Magnitude Systems

What is the absolute magnitude of the Sun? Apparent magnitude of the sun as seen from Earth is $m_\odot = -26.8^m$.

Using Pogson’s equation,

$$M_\odot = m_\odot - 5 \log \frac{1 \, \mathrm{AU}}{10 \, \mathrm{pc}} = \boxed{4.8^m}$$

Albireo is a star in the constellation of Cygnus. It is a double and the system has a total magnitude of $3.0^m$. The fainter companion (Albireo B) has a magnitude of $5.1^m$. Find out the apparent magnitude of the brighter component of the double, Albireo A.

The light which we see coming from a double is the sum of the light from both components. Let $F_0$ be the flux which corresponds to a magnitude of 0. Let $L_A$ and $L_B$ be the luminosities of Albireo A and B, respectively. The total luminosity of the system is thus $L_A + L_B$.

The luminosity of Albireo B can be found using

$$m_B = -2.5 \log \frac{L_B}{F_0} \implies L_B = F_0 \cdot 10^{-0.4 m_B}$$

The total luminosity of the system can be found using

$$m_\text{tot} = -2.5 \log \frac{L_A + L_B}{F_0} \implies L_A + L_B = F_0 \cdot 10^{-0.4 m_\text{tot}}$$

Hence the luminosity of Albireo A is

$$L_A = F_0 \cdot 10^{-0.4 m_\text{tot}} - F_0 \cdot 10^{-0.4 m_B}$$

The apparent magnitude of Albireo A is given by Pogson’s equation as

$$m_A = -2.5 \log \frac{L_A}{F_0} = -2.5 \log \left(10^{-0.4 m_\text{tot}} - 10^{-0.4 m_B} \right)$$

Therefore we get that $\boxed{m_A = 3.2^m}$

Find an expression for the bolometric correction, assuming the visual sensitivity function is given by

$$S_V(\lambda) = \begin{cases} 1 & \text{if } \lambda_1 < \lambda < \lambda_2 \\ 0 & \text{otherwise} \end{cases}$$

Where $\lambda_1 = 551 - 44$ nm and $\lambda_2 = 551 + 44$ nm.

Consider the zero point of both the visual and bolometric magnitudes to be the same, i.e. $C_V = C_\text{bol}$. Assume the emitting body emits like a perfect blackbody of temperature $5800 \, \mathrm{K}$.

The visual magnitude is given by

$$m_V = -2.5 \log \int_0^\infty F_\lambda S_V(\lambda)\, d\lambda + C_V$$

The bolometric magnitude is given by

$$m_\text{bol} = -2.5 \log \int_0^\infty F_\lambda\, d\lambda + C_\text{bol}$$

Thus the bolometric correction is given by

$$BC = m_\text{bol} - m_V = -2.5 \log \left( \frac{\int_0^\infty F_\lambda\, d\lambda}{\int_0^\infty F_\lambda S_V(\lambda)\, d\lambda} \right)$$

Since $F_\lambda$ follows a blackbody spectrum, we can write

$$F_\lambda = \pi \frac{R^2}{D^2} \cdot \frac{2 \pi h c^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k T}} - 1}$$

Where $T$ is the temperature of the object, $R$ is the radius of the object and $D$ is the distance to the object. Defining $x = \frac{hc}{\lambda kT}$, and doing some algebra, we get that the expression for the bolometric correction simplifies to

$$BC = 2.5 \log \left( \frac{15}{\pi^4} \int_{x_2}^{x_1} \frac{x^3}{e^x - 1} \,dx \right)$$

where $x_1 = \frac{hc}{\lambda_1 k T} \approx 4.901$ and $x_2 = \frac{hc}{\lambda_2 k T} \approx 4.176$. Numerical integration gives

$$\boxed{BC \approx -2.37^m}$$

The bolometric correction of the sun, which too has a temperature of about $5800 \, \mathrm{K}$, is about $-0.7^m$. This large difference is due to the assumptions we made: Sun does not emit like a perfect blackbody, the zero points of the visual and bolometric magnitudes are not the same, and the sensitivity function is not a perfect rectangle, but rather shaped like a bell curve.

