Stefan-Boltzmann Luminosity Law

\[L = 4\pi R^2 \sigma T_\text{surface}^4\quad\text{[Eq. (6.16)]}\]

Note: If you feel uncomfortable with working with exponents, variables, unit conversions, or percentages you should review the following tutorials before working through this tutorial:

This tutorial looks a bit more complex and perhaps a little scary for some students. Don’t let it intimidate you. The Stefan-Boltzmann luminosity law is the most challenging equation in the textbook, but it really only comes down to multiplication, division (in other forms), and simple exponents (and roots, again in other forms). If you can do \(2^2 = 2\times2=4\), or \(3^4 = 3\times3\times3\times3 = 81\), you’re in business. If you follow along carefully, the tutorial should help build your confidence. If you need extra help, take a look at the tutorials listed above, especially the exponents tutorial. Getting good at something requires persistence and practice; mathematics is no different.

It is important to remember that variables, like \(x\), \(y\), \(L\), \(R\), and \(T_\text{surface}\) are nothing more than symbolic placeholders for actual numbers. For example, \(L\) is waiting to be replaced by the value for the luminosity of a star, either by calculating it using the Stefan-Boltzmann luminosity law, or by measuring it. All algebra done with variables can also be done through arithmetic with numbers. Constants like \(4\), \(\pi\), and \(\sigma\) (sigma) have permanently fixed values and cannot be changed. These particular constants are universal in time and space; \(4\) will always be \(4\), \(\pi\) will always be \(\pi\), and \(\sigma\) will always be \(\sigma\).

Sections

Introduction

As the name of the Stefan-Boltzmann luminosity law (for a spherical blackbody) indicates, it describes the luminosity produced (the energy radiated every second) by a blackbody radiator, such as a star or a planet. The name of the law is often (usually) more simply the Stefan-Boltzmann law, but it is still critically important to remember that it is used to determine the luminosity of a perfect blackbody. The law also has another restriction; it applies only to spherical objects. You may recognize the restriction by the first part of the equation, \(4\pi R^2\), which is the surface area of a sphere of radius \(R\). The second half of the equation, \(\sigma T_\text{surface}^4\), is the flux [Stefan’s law, Equation (6.15)] from an area of the object’s surface that has a temperature, \(T_\text{surface}\). (Flux is the amount of energy every second that is radiated from a specific amount of area of the blackbody. )

The first half of the Stefan-Boltzmann law has units of meters squared (\(\text{m}^2\)) and the second half of the equation has units of joules (energy) per second per meter squared (\(\text{J}/\text{s}/\text{m}^2\)). Since a joule per second is defined as a unit of power known as a watt (W), the second half of the equation has units of watts per meter squared (\(\text{W}/\text{m}^2\)). Multiplying the units of the first half of the Stefan-Boltzmann law by the units of the second half gives \[\text{m}^2 \times \frac{\text{W}}{\text{m}^2} = \text{W}.\]

Canceling Constants

Many times throughout Astronomy: The Human Quest for Understanding we use a “trick” to eliminate the need for constants in equations [think of Kepler’s third law, Equation (5.10)]. We can do the same thing with the Stefan-Boltzmann equation by comparing one object with another, such as a star with the Sun, or a planet with Earth. For a star assume that its luminosity is \(L_\text{star}\), its radius is \(R_\text{star}\), and its surface temperature is \(T_\text{star, surface}\). Next, divide the star’s version of the Stefan-Boltzmann equation by the Sun’s values:

\[\frac{L_\text{star}}{L_\text{Sun’s value}} = \frac{4\pi R_\text{star}^2\,\sigma T_\text{star, surface}^4}{4\pi R_\text{Sun’s value}^2\,\sigma T_\text{Sun’s value}^4}\].
Since the constants \(4\), \(\pi\), and \(\sigma\) are in the numerator and the denominator, they all cancel, which leaves
\[\frac{L_\text{star}}{L_\text{Sun’s value}} = \left(\frac{R_\text{star}}{R_\text{Sun’s value}}\right)^2\,\left(\frac{T_\text{star, surface}}{T_\text{Sun’s value}}\right)^4\]. To make the equation look less complicated, let \(\frac{L_\text{star}}{L_\text{Sun’s value}}\) be replaced by \(L_\text{Sun}\) and similarly for the other fractions. Finally, we arrive at \[L_\text{Sun} = R_\text{Sun}^2 T_\text{surface, Sun}^4\qquad\qquad\text{Equation (6.17)}\].

