1) A Matter-only Model of the Universe in Newtonian approach
In this exercise, we will derive the first and second Friedmann equations of a homogeneous, isotropic and matter-only universe. We use the Newtonian approach
Consider a universe filled with matter which has a mass density $\rho (t)$. Note that as the universe expands or contracts, the density of the matter changes with time, which is why it is a function of time $t$.
Now consider a mass shell of radius $R$ within this universe. The total mass of the matter enclosed by this shell is $M$. In the case we consider (homogeneous and isotropic universe), there is no shell crossing, so $M$ is a constant.
(a) What is the acceleration of this shell? Express the acceleration as the time derivative of velocity, $\dot{v}$ (pronounced $v$-dot) to avoid confusion with the scale factor $a$ (which you learned about last week).
\begin{align*}
F_g &= m\dot{v}\\
-\frac{Gm^2}{R^2} &= m\dot{v}\\
\dot{v} &= -\frac{Gm}{R^2}
\end{align*}
(b) To derive an energy equation, it is a common trick to multiply both sides of your acceleration equation by $v$. Turn your velocity, $v$, into $\frac{dR}{dt}$, cancel $dt$, and integrate both sides of your equation, This is an indefinite integral, so you will have a constant integration; combine these integration constants call their sum $C$. You should arrive at the following equation:
\begin{align*}
\frac{1}{2}\dot{R}^2 - \frac{GM}{R} = C
\end{align*}
Convince yourself the equation you've written down has units of energy per unit mass .
\begin{align*}
\dot{v} &= -\frac{Gm}{R^2}\\
\dot{v}v &= -\frac{Gm}{R^2}v\\
\dot{v}\frac{dR}{dt}&= -\frac{Gm}{R^2}\frac{dR}{dt}\\
\dot{v}dR &= -\frac{Gm}{R^2}dR\\
\int \ddot{Rdr} &= \int -\frac{Gm}{R^2}dR\\
\frac{1}{2}\dot{R}^2 + C_1 &= \frac{Gm}{R} + C_2\\
C &= \frac{1}{2}\dot{R}^2 - \frac{Gm}{R}
\end{align*}
(c) Express the total mass, $M$ using the mass density, and plug it into the above equation. Rearrange your equation to give an expression for $\left(\frac{\dot{R}}{R}\right)^2$, where $\dot{R}$ is equal to $\frac{dR}{dt}$.
The total mass, $M$ in terms of density is:
\begin{align*}
M = \frac{4}{3}\pi R^3 \rho(t)
\end{align*}
We can plug in this mass, $M$, to the equation in part (b) as follows:
\begin{align*}
C &= \frac{1}{2}\dot{R}^2 - \frac{Gm}{R}\\
C &= \frac{1}{2}\dot{R}^2 - \frac{G\left(\frac{4}{3}\pi R^3 \rho(t)\right)}{R}\\
2C &= \dot{R}^2 - G\left(\frac{8}{3}\pi R^2 \rho(t)\right)\\
\dot{R}^2 &= 2C + G\left(\frac{8}{3}\pi R^2 \rho(t)\right)\\
\left(\frac{\dot{R}}{R}\right)^2 &= \frac{2C}{R^2} + \frac{8 G \pi \rho(t)}{3}
\end{align*}
(d) $R$ is the physical radius of the sphere. It is often convenient to express $R$ as $R = a(t)r$, where $r$ is the comoving radius of the sphere. The comoving coordinate for a fixed shell remains constant in time. The time dependence of $R$ is captured by the scale factor $a(t)$. The comoving radius equals to the physical radius at the epoch when $a(t) =1$. Rewrite your equation in terms of the comoving radius, $R$, and the scale factor , $a(t)$.
We know that $R = a(t)r$. We also need to know what $\dot{R}$ is. In order to do that, we need to find the derivative of $R = a(t)r$, which is $\dot{R} = \dot{a}(t)r$. We can substitute these values for $R$ and $\dot{R}$ in the equation we derived in part (c) as follows:
\begin{align*}
\left(\frac{\dot{a}(t)r}{a(t)r}\right)^2 &= \frac{2C}{(a(t)r)^2} + \frac{8 G \pi \rho(t)}{3}
\end{align*}
(e) Rewrite the above expression so that $\left(\frac{\dot{a}}{a}\right)^2$ appears alone on the left side of the equation.
Rewriting the above equation in terms of $\left(\frac{\dot{a}}{a}\right)^2$, we get:
\begin{align*}
\left(\frac{\dot{a}}{a}\right)^2 &= \frac{2C}{a^2 r^2} + \frac{8 G \pi \rho(t)}{3}
\end{align*}
(f) Derive the first Friedmann Equation: From the previous worksheet, we know that $H(t) = \frac{\dot{a}}{a}$. Plugging this relation into your above result and identifying the constant $\frac{2C}{r^2} = -kc^2$, where $k$ is the "curvature" parameter, you will get the first Friedmann equation. The Friedmann equation tells us about how the shell expands or contracts; in other words, it tells us about the Hubble expansion (or contraction) rate of the universe.
Substituting $H(t) = \frac{\dot{a}}{a}$ and $\frac{2C}{r^2} = -kc^2$ to the equation in part (e), we get the first Friedmann Equation:
\begin{align*}
H^2(t) &= \frac{8 G \pi \rho(t)}{3} - \frac{kc^2}{a^2}
\end{align*}
(g) Derive the second Friedmann Equation: Now express the acceleration of the shell in terms of the density of the universe, and replace $R$ with $R = a(t)r$. You should see that $\frac{\dot{a}}{a} = -\frac{4 \pi}{3}Gp$, which is known as the second Friedmann equation.
The more complete second Friedmann equation has another term involving the pressure following from Einstein's general relativity (GR), which is not captured in the Newtonian derivation.
If the matter is cold, its pressure is zero. Otherwise, if it is warm or hot we will need to consider the effect of the pressure.
We follow the same basic steps as we did to get the first Friedmann equation, and start with the equation we derived in part (a):
\begin{align*}
\dot{v} &= -\frac{Gm}{R^2}
\end{align*}
We know that acceleration, represented as the first derivative of velocity, $\dot{v}$, can also be represented by the second derivative of distance, as $\ddot{r}$. Since we know that $R = a(t)r$, we can say that the second derivative of a distance can be represented as, $\ddot{r} = \ddot{a}(t)r$. Therefore :
\begin{align*}
\dot{v} = \ddot{r} = \ddot{a}(t)r
\end{align*}
This can be substituted in the equation from part (a) to get the equation:
\begin{align*}
\ddot{a}(t)r &= -\frac{Gm}{\left(a(t)r\right)^2}\\
\frac{\ddot{a}(t)r}{a(t)} &= -\frac{Gm}{r^2}\\
\frac{\ddot{a}}{a} &= -\frac{Gm}{r^3}\\
\end{align*}
Finally, we can substitute the mass for the mass density, $m = \frac{4}{3}\pi r^3 \rho(t)$ to derive the Second Friedmann Equation:
\begin{align*}
\frac{\ddot{a}}{a} &= -\frac{Gm}{r^3}\\
\frac{\ddot{a}}{a} &= -\frac{G\left(\frac{4}{3}\pi r^3 \rho(t)\right)}{r^3}\\
\frac{\ddot{a}}{a} &= -\frac{4}{3} G \pi \rho(t)
\end{align*}
Monday, February 8, 2016
Blog #27: Size of the Universe
3) It is not strictly correct to associate this ubiquitous distance-dependent redshift we observe with teh velocity of the galaxies (at very large separations, Hubble's Law gives 'velocities' that exceeds the speed of light and becomes poorly defined). What we have measured is the cosmological redshift, which is actually due to the overall expansion of the Universe itself. This phenomenon is dubbed the Hubble Flow, and it is due to space itself being stretched in an expanding Universe.
Since everything seems to be getting away from us, you might be tempted to imagine we are located at the center of this expansion. But, as you explored in the opening thought experiment, in actuality, everything is rushing away from everything else, everywhere in the universe, in the same way. So, an alien astronomer observing the motion of the galaxies in its locality would arrive at the same conclusions we do.
In cosmology, the scale factor, a(t), is a dimensionless parameter that characterizes the size of the universe and the small amount of space in between grid points in the universe at time $t$. In the current epoch, $t = t_0$ and $a(t_0) \equiv 1$. $a(t)$ is a function of time. It changes over time, and it was smaller in the past (since the universe is expanding). This means that two galaxies in the Hubble Flow separated by distance $d_0 = d(t_0)$ in the present were $d(t) = a(t)(d_0)$ apart at time $t$.
The Hubble Constant is also a function of time, and is defined so as to characterize the fractional rate of change of the scale factor:
\begin{align*}
H(t) = \frac{1}{a(t)}\frac{da}{dt} \Big|_t
\end{align*}
and the Hubble Law is locally valid for any $t$:
\begin{align*}
v = H(t)d
\end{align*}
where $v$ is the relative recessional velocity between two points and $d$ the distance that separates them.
(a) Assume the rate of expansion, $\dot{a} \equiv \frac{da}{dt}$, has been constant for all time. How long ago was the Big Bang (i.e. when $a(t=0) = 0$)? How does this compare with the age of the oldest globular clusters (~ 12 Gyr)? What you will calculate is known as the Hubble Time.
In order to solve for the moment when the Big Bang occurred, we need to solve for $t_0$.
We can start with the equation given to us:
\begin{align*}
H(t) = \frac{1}{a(t)}\frac{da}{dt} \Big|_t
\end{align*}
We know that $\dot{a} \equiv \frac{da}{dt}$ and $a(t_0) = 1$, which we can substitute in this equation:
\begin{align*}
H(t) &= \frac{1}{a(t)}\frac{da}{dt} \Big|_t\\
H(t_0) &= \frac{1}{a(t_0)}\frac{da}{dt}\\
H_0 &=\frac{da}{dt}\\
H_0 dt &= da\\
\int_0^{t_0} H_0dt &= \int_0^{a(t_0) = 1} da\\
H_0 t_0 &= 1\\
t_0 = \frac{1}{H_0}
\end{align*}
A quick Google search shows that the value for $H_0$ is about $67.8 \frac{\frac{km}{s}}{Mpc}, which converted into seconds is: $H_0 = 2.3 \times 10^{-18} \frac{1}{s}. Using this information, we can solve for $t_0$ as follows:
\begin{align*}
t_0 &= \frac{1}{H_0}\\
t_0 &= \frac{1}{2.3 \times 10^{-18} \frac{1}{s}}\\
t_0 &\approx 4.4 \times 10^{17}\text{ seconds}\\
t_0 &\approx 14 \text{ Gyr}
\end{align*}
This shows that the Hubble Time, which is the time at the beginning of the Universe, is about 14 billion years, which is about 2 billion years earlier than the earliest globular clusters.
(b) What is the size of the observable universe? What you will calculate is known as the Hubble Length.
Distance is measured by rearranging the equation for velocity as follows:
\begin{align*}
d = vt
\end{align*}
Since we know that the time in this equation is the Hubble Time, and $v$ is the speed of light, $c$, which gives us:
\begin{align*}
d &= vt\\
d &= H_0c\\
d &= (4.4 \times 10^{17}) \times (3 \times 10^{10})\\
d &= 1.32 \times 10^{28} \text{ cm}\\
d &= 1.4 \times 10^{10} \text{ light years}
\end{align*}
This shows that the size of the observable universe is $1.4 \times 10^{10}$ light years, which is also known as the Hubble Length.
Since everything seems to be getting away from us, you might be tempted to imagine we are located at the center of this expansion. But, as you explored in the opening thought experiment, in actuality, everything is rushing away from everything else, everywhere in the universe, in the same way. So, an alien astronomer observing the motion of the galaxies in its locality would arrive at the same conclusions we do.
In cosmology, the scale factor, a(t), is a dimensionless parameter that characterizes the size of the universe and the small amount of space in between grid points in the universe at time $t$. In the current epoch, $t = t_0$ and $a(t_0) \equiv 1$. $a(t)$ is a function of time. It changes over time, and it was smaller in the past (since the universe is expanding). This means that two galaxies in the Hubble Flow separated by distance $d_0 = d(t_0)$ in the present were $d(t) = a(t)(d_0)$ apart at time $t$.
The Hubble Constant is also a function of time, and is defined so as to characterize the fractional rate of change of the scale factor:
\begin{align*}
H(t) = \frac{1}{a(t)}\frac{da}{dt} \Big|_t
\end{align*}
and the Hubble Law is locally valid for any $t$:
\begin{align*}
v = H(t)d
\end{align*}
where $v$ is the relative recessional velocity between two points and $d$ the distance that separates them.
(a) Assume the rate of expansion, $\dot{a} \equiv \frac{da}{dt}$, has been constant for all time. How long ago was the Big Bang (i.e. when $a(t=0) = 0$)? How does this compare with the age of the oldest globular clusters (~ 12 Gyr)? What you will calculate is known as the Hubble Time.
In order to solve for the moment when the Big Bang occurred, we need to solve for $t_0$.
We can start with the equation given to us:
\begin{align*}
H(t) = \frac{1}{a(t)}\frac{da}{dt} \Big|_t
\end{align*}
We know that $\dot{a} \equiv \frac{da}{dt}$ and $a(t_0) = 1$, which we can substitute in this equation:
\begin{align*}
H(t) &= \frac{1}{a(t)}\frac{da}{dt} \Big|_t\\
H(t_0) &= \frac{1}{a(t_0)}\frac{da}{dt}\\
H_0 &=\frac{da}{dt}\\
H_0 dt &= da\\
\int_0^{t_0} H_0dt &= \int_0^{a(t_0) = 1} da\\
H_0 t_0 &= 1\\
t_0 = \frac{1}{H_0}
\end{align*}
A quick Google search shows that the value for $H_0$ is about $67.8 \frac{\frac{km}{s}}{Mpc}, which converted into seconds is: $H_0 = 2.3 \times 10^{-18} \frac{1}{s}. Using this information, we can solve for $t_0$ as follows:
\begin{align*}
t_0 &= \frac{1}{H_0}\\
t_0 &= \frac{1}{2.3 \times 10^{-18} \frac{1}{s}}\\
t_0 &\approx 4.4 \times 10^{17}\text{ seconds}\\
t_0 &\approx 14 \text{ Gyr}
\end{align*}
This shows that the Hubble Time, which is the time at the beginning of the Universe, is about 14 billion years, which is about 2 billion years earlier than the earliest globular clusters.