A star’s apparent magnitude in the V band is $15.1^m$ and absolute magnitude is equal to $1.3^m$. Extinction of the interstellar medium per kpc is $1^m$.

a) Find the distance to the star.
b) If the star’s observed color index is $(B-V) = 1.6^m$, find its intrinsic color.
c) If $BC = -0.6^m$ for the star, find its luminosity.

a) The distance modulus is given by

$$m - M = 5 \log \frac{r}{10 \times 10^{-3}} + ar$$

Where $r$ is measured in kpc, and $a = 1^m \, \mathrm{kpc^{-1}}$. Substituting the values, we get

$$3.8 = 5 \log r + r$$

Solving it numerically, we get $\boxed{r \approx 2.14 \, \mathrm{kpc}}$.

b) The intrinsic color index is given by

$$(B-V)_0 = (B-V) - E_{B-V}$$

We can use the relation between $A_V$ and $E_{B-V}$ to find the color excess. The extinction in visual band is

$$A_V = ar = 2.14^m$$

Hence the color excess is

$$E_{B-V} = \frac{A_V}{R} \approx 0.69^m$$

Thus the intrinsic color index of the star is

$$\boxed{(B-V)_0 = 1.6^m - 0.69^m = 0.91^m}$$

c) The bolometric magnitude of the star is

$$M_\text{bol} = M_V + BC = 1.3^m - 0.6^m = 0.7^m$$

Comparing with the bolometric magnitude of the Sun gives the luminosity of the star

$$M_\text{bol} - M_{\odot, \text{bol}} = -2.5 \log \frac{L}{L_\odot}$$

$$\implies L = L_\odot \cdot 10^{-0.4(M_\text{bol} - M_{\odot, \text{bol}})}$$

The absolute bolometric magnitude of the Sun is $4.75^m$. Thus we get that the luminosity of the star is

$$\boxed{L \approx 41.7 \, L_\odot}$$

Vega is known to have an apparent magnitude of about $0$ (in reality it is about $0.03$ and fluctuates). Given that it has an absolute magnitude of about $0.58$, find its distance.

Using Pogson’s equation, we have

$$m - M = 5 \log \frac{r}{10 \, \mathrm{pc}}$$

Substituting the values, we get

$$0 - 0.58 = 5 \log \frac{r}{10}$$

$$\implies r = 10 \cdot 10^{-0.116} \approx 7.7 \, \mathrm{pc}$$

Hence the distance to Vega is approximately $\boxed{7.7 \, \mathrm{pc}}$.

Note that since Vega is so close to us, the extinction effects are negligible, and we can ignore them in this case.

What is the optical thickness of a layer of fog, if the Sun seen through the fog seems as bright as a full moon in a cloudless sky?

The apparent magnitudes of the Sun and the Moon are $−26.8$ and $−12.5$, respectively. Thus the total extinction in the cloud must be $A = 14.3$. Since

$$A = (2.5 \log e) \tau$$

where $\tau$ is the optical thickness of the fog, we can rearrange to get

$$\tau = \frac{A}{2.5 \log e} = \frac{14.3}{1.086} \approx \boxed{13.2}$$

Assume that all stars have the same absolute magnitude and stars are evenly distributed in space. Let $N(m)$ be the number of stars brighter than m magnitudes. Find the ratio $N(m + 1)/N(m)$.

Let the number density of stars be $n$, and their absolute magnitude be $M$. Let the apparent magnitude of the star at a distance $r$ pc from the observer be $m$. $m$ and $r$ are related by the equation

$$m - M = 5 \log r - 5$$

Rearranging gives

$$r = 10^{0.2(m - M) + 1}$$

All the stars brighter than magnitude $m$ will lies within $r$ pc distance to the observer. Thus the volume of the sphere upto radius $r$ is

$$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left( 10^{0.2(m - M) + 1} \right)^3$$

The number of stars brighter than magnitude $m$ is given by

$$N(m) = n V = n \frac{4}{3} \pi \left( 10^{0.2(m - M) + 1} \right)^3$$

The number of stars brighter than magnitude $m + 1$ is given by

$$N(m + 1) = n \frac{4}{3} \pi \left( 10^{0.2(m + 1 - M) + 1} \right)^3$$

The ratio of the two is given by

$$\frac{N(m + 1)}{N(m)} = \left( \frac{10^{0.2(m + 1 - M) + 1}}{10^{0.2(m - M) + 1}} \right)^3 = 10^{0.6} \approx \boxed{3.98}$$

An old planetary nebula, with a white dwarf (WD) in its center, is located 50 pc away from Earth. Exactly in the same direction, but behind the nebula, lies another WD, identical to the frist, but located at 150 pc from the Earth. Consider that the two WDs have absolute bolometric magnitude +14.2 and intrinsic color indexes $B - V = 0.300$ and $U - V = 0.330$. Extinction occurs in the interstellar medium and in the planetary nebula.

