3. Baryon-to-photon ratio of our universe.
a) Despite the fact that the CMB has a very low temperature, the number of photons is enormous. Let us estimate what that number is. Each photon has energy \(h\nu\). From the Planck spectrum, figure out the number density, \(n_{\nu}\), of the photon per frequency interval \(d\nu\). Integrate over \(d\nu\) to get an expression for total number density of photon given temperature T. Keep all factors and use the fact that \(\int^{\infty}_0\frac{x^2}{e^x-1}dx\approx2.4\).
The number density is the energy density (\(u_{\nu}d\nu\)) divided by the energy per photon (\(h\nu\).
\(n_{\nu}=\int^{\infty}_0\frac{8\pi h\nu^3}{c^3}\frac{1}{e^{\frac{h_P\nu}{k_BT}}-1}\frac{1}{h\nu}d\nu=\frac{8\pi}{c^3}\int^{\infty}_0\frac{\nu^2}{e^{\frac{h_P\nu}{k_BT}}-1}d\nu\)
By doing u-substitution and using the fact provided above, we can finish the integration.
\(u=\frac{h_P\nu}{k_BT}\)
\(n_{\nu}=\frac{8\pi}{c^3}(\frac{k_BT}{h_P})^3\int^{\infty}_0\frac{u^2}{e^u-1}du=\frac{8\pi}{c^3}(\frac{k_BT}{h_P})^3 2.4\)
b) Using the following values for the constants: \(k_B=1.38\times10^{-16}erg/K\), \(c=3.00\times10^{10}cm/s\), \(h_P=6.62\times10^{-27}erg\cdot s\), and use the temperature of the CMB today that you have computed to calculate the number density of photons in our universe today.
\(\frac{8\pi}{(3\times10^{10}cm/s)^3}[\frac{(1.38\times10^{-16}erg/K)(2.72K)}{6.62\times10^{-27}erg\cdot s}]^32.4=407/cm^3\)
c) Let us calculate the average baryon number density today. In general, baryons refer to protons or neutrons. The present-day density (matter + radiation + dark energy) of our universe is \(9.2\times10^{-30}g/cm^3\). The baryon density is about 4% of it. The masses of proton and neutron are very similar (\(\approx1.7\times10^{-24}g\)). What is the number density of baryons?
\(\frac{0.04(9.2\times10^{-30}g)}{1cm^3}\frac{1 baryon}{1.7\times10^{-24}g}=2.16\times10^{-7} baryons/cm^3\)
d) Divide the above two numbers to get the baryon-to-photon ratio. As you can see, our universe contains many more photons than baryons.
\(\frac{2.16\times10^{-7}baryons/cm^3}{407photons/cm^3}=5.32\times10^{-10}baryons/photon\)
Monday, November 30, 2015
Blog #34: WS11.1, #2
2. Cosmic microwave background. One of the successful predictions of the Big Bang model is the cosmic microwave background (CMB) existing today. In this exercise let us figure out the spectrum and temperature of the CMB today.
In the Big Bang model, the universe started with a hot radiation-dominated soup in thermal equilibrium. In particular, the spectrum of the electromagnetic radiation satisfies the Planck spectrum \[u_{\nu}d\nu=\frac{8\pi h_P\nu^3}{c^3}\frac{1}{e^{\frac{h_P\nu}{k_BT}}-1}d\nu\]At about the redshift \(z\approx 1100\) when the universe had the temperature \(T\approx 3000K\), almost all the electrons and protons in our universe are combined and the universe becomes electromagnetically neutral. So the electromagnetic waves (photons) no longer get absorbed or scattered by the rest of the contents of the universe. They started to propagate freely in the universe until reaching our detectors. Interestingly, even though the photons are no longer in equilibrium with their environment, as we will see, the spectrum still maintains an identical form to the Planck spectrum, albeit characterized by a different temperature.
a) If the photon was emitted at redshift z with frequency \(\nu\), what is its frequency \(\nu'\) today?
Based on the redshift equation, we can find that \(z=\frac{\nu-\nu'}{\nu'}=\frac{\nu}{\nu'}-1\)
Solving for \(\nu'\), we get \(\nu'=\frac{\nu}{z+1}\).
b) If a photon at redshift z had the energy density \(u_{\nu}d\nu\), what is its energy density \(u_{\nu'}d\nu'\) today?
The energy of the photon decreases as its wavelength gets stretched with the expanding of space. This is quantified by the changing frequency calculated above: the energy decreases by a factor of \(\frac{1}{z+1}\). The number density of photons also decreases as space expands. If we imagine a region of space expanding:
\(v\propto r^3\propto a^3r^3\propto(\frac{1}{z+1})^3r^3\), since the volume depends on the scale factor a, which is itself also proportional to \(\frac{1}{z+1}\).
The combined effect means that \(u_{\nu'}d\nu'=u_{\nu}d\nu(\frac{1}{z+1})^4\).
c) Plug in the relation between \(\nu\) and \(\nu'\) into the Planck spectrum and multiply it with the overall energy density dilution factor that you have just figured out to get the energy density today. Write the final expression as the form \(u_{\nu'}d\nu'\. What is \(u_{\nu'}\)? This is the spectrum we observe today. Show that it is exactly the same as the Planck spectrum, except that the temperature is now \(T'=T(1+z)^{-1}\).
We know the following:
\(T=T'(z+1)\)
\(u_{\nu}d\nu=u_{\nu'}d\nu'(z+1)^4\)
\(\nu=\nu'(z+1)\), which means that \(d\nu=d\nu'(z+1)\)
We can then plug these into the Planck equation.
\(u_{\nu'}d\nu'(z+1)^4=\frac{8\pi h_P[\nu'(z+1)]^3}{c^3}\frac{1}{e^{\frac{h_P\nu'(z+1)}{k_BT'(z+1)}}}d\nu'(z+1)\)
\(u_{\nu'}d\nu'=\frac{8\pi h_P\nu'^3}{c^3}\frac{1}{e^{\frac{h_P\nu'}{k_BT'}}}d\nu'\)
This is the same form as the original Planck spectrum.
\(u_{\nu'}=\frac{8\pi h_P\nu'^3}{c^3}\frac{1}{e^{\frac{h_P\nu'}{k_BT'}}}\)
d) According to the Big Bang model, we should observe a black body radiation with temperature T' filled in the entire universe. This is the CMB. Using the information given at the beginning of this problem, what is the temperature T' today?
\(T'=T(z+1)^{-1}\)
\(T'=\frac{3000K}{1100+1}=2.72K\)
In the Big Bang model, the universe started with a hot radiation-dominated soup in thermal equilibrium. In particular, the spectrum of the electromagnetic radiation satisfies the Planck spectrum \[u_{\nu}d\nu=\frac{8\pi h_P\nu^3}{c^3}\frac{1}{e^{\frac{h_P\nu}{k_BT}}-1}d\nu\]At about the redshift \(z\approx 1100\) when the universe had the temperature \(T\approx 3000K\), almost all the electrons and protons in our universe are combined and the universe becomes electromagnetically neutral. So the electromagnetic waves (photons) no longer get absorbed or scattered by the rest of the contents of the universe. They started to propagate freely in the universe until reaching our detectors. Interestingly, even though the photons are no longer in equilibrium with their environment, as we will see, the spectrum still maintains an identical form to the Planck spectrum, albeit characterized by a different temperature.
a) If the photon was emitted at redshift z with frequency \(\nu\), what is its frequency \(\nu'\) today?