(b) What is the size of the observable universe? What you will calculate is known as the Hubble Length.
Distance is measured by rearranging the equation for velocity as follows:
\begin{align*}
d = vt
\end{align*}
Since we know that the time in this equation is the Hubble Time, and $v$ is the speed of light, $c$, which gives us:
\begin{align*}
d &= vt\\
d &= H_0c\\
d &= (4.4 \times 10^{17}) \times (3 \times 10^{10})\\
d &= 1.32 \times 10^{28} \text{ cm}\\
d &= 1.4 \times 10^{10} \text{ light years}
\end{align*}
This shows that the size of the observable universe is $1.4 \times 10^{10}$ light years, which is also known as the Hubble Length.
Blog #26: Distance and Velocity at a Frame of Reference
1) Before we dive into the Hubble Flow, let's do a thought experiment. Pretend that there is an infinitely long series of balls sitting in a row. Imagine that during a time interval $\Delta t$ the space between each ball increases by $\Delta x$.
(a) Look at the shadd ball, Ball C, in the figure above. Imagine that Ball C is sitting still (so we are in the reference frame of Ball C). What is the distance to Ball D after time $\Delta t$? What about Ball B?
Based on the picture above, Ball B and Ball D are moving away from Ball C.
The distance between Ball B and Ball C at time $t = 0$ is $X_{{CB}_0}$.
The distance between Ball C and Ball D at time $t = 0$ is $X_{{CD}_0}$
The distance between Ball B and Ball C at time $t = \Delta t$ is:
\begin{align*}
X_{CB}(\Delta t) = X_{{CB}_0} + \Delta x
\end{align*}
The distance between Ball C and Ball D at time $t = \Delta t$ is:
\begin{align*}
X_{CD}(\Delta t) = X_{{CD}_0} + \Delta x
\end{align*}
(b) What are the distances from Ball C to Ball A and Ball E?
The distance between Ball C and Ball A in the time span $\Delta t$ is the same as the distance between Ball C to Ball B and Ball B to Ball A, as shown below:
\begin{align*}
X_{CA}(\Delta t) &= X_{CB}(\Delta t) + X_(BA)(\Delta t)\\
X_{CA}(\Delta t) &= (X_{{CB}_0} + \Delta x) + (X_{{BA}_0} + \Delta x)\\
X_{CA}(\Delta t) &= X_{{CA}_0} + 2 \Delta x
\end{align*}
The same logic can be used for the distance between Ball A and Ball E, as follows:
\begin{align*}
X_{CE}(\Delta t) &= X_{CD}(\Delta t) + X_(DE)(\Delta t)\\
X_{CE}(\Delta t) &= (X_{{CD}_0} + \Delta x) + (X_{{DE}_0} + \Delta x)\\
X_{CE}(\Delta t) &= X_{{CE}_0} + 2 \Delta x
\end{align*}
(c) Write a general expression for the distance to a ball $N$ balls away from Ball C after time $\Delta t$. Interpret your findings.
Based on the answer to part (b), we can see that the distance between Ball C and a ball N balls away is the sum of the distance between the individual balls between the two balls. Therefore, we can generalize the distance between Ball C and another ball $N$ balls away during time $\Delta t$ as follows:
\begin{align*}
X_{CN}(\Delta t) = X_{{CN}_0} + N\Delta x
\end{align*}
(d) Write the velocity of a ball $N$ balls away from Ball C during $\Delta t$. Interpret your finding.
Velocity is described as a change in distance over a time, $t$, as described by the equation $v(t) = \frac{\Delta x}{t}$. The change in distance between two balls over a time $\Delta t$ is given by the answer in part (b) as $N\Delta x$. Therefore, the velocity of a ball $N$ balls away from Ball C is:
\begin{align*}
v(t) &= \frac{\Delta x}{t}\\\
v(\Delta t) &= \frac{N\Delta x}{\Delta t}
\end{align*}
This shows that balls that are further away from Ball C are moving faster in the frame of reference for Ball C than balls that are closer to Ball C.
(a) Look at the shadd ball, Ball C, in the figure above. Imagine that Ball C is sitting still (so we are in the reference frame of Ball C). What is the distance to Ball D after time $\Delta t$? What about Ball B?
Based on the picture above, Ball B and Ball D are moving away from Ball C.
The distance between Ball B and Ball C at time $t = 0$ is $X_{{CB}_0}$.
The distance between Ball C and Ball D at time $t = 0$ is $X_{{CD}_0}$
The distance between Ball B and Ball C at time $t = \Delta t$ is:
\begin{align*}
X_{CB}(\Delta t) = X_{{CB}_0} + \Delta x
\end{align*}
The distance between Ball C and Ball D at time $t = \Delta t$ is:
\begin{align*}
X_{CD}(\Delta t) = X_{{CD}_0} + \Delta x
\end{align*}
(b) What are the distances from Ball C to Ball A and Ball E?
The distance between Ball C and Ball A in the time span $\Delta t$ is the same as the distance between Ball C to Ball B and Ball B to Ball A, as shown below:
\begin{align*}
X_{CA}(\Delta t) &= X_{CB}(\Delta t) + X_(BA)(\Delta t)\\
X_{CA}(\Delta t) &= (X_{{CB}_0} + \Delta x) + (X_{{BA}_0} + \Delta x)\\
X_{CA}(\Delta t) &= X_{{CA}_0} + 2 \Delta x
\end{align*}
The same logic can be used for the distance between Ball A and Ball E, as follows:
\begin{align*}
X_{CE}(\Delta t) &= X_{CD}(\Delta t) + X_(DE)(\Delta t)\\
X_{CE}(\Delta t) &= (X_{{CD}_0} + \Delta x) + (X_{{DE}_0} + \Delta x)\\
X_{CE}(\Delta t) &= X_{{CE}_0} + 2 \Delta x
\end{align*}
(c) Write a general expression for the distance to a ball $N$ balls away from Ball C after time $\Delta t$. Interpret your findings.
Based on the answer to part (b), we can see that the distance between Ball C and a ball N balls away is the sum of the distance between the individual balls between the two balls. Therefore, we can generalize the distance between Ball C and another ball $N$ balls away during time $\Delta t$ as follows:
\begin{align*}
X_{CN}(\Delta t) = X_{{CN}_0} + N\Delta x
\end{align*}
(d) Write the velocity of a ball $N$ balls away from Ball C during $\Delta t$. Interpret your finding.
Velocity is described as a change in distance over a time, $t$, as described by the equation $v(t) = \frac{\Delta x}{t}$. The change in distance between two balls over a time $\Delta t$ is given by the answer in part (b) as $N\Delta x$. Therefore, the velocity of a ball $N$ balls away from Ball C is:
\begin{align*}
v(t) &= \frac{\Delta x}{t}\\\
v(\Delta t) &= \frac{N\Delta x}{\Delta t}
\end{align*}
This shows that balls that are further away from Ball C are moving faster in the frame of reference for Ball C than balls that are closer to Ball C.
Wednesday, January 27, 2016
Blog #25: Quasars and their Black Holes
4) One feature you surely noticed was the strong, broad emission lines. Here is a closer look at the strongest emission line in the spectrum:
This feature arises from hydrogen gas in the accretion disk. The photons radiated during the accretion process are constantly ionizing nearby hydrogen atoms. So there are many free protons and electrons in the disk. When one of these protons comes close enough to an electron, they recombine into a new hydrogen atom, and the electron will lose energy until it reaches the lowest allowed energy state, labeled $n=1$ in the model of the hydrogen atom shown below (and called the ground state):
On its way to the ground state, the electron passes through other allowed energy states (called excited states). Technically speaking, atoms have an infinite number of allowed energy states, but electrons spend most of their time occupying those of lowest energies, and so only the $n = 3$ and $n=3$ excited states are shown above for simplicity.
Because the difference in energy between e.g., the $n =2$ and $n= 1$ states are always the same, the electron always loses the same amount of energy when it passes between them. Thus, the photon it emits during this process will always have the same wavelength. For the hydrogen atom, the energy difference between the $n=2$ and $n=1$ energy levels 10.19 eV, corresponding to a photon wavelength of $\lambda = 1215.67$ Angstroms. This is the most commonly-observed atomic transition in all of astronomy, as hydrogen is by far the most abundant element in the Universe. It is referred to as the Lyman $\alpha$ transition (or Ly$\alpha$ for short).
It turns out that that strongest emission feature you observed in the quasar spectrum aboves arises from Ly$\alpha$ emission from material orbiting around the central black hole.
(a) Recall the Doppler equation:
\begin{align*}
\frac{\lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}} = z \approx \frac{v}{c}
\end{align*}
Using the data provided, calculate the redshift of this quasar.
Looking at the zoomed-in graph of the spectrum above, the peak of the emission line is at 1410 Angstroms, which shows that the $\lambda_{observed} = 1410$ Angstrom. We know that the actual emitted emission for the Ly$\alpha$ transition is $\lambda_{emitted} = 1215.67$. Using the equation above, we can solve for the redshift as follows:
\begin{align*}
z &= \frac{\lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}}\\
z &= \frac{1410 - 1215.67}{1215.67}\\
z &\approx 0.16
\end{align*}
Therefore, the redshift of this quasar is 0.16.
(b) Again using the data provided, along with the Virial Theorem, estimate the mass of the black hole in this quasar. It would help to know that the typical accretion disk around a $10^8 M_{\odot}$ black hole extends to a radius of $r = 10^{15}$ m.
The velocity at which this quasar is moving away from us can be found by looking at the width of the broadened peak. The broadened peak is about 15 Angstroms, so the redshift between these two peaks would be $z = \frac{15}{1215} \approx 0.012$.
However, we only use half that value for the actual velocity, since half of the quasar is coming towards us, and the other half is going away from us. Therefore, $z = 0.006$. Using the equation above we can calculate how fast the quasar is going away from us:
\begin{align*}
z &= \frac{v}{c}\\
v&= zc\\
v &= 0.006c
\end{align*}
Now that we have the speed at which the quasar is moving away from us, we can use the Virial Theorem to solve for the mass of the black hole, $M$ as follows:
\begin{align*}
K &= -\frac{1}{2}U\\
\frac{1}{2}mv^2 &= \frac{1}{2}\frac{GMm}{r}\\
v^2 &= \frac{GM}{r}\\
M &= \frac{v^2r}{G}\\
M &= \frac{(0.006 \times 3 \times 10^{10})^2(10^{17})}{6.7 \times 10^{-8}}\\
M &= 4.8 \times 10^{40} \text{ grams}\\
M &= 2.4 \times 10^7 M_{\odot}
\end{align*}
The mass of the black hole is $2.4 \times 10^7 M_{\odot}$.
This feature arises from hydrogen gas in the accretion disk. The photons radiated during the accretion process are constantly ionizing nearby hydrogen atoms. So there are many free protons and electrons in the disk. When one of these protons comes close enough to an electron, they recombine into a new hydrogen atom, and the electron will lose energy until it reaches the lowest allowed energy state, labeled $n=1$ in the model of the hydrogen atom shown below (and called the ground state):
On its way to the ground state, the electron passes through other allowed energy states (called excited states). Technically speaking, atoms have an infinite number of allowed energy states, but electrons spend most of their time occupying those of lowest energies, and so only the $n = 3$ and $n=3$ excited states are shown above for simplicity.
Because the difference in energy between e.g., the $n =2$ and $n= 1$ states are always the same, the electron always loses the same amount of energy when it passes between them. Thus, the photon it emits during this process will always have the same wavelength. For the hydrogen atom, the energy difference between the $n=2$ and $n=1$ energy levels 10.19 eV, corresponding to a photon wavelength of $\lambda = 1215.67$ Angstroms. This is the most commonly-observed atomic transition in all of astronomy, as hydrogen is by far the most abundant element in the Universe. It is referred to as the Lyman $\alpha$ transition (or Ly$\alpha$ for short).
It turns out that that strongest emission feature you observed in the quasar spectrum aboves arises from Ly$\alpha$ emission from material orbiting around the central black hole.
(a) Recall the Doppler equation:
\begin{align*}
\frac{\lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}} = z \approx \frac{v}{c}
\end{align*}
Using the data provided, calculate the redshift of this quasar.
Looking at the zoomed-in graph of the spectrum above, the peak of the emission line is at 1410 Angstroms, which shows that the $\lambda_{observed} = 1410$ Angstrom. We know that the actual emitted emission for the Ly$\alpha$ transition is $\lambda_{emitted} = 1215.67$. Using the equation above, we can solve for the redshift as follows:
\begin{align*}
z &= \frac{\lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}}\\
z &= \frac{1410 - 1215.67}{1215.67}\\
z &\approx 0.16
\end{align*}
Therefore, the redshift of this quasar is 0.16.
(b) Again using the data provided, along with the Virial Theorem, estimate the mass of the black hole in this quasar. It would help to know that the typical accretion disk around a $10^8 M_{\odot}$ black hole extends to a radius of $r = 10^{15}$ m.
The velocity at which this quasar is moving away from us can be found by looking at the width of the broadened peak. The broadened peak is about 15 Angstroms, so the redshift between these two peaks would be $z = \frac{15}{1215} \approx 0.012$.