When we measure the color indices for the closer WD (the one who lies at the center of the nebula), we find the values $B - V = 0.327$ and $U - B = 0.038$. In this part of the Galaxy, the interstellar extinction rates are 1.50, 1.23 and 1.00 magnitudes per kiloparsec for the filters U, B and V, respectively. Calculate the color indices as they would be measured for the second star.

Let $B_0$, $V_0$, $U_0$ be the intrinsic absolute magnitudes of the white dwarf. Let $a_B$, $a_V$, $a_U$ be the interstellar extinction coefficients per kpc. Let $A_{neb,B}$, $A_{neb,V}$, $A_{neb,U}$ be the total extinction for the planetary nebula around the first WD, for light travelling from the centre of the nebula to the edge.

$$(U-B)_0 = (U-V)_0 - (B-V)_0 = 0.330 - 0.300 = 0.030$$

For the closer WD,

$$(U-V)_1 = (U-B)_1 + (B_V)_1 = 0.038 + 0.327 = 0.365$$

$$(B-V)_1 = (B-V)_0 + (a_B - a_V) r + (A_{neb,B} - A_{neb,V})$$

$$\begin{align*} \therefore (A_{neb,B} - A_{neb,V}) &= (B-V)_1 - (B-V)_0 - (a_B - a_V) r \\ &= 0.0155 \end{align*}$$

Similarily,

$$\begin{align*} (A_{neb,U} - A_{neb,V}) &= (U-V)_1 - (U-V)_0 - (a_U - a_V) r \\ &= 0.0100 \end{align*}$$

$$\begin{align*} (A_{neb,U} - A_{neb,B}) &= (U-B)_1 - (U-B)_0 - (a_U - a_B) r \\ &= -0.0055 \end{align*}$$

For second white dwarf, the extinction in planetary nebula will be doubled as extinction the light from this pulsar has to cross the entire diameter of the nebula. Overall distance of this pulsar is 3 times larger.

$$(B-V)_2 = (B-V)_0 + 3(a_B-a_V)r + 2(A_{neb,B} - A_{neb,V})$$

$$\implies \boxed{(B-V)_2 \approx 0.366}$$

$$(U-V)_2 = (U-V)_0 + 3(a_U-a_V)r + 2(A_{neb,U} - A_{neb,V})$$

$$\implies \boxed{(U-V)_2 \approx 0.425}$$

$$(U-B)_2 = (U-B)_0 + 3(a_U-a_B)r + 2(A_{neb,U} - A_{neb,B})$$

$$\implies \boxed{(U-B)_2 \approx 0.060}$$

A UBV photometric observation of a star gives $U = 8.15$, $B = 8.50$, and $V = 8.14$. Based on the spectral class, one gets the intrinsic colour $(U − B)_0 = −0.45$. If the star is known to have radius of $2.3R_\odot$, absolute bolometric magnitude of $−0.25$, and bolometric correction (BC) of $−0.15$, determine:

a) the intrinsic magnitudes U, B, and V of the star
b) the effective temperature of the star
c) the distance to the star

Take the ratio of total to selective extinction $R_V = 3.2$ and the colour excess in $(B − V)$ to be about 72% of the colour excess in $(U − B)$.

a) Let intrinsic magnitudes of the star be $U_0$, $B_0$, and $V_0$ respectively.

