Download C:\Documents and Settings\Administrator\Desktop\Lecture 15.wpd

Review Lecture 15
The Central Problem in astronomy is distance. What we see is basically a twodimensional picture of the sky. To interpret many pieces of information available
to the astronomer we need to know how far away a star or galaxy is.
Example: If you look at the sky, Sirius is brighter than Betelgeuse. But Betelgeuse
is actually 300 times more luminous or brighter than Sirius. That is, Betelgeuse
actually emits far more energy than Sirius. The reason is that Betelgeuse is just
much further away from us than Sirius; thus, apparently less bright or dimmer. To
get at these quantities we need three definitions.
Luminosity = L, measured in Watts, is the power output(at all wavelengths) of the
star,
L = σAT4
where σ = 5.67 x10
−8
in Watts
W
2
m K
4
, A is the surface area of the star, and T is the
temperature measured in Kelvin.
This power is spread out into the area of a sphere, where spherical area = 4 πd
according to the Inverse Square Law.
2
Apparent brightness and the Inverse Square Law, measured in Watts/meter2, is
what we appear to see.
luminosity
L
brightness = b =
=
spherical area 4 π d 2
(1)
Thus, we can only find the luminosity of the star if we know the apparent
brightness of the star and the distance to the star, d. The brightness at the surface of
a star is called the flux
lu min osity
L
flux = σT4 =
=
in W/m2.
surface area 4 πd 2
[Note: This looks like the Stefan-Boltzmann Law and is similar. If you will, flux is
the brightness of a star measured at the surface of the star.]
Brightness - unfortunately, Hipparchus (130 B.C.) measured brightness in terms of
apparent magnitude, m, with a scale such that
m = 1 the brightest star in the sky visible to the eye
m = 6 the dimmest “
”
∆ m = 5 represents 100 times brighter by instrument in a logarithmic fashion.
Thus, if x represents the increase in brightness for ∆ m = 1, then
x5 = 100
⇒ x = 2.512 ≅ 2.5
So, going from the brightest to the dimmest
brightness
magnitude
5
(2.5) = 100
m=1
brightest
4
(2.5) = 39.1
m=2
3
(2.5) = 12.17 m = 3
(2.5)2 = 6.25
m=4
2.5
m=5
1
m=6
dimmest
Note: The eye has a different response to light than mechanical devices. Thus:
mV = visual apparent magnitude
mB = bolometric apparent magnitude (magnitude including all wavelengths)
Relationship between apparent magnitude and apparent brightness
What this says is that if I have two stars with magnitudes m1 and m2, where 1 is
brighter than 2, then the ratio of brightnesses is given by
( 2.5) m − m = ( 2.5) ∆m =
2
1
b1
b2
(2)
This can be rewritten to solve for the difference in magnitudes by taking the log of
both sides, i.e.
⎛b ⎞
log10 2.5∆m = ∆ m log10 2.5 = 0.4 ∆ m = log10 ⎜ 1 ⎟ or
⎝ b2 ⎠
(3)
⎛ b1 ⎞
∆ m = 2.5 log10 ⎜ ⎟
⎝ b2 ⎠
As an example, consider α Centauri(m1 = 0) and Bernard’s Star(m2 = 10), then
∆ m = 10 and the ratio of brightnesses is (2.5)10 = (2.5)5(2.5)5 = 1002 = 10,000.
Thus α Centauri has an apparent brightness 10,000 greater than Bernard’s Star.
The inverse square law of equation 1 says that equation 3 can be rewritten as
2
⎛b ⎞
⎛d ⎞
⎛d ⎞
∆ m = 2.5 log10 ⎜ 1 ⎟ = 2.5 log10 ⎜ 2 ⎟ = 5 log10 ⎜ 2 ⎟
⎝ b2 ⎠
⎝ d1 ⎠
⎝ d1 ⎠
(4)
Absolute magnitude measures the apparent brightness of a star at 10 parsecs.
By implication, a star at a distance of 10 parsecs would have the same apparent
magnitude as absolute magnitude. Let the standard distance then be
dstandard = 10 parsecs = 3 x 1017 m
Example: The Sun
mSun = -26.5, the Sun is the brightest star in the sky
dsun = 1 AU = 1.5 x 1011 m
The Inverse Square Law says that brightness ~
⇒
b sun
b s tan dard
1
dis tan ce 2
=
1
d2
2
d 2s tan dard ⎛ 3 x 1017 m ⎞
6 2
⎜
⎟
2
x
10
=
=
=
= 4 x 1012
(
)
2
11
d sun
. x 10 m ⎠
⎝ 15
This says the ratio of brightness is inversely proportional to the squared distances.
