Apparent Magnitude, Absolute Magnitude and Distance

By , 22nd April 2008 in Astronomy

Apparent magnitude and Absolute magnitude are two ways of comparing an object's brightness. In this example we look at the relationship between absolute magnitude, apparent magnitude and luminosity.

There are two main types of magnitude commonly used in astronomy. The first of these, apparent magnitude, is the brightness of the object as seen by an observer on the Earth. The apparent magnitude of a star is dependent on two factors:

  1. The luminosity of the star (total energy per second radiated)
  2. The distance of the star from Earth

The second, absolute magnitude, is dependent solely on the stars luminosity and can be regarded as an intrinsic property of the star. Absolute magnitude is defined as the apparent magnitude of an object if it were a standard distance from the Earth. The standard distance is 10 parsecs. Since distance is always equal when comparing absolute magnitudes, it can be removed as a factor in the stars brightness which is why it can be regarded as an intrinsic property.

Absolute magnitude and Luminosity

A stars luminosity, L, is the total amount of energy radiated per unit time. The absolute magnitude of a star is related to its luminosity in the same way as apparent magnitude is related to flux. If we compare the ratio of the brightness of two stars, expressed in terms of their luminosities, then we obtain a relation for the difference in their absolute magnitudes.

Equation 23 - Absolute Magnitude Relation

Capital letters are used to indicate absolute magnitudes and lower case letters are used to identify apparent magnitudes.

As we have previously stated, absolute magnitude is the apparent magnitude of an object if it was a distance of 10 parsecs from the Earth.

It is clear from this definition that a star located at 10 parsecs from the Earth will have the same apparent and absolute magnitude. A star that is further away than 10 parsecs will have a fainter apparent magnitude than absolute magnitude and a star that is closer than 10 parsecs will have a brighter apparent magnitude than absolute magnitude.

How do we know a stars absolute magnitude? We could travel to every star and measure the apparent brightness from a distance of 10 parsecs, but at the moment that really isn't a practical solution. Luckily for us however, the apparent and absolute magnitudes are related by a very important formula.

Distance Modulus

Distance Modulus is the difference between the apparent and absolute magnitudes. This can be obtained by combining the definition of absolute magnitude with an expression for the inverse square law and Pogson's relation. Using the distance modulus it is possible to establish a relationship between the absolute magnitude, M, of a star, its apparent magnitude, m, and its distance, d.

The inverse square law tells us that for a star at distance d (parsecs), with observed flux Fm, then its flux FM at 10 parsecs would be given by:

Equation 24 - Inverse Square Law for Flux

We can combine this with equation 23 above to give the distance modulus equation.

Equation 25 - Distance Modulus

If we measure a stars apparent magnitude, and its distance in parsecs is known, then we can determine the absolute magnitude and hence the luminosity of the star. If we know the stars absolute and apparent magnitudes we can use distance modulus to calculate the distance to the star. This equation is very powerful and will be used a great many times in upcoming tutorials.

The formula for calculating Absolute Magnitude within our galaxy is:

Equation 31 - Absolute Magnitude

Where D is the distance to the star in parsecs.


Barnard's Star lays 1.82 parsecs away and has an observed (apparent) magnitude of 9.54.

m - M = 5((log10 D)-1)
M = 9.54 * 5((log10 1.82)-1)
M = 9.54 - (-3.7)
M = 13.24

If Barnard's Star were to be moved to a distance of 10 parsecs from the Earth it would then have a magnitude of 13.24.

If we already know both Apparent and Absolute magnitudes, it is possible to calculate the distance to the star:

d = 100.2(m - M + 5)

Using Barnard's Star again,

d = 100.2(9.54-13.24+5)
d = 100.26
d = 1.82 parsecs

Bolometric Magnitude

Another type of magnitude of interest to astronomers is the bolometric magnitude. So far the absolute and apparent magnitudes are based on the total visible energy radiated from the star. We know that not all of that energy is received on Earth since it is filtered out by our atmosphere.

Bolometric magnitude is a based on the flux throughout the entire electromagnetic spectrum. The term absolute bolometric magnitude is based specifically on the luminosity (or total rate of energy output) of the star. We will look at bolometric magnitudes in a later tutorial.

Further Reading
  1. Danyale

    is absolute magnitude or apparent magnitude relative to the distance from Earth?

  2. Chris

    Why does the size of a star correlate with bolometric luminosity? why cant a star be relatively small but bolometrically bright, or large but bolometrically dim?

  3. Phileas Fogg
    Phileas Fogg

    Does anybody know how to "unaccount" for the atmosphere when measuring absolute magnitude?

  4. tiffany forshee
    tiffany forshee

    okay so what is the difference between absolute and apparent magnitude?

    1. Palustec

      Apparent Magnitude - is how bright something appears to you from were you are.
      Absolute Magnitude - is how bright something looks from a set distance of 10 parsecs (32.6 lightyears).

      A dim star will still be very bright if you are really close to it. Conversely, a really bright star can be seen from very far away.

  5. AB Atwater
    AB Atwater

    I am very interested in astronomy, but I suck at mathematic calculations (go figure...). I am taking a class currently, and a question posed is as follows:

    Polaris is a second-magnitude star. Phi Pegasi is about sixteen times fainter than Polaris. What is the approximate magnitude of Phi Pegasi?

    I have been given the choices of 18, -14, 3, -3, and 5. I do not know the calculation/equation that I need to use to reach an answer. I am not asking that you answer this for me, but I would like to know / to understand the appropriate equation. Please respond via e-mail, as soon as possible. Thank you!

    1. Palustec

      The answer is 5th magnitude (apparent).

      each magnitude is 2.512 brighter or fainter then the next increment. So a 1st magnitude star is 2.512 times brighter then a 2nd magnitude star. Therefore, 2.512 * 2.512 * 2.512 = 15.85 times brighter or fainter - notice that 15.85 is close to 16). So that means there are 3 magnitudes of difference between Polaris and Phi Pegasi - so 2 (mag of Polaris) + 3 (mag diff) = 5 th mag for Phi Pegasi.

  6. Jerry Abbott
    Jerry Abbott

    Does anybody know the equation relating all-spectrum flux with bolometric apparent magnitude?

    If we know that a certain star has a bolometric apparent magnitude of EXACTLY zero, then how many watts per meter squared do you get from it when you integrate over the entire spectrum?

  7. carolina

    omg this calculation is so difficult

  8. Su

    (m-M) = 5logd-5. Great but how do I get d = ?????
    I cannot see how to rearrange can you help
    Many thanks

    1. Palustec

      d = 10^(1 + (m-M)/5)

      10 raised to the power of 1 plus the quantity of apparent mag minus absolute mag divided by 5.

      a quick sanity check: if m = M then 10 raised to the 1st power is 10. which is 10 parsecs.

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