However, we only use half that value for the actual velocity, since half of the quasar is coming towards us, and the other half is going away from us. Therefore, $z = 0.006$. Using the equation above we can calculate how fast the quasar is going away from us:
\begin{align*}
z &= \frac{v}{c}\\
v&= zc\\
v &= 0.006c
\end{align*}
Now that we have the speed at which the quasar is moving away from us, we can use the Virial Theorem to solve for the mass of the black hole, $M$ as follows:
\begin{align*}
K &= -\frac{1}{2}U\\
\frac{1}{2}mv^2 &= \frac{1}{2}\frac{GMm}{r}\\
v^2 &= \frac{GM}{r}\\
M &= \frac{v^2r}{G}\\
M &= \frac{(0.006 \times 3 \times 10^{10})^2(10^{17})}{6.7 \times 10^{-8}}\\
M &= 4.8 \times 10^{40} \text{ grams}\\
M &= 2.4 \times 10^7 M_{\odot}
\end{align*}
The mass of the black hole is $2.4 \times 10^7 M_{\odot}$.
Blog #24: Analyzing a Quasar's Spectrum
3) Such bright objects, known as quasars, can be easily observed at great distances, and astronomers started taking spectra of them back in the 1960's. Here's a spectrum of the first quasar ever discovered, called 3C 273:
What are the main features you see in this spectrum (ignoring the gap in the data at around 1625 A)?
The first noticeable thing in this plot is that as you get to larger wavelengths, the flux drops, as is evident by the negative slope of the data.
The second noticeable thing is that the graph has 3 peaks that signify spectral emissions. The three peaks are at around 1400 Angstroms, 1800 Angstroms, and 2200 Angstroms.
The last noticeable thing in this graph is that there are several small dips in the graph, showing the absorption lines in the spectrum.
What are the main features you see in this spectrum (ignoring the gap in the data at around 1625 A)?
The first noticeable thing in this plot is that as you get to larger wavelengths, the flux drops, as is evident by the negative slope of the data.
The second noticeable thing is that the graph has 3 peaks that signify spectral emissions. The three peaks are at around 1400 Angstroms, 1800 Angstroms, and 2200 Angstroms.
The last noticeable thing in this graph is that there are several small dips in the graph, showing the absorption lines in the spectrum.
Blog #23: Astronomy + X = The Martian
This past month, I went to see the critically acclaimed film, The Martian. I was absolutely. blown. away (just like Mark Watney). I think this is the best film, bar none, that I have ever seen. It meshes the fields of science fiction, and actual science, in a way that it brings the literal future of human space travel at the forefront of American cinema.
The plot of the movie was very straight forward. 6 astronauts were on the Ares IV manned mission to Mars. A dust storm hit the crew, which caused Mark Watney, a botanist, to be impaled by a satellite dish and flown away from his crew members. The crew received a signal from Watney's suit saying he died, and the crew boarded the Hermes spacecraft and left without him. However, it turns out that the Mark Watney was still alive. The rest of the story is about how Watney tries to survive on a barren planet by growing his own food, communicating with NASA through the decommissioned Pathfinder mission, and eventually preparing to leave Mars. Along the way, the harsh Martian environment hampers his plans and make it nearly impossible for Watney to survive.
 |
| The first images of the Martian surface taken by Mariner 4 in 1965. |
 |
| Images by the Mariner 9 taken in 1971 that shocked the world to see craters and valleys on the Martian surface |
 |
| First color photo taken on the surface of Mars by Spirit in 2003 |
6 years later, the Mariner 9 mission in 1971 shocked the scientific community when they discovered that Mars had craters and valleys. Fast forwarding to 2003, the Mars Exploration Rovers, Spirit and Opportunity, took the first high definition colored photos from the surface of Mars that showed Mars' rugged surface in detail. Finally, right now, the Mars Science Laboratory, known as Curiosity, has sent back high def photos of Mars that show the same level of detail that was seen in the movie, The Martian. In fact, when looking at the pictures of Mars taken from Curiosity and those depicted in The Martian, the similarities are uncanny.
 |
| Images from Mars Science Laboratory, Curiosity |
 |
| Images of the surface of Mars depicted in The Martian. |
The other amazing feat that the Martian was able to pull off was that all the science explained in the movie was actual science, including the technologies being used in the movie. Every technology used in The Martian was the technology we currently have to take people to Mars, so no technology was fabricated.
So that begs the question, if the technology in the movie was real, and the surface of Mars was depicted as scientifically accurate, is it even fair to call this movie science fiction? Well that dust storm in the beginning of the movie was probably the most unrealistic aspect of the whole movie, since Mars doesn't have an atmosphere thick enough to sustain such forceful winds. However, it is very much possible for people to go to Mars, right now, if given enough funding to build everything necessary to send a manned mission. This film gave NASA and the scientific community the public relations boost it needed to re-ignite the interest in space travel to average population. Right now, NASA just started recruiting the next generation of astronauts, because a real manned mission to Mars is being planned for the 2030's. Hopefully in the real manned missions, NASA doesn't leave anyone behind, and if they do, at least that astronaut has a decent shot at survival if they simply watch The Martian.
Blog #22: Sound-Crossing Timescale - White Dwarf Goes Supernova
3) A white dwarf that exceeds the Chandrasekhar mass will start to fuse carbon in its interior, which releases a great deal of heat, which increases the internal pressure of the white dwarf. However, because the white dwarf is "trying" to support itself using degeneracy pressure, and increasing this pressure doesn't change the star's radius, the increasing temperature leads to more fusion, more energy, and a run-away fusion process is initiated.
Once the run-away fusion inside the white dwarf is "ignited", it propagates as a wave travelling outward at the speed of sound $c_s$. How much time does it take the flame to sweep outward across the radius of the white dwarf? This is also known as the "sound-crossing timescale."
How does this time scale relate to the density of the white dwarf?
In problem 2, the speed of sound, in terms of mass, and radius, is determined to be:
\begin{align*}
c_s = \left(\frac{GM}{5R}\right)^{\frac{1}{2}}
\end{align*}
We also know that speed is defined as a distance over time, which can rearranged to solve for time. The equation for time, therefore, is:
\begin{align*}
t_{sc} = \frac{R}{c_{s}}
\end{align*}
where $c_s$ is the speed of sound, $t_{sc}$ is the "sound-crossing timescale", and $R$ is the radius of the white dwarf. We can substitute the speed of sound $c_s = \left(\frac{GM}{5R}\right)^{\frac{1}{2}}$ and solve for the "sound-crossing timescale" as follows:
\begin{align*}
t_{sc} &= \frac{R}{c_{s}}\\
t_{sc} &= \frac{R}{ \left(\frac{GM}{5R}\right)^{\frac{1}{2}}}\\
t_{sc} &= \left(\frac{5R^3}{GM}\right)^{\frac{1}{2}}
\end{align*}
In order to see how the sound-crossing timescale relates to density, we can use dimensional analysis:
\begin{align*}
t_{sc} &= \left(\frac{5R^3}{GM}\right)^{\frac{1}{2}}\\
t_{sc} &= \left[\frac{m^3}{g}\right]^{\frac{1}{2}}\\
t_{sc} &\propto \sqrt{\frac{1}{\rho}}
\end{align*}
This shows that the sound-crossing timescale is inversely proportional to the square root of the density.
Once the run-away fusion inside the white dwarf is "ignited", it propagates as a wave travelling outward at the speed of sound $c_s$. How much time does it take the flame to sweep outward across the radius of the white dwarf? This is also known as the "sound-crossing timescale."
How does this time scale relate to the density of the white dwarf?
In problem 2, the speed of sound, in terms of mass, and radius, is determined to be:
\begin{align*}
c_s = \left(\frac{GM}{5R}\right)^{\frac{1}{2}}
\end{align*}
We also know that speed is defined as a distance over time, which can rearranged to solve for time. The equation for time, therefore, is:
\begin{align*}
t_{sc} = \frac{R}{c_{s}}
\end{align*}
where $c_s$ is the speed of sound, $t_{sc}$ is the "sound-crossing timescale", and $R$ is the radius of the white dwarf. We can substitute the speed of sound $c_s = \left(\frac{GM}{5R}\right)^{\frac{1}{2}}$ and solve for the "sound-crossing timescale" as follows:
\begin{align*}
t_{sc} &= \frac{R}{c_{s}}\\
t_{sc} &= \frac{R}{ \left(\frac{GM}{5R}\right)^{\frac{1}{2}}}\\
t_{sc} &= \left(\frac{5R^3}{GM}\right)^{\frac{1}{2}}
\end{align*}
In order to see how the sound-crossing timescale relates to density, we can use dimensional analysis:
\begin{align*}
t_{sc} &= \left(\frac{5R^3}{GM}\right)^{\frac{1}{2}}\\
t_{sc} &= \left[\frac{m^3}{g}\right]^{\frac{1}{2}}\\
t_{sc} &\propto \sqrt{\frac{1}{\rho}}
\end{align*}
This shows that the sound-crossing timescale is inversely proportional to the square root of the density.
Blog #21: White Dwarf Goes Type 1a Supernova
1) White dwarfs are supported internally against the force of gravity by "electron degeneracy" pressure (encountered in Astronomy 16). The maximum mass that can be supported by this exotic form of pressure is the 1.4 $M_{\odot}$ (also known as the Chandrasekhar Mass). The radius of our white dwarf is approximately twice the radius of the Earth, or ~ $12 \times 10^8$ cm.
Given the mass, M, and radius, R, derive an algebraic expression for the internal pressure of a white dwarf with these properties. Start with the Virial theorem, recall that the internal kinetic energy per particle is $\frac{3}{2}kT$, where $k = 1.4 \times 10^{-16}$ erg $K^{-1}$ is the Boltzmann constant. You can also assume the interior of the white dwarf is an ideal gas, and its mass is uniformly distributed.
The internal kinetic energy of a single particle is $\frac{3}{2}kT$, so the kinetic energy of the total system of $N$ particles is going to be:
\begin{align*}
K = \frac{3}{2}NkT
\end{align*}
Since the interior of a star can be considered an ideal gas, we can use the ideal gas relation, $PV = NkT$, where $V = \frac{4}{3}\pi R^3$ is the volume of the star. Using this equation, we can solve for $K$ in terms of pressure and volume as follows:
\begin{align*}
K &= \frac{3}{2}NkT\\
K &= \frac{3}{2}PV\\
K &= \frac{3}{2}P \left(\frac{4}{3}\pi R^3\right)\\
K &= 2\pi PR^3
\end{align*}
We also know that the total potential energy in the system from the previous worksheet is:
\begin{align*}
U = -\frac{3GM^2}{5R}
\end{align*}
Using the Virial Theorem, we can solve for the internal pressure as follows:
\begin{align*}
K &= -\frac{1}{2}U\\
2\pi PR^3 &= -\frac{1}{2}\left(-\frac{3GM^2}{5R}\right)\\
P &= \frac{\frac{3GM^2}{10R}}{2\pi R^3}\\
P &= \frac{3GM^2}{20\pi R^4}
\end{align*}
The internal pressure inside the white dwarf star is: $P = \frac{3GM^2}{20\pi R^4}$.
Given the mass, M, and radius, R, derive an algebraic expression for the internal pressure of a white dwarf with these properties. Start with the Virial theorem, recall that the internal kinetic energy per particle is $\frac{3}{2}kT$, where $k = 1.4 \times 10^{-16}$ erg $K^{-1}$ is the Boltzmann constant. You can also assume the interior of the white dwarf is an ideal gas, and its mass is uniformly distributed.
The internal kinetic energy of a single particle is $\frac{3}{2}kT$, so the kinetic energy of the total system of $N$ particles is going to be:
\begin{align*}
K = \frac{3}{2}NkT
\end{align*}
Since the interior of a star can be considered an ideal gas, we can use the ideal gas relation, $PV = NkT$, where $V = \frac{4}{3}\pi R^3$ is the volume of the star. Using this equation, we can solve for $K$ in terms of pressure and volume as follows:
\begin{align*}
K &= \frac{3}{2}NkT\\
K &= \frac{3}{2}PV\\
K &= \frac{3}{2}P \left(\frac{4}{3}\pi R^3\right)\\
K &= 2\pi PR^3
\end{align*}
We also know that the total potential energy in the system from the previous worksheet is:
\begin{align*}
U = -\frac{3GM^2}{5R}
\end{align*}
Using the Virial Theorem, we can solve for the internal pressure as follows:
\begin{align*}
K &= -\frac{1}{2}U\\
2\pi PR^3 &= -\frac{1}{2}\left(-\frac{3GM^2}{5R}\right)\\
P &= \frac{\frac{3GM^2}{10R}}{2\pi R^3}\\
P &= \frac{3GM^2}{20\pi R^4}
\end{align*}
The internal pressure inside the white dwarf star is: $P = \frac{3GM^2}{20\pi R^4}$.
Blog #20: The Hubble Tuning Fork
In 1926, Edwin Hubble, for whom the Hubble Space Telescope is named after, saw that the different galaxies seen in the night sky had different shapes and structures. Therefore, he created a classification system for galaxies that split these galaxies into 3 categories, namely: (1) Elliptical galaxies, (2) Spiral galaxies, and (3) Irregular galaxies. Each of these three classifications had further sub-divisions, which ended up creating a system of galaxy classification known as the Hubble Tuning Fork. The Hubble Tuning Fork shows how the different categories of galaxies and their respective subdivisions are all related to each other in terms of shape and structure. In order to understand the Tuning Fork, let's first discuss the subdivisions of the Elliptical, Spiral, and Irregular galaxies.
