Homework Statement

A variable star changes in brightness by a factor of 4. What is the change in magnitude?

Homework Equations

m and n represent two stars of magnitude
m-n=2.5 log(f(n)/f(m))
log (f(n)/f(m))=.4*(m-n)

The Attempt at a Solution

I think m-n is suppose to represent a change in magnitude.perhaps a factor of 4 = log(f(n)/f(m) . Therefore , m-n=2.5*4=10 ?

Magnitude is a logarithmic scale, whereas brightness is measured on a geometric scale. So a star that is 4 x brighter represents a smaller change in magnitude. I get a 1.5 change in magnitude.

Let x = the variation in magnitude and let b = brightness. According to the rather arcane method of determining stellar magnitudes (see: http://en.wikipedia.org/wiki/Apparent_magnitude ) this is the relationship between x and b:

\log_{10}(100^{.2})^x = 2.512^x = b

AM

 

[Benzoate]

Magnitude is a logarithmic scale, whereas brightness is measured on a geometric scale. So a star that is 4 x brighter represents a smaller change in magnitude. I get a 1.5 change in magnitude.

Let x = the variation in magnitude and let b = brightness. According to the rather arcane method of determining stellar magnitudes (see: http://en.wikipedia.org/wiki/Apparent_magnitude ) this is the relationship between x and b:

\log_{10}(100^{.2})^x = 2.512^x = b
AM

I don't understand. so factor 4 can also represent the brightness of the star? Also when you say let x be a variation in magnitude, do you mean let x be the change in magnitude?

 

[Kurdt]

Using the original equation you had you know the ratio between the two extremes of brightness is 4. Therefore f(n)/f(m)=4. The method Andrew has used is the inverse. For a difference of 1 in the apparent magnitude the ratio of the apparent brightness is 2.512. for a difference of 2 in the apparent magnitude the difference in apparent brightness is (2.512)² ~ 6.31. etc.

 

[dynamicsolo]

For the sake of clarifying the basis of this system (for which Hipparchus gets the initial blame), a difference of five magnitudes corresponds to a ratio of 100 in brightness, intensity or power. (The decibel scale for sound works similarly, a difference of 10 decibels corresponds to a ratio of 10 in intensity.) So the relation between magnitudes and intensity is

( I1/I2 ) = (100)^[(m2 - m1)/5] ,

which is the basis of the equation Andrew Mason gives: \log_{10}(100^{.2})^x = 2.512^x = b . [Revision: I forgot to reverse the magnitudes; the brighter the star, the lower the magnitude -- fixed now. Thanksalot, Hipparchus...]

You can also take the common logarithm of both sides and rearrange it to get

log10 ( I1/I2 ) = (2/5)·(m2 - m1) or

(m1 - m2) = -2.5 · log10( I1/I2 ) , the form you have in your first post.

In your question, it would be I1/I2 = 4.

Systems such as the magnitude scale or the decibel (dB) scale exist because physiological sensory systems (like eyes and ears) and many artificial detection devices have such a logarithmic "response" over a large range of intensity levels.