Color in Astrophysics: Worked Examples and Complete Solutions

(Example)

Color in Astrophysics: Worked Examples and Complete Solutions

This companion develops practical skill with photometric colors, flux ratios, reddening, blackbody color, uncertainty propagation, and unresolved binary systems.

Unless a problem explicitly names a standard photometric system, idealized examples that convert color directly to a flux-density ratio should be interpreted as AB-like comparisons in which the same reference flux density applies to both bands.

Part I: Problems

Problem 1: calculate a color index

A star has

m_B	= 10.43,	(1)
m_V	= 9.78.	(2)

Calculate B − V .

State whether the source is redder or bluer than an object with B − V = 0.20.

Problem 2: color to flux-density ratio

Two idealized AB bands X and Y give

mX − mY = 0.65.

(3)

Find

⟨f-ν⟩X-. ⟨fν⟩Y

(4)

Problem 3: flux-density ratio to color

An idealized AB observation gives

⟨fν⟩X ------= 2.50. ⟨fν⟩Y

(5)

Find m_X − m_Y.

Problem 4: intrinsic color and color excess

A star has

(B − V )_obs	= 0.62,	(6)
(B − V )₀	= 0.30.	(7)

Find E(B − V ).

If an extinction law with

RV = 3.1

(8)

is assumed for this example, estimate A_V and A_B.

Problem 5: deredden a measured color

A source has

g − r	= 1.10,	(9)
E(g − r)	= 0.24.	(10)

Find the intrinsic color (g − r)₀.

Problem 6: idealized blackbody color

Treat two narrow AB-like bands as monochromatic measurements at

λ_X	= 450 nm,	(11)
λ_Y	= 650 nm.	(12)

For a blackbody at

T = 6000 K,

(13)

use the Planck function per unit frequency,

2hν3 1 B ν(T) = --2-----------------, c exp(hν∕kT ) − 1

(14)

to estimate the idealized color

( ) m − m = − 2.5log B-ν,X- . X Y 10 B ν,Y

(15)

Problem 7: compare two blackbody temperatures

Repeat Problem 6 for

T = 10000 K.

(16)

Which blackbody is bluer?

Problem 8: color as a spectral slope

A source has

m − m = 0.40 450 650

(17)

in two idealized AB bands centered at 450 nm and 650 nm.

Assume

α fν ∝ ν .

(18)

Estimate α.

Problem 9: color uncertainty

A source has

m_X	= 15.42 ± 0.02,	(19)
m_Y	= 14.88 ± 0.03.	(20)

Assuming independent errors, calculate the color and its uncertainty.

Problem 10: unresolved binary color

In two idealized bands, star 1 has flux densities

f_X,1	= 1.00,	(21)
f_Y,1	= 1.00,	(22)

while star 2 has

f_X,2	= 0.20,	(23)
f_Y,2	= 0.50.	(24)

All flux densities are in the same arbitrary units.

Calculate:

the color of star 1;
the color of star 2;
the unresolved combined color.

Show that the combined color is not the arithmetic average of the two component colors.

Problem 11: equal-component binary in a color-magnitude diagram

Two identical stars are unresolved.

Each component has color

X − Y = 0.70

(25)

and absolute magnitude

MY = 5.00.

(26)

Find the unresolved system’s:

color;
Y -band absolute magnitude.

Problem 12: color change during eclipse

An unresolved binary has out-of-eclipse fluxes

F_X	= 10.0,	(27)
F_Y	= 12.0.	(28)

During primary eclipse, the hotter component is partially hidden and the fluxes become

F_X	= 6.0,	(29)
F_Y	= 9.0.	(30)

Assume identical AB-like zero points.

Calculate the color before and during eclipse.

Does the system become bluer or redder?

Figure 1. Binary star colors must be calculated after adding the component fluxes in each band.

Part II: Complete Solutions

Solution 1

By definition,

B − V	= m_B − m_V	(31)
	= 10.43 − 9.78	(32)
	= 0.65.	(33)

Compared with

B − V = 0.20,

(34)

the source with B − V = 0.65 has the larger blue-minus-visual color and is therefore redder.

Solution 2

For an idealized AB color,

( ) m − m = − 2.5log fX- . X Y 10 fY

(35)

Therefore

	= 10^−0.4(0.65)	(36)
	≈ 0.550.	(37)

The source has only about 55 percent as much AB flux density in X as in Y .

Figure 2. A positive short-minus-long wavelength AB color corresponds to a smaller flux density in the shorter-wavelength band.

Solution 3

Use

m_X − m_Y	= −2.5 log ₁₀(2.50)	(38)
	≈−0.995.	(39)

The negative color indicates that the source is relatively strong in band X.

Solution 4

The color excess is

E(B − V )	= (B − V )_obs − (B − V )₀	(40)
	= 0.62 − 0.30	(41)
	= 0.32.	(42)

With the assumed illustrative value

RV = 3.1,

(43)

A_V	= R_VE(B − V )	(44)
	= 3.1(0.32)	(45)
	= 0.992 mag.	(46)

Since

E (B − V) = AB − AV ,

(47)

we obtain

A_B	= A_V + E(B − V )	(48)
	= 0.992 + 0.32	(49)
	= 1.312 mag.	(50)

The larger blue-band extinction is what makes the observed color redder.

Solution 5

The intrinsic color is

(g − r)₀	= (g − r) − E(g − r)	(51)
	= 1.10 − 0.24	(52)
	= 0.86.	(53)

Figure 3. Dereddening subtracts the appropriate color excess and moves the observed color back toward the intrinsic spectral energy distribution.