Elliptical Galaxies:
Elliptical galaxies get their name for their uniform, elliptical shape. The elliptical galaxies are classified by the eccentricity of the ellipse, where E0 is an elliptical galaxy with no eccentricity, and is therefore perfectly round, to E7, where the eccentricity is very high. The "E" in the E0 - E7 range stands for "elliptical", specifying only elliptical galaxies. Look at the diagram below to get a sense of the categories of Elliptical galaxies:
Spiral Galaxies:
Spiral galaxies get their name for their characteristic bands, or "arms" of gas and dust that swirl around the center of the galaxy. Spiral galaxies come in two subcategories: (1) without a central bar $S$ (top photo), (2) with a central bar $SB$ (bottom photo).
These two categories are further divided by the tightness of their spiral arms, where spiral galaxies with very tight spirals are in category "a", galaxies with medium spirals in category "b", and galaxies with loose spirals are in category "c". See below for the differences between these galaxies:
There is also another category of spiral galaxies, where there are no spirals at all! This category, S0, is characterized by the fact that there is a bulge, and a distinct disk, unlike elliptical galaxies, but the disk doesn't have any arms. An example of a S0 galaxy is the Sombrero Galaxy:
Irregular Galaxies:
Irregular galaxies are those that do not belong to either spiral or elliptical galaxies, and can take on a range of shapes, from shapeless blobs, to distorted spirals.
Hubble's Tuning Fork:
All these different types of galaxies can be rearranged to show how their shapes relate to each other in the form of a tuning fork. Here's a tuning fork I made:
Elliptical Galaxies:
Elliptical galaxies get their name for their uniform, elliptical shape. The elliptical galaxies are classified by the eccentricity of the ellipse, where E0 is an elliptical galaxy with no eccentricity, and is therefore perfectly round, to E7, where the eccentricity is very high. The "E" in the E0 - E7 range stands for "elliptical", specifying only elliptical galaxies. Look at the diagram below to get a sense of the categories of Elliptical galaxies:
Spiral Galaxies:
 |
 |
Spiral galaxies get their name for their characteristic bands, or "arms" of gas and dust that swirl around the center of the galaxy. Spiral galaxies come in two subcategories: (1) without a central bar $S$ (top photo), (2) with a central bar $SB$ (bottom photo).
These two categories are further divided by the tightness of their spiral arms, where spiral galaxies with very tight spirals are in category "a", galaxies with medium spirals in category "b", and galaxies with loose spirals are in category "c". See below for the differences between these galaxies:
There is also another category of spiral galaxies, where there are no spirals at all! This category, S0, is characterized by the fact that there is a bulge, and a distinct disk, unlike elliptical galaxies, but the disk doesn't have any arms. An example of a S0 galaxy is the Sombrero Galaxy:
Irregular Galaxies:
Irregular galaxies are those that do not belong to either spiral or elliptical galaxies, and can take on a range of shapes, from shapeless blobs, to distorted spirals.
Hubble's Tuning Fork:
All these different types of galaxies can be rearranged to show how their shapes relate to each other in the form of a tuning fork. Here's a tuning fork I made:
Blog #19: An Explosive Signature of Explosive Galaxies
We learned in Astro 16 the conditions that allow a star to form. Ideally, stars form in large clouds of gas and dust.
We learned in Astro 17 that galaxies can collide. For example, the Milky Way Galaxy and the Andromeda Galaxy are in a direct collision course with each other.
So what does star formation and galaxy mergers have in common? The answer lies in what is known as Gamma-Ray Bursts (GRBs). Gamma-Ray Bursts are narrow beams of intense radiation that are the most powerful event in the Universe aside from the Big Bang itself. They occur when a truly massive star, with a mass of at least 15 $M_{\odot}$ dies and collapses in on itself. When a star of this size runs out of fuel and dies, it explodes in what is known as a "hypernova", which are considered to be substantially more energetic explosions than standard supernovae. In addition to the hypernova, the collapsing star also releases Gamma Ray Bursts.
Another important thing to note is that stars this massive have relatively short life-spans since they use up all of their fuel much faster than smaller stars. Therefore, galaxies that seem to have a lot of GRB's signal that the galaxy is forming new stars. In galaxies, there are two types of gasses: (1) hot, ionized gas, and (2) cold, neutral gas. Star formation, especially for the really massive stars, happens in regions of cold, neutral gas, which is hard to detect. This article shows how researchers observed this cold, neutral gas in galaxies within a 100 million light years away, which is about 50 times further than the Andromeda Galaxy. Because of their relative close distance to us, researchers were able to use the radio frequency on the 21 cm line to detect this cold neutral gas. This region of the galaxy had a lot of GRBs.
Looking at a density map of the cold, neutral gas, it seemed that the gas was disturbed, and was outside the main disk of the galaxy. This suggests that the galaxy had collided with another, smaller galaxy, which resulted in the scattered cold, neutral gas. The other important conclusion was that the collision could have "shock-compressed" the gas, which sparked the formation of massive stars. Since these massive stars give off GMRs when they die, the presence of GMRs could mean that a galaxy is the result of a galaxy collision and merger.
Unfortunately, this was observed only in galaxies that are within a 100 million miles from us. Since the cold, neutral gas is hard to detect because the 21 cm emission line is harder to see further out, it will be harder to conclude that the presence of GRBs are predictive of galaxy mergers. However, this research is on the path to show not only that GMRs are indicative of galaxies that are forming new stars, but the composition and history of the galaxy in which they reside.
We learned in Astro 17 that galaxies can collide. For example, the Milky Way Galaxy and the Andromeda Galaxy are in a direct collision course with each other.
So what does star formation and galaxy mergers have in common? The answer lies in what is known as Gamma-Ray Bursts (GRBs). Gamma-Ray Bursts are narrow beams of intense radiation that are the most powerful event in the Universe aside from the Big Bang itself. They occur when a truly massive star, with a mass of at least 15 $M_{\odot}$ dies and collapses in on itself. When a star of this size runs out of fuel and dies, it explodes in what is known as a "hypernova", which are considered to be substantially more energetic explosions than standard supernovae. In addition to the hypernova, the collapsing star also releases Gamma Ray Bursts.
Another important thing to note is that stars this massive have relatively short life-spans since they use up all of their fuel much faster than smaller stars. Therefore, galaxies that seem to have a lot of GRB's signal that the galaxy is forming new stars. In galaxies, there are two types of gasses: (1) hot, ionized gas, and (2) cold, neutral gas. Star formation, especially for the really massive stars, happens in regions of cold, neutral gas, which is hard to detect. This article shows how researchers observed this cold, neutral gas in galaxies within a 100 million light years away, which is about 50 times further than the Andromeda Galaxy. Because of their relative close distance to us, researchers were able to use the radio frequency on the 21 cm line to detect this cold neutral gas. This region of the galaxy had a lot of GRBs.
Looking at a density map of the cold, neutral gas, it seemed that the gas was disturbed, and was outside the main disk of the galaxy. This suggests that the galaxy had collided with another, smaller galaxy, which resulted in the scattered cold, neutral gas. The other important conclusion was that the collision could have "shock-compressed" the gas, which sparked the formation of massive stars. Since these massive stars give off GMRs when they die, the presence of GMRs could mean that a galaxy is the result of a galaxy collision and merger.
Unfortunately, this was observed only in galaxies that are within a 100 million miles from us. Since the cold, neutral gas is hard to detect because the 21 cm emission line is harder to see further out, it will be harder to conclude that the presence of GRBs are predictive of galaxy mergers. However, this research is on the path to show not only that GMRs are indicative of galaxies that are forming new stars, but the composition and history of the galaxy in which they reside.
Monday, October 19, 2015
Blog 18: Normal Galaxies and the Tully Fisher Relation
4) Over time, from measurements of the photometric and kinematic properties of normal galaxies, it became apparent that there exist correlations between the amount of motion of objects in the galaxy and the galaxy's luminosity. In this problem, we'll explore one of these relationships.
Spiral galaxies obey the Tully-Fisher Relation:
\begin{align}
L \sim v_{max}^4
\end{align}
where $L$ is total luminosity, and $v_{max}$ is the maximum observed rotational velocity. This relation was initially discovered observationally, but it is not hard to derive in a crude way:
(a) Assume that $v_{max} \sim \sigma$ (is this a good approximation?). Given what you know about the Virial Theorem, how should $v_{max}$ relate to the mass and radius of the Galaxy?
Using the Virial Theorem, we derived in problem 3 that the mass, radius, and stellar velocity can be related with the equation:
\begin{align}
M \approx \frac{\sigma^2 R}{G}
\end{align}
We can rearrange this equation to solve for $\sigma^2$ as follows:
\begin{align}
M \approx \frac{\sigma^2 R}{G}\\
\sigma^2 = \frac{M G}{R}
\end{align}
Let's assume that $v_{max} \sim \sigma$, which is a good assumption because we can think of $v_{max}$ as the average velocity of stars in the galaxy, and that they don't vary in speed too mych. Using this assumption, we can use the equation above to show the relationship between $v_{max}$ to the mass and radius of the galaxy as follows:
\begin{align}
v_{max}^2 \sim \frac{M}{R}
\end{align}
(b and c) To proceed from here, you need some handy observational facts. First, all spiral galaxies have a similar disk surface brightnesses ($\langle I \rangle = \frac{L}{R^2}$) (Freeman's Law). Second, they also have similar total mass-to-light ratios $(\frac{M}{L})$.
Use some squiggly math (drop the constants and use $\sim$ instead of $=$) to find the Tully-Fisher relationship.
Based on the Freeman's Law, we know that $\langle I \rangle = \frac{L}{R^2}$, which can be re-written as:
\begin{align}
I &= \frac{L}{R^2}\\
IR^2 &= L\\
R &\sim \sqrt{L}\\
\end{align}
Since we also know the mass-luminosity ratio as $\frac{M}{L}$, we can say that $M \sim L$.
We can plug these values for the $v_{max}$ derived in part (a) to get the Tully-Fisher Relation:
\begin{align}
v_{max}^2 &\sim \frac{M}{R} &\\
v_{max}^2 &\sim \frac{M}{\sqrt{L}} \rightarrow \text{Substituting for Freeman's Law}&\\
v_{max}^2 &\sim \frac{L}{\sqrt{L}} \rightarrow \text{Substituting for mass-ratio relation}&\\
v_{max}^2 &\sim \sqrt{L} &\\
v_{max}^4 &\sim L &\\
\end{align}
This is the Tully-Fisher Relation!
(d) It turns out the Tully-Fisher Relation is so well-obeyed that it can be used as a standard candle, just like the Cepheids and Supernova Ia seen previously. In the B-band ($\lambda_{cen} \sim 445$ nm, blue light), this relation is approximately:
\begin{align}
M_B = -10 \log \left(\frac{v_{max}}{\text{km/s}}\right) + 3
\end{align}
Suppose you observe a spiral galaxy with apparent, extinction-corrected magnitude B = 13 mag. You perform longslit optical spectroscopy, obtaining a maximum rotational velocity of 400 km/s for this galaxy. How distant do you infer this spiral galaxy to be?
We can calculate the distance of the spiral galaxy using its apparent and absolute magnitudes with the distance modulus equation:
\begin{align}
d = 10^{(\frac{m - M_B}{5} + 1)}
\end{align}
where $d$ is the distance to the spiral galaxy, $M_B$ is the absolute magnitude of the galaxy, and $m$ is the apparent magnitude of the galaxy. We are told that the apparent, extinction corrected magnitude is $m = B = 13$. Therefore, we just need to find the absolute magnitude of the galaxy.
We can solve for the absolute magnitude of the galaxy using the relation:
\begin{align}
M_B = -10 \log \left(\frac{v_{max}}{\text{km/s}}\right) + 3
\end{align}
where $M_B$ is the absolute magnitude of the galaxy, and $v_{max}$ is the maximum rotational velocity of this galaxy, which is told to be $v_{max}= 400$ km/s. We can plug in maximum rotational velocity into this equation to solve for the absolute magnitude, like so:
\begin{align}
M_B &= -10 \log \left(\frac{v_{max}}{\text{km/s}}\right) + 3\\
M_B &= -10 \log \left(\frac{400 \text{ km/s}}{\text{km/s}}\right) + 3\\
M_B &= -23
\end{align}
Now that we have both the apparent magnitude, $m = 13$ and the absolute magnitude, $M_B = -23$, we can plug these magnitudes into the distance modulus and solve for the distance, $d$ as follows:
\begin{align}
d &= 10^{(\frac{m - M_B}{5} + 1)}\\
d &= 10^{(\frac{13 + 23}{5} + 1)}\\
d &= 1.6 \times 10^8 \text{ pc}
\end{align}
Therefore, the spiral galaxy is $1.6 \times 10^8$ pc away!
Spiral galaxies obey the Tully-Fisher Relation:
\begin{align}
L \sim v_{max}^4
\end{align}
where $L$ is total luminosity, and $v_{max}$ is the maximum observed rotational velocity. This relation was initially discovered observationally, but it is not hard to derive in a crude way:
(a) Assume that $v_{max} \sim \sigma$ (is this a good approximation?). Given what you know about the Virial Theorem, how should $v_{max}$ relate to the mass and radius of the Galaxy?
Using the Virial Theorem, we derived in problem 3 that the mass, radius, and stellar velocity can be related with the equation:
\begin{align}
M \approx \frac{\sigma^2 R}{G}
\end{align}
We can rearrange this equation to solve for $\sigma^2$ as follows:
\begin{align}
M \approx \frac{\sigma^2 R}{G}\\
\sigma^2 = \frac{M G}{R}
\end{align}
Let's assume that $v_{max} \sim \sigma$, which is a good assumption because we can think of $v_{max}$ as the average velocity of stars in the galaxy, and that they don't vary in speed too mych. Using this assumption, we can use the equation above to show the relationship between $v_{max}$ to the mass and radius of the galaxy as follows:
\begin{align}
v_{max}^2 \sim \frac{M}{R}
\end{align}
(b and c) To proceed from here, you need some handy observational facts. First, all spiral galaxies have a similar disk surface brightnesses ($\langle I \rangle = \frac{L}{R^2}$) (Freeman's Law). Second, they also have similar total mass-to-light ratios $(\frac{M}{L})$.
