Magnitude & Distance
|In this EBook we discuss how brightness of celestial objects can relate to their distance.|
|Tags||Astronomy, Stars, Galaxies, Hipparchus, brightness, luminosity, flux, apparent magnitude, absolute magnitude, distance modulus, standard candles.|
|First Published||October 2008|
|This Edition - 3.0||December 2018|
Units of Distance in Astronomy
The standard unit of length (or distance) is the metre (symbol: m).
In Astronomy we work with very large distances as compared to distances on Earth. Therefore we commonly use other units for distance in Astronomy.
- Astronomical Unit (AU) - average distance Earth - Sun
- Light year (ly) – the distance light travels in one year
- Parsec (pc)
The Astronomical Unit is used to express distances across the Solar System.
1 astronomical unit (AU) = average distance from the Earth to the Sun;
1 AU = 149,597,870,691 m = about 1.5 x 1011 m.
The Light Year is commonly used to express distances to objects, e.g. to stars. Note that the lightyear is not a measure of time!
1 light year (ly) = distance that light travels in one year
1 ly = 9.461 x 1015 m = 9.461 x 1012 km
1 ly = 63,240 AU.
To get an idea about how much a light year is, remember that light travels at about 3x108 m/s; this is about 7½ times around the Earth in one second.
For relatively short distances we use e.g.:
Distance to the Moon is in average 384,402 km or 1.282 Light second;
Distance to the Sun is 1 AU or about 8.3 Light minutes.
This unit also indicates how long ago the light that we see departed from the source.
The Parsec is a unit used to express distances outside our Solar system to stars and galaxies, etc.
1 parsec = 1 pc = 206,265 AU = 3.086×1016 m = 3.262 ly
This unit originates from determining the distance to a star by measuring parallax. It is the distance for which a star has a parallax of one arc second when the Earth moves 1 AU. We discuss this below.
An isosceles triangle is determined by the length of the base and the opposite angle.
If the base is fixed, the height (or distance) can be found by measuring the angle. In this geometric sense, the top angle is called the parallax.
The most common method of parallax to measure distance to stars is the heliocentric parallax, where we use the distance from Earth to the Sun as the base of the parallax triangle. Take a photo of the night sky in two different seasons, about half a year apart. The orbit of the earth around the sun is used as the base.
Nearby stars appear to move more than stars further away. The baseline can be up to 2 AU.
The green star is closest because it has the largest parallactic shift.
The blue star is further away, but still a lot closer than the red background stars, who do not appear to have any shift at all.
The animation below shows how parallax affects our perception of the apparent position of a star near to us in the course of a year. It's position relative to other, farther away stars seems to change because our viewing position shifts as the earth moves around the sun.
Drag the slider on the illustration to learn how we can deduce the distance to a star
by measuring the stellar parallax value.
Calculating astronomical parallax
The angles involved are very small, typically less than 1 arc second. (Remember that 1 arc second = 1/3600 of a degree). To determine the distance to a star we can write (for small angles):
d = b / 2p = ½ b / p
where d is the distance to the star, p is the parallax angle expressed in radians (see diagram), and b is the baseline. In the case of heliocentric parallax, the base b is equal to 2 Astronomical Unit (AU) -- the diameter of the Earth's orbit.
Since there are 206,265 arc seconds per radian, the formula can be re-written as:
d (in AU) = 206,265 / p
with p now measured in arc seconds.
Or, if we write the distance of one parsec as 206,265 AU, we get:
d (in parsecs) = 1 / p (in arc seconds)
Thus one parsec is the distance for which the heliocentric parallax is 1 arcsecond.
This is the distance unit astronomers use most frequently, and it is equivalent to 3.26 light years.
This unit was chosen because it makes the parallactic formula very simple.
A star is very nearly a blackbody, therefore its brightness is directly related to the total amount of energy radiated from its surface per unit time.
In astronomy this is called Luminosity which is the amount of energy emitted per second.
This is exactly the same as Power which is energy per unit time (Joule per second).
We all know this when we buy a light bulb that has a certain number of Watt.
The unit for power is Watt (W) or Joule (energy) per second or W = J s-1.
In astronomy luminosity is not expressed in Watt but in multiples of the Sun’s power (i.e. the Sun’s luminosity),
LSun = 3.83x1026 W.
We can measure the power received on Earth on every square metre. This is called the Solar Constant expressed in W/m2.
Multiply this with the surface area of a sphere of radius 1 AU (distance Sun - Earth) and you get LSun.
Flux and the Inverse square law
Flux is the amount of power going through a unit area.
This diminishes with the square of the distance.
where L is the Luminosity and r is the distance to the star.
So the amount of flux decreases with the square of the distance.
Let us look at a practical example
If someone takes two torches, say one of 20 Watt and the other one of 10 Watt and walks to a distance of say 50 m, then we will see the bright torch still brighter than the less bright one, but both are less bright than at a closer distance.
If you now leave the one of 10 Watt at that distance and walk further say to 100 m with the one of 20 Watt. That more luminous one may now look fainter than the less luminous one that is still at a distance of 50 m.
This shows that if we do not know the actual luminosity of an object (say a star), the brightness (or flux) that we see does not tell us anything about the distance of the star. Brighter stars can be further away than fainter ones or the other way around. We cannot know.
Inverse Square Law
We saw that the flux decreases with the square of the distance. This means that if the distance increases by a factor of two, the flux diminishes by a factor of two-squared or four. If the distance is tripled, the flux decreases with a factor of nine, etc.
This relationship is well known in Physics and is called the Inverse Square Law.
The flux is an indicator for the brightness that we see.
So how can we find the distance to a star with the brightness that we observe?
