Acidity and basicity are ways of measuring the concentration of protons in a solution--protons are acidic, so an acidic solution is characterized by having a high [H+]. One equation that is commonly used in biology and chemistry is the Henderson-Hasselbalch equation (which I primarily chose to write on because of its cool name). This equation relates the pH of a solution (the negative log of [H+]) with the chemical's acidity (pKa--an inherent property of the chemical) and its concentrations in both its acidic ([HA]) and basic ([A-]) forms: \[pH=pK_a+log(\frac{[A^-]}{[HA]})\]If we imagine a solution of a chemical with a defined pKa and a known ratio of \(\frac{[A^-]}{[HA]}\), we can easily calculate the pH of that solution using this equation. This equation is especially useful in buffers--solutions whose pH's change very little even when strong acids or strong bases are added.
Sources:
https://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation
https://en.wikipedia.org/wiki/Buffer_solution
Friday, January 29, 2016
Week 1: Introduction
Hi! My name is Kaelyn, and I'm a junior concentrating in Molecular and Cellular Biology (aka the best concentration). On top of that, I'm interested in pursuing Astrophysics as a secondary, which is why I'll be taking Astro 16. I had a great time in Astro 17 last semester, but am even more excited for this semester. As I have much more experience with the life sciences than I do physics, I'm especially excited to learn a bit more about astrochemistry and astrobiology. My ultimate goal is to go to medical school and become a surgeon, but after that I wouldn't turn down an astronaut job!
Wednesday, December 9, 2015
Blog #37: Illustris Simulation
The Halo Mass Function
When I zoomed in on a random region of high density, this is what I found:
Because I am technologically inept, here is a hand-drawn histogram of the masses of the subhaloes that I zoomed in. It appears that low-mass haloes are more common than high-mass haloes.
Because I'm only mostly technologically inept, I was able to use Excel to calculate that approximately 91% on average of the mass of the subhaloes (or at least the subhaloes that Illustris provided me with) was stellar mass.
Exploring Structure and Reionization in the Illustris Simulation
Large scale:
Small scale:
The dark matter appears to be more confined to the filamentary structure than the gas is. This could be because it is denser than the gas and therefore it is more favorable for it to aggregate along the filaments, whereas gas is less affected by gravity and therefore more free to diffuse throughout space. On the smaller scale, however, the gas appears to aggregate more tightly than the dark matter does.
In medium to large galaxies, the gas appears to be densest near the center of the disk. This makes sense given what we have learned about the structure of the Milky Way, that a higher density of matter, including gas, towards its center.
The most massive galaxies tend to cluster, which makes sense given their gravitational attraction to each other.
When I zoomed in on a random region of high density, this is what I found:
Because I am technologically inept, here is a hand-drawn histogram of the masses of the subhaloes that I zoomed in. It appears that low-mass haloes are more common than high-mass haloes.
Exploring Structure and Reionization in the Illustris Simulation
Large scale:
 |
| Dark matter density |
 |
| Gas density |
Small scale:
 |
| Dark matter density |
 |
| Gas density |
In medium to large galaxies, the gas appears to be densest near the center of the disk. This makes sense given what we have learned about the structure of the Milky Way, that a higher density of matter, including gas, towards its center.
 |
| Gas density (blue is denser) relative to galaxies (circles) |
 |
| The clustering of galaxies (circles) |
After watching the video, it seems clear that the dark matter is leading the structure formation. The formation of filaments is apparent much earlier in the dark energy box.
This is only a bit after the beginning of the "Epoch of Reionization," when hydrogen atoms become ionized (indicated by the color that is starting to appear in the gas box). According to the video, it appeared to start around a redshift of 7.7, or 0.7 billion years.
One of the fastest rates of star formation appears to be at redshift of between 3.50-2.90 or so, although it seems that star formation goes through cycles of speeding up and slowing down. It also appears that new objects are formed when large objects break up, as there are explosions occurring in the gas box, which will eject matter away to form smaller objects later.
Structures probably form along these filaments because that is where the dark matter is. The dark matter gravitationally attracts normal matter, which makes it more favorable for structures to form there instead of in random regions of space.
Blog #36: WS12.1, #1&2d
1. Linear perturbation theory. In the early universe, the matter/radiation distribution of the universe is very homogenous and isotropic. At any given time, let us denote the average density of the universe as \(\bar{\rho}(t)\). Nonetheless, there are some tiny fluctuations and not everywhere exactly the same. So let us define the density at comoving position r and time t as \(\rho(x,t)\) and the relative density contrast as \[\delta(r,t)\equiv\frac{\rho(r,t)-\bar{\rho}(t)}{\bar{\rho}(t)}.\]In this exercise we focus on the linear theory, namely, the density contrast in the problem remains small enough so we only need to consider terms linear in \(\delta\). We assume that cold dark matter, which behaves like dust (that is, it is pressureless) dominates the content of the universe at the early epoch. The absence of pressure simplifies the fluid dynamics equations used to characterize the problem.
a) In the linear theory, it turns out that the fluid equations simplify such that the density contrast \(\delta\) satisfies the following second-order differential equation: \[\frac{d^2\delta}{dt^2}+\frac{2\dot{a}}{a}\frac{d\delta}{dt}=4\pi G\bar{\rho}\delta,\]where a(t) is the scale factor of the universe. Notice that remarkably in the linear theory this equation does not contain spatial derivatives. Show that this means that the spatial shape of the density fluctuations is frozen in comoving coordinates and only their amplitude changes. Namely this means that we can factorize \[\delta(x,t)=D(t)\tilde{\delta}(x),\]where \(\tilde{\delta}(x)\) is arbitrary and independent of time, and D(t) is a function of time and valid for all x. D(t) is not arbitrary and must satisfy a differential equation. Derive this differential equation.
