Cepheids and Supernovae as Indicators of Distance
Because their periods are directly linked to their luminosities, Cepheid variables are one of the most reliable tools astronomers have for determining the distances to galaxies. To this day, astronomers use this link—much as Hubble did back in the 1920s—to measure intergalactic distances. More recently, they have begun to use Type Ia supernovae, which are far more luminous and thus can be seen much farther away, to determine the distances to very remote galaxies.
EXAMPLE: In 1992 a team of astronomers observed Cepheid variables in a galaxy called IC 4182 to deduce this galaxy’s distance from Earth. One such Cepheid has a period of 42.0 days and an average apparent magnitude (m) of +22.0. (See Box 17-3 for an explanation of the apparent magnitude scale.) By comparison, the dimmest star you can see with the naked eye has m = +6; this Cepheid in IC 4182 appears less than one one-millionth as bright.
According to the period-luminosity relation shown in Figure 19-20, such a Cepheid with a period of 42.0 days has an average luminosity of 33,000 L⊙. An equivalent statement is that this Cepheid has an average absolute magnitude (M) of −6.5. (This compares to M = +4.8 for the Sun.) Use this information to determine the distance to IC 4182.
Situation: We are given the apparent magnitude m = +22.0 and the absolute magnitude M = −6.5 of the Cepheid variable star in IC 4182. Our goal is to calculate the distance to this star, and hence the distance to the galaxy of which it is part.
Tools: We use the relationship between apparent magnitude, absolute magnitude, and distance given in Box 17-3.
Answer: In Box 17-3, we saw that the apparent magnitude of a star is related to its absolute magnitude and distance in parsecs (d) by
m − M = 5 log d − 5
This equation can be rewritten as
We have m − M = (+22.0) − (−6.5) = 22.0 + 6.5 = 28.5. (Recall from Box 17-3 that m − M is called the distance modulus.) Hence, our equation becomes
d = 10(28.5 + 5)/5 parsecs = 106.7 parsecs = 5 × 106 parsecs
(A calculator is needed to calculate the quantity 106.7.)
Review: Our result tells us that the galaxy is 5 million parcsecs, or 5 Mpc, from Earth (1 Mpc = 106 pc). This distance can also be expressed as 16 million light-years.
EXAMPLE: Astronomers are interested in IC 4182 because a Type Ia supernova was observed there in 1937. All Type Ia supernovae are exploding white dwarfs that reach nearly the same maximum brightness at the peak of their outburst (see Section 20-9). Once astronomers know the peak absolute magnitude of Type Ia supernovae, they can use these supernovae as distance indicators. Because the distance to IC 4182 is known from its Cepheids, the 1937 observations of the supernova in that galaxy help us calibrate Type Ia supernovae as distance indicators.
At maximum brightness, the 1937 supernova reached an apparent magnitude of m = +8.6. What was its absolute magnitude at maximum brightness?
Situation: We are given the supernova’s apparent magnitude m, and we know its distance from the previous example. Our goal is to calculate its absolute magnitude M.
Tools: We again use the relationship m − M = 5 log d − 5.
Answer: We could plug in the value of d found in the previous example. But it is simpler to note that the distance modulus m − M has the same value no matter whether it refers to a Cepheid, a supernova, or any other object, just so it is at the same distance d. From the Cepheid example we have m − M = 28.5 for IC 4182, so
M = m − (m − M) = 8.6 − (28.5) = −19.9
This absolute magnitude corresponds to a remarkable peak luminosity of 1010 L⊙.
Review: Whenever astronomers find a Type Ia supernova in a remote galaxy, they can combine this absolute magnitude with the observed maximum apparent magnitude to get the galaxy’s distance modulus, from which the galaxy’s distance can be easily calculated (just as we did above for the Cepheids in IC 4182). This technique has been used to determine the distances to galaxies hundreds of millions of parsecs away (see Section 25-4).