The method of finding distance we are going to discuss is based on assumptions about the luminosity of the star. Suppose we know the luminosity (L in the image), then with the Flux (brightness that we observe) we are able to calculate the distance with the inverse square law.
In astronomy it is common to express brightness not in Flux but in Magnitude.
The magnitude scale
Magnitude is a historic concept. Hipparchus, about 190-120 BCE, categorized stars of the 1st magnitude (for the typical brightest stars) through to stars of the 5th magnitude (faintest to the eye). This describes the brightness of a star without any reference to how far away it is and is denoted with a lowercase m.
Hipparchus thus classified stars by their relative brightness and began the magnitude scale. This classification was based on visual observations because in those days they could not measure flux or brightness with photometers like we do today.
Hipparchus defined a magnitude +2 star as half as bright as a magnitude +1 star. The total range of his scale was from +1 (brightest star) to +6 (dimmest visible to naked eye). In effect this is a logarithmic scale.
Based on measurements in the 19th century the magnitude scale was adjusted so that first magnitude stars were actually 100 times brighter (in terms of flux) than sixth magnitude stars. So the scale has been revised so that a first (+1) magnitude star is exactly 100 times brighter than a sixth (+6) magnitude star. So an interval of 5 magnitudes is the same as a factor of 100 in measured flux.
This means that for every unit increase in magnitude, the brightness increases by a factor of 2.512 because
2.512 x 2.512 x 2.512 x 2.512 x 2.512 = (2.512)5 = 100
This factor of 2.512 has of course been calculated from 1001/5 = 2.512
This was expressed by Pogson in 1856 as the “fifth root of hundred” ratio, or the Pogson’s ratio.
The observed magnitude is since referred to as Apparent Magnitude with symbol (m) and is still widely in use today.
In order to have a similar measure for a stars luminosity, the concept of Absolute Magnitude (symbol M) has been defined as the apparent magnitude that a star would have at a standard distance of 10 parsec (pc).
As an example the Sun has an apparent magnitude m = -26.73 but it has an absolute magnitude of +4.75. That is the apparent magnitude we should see if the Sun was at a distance of 10 pc.
With refined methods to measure brightness, it has been found that there are stars actually brighter than magnitude +1 stars. These are given a negative number for their magnitude.
Sirius, the brightest star, has an apparent magnitude -1.44. The scale can also be used for other celestial objects brighter than the brightest stars and dimmer than the stars we can see with the naked eye.
Examples of objects with various apparent magnitudes
|−4.4||Maximum brightness of Venus|
|−4.0||Faintest objects observable during the day with naked eye|
|−2.8||Maximum brightness of Mars|
|−1.5||Brightest star at visible wavelengths: Sirius|
|−0.7||Second brightest star: Canopus|
|0||The zero point by definition: Close to Vega’s apparent magnitude|
|~3||Faintest stars visible in an urban neighbourhood|
|~6||Faintest stars observable with naked eye|
|27||Faintest objects observable in visible light with 8m ground-based telescopes|
|30||Faintest objects observable in visible light with Hubble Space Telescope|
Note that the star Vega is the reference star for the magnitude scale (m=0).
Summary apparent magnitude
- Apparent magnitude (symbol m)
- indicates the brightness of a star as we see it
- is a scale that runs opposite to brightness
- is centred at the apparent magnitude of Vega (m = 0)
- allows negative values for bright stars
- one unit in magnitude corresponds to a factor of 2.512 in brightness, or 5 units in magnitude correspond to a factor of 100 in brightness.
Which of the two is brighter and by how much?
The magnitude difference is 1.64 - 0.61 = 1.03.
This corresponds to a difference in brightness of (2.512)1.03 = 2.58.
Therefore Beta Centauri is 2.58 times brighter than Gamma Crucis.
This page requires working with logarithms.
Now we need to find out how to calculate distance from absolute magnitude (M) and apparent magnitude (m). We assume that we will have a way to find the absolute magnitude of stars. We will come back to that later.
Absolute magnitude is defined as the apparent magnitude at a distance of 10 pc. So the flux ratio between the flux (FD) we receive at our distance D and the flux at the standard distance of 10 pc (F10) is
(this derives from Pogson’s ratio, or the definition that 5 units in magnitude correspond with a factor of 100 in flux).
Each flux can be expressed in Luminosity (L) from the Inverse Square Law as (units for distance are parsec)
Substitution into the first formula gives
Taking the logarithm and re-arranging gives
with D in pc.
This is the Distance Modulus , the relationship between apparent and absolute magnitude and distance.
Apparent magnitude m is what we observe. Provided that we have a way to find absolute magnitude, we can calculate the distance from the Distance Modulus.
Calculate an example
A star has an absolute magnitude of -1.0 and an apparent magnitude of +14.0. What is the distance to that star?
Take power of 10 on both sides
Hence the star is at a distance of 10,000 parsec.
The good news is that in astronomy there are methods to find the luminosity or absolute magnitude of a star, which will enable us to calculate the distance. This is why we did develop the basic method above.
Celestial objects of which we (think we) know the luminosity or absolute magnitude are termed Standard Candles.
Any object that can be considered as a Standard Candle in astronomy needs to have the following properties:
- it must be easy to identify and not being confused with a different type of object
- it must have a known luminosity related to some physical property that we can measure
- it should preferably be very bright so we can use it to large distances.
Fortunately there are several ways in which we can find estimates for a star’s luminosity. These methods are generally deducted from particular physical processes in a star (or entire galaxy) that are indicative for the object’s luminosity. These techniques allow us to estimate distances, even at inter-galactic scale.
In our EBook Stellar Distance these and other methods in Astronomy are discussed in detail.
The particular methods that use the Distance Modulus method described above
are Spectroscopic Parallax and Main Sequence Fitting .