Important note: Don’t forget that the subscript “Sun” is only there to remind you that the luminosity, \(L_\text{Sun}\), radius, \(R_\text{Sun}\), and surface temperature, \(T_\text{surface, Sun}\), are to be written only in terms of the Sun’s values (they are ratios with the Sun’s values and so they don’t actually have units; they are pure numbers only).

Stars of the Same Temperature But Different Radii

Suppose that Star A has the same surface temperature as the Sun, but it has three times the radius. How would its luminosity compare to the luminosity of the Sun?

First of all, if \(T_\text{star, surface} = T_\text{Sun}\) then \(T_\text{surface, Sun} = 1\), which means that the luminosity equation above becomes simpler:

\[L_\text{Sun} = R_\text{Sun}^2 T_\text{surface, Sun}^4 = R_\text{Sun}^2 (1)^4\qquad\qquad\text{or}\qquad\qquad L_\text{Sun} = R_\text{Sun}^2..\]

What this equation tells us if that a star has three times the radius of the Sun but the same surface temperature, its luminosity is \(L_\text{Sun} = 3^2 = 3\times 3 = 9\) times the Sun’s luminosity. If the star has five times the radius of the Sun, its luminosity is \(L_\text{Sun} = 5^2 = 5\times 5=25\) times the luminosity of the Sun. And if its radius is 100 times the Sun’s radius, its luminosity is \(L_\text{Sun} = 100^2 = 10,000\) times the luminosity of the Sun.

The equation also tells us what happens when a star is smaller than the Sun but has the same surface temperature as the Sun. If its radius is one-half of the Sun’s radius, its luminosity is \(L_\text{Sun} = \left(\frac{1}{2}\right)^2 = \frac{1}{2^2} = \frac{1}{4}\) of the Sun’s luminosity (or 0.25 times the Sun’s luminosity; \((25\%)\). If it is one-tenth of the size of the Sun, then its luminosity is just \(L_\text{Sun} = \left(\frac{1}{10}\right)^2 = \frac{1}{10^2} = \frac{1}{100} = 0.01 = 1\%\) and so on.

Stars of the Same Radii But Different Temperatures

The same method of calculating a star’s luminosity can be used if its radius is identical to the radius of the Sun, but it has a different surface temperature.

In this case \(R_\text{Sun}^2 = 1\), and so the equation for the star’s luminosity becomes \(L_\text{Sun} = T_\text{surface}^4\).

Suppose that a star has the same radius as the Sun, but is three times as hot at its surface. This means that its luminosity relative to the Sun’s luminosity is \(L_\text{Sun} = 3^4 = 3 \times 3 \times 3 \times 3 \times 3 = 81\) times greater. If the star’s surface temperature is five times higher, then its luminosity is \(5^4 = 625\) times greater, and if it is ten times hotter, its luminosity is \(10^4 = \text{10,000}\) times greater.

If the star’s surface temperature is one-half the Sun’s surface temperature, but the star has the same radius as the Sun, then its luminosity is \(L_\text{Sun} = \left(\frac{1}{2}\right)^4 = \frac{1}{2^4} = \frac{1}{16} = 0.0625 = 6.25\%\) of the Sun’s luminosity. This calculation could also have been done by replacing \(1/2\) by \(0.5\), in which case \(L_\text{Sun} = 0.5^4 = 0.5 \times 0.5 \times 0.5 \times 0.5 = 0.0625\); we arrive at the same answer (thank goodness!).

Luminosity changes rapidly with surface temperature.

The Sun’s surface temperature is \(T_\text{Sun, surface} = 5772~\text{K}\), so another way to write the equation would be \(L_\text{Sun} = \left(\frac{T_\text{surface}}{5772~\text{K}}\right)^4\), assuming that the star’s radius is identical to the Sun’s radius. If a star’s surface temperature is 7000 K and it has the Sun’s radius, then its luminosity would be \(\left(\frac{7000~\text{K}}{5772~\text{K}}\right)^4 = 1.213^4 = 2.16\) greater than the Sun’s luminosity. If its temperature is 3000 K, then its luminosity would be \(\left(\frac{3000~\text{K}}{5772~\text{K}}\right)^4 = 0.5198^4 = 0.073 = 7.3\%\) of the Sun’s luminosity.

Putting the Equation Together

Now that you know how to work with either half of the right-hand side of Equation (6.17), it is straightforward to use both halves together; the complete Stefan-Boltzmann equation: \[L_\text{Sun} = R_\text{Sun}^2 T_\text{surface, Sun}^4\].

Imagine you discover a star that is 120 times the Sun’s radius and it has a surface temperature of 3500 K. What would the star’s luminosity be?

