(week1-beerslaw)= # Beers and inverse squared laws Comments/derivations for the material in Stull p. 38-39 (differences)= ## Differences, differentials and derivatives 1. What is a derivative? - According to [wikipedia](https://en.wikipedia.org/wiki/Derivative), a derivative is defined as: The derivative of a function of a real variable measures the sensitivity to change of a quantity (a function value or dependent variable) which is determined by another quantity (the independent variable). 2. What is the derivative of $F = \sigma T^4$ with respect to T? $$ \frac{dF}{dT} = \frac{d \sigma T^4}{dT} = 4 \sigma T^3 $$ Does that mean that you are free to treat dF/dT as a fraction, and cancel dT in the denominator to write: $$ dF = \frac{d \sigma T^4}{dT} dT = 4 \sigma T^3 dT $$ Scientists (and mathematicians) do this all the time, but that isn't how derivatives work -- dF/dT is a function, not a fraction. So what is really going on? 3. Instead, we need to treat the differentials dF and dT as limits, not numbers. We know that the following is true: given a small temperature difference $\Delta T$ the flux difference $\Delta F$ is aproximately: $$ \Delta F \approx \frac{dF}{dT} \Delta T = 4 \sigma T^3 \Delta T $$ i.e. -- change in x times slope = change in y. This equation in words is called a "first order Taylor series expansion" of the function $F = \sigma T^4$. It is not exact, but it gets better as $\Delta T$ gets smaller. We can make it exact by writing out the whole Taylor series expansion which is a power series with an infinite number of terms. (See the Wikipedia entry [here](https://en.wikipedia.org/wiki/Taylor_series)) 4. What is a differential? Roughly, the differentials dF and dT represent limits that are evaluated by integration: $dT = \lim{\ \Delta T \to 0}$, so that $\int dT = T$ and $$ dF = \lim_{ \Delta T \to 0} \Delta F $$ So if we take the $\lim{\ \Delta T \to 0}$ we can write $$ dF = \frac{d \sigma T^4}{dT} dT = 4 \sigma T^3 dT $$ Note that this is just doing the steps in Stull p. 38 in the opposite order. He starts with my last equation and winds up with 3). The advantage of the alternative approach we're using here is that we don't need to treat the function $\frac{dF}{dT}$ as if it can be ripped apart to get $dF$ and $dT$. If you really care, I recommend reading [this answer](http://math.stackexchange.com/questions/23902/what-is-the-practical-difference-between-a-differential-and-a-derivative) to a question about differentials. This is also closely related to the [chain rule](https://en.wikipedia.org/wiki/Chain_rule) for derivatives of functions. By taking the limit as $\Delta T$ approaches 0 we've turned a *difference equation* for the number $\Delta F$ (which we can solve in python) into a *differential equation* for the differential $dF$ (which we can integrate using calculus). (beers-law-diff)= ## Beers law -- using differentials Both physics (Maxwell's equations) and observations show that when a narrow beam of photons are travelling through a material (like air or water) of constant composition, the flux obeys "Beer's law": $$ F(s) = F_i \exp (-n b s) $$ where $s\ (m)$ is the distance travelled (the path length), $n\ (\#/m^3)$ is the number denstiy of reflecting/absorbing particles and $b\ (m^2)$ is the extinction cross section due to both absorption and scattering. You can think of $b$ as the *target size* of the particles (molecules, smoke particles etc.) which may be much different than the physical size of the particle because of quantum mechanical and wave interference effects. To get the differential version of Beers law, take the derivitive and use the limit argument we went through above and get: $$ dF = F_i (-nb) \exp(-nbs) ds = (-nb) F ds $$ (beers-flux) or $$ \frac{dF}{F} = -n b ds $$ (eq:difflux) If we define the differential optical depth as: $$ d\tau = n b ds $$ Then we get: $$ \frac{dF^\prime}{F^\prime} = d \ln F^\prime = -d \tau^\prime $$ (eq:diffbeers) and integrating from $\tau^\prime=0,\ F^\prime=F_i$ to $\tau^\prime = \tau,F^\prime = F$ gives: $$ \int_{F_i}^F d \ln F^\prime = -\int_0^\tau d\tau^\prime $$ $$ \ln \left ( \frac{F}{F_i} \right ) = - \tau $$ $$ F = F_i \exp (-\tau) $$ which is stull 2.31c. An important point is that the extinction cross section $b$ can vary enormously over small wavlength ranges (called "absorption bands") where the cross section can increase by a factor of 100,000. This is why small concentrations of carbon dioxide have such large impacts on climate. Also note that now that we have the differential form of Beer's law, we don't have to assume that $n$ or $b$ are constant, we can make them depend on position and just do the (more complicated) integral to get $\tau$ ## Inverse Square Law: In-class problem in energy conservation On page 39, Stull asserts the inverse square law: $$ F_2 = F_1^* \left ( \frac{R_1^2}{R_2^2} \right ) $$ 1. Prove this using conservation of energy (i.e. conservation of Joules) 2. Suppose a 10 cm x 10 cm piece of white paper with a visible reflectivity of 80% is pinned to a wall and illuminated by visible light with a flux of 100 $W\,m^{-2}$. If the paper reflects evenly in all directions (isotropic, not glossy), what is the flux from the paper 3 meters from the wall? What about 6 meters from the wall? 3. Stull defines the direct beam transmissivity as: $$ t = \frac{F}{F_{i}} = \exp(-\tau) $$ Suppose I have two pieces of translucent glass that absorb but don't reflect light, and their indvidual transmissivities are $t_1$ and $t_2$. Use conservation of energy to prove that if we stack the two pieces, the combined transmissivity will be $t_1 \times t_2$, which means that the combined optical depth will be $\tau_1 + \tau_2$ (i.e. optical depths add). Note that I had to assume no reflection so that photons wouldn't bounce back and forth between the two plates.