We can check that this is a solution by plugging the third equation into the second.
\(\frac{d\delta}{dt}=\dot{D}(t)\tilde{\delta}(x)\)
\(\frac{d^2\delta}{dt^2}=\ddot{D}(t)\tilde{\delta}(x)\)
\(\ddot{D}(t)\tilde{\delta}(x)+\frac{2\dot{a}}{a}\dot{D}(t)\tilde{\delta}(x)=4\pi G\bar{\rho}D(t)\tilde{\delta}(x)\)
\(\ddot{D}(t)+\frac{2\dot{a}}{a}\dot{D}(t)=4\pi G\bar{\rho}D(t)\)
This is a solution to the differential equation.
b) Now let us consider a matter-dominated flat universe, so that \(\bar{\rho}(t)=a^{-3}\rho_{c,0}\) where \(\rho_{c,0}\) is the critical density today, \(3H^2_0/8\pi G\) as in WS11.1. Recall that the behavior of the scale factor of this universe can be written \(a(t)=(3H_0t/2)^{2/3}\), which you learned in previous worksheets, and solve the differential equation for D(t). Hint: you can use the ansatz \(D(t)\propto t^q\) and plug it into the equation that you derived above; you will end up with a quadratic equation for q, There are two solutions, and the general solution for D is a linear combination of two components: one gives you a growing function in t, denoting it as \(D_+(t)\); another decreasing function in t, denoting it as \(D_-(t)\).
By this ansatz, we can say:
\(D(t)=t^q\)
\(\dot{D}(t)=qt^{q-1}\)
\(\ddot{D}(t)=q(q-1)t^{q-2}\)
We can then plug into the differential equation.
\((q)(q-1)t^{q-2}+\frac{2\dot{a}}{a}qt^{q-1}=4\pi G\bar{\rho}t^q\)
We can multiply by \(t^{2-q}\) and substitute in today's critical density.
\((q)(q-1)+\frac{2\dot{a}}{a}qt-4\pi Ga^{-3}\rho_{c,0}t^2=0\)
We plug in the critical density and \(\frac{2}{3t}\) for \(\frac{\dot{a}}{a}\).
\(q^2+\frac{1}{3}q-\frac{2}{3}=0\)
\(q=\frac{2}{3}, -1\)
c) Explain why the \(D_+\) component is generically the dominant one in structure formation, and show that in the Einstein-de Sitter model, \(D_+(t)\propto a(t)\).
Since this is a linear combination, it means that \(D(t)=D_+(t)+D_-(t)=At^{2/3}+Bt^{-1}\).
\(D_+(t)\propto t^{2/3}\propto a(t)\). This term will grow as time increases, since t's exponent is positive.
\(D_-(t)\propto t^{-1}\). This term will decrease as time increases, since t's exponent is negative.
As a result, \(D_+(t)\) will come to dominate as time moves forward.
2. Spherical collapse. Gravitational instability makes initial small density contrasts grow in time. When the density perturbation grows large enough, the linear theory, such as the one presented in the above exercise, breaks down. Generically speaking, non-linear and non-perturbative evolution of the density contrast have to be dealt with in numerical calculations. However, in some very special situations, analytical treatment is possible and provides some insights to some important natures of gravitational collapse. In this exercise we study such an example.
d) Plot r as a function of t for all three cases (open, closed, and flat universe), and show that in the closed case, the particle turns around and collapses; in the open case, the particle keeps expanding with some asymptotically positive velocity; and in the flat case, the particle reaches an infinite radius but with a velocity that approaches zero.
Closed case: blue
\(r=A(1-cos\eta)\)
\(t=B(\eta-sin\eta)\)
The particle's radius increases, but then decreases until the particle collapses to zero again.
Open case: red
\(r=A(cosh\eta-1)\)
\(t=B(sinh\eta-\eta)\)
The particle's radius appears to continue to increase.
Flat case: green
\(r=A\eta^2/2\)
\(t=B\eta^3/6\)
The particle's radius appears to increase with a decreasing velocity.
