Photometry
Atmosphere & Standardization
ASTR 3010
Lecture 13
Textbook 10.6 & 10.7
Extinction by Atmosphere
• Observing the incoming radiation at depth H in
the atmosphere.
Measured spectrum φA(λ)
mPA = mOP + Aatm + Aism + Aexg + CPZ
= mP + Aatm
minside = moutside
jA (l) = j(l) · e -t( l ,H ) X
where optical depth τ
t (l,H) =
ò
and X is air mass.
H
0
a (h)dh, a is absorption coeff
jA ( l )
j(l)
ìï ò T f S dl üï
P l atm
ý
- 2.5logí
ïî ò TP f l dl ïþ
Different notations
mlA = ml + k( l) X Þ ml,measured = ml ,corrected + k( l) X
mlSTD
= ml1 + a12 (color index)12 + a1
S1
B = b + a b (B -V) + zb
Bouguer’s Law
jA (l) = j(l) · e
-t ( l ,H ) X
m = -2.5log(F)
mlA = -2.5log(jA ( l)) = -2.5log(j( l)) + 2.5 t ( l) X log(e)
mlA = ml +1.086 t ( l) X
mlA = ml + k ( l) X where k(l ) º 1.086 t ( l) is monochromatic extinction coeff
Take multiple measurements of
non-varying object at several
different airmasses!
one can get a mean extinction
coeff from the slope
with known airmass, one can
recover mλ for any other stars!
mlA
slope = k
ml
X
0
1
2
3
Sources of extinction
1.
2.
Rayleigh scattering
Absorption by Ozone
stable over long time
3.
4.
Scattering by Aerosols
Molecular-band absorption
variable due to a weather system
Photometric Condition
To be able to use Bouguer’s Law, we need two conditions
1.
2.
k is stationary
k is isotropic
when these two conditions
are met, the night is called
“photometric”
A
ml
slope = k
ml
0
1
X
2
3
Example of non-stationary extinction during the obs.
Measuring monochromatic extinction
Assume  use observatory’s value
1.
2.
Use a reference  observe a star with known mλ
3.
From the Bouguer line of your measurements
4.
Variable extinction / multi-night data
5.
mlA
ml
0
o
measure two standard stars at a given time at different airmass
o
repeat the pair observation several times per night
Use all data
1 X 2
3
Heterochromatic extinction
• Apparent magnitudes versus airmass different slopes for different colors
ìï ò T f S dl üï
P l atm
ý
m - mP º kP X = -2.5logí
ïî ò TP f l dl ïþ
A
P
Forbes Effect
= spectrum of a star changes with airmass
2nd order extinction coefficients
• Taylor Expand kP (or parameterize kP)
ìï ò T f S dl üï
P l atm
ý
m - mP º kP X = -2.5logí
ïî ò TP f l dl ïþ
A
P
kP » kP0 + k1P × (spectral shape) + k1P × (something else)
• For example, (B-V) color can be used to indicate the spectral shape.
kP = kP0 + k1P × ( B -V )
• This color-dependent term is not changing rapidly and takes many data to
measure one can use observatory’s value
Transformation to a standard system
• instrumental (outside the atmosphere) magnitudes measured with two
filters at λ1 and λ2 where standard wavelengths are λS1 and λS2.
From m = -2.5log( F ) + C
we get
ml1
Then,
( )
= -2.5log( F ) + C
STD
mlSTD
=
-2.5log
F
+
C
l
1
S1
S1
l1
1
[ ( )
( )]
STD
mlSTD
m
=
-2.5
log
F
log
F
+
C
- C1
l1
l S1
l1
1
S1
color term
color coefficient
efficiency term
zero-point constant
mlSTD
= ml1 + a12 (color index)12 + a1
S1
Transformation to a standard system
• In practice, you measure mλ1 and (color index)12 or mλ1 and mλ2
then plot mlSTD
- ml1 versus (color index)12
S1
y = m STD - m
-1
y = a0 + a1x + a2 x
0
+1
X = Color Index
2
+2
Example (Homework)
An observer used B and V filters to obtain four exposures of the same field
at different air masses: two B exposures at air masses 1.05 and 2.13, and
two V exposures at airmasses 1.10 and 2.48. Four stars in this field are
photometric standards. Their measured magnitudes are given below.
(B-V)
V
Airmass
b(1)
b(2)
v(1)
v(2)
1.05
2.13
1.10
2.48
Star A
-0.07
12.01
9.853
10.687
8.778
9.427
Star B
0.36
12.44
10.693
11.479
9.160
9.739
Star C
0.69
12.19
10.759
11.462
8.873
9.425
Star D
1.15
12.89
11.898
12.547
9.522
10.001
Example (Homework)
(B-V)
V
Airmass
b(1)
b(2)
v(1)
v(2)
1.05
2.13
1.10
2.48
Star A
-0.07
12.01
9.853
10.687
8.778
9.427
Star B
0.36
12.44
10.693
11.479
9.160
9.739
Star C
0.69
12.19
10.759
11.462
8.873
9.425
Star D
1.15
12.89
11.898
12.547
9.522
10.001
1. Calculate extinction coefficients for the instrumental system for B and
V bands.
2. Compute the standard transformation coefficients αV and αB-V (or αB)
3. Calculate standard magnitudes of Obj1 (i.e., V and B-V) whose
instrumental magnitudes are v=9.850 and b=10.899 taken at airmass=1.50
mlA = ml + k( l) X
mlSTD
= ml1 + a12 (color index)12 + a1
S1
Example (Homework)
(B-V)
V
Airmass
b(1)
b(2)
v(1)
v(2)
1.05
2.13
1.10
2.48
Star A
-0.07
12.01
9.853
10.687
8.778
9.427
Star B
0.36
12.44
10.693
11.479
9.160
9.739
Star C
0.69
12.19
10.759
11.462
8.873
9.425
Star D
1.15
12.89
11.898
12.547
9.522
10.001
1. Calculate extinction coefficients for the instrumental system for B and
V bands.
mlA = ml + k ( l ) X @ ml + ( k0 + k1 (B -V )) X
unknown
unknown
Plot b(2)-b(1)/(X2-X1) and measure the slope for k1(B-V)
Example (Homework)
(B-V)
V
Airmass
b(1)
b(2)
v(1)
v(2)
1.05
2.13
1.10
2.48
Star A
-0.07
12.01
9.853
10.687
8.778
9.427
Star B
0.36
12.44
10.693
11.479
9.160
9.739
Star C
0.69
12.19
10.759
11.462
8.873
9.425
Star D
1.15
12.89
11.898
12.547
9.522
10.001
2. Compute the standard transformation coefficients αV and αB-V (or αB)
mlSTD
= ml1 + a12 (color index)12 + a1
S1
Plot m STD - m as a function
lS1
l1
of color
index
(e.g., B-V)
 Slope
= α12
 y-intercept= α1
Example (Homework)
3. Calculate standard magnitudes of Obj1 (i.e., V and B-V) whose
instrumental magnitudes are v=9.850 and b=10.899 taken at airmass=1.50
mb = mbA - éëkb0 + kb1 (B -V )ùû X
mv = mvA - éëkv0 + kv1 (B -V )ùû X
B = mbSTD = mb + a bB-V (B -V ) + a b0
V = mvSTD = mv + a vB-V (B -V ) + a v0
B -V =
mb - mv + (a b0 - a v0 )
1- (a bB-V - a vB-V )
In summary…
Important Concepts
Important Terms
• Bouguer’s Law
• Photometric condition
• Standard Transformation
• Extinction coefficient
• Forbes effect
Chapter/sections covered in this lecture : 10.6 & 10.7