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.
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