Based on the redshift equation, we can find that \(z=\frac{\nu-\nu'}{\nu'}=\frac{\nu}{\nu'}-1\)
Solving for \(\nu'\), we get \(\nu'=\frac{\nu}{z+1}\).
b) If a photon at redshift z had the energy density \(u_{\nu}d\nu\), what is its energy density \(u_{\nu'}d\nu'\) today?
The energy of the photon decreases as its wavelength gets stretched with the expanding of space. This is quantified by the changing frequency calculated above: the energy decreases by a factor of \(\frac{1}{z+1}\). The number density of photons also decreases as space expands. If we imagine a region of space expanding:
\(v\propto r^3\propto a^3r^3\propto(\frac{1}{z+1})^3r^3\), since the volume depends on the scale factor a, which is itself also proportional to \(\frac{1}{z+1}\).
The combined effect means that \(u_{\nu'}d\nu'=u_{\nu}d\nu(\frac{1}{z+1})^4\).
c) Plug in the relation between \(\nu\) and \(\nu'\) into the Planck spectrum and multiply it with the overall energy density dilution factor that you have just figured out to get the energy density today. Write the final expression as the form \(u_{\nu'}d\nu'\. What is \(u_{\nu'}\)? This is the spectrum we observe today. Show that it is exactly the same as the Planck spectrum, except that the temperature is now \(T'=T(1+z)^{-1}\).
We know the following:
\(T=T'(z+1)\)
\(u_{\nu}d\nu=u_{\nu'}d\nu'(z+1)^4\)
\(\nu=\nu'(z+1)\), which means that \(d\nu=d\nu'(z+1)\)
We can then plug these into the Planck equation.
\(u_{\nu'}d\nu'(z+1)^4=\frac{8\pi h_P[\nu'(z+1)]^3}{c^3}\frac{1}{e^{\frac{h_P\nu'(z+1)}{k_BT'(z+1)}}}d\nu'(z+1)\)
\(u_{\nu'}d\nu'=\frac{8\pi h_P\nu'^3}{c^3}\frac{1}{e^{\frac{h_P\nu'}{k_BT'}}}d\nu'\)
This is the same form as the original Planck spectrum.
\(u_{\nu'}=\frac{8\pi h_P\nu'^3}{c^3}\frac{1}{e^{\frac{h_P\nu'}{k_BT'}}}\)
d) According to the Big Bang model, we should observe a black body radiation with temperature T' filled in the entire universe. This is the CMB. Using the information given at the beginning of this problem, what is the temperature T' today?
\(T'=T(z+1)^{-1}\)
\(T'=\frac{3000K}{1100+1}=2.72K\)
Monday, November 23, 2015
Blog #33: Why Thanksgiving is the Most Underrated Holiday
We all know those people who, as soon as Halloween is over, break out the Christmas music (not guilty) and the winter-themed socks (definitely guilty). And it's not like they're being discouraged from doing so; for example, Spotify has been giving me aggressive Christmas-themed ads for the past couple weeks at least. Don't get me wrong, I love Christmas carols, Elf, and candy canes just as much as the next girl (plus I wrote a Christmas-themed blog post in October so that might make me a worse offender than the Christmas-songs-in-November crowd), but I don't like it when people forget about the major holiday that exists between Halloween and Christmas--and I'm not talking about Black Friday.
Thanksgiving! There are so many wonderful things about Thanksgiving that I don't even know where to start, so I'll just repeat: Thanksgiving!!! When most people think about Thanksgiving, probably the first thing they think about is food (note: I just proved this scientifically by asking my roommate). I love food. Probably my only weakness is my inability to resist delicious food (along with my crappy sense of direction).
My dad is an awesome cook and he spends the entire day and a half before Thanksgiving cooking. Bacon-covered turkey is a prime example and one of my favorites, but our guests are from all over the world so they always bring really interesting food too.
Another great thing about Thanksgiving is getting to spend time with family and friends. This is something I've definitely come to appreciate more since coming to college. I haven't seen my family since August, and it'll be great to be able to annoy my parents again in person and bug my little brothers about the presumably-directionless lives they've been living without my guidance. I also get to be reunited with some of my best friends from home, my pickup truck (which I probably like more than I like most people), and, perhaps most importantly, my cat (who I definitely like more than I like most people).
But wait!--you say. Both of those things are true of Christmas too!
Yes, but there are two other things that make Thanksgiving so amazing. First, the timing. It's getting to the point in the semester where I really just need a break (actually we had probably gotten to that point about 4 or 5 weeks ago, but I've made it since then with only a few minor crises). I tend to get really sick at the end of each spring semester, and I think Thanksgiving break is what rejuvenates me enough to not die at the end of each fall semester.
The second thing is that Thanksgiving focuses on being happy about what you already have, whereas Christmas, in all of its consumerist glory, kind of ends up feeling the opposite. I always spend the few weeks before Christmas panicking about getting gifts for other people and panicking whenever anyone asks me what I want for Christmas, because I never have any idea. Then I spend the week after Christmas feeling vaguely annoyed and dissatisfied with whatever I've gotten, regardless of the fact that I didn't know what I wanted in the first place so why would anybody else know what to get me? (I apologize to anyone who has ever had the misfortune of having to buy me a gift.)
Anyway, I'm not saying that Thanksgiving is necessarily better than Christmas--I really love Christmas too. Basically the bottom line (so that I don't end on a cliché) is that people need to stop under-appreciating Thanksgiving because it's a pretty awesome holiday, and I'm super pumped to be on break in less than 24 hours (shoutout to my Tuesday classes for canceling lecture!).
Happy Thanksgiving from me and my dorky brothers!
Thanksgiving! There are so many wonderful things about Thanksgiving that I don't even know where to start, so I'll just repeat: Thanksgiving!!! When most people think about Thanksgiving, probably the first thing they think about is food (note: I just proved this scientifically by asking my roommate). I love food. Probably my only weakness is my inability to resist delicious food (along with my crappy sense of direction).
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| People always used to compare me to Pam but I didn't understand until seeing this. |
Another great thing about Thanksgiving is getting to spend time with family and friends. This is something I've definitely come to appreciate more since coming to college. I haven't seen my family since August, and it'll be great to be able to annoy my parents again in person and bug my little brothers about the presumably-directionless lives they've been living without my guidance. I also get to be reunited with some of my best friends from home, my pickup truck (which I probably like more than I like most people), and, perhaps most importantly, my cat (who I definitely like more than I like most people).
But wait!--you say. Both of those things are true of Christmas too!
Yes, but there are two other things that make Thanksgiving so amazing. First, the timing. It's getting to the point in the semester where I really just need a break (actually we had probably gotten to that point about 4 or 5 weeks ago, but I've made it since then with only a few minor crises). I tend to get really sick at the end of each spring semester, and I think Thanksgiving break is what rejuvenates me enough to not die at the end of each fall semester.
