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.
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| The first images of the Martian surface taken by Mariner 4 in 1965. |
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| Images by the Mariner 9 taken in 1971 that shocked the world to see craters and valleys on the Martian surface |
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| 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.
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| Images from Mars Science Laboratory, Curiosity |
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| 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.
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