You already know that \(R_\text{Sun} = 120\), but you still need to calculate \(T_\text{surface, Sun}\). Recalling that \(T_\text{surface, Sun}\) is just the ratio of the star’s surface temperature to the Sun’s surface temperature, we find that \(T_\text{surface, Sun} = 3500~\text{K}/5772~\text{K} = 0.6064\). Putting the numbers into Equation (6.17) gives \[L_\text{Sun} = 120^2\times 0.6064^4 = \text{14,400}\times 0.1352 = 1950\]. Your star is 1950 times more luminous than our Sun.

Solving For Either \(R_\text{Sun}\) or \(T_\text{surface, Sun}\)

The Stefan-Boltzmann law is more flexible that just calculating a star’s luminosity. If fact, if you know any two of \(L_\text{Sun}\), \(R_\text{Sun}\), or \(T_\text{surface, Sun}\), you can calculate the third value. A little bit of algebra leads to \[R_\text{Sun} = \frac{\sqrt{L_\text{Sun}}}{\sqrt{T_\text{surface, Sun}^4}} = \frac{\sqrt{L_\text{Sun}}}{T_\text{surface, Sun}^2} = \frac{L_\text{Sun}^{1/2}}{T_\text{surface, Sun}^2}\] and \[T_\text{surface, Sun} = \frac{\sqrt[4]{L_\text{Sun}}}{\sqrt[4]{R_\text{Sun}^2}} = \frac{\sqrt[4]{L_\text{Sun}}}{\sqrt{R_\text{Sun}}} = \frac{L_\text{Sun}^{1/4}}{R_\text{Sun}^{1/2}} .\]

The “square root” of a number is the flip side of squaring a number. In mathematical terms, the “square root” is the inverse function of the square. What that simply means is that if you take the square root of a number, say \(\sqrt{4}\), you get a number that, when multiplied by itself, gets the original number back; in this case \(\sqrt{4} = 2\) and \(2 \times 2 = 2^2 = 4\). As another example, \(\sqrt{16} = 4\) and \(4\times 4 = 4^2 = 16\). Fancy ways of writing this with algebra: \(\sqrt{x} = y\) and \(y \times y = y^2 = x\). It is also true that \(\sqrt{x^2} = x\) and \(\sqrt{x} \times \sqrt{x} = \left(\sqrt{x}\right)^2 = x\). Again, let’s say that \(x=2\), then \(\sqrt{2^2} = \sqrt{4} = 2\), and \(\sqrt{2} \times \sqrt{2} = 1.414213 \times 1.414213 = 1.414213^2 = 2\). (We won’t worry about taking square roots of negative numbers here, although there are very good, but imaginary, reasons in mathematics and physics for doing so!)

Square roots undo squares and squares undo square roots; this is what is meant by being inverse functions of each other. This same behavior works for any power and its inverse; square root is special only because it has a fancy symbol, \(\sqrt{x}\) to express it. It is possible to express cubed roots, fourth roots, and so on by extensions of the \(\sqrt{x}\) symbol, such as \(\sqrt[3]{x}\), but that notation is very clunky and hard to work with in general. It is much easier to use exponents for powers and their roots. As an example, \(\sqrt{2} = 2^{1/2} = 2^{0.5}\). Give it a try on your calculator, you should find that \(\sqrt{2} = 1.414213\), \(2^{1/2} = 1.414213\), and \(2^{0.5} = 1.414213\).

Powers also make understanding inverses easy. To show that \(\left(\sqrt{2}\right)^2 = 2\), write the square root as a power: \(\left(2^{1/2}\right)^2 = 2^{1/2} \times 2^{1/2} = 2^{(1/2 + 1/2)} = 2^1 = 2.\) The square of the square root of 2 equals 2. If this is a bit confusing you can always review exponents.

The reason for this little stroll down exponent lane is that the two equations for \(R_\text{Sun}\) and \(T_\text{surface, Sun}\) contain square roots, (\(\sqrt{L_\text{Sun}}\) or \(L_\text{Sun}^{1/2}\); \(\sqrt{R_\text{Sun}}\) or \(R_\text{Sun}^{1/2}\)), and fourth roots, (\(\sqrt[4]{L_\text{Sun}}\) or \(L_\text{Sun}^{1/4}\)).

Let’s try a couple of examples. First, Star A has a luminosity that is 25 times greater than the luminosity of the Sun, and a surface temperature that is twice the surface temperature of the Sun. How does its size compare to the size of the Sun?

