Stellar distances

Stellar Parallax

Parallax angle (Ɵ) 1 a.u.

Parallax Method January very distant reference star Sun ‘near’ star   Sun   ‘near’ star   very distant reference star June  

Very small angles

Parallax and Distance The distance to a star is inversely related to its parallax, i.e. nearer stars have larger parallaxes. The relation between parallax  and distance d is: d = 1/ Where, if  is measured in arcseconds, d is then determined in parsecs. Thus, a star whose parallax is 1.0 arcseconds has a distance of 1.0 parsecs; one with  = 0.01 seconds of arc has a distance of 1/0.01 = 100 parsecs, etc. One parsec equals 3.26 light years (the distance light travels in one year) or about 2x1013 miles! That is more than 200,000 AU.

Parsec If the parallax angle is one arcsecond (1/3600 degree in radians) the distance to the star is called a parsec (pc)

Problems 1) The star Proxima Centauri has a parallax of 0.762 arcsecs (0.762”). Calculate the distance to the star in i) parsecs (pc), ii) light years (ly) 2) Friedrich Bessel is credited as the first person to measure the distance to a star using trigonometric parallax. He measured the value of p=8.06 x 10-5 degrees for the star 61 Cygnus. Calculate the distance to 61 Cygnus in i) parsecs (pc), ii) light years (ly) 1 parsec = 3.26 ly

Some of the Brightest Stars The Five Nearest Stars Star Parallax Distance arcsec pc a Centauri 0.76 1.31 Barnard’s Star 0.54 1.83 Wolf 359 0.43 2.35 Lalande 21185 0.40 2.49 Sirius 0.37 2.67 Some of the Brightest Stars Star Parallax Distance arcsec pc Sirius 0.37 2.67 Canopus 0.03 30 a Centauri 0.76 1.31 Arcturus 0.09 11 Vega 0.13 8 Capella 0.07 14 Betelgeuse 0.006 150 Deneb 0.002 430 Note that the brightest stars are not necessarily the nearest. What does that mean?

Apparent and absolute magnitudes

Hipparchus Greek astronomer Lived 2000 years ago

Hipparchus compared the relative brightness of stars (as seen from earth) Brightest star – magnitude 1 Faintest star – magnitude 6

Apparent magnitude and brightness Magnitude 1 star is 100 times brighter than a magnitude 6 star

The difference between a magnitude 1 star and a magnitude 6 star is ‘5 steps’ on the magnitude scale and the scale is logarithmic. This means that each ‘step’ equated to a brightness decrease of 2.512 since (2.512)5=100 Magnitude 1 Magnitude 2 Magnitude 3 Magnitude 4 r5 = 100 Magnitude 5 r = 2.512 Magnitude 6

* Under what conditions? Clear sky When viewed from earth As visible to the naked eye

Can a star have a magnitude greater than 6?

Can a star have a magnitude greater than 6? Yes, but these stars are only visible through a telescope

Negative apparent magnitude? They are very bright!!

Guess the apparent magnitude of Sun It is -26.7

This is Pogson’s Equation Apparent magnitude The apparent magnitude m, of a star of apparent brightness I is defined by m = -2.5log10I+K where K is set by assigning a magnitude to a specific reference star. This is Pogson’s Equation

Differences in intensity One type of question often encountered is to compare the brightness of two stars, given their apparent magnitudes

Example Apparent magnitude of Sun is -26.7 and that of Betelgeuse is 0.5. How much brighter is Sun than Betelgeuse?

Difference in magnitudes is 0.5 - -26.7 = 27.2 Apparent magnitude of Sun is -26.7 and that of Betelgeuse is 0.5. How much brighter is Sun than Betelgeuse? Difference in magnitudes is 0.5 - -26.7 = 27.2 Each difference in magnitude is a difference of 2.512 in brightness Therefore the difference in brightness = 2.51227.2 = 7.6 x 1010

Sun is 76 billion times brighter than Betelgeuse

Question Apparent magnitudes of Andromeda galaxy and Crab nebula are 4.8 and 8.4 respectively. Which of these is brightest? By what factor?

