LUMINOSITY & RADIANT FLUX
Defining Luminosity
- Luminosity $L$ is defined as:
The total power output of radiation emitted by a star - It is measured in units of Watts (W)
- Radiant flux intensity $F$ is defined as:
The observed amount of intensity, or the radiant power transmitted normally though a surface per unit of area, of radiation measured on Earth - The best way to picture this is:
- The luminosity is the total radiation that leaves the star
- The radiant flux intensity is the amount of radiation measured on Earth
- By the time the radiation reaches the Earth, it will have spread out a great deal, therefore, it will only be a fraction of the value of the luminosity
Inverse Square Law of Flux
- Light sources which are further away appear fainter because the light it emits is spread out over a greater area
- The moment the light leaves the surface of the star, it begins to spread out uniformly through a spherical shell
- The surface area of a sphere is equal to $4 \pi r^2$
- The radius $r$ of this sphere is equal to the distance $d$ between the star and the Earth
- By the time the radiation reaches the Earth, it has been spread over an area of $4 \pi d^2$
- The inverse square law of flux can therefore be calculated using:
$
F=\frac{L}{4 \pi d^2}
$
- Where:
- $\mathrm{F}=$ radiant flux intensity, or observed intensity on Earth $\left(\mathrm{W} \mathrm{m}^{-2}\right)$
- $\mathrm{L}$ = luminosity of the source (W)
- $\mathrm{d}=$ distance between the star and the Earth $(\mathrm{m})$
- This equation assumes:
- The power from the star radiates uniformly through space
- No radiation is absorbed between the star and the Earth
- This equation tells us:
- For a given star, the luminosity is constant
- The radiant flux follows an inverse square law
- The greater the radiant flux (larger F) measured, the closer the star is to the Earth (smaller d)
Worked example: Inverse square law of flux
A star has a luminosity that is known to be $4.8 \times 10^{29} \mathrm{~W}$. A scientist observing this star finds that the radiant flux intensity of light received on Earth from the star is $2.6 \mathrm{nW} \mathrm{m}^{-2}$. Determine the distance of the star from Earth.
Answer/Explanation
Step 1:
Write down the known quantities
Luminosity, $L=4.8 \times 10^{29} \mathrm{~W}$
Radiant flux intensity, $F=2.6 \mathrm{nW} \mathrm{m}^{-2}=2.6 \times 10^{-9} \mathrm{~W} \mathrm{~m}^{-2}$
Step 2:
Write down the inverse square law of flux
$
\mathrm{F}=\frac{L}{4 \pi d^2}
$
Step 3:
Rearrange for distance $d$, and calculate
$
d=\sqrt{\frac{L}{4 \pi F}}=\sqrt{\frac{4.8 \times 10^{29}}{4 \pi \times\left(2.6 \times 10^{-9}\right)}}=3.8 \times 10^{18} \mathrm{~m}
$
STANDARD CANDLES $\&$ STELLAR DISTANCES
Standard Candles
- A standard candle is defined as:
An astronomical object which has a known luminosity due to a characteristic quality possessed by that class of object - Examples of standard candles are:
- Cepheid variable stars
- A type of pulsating star which increases and decreases in brightness over a set time period
- This variation has a well defined relationship to the luminosity
- Type la supernovae
- A supernova explosion involving a white dwarf
- The luminosity at the time of the explosion is always the same
Using Standard Candles as a Distance Indicator
- Measuring astronomical distances accurately is an extremely difficult task
- A direct distance measurement is only possible if the object is close enough to the Earth
- For more distant objects, indirect methods must be used – this is where standard candles come in useful
- If the luminosity of a source is known, then the distance can be estimated based on how bright it appears from Earth
- Astronomers measure the radiant flux intensity, of the electromagnetic radiation arriving at the Earth
- Since the luminosity is known (as the object is a standard candle), the distance can be calculated using the inverse square law of flux
- Each standard candle method can measure distances within a certain range
- Collating the data and measurements from each method allows astronomers to build up a larger picture of the scale of the universe
- This is known as the cosmic distance ladder
WIEN’S DISPLACEMENT LAW
- Wien’s displacement law relates the observed wavelength of light from a star to its surface temperature, it states:
The black body radiation curve for different temperatures peaks at a wavelength which is inversely proportional to the temperature
- This relation can be written as:
$
\lambda_{\max } \propto \frac{1}{T}
$
- $\lambda_{\max }$ is the maximum wavelength emitted by the star at the peak intensity
- A black-body is an object which:
- Absorbs all the radiation that falls on it, and is also a good emitter
- Does not reflect or transmit any radiation
- A black-body is a theoretical object, however, stars are the best approximation there is
- The radiation emitted from a black-body has a characteristic spectrum that is determined by the temperature alone
- The full equation for Wien’s Law is given by
$
\lambda_{\max } \mathrm{T}=2.9 \times 10^{-3} \mathrm{~m} \mathrm{~K}
$
- Where:
- $\lambda_{\max }=$ peak wavelength of the star $(m)$
- $T=$ thermodynamic temperature at the surface of the star $(K)$
- This equation tells us the higher the temperature of a body:
- The shorter the wavelength at the peak intensity, so hotter stars tend to be white or blue and cooler stars tend to be red or yellow
- The greater the intensity of the radiation at each wavelength
Worked example: Wien’s displacement law
The spectrum of the star Rigel in the constellation of Orion peaks at a wavelength of $263 \mathrm{~nm}$, while the spectrum of the star Betelgeuse peaks at a wavelength of $828 \mathrm{~nm}$. Which of these two stars is cooler, Betelgeuse or Rigel?
Answer/Explanation
Step 1:
Write down Wien’s displacement law
$
\lambda_{\max } \mathrm{T}=2.9 \times 10^{-3} \mathrm{~m} \mathrm{~K}
$
Step 2:
Rearrange for temperature $\mathrm{T}$
$
T=\frac{2.9 \times 10^{-3}}{\lambda_{\max }}
$
Step 3: $\quad$ Calculate the surface temperature of each star
Rigel: $T=\frac{2.9 \times 10^{-3}}{\lambda_{\max }}=\frac{2.9 \times 10^{-3}}{263 \times 10^{-9}}=11026=11000 \mathrm{~K}$
Betelguese: $\quad \mathrm{T}=\frac{2.9 \times 10^{-3}}{\lambda_{\max }}=\frac{2.9 \times 10^{-3}}{828 \times 10^{-9}}=3502=3500 \mathbf{K}$
Step 4: Write a concluding sentence
Betelguese has a surface temperature of $3500 \mathrm{~K}$, therefore, it is much cooler than Rigel
Exam Tip
Note that the temperature used in Wien’s Law is in Kelvin (K). Remember to convert from ${ }^{\circ} \mathrm{C}$ if the temperature is given in degrees in the question before using the Wien’s Law equation.