Use some squiggly math (drop the constants and use $\sim$ instead of $=$) to find the Tully-Fisher relationship.
Based on the Freeman's Law, we know that $\langle I \rangle = \frac{L}{R^2}$, which can be re-written as:
\begin{align}
I &= \frac{L}{R^2}\\
IR^2 &= L\\
R &\sim \sqrt{L}\\
\end{align}
Since we also know the mass-luminosity ratio as $\frac{M}{L}$, we can say that $M \sim L$.
We can plug these values for the $v_{max}$ derived in part (a) to get the Tully-Fisher Relation:
\begin{align}
v_{max}^2 &\sim \frac{M}{R} &\\
v_{max}^2 &\sim \frac{M}{\sqrt{L}} \rightarrow \text{Substituting for Freeman's Law}&\\
v_{max}^2 &\sim \frac{L}{\sqrt{L}} \rightarrow \text{Substituting for mass-ratio relation}&\\
v_{max}^2 &\sim \sqrt{L} &\\
v_{max}^4 &\sim L &\\
\end{align}
This is the Tully-Fisher Relation!
(d) It turns out the Tully-Fisher Relation is so well-obeyed that it can be used as a standard candle, just like the Cepheids and Supernova Ia seen previously. In the B-band ($\lambda_{cen} \sim 445$ nm, blue light), this relation is approximately:
\begin{align}
M_B = -10 \log \left(\frac{v_{max}}{\text{km/s}}\right) + 3
\end{align}
Suppose you observe a spiral galaxy with apparent, extinction-corrected magnitude B = 13 mag. You perform longslit optical spectroscopy, obtaining a maximum rotational velocity of 400 km/s for this galaxy. How distant do you infer this spiral galaxy to be?
We can calculate the distance of the spiral galaxy using its apparent and absolute magnitudes with the distance modulus equation:
\begin{align}
d = 10^{(\frac{m - M_B}{5} + 1)}
\end{align}
where $d$ is the distance to the spiral galaxy, $M_B$ is the absolute magnitude of the galaxy, and $m$ is the apparent magnitude of the galaxy. We are told that the apparent, extinction corrected magnitude is $m = B = 13$. Therefore, we just need to find the absolute magnitude of the galaxy.
We can solve for the absolute magnitude of the galaxy using the relation:
\begin{align}
M_B = -10 \log \left(\frac{v_{max}}{\text{km/s}}\right) + 3
\end{align}
where $M_B$ is the absolute magnitude of the galaxy, and $v_{max}$ is the maximum rotational velocity of this galaxy, which is told to be $v_{max}= 400$ km/s. We can plug in maximum rotational velocity into this equation to solve for the absolute magnitude, like so:
\begin{align}
M_B &= -10 \log \left(\frac{v_{max}}{\text{km/s}}\right) + 3\\
M_B &= -10 \log \left(\frac{400 \text{ km/s}}{\text{km/s}}\right) + 3\\
M_B &= -23
\end{align}
Now that we have both the apparent magnitude, $m = 13$ and the absolute magnitude, $M_B = -23$, we can plug these magnitudes into the distance modulus and solve for the distance, $d$ as follows:
\begin{align}
d &= 10^{(\frac{m - M_B}{5} + 1)}\\
d &= 10^{(\frac{13 + 23}{5} + 1)}\\
d &= 1.6 \times 10^8 \text{ pc}
\end{align}
Therefore, the spiral galaxy is $1.6 \times 10^8$ pc away!
Blog 17: Modeling Normal Galaxies Using the Virial Theorem
3) One of the most useful equations in astronomy is an extremely simple relationship known as the Virial Theorem. It can be used to derive Kepler's Third Law, measure the mass of a cluster of stars, or the temperature and brightness of a newly-formed planet. The Virial Theorem applies to a system of particles in equilibrium that are bound by a force that is defined by an inverse central-force law $(F \propto \frac{1}{r^\alpha})$. It relates the kinetic (or thermal) energy of a system $K$, to the potential energy, $U$, giving:
\begin{align}
k = -\frac{1}{2}U
\end{align}
(a) Consider a spherical distribution of $N$ particles, each with a mass $m$. The distribution has total mass $M$ and a total radius, $R$. Convince yourself that the total potential energy, $U$, is approximately
\begin{align}
U \approx -\frac{GM^2}{R}
\end{align}
You can derive or look up the actual numerical constant out front. But in general astronomy you don't need this prefactor, which is of order unity.
The total potential energy of the system can be modeled by the following equation:
\begin{align}
U = -\frac{G(\sum_{i = 1}^{N} m_i) (\sum_{j = 1}^{N} m_j)}{ r_{i,j}}
\end{align}
where $G$ is the gravitational constant, $m$ is the mass of a single particle in the spherical distribution, and $r$ is the radius between two particles, $i, j$ within a distribution of $N$ particles. Since there are $N$ particles, and the radius between two particles can be averaged out to the radius of the entire sphere, $R$, the equation above can be simplified to:
\begin{align}
U &= -\frac{G (N \times m) (N \times m)}{ R}\\
U &= -\frac{G N^2 m^2}{ R}\\
\end{align}
The total mass of the distribution, $M$ is the equivalent of the mass of a single particle, $m$, multiplied by the total number number of particles, $N$, given by the formula, $M = Nm$. We can substitute this into the previous equation to solve for the total potential energy as:
\begin{align}
U &= -\frac{G M^2}{ R}
\end{align}
However, the problem says that there should be a prefactor constant that is not included. Since it is a unity factor, it can be ignored, so the true potential energy can be modeled as:
\begin{align}
U \approx -\frac{G M^2}{ R}
\end{align}
(b) Now let's figure out what $K$ is equal to. Consider a bound spherical distribution of $N$ particles (perhaps stars in a globular cluster), each of mass $m$, and each moving away with a velocity of $v_i$ with respect to the center of mass. If these stars are far away in space, their individual velocity vectors are very difficult to measure directly. Generally, it is much easier to measure the scatter around the mean velocity if the system along our line of sight, the velocity scatter $\sigma^2$. Show that the kinetic energy of the system is:
\begin{align}
K = N \frac{3}{2}m \sigma^2
\end{align}
The equation for the total kinetic energy of a systemis:
\begin{align}
K = \frac{1}{2}M v^2
\end{align}
However, since the total mass $M$ is composite of the individual particle mass, $m$, times the total number of particles there are, $N$, $M = Nm$, and the equation for total kinetic energy can be written as:
\begin{align}
K = N \frac{1}{2}m v^2
\end{align}
Since the velocity scattering happens in a three dimensional frame, the velocity-squared of the system can be represented as:
\begin{align}
v^2 = 3 \sigma^2
\end{align}
Substituting this into the previous equation, you get a total kinetic energy of the system as:
\begin{align}
K = N \frac{3}{2}m \sigma^2
\end{align}
(c) Use the Virial Theorem to show that the total mass of, say, a globular cluster of radius $R$ and stellar velocity dispersion $\sigma$ is (to some prefactor of order unity):
\begin{align}
M \approx \frac{\sigma^2 R}{G}
\end{align}
Let's keep track of everything we know:
(1) The Virial Theorem:
\begin{align}
K = -\frac{1}{2}U
\end{align}
(2) The total kinetic energy of a system:
\begin{align}
K = N \frac{3}{2}m \sigma^2
\end{align}
(3) The total potential energy of a system:
\begin{align}
U \approx -\frac{G M^2}{ R}
\end{align}
(4) The relationship between the total mass, $M$, of a system, and the mass of a single particle of a system, $m$ for $N$ particles:
\begin{align}
M = Nm
\end{align}
Using this information, we can solve for the mass of the globular cluster, $M$, using the Virial Theorem, as follows:
\begin{align}
K &= -\frac{1}{2}U\\
N \frac{3}{2}m \sigma^2 &\approx -\frac{1}{2}\left(-\frac{G M^2}{ R}\right)\\
3 M \sigma^2 &\approx \frac{G M^2}{ R}\\
3 \sigma^2 &\approx \frac{G M}{ R}\\
\frac{3 \sigma^2 R}{G} &\approx M\\
M &\approx \frac{3 \sigma^2 R}{G}
\end{align}
We have derived an approximation for $M$ in terms of the velocity scatter $\sigma^2$, the radius $R$ of the globular cluster, and the gravitational constant $G$ as was intended. However, there is the coefficient 3 included that signifies the 3 dimensions of space. We can remove the coefficient to show the relationship as:
\begin{align}
M &\approx \frac{\sigma^2 R}{G}
\end{align}
Wednesday, October 7, 2015
Blog #16: The Great Debate (Shapley-Curtis Debate)
 |
| Harlow Shapley (left) and Heber Curtis (right) Source: http://education.ezinemark.com/top-10-science-debates-in-history-773690137c70.html |
At the heart of this debate was the question of the true scale of the Universe. Shapley asserted that Andromeda could not have been a separate galaxy because otherwise, by its apparent size and magnitude, it would have to be $10^8$ light years away. That distance seemed outlandish because it was further away than any other object in the Universe (based on the understanding at the time). Furthermore, astronomer Adriaan vas Maanen claimed that he measured what is now known as the Pinwheel Galaxy to be rotating within a timescale of years. Extrapolating the rate of rotation in the circumstance that the Pinwheel Galaxy was a separate galaxy would require the orbital velocity to be faster than the speed of light, thus defying the laws of physics. Finally, Shapley argued that he had observed a nova in the Andromeda "nebula" that outshone the entire nebula. Considering its brightness relative to all of Andromeda, if Andromeda was a separate galaxy, the amounts of energy release in the nova would be unimaginable.
Heber Curtis, however argued that it was strange that there were more "novas" inside the region marked as Andromeda, than at any other part of the Milky Way. This would suggest that Andromeda must be it's own galaxy far away. Curtis also cited dust clouds found in other galaxies similar to those in the Milky Way. Finally, he observed massive doppler shifts in objects he thought were galaxies, thus giving these galaxies a signature age that was different from that of the Milky Way.
However, the argument presented by Adriaan vas Maanen regarding the observation of rotating pinwheels in the Pinwheel Galaxy seemed to be the arbiter of the debate, since Curtis agreed that if the Pinwheel Galaxy was indeed rotating within an observational scale of years, then it would suggest that the orbital velocity of the galaxy was faster than the speed of light, and therefore his hypothesis would be wrong. It was later observed that, in fact, the rotation of the Pinwheel Galaxy could not be observed within the lifespan of humans, thus giving credibility to Curtis' argument.
Curtis' arguments were reconsidered when Edwin Hubble discovered Cepheid Variables in Andromeda and other "nebulae", which were calculated to be much further away than the Milky Way, suggesting that the Universe was, in fact, comprised of multiple galaxies, not just the Milky Way. Shapley's argument regarding the massively bright nova in Andromeda as having too much energy to be feasible was debunked when the existence of supernova was proven. Supernova do outside their respective galaxy, and are high-energy events measured in scales unimaginable in the early days of astronomy.
This debate provided the platform to discuss the "Scale of the Universe" to get a better insight of the immensity of the Universe we reside in. This debate showed how different points of view in astronomy were formulated based on the best information available at the time. Now we know that the Universe is filled with hundreds of billions of galaxies, and the Milky Way is just one of them.
Source: https://en.wikipedia.org/wiki/Great_Debate_(astronomy)
Monday, October 5, 2015
Blog #15: Worksheet 5.2- Cepheid Relations
Let's analyze real data about Cepheid Relations using Henrietta Swan Leavitt's actual date from 1908!
Included was a CSV file with the data.
(a) The data file, "Cepheid_variables.csv", contains data for 25 Cepheid variables located in the Small Magellanic Cloud (SMC). Each line contains a specific Cepheid's (1) ID number, (2) Maximum apparent magnitude, (3) Minimum apparent magnitude and (4) Period. Calculate the mean apparent magnitude for each Cepheid.
In order to find the mean apparent magnitude, I simply took the minimum and maximum apparent magnitudes of each Cepheid and averaged the two.
Since there were many individual Cepheids, I decided to get the mean apparent magnitude using Excel's "AVERAGE" function using the maximum and minimum apparent magnitudes as the input. The resulting mean apparent magnitude for each Cepheid is shown in the blue column in Table 1.
(b) The distance to the SMC is about 60 kpc, where kpc = 1000 pc. Convert your mean apparent magnitudes into mean absolute magnitudes. Plot the Cepheid mean absolute magnitudes as a function of period. This plot should look exponential.
In order to get the mean absolute magnitude, I had to convert the mean apparent magnitude into the mean absolute magnitude using the equation:
\begin{align}
M = m - 5 \log {\left(\frac{d}{10 \text{ pc}}\right)}
\end{align}
where $m$ is the mean apparent magnitude, and $d$ is the distance to the SMC, which is 6000 pc. The values of this conversion for each Cepheid to its mean absolute magnitude are in Table 1 in the orange column.
The plot looks as follows:
It does indeed look exponential.
(3) It is often handy to plot exponential (or power-law) functions with one or more logarithmic axes, which "straighten out" the data. Magnitudes are already exponential, so we don't need to adjust that axis. Plot the Cepheid mean absolute magnitudes as a function of $\log_{10}$(Period). Verify that the plot now looks linear.
In order to plot the mean absolute magnitude against a linear axis, I converted the period of each Cepheid into the log of the period, as shown in the purple column in Table 1, and got the following chart:
(4) and (5) Now that the data looks linear, we can estimate the parameters of the linear relation, $M_v(P) = A \log_{10}(\text{Period})+ B$. $A$ and $B$ are "free parameters" that allow the function to match the data.