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| Me at the end of every semester. |
The second thing is that Thanksgiving focuses on being happy about what you already have, whereas Christmas, in all of its consumerist glory, kind of ends up feeling the opposite. I always spend the few weeks before Christmas panicking about getting gifts for other people and panicking whenever anyone asks me what I want for Christmas, because I never have any idea. Then I spend the week after Christmas feeling vaguely annoyed and dissatisfied with whatever I've gotten, regardless of the fact that I didn't know what I wanted in the first place so why would anybody else know what to get me? (I apologize to anyone who has ever had the misfortune of having to buy me a gift.)
Anyway, I'm not saying that Thanksgiving is necessarily better than Christmas--I really love Christmas too. Basically the bottom line (so that I don't end on a cliché) is that people need to stop under-appreciating Thanksgiving because it's a pretty awesome holiday, and I'm super pumped to be on break in less than 24 hours (shoutout to my Tuesday classes for canceling lecture!).
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| Confession: I was actually ready for Thanksgiving in August. |
Happy Thanksgiving from me and my dorky brothers!
Sunday, November 22, 2015
Blog #32: WS10.1, #3
3. Observable Universe. It is important to realize that in our Big Bang universe, at any given time, the size of the observable universe is finite. The limit of this observable part of the universe is called the horizon (or particle horizon to be precise, since there are other definitions of horizon). In this problem, we'll compute the horizon size is a matter dominated universe in co-moving coordinates.
a) First of all, why do we use the light to figure out the horizon size?
Since light has a finite speed and finite time has passed since the Big Bang, the part of the universe that is observable to us can be no larger than the distance that light can travel in the time interval since the Big Bang.
b) Light satisfies the statement that \(ds^2=0\). Using the FRW metric, write down the differential equation that describes the path light takes. We call this path the geodesic for a photon. Choosing convenient coordinates in which the light travels in the radial direction so that we can set \(d\theta=d\phi=0\), find the differential equation in terms of the coordinates t and r only.
The FRW metric is:
\(ds^2=-c^2dt^2+a^2(t)[\frac{dr^2}{1-kr^2}+r^2(d\theta^2+sin^2\theta d\phi^2)]\)
Since we know that \(ds^2=d\theta=d\phi=0\), it becomes:
\(0=-c^2dt^2+a^2(t)[\frac{dr^2}{1-kr^2}]\)
\(c^2dt^2=a^2(t)[\frac{dr^2}{1-kr^2}]\)
\(\frac{dr}{dt}=\frac{c}{a(t)}(\frac{1}{1-kr^2})^{-1/2}\)
c) Suppose we consider a flat universe. Let's consider a matter dominated universe so that a(t) as a function of time is known. Find the radius of the horizon today (at \(t=t_0\)).
Move all terms with variable r to the LHS and t to the RHS. Integrate both sides, namely r from 0 to \(r_{horizon}\) and t from 0 to \(t_0\).
\(\int^{t_0}_0\frac{cdt}{a(t)}=\int^{r_{horizon}}_0(\frac{1}{1-kr^2})^{1/2}dr\)
We know that k = 0, since this is a flat universe. We also know that \(a\propto t^{2/3}\) in a matter-dominated universe.
\(c\int^{t_0}_0t^{-2/3}=\int^{r_{horizon}}_0(\frac{1}{1})^{1/2}dr\)
\(3ct_0^{1/2}\propto r_{horizon}\)
a) First of all, why do we use the light to figure out the horizon size?
Since light has a finite speed and finite time has passed since the Big Bang, the part of the universe that is observable to us can be no larger than the distance that light can travel in the time interval since the Big Bang.
b) Light satisfies the statement that \(ds^2=0\). Using the FRW metric, write down the differential equation that describes the path light takes. We call this path the geodesic for a photon. Choosing convenient coordinates in which the light travels in the radial direction so that we can set \(d\theta=d\phi=0\), find the differential equation in terms of the coordinates t and r only.
The FRW metric is:
\(ds^2=-c^2dt^2+a^2(t)[\frac{dr^2}{1-kr^2}+r^2(d\theta^2+sin^2\theta d\phi^2)]\)
Since we know that \(ds^2=d\theta=d\phi=0\), it becomes:
\(0=-c^2dt^2+a^2(t)[\frac{dr^2}{1-kr^2}]\)
\(c^2dt^2=a^2(t)[\frac{dr^2}{1-kr^2}]\)
\(\frac{dr}{dt}=\frac{c}{a(t)}(\frac{1}{1-kr^2})^{-1/2}\)
c) Suppose we consider a flat universe. Let's consider a matter dominated universe so that a(t) as a function of time is known. Find the radius of the horizon today (at \(t=t_0\)).
Move all terms with variable r to the LHS and t to the RHS. Integrate both sides, namely r from 0 to \(r_{horizon}\) and t from 0 to \(t_0\).
\(\int^{t_0}_0\frac{cdt}{a(t)}=\int^{r_{horizon}}_0(\frac{1}{1-kr^2})^{1/2}dr\)
We know that k = 0, since this is a flat universe. We also know that \(a\propto t^{2/3}\) in a matter-dominated universe.
\(c\int^{t_0}_0t^{-2/3}=\int^{r_{horizon}}_0(\frac{1}{1})^{1/2}dr\)
\(3ct_0^{1/2}\propto r_{horizon}\)
Blog #31: WS10.1, #2
2. Ratio of circumference to radius. Let's continue to study the difference between closed, flat, and open geometries by computing the ratio between the circumference and radius of a circle.
a) To compute the radius and circumference of a circle, we look at the spatial part of the metric and concentrate on the two-dimensional part by setting \(d\phi=0\) because a circle encloses a two-dimensional surface. For the flat case, this part is just \[ds^2_{2d}=dr^2+r^2d\theta^2.\]The circumference is found by fixing the radial coordinate (r = R and dr = 0) and both sides of the equation (note that \(\theta\) is integrated from 0 to 2π).
The radius is found by fixing the angular coordinate (\(\theta, d\theta=0\)) and integrating both sides (note that dr is integrated from 0 to R).
Compute the circumference and radius to reproduce the famous Euclidean ratio 2π.
Since dr = 0 and r = R, our equation becomes \(ds^2_{2d}=R^2d\theta^2\).
\(ds_{2d}=Rd\theta\)
We can then integrate. Since we want the equation for circumference, we integrate ds from 0 to the full circumference.
\(\int^{circ}_0ds_{2d}=R\int^{2\pi}_0d\theta\)
\(circumference=2\pi R\), as we know.
The ratio of circumference to radius would therefore be \(\frac{2\pi R}{R}=2\pi\).
b) For a closed geometry, we calculated the analogous two-dimensional part of the metric in problem 1. This can be written as: \[ds^2_{2d}=d\xi^2+sin^2\xi d\theta^2.\] Repeat the same calculation above and derive the ratio for the closed geometry.
Compare your results to the flat (Euclidean) case; which ratio is larger?
Since \(\xi\) essentially represents the radius of the circle in this geometry, we can again fix it such that \(d\xi=0\). The process is then very similar to that of part a:
\(ds^2_{2d}=sin^2\xi d\theta^2\)
\(ds_{2d}=sin\xi d\theta\)
\(\int^{circ}_0ds_{2s}=\int^{2\pi}_0sin\xi d\theta\)
\(circumference=2\pi sin\xi\)
The ratio of circumference to radius would be \(\frac{2\pi sin\xi}{\xi}\). This would be smaller than the Euclidean ratio, since \(\xi\), the denominator, would increase faster than \(\sin\xi\), the numerator, would.
c) Repeat the same analyses for the open geometry, and comparing to the flat case.