$$E_{U-B} = (U-B) - (U - B)_0 = (8.15 - 8.50) - (-0.45) = 0.1$$

Since $E_{B-V} = 0.72 \, E_{U-B}$,

$$E_{B-V} = 0.72 \cdot 0.1 = 0.072$$

Using the relation $A_V = R_V \, E_{B-V}$, we get

$$A_V = 3.2 \cdot 0.072 \approx 0.23$$

$$\therefore \boxed{V_0 = V - A_V = 8.14 - 0.23 = 7.91}$$

The intrinsic $B_0$ and $U_0$ color can be found using

$$(B-V)_0 = (B-V) - E_{B-V} = (8.50 - 8.14) - 0.072 = 0.288$$

$$\begin{align*} \implies &\boxed{B_0 = V_0 + (B-V)_0 = 8.20} \\ &\boxed{U_0 = B_0 + (U-B)_0 = 7.75} \end{align*}$$

b) Comparing the star’s bolometric magnitude to that of the Sun,

$$\begin{align*} &M_\text{bol} - M_{\odot, \text{bol}} = -2.5 \log \frac{L}{L_\odot} \\ \implies &M_{\odot, \text{bol}} - M_\text{bol} = 5 \log \frac{R}{R_\odot} + 10 \log \frac{T}{T_\odot} \end{align*}$$

Substituting the values ($M_{\odot, \text{bol}} \approx 4.75$), we get

$$4.75 - (-0.25) = 5 \log 2.3 + 10 \log \frac{T}{T_\odot}$$

$$\implies \boxed{T = 2.08 T_\odot \approx 12000 \, \mathrm{K}}$$

c) The absolute visual magnitude can be found using the bolometric correction

$$M_V = M_\text{bol} - BC = -0.25 - (-0.15) = -0.10$$

Hence the distance can be found with $V_0$ (intrinsic apparent magnitude)

$$V_0 - M_V = 5 \log \frac{r}{10 \, \mathrm{pc}}$$

$$\implies 7.91 - (-0.10) = 5 \log \frac{r}{10}$$

$$\implies \boxed{r \approx 400 \, \mathrm{pc}}$$

A pulsar, located 1000 pc far from Earth and 10,000 times more luminous than our Sun, emits radiation only from its two opposite poles, creating a homogeneous emission beam shaped as double cone with opening angle $\alpha = 4^\circ$. Assuming the angle between the rotation axis and the emission axis is $\theta = 30^\circ$, and assuming a random orientation of the pulsar beams in relation to an observer on Earth, what is the probability of detecting the pulses? In case we can see it, what is the apparent bolometric magnitude of the pulsar?

As the pulsar rotates, its beam sweeps out a conical surface. The solid angle into which radiation is emitted (by both poles) in one rotation is

$$\begin{align*} \Omega &= 2 \cdot \left[ 2 \pi (1 - \cos (\theta + \frac{\alpha}{2})) - 2 \pi (1 - \cos (\theta - \frac{\alpha}{2})) \right] \\ &= 8 \pi \sin \theta \sin \frac{\alpha}{2} \end{align*}$$

To detect the pulsar, Earth must lie inside this solid angle. Hence the probability is

$$p = \frac{\Omega}{4 \pi} = 2 \sin \theta \sin \frac{\alpha}{2} \approx \boxed{0.035}$$

The pulsar emits radiation into a cone having solid angle

$$\Omega = 2 \cdot 4 \pi \sin^2 \frac{\alpha}{4}$$

Thus the flux recieved on Earth is

$$F = \frac{L}{A} = \frac{L}{\omega r^2} = \frac{L}{8 \pi r^2} \frac{1}{\sin^2 \frac{\alpha}{4}}$$

Comparing the flux with the solar flux gives the apparent magnitude of the pulsar

$$\begin{align*} m &= M_\odot - 2.5 \log \frac{F}{F_\odot} \\ &= M_\odot - 2.5 \log \frac{L}{8 \pi r^2 \sin^2 \frac{\alpha}{4}} \frac{4 \pi r_0^2}{L_\odot} \\ &= M_\odot -2.5 \log \frac{L}{L_\odot} + 5 \log \frac{r}{10 \, \mathrm{pc}} + 5 \log (\sqrt{2} \sin \frac{\alpha}{4}) \\ &= 4.72 - 8.04 = \boxed{-3.32} \end{align*}$$

A star has an apparent magnitude $m_U = 15.0$ in the U-band. The U-band filter is ideal, i.e., it has perfect (100%) transmission within the band and is completely opaque (0% transmission) outside the band. The filter is centered at 360 nm, and has a width of 80 nm. It is assumed that the star also has a flat energy spectrum with respect to frequency. The conversion between magnitude, $m$, in any band and flux density, $f$, of a star in Jansky is given by

$$f = 3631 \times 10^{-0.4m} \, \mathrm{Jy}$$

a) Approximately how many U-band photons, $N_0$, from this star will be incident normally on a $1 \, \mathrm{m^2}$ area at the top of the Earth’s atmosphere every second?