You can determine the precise values of A and B by minimizing the difference between the observed points and model using the metric:
\begin{align}
\chi^2 \equiv \sum_{i=1}^{25}(O_i - C_i)^2
\end{align}
where $O_i$ is the observed value and $C_i$ is the predicted (model) value.
In this case, $A = -2.0332$ and $B = -2.7276$.
Included was a CSV file with the data.
(a) The data file, "Cepheid_variables.csv", contains data for 25 Cepheid variables located in the Small Magellanic Cloud (SMC). Each line contains a specific Cepheid's (1) ID number, (2) Maximum apparent magnitude, (3) Minimum apparent magnitude and (4) Period. Calculate the mean apparent magnitude for each Cepheid.
In order to find the mean apparent magnitude, I simply took the minimum and maximum apparent magnitudes of each Cepheid and averaged the two.
 |
| Table 1: Cepheid Relations Chart |
(b) The distance to the SMC is about 60 kpc, where kpc = 1000 pc. Convert your mean apparent magnitudes into mean absolute magnitudes. Plot the Cepheid mean absolute magnitudes as a function of period. This plot should look exponential.
In order to get the mean absolute magnitude, I had to convert the mean apparent magnitude into the mean absolute magnitude using the equation:
\begin{align}
M = m - 5 \log {\left(\frac{d}{10 \text{ pc}}\right)}
\end{align}
where $m$ is the mean apparent magnitude, and $d$ is the distance to the SMC, which is 6000 pc. The values of this conversion for each Cepheid to its mean absolute magnitude are in Table 1 in the orange column.
The plot looks as follows:
 |
| Figure 1: Mean Absolute Magnitude vs. Period (days) |
It does indeed look exponential.
(3) It is often handy to plot exponential (or power-law) functions with one or more logarithmic axes, which "straighten out" the data. Magnitudes are already exponential, so we don't need to adjust that axis. Plot the Cepheid mean absolute magnitudes as a function of $\log_{10}$(Period). Verify that the plot now looks linear.
In order to plot the mean absolute magnitude against a linear axis, I converted the period of each Cepheid into the log of the period, as shown in the purple column in Table 1, and got the following chart:
 |
| Figure 2: Mean Absolute Magnitude vs. log(Period) in Days |
(4) and (5) Now that the data looks linear, we can estimate the parameters of the linear relation, $M_v(P) = A \log_{10}(\text{Period})+ B$. $A$ and $B$ are "free parameters" that allow the function to match the data.
You can determine the precise values of A and B by minimizing the difference between the observed points and model using the metric:
\begin{align}
\chi^2 \equiv \sum_{i=1}^{25}(O_i - C_i)^2
\end{align}
where $O_i$ is the observed value and $C_i$ is the predicted (model) value.
In this case, $A = -2.0332$ and $B = -2.7276$.
Blog #14: Worksheet 5.1 - Extragalactic Distance Ladder
(a) Suppose you are observing two stars, Star A and Star B. Star A is 3 magnitudes fainter than Star B. How much longer do you need to observe Star A to collect the same amount of energy in your detector as you do for Star B?
In order to relate the magnitudes of a star to their observation time, we need to look at how much light we are receiving per unit of time for each star. Flux is the measure of energy per time per area, and in this case, the "per area" is the same, since we are looking at the perceived flux reaching our eyes (or telescope if we are observing with a telescope). Since we are looking to see how much longer it will take for the light of Star A to reach us, we can use the proportion of fluxes of stars A and B using the relation:
\begin{align}
\frac{F_{B}}{F_{A}} \approx 2.5^{(m_A - m_B)}
\end{align}
where in this case, the factor $(m_A - m_B)$ is the difference in magnitudes between Star A and Star B, which we know to be 3. Therefore, the ratio of fluxes of Star A and Star B is:
\begin{align}
\frac{F_{B}}{F_{A}} \approx 2.5^{(m_A - m_B)} \approx 16
\end{align}
Since flux is related linearly with time, since the flux of Star A is 16 times less than the flux of Star B, it will take 16 times longer for Star A to collect the same amount of energy as Star B.
(b) Stars have both an apparent magnitude, $m$, which is how bright they appear from Earth. They also have an absolute magnitude, $\mathcal{M}$, which is the apparent magnitude a star would have at $d = 10$ pc. How does the apparent magnitude, $m$, of a star with absolute magnitude $\mathcal{M}$, depend on its distance, $d$ away from you?
Let's consider a star with an apparent magnitude, $m$, and an absolute magnitude, $\mathcal{M}$. The flux of the star is determined by the equation :
\begin{align}
F = \frac{\text{Luminosity}}{\text{Area}} = \frac{L}{\pi d^2}
\end{align}
where, L is the inherent luminosity of the star, and $d$ is the distance the star is away from you.
Knowing this, we can use the equation for the ratio of fluxes to determine the relationship between the apparent magnitude of a star, $m$ at a distance $d$, and the apparent magnitude of the same star if it were located $d = 10$ pc away.
The flux of the star at any given distance is:
\begin{align}
F_m = \frac{L}{4\pi d ^2}
\end{align}
The flux of a star at exactly $d = 10$ pc is as follows:
\begin{align}
F_M = \frac{L}{4\pi (10 \text{ pc}) ^2}
\end{align}
To solve for the relationship between the the apparent magnitude, $m$, and the absolute magnitude, $\mathcal{M}$ the ratio of the fluxes of these two values are as follows:
\begin{align}
\frac{F_{m}}{F_{M}} &= 10^{0.4(M - m)}\\
\frac{\frac{L}{4\pi d ^2}}{\frac{L}{4\pi (10 \text{ pc}) ^2}} &= 10^{0.4(M - m)}\\
\left(\frac{10 \text{ pc}}{d}\right)^2 &= 10^{0.4(M - m)}\\
\log{\left(\frac{10 \text{ pc}}{d}\right)^2} &= 0.4(M - m)\\
\frac{\log{\left(\frac{10 \text{ pc}}{d}\right)^2}}{0.4} &= M - m\\
m &= M - \frac{2 \log{\left(\frac{10 \text{ pc}}{d}\right)}}{0.4} \\
m &= M - 5 \log{\left(\frac{10 \text{ pc}}{d}\right)}
\end{align}
This equation is known as the distance module, and the distance needs to be measured in parsecs.
(c) What is the star's parallax in terms of its apparent and absolute magnitudes?
The star's parallax with respect to its distance is given by the equation:
\begin{align}
\theta = \frac{1 \text{AU}}{d}
\end{align},
where, $theta$ is the parallax angle, and $d$ is the distance to the star, measured in parsecs.
We can use the equation derived in part (b) to solve for the distance, $d$, as follows:
\begin{align}
m &= M - 5 \log{\left(\frac{10 \text{ pc}}{d}\right)}\\
M - m &= 5 \log{\left(\frac{10 \text{ pc}}{d}\right)}\\
\frac{M - m}{5} &= \log{\left(\frac{10 \text{ pc}}{d}\right)}\\
10^{\frac{M - m}{5}} &= \frac{10 \text{ pc}}{d}\\
d &= \frac{10^{0.2(M - m)}}{10 \text{ pc}}\\
\end{align}
Now that we have $d$ in terms of magnitudes, we can substitute it into the parallax equation above as follows:
\begin{align}
\theta &= \frac{1 \text{AU}}{d}\\
\theta &= \frac{1 \text{AU}}{\frac{10^{0.2(M - m)}}{10 \text{ pc}}}\\
\theta &= \frac{1 \text{AU}}{10^{0.2(M - m)}} \times 10 \text{ pc}
\end{align},
In order to relate the magnitudes of a star to their observation time, we need to look at how much light we are receiving per unit of time for each star. Flux is the measure of energy per time per area, and in this case, the "per area" is the same, since we are looking at the perceived flux reaching our eyes (or telescope if we are observing with a telescope). Since we are looking to see how much longer it will take for the light of Star A to reach us, we can use the proportion of fluxes of stars A and B using the relation:
\begin{align}
\frac{F_{B}}{F_{A}} \approx 2.5^{(m_A - m_B)}
\end{align}
where in this case, the factor $(m_A - m_B)$ is the difference in magnitudes between Star A and Star B, which we know to be 3. Therefore, the ratio of fluxes of Star A and Star B is:
\begin{align}
\frac{F_{B}}{F_{A}} \approx 2.5^{(m_A - m_B)} \approx 16
\end{align}
Since flux is related linearly with time, since the flux of Star A is 16 times less than the flux of Star B, it will take 16 times longer for Star A to collect the same amount of energy as Star B.
(b) Stars have both an apparent magnitude, $m$, which is how bright they appear from Earth. They also have an absolute magnitude, $\mathcal{M}$, which is the apparent magnitude a star would have at $d = 10$ pc. How does the apparent magnitude, $m$, of a star with absolute magnitude $\mathcal{M}$, depend on its distance, $d$ away from you?
Let's consider a star with an apparent magnitude, $m$, and an absolute magnitude, $\mathcal{M}$. The flux of the star is determined by the equation :
\begin{align}
F = \frac{\text{Luminosity}}{\text{Area}} = \frac{L}{\pi d^2}
\end{align}
where, L is the inherent luminosity of the star, and $d$ is the distance the star is away from you.
Knowing this, we can use the equation for the ratio of fluxes to determine the relationship between the apparent magnitude of a star, $m$ at a distance $d$, and the apparent magnitude of the same star if it were located $d = 10$ pc away.
The flux of the star at any given distance is:
\begin{align}
F_m = \frac{L}{4\pi d ^2}
\end{align}
The flux of a star at exactly $d = 10$ pc is as follows:
\begin{align}
F_M = \frac{L}{4\pi (10 \text{ pc}) ^2}
\end{align}
To solve for the relationship between the the apparent magnitude, $m$, and the absolute magnitude, $\mathcal{M}$ the ratio of the fluxes of these two values are as follows:
\begin{align}
\frac{F_{m}}{F_{M}} &= 10^{0.4(M - m)}\\
\frac{\frac{L}{4\pi d ^2}}{\frac{L}{4\pi (10 \text{ pc}) ^2}} &= 10^{0.4(M - m)}\\
\left(\frac{10 \text{ pc}}{d}\right)^2 &= 10^{0.4(M - m)}\\
\log{\left(\frac{10 \text{ pc}}{d}\right)^2} &= 0.4(M - m)\\
\frac{\log{\left(\frac{10 \text{ pc}}{d}\right)^2}}{0.4} &= M - m\\
m &= M - \frac{2 \log{\left(\frac{10 \text{ pc}}{d}\right)}}{0.4} \\
m &= M - 5 \log{\left(\frac{10 \text{ pc}}{d}\right)}
\end{align}
This equation is known as the distance module, and the distance needs to be measured in parsecs.
(c) What is the star's parallax in terms of its apparent and absolute magnitudes?
The star's parallax with respect to its distance is given by the equation:
\begin{align}
\theta = \frac{1 \text{AU}}{d}
\end{align},
where, $theta$ is the parallax angle, and $d$ is the distance to the star, measured in parsecs.
We can use the equation derived in part (b) to solve for the distance, $d$, as follows:
\begin{align}
m &= M - 5 \log{\left(\frac{10 \text{ pc}}{d}\right)}\\
M - m &= 5 \log{\left(\frac{10 \text{ pc}}{d}\right)}\\
\frac{M - m}{5} &= \log{\left(\frac{10 \text{ pc}}{d}\right)}\\
10^{\frac{M - m}{5}} &= \frac{10 \text{ pc}}{d}\\
d &= \frac{10^{0.2(M - m)}}{10 \text{ pc}}\\
\end{align}
Now that we have $d$ in terms of magnitudes, we can substitute it into the parallax equation above as follows:
\begin{align}
\theta &= \frac{1 \text{AU}}{d}\\
\theta &= \frac{1 \text{AU}}{\frac{10^{0.2(M - m)}}{10 \text{ pc}}}\\
\theta &= \frac{1 \text{AU}}{10^{0.2(M - m)}} \times 10 \text{ pc}
\end{align},
Monday, September 28, 2015
Blog #13: Nathalie Cabrol, the Director for SETI
Nathalie Cabrol is an astrobiologist from France who studied at University of Paris in Nanterre as part of the Sorbonne, where she received her Ph.D in the field that is now known as planetary geology. Cabrol's work lies at the intersection of astronomy, geology, and biology as she attempts to intertwine these fields into the field of astrobiology. Cabrol has led the way in exploring uncharted territories in the search for extraterrestrial life by finding the limits of life on Earth. By exploring the harshest conditions on Earth, Cabrol has studied the extreme conditions in which life can ultimately survive.
As part of her studies as a planetary geologist, Cabrol's research focused on Gusev Crater on Mars. Through her expertise on Gusev Crater, one of Cabrol's signature achievements was to argue before NASA to choose Gusev Crater as a possible landing site for the Mars Exploration Rovers. In 2004, Gusev Crater was successfully chosen as the landing site for the rover Spirit.
After leaving NASA, Cabrol became the Principal Investigator and the expedition leader in the High Lakes Project. The High Lakes Project is an undertaking by SETI to understand the impact of extreme environmental stress on lake habitats and the effects of climate change in these habitats. As part of the project, Cabrol studied organisms that are prone to living in harsh conditions, called extremophiles, and saw that these organisms were thriving in conditions such as volcanic thermal vents, high irradiance and ultraviolet exposure, and dramatic pH changes due to the variable environment. Cabrol's research showed that regardless of how extreme the environment was, life seems to have evolved to adapt to any and all conditions. This is a profound discovery as it pertains to the possibility of life existing in other parts of the solar system, and the Universe as a whole.