Since the simplified metric is not provided, we have to start from the FRW metric. Since this is an open, hyperbolic geometry, we know that k = -1, \(r=sinh\xi\), and therefore \(dr=cosh\xi\). Since this is the 2-dimensional case, we can again set \(d\phi=0\).
\(ds^2_{3d}=\frac{dr^2}{1-kr^2}+r^2(d\theta^2+sin^2\theta d\phi^2)\)
\(ds^2_{2d}=\frac{cosh^2\xi}{1+sinh^2\xi}d\xi^2+sinh^2\xi d\theta^2\)
\(ds^2_{2d}=d\xi^2+sinh^2\xi d\theta^2\)
Since we are working with a constant radius, \(d\xi=0\).
\(ds^2_{2d}=sinh^2\xi d\theta^2\)
a) To compute the radius and circumference of a circle, we look at the spatial part of the metric and concentrate on the two-dimensional part by setting \(d\phi=0\) because a circle encloses a two-dimensional surface. For the flat case, this part is just \[ds^2_{2d}=dr^2+r^2d\theta^2.\]The circumference is found by fixing the radial coordinate (r = R and dr = 0) and both sides of the equation (note that \(\theta\) is integrated from 0 to 2π).
The radius is found by fixing the angular coordinate (\(\theta, d\theta=0\)) and integrating both sides (note that dr is integrated from 0 to R).
Compute the circumference and radius to reproduce the famous Euclidean ratio 2π.
Since dr = 0 and r = R, our equation becomes \(ds^2_{2d}=R^2d\theta^2\).
\(ds_{2d}=Rd\theta\)
We can then integrate. Since we want the equation for circumference, we integrate ds from 0 to the full circumference.
\(\int^{circ}_0ds_{2d}=R\int^{2\pi}_0d\theta\)
\(circumference=2\pi R\), as we know.
The ratio of circumference to radius would therefore be \(\frac{2\pi R}{R}=2\pi\).
b) For a closed geometry, we calculated the analogous two-dimensional part of the metric in problem 1. This can be written as: \[ds^2_{2d}=d\xi^2+sin^2\xi d\theta^2.\] Repeat the same calculation above and derive the ratio for the closed geometry.
Compare your results to the flat (Euclidean) case; which ratio is larger?
Since \(\xi\) essentially represents the radius of the circle in this geometry, we can again fix it such that \(d\xi=0\). The process is then very similar to that of part a:
\(ds^2_{2d}=sin^2\xi d\theta^2\)
\(ds_{2d}=sin\xi d\theta\)
\(\int^{circ}_0ds_{2s}=\int^{2\pi}_0sin\xi d\theta\)
\(circumference=2\pi sin\xi\)
The ratio of circumference to radius would be \(\frac{2\pi sin\xi}{\xi}\). This would be smaller than the Euclidean ratio, since \(\xi\), the denominator, would increase faster than \(\sin\xi\), the numerator, would.
c) Repeat the same analyses for the open geometry, and comparing to the flat case.
Since the simplified metric is not provided, we have to start from the FRW metric. Since this is an open, hyperbolic geometry, we know that k = -1, \(r=sinh\xi\), and therefore \(dr=cosh\xi\). Since this is the 2-dimensional case, we can again set \(d\phi=0\).
\(ds^2_{3d}=\frac{dr^2}{1-kr^2}+r^2(d\theta^2+sin^2\theta d\phi^2)\)
\(ds^2_{2d}=\frac{cosh^2\xi}{1+sinh^2\xi}d\xi^2+sinh^2\xi d\theta^2\)
\(ds^2_{2d}=d\xi^2+sinh^2\xi d\theta^2\)
Since we are working with a constant radius, \(d\xi=0\).
\(ds^2_{2d}=sinh^2\xi d\theta^2\)
\(\int^{circ}_0ds_{2d}=\int^{2\pi}_0sinh\xi d\theta\)
\(circumference=2\pi sinh\xi\)
The ratio of circumference to radius would be \(\frac{2\pi sinh\xi}{\xi}\). This would be larger than the Euclidean case.
d) You may have noticed that, except for the flat case, this ratio is not a constant value. However, in both the open and closed case, there is a limit where the ratio approaches the flat case. Which limit is that?
At small radii, both the closed and open geometries will have ratios close to that of the flat geometry because \(\frac{sin(x)}{x}\) and \(\frac{sinh(x)}{x}\) both tend towards 1 as x approaches 0. This makes sense, because a circle drawn on a sphere, if much smaller than the sphere, will appear to be flat. The Earth is a good example--if you draw a circle on the floor, it will look flat even though it's technically on a sphere.
Monday, November 9, 2015
Blog #30: Lab Intro
Giant molecular clouds (GMCs) are some of the largest structures in the Galaxy, composed mostly of hydrogen and helium atoms, though other atoms and molecules are present as well. GMCs are where matter comes together to form new planets and stars.
In this lab, our goal is to determine the radial velocities of various GMCs in our Galaxy to see how the Galaxy's rotation speed varies by radius. We'll be doing this by measuring the Doppler shifts (specifically by using the emission spectrum of CO) of the GMCs in the radio frequency. This will allow us to see past the gas and dust that would obscure the GMCs were we to look for them using the visible spectrum. We will be observing in the Galactic plane, at Galactic longitudes from 10˚-70˚.
The GMC moving fastest from us will be those whose orbit is tangential to our line of sight. It will be the one with the tightest orbit, and if it is at the tangent point, its entire velocity (\(V_r(max)\)) will be directly away from us. However, we also need to take into account that the Sun, because of its own rotation, will be moving towards that GMC. Therefore, the total circular velocity \(V_{cir}=V_r(max)+V_{\odot}sinl\), where the sin(l) term corrects for the direction of the Sun's rotation.
Our ultimate goal with this lab is to create a Galactic rotation curve. We have talked a bit about the Galactic rotation curve earlier in the class--specifically in the context of how we know about the existence of dark matter. Along with that, we'll calculate/plot the orbital period, the total mass enclosed within each GMC's radius, the number of stars interior to the Sun, and the Sun's Galactic age.
Sources:
http://coolcosmos.ipac.caltech.edu/cosmic_classroom/cosmic_reference/molecular_clouds.html
http://coolcosmos.ipac.caltech.edu/cosmic_classroom/cosmic_reference/images/horsehead.jpg
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| An example of a GMC |
In this lab, our goal is to determine the radial velocities of various GMCs in our Galaxy to see how the Galaxy's rotation speed varies by radius. We'll be doing this by measuring the Doppler shifts (specifically by using the emission spectrum of CO) of the GMCs in the radio frequency. This will allow us to see past the gas and dust that would obscure the GMCs were we to look for them using the visible spectrum. We will be observing in the Galactic plane, at Galactic longitudes from 10˚-70˚.