This star is being observed in the U-band using a ground based telescope, whose primary mirror has a diameter of $2.0 \, \mathrm{m}$. Atmospheric extinction in U-band during the observation is 50%. You may assume that the seeing is diffraction limited. Average surface brightness of night sky in U-band was measured to be $22.0 \, \mathrm{mag/arcsec^2}$.

b) What is the ratio, $R$, of number of photons received per second from the star to that received from the sky, when measured over a circular aperture of diameter $2''$?
c) In practice, only 20% of U-band photons falling on the primary mirror are detected. How many photons, $N_t$, from the star are detected per second?

a) The U-band is defined as ($360 \pm 40$) nm. Thus, the minimum, maximum and average frequencies of the band are

$$\begin{align*} \nu_\text{min} &= \frac{c}{\lambda_\text{max}} = 7.495 \times 10^{14} \, \mathrm{Hz} \\ \nu_\text{max} &= \frac{c}{\lambda_\text{min}} = 9.369 \times 10^{14} \, \mathrm{Hz} \\ \nu_\text{avg} &= \frac{\nu_\text{min} + \nu_\text{max}}{2} = 8.432 \times 10^{14} \, \mathrm{Hz} \\ \Delta \nu &= \nu_\text{max} - \nu_\text{min} = 1.874 \times 10^{14} \, \mathrm{Hz} \end{align*}$$$$\therefore f = 3631 \times 10^{-0.4 \cdot 15.0} \, \mathrm{Jy} = 3.631 \times 10^{-29} \, \mathrm{W \, Hz^{-1} \, m^{-2}}$$

Now, $N_0 \times h \nu_\text{avg} = \Delta \nu \cdot f \cdot A \cdot \Delta t$, where $A = 1 \, \mathrm{m^2}$ and $\Delta t = 1 \, \mathrm{s}$.

$$\therefore N_0 = \frac{f A \Delta \nu \Delta t}{h \nu_\text{avg}} \approx \boxed{12180}$$

b) Let us call sky flux per square arcsec as $\Phi$ and total sky flux for the given aperture as $\phi_\text{sky}$. Let total star flux be $\phi_\text{st}$.

$$\phi_\text{sky} = \Phi A = \pi (1'')^2 \Phi$$

Using the formula for surface brightness, we get that the magnitude corresponding to this flux is

$$m_{sky} = 22.0 + 2.5 \log \frac{\Phi}{\phi_\text{sky}} = 20.76^m$$

As extinction is 50%,

$$R = \frac{0.5 \phi_\text{st}}{\phi_\text{sky}} = 0.5 \times 10^{-0.4(15 - 20.76)} \approx \boxed{100}$$

c) Equating the number of photons,

$$N_T \times 1 \, \mathrm{m^2} = N_0 \times 0.5 \times 0.2 \times \pi \left( \frac{2.0}{2} \right)^2 \mathrm{m^2}$$

$$\implies N_T = 12180 \times 0.5 \times 0.2 \times \pi \approx \boxed{3826}$$

The star $\beta$-Doradus is a Cepheid variable star with a pulsation period of 9.84 days. We make a simplifying assumption that the star is brightest when it is most contracted (radius being $R_1$) and it is faintest when it is most expanded (radius being $R_2$). For simplicity, assume that the star maintains its spherical shape and behaves as a perfect black body at every instant during the entire cycle. The bolometric magnitude of the star varies from $3.46$ to $4.08$. From Doppler measurements, we know that during pulsation the stellar surface expands or contracts at an average radial speed of $12.8 \, \mathrm{km \, s^{-1}}$. Over the period of pulsation, the peak of thermal radiation (intrinsic) of the star varies from 531.0 nm to 649.1 nm.

a) Find the ratio of radii of the star in its most contracted and most expanded states ($R_1/R_2$).
b) Find the radii of the star (in metres) in its most contracted and most expanded states ($R_1$ and $R_2$).
c) Calculate the flux of the star, $F_2$, when it is in its most expanded state.
d) Find the distance to the star, $D_\text{star}$, in parsecs.

a) The ratio of the fluxes is

$$m_1 - m_2 = -2.5 \log \frac{F_1}{F_2} \implies \frac{F_1}{F_2} = 10^{-0.4(m_1 - m_2)} = 1.77$$

The flux is given by $F = \frac{4 \pi R^2 \sigma T^4}{4 \pi D^2}$, if the star is at a distance $D$ from the observer.