Apparent Magnitude Apparent magnitude is how bright an object appears to us This will be affected by 1) how bright the objects actually is (absolute magnitude) and 2) how far away the object is. E.g. a dim object close to us could have the same apparent magnitude as a bright object far away.

Absolute magnitude is the apparent magnitude of a star when viewed from a distance of 10 parsecs.

Absolute magnitude

Absolute magnitude M and apparent magnitude m m – M = 5 log (d/10) d is in parsecs!

Question Calculate the absolute magnitude of Sun. Apparent magnitude = -26.7 Distance from earth = 4.9 x 10-6 pc

m – M = 5 log(d/10) -26.7- M = 5 log (4.9 x 10-6/10) M =-26.7 – 5log(4.9 x 10-7) M = 4.85

M = 4.85 This means at a standard distance of 10 parsecs the sun would appear to be a dim star.

Can absolute magnitude be Positive ? Negative ? Any value?

Let’s try some questions! ‘Apparent and absolute magnitude questions’

Spectroscopic parallax

Spectroscopic parallax This refers to the method of finding the distance to a star given the star’s luminosity (total power output of star in W) and apparent brightness (intensity as seen at the Earth in Wm-2). It doesn’t use parallax! Limited to distances less than 10 Mpc We know that b = L/(4πd2) so d = (L/(4 πb))½

Spectroscopic parallax - Example A main sequence star emits most of its energy at λ = 2.4 x 10-7 m. Its apparent brightness is measured to be 4.3 x 10-9 W.m-2. How far away is the star? λ 0T = 2.9 x 10-3 Km T = 2.9 x 10-3 / 4.3 x 10-9 = 12000K

T = 12000K. From an HR diagram we can see this corresponds to a brightness of about 100x that of the sun (= 100 x 3.9 x 1026 = 3.9 x 1028 W)

Spectroscopic parallax - Example Thus d = (L/(4 πb))½ d = (3.9 x 1028/(4 x π x 4.3 x 10-9))½ d = 8.5 x 1017 m = 90 ly = 28 pc

Question The peak wavelength of a main sequence star is 720nm. Its apparent brightness is measured as 2 x 10-9Wm-2. What is: a) its surface temperature? b) its luminosity? c) its distance from the Earth?

The same technique can be used for magnitudes The light from a main sequence star has a peak wavelength of 100nm, and an apparent magnitude of 5 What is: a) its surface temperature? b) its distance from the Earth in parsecs? c) its radius?

Using cepheids to measure distance

Cepheid variables At distances greater than Mpc, neither parallax nor spectroscopic parallax can be relied upon to measure the distance to a star. When we observe another galaxy, all of the stars in that galaxy are approximately the same distance away from the earth. What we really need is a light source of known luminosity in the galaxy. If we had this then we could make comparisons with the other stars and judge their luminosities. In other words we need a ‘standard candle’ –that is a star of known luminosity. The outer layers of Cepheid variable stars undergo periodic expansion and contraction, producing a periodic variation in its luminosity.

Cepheid variable stars are useful to astronomers because of the period of their variation in luminosity turns out to be related to the average absolute magnitude of the Cepheid. Thus the luminosity of the Cepheid can be calculated by observing the variation in brightness.

The process of estimating the distance to a galaxy (in which the individual stars can be imagined) might be as follows: Locate a Cepheid variable in the galaxy Measure the variation in brightness over a given period of time. Use the luminosity-period relationship for Cepheids to estimate the average luminosity. Use the average luminosity, the average brightness and the inverse square law to estimate the distance to the star.

Cepheid calculation - Example

From the left-hand graph we can see that the period of the cepheid is 5.4 days. From the second graph we can see that this corresponds to a luminosity of about 103 suns (3.9 x 1029 W).

From the left hand graph we can see the peak apparent magnitude is 3 From the left hand graph we can see the peak apparent magnitude is 3.6 which means we can find the apparent brightness from b/b0 = 2.512-m b = 2.52 x 10-8 x 2.512-3.6 = 9.15 x 10-10 W.m-2

Now using the relationship between apparent brightness, luminosity and distance d = (L/(4πb))½ d = (3.9 x 1029/(4 x π x 9.15 x 10-10))½ d = 5.8 x 1018 m = 615 ly = 189 pc