The GMC moving fastest from us will be those whose orbit is tangential to our line of sight. It will be the one with the tightest orbit, and if it is at the tangent point, its entire velocity (\(V_r(max)\)) will be directly away from us. However, we also need to take into account that the Sun, because of its own rotation, will be moving towards that GMC. Therefore, the total circular velocity \(V_{cir}=V_r(max)+V_{\odot}sinl\), where the sin(l) term corrects for the direction of the Sun's rotation.
Our ultimate goal with this lab is to create a Galactic rotation curve. We have talked a bit about the Galactic rotation curve earlier in the class--specifically in the context of how we know about the existence of dark matter. Along with that, we'll calculate/plot the orbital period, the total mass enclosed within each GMC's radius, the number of stars interior to the Sun, and the Sun's Galactic age.
 |
| In this diagram, GMC (b) is the one moving fastest from us, so it corresponds to the rightmost peak in the emissions spectrum. (From WS9.2) |
Sources:
http://coolcosmos.ipac.caltech.edu/cosmic_classroom/cosmic_reference/molecular_clouds.html
http://coolcosmos.ipac.caltech.edu/cosmic_classroom/cosmic_reference/images/horsehead.jpg
Blog #29: WS9.1, #2
2. In Question 1, you derived the Friedmann Equation in a matter-only universe in the Newtonian approach. That is, you now have an equation that describes the rate of change of the size of the universe, should the universe be made of matter (this includes stars, gas, and dark matter) and nothing else. Of course, the universe is not quite so simple. In this question we'll introduce the full Friedmann Equation which describes a universe that contains matter, radiation, and/or dark energy. We will also see some correction terms to the Newtonian derivation.
a) The full Friedmann equations follow from Einstein's GR, which we will not go through in this course. Analogous to the equations that we derived in Question 1, the full Friedmann equations express the expansion/contraction rate of the scale factor of the universe in terms of the properties of the content in the universe, such as the density, pressure, and cosmological constant. The first Friedmann equation (1) is:
\[(\frac{\dot{a}}{a})^2=\frac{8\pi}{3}G\rho-\frac{kc^2}{a^2}+\frac{\Lambda}{3}\]
The second Friedmann equation (2) is:
\[\frac{\ddot{a}}{a}=\frac{-4\pi G}{3c^2}(\rho c^2+3P)+\frac{\Lambda}{3}\]
In these equations, \(\rho\) and P are the density and pressure of the content, respectively. k is the curvature parameter; k = -1, 0, 1 for open, flat, and closed universe, respectively. \(\Lambda\) is the cosmological constant. Note that in GR, not only density but also pressure are the sources of energy.
Starting from these two equations, derive the third Friedmann equation (3), which governs the way average density in the universe changes with time.
\[\dot{\rho}c^2=-3\frac{\dot{a}}{a}(\rho c^2+P)\]
To derive this equation, first multiply \(a^2\) on both sides of (1) and then take time derivative on both sides; plug (2) into your expression to eliminate \(\ddot{a}\).
\(\dot{a}^2=\frac{8\pi}{3}G\rho a^2-kc^2+\frac{\Lambda}{3}a^2\)
\(2\dot{a}\ddot{a}=\frac{8\pi}{3}G[\dot{\rho}a^2+2\rho a\dot{a}]+\frac{2\Lambda}{3}a\dot{a}\)
\(2\dot{a}\frac{\ddot{a}}{a}=\frac{8\pi}{3}G[\dot{\rho}a+2\rho\dot{a}]+\frac{2\Lambda}{3}\dot{a}\)
\(2\dot{a}(\frac{-4\pi G}{3c^2}[\rho c^2+3P]+\frac{\Lambda}{3})=\frac{8\pi}{3}G[\dot{\rho}a+2\rho\dot{a}]+\frac{2\Lambda}{3}\dot{a}\)
Simplify to \(\dot{\rho}c^2=-3\frac{\dot{a}}{a}(\rho c^2+P)\)
a) The full Friedmann equations follow from Einstein's GR, which we will not go through in this course. Analogous to the equations that we derived in Question 1, the full Friedmann equations express the expansion/contraction rate of the scale factor of the universe in terms of the properties of the content in the universe, such as the density, pressure, and cosmological constant. The first Friedmann equation (1) is:
\[(\frac{\dot{a}}{a})^2=\frac{8\pi}{3}G\rho-\frac{kc^2}{a^2}+\frac{\Lambda}{3}\]
The second Friedmann equation (2) is:
\[\frac{\ddot{a}}{a}=\frac{-4\pi G}{3c^2}(\rho c^2+3P)+\frac{\Lambda}{3}\]
In these equations, \(\rho\) and P are the density and pressure of the content, respectively. k is the curvature parameter; k = -1, 0, 1 for open, flat, and closed universe, respectively. \(\Lambda\) is the cosmological constant. Note that in GR, not only density but also pressure are the sources of energy.
Starting from these two equations, derive the third Friedmann equation (3), which governs the way average density in the universe changes with time.
\[\dot{\rho}c^2=-3\frac{\dot{a}}{a}(\rho c^2+P)\]
To derive this equation, first multiply \(a^2\) on both sides of (1) and then take time derivative on both sides; plug (2) into your expression to eliminate \(\ddot{a}\).
\(\dot{a}^2=\frac{8\pi}{3}G\rho a^2-kc^2+\frac{\Lambda}{3}a^2\)
\(2\dot{a}\ddot{a}=\frac{8\pi}{3}G[\dot{\rho}a^2+2\rho a\dot{a}]+\frac{2\Lambda}{3}a\dot{a}\)
\(2\dot{a}\frac{\ddot{a}}{a}=\frac{8\pi}{3}G[\dot{\rho}a+2\rho\dot{a}]+\frac{2\Lambda}{3}\dot{a}\)
\(2\dot{a}(\frac{-4\pi G}{3c^2}[\rho c^2+3P]+\frac{\Lambda}{3})=\frac{8\pi}{3}G[\dot{\rho}a+2\rho\dot{a}]+\frac{2\Lambda}{3}\dot{a}\)
Simplify to \(\dot{\rho}c^2=-3\frac{\dot{a}}{a}(\rho c^2+P)\)
b) If the matter is cold, its pressure P = 0, and the cosmological constant \(\Lambda\) = 0. Use the third Friedmann equation to derive the evolution of the density of the matter \(\rho\) as a function of the scale factor of the universe a. You can leave this equation in terms of \(\rho\), \(\rho_0\), \(a\) and \(a_0\), where \(\rho_0\) and \(a_0\) are current values of the mass density and scale factor.
The result you got has the following simple interpretation. The cold matter behaves like “cosmological dust” and it is pressureless (not to be confused with warm/hot dust in the interstellar medium!). As the universe expands, the mass of each dust particle is fixed, but the number density of the dust is diluted - inversely proportional to the volume.
Using the relation between \(\rho\) and a that you just derived and the first Friedmann equation, derive the differential equation for the scale factor a for the matter dominated universe. Solve this differentiation equation to show that \(a(t)\propto t^{2/3}\). This is the characteristic expansion history of the universe if it is dominated by matter.