The x-axis is time and the y-axis is radius. A and B were arbitrarily assigned values of 1, but in practice these values depend on the mass of the patch of over-density.
a) In the linear theory, it turns out that the fluid equations simplify such that the density contrast \(\delta\) satisfies the following second-order differential equation: \[\frac{d^2\delta}{dt^2}+\frac{2\dot{a}}{a}\frac{d\delta}{dt}=4\pi G\bar{\rho}\delta,\]where a(t) is the scale factor of the universe. Notice that remarkably in the linear theory this equation does not contain spatial derivatives. Show that this means that the spatial shape of the density fluctuations is frozen in comoving coordinates and only their amplitude changes. Namely this means that we can factorize \[\delta(x,t)=D(t)\tilde{\delta}(x),\]where \(\tilde{\delta}(x)\) is arbitrary and independent of time, and D(t) is a function of time and valid for all x. D(t) is not arbitrary and must satisfy a differential equation. Derive this differential equation.
We can check that this is a solution by plugging the third equation into the second.
\(\frac{d\delta}{dt}=\dot{D}(t)\tilde{\delta}(x)\)
\(\frac{d^2\delta}{dt^2}=\ddot{D}(t)\tilde{\delta}(x)\)
\(\ddot{D}(t)\tilde{\delta}(x)+\frac{2\dot{a}}{a}\dot{D}(t)\tilde{\delta}(x)=4\pi G\bar{\rho}D(t)\tilde{\delta}(x)\)
\(\ddot{D}(t)+\frac{2\dot{a}}{a}\dot{D}(t)=4\pi G\bar{\rho}D(t)\)
This is a solution to the differential equation.
b) Now let us consider a matter-dominated flat universe, so that \(\bar{\rho}(t)=a^{-3}\rho_{c,0}\) where \(\rho_{c,0}\) is the critical density today, \(3H^2_0/8\pi G\) as in WS11.1. Recall that the behavior of the scale factor of this universe can be written \(a(t)=(3H_0t/2)^{2/3}\), which you learned in previous worksheets, and solve the differential equation for D(t). Hint: you can use the ansatz \(D(t)\propto t^q\) and plug it into the equation that you derived above; you will end up with a quadratic equation for q, There are two solutions, and the general solution for D is a linear combination of two components: one gives you a growing function in t, denoting it as \(D_+(t)\); another decreasing function in t, denoting it as \(D_-(t)\).
By this ansatz, we can say:
\(D(t)=t^q\)
\(\dot{D}(t)=qt^{q-1}\)
\(\ddot{D}(t)=q(q-1)t^{q-2}\)
We can then plug into the differential equation.
\((q)(q-1)t^{q-2}+\frac{2\dot{a}}{a}qt^{q-1}=4\pi G\bar{\rho}t^q\)
We can multiply by \(t^{2-q}\) and substitute in today's critical density.
\((q)(q-1)+\frac{2\dot{a}}{a}qt-4\pi Ga^{-3}\rho_{c,0}t^2=0\)
We plug in the critical density and \(\frac{2}{3t}\) for \(\frac{\dot{a}}{a}\).
\(q^2+\frac{1}{3}q-\frac{2}{3}=0\)
\(q=\frac{2}{3}, -1\)
c) Explain why the \(D_+\) component is generically the dominant one in structure formation, and show that in the Einstein-de Sitter model, \(D_+(t)\propto a(t)\).
Since this is a linear combination, it means that \(D(t)=D_+(t)+D_-(t)=At^{2/3}+Bt^{-1}\).
\(D_+(t)\propto t^{2/3}\propto a(t)\). This term will grow as time increases, since t's exponent is positive.
\(D_-(t)\propto t^{-1}\). This term will decrease as time increases, since t's exponent is negative.
As a result, \(D_+(t)\) will come to dominate as time moves forward.
2. Spherical collapse. Gravitational instability makes initial small density contrasts grow in time. When the density perturbation grows large enough, the linear theory, such as the one presented in the above exercise, breaks down. Generically speaking, non-linear and non-perturbative evolution of the density contrast have to be dealt with in numerical calculations. However, in some very special situations, analytical treatment is possible and provides some insights to some important natures of gravitational collapse. In this exercise we study such an example.
d) Plot r as a function of t for all three cases (open, closed, and flat universe), and show that in the closed case, the particle turns around and collapses; in the open case, the particle keeps expanding with some asymptotically positive velocity; and in the flat case, the particle reaches an infinite radius but with a velocity that approaches zero.
Closed case: blue
\(r=A(1-cos\eta)\)
\(t=B(\eta-sin\eta)\)
The particle's radius increases, but then decreases until the particle collapses to zero again.
Open case: red
\(r=A(cosh\eta-1)\)
\(t=B(sinh\eta-\eta)\)
The particle's radius appears to continue to increase.
Flat case: green
\(r=A\eta^2/2\)
\(t=B\eta^3/6\)
The particle's radius appears to increase with a decreasing velocity.
 |
| Zoomed in |
 |
| Zoomed out |
Monday, November 30, 2015
Blog #35: WS11.1, #3
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.
\(\frac{2.16\times10^{-7}baryons/cm^3}{407photons/cm^3}=5.32\times10^{-10}baryons/photon\)
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.
\(\frac{2.16\times10^{-7}baryons/cm^3}{407photons/cm^3}=5.32\times10^{-10}baryons/photon\)
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.
 |
| 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!).
 |
| Confession: I was actually ready for Thanksgiving in August. |
Happy Thanksgiving from me and my dorky brothers!
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