Solution 6

For

λ_X	= 450 nm,	(54)
λ_Y	= 650 nm,	(55)

the frequencies are

ν = c. λ

(56)

Using the Planck function at

T = 6000 K,

(57)

the ratio is approximately

B-ν,X-≈ 0.573. B ν,Y

(58)

Therefore

m_X − m_Y	= −2.5 log ₁₀(0.573)	(59)
	≈ 0.605.	(60)

This is an idealized monochromatic AB-like color, not a calibrated Johnson, SDSS, or Gaia color.

Figure 4. Increasing blackbody temperature shifts the short-to-long wavelength flux ratio and therefore changes the measured color.

Solution 7

T = 10000 K,

(61)

the same two wavelengths give approximately

B-ν,X- B ν,Y ≈ 1.046.

(62)

Thus

m_X − m_Y	= −2.5 log ₁₀(1.046)	(63)
	≈−0.049.	(64)

The 10000 K blackbody has the smaller, more negative color and is therefore bluer than the 6000 K blackbody.

Solution 8

For a power law,

mX − mY α = − ---------------- . 2.5 log10(νX ∕νY)

(65)

Since

νX- = λY-= 650, νY λX 450

(66)

we obtain

α	= −	(67)
	≈−1.00.	(68)

Thus the flux density behaves approximately as

fν ∝ ν− 1.

(69)

Solution 9

The color is

m_X − m_Y	= 15.42 − 14.88	(70)
	= 0.54.	(71)

For independent errors,

σ_C	=	(72)
	≈ 0.036.	(73)

Therefore

|------------------------| |m − m = 0.54 ± 0.04 | ---X-----Y---------------

(74)

to two significant figures in the uncertainty.

Solution 10

For star 1,

C₁	= −2.5 log ₁₀	(75)
	= 0.	(76)

For star 2,

C₂	= −2.5 log ₁₀	(77)
	= −2.5 log ₁₀(0.40)	(78)
	≈ 0.995.	(79)

The unresolved fluxes are

F_X,tot	= 1.20,	(80)
F_Y,tot	= 1.50.	(81)

Therefore

C_tot	= −2.5 log ₁₀	(82)
	= −2.5 log ₁₀(0.80)	(83)
	≈ 0.242.	(84)

The arithmetic average of the component colors would be

0 + 0.995 --------- ≈ 0.498, 2

(85)

which is not the correct combined color.

Fluxes must be added first.

Solution 11

Because the two stars are identical, doubling both band fluxes leaves their ratio unchanged.

Therefore

|------------------| |(X − Y ) = 0.70.| ---------tot---------

(86)

The total flux in band Y doubles, so the magnitude changes by

ΔM_Y	= −2.5 log ₁₀2	(87)
	≈−0.753.	(88)

Thus

M_Y,tot	= 5.00 − 0.753	(89)
	= 4.247.	(90)

The unresolved equal-component binary moves vertically upward in a color-magnitude diagram while retaining the same color.

Solution 12

Out of eclipse,

C_out	= −2.5 log ₁₀	(91)
	≈ 0.198.	(92)

During eclipse,

C_ecl	= −2.5 log ₁₀	(93)
	≈ 0.440.	(94)

The color becomes more positive:

ΔC ≈ 0.242.

(95)

Therefore the system becomes

|-------------------------| redder during the eclipse. | ---------------------------

(96)

This is consistent with preferentially removing Light from a hotter, bluer component.

Part III: Additional conceptual checks

Check 1: does distance change color?

If there is no wavelength-dependent extinction and the same source is simply moved farther away, both band fluxes decrease by the same inverse-square factor.

The flux ratio is unchanged.

Therefore the color is unchanged.

Check 2: can two stars have the same color but different luminosities?

Yes.

Two stars can have similar spectral shapes and therefore similar colors while having different radii and total luminosities.

Color is primarily a spectral-shape measurement, not a total-power measurement.

Check 3: can reddening mimic a cooler star?

Yes.

Dust can make a hot source appear redder.

This creates a temperature-reddening degeneracy that often requires additional colors, spectra, or distance information to break.

Check 4: why use several colors?

One color gives one coarse spectral slope.

Several independent colors sample more of the spectral energy distribution and can help separate temperature, extinction, metallicity, and unusual spectral features.

Summary of useful formulas

|------------------| |CXY = mX − mY | -------------------

(97)

|--------------------(-------)-| | ⟨fν⟩X- | |CXY,AB = − 2.5log10 ⟨f ⟩ | -------------------------ν-Y---

(98)

|----------------------------------------------| -E-(X--−-Y-)-=-(X-−--Y)-−-(X--−-Y-)0-=-AX--−-AY--|

(99)

|----------------------------------| -(X-−--Y)0-=-(X--−-Y-) −-E-(X-−-Y-)|

(100)

|-----∘----------| | 2 2 | -σC-=---σ-X-+-σY--

(101)

for independent magnitude errors, and

|--------∑--------------------∑-------| FX,tot = FX,i, FY,tot = FY,i | ----------i--------------------i-------

(102)

before calculating the color of an unresolved multiple system.

References

[1] B. W. Carroll and D. A. Ostlie, An Introduction to Modern Astrophysics, 2nd ed., Cambridge University Press, 2017.

[2] M. S. Bessell, Standard Photometric Systems, Annual Review of Astronomy and Astrophysics, 43, 293–336, 2005.

[3] M. S. Bessell, F. Castelli, and B. Plez, Model atmospheres broad-band colors, bolometric corrections and temperature calibrations for O–M stars, Astronomy and Astrophysics, 333, 231–250, 1998.

"Color in Astrophysics: Worked Examples and Complete Solutions" is owned by bloftin.
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Color in Astrophysics: Worked Examples and Complete Solutions