Determining the density of matter as a function of the scale factor:
\(\dot{\rho}c^2=-3\frac{\dot{a}}{a}\rho c^2\)
\(\frac{d\rho}{dt}=\frac{-3}{a}\frac{da}{dt}\rho\)
\(\int^{\rho}_{\rho_0}\frac{d\rho'}{\rho'}=\int^{a}_{a_0}\frac{-3da'}{a'}\)
\(ln(\frac{\rho}{\rho_0})=-3ln(\frac{a}{a_0})\)
Simplify to \(\rho(a)=\rho_0(\frac{a_0}{a})^3\)
Determining the scale factor for the matter-dominated universe as a function of time:
\((\frac{\dot{a}}{a})^2=\frac{8\pi}{3}G\rho-\frac{kc^2}{a^2}+\frac{\Lambda}{3}\)
\((\frac{\dot{a}}{a})^2=\frac{8\pi}{3}G\rho\) since \(\Lambda\) and k = 0 in the matter-dominated universe.
\(\frac{da}{dt}=(\frac{8\pi}{3}G\rho_0\frac{a_0^3}{a})^{1/2}\)
The right side of the above equation is a constant.
\(\int^a_0a'^{1/2}da'=\int^t_0Cdt'\)
\(\frac{2}{3}a^{2/3}=Ct\) so \(a(t)\propto t^{2/3}\)
c) Let us repeat the above exercise for a universe filled with radiation only. For radiation, \(P=\frac{1}{3}\rho c^2\) and \(\Lambda=0\). Again, use the third Friedmann equation to see how the density of the radiation changes as a function of scale factor.
The result also has a simple interpretation. Imagine the radiation being a collection of photons. Similar to the matter case, the number density of the photon is diluted, inversely proportional to the volume. Now the difference is that, in contrast to the dust particle, each photon can be thought of as wave. As you learned last week, the wavelength of the photon is also stretched as the universe expands, proportional to the scale factor of the universe. According to quantum mechanics, the energy of each photon is inversely proportional to its wavelength: E “ hν. Unlike the dust case where each particle has a fixed energy. So in an expanding universe, the energy of each photon is decreasing inversely proportional to the scale factor. Check that this understanding is consistent with the result you got.
Again using the relation between \(\rho\) and a and the first Friedmann equation to show that \(a(t)\propto t^{1/2}\) for the radiation only universe.
Again using the relation between \(\rho\) and a and the first Friedmann equation to show that \(a(t)\propto t^{1/2}\) for the radiation only universe.
Determining the density of radiation as a function of the scale factor:
\(\dot{\rho}c^2=-3\frac{\dot{a}}{a}(\rho c^2+\frac{1}{3}\rho c^2\)
\(\dot{\rho}=-4\frac{\dot{a}}{a}\rho\)
\(\frac{d\rho}{dt}=\frac{-4}{a}\frac{da}{dt}\rho\)
\(\int^{\rho}_{\rho_0}\frac{d\rho'}{\rho'}=\int^{a}_{a_0}\frac{-4da'}{a'}\)
Simplify to \(\rho(a)=\rho_0(\frac{a_0}{a})^4\)
Determining the scale factor for the radiation-dominated universe as a function of time:
\((\frac{\dot{a}}{a})^2=\frac{8\pi}{3}G\rho_0(\frac{a_0}{a})^4\)
\(\dot{a}^2=\frac{8\pi}{3}G\rho_0\frac{a_0^4}{a^2}\)
\(\frac{da}{dt}=(\frac{8\pi}{3}G\rho_0)^{1/2}\frac{a_0^2}{a}\)
\(\int^a_0a'da'=\int^t_0Cdt'\)
\(\frac{1}{2}a^2=Ct\) so \(a(t)\propto t^{1/2}\)
d) Cosmological constant/dark energy dominated universe. Imagine a universe dominated by the cosmological-constant-like term. Namely in the Friedmann equation, we can set \(\rho=0\) and P = 0 and only keep \(\Lambda\) nonzero.
As a digression, notice that we said “cosmological-constant-like” term. This is because the effect of the cosmological constant may be mimicked by a special content of the universe which has a negative pressure \(P=-\rho c^2\). Check that the effect of this content on the right-hand-side of third Friedmann equation is exactly like that of the cosmological constant. To be general we call this content the Dark Energy. How does the energy density of the dark energy change in time?
As a digression, notice that we said “cosmological-constant-like” term. This is because the effect of the cosmological constant may be mimicked by a special content of the universe which has a negative pressure \(P=-\rho c^2\). Check that the effect of this content on the right-hand-side of third Friedmann equation is exactly like that of the cosmological constant. To be general we call this content the Dark Energy. How does the energy density of the dark energy change in time?
Show that the scale factor of the cosmological-constant-dominated universe expands exponentially in time. What is the Hubble parameter of this universe?
Hint: While calculating the scale factor as a function of time, you will find that setting \(a(0)=0\) leads to a negative infinity. Feel free to ignore this term to show the dependence.
Hint: While calculating the scale factor as a function of time, you will find that setting \(a(0)=0\) leads to a negative infinity. Feel free to ignore this term to show the dependence.
The dark energy is a property of spacetime itself, so it remains constant through time.
Determining the scale factor for the dark energy-dominated universe as a function of time:
\((\frac{\dot{a}}{a})^2=\frac{8\pi}{3}G\rho-\frac{kc^2}{a^2}+\frac{\Lambda}{3}\)
\((\frac{\dot{a}}{a})^2=\frac{\Lambda}{3}\) since \(\rho\) and k = 0 in the dark energy-dominated universe.
\(\dot{a}=a(\frac{\Lambda}{3})^{1/2}\)
\(\int^a_0\frac{da'}{a'}=\int^t_0(\frac{\Lambda}{3})^{1/2}dt'\)
\(a(t)=exp(t(\frac{\Lambda}{3})^{1/2})\)
Determining the Hubble parameter:
\(H(t)=\frac{\dot{a}}{a}\) so \(H(t)=(\frac{\Lambda}{3})^{1/2}\)
(e) Suppose the energy density of a universe at its very early time is dominated by half matter and half radiation. (This is in fact the case for our universe 13.7 billion years ago and only 60 thousand years after the Big Bang.) As the universe keeps expanding, which content, radiation or matter, will become the dominant component? Why?
Matter will become the dominant component in this universe as time goes on. The radiation component loses energy not only because the number density decreases as the volume increases, but also because the energy of each individual photon decreases as its wavelength is stretched. This can also be seen mathematically, since at large values for t, the value of the scale factor in the matter-dominated universe (\(a(t)\propto t^{2/3}\)) will be larger than that of the radiation-dominated universe (\(a(t)\propto t^{1/2}\)).
(f) Suppose the energy density of a universe is dominated by similar amount of matter and dark energy. (This is the case for our universe today. Today our universe is roughly 68% in dark energy and 32% in matter, including 28% dark matter and 5% usual matter, which is why it is acceleratedly expanding today.) As the universe keeps expanding, which content, matter or the dark energy, will become the dominant component? Why? What is the fate of our universe?
The energy of dark energy remains constant as the universe expands, whereas the energy of matter decreases as volume increases (the matter gets diluted). Therefore, as the universe expands and matter gets increasingly diluted, dark energy will come to dominate. Our universe will eventually come to be dominated by dark energy as the universe expands enough to make the contributions from matter and radiation negligible.
Blog #28: WS9.1, #1
1. In this exercise, we will derive the first and second Friedmann equations of a homogenous, 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's a function of time.
