Laser Gun Recoil: Follow-up

Matt Springer over at Built on Facts has a very nice post following up on my earlier analysis of whether or not a laser gun would recoil. In my analysis I came to the conclusion that the momentum delivered was much less than that of a conventional gun, but that the impulse, and hence the force delivered, was about the same. However, that force was delivered over a \~30ns timescale, which left open the question of whether or not the wielder would be able to feel that recoil. While I had thought to possibly return to this issue at some point, Matt has beat me to it. This is fortuitous, because I wasn't sure quite where to start with the question, while he approaches it in a very sensible, clear manner. The gist of his post is that he compares our response to short time scale forces to our ability to sense sound. I won't go through his calculations, you can check those out for yourself, but he concludes that it is very unlikely that we could feel the recoil of our laser gun. Case closed. Though, like any good scientist, we may choose to reopen it later if more evidence comes to light.

Q Factors

When I walk in my door when I get home, I hook my keys, which I keep on a carabiner, onto a binder clip that I've clipped onto my window sill. Its a great way to never lose your keys. But one thing I always notice is that when I hook it on, it swings, and every time it swings it makes a click. This you might expect. What always surprises me is how long the keys keep swinging. They seem to swing for a surprisingly long time, minutes. It always catches me off guard. In order to explain why, I get to talk about Q Factors The Q factor stands for quality factor. Its a nondimensional parameter (my favorite kind) that tells you how pure your oscillator is. Lets back up a step. Lots of things in the world oscillate. Think about a swing. If you get going on the swing and then stop rocking, you swing back and forth, back and forth, but eventually you come to a stop. Imagine swinging on a rusted old swing set. Now give the joint where the swing swings from a nice shot of WD-40. You can imagine that if you repeated the experiment (get swinging to some height and then stop pumping), you'd continue to swing longer. Why? Because the Q factor has increased. You're swinging on a higher quality swing. Mathematically its defined to be $$Q = 2 \pi \times \frac{ U }{ \Delta U }$$ or 2 pi times the total energy stored in the oscillator divided by the energy lost in a cycle. But, another way to gauge the Q factor is the fact that it tells you something about how the oscillators get damped each period. As a number it tells you how many periods need to go by for the amplitude of the oscillations to be damped by $$\frac{1}{e^{2\pi}} \sim \frac{1}{535}$$ This allows you to estimate Q factors for everyday objects. A factor of 1/535 is pretty near to my threshold for observing a lot of things. What does a factor of 535 mean in terms of sound, one of the most common ways I interact with things around me? Well, sound is measured in decibels, which is a logarithmic scale, where a factor of 535 in the power output by something corresponds to a change in the decibels of $$dB = 10 \log_{10} \frac{1}{535} \sim -27$$ What is a decibel change of 27 mean? Well, wikipedia tells me that a calm room is somewhere between 20 and 30 decibels, where as a TV set about a meter away is at about 60 dB. So that tells me that if something like my keys start off making a sound comparable to the volume I set my TV at, I can listen to it until it just gets drowned out by the room and that should give me some estimate for the Q of my keys. I'll keep you in suspense just a bit longer. I said I was surprised how long the keys swing. In order to put the Q that I measured in context, I'll tell you about a few other Qs of things you might have some experience with. Most swinging things that I seem to remember coming in contact with have quality factors of about 10 or so. Swings, or things letting a meter stick swing, stuff like that. Tuning forks, which are built to be accurate resonators will have quality factors of about a thousand or so. The quartz crystal in your watch, which is really supposed to be a good oscillator has a quality factor of 10 thousand or so. One of the best Q factors achieved by man is 10^14. So, what was the Q factor of my keys? I counted the times I could hear them swinging and got a count of 435. This number isn't to be taken too seriously, but it indicated that my swinging keys have a quality factor of something between 400 and 500, which is pretty darn good for something that wasn't engineered. That explains why it always surprises me, the keys always seem to swing much longer than I would anticipate.