From the data presented, \(L_\text{star} = 25\) and \(T_\text{surface, Sun} = 2\). From the equation for radius we can insert these numbers for their placeholders: \[R_\text{Sun} = \frac{\sqrt{L_\text{Sun}}}{T_\text{surface, Sun}^2} = \frac{\sqrt{25}}{2^2} = \frac{5}{4} = 1.2\]; the star’s radius is 1.2 times greater than the radius of the Sun.

As another example, Star B has a radius that is five times the radius of the Sun, and its luminosity is 100 times the Sun’s luminosity. Therefore, \(R_\text{Sun} = 5\) and \(L_\text{Sun} = 100\). From the information given, we can calculate the star’s surface temperature: \[T_\text{surface, Sun} = \frac{L_\text{Sun}^{1/4}}{R_\text{Sun}^{1/2}} = \frac{100^{1/4}}{5^{1/2}} = \frac{3.162}{2.236} = 1.414.\]

Perhaps you noticed in the last example, something that could have simplified your calculations slightly: since the fourth root of a number is the square root of its square root (think about it :), \(\frac{100^{1/4}}{5^{1/2}} = \frac{10^{1/2}}{5^{1/2}} = \left(\frac{10}{5}\right)^{1/2} = 2^{1/2} = \sqrt{2} = 1.414\).

Quiz Yourself

Try the following exercises to make sure that you really do understand the material just presented. By the way, computers and many calculators write scientific notation in “computer-speak,” where \(10\) raised to an integer (either positive or negative) replaces \(\times 10^\text{exponent}\) with \({\rm E(exponent)}\) or \({\rm e(exponent)}\); in other words, \(3.14159 \times 10^9\) is represented by \(3.14159\text{E}9\). If your calculator has the \(\Large{\hat{ }}\) symbol, then you would enter the number as \(3.14159 \times 10{\Large{\hat{ }}}9\). If your calculator has a \(10^x\) button then you would enter the number as \(3.14159\times9\,\,10^x\). If you have having problems with entering numbers in scientific notation, read the manual or help screens, or ask your instructor.

To enter squares and square roots, calculators usually have the keys \(x^2\) and \(\sqrt{x}\), respectively. You enter a number and hit the appropriate key to carry out the operation. For more general exponent calculations, there is usually a key similar to \(x^y\); in this case you first enter the \(x\) number, then hit the exponent function key, followed by the value for \(y\) and then the \(=\) sign to complete the operation.

(Answers are available below.)

Try some of these problems to make sure that you understand the Stefan-Boltzmann luminosity law:

  1. If a star has \(R_\text{Sun} = 25\) but has the same surface temperature as the Sun, what is the star’s luminosity compared to the Sun’s luminosity?
  2. The radius of a star is 500 times the radius of the Sun, but it has the Sun’s surface temperature, what is its luminosity in terms of the Sun’s luminosity?
  3. An unrealistic star’s radius is only \(1/10\) of the Sun’s radius even though they have the same surface temperature. What is the unrealistic star’s luminosity?
  4. If a star has a radius that is \(75\%\) of the Sun’s radius, and it has the Sun’s surface temperature. What is its luminosity compared to the luminosity of the Sun?
  5. A star with the Sun’s radius has a surface temperature of \(\text{12,000 K}\). What is the star’s luminosity?
  6. If a star has a surface temperature of \(\text{2500 K}\) with the Sun’s radius, what is its luminosity in terms of the luminosity of the Sun?
  7. A star is discovered that has a radius that is only \(0.84\%\) of the Sun’s radius and has a temperature of \(\text{25,000 K}\). What is its luminosity? (This tiny, but hot star is Sirius B, a white dwarf star.)
  8. A star has a radius that is \(350\) times the radius of the Sun, with a temperature of \(\text{2900 K}\). Compare its luminosity to that of the Sun.
  9. A star with a luminosity that is \(1\%\) of the Sun’s luminosity and has a radius that is \(50\%\) of the Sun’s radius. What is its surface temperature?
  10. A very massive star is found in a gas and dust cloud that has a luminosity that is 4.7 million times the luminosity of the Sun \(4.7\times 10^6\) and a surface temperature of \(\text{46,000 K}\). What is the star’s radius?

Answers

  1. \(625\) times greater
  2. \(\text{250,000}\) times greater
  3. \(1/100\) of the Sun’s luminosity
  4. \(0.5625\) times the luminosity of the Sun
  5. \(18.68\) times the Sun’s luminosity
  6. \(0.035\) times the Sun’s luminosity
  7. \(0.0248\) times the luminosity of the Sun, or \(2.48\%\)
  8. \(7800\) times the luminosity of the Sun
  9. \(\text{2580 K}\)
  10. \(34\) times the Sun’s radius
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