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 (homogenous 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 \(\dot{v}\) to avoid confusion with the scale factor a.
Since F=ma, we can also say that \(F=m\dot{v}\). The only force in this system is gravity, so we can equate this with the gravitational force.
\(\frac{-GMm}{R^2}=m\dot{v}\)
\(\dot{v}=\frac{-GM}{R^2}\)
b) To derive an energy equation, it is a common trick to multiply both sides of your acceleration equation by v. Turn your velocity into \(\frac{dR}{dt}\), cancel dt, and integrate both sides of your equation. This is an indefinite integral, so you will have a constant of integration; combine these integration constants (C). You should arrive at the following equation:
\[\frac{1}{2}\dot{R}^2-\frac{GM}{R}=C\]
This has units of energy per unit mass.
\(\dot{v}\frac{dR}{dt}=\frac{-GM}{R^2}\frac{dR}{dt}\)
We can replace \(\dot{v}\) with \(\ddot{R}\) since velocity is the time derivative of the radius:
\(\int\ddot{R}d\dot{R}=-GM\int\frac{1}{R^2}dR\)
\(\frac{1}{2}\dot{R}^2+C=\frac{GM}{R}+C\)
\(\frac{1}{2}\dot{R}^2-\frac{GM}{R}=C\)
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 \((\frac{\dot{R}}{R})^2\), where \(\dot{R}\) is equal to \(\frac{dR}{dt}\).
The mass of the sphere can be expressed as \(M=\frac{4}{3}\pi R^3\rho(t)\)
\(\frac{1}{2}\dot{R}^2-\frac{4G\pi R^2\rho(t)}{3}=C\)
\((\frac{\dot{R}}{R})^2=\frac{8G\pi R^2\rho(t)}{3}+\frac{2C}{R^2}\)
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 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).
If \(R=a(t)r\), then \(\dot{R}=\dot{a}(t)r\).
\((\frac{\dot{a}(t)}{a(t)})^2=\frac{8G\pi\rho(t)}{3}+\frac{2C}{a^2(t)r^2}\)
e) Rewrite the above expression so that \((\frac{\dot{a}}{a})^2\) appears on the left side of the equation.
\((\frac{\dot{a}}{a})^2=\frac{8G\pi\rho(t)}{3}+\frac{2C}{a^2(t)r^2}\)
f) Derive the first Friedmann Equation: We know that \(H(t)=\frac{\dot{a}}{a}\). Plugging this relation into your above result and identifying the constant \(2C/r^2=-kc^2\) 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.
\(H^2(t)=\frac{8G\pi\rho(t)}{3}-\frac{kc^2}{a^2(t)}\)
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 \(\frac{\ddot{a}}{a}=-\frac{4\pi}{3}G\rho\), which is known as the second Friedmann equation.
\(\dot{v}=\frac{-GM}{R}=\frac{-4}{3}G\pi R\rho(t)\)
Since \(\dot{v}=\ddot{R}\) and \(\dot{R}=\dot{a}(t)r\), then \(\dot{v}=\ddot{a}(t)\).
\(\ddot{a}(t)=\frac{-4}{3}G\pi\rho(t)a(t)\)
\(\frac{\ddot{a}}{a}=\frac{-4}{3}G\pi\rho(t)\)
The more complete second Friedmann equation actually has another term involving the pressure following from Einstein's general relative, 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.
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's a function of time.
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 (homogenous 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 \(\dot{v}\) to avoid confusion with the scale factor a.
Since F=ma, we can also say that \(F=m\dot{v}\). The only force in this system is gravity, so we can equate this with the gravitational force.
\(\frac{-GMm}{R^2}=m\dot{v}\)
\(\dot{v}=\frac{-GM}{R^2}\)
b) To derive an energy equation, it is a common trick to multiply both sides of your acceleration equation by v. Turn your velocity into \(\frac{dR}{dt}\), cancel dt, and integrate both sides of your equation. This is an indefinite integral, so you will have a constant of integration; combine these integration constants (C). You should arrive at the following equation:
\[\frac{1}{2}\dot{R}^2-\frac{GM}{R}=C\]
This has units of energy per unit mass.
\(\dot{v}\frac{dR}{dt}=\frac{-GM}{R^2}\frac{dR}{dt}\)
We can replace \(\dot{v}\) with \(\ddot{R}\) since velocity is the time derivative of the radius:
\(\int\ddot{R}d\dot{R}=-GM\int\frac{1}{R^2}dR\)
\(\frac{1}{2}\dot{R}^2+C=\frac{GM}{R}+C\)
\(\frac{1}{2}\dot{R}^2-\frac{GM}{R}=C\)
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 \((\frac{\dot{R}}{R})^2\), where \(\dot{R}\) is equal to \(\frac{dR}{dt}\).
The mass of the sphere can be expressed as \(M=\frac{4}{3}\pi R^3\rho(t)\)
\(\frac{1}{2}\dot{R}^2-\frac{4G\pi R^2\rho(t)}{3}=C\)
\((\frac{\dot{R}}{R})^2=\frac{8G\pi R^2\rho(t)}{3}+\frac{2C}{R^2}\)
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 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).
If \(R=a(t)r\), then \(\dot{R}=\dot{a}(t)r\).
\((\frac{\dot{a}(t)}{a(t)})^2=\frac{8G\pi\rho(t)}{3}+\frac{2C}{a^2(t)r^2}\)
e) Rewrite the above expression so that \((\frac{\dot{a}}{a})^2\) appears on the left side of the equation.
\((\frac{\dot{a}}{a})^2=\frac{8G\pi\rho(t)}{3}+\frac{2C}{a^2(t)r^2}\)
f) Derive the first Friedmann Equation: We know that \(H(t)=\frac{\dot{a}}{a}\). Plugging this relation into your above result and identifying the constant \(2C/r^2=-kc^2\) 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.
\(H^2(t)=\frac{8G\pi\rho(t)}{3}-\frac{kc^2}{a^2(t)}\)
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 \(\frac{\ddot{a}}{a}=-\frac{4\pi}{3}G\rho\), which is known as the second Friedmann equation.
\(\dot{v}=\frac{-GM}{R}=\frac{-4}{3}G\pi R\rho(t)\)
Since \(\dot{v}=\ddot{R}\) and \(\dot{R}=\dot{a}(t)r\), then \(\dot{v}=\ddot{a}(t)\).
\(\ddot{a}(t)=\frac{-4}{3}G\pi\rho(t)a(t)\)
\(\frac{\ddot{a}}{a}=\frac{-4}{3}G\pi\rho(t)\)
The more complete second Friedmann equation actually has another term involving the pressure following from Einstein's general relative, 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.
Monday, November 2, 2015
Blog #27: WS8.1, #3
3. It is not strictly correct to associate this ubiquitous distance-dependent redshift we observe with the velocity of galaxies (at very large separations, Hubble's Law gives 'velocities' that exceed 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, 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 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 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:
\[H(t)=\frac{1}{a(t)}\frac{da}{dt}|_t\]
and the Hubble Law is locally valid for any t:
\[v=H(t)d\]
where v is the relative recessional velocity between two points and d the distance that separates them.
a) Assume the rate of expansion, \(\dot{a}\equiv da/dt\), 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.