The End of Earth Physics I

I was reading the Wikipedia page for the Hitchhiker's Guide books the other day and found that it started as a series of radio shows called "The Ends of Earth." At the end of each episode, the Earth would be destroyed. Since I feel like this is the best way to end any TV show/movie/book/news broadcast/Mayan calendar, I will shamelessly steal the idea. Since this is the first End of the Earth post, we will start small and just consider the boiling off of all the world's oceans. To be precise, we will consider how much energy it would take to turn all the world's water at 0 degrees Celsius to water vapor at 100 degrees Celsius. First we need to estimate the mass of all the water on earth. I will make the assumptions that all water is fresh water and that the oceans account for all water on earth. These are obviously not exactly true, but will give the proper order of magnitude. The volume of a spherical shell is given by $$\text{V} = 4 \pi R^{2} \Delta R$$ Taking the radius of the earth to be 6000 km, the ocean depth to be 1 km, and the fraction of earth covered by water to be 7/10 , we see that $$V_\text{water} = 4 \pi {R_{\text{earth}}^{2} \times \text{height} \times \text{fraction}$$ $$V_\text{water} = 4 \pi (6 \times 10^{6} \text{m})^{2} \times 10^{3} \text{m} \times (\frac{7}{10}) = 3 \times 10^{17} \text{m}^{3}$$ Now we have the volume of the ocean. Now, since $$\text{Density} = \frac{Mass}{Volume}$$ we can calculate the mass of the oceans using the density of water to be 1000 kg / m^3. $$\text{Mass} = (\rho_\text{water})(V_\text{water}) = (1000 \frac{\text{kg}}{\text{m}^{3}})(3 \times 10^{17} \text{m}^{3}) = 3 \times 10^{20} \text{kg}$$ FUN FACT: The mass of all the water on earth is about 2% the mass of that celestial punching bag known as Pluto. Alright, now that we have the mass of the earth, we can start calculating how much energy it would take to heat it up and boil it away. The amount of heat Q required to raise the temperature of a given material is $$Q_\text{heat} = \text{mc}\Delta\text{T}$$ where m is the mass of the thing we are heating, c is the specific heat ( the amount of energy required to heat our material up 1 degree Celcius) and delta T is our change in temperature. This equation only holds when there are no phase transitions, so we can use it to calculate how much energy is needed to heat up the oceans from 0 degrees to 100 degrees: $$Q_\text{heat} = (3 \times 10^{20} \text{kg}) \times (4000 \frac{\text{J}}{\text{kg C}}) \times (100 \text{deg C}) = 10^{26} \text{J}$$ where we have used the fact that the specific heat of water is about 4000 J/kg deg C. So now we know how much energy it takes to bring water to its boiling point, but this is not the same as boiling it off. To find that, we need to calculate how much energy we have to add to liquid water at a constant 100 degrees to turn it into water vapor. This is given by $$Q_{\text{boil}} = \text{mL}$$ where m is the mass again and L is the "latent heat of vaporization." It tells us how much energy we need to add to turn a kilogram of liquid water at 100 degrees to a kilogram of water vapor at 100 degrees. For water, L is about 2*10^6 J/kg, so $$Q_{\text{boil}} = ( 3 \times 10^{20} \text{kg} ) \times (2 \times 10^{6} \frac{\text{J}}{\text{kg}}) = 6 \times 10^{26} \text{J}$$ Comparing this to the energy calculated before, we see that amount of energy needed to vaporize water at 100 degrees is about six times larger than the amount of energy needed to heat up the water by 100 degrees! Now we can calculate the total amount of energy to boil way the oceans: $$Q_{\text{total}} = Q_{\text{heat}} + Q_{\text{boil}} = 7 \times 10^{26} \text{J}$$ So how much energy is that really? Well the total power of the sun is $$L_{\text{sun}} = 4 \times 10^{26} \frac{\text{J}}{\text{s}}$$ Thus, it would take the entire energy output of the sun for about 2 seconds to boil away all the earth's oceans. Now thats a lot of energy. To put this into the conventional unit of destruction, this is equivalent to 10^11 Megatons of TNT. Typical hydrogen bombs are 10 Megatons. So if you want to boil away the oceans with H-bombs, you'd need about ten billion hydrogen bombs. Luckily, short of constructing a reflecting Dyson sphere with a hole that directs all the sun's radiation at the earth, these energies are unobtainable. So don't panic.