Hence we get that

$$\frac{F_1}{F_2} = \frac{R_1^2 T_1^4}{R_2^2 T_2^4} = 1.77$$

Using Wien’s displacement law, $\frac{T_1}{T_2} = \frac{\lambda_2}{\lambda_1}$.

$$\implies \frac{R_1}{R_2} = \sqrt{\frac{F_1}{F_2}} \left( \frac{\lambda_1}{\lambda_2} \right)^2 = \boxed{0.890}$$

b) The radial velocity causes the change in radius of the star

$$R_2 - R_1 = v \cdot \frac{P}{2} = 5.441 \times 10^9 \, \mathrm{m}$$

$$(1-0.890)R_2 = 5.441 \times 10^9 \, \mathrm{m}$$

$$\begin{align*} \therefore \: &\boxed{R_2 = 4.95 \times 10^10 \, \mathrm{m}} \\ &\boxed{R_1 = 4.40 \times 10^{10} \, \mathrm{m}} \end{align*}$$

c) To get the absolute value of flux ($F_2$) we must compare it with observed flux of the Sun.

$$m_2 - m_\odot = -2.5 \log \frac{F_2}{F_\odot}$$

$$\implies F_2 = F_\odot \cdot 10^{-0.4(m_2 - m_\odot)}$$$$F_2 = 1361 \, \mathrm{W \, m^{-2}} \cdot 10^{-0.4(4.08 + 26.72)} = \boxed{6.51 \times 10^{-10} \, \mathrm{W \, m^{-2}}}$$

d) The temperature of the star when it is fully expanded is

$$T_2 = \frac{b}{\lambda_2} = 4465 \, \mathrm{K}$$

The distance can be found using the observed flux

$$F_2 = \frac{4 \pi R_2^2 \sigma T_2^4}{4 \pi D_\text{star}^2}$$

$$\implies D_\text{star} = R_2 T_2^2 \sqrt{\frac{\sigma}{F_2}} = \boxed{298 \, \mathrm{pc}}$$

Given the intrinsic $(B-V)$ color index of a star, find its color temperature $T_c$.

Assuming the star to be a blackbody, the spectral flux density in the Wien approximation of the star is propotional to

$$F_\lambda \propto \lambda^{-5} e^{-hc/\lambda kT_c}$$

The color index is defined as

$$(B-V) = -2.5 \log \left( \frac{F_B}{F_V} \right) + C$$

$$\implies (B-V) = -2.5 \log \left( \frac{\lambda_B^{-5} e^{-hc/\lambda_B kT_c}}{\lambda_V^{-5} e^{-hc/\lambda_V kT_c}} \right) + C$$

$$\implies (B-V) = 2.5 \log \left( \frac{\lambda_B}{\lambda_V} \right)^5 + 2.5 \frac{hc}{kT_c} \log e \left( \frac{1}{\lambda_B} - \frac{1}{\lambda_V} \right) + C$$

$$\implies T_c = \frac{hc}{k} \log e \cdot \frac{2.5 \left( \frac{1}{\lambda_B} - \frac{1}{\lambda_V} \right)}{(B-V) - 12.5 \log \left( \frac{\lambda_B}{\lambda_V} \right) - C}$$

Using the given wavelengths $\lambda_B = 440 \, \mathrm{nm}$ and $\lambda_V = 550 \, \mathrm{nm}$, the term $-12.5 \log (\lambda_B/\lambda_V) \approx 1.21$. Assuming standard zero-point calibration, we get the approximate relation:

$$\boxed{T_c \approx \frac{7000 \mathrm{K}}{(B-V) + 0.47}}$$