Luminosity and how far away things are
In this class, we will describe
how bright a star or galaxy really is by its luminosity.

The luminosity is how much energy is coming from the per second. The units
are watts (W).

Astronomers often use another measure, absolute magnitude.
Absolute magnitude is based on a ratio scale, like apparent magnitued. I think
it is confusing.

The luminosity of the Sun is

Lsun = 3.9 x 1026 W We will often measure
luminosities of stars in units of the luminosity. That is, we might say for a
certain star Lstar = 5.2 x Lsun, meaning that the star has
5.2 times the energy output per second of the Sun.

Apparent brightness
In this class, we will describe how bright a star
seems as seen from Earth by its apparent brightness. This is often
called the intensity of the starlight. Sometimes it is called the
flux of light.

The apparent brightness is how much energy is coming from the star per square
meter per second, as measured on Earth. The units are watts per square meter
(W/m2).

Astronomers usually use another measure, magnitude. (Our book
calls it apparent magnitude.) Since magnitude is so commonly used,
we need to understand a little about it too. The magnitude system stems from
ancient Greece. A very bright star was called ``first magnitude,'' a pretty
bright star is ``second magnitude,''... a barely visible star is ``sixth
magnitude.''

Measuring luminosity
Since we can't go to a star to measure its
luminosity, we have to be clever. If we know the distance to the star we can do
it, because there is a simple relation between

the distance d to the star,

the apparent brightness b of the star, and

the luminosity L of the star.

The relation is

This relation is based on conservation of energy. Consider an observer a
distance d from the star. How much energy crosses a square meter
detector that this observer has?

All of the energy produced by the star per second must cross a sphere of
radius d.

The study of geometry tells us that area of this sphere is 4 Pi
d2.

a 1 m2 detector gets a fraction

(1 m2)/(total area)

of this.

Note that the brightness decreases proportionally to the square of the
distance as one moves away from the star. This can be demonstrated
experimentally in the laboratory.

We can write the relation so that it gives L in terms of b
and d.

It is easy to measure the apparent brightness b (e.g.
with a telescope and a CCD).

It is not so easy to measure the distance, but we have already seen
something about how that can be done.

Putting these toghther, we get L.

Luminosities of stars
One finds

There are a few stars that are more luminous than the Sun.

For instance, Betelgeuse has L ~ 14000 x Lsun.

Most stars are less luminous than the Sun.

For instance, Lstar ~ 0.001 x Lsun is not unusual.

There are lots more low luminosity stars than high luminosity stars.

Using brightness and luminosity to get distance

It is easy to measure the apparent brightness of a star, a galaxy, a
supernova, ...

If somehow we know the luminosity of such an object, then we can
compute its distance from us.

This is the essence of how we learn the distances to things.

Example: You have a 100 W lightbulb in your laboratory. Standing a
distance d from the lightbulb, you measure apparent brightness of the
lightbulb to be 0.1 W/m2. How can you use this information to
determine the distance from you to the lightbulb? (An approximate answer is
OK. In particular, you can approximate 4 Pi = 12.566 by 10.)

The luminosity of the lightbulb is L = 100 W.

The brightness is b = 0.1 W/m2.

So the distance is given by d2 = (100 W)/(4 Pi x 0.1
W/m2).

Since 4 Pi is approximately 10, this is d2 = (100 / 1)
m2.

Thus d2 = 100 m2.

We now know what d2 is. We want to know what d is. So we take
the square root.

So d = 10 m.

Example: You see a 1000 W streetlamp on a distant hill. You measure the
apparent brightness of the streetlamp to be 0.000001 W/m2. How can
you use this information to determine the distance to the hill? (An
approximate answer is OK. In particular, you can approximate 4 Pi = 12.566 by
10.))

The luminosity of the streetlamp is L = 1000 W = 103 W.

The brightness is b = 0.000001 W/m2 = 10-6 =
W/m2.

So the distance is given by d2 = (103 W)/(4 Pi x
10-6 W/m2).

Since 4 Pi is approximately 10, this is d2 = (103
/(10 x 10-6)) m2.

Thus d2 = (103 / 10-5) m2.

So d2 = 108 m2

We now know what d2 is. We want to know what d is. So we take
the square root.

So d = 104 m = 10 km.

Davison E. Soper, Institute of Theoretical Science, University of
Oregon, Eugene OR 97403 USA [email-masked]