Onsager's Tour de Force

In 1943 in a tour de force of mathematical physics, Lars Onsager solved the 2D Ising Model. His solution has proved crucial in furthering statistical mechanics, allowing theorists to check all of there approximation schemes against analytical results. I call his effort a tour de force because it was a huge mathematical exercise, his solution spanning 33 pages. I also call it a 'tour de force' because I have seen it referenced as such in no less than 5 different sources, as well as numerous times in speech. This got me wondering, just how many times is Onsager's solution called a tour de force... So, first I started with a Google books search, and turned up 39 Books, among the one's that Google has indexed, which surely represent only the tip of the iceburg. The books search turns up some of the more popular statistical mechanics books including Kadanoff and Goldenfeld. And I happen to know its also called a tour de force in Sethna's book and Cardy's. Interestingly, the earliest mention in the book search is Magnetism, Volume 2, Part 1 By George Tibor Rado, Harry Suhl, from 1963, 19 years after Onsager's paper. Next I used Google Scholar to try and turn up some references in papers as well. I got 73 results, the earliest of which I have access to is: Field theory of the two-dimensional Ising model: Equivalence to the free particle one-dimensional Dirac equation, Ferrell, Richard A. In fact, making a histogram of the appearances, it looks like the term's usage is only increasing. I'm not sure what's more impressive. The fact that Onsager was able to solve the 2D Ising model, or the fact that his solution was so impressive that it has become almost necessary to refer to it as a tour de force. If you feel like a challenge, we're still waiting on a solution to the 3D Ising model. It appears to be a hard problem, in fact it looks as though its NP Complete. So get to work! (Note: the last document doesn't refer to Onsager's solution by its proper name, instead calling it a breakthrough, tsk tsk)