Cabrol's research as a planetary geologist who studied planetary formations on Mars that could have been potential lake beds, along with her research on extremophiles in the High Lakes Project, make her uniquely suited to find analogues of conditions in the Universe that are hospitable to life with their Earth counterparts. As a result, in 2015, Cabrol became the director of the Carl Sagan Center at SETI, an organization aimed to answer the fundamental questions of the origin of life, and the possibility of life in other parts of the Universe.
Sources:
http://www.seti.org/users/nathalie-cabrol
https://astrobiology.nasa.gov/nai/reports/annual-reports/2008/seti/the-high-lakes-project-hlp/
http://www.seti.org/mos/nathalie-cabrol
Sunday, September 27, 2015
Blog #12: Say Cheese! Photographing light as a particle and a wave
We have all heard about the weird nature of elementary particles in the quantum world, especially the dual nature of light as both a particle and a wave.. Einstein's work on the photoelectric effect showed that light acts as discrete particles of light, called photons. However, other experiments, such as the double slit-experiment, shows light acting like, and having all the properties of a wave. Although at first this wave-particle duality was met with skepticism, today, it is accepted as fact.
However, although we now know that light acts as both a particle and a wave, the weird nature of quantum physics makes it so that the act of observing light determines whether or not you see its particulate properties, or its wave-like properties. We have never been able to see light act as both a particle and a wave at the same time!
This past year, however, scientists at the Swiss Federal Institute of Technology in Lausanne made headlines around the world after successfully capturing the first photograph of light acting as both a particle and a wave at the same time.
This image captures the wave nature of light as a $sinc()$ function. However, the shadow on the bottom of the picture shows discrete particles.
So how did these scientists manage to capture a photograph of light acting as both a particle and a wave, especially since you need light to take a photo in the first place?
Well, they used a classic imaging technique of using a beam of electrons and it's interaction with the object it is imaging, to take a picture of light. This technique is most notably used in electron microscopes, which give us the scarily up-close images of the microscopic world that are light microscopes are not able to see.
Okay, well things underneath an electron microscope are not moving, so we can take pictures of them. But light is always moving, so how do we keep it in one place to take a picture of it?
The scientists in Lausanne shot a beam of ultraviolet light onto a nano-wire. The UV light increased the energy of the wire, and caused a stream of photons to travel in opposite directions, creating a standing wave. In order to capture an image of this standing wave, the scientists shot a beam of electrons to see the light particles interact with the electrons, which was recorded to produce the first picture of light as both a particle and a wave!
This imaging technology can prove to be revolutionary, for it would allow scientists to image and even record the quantum world and it's effects directly. It could provide breakthroughs in fields like quantum computing, as well as help bridge the gap between the macro-world of stellar astronomy, and the quantum world.
For more information about this, check out this video:
Citation:
http://www.spacedaily.com/reports/The_first_ever_photograph_of_light_as_a_particle_and_a_wave_999.html
However, although we now know that light acts as both a particle and a wave, the weird nature of quantum physics makes it so that the act of observing light determines whether or not you see its particulate properties, or its wave-like properties. We have never been able to see light act as both a particle and a wave at the same time!
This past year, however, scientists at the Swiss Federal Institute of Technology in Lausanne made headlines around the world after successfully capturing the first photograph of light acting as both a particle and a wave at the same time.
 |
| This is the first image captured that shows light to be both a particle and a wave. Source: http://www.extremetech.com/extreme/200277-scientists-capture-first-ever-image-of-light-acting-as-a-particle-and-a-wave |
This image captures the wave nature of light as a $sinc()$ function. However, the shadow on the bottom of the picture shows discrete particles.
So how did these scientists manage to capture a photograph of light acting as both a particle and a wave, especially since you need light to take a photo in the first place?
 |
| This is how the avian flu virus looks like under an electron microscope. Source: http://blog.silive.com/health/2008/10/avian-flu-virus.jpg |
Well, they used a classic imaging technique of using a beam of electrons and it's interaction with the object it is imaging, to take a picture of light. This technique is most notably used in electron microscopes, which give us the scarily up-close images of the microscopic world that are light microscopes are not able to see.
Okay, well things underneath an electron microscope are not moving, so we can take pictures of them. But light is always moving, so how do we keep it in one place to take a picture of it?
The scientists in Lausanne shot a beam of ultraviolet light onto a nano-wire. The UV light increased the energy of the wire, and caused a stream of photons to travel in opposite directions, creating a standing wave. In order to capture an image of this standing wave, the scientists shot a beam of electrons to see the light particles interact with the electrons, which was recorded to produce the first picture of light as both a particle and a wave!
This imaging technology can prove to be revolutionary, for it would allow scientists to image and even record the quantum world and it's effects directly. It could provide breakthroughs in fields like quantum computing, as well as help bridge the gap between the macro-world of stellar astronomy, and the quantum world.
For more information about this, check out this video:
Citation:
http://www.spacedaily.com/reports/The_first_ever_photograph_of_light_as_a_particle_and_a_wave_999.html
Blog #11: Microlensing
3) When speaking about microlensing, it is often easier to refer to angular quantities in units of $\theta_E$. Let's define $u \equiv \frac{\beta}{\theta_E}$ and $y \equiv \frac{\theta}{\theta_E}$.
(a) Show that the lens equation can be written as:
\begin{align}
u \equiv y - y^{-1}
\end{align}
Let's recall that the lens equation was obtained from Problem 2 and was defined as follows:
\begin{align} \beta = \theta - \alpha \end{align} This equation was later proved to be rewritten in part 2(d) as:
\begin{align}
\beta = \theta - \frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)
\end{align}
Therefore, we know that:
\begin{align}
\alpha' = \frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)
\end{align}
All of this information will come in handy in just a minute. First, let's try to rewrite the lens equation in terms of $\theta_E$:
\begin{align}
\beta &= \theta - \alpha'\\
\frac{\beta}{\theta_E} &= \frac{\theta}{\theta_E} - \frac{\alpha'}{\theta_E}\\
\end{align}
In order to get the lens equation in the form $u = y - y^{-1}$, the following three conditions need to be true:
\begin{align}
(1) && u = \frac{\beta}{\theta_E}\\
(2) && y = \frac{\theta}{\theta_E}\\
(3) && y^{-1} = \frac{\alpha'}{\theta_E}
\end{align}
Since we don't know what $\theta_E$ is, we can't be sure if this relation will hold true if the three conditions above are met. So let's solve for $\theta_t$, given that $y^{-1} = \frac{\theta_E}{\theta}$ and $\alpha = \frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)$ :
\begin{align}
y^{-1} &= \frac{\alpha'}{\theta_E}\\
\frac{\theta_E}{\theta} &= \frac{\alpha'}{\theta_E}\\
\theta_E^2 &= \alpha' \theta\\
\theta_E^2 &= \frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right) \times \theta\\
\theta_E &= \left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}
\end{align}
In order to solve prove that the lens equation can be rewritten as $u = y-y^{-1}$, let's substitute this equation with the lens equation and see if it holds true:
\begin{align}
u &= y - y^{-1}\\
\frac{\beta}{\theta_E} &= \frac{\theta}{\theta_E} - \frac{\alpha'}{\theta_E}\\
\frac{\theta - \frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)}{\left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}} &= \frac{\theta}{\left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}} - \frac{\frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)}{\left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}}\\
\frac{\theta - \frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)}{\left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}} &= \frac{\theta - \frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)}{\left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}}
\end{align}
Since both sides equal to each other means that the three conditions above held true, and the lens equation can be re-written as $u = y - y^{-1}$.
(b) Solve for the roots of $y(u)$ in terms of $u$. These equations prescribe the angular position of the images as a function of the (mis)alignment between the source and lens. For the situation given in question 2(f) and a lens-source angular separation of 100 $\mu as$ (micro-arcseconds), indicate the position of the image in a drawing.
In order to solve for the roots of $y(u)$ in terms of $u$, we can look at the equation $u = y - y^{-1}$ and rewrite it and modify it to look like a polynomial equation as follows:
\begin{align}
u &= y - y^{-1}\\
0 &= y - y^{-1} - u\\
0(y) &= (y - y^{-1} - u) (y)\\
0 &= y^2 - uy - 1\\
\end{align}
Having the equation written in a polynomial form, we can solve for the roots of $y(u)$ in terms of $u$ using the quadratic formula as follows:
\begin{align}
y = \frac{u \pm \sqrt{u^2 + 4}}{2}
\end{align}
Now that we have an equation to get the two roots of $y$ in terms of $u$, let's try to calculate $u$. But before we try to calculate $u$, let's see what is all the information given to us:
We know from part (a) that $u = \frac{\beta}{\theta_E}$, and $\theta_E = \left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}$, which plug in the known values to solve for $u$:
\begin{align}
u &= \frac{\beta}{\theta_E}\\
u &= \frac{\beta}{\left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}}\\
u &= \frac{100 \mu as}{\left[ \frac{4(4.3 \times 10^{-3} \frac{\text{pc} \cdot \text{km}^2}{M_\odot \cdot s^2})(0.3 M_{\odot})}{(3 \times 10^5 \frac{\text{km}}{s})^2}\left(\frac{8000 \text{ pc} - 4000 \text{ pc}}{(8000 \text{ pc}) (4000 \text{ pc})}\right)\right]^{\frac{1}{2}}}\\
u &= \frac{100 \mu as}{\left[ \frac{4(4.3 \times 10^{-3} \frac{\text{pc} \cdot \text{km}^2}{M_\odot \cdot s^2})(0.3 M_{\odot})}{(3 \times 10^5 \frac{\text{km}}{s})^2}\left(\frac{8000 \text{ pc} - 4000 \text{ pc}}{(8000 \text{ pc}) (4000 \text{ pc})}\right)\right]^{\frac{1}{2}}}\\
u &= \frac{0.1 \text{ milli -arcseconds}}{2.7 \times 10^{-9} \text{ radians}}\\
u &= \frac{0.1 \text{ milli -arcseconds}}{0.6 \text{ milli-arcseconds}}\\
u &= 0.17 \text{ milli-arcseconds}
\end{align}
Now that we have solved for $u$, we can solve for the roots of $y(u)$ as follows:
\begin{align}
y &= \frac{u \pm \sqrt{u^2 + 4}}{2}\\
y &= \frac{0.17 \text{ milli-arcseconds}\pm \sqrt{(0.17 \text{ milli-arcseconds})^2 + 4}}{2}\\
y &= 1.09 \text{ milli-arcseconds}\\
&= -0.92 \text{ milli-arcseconds}
\end{align}
Okay, so now we have the 2 roots of $y$. One of them is a positive value and the other is a negative value. These two values have a real physical impact in how microlensing works. Look at the image below:
The positive value correlates to the larger, right side of the distorted image in the Einstein ring, and the negative value correlates to the smaller, left side of the distorted image in the Einstein ring. The positive and negatives indicates that the two images are mirror images of each other, with the positive value image being larger than the negative value image.
(a) Show that the lens equation can be written as:
\begin{align}
u \equiv y - y^{-1}
\end{align}
Let's recall that the lens equation was obtained from Problem 2 and was defined as follows:
\begin{align} \beta = \theta - \alpha \end{align} This equation was later proved to be rewritten in part 2(d) as:
\begin{align}
\beta = \theta - \frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)
\end{align}
Therefore, we know that:
\begin{align}
\alpha' = \frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)
\end{align}
All of this information will come in handy in just a minute. First, let's try to rewrite the lens equation in terms of $\theta_E$:
\begin{align}
\beta &= \theta - \alpha'\\
\frac{\beta}{\theta_E} &= \frac{\theta}{\theta_E} - \frac{\alpha'}{\theta_E}\\
\end{align}
In order to get the lens equation in the form $u = y - y^{-1}$, the following three conditions need to be true:
\begin{align}
(1) && u = \frac{\beta}{\theta_E}\\
(2) && y = \frac{\theta}{\theta_E}\\
(3) && y^{-1} = \frac{\alpha'}{\theta_E}
\end{align}
Since we don't know what $\theta_E$ is, we can't be sure if this relation will hold true if the three conditions above are met. So let's solve for $\theta_t$, given that $y^{-1} = \frac{\theta_E}{\theta}$ and $\alpha = \frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)$ :
\begin{align}
y^{-1} &= \frac{\alpha'}{\theta_E}\\
\frac{\theta_E}{\theta} &= \frac{\alpha'}{\theta_E}\\
\theta_E^2 &= \alpha' \theta\\
\theta_E^2 &= \frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right) \times \theta\\
\theta_E &= \left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}
\end{align}
In order to solve prove that the lens equation can be rewritten as $u = y-y^{-1}$, let's substitute this equation with the lens equation and see if it holds true:
\begin{align}
u &= y - y^{-1}\\
\frac{\beta}{\theta_E} &= \frac{\theta}{\theta_E} - \frac{\alpha'}{\theta_E}\\
\frac{\theta - \frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)}{\left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}} &= \frac{\theta}{\left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}} - \frac{\frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)}{\left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}}\\
\frac{\theta - \frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)}{\left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}} &= \frac{\theta - \frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)}{\left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}}
\end{align}
Since both sides equal to each other means that the three conditions above held true, and the lens equation can be re-written as $u = y - y^{-1}$.
(b) Solve for the roots of $y(u)$ in terms of $u$. These equations prescribe the angular position of the images as a function of the (mis)alignment between the source and lens. For the situation given in question 2(f) and a lens-source angular separation of 100 $\mu as$ (micro-arcseconds), indicate the position of the image in a drawing.