Thus, the Sun is 4 x 1012 times brighter where it is, as seen from the Earth, than at
10 parsecs. What is the absolute magnitude of the Sun at 10 parsecs. Note:
increased distance ⇒ decreased intensity ⇒ ∆ m positive. According to equation
∆m
= 4 x101 2
∆ m1
( )
1, we need to determine ∆ m such that ( 2.5)
Two methods:
12
1) Estimate 4 x 10 = 4 x (100)
(2.5)1 = 2.5
(2.5)2 = 6.2
6
= 2.5
x 2.5
5 ∆ m2
= (2.5)
∆ m1 + 5 ∆ m2
4 is somewhere between 2.5 and 6.2. Guess ∆m1 = 1.5
( )
100 = 2.5
6
5 6
= (2.5)
30
⇒ ∆m2 = 30
∆m = ∆m1 + 5 ∆m2 = 15
. + 30 = +315
.
2) Calculate by taking logarithms of both sides
( 2.5) ∆m = 4 x 1012
log(4 x 1012 ) = log(2.5m ) = m log( 2.5)
log(4 x 1012 ) 12.6
∆m =
=
= + 317
. ≅ 315
. estimate
log( 2.5)
.4
⇒ The Absolute Visual Magnitude of the Sun is
MSun = -26.5 + 31.5 = +5
The Sun would be a dim star just visible in Cortland at 10 parsecs.
Distance
Distance can only be determined directly
from a local group of stars near enough to have
their parallax measured.
1
d=
(in par sec s)
p(in arc sec onds)
From Earth the smallest parallax measurable is
about .01 arcseconds corresponding to a distance
of 100 pc = 326 light years. There are only about
2000 stars within this range. Satellite
measurements extend this distance to about 1000
pc and 200,000 stars.
In comparison to the Sun, equation (1) can be
written as
L
LSun
2
⎛ d ⎞ b
=⎜
⎟
⎝ d Sun ⎠ bSun
Absolute Magnitude from distances
Equation 3 can be rewritten as
⎛ D⎞
∆ m = M − m = 5 log⎜ ⎟ = 5 log D − 5 log d
⎝ d⎠
= 5 log 10 − 5 log d = 5 − 5 log d
(5)
This can be rearranged to say that the absolute magnitude can be determined from
the apparent magnitude and the distance from the observer, i.e.
(6)
M = m + 5 − 5 log10 d
This equation can be used to produce the log-log plot shown below.
To use the plot you need two of the three quantities: relative magnitude, distance,
or absolute magnitude. In the figure below, Sirius has a bolometric apparent
magnitude of m = -1.4 (the brightest star in the sky other than the Sun has a visual
magnitude of 1), and from parallax measurements d = 2.7 parsecs away. Finding
this point on the plot and extrapolating over we find that Sirius has an absolute
magnitude of MSirius = +1 and a luminosity relative to the Sun of LSirius = 23Lsun.
Conversely, the Sun has an absolute magnitude of +5; thus, at 10 parsecs the
relative magnitude would be +5.
Stars Within 4 Parsecs
Distance
(pc)
Name
Part
Apparent
Magnitude
Absolute
Magnitude
Spectral
Type
Mass (MSun)
Radius (RSun)
0.0
Sun
Jupiter
A
B
-26.6
-2.6
5
26
G2
1.0
0.001
1.0
0.1
1.3
Alpha Centauri
A
B
C
0
1
11
4
6
15
G2
K0
M5
1.1
0.9
0.1
1.2
0.9
?
1.8
Bernard’s Star
10
13
M5
?
?
2.3
Wolf 359
14
17
M8
?
?
8
10
M2
0.35
?
o
2.5
BD +36 2147
2.7
L726-8
A
B
12
13
15
16
M6
M6
0.11
0.11
?
?
2.9
Sirius
A
B
-2
8
1
11
A1
A5 Wh Df
2.3
1.0
1.8
0.02
2.9
Ross 154
11
13
M5
?
?
3.1
Ross 248
12
15
M6
?
?
3.2
L789-6
12
15
M6
?
?
3.3
ε Eridani
4
6
K2
0.9
?
3.3
Ross 128
11
14
M5
?
?
3.3
61 Cygni
5
6
8
8
K5
K7
0.63
0.6
?
?
3.4
ε Indi
5
7
K5
?
?
3.5
Procyon
A
B
0
11
3
13
F5
Wh Dwarf
1.8
0.6
1.7
0.01
3.5
Σ
2398
A
B
9
10
11
12
M4
M5
0.4
0.4
?
?
3.5
BD +43o44
A
B
C
8
11
?
10
13
?
M1
M6
K?
?
?
?
?
?
?
3.6
τ Ceti
4
6
G8
?
1.0
7
10
M2
?
?
10
?
12
?
M4
?
?
?
?
?
A
B
o
3.6
CD -36 15693
3.7
BD +5o1668
3.7
G51-15
15
17
?
?
?
3.8
L725-32
12
14
M5
?
?
3.8
o
CD -39 14192
7
9
M0
?
?
3.9
Kapteyn’s Star
9
11
M0
?
?
4.0
Kruger 60
A
B
C
10
11
?
12
13
?
M4
M6
?
0.27
0.16
0.01
0.51
?
?
4.0
Ross 614
A
B
11
15
13
17
M5
?
0.14
0.08
?
?
4.0
BD -12o4523
10
12
M5
?
?
A
B