Since we know the rate of expansion is constant, we can figure out a as a function of t.
\(\frac{da}{dt}=c\)
\(\int_0^{a(t)}da=\int_0^tdt'\)
\(a(t)=ct\)
We can then plug this into the equation for the Hubble Constant, with \(\frac{da}{dt}\) simplified to a constant.
\(H(t)=\frac{1}{ct}c\)
\(H(t)=\frac{1}{t}\)
Now we just need to plug in the Hubble Constant (68.6 (km/s)/Mpc, calculated in #2) in for H(t) to calculate the Hubble Time.
\(68.6\frac{km/s}{Mpc}=\frac{1}{t}\)
\(t=0.0146\frac{Mpc}{km/s}\rightarrow14.28Gyr\)
The universe is about 2Gyr older than the oldest globular clusters.
b) What is the size of the observable universe? What you will calculate is known as the Hubble Length.
The radius of the observable universe should be the distance that light can travel over the Hubble Time, so we can just multiply the Hubble Time by the speed of light.
\(r=(0.0146\frac{Mpc}{km/s})(3.0*10^5km/s)=4380Mpc\)
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, 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 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 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:
\[H(t)=\frac{1}{a(t)}\frac{da}{dt}|_t\]
and the Hubble Law is locally valid for any t:
\[v=H(t)d\]
where v is the relative recessional velocity between two points and d the distance that separates them.
a) Assume the rate of expansion, \(\dot{a}\equiv da/dt\), 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.
Since we know the rate of expansion is constant, we can figure out a as a function of t.
\(\frac{da}{dt}=c\)
\(\int_0^{a(t)}da=\int_0^tdt'\)
\(a(t)=ct\)
We can then plug this into the equation for the Hubble Constant, with \(\frac{da}{dt}\) simplified to a constant.
\(H(t)=\frac{1}{ct}c\)
\(H(t)=\frac{1}{t}\)
Now we just need to plug in the Hubble Constant (68.6 (km/s)/Mpc, calculated in #2) in for H(t) to calculate the Hubble Time.
\(68.6\frac{km/s}{Mpc}=\frac{1}{t}\)
\(t=0.0146\frac{Mpc}{km/s}\rightarrow14.28Gyr\)
The universe is about 2Gyr older than the oldest globular clusters.
b) What is the size of the observable universe? What you will calculate is known as the Hubble Length.
The radius of the observable universe should be the distance that light can travel over the Hubble Time, so we can just multiply the Hubble Time by the speed of light.
\(r=(0.0146\frac{Mpc}{km/s})(3.0*10^5km/s)=4380Mpc\)
Blog #26: WS8.1, #1
1. 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 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?
If the initial distance between Ball C and Ball B (or D) is \(x_0\), the distance after the time interval should be \(d=x_0+\Delta x\).
b) What are the distances from Ball C to Ball A and Ball E?
If we assume that the balls are evenly spaced to begin with, the distance to Ball A should be \(d=2(x_0+\Delta x)\), to account for both the distance that Ball B moves from Ball C and the distance that Ball A moves from Ball B. The same is true for Ball E.
c) Write a general expression for the distance to a ball N balls away from Ball C after time \(\Delta t\). Interpret your finding.
\(d=N(x_0+\Delta x)\)
\(d(t)=N\frac{\Delta x}{\Delta t}\Delta t+Nx_0\)
This means that the distance between a ball N balls away from Ball C increases linearly with time.
d) Write the velocity of a ball N balls away from Ball C during \(\Delta t\). Interpret your finding.
The velocity would be the slope of the above equation: \(N\frac{\Delta x}{\Delta t}\). This means that the velocity is not only dependent on the rate of change of the distance, but also on how many balls are between that ball and Ball C. Farther balls will move away from Ball C at higher velocities than balls close to Ball C.
Blog #25: WS7.2, #5
5. You may have also noticed some weak "dips" (or absorption features) in the spectrum:
a) Suggest some plausible origins for these features. By way of inspiration, you may want to consider what might occur if the bright light from this quasar's accretion disk encounters some gaseous material on its way to Earth. That gaseous material will definitely contain hydrogen, and those hydrogen atoms will probably have electrons occupying the lowest allowed energy state.
These dips are probably due to the absorption of photons by the gaseous material between the quasar and the Earth. The wavelengths probably correspond to the photons at appropriate energies to excite electrons from the ground state to the second, third, or higher energy levels. There is decreased flux at these wavelengths because some of those photons will be absorbed by the gas and not reach the Earth.
b) A spectrum of a different quasar is shown below. Assuming the strongest emission line you see here is due to Lyα, what is the approximate redshift of the object?
The observed wavelength appears to be about 5600Å.
\(z=\frac{5600Å-1216Å}{1216Å}=3.61\)
c) What is the most noticeable difference between this spectrum and the spectrum of 3C 273 (below)?
What conclusion might we draw regarding the incidence of gas in the early Universe as compared to the nearby Universe?
One noticeable feature is that the other spectrum is located at much greater wavelengths than the 3C 273 spectrum. This means that the redshift is much greater (as we calculated in part b), and therefore the object is farther away than 3C 273. We also notice all of the dips to the left of the Lyα peak. These are probably caused by the absorption of the quasar's radiation by dust and gas. This implies that gas, specifically neutral Hydrogen gas, was very prevalent in the early Universe.
Blog #24: WS7.2, #4
4. One prominent feature of quasar spectra are the strong, broad emission lines. Here's a close-up of one:
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 (called the ground state), labeled n = 1.
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.
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 is 10.19eV, 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 α transition (or Lyα for short).
It turns out that the strongest emission feature you observed in the quasar spectrum above arises from Lyα emission from material orbiting around the central black hole.
a) Recall the Doppler equation:
\[\frac{\lambda_{observed}-\lambda_{emitted}}{\lambda_{emitted}}=z\approx\frac{v}{c}\]
Using the data provided, calculate the redshift of this quasar.
Looking at the spectrum, the peak appears to be at about 1405 Angstroms. Since we know the actual wavelength, we can just plug the numbers in to calculate the redshift.
\(z=\frac{1405Å-1216Å}{1216Å}=0.155\)
b) Again using the data provided, along with the Virial Theorem, estimate the mass of the black hole in this quasar. It will help to know that the typical accretion disk around a \(10^8M_\odot\) black hole extends to a radius of \(r=10^{15}m\).
The Virial Theorem can be used to show that \(M\approx\frac{\sigma^2R}{G}\). We then need to find \(\sigma\) (the stellar velocity dispersion) so that we can calculate the mass. We can do this by looking at the wavelength at half the peak flux (which appears to be about 1400Å and 1415Å). Taking the average difference between these value and the peak wavelength (1405Å) gives us 7.5Å. We can divide this wavelength by the peak wavelength to calculate the dispersion.
\(\frac{7.5Å}{1405Å}=0.0053\approx\frac{\sigma}{c}\)
\(\sigma=0.0053*c=1.59*10^6m/s\)
Finally, we can plug this into the equation, using the typical radius.
\(M=\frac{(1.59*10^6m/s)^2(10^{15}m)}{6.674*10^{-14}m^3g^{-1}s^{-2}}=3.79*10^{40}g\rightarrow1.90*10^7M_\odot\)
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