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\title{PHYS 2380 - Assignment 1}
\author{Due February 8, 2017}
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\section{Application: The Planck formula}

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The Planck formula we derived in class is
\begin{equation}
u(\nu) = \frac{8 \pi h \nu^3}{c^3 \left( e^\frac{h \nu}{k_b T}-1 \right)}.
\end{equation}
Write down the corresponding relationship that depends on wavelength, $u(\lambda)$. 

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This is a variation of question 3-14 in Tipler \& Llewellyn.

\section{Application: Power in the cosmic microwave background}

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The oldest radiation in the Universe is the cosmic microwave background. This relic radiation is the afterglow of the big bang, emitted when the Universe was only $3\times 10^5$ years old. At this time all matter was in a compact, hot and dense state. The radiation emitted by this cosmic furnace can still be observed today, and the spectrum of the radiation is very accurately fit by a blackbody with temperature of $2.72548$ K.

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a.) Using this temperature, find the wavelength at the maximum intensity of the blackbody spectrum.  

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b.) What is the frequency that corresponds to the wavelength in part a? 

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c.) Find the total power delivered to the Earth from the cosmic microwave background, given the radius of the Earth, $R_\oplus = 6.38 \times 10^6$ m. 

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This is question 3-15 in Tipler \& Llewellyn. 

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\section{Application: Power radiated by the human body}

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The body temperature of a healthy human being is $37^\circ$ C. 
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a.) Find the power radiated by the human body per unit area.

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b.) Besides emitting heat energy, humans also absorb energy from the surrounding environment. If the ambient temperature is $27^\circ$ C on a nice summer day, find the power absorbed and calculate the net power lost using the result from part a. 

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c.) If the typical surface area of a human is $1.73$ m$^2$, find the total power radiated per day. Also, convert the energy to food calories if $1$ Cal = $4180$ J. 

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\section{Theory: Classical mean energy}

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In this exercise we will derive the relationship discussed in class for the mean energy of an individual object in a large group at thermal equilibrium, $\bar{E}=k_bT$, where $k_b$ is the Boltzmann constant. 

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a.) The Maxwell-Boltzmann distribution describes how a fixed amount of energy is partitioned among many identical individuals in thermal equilibrium, such as oscillators in the walls of a blackbody cavity, or the modes of the electromagnetic field inside the cavity. This is called the equipartition of energy in classical physics. The Maxwell-Boltzmann distribution is: 
\begin{equation}
f(E) = A e^{-E/k_b T}
\end{equation}
where $A$ is a constant. Find the value of this constant by calculating
\begin{equation}
\int_0^\infty f(E) dE = 1.
\end{equation}

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b.) Find the mean energy $\bar{E}$ using
\begin{equation}
\bar{E}=\frac{\int_0^\infty E f(E) dE}{\int_0^\infty f(E) dE}.
\end{equation}

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\section{Theory: Quantum mean energy}

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Now let's evaluate the expression for the mean energy using Planck's formula $E=nh\nu$. 
We write the quantum version of the mean energy as 
\begin{equation}
\bar{E}=\frac{ \sum_{n=0}^\infty nh\nu e^{-\frac{nh\nu}{k_bT}} }{\sum_{n=0}^\infty e^{-\frac{nh\nu}{k_bT}} }
\end{equation} 

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a.) Simplify the expression for $\bar{E}$ above using the identity
\begin{equation}
\frac{ \sum_{n=0}^\infty ne^{-n \alpha} }{ \sum_{n=0}^\infty e^{-n \alpha} } = - \frac{\text{d}}{\text{d} \alpha} \left[ \ln\left( \frac{1}{1-e^{-\alpha}} \right) \right]
\end{equation}
You do not have to prove this complicated identity! Just use it to simplify the expression for $\bar{E}$ above.  

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b.) Evaluate the right hand side of the identity and use the result of part a to show that the quantum estimate is 
\begin{equation}
\bar{E}=\frac{ h \nu }{\left( e^\frac{h\nu}{k_bT}-1 \right)}
\end{equation}
which was written down in class. 

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