In order to solve for the roots of $y(u)$ in terms of $u$, we can look at the equation $u = y - y^{-1}$ and rewrite it and modify it to look like a polynomial equation as follows:
\begin{align}
u &= y - y^{-1}\\
0 &= y - y^{-1} - u\\
0(y) &= (y - y^{-1} - u) (y)\\
0 &= y^2 - uy - 1\\
\end{align}
Having the equation written in a polynomial form, we can solve for the roots of $y(u)$ in terms of $u$ using the quadratic formula as follows:
\begin{align}
y = \frac{u \pm \sqrt{u^2 + 4}}{2}
\end{align}
Now that we have an equation to get the two roots of $y$ in terms of $u$, let's try to calculate $u$. But before we try to calculate $u$, let's see what is all the information given to us:
- $\beta = 100 \mu as = 0.1$ milli-arcseconds
- $M_L = 0.3 M_{\odot}$
- $D_L = 4 kpc = 4000 \text{ pc}$
- $D_S = 8 kpc = 8000 \text{ pc}$
- $G = 4.3 \times 10^{-3} \frac{\text{pc} \cdot \text{km}^2}{M_\odot \cdot s^2}$
- $c = 3 \times 10^5 \frac{\text{km}}{s}$
We know from part (a) that $u = \frac{\beta}{\theta_E}$, and $\theta_E = \left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}$, which plug in the known values to solve for $u$:
\begin{align}
u &= \frac{\beta}{\theta_E}\\
u &= \frac{\beta}{\left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}}\\
u &= \frac{100 \mu as}{\left[ \frac{4(4.3 \times 10^{-3} \frac{\text{pc} \cdot \text{km}^2}{M_\odot \cdot s^2})(0.3 M_{\odot})}{(3 \times 10^5 \frac{\text{km}}{s})^2}\left(\frac{8000 \text{ pc} - 4000 \text{ pc}}{(8000 \text{ pc}) (4000 \text{ pc})}\right)\right]^{\frac{1}{2}}}\\
u &= \frac{100 \mu as}{\left[ \frac{4(4.3 \times 10^{-3} \frac{\text{pc} \cdot \text{km}^2}{M_\odot \cdot s^2})(0.3 M_{\odot})}{(3 \times 10^5 \frac{\text{km}}{s})^2}\left(\frac{8000 \text{ pc} - 4000 \text{ pc}}{(8000 \text{ pc}) (4000 \text{ pc})}\right)\right]^{\frac{1}{2}}}\\
u &= \frac{0.1 \text{ milli -arcseconds}}{2.7 \times 10^{-9} \text{ radians}}\\
u &= \frac{0.1 \text{ milli -arcseconds}}{0.6 \text{ milli-arcseconds}}\\
u &= 0.17 \text{ milli-arcseconds}
\end{align}
Now that we have solved for $u$, we can solve for the roots of $y(u)$ as follows:
\begin{align}
y &= \frac{u \pm \sqrt{u^2 + 4}}{2}\\
y &= \frac{0.17 \text{ milli-arcseconds}\pm \sqrt{(0.17 \text{ milli-arcseconds})^2 + 4}}{2}\\
y &= 1.09 \text{ milli-arcseconds}\\
&= -0.92 \text{ milli-arcseconds}
\end{align}
Okay, so now we have the 2 roots of $y$. One of them is a positive value and the other is a negative value. These two values have a real physical impact in how microlensing works. Look at the image below:
The positive value correlates to the larger, right side of the distorted image in the Einstein ring, and the negative value correlates to the smaller, left side of the distorted image in the Einstein ring. The positive and negatives indicates that the two images are mirror images of each other, with the positive value image being larger than the negative value image.
Blog #10: Microlensing
1) Mass bends space-time! This is a prediction of general relativity, but fortunately we can heuristically derive the effect (up to a factor of 2) using Newtonian mechanics and some simplifying assumptions,
Consider a photon of "mass" $m_\gamma$ passing near an object of mass $M_L$; we'll call this object a "lens" (the 'L' in $M_L$ stands for "lens", which is the object doing the bending). The closest approach ($b$) of the photon is known as the impact parameter. We can imagine that the photon feels a gravitational acceleration from this lens, which we imagine is vertical (see diagram below:
(a) Give an expression for the gravitational acceleration in the vertical direction in terms of $M$, $b$, and $G$.
Okay, so we know that we are using basic Newtonian physics to solve for the distortion of space-time. Therefore, we can use basic Newtonian equations to solve for the gravitational acceleration in the vertical direction.
Recall that the relationship between force and acceleration can be described by Newton's Second Law of Motion:
\begin{align}
F = ma
\end{align}
Well in this case, we can use the force of gravitational acceleration, $F_g = \frac{GM_1M_2}{r^2}$, where the two masses, $M_1$ and $M_2$ are the mass of the "lens", $M_L$ and the "mass" of the photon, $m_{\gamma}$, respectively. The radius between the two bodies, $r$ is described by the variable $b$. Knowing this, we can rearrange Newton's second law and substitute in these variables for gravitational force, $F_g = \frac{G M_L M_\gamma}{b^2}$ in order to solve for the gravitational acceleration, $a$ as follows:
\begin{align}
F_g &= m_\gamma a\\
a &= \frac{F_g}{m}\\
a &= \frac{\frac{G M_L M_\gamma}{b^2}}{m_\gamma}\\
a &= \frac{G M_L}{b^2}\\
\end{align}
(b) Consider the time of interaction $\Delta t$. Assume that most of the influence the photon feels occurs in a horizontal distance $2b$. Express $\Delta t$ in terms of $b$ and the speed of the photon.
In order to solve for the "time of interaction", $\Delta t$, let's think about the direction that the photon was travelling. Since the photon was traveling in the horizontal direction, it feels the effects of gravity of $M_L$ the most over the horizontal distance, $2b$.
Since we know that the photon is traveling a distance of $2b$, and we are solving for the $\Delta t$, we can use the rudimentary knowledge of classical mechanics to relate the distance and time using the equation for velocity. Since velocity is measured as the distance traveled over a specified time, and we know that a photon travels at the speed of light, $c$, we can use the velocity equation to solve for $\Delta t$ as follows:
\begin{align}
v_{photon} &= \frac{\text{distance}}{\Delta t}\\
\Delta t &= \frac{\text{distance}}{v_{photon}}\\
\Delta t &= \frac{2b}{c}
\end{align}
(c) Solve for the change in velocity, $\Delta v$, in the direction perpendicular to the original photon path, over this time of interaction.
We are solving for the "change in velocity... over time of interaction", which sounds very much like the definition of acceleration ($a = \frac{\Delta v}{\Delta t}$)! Since the change in velocity, $\Delta v$, is happening over the distance perpendicular to the original photon path, we can use the acceleration derived in part (a) to solve for the change in velocity $\Delta v$. And since we are only measuring the "change in velocity", $\Delta v$ over the "time of interaction", $\Delta t$, we have all the necessary information to use the acceleration equation.
Recall that the acceleration in part (a) was $a = \frac{G M_L}{b^2}$, and the time of interaction, $\Delta t$ calculated in part (b) was $\Delta t = \frac{2b}{c}$. Using this information, we can rearrange the acceleration equation to solve for $\Delta v$ as follows:
\begin{align}
a &= \frac{\Delta v}{\Delta t}\\
\Delta v &= a \Delta t\\
\Delta v &= \frac{G M_L}{b^2} \times \frac{2b}{c}\\
\Delta v &= \frac{2G M_L}{bc}
\end{align}
(d) Now solve for the deflection angle ($\alpha$) in terms of $G$, $M_L$, $b$, and $c$ using your answer from part (a), (b), and (c). This result is a factor of 2 smaller than the correct, relativistic result.
According to the diagram above, the deflection angle $\alpha$ seems to create a right triangle with the horizontal and vertical components of the velocity of the path the photon is trying to take. Therefore, we can model the deflection angle as follows:
Using trigonometry, we can determine the following:
\begin{align}
\tan \alpha = \frac{\text{opposite}}{\text{adjacent}} = \frac{\Delta v}{c}
\end{align}
However, since the change in angle is very minute, we can use the small angle approximation to say that:
\begin{align}
\tan \alpha \approx \alpha = \frac{\Delta v}{c}
\end{align}
Since we can ignore the tangent function, we can solve for $\alpha$ using the equation above and the value calculated for $\Delta v = \frac{2G M_L}{bc}$ as follows:
\begin{align}
\alpha &= \frac{\Delta v}{c} \\
\alpha &= \frac{\frac{2G M_L}{bc}}{c} \\
\alpha &= \frac{2G M_L}{bc^2}
\end{align}
So now, we have the deflection angle, $\alpha$ using classical Newtonian mechanics. However, as the problem states, this answer is smaller than the actual answer obtained by general relativity by a factor of 2. Therefore, the correct relativistic deflective angle $\alpha$ is:
\begin{align}
\alpha = \frac{4G M_L}{bc^2}
\end{align}
Consider a photon of "mass" $m_\gamma$ passing near an object of mass $M_L$; we'll call this object a "lens" (the 'L' in $M_L$ stands for "lens", which is the object doing the bending). The closest approach ($b$) of the photon is known as the impact parameter. We can imagine that the photon feels a gravitational acceleration from this lens, which we imagine is vertical (see diagram below:
(a) Give an expression for the gravitational acceleration in the vertical direction in terms of $M$, $b$, and $G$.
Okay, so we know that we are using basic Newtonian physics to solve for the distortion of space-time. Therefore, we can use basic Newtonian equations to solve for the gravitational acceleration in the vertical direction.
Recall that the relationship between force and acceleration can be described by Newton's Second Law of Motion:
\begin{align}
F = ma
\end{align}
Well in this case, we can use the force of gravitational acceleration, $F_g = \frac{GM_1M_2}{r^2}$, where the two masses, $M_1$ and $M_2$ are the mass of the "lens", $M_L$ and the "mass" of the photon, $m_{\gamma}$, respectively. The radius between the two bodies, $r$ is described by the variable $b$. Knowing this, we can rearrange Newton's second law and substitute in these variables for gravitational force, $F_g = \frac{G M_L M_\gamma}{b^2}$ in order to solve for the gravitational acceleration, $a$ as follows:
\begin{align}
F_g &= m_\gamma a\\
a &= \frac{F_g}{m}\\
a &= \frac{\frac{G M_L M_\gamma}{b^2}}{m_\gamma}\\
a &= \frac{G M_L}{b^2}\\
\end{align}
(b) Consider the time of interaction $\Delta t$. Assume that most of the influence the photon feels occurs in a horizontal distance $2b$. Express $\Delta t$ in terms of $b$ and the speed of the photon.
In order to solve for the "time of interaction", $\Delta t$, let's think about the direction that the photon was travelling. Since the photon was traveling in the horizontal direction, it feels the effects of gravity of $M_L$ the most over the horizontal distance, $2b$.
Since we know that the photon is traveling a distance of $2b$, and we are solving for the $\Delta t$, we can use the rudimentary knowledge of classical mechanics to relate the distance and time using the equation for velocity. Since velocity is measured as the distance traveled over a specified time, and we know that a photon travels at the speed of light, $c$, we can use the velocity equation to solve for $\Delta t$ as follows:
\begin{align}
v_{photon} &= \frac{\text{distance}}{\Delta t}\\
\Delta t &= \frac{\text{distance}}{v_{photon}}\\
\Delta t &= \frac{2b}{c}
\end{align}
(c) Solve for the change in velocity, $\Delta v$, in the direction perpendicular to the original photon path, over this time of interaction.
We are solving for the "change in velocity... over time of interaction", which sounds very much like the definition of acceleration ($a = \frac{\Delta v}{\Delta t}$)! Since the change in velocity, $\Delta v$, is happening over the distance perpendicular to the original photon path, we can use the acceleration derived in part (a) to solve for the change in velocity $\Delta v$. And since we are only measuring the "change in velocity", $\Delta v$ over the "time of interaction", $\Delta t$, we have all the necessary information to use the acceleration equation.
Recall that the acceleration in part (a) was $a = \frac{G M_L}{b^2}$, and the time of interaction, $\Delta t$ calculated in part (b) was $\Delta t = \frac{2b}{c}$. Using this information, we can rearrange the acceleration equation to solve for $\Delta v$ as follows:
\begin{align}
a &= \frac{\Delta v}{\Delta t}\\
\Delta v &= a \Delta t\\
\Delta v &= \frac{G M_L}{b^2} \times \frac{2b}{c}\\
\Delta v &= \frac{2G M_L}{bc}
\end{align}
(d) Now solve for the deflection angle ($\alpha$) in terms of $G$, $M_L$, $b$, and $c$ using your answer from part (a), (b), and (c). This result is a factor of 2 smaller than the correct, relativistic result.
According to the diagram above, the deflection angle $\alpha$ seems to create a right triangle with the horizontal and vertical components of the velocity of the path the photon is trying to take. Therefore, we can model the deflection angle as follows:
Using trigonometry, we can determine the following:
\begin{align}
\tan \alpha = \frac{\text{opposite}}{\text{adjacent}} = \frac{\Delta v}{c}
\end{align}
However, since the change in angle is very minute, we can use the small angle approximation to say that:
\begin{align}
\tan \alpha \approx \alpha = \frac{\Delta v}{c}
\end{align}
Since we can ignore the tangent function, we can solve for $\alpha$ using the equation above and the value calculated for $\Delta v = \frac{2G M_L}{bc}$ as follows:
\begin{align}
\alpha &= \frac{\Delta v}{c} \\
\alpha &= \frac{\frac{2G M_L}{bc}}{c} \\
\alpha &= \frac{2G M_L}{bc^2}
\end{align}
So now, we have the deflection angle, $\alpha$ using classical Newtonian mechanics. However, as the problem states, this answer is smaller than the actual answer obtained by general relativity by a factor of 2. Therefore, the correct relativistic deflective angle $\alpha$ is:
\begin{align}
\alpha = \frac{4G M_L}{bc^2}
\end{align}
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