Another Reason Why The Core is Stupid

I assume everyone has heard of The Core, the terrible scifi movie from 2003. If you haven't you're missing out on what appears to be, according to Discover magazine, the worst sci-fi film ever. There are already numerous sites that discuss the bad science in the core (here, or over at Bad Astronomy), but they all seem to ignore another fundamental problem with the plot. I don't think I'll give too much away if I tell you that the basic premise of the movie is that the earth's core has stopped rotating, and so the earth's magnetic field is collapsing, which they claim will mean that all of the previously deflected microwaves (note: EM radiation is not bent by a magnetic field) will cook us all. Now, a lot of people have focused on the microwaves bit, which while bad science, one could argue that we would still have some bad effects from loosing our magnetic field. The problem I have is that the Earth's magnetic field cannot change that abruptly. I'm currently teaching undergraduate honors E&M, and we're working out of the fantastic texbook (unfortunately now out of print) by Purcell. And in the chapter on electromagnetic induction he has an illuminating exercise. Lets try and estimate how quickly the magnetic field of the earth can change. Well, lets sort of work backwards. We know that if we have a conducting ring with a current flowing through it, this will create a magnetic field. So, if we can try and model some sort of circuit that approximates the earth, and then look at how quickly energy is dissipated in that circuit, we can estimate how fast the magnetic field decays. So lets imagine a thick torus with height and width a. Flowing around this torus is some current I, distributed in a complicated way. The torus is made out of a material with some conductivity sigma. Now, we know that for a wire made out of some material with a conductivity sigma, we can estimate its resistance as $$R = \frac{ L }{ \sigma A }$$ where L is the length, and A is the cross sectional area of the wire. Lets do that with our torus, calling $$A = a^2 \qquad L = 2 \pi a$$ giving us a resistance $$R \sim \frac{ 2 \pi }{ \sigma a }$$ To estimate the magnetic field of this torus, lets just take the magnetic field of a loop with radius a/2. I.e. $$B = \frac{ \mu_0 I }{2 \pi (a/2) }$$ Now we know that the energy stored in the magnetic field is $$U = \frac{1}{2 \mu_0 } \int B^2 \ dV \sim \frac{1}{2 \mu_0} B V$$ where we take the magnetic field to be the magnetic field of the simple loop and V to be the volume of a fat cylinder or so, i.e. $$V \sim \pi a^2 \times a$$ Now if we have a circuit with a known resistance we know that the energy is dissipated through the resistor $$\frac{dU}{dt} = - I^2 R$$ so if we just want an order of magnitude estimate for the characteristic decay time, we can take $$\tau \sim \frac{ U }{ I^2 R }$$ Putting in our approximations from above we have $$\tau \sim \frac{ \frac{1}{2\mu_0 } B^2 V }{ I^2 \frac{2 \pi }{ \sigma a } } = \frac{ \frac{1}{2 \mu_0 } ( \pi a^3 ) \left( \frac{ \mu_0 I }{ 2 \pi (a/2) } \right)^2 }{ I^2 \frac{ 2 \pi }{\sigma a} } = \frac{ \mu_0 }{4 \pi^2 } \sigma a^2$$ where we know $$\mu_0 = 4 \pi \times 10^{-7} N/A^2$$ we obtain roughly (i.e. ignoring the other pi) $$\tau \sim \sigma a^2 \times 10^{-7} (s)$$ Now, lets take the radius of the core to be about half the radius of the earth, or 3000 km or so, and take the conductivity of the core to be about a tenth of that of iron at room temperature (iron becomes a worse conductor when its heated), i.e. $$a \sim 3000 (km) \qquad \sigma \sim 10^6 (S/m)$$ We obtain $$\tau \sim 10^12 (s) = 300 (centuries)$$ So, even if you could magically make the core of the earth stop spinning, the magnetic field is not going to change instantaneously, in fact it would only be able to change on the order of 300 centuries or so. This is really short on geologic time scales, but nothing like the week or so that the movie The Core takes place over. Just one more reason why one of the worst sci-fi movies of all time is bad.

Locating the Sun Photo

This photo has been making the rounds on the internet lately:

Originally from here. Its a time lapse photo of the sun taken from a pinhole camera from June until December. The real question is: Where was this photo taken?

Wandering Sun

Well, first lets think a second about why the sun would wander in the sky. As you may already know, the earth's pole is not aligned with our solar system's axis. The earth's axis has what is called a 23.5 degree axial tilt, or obliquity. As a result, if you sit at a fixed latitude on the earth, then as the earth revolves around the sun and the year goes by, the angle the sun makes with the horizon changes. We need to estimate how much the sun's angle in the sky changes. Consider the following diagram

You'll notice that the angle that the dashed line towards the sun makes with respect to the dotted horizon line, changes when the earth is on either side of the sun. But how much does it change? We can estimate it by realizing that the angle of that triangle centered at the sun is very small. Very small indeed. We could estimate its size by: $$\arctan \frac{R_{\text{Earth}}}{D_{\text{Earth-Sun}}} \approx 0.002^{\circ}$$ Very small. So, knowing that the one angle is very tiny, and remembering that the sum of all of the angles in a triangle must add up to 180 degrees, we can calculate how much that angle in the sky changes. Notice that the angle between the line pointing to our latitude of interest and the equator has a known angle, namely the angle there is our latitude +/- 23.5 degrees. So if that is the angle our observation point makes with the horizontal, by the reasoning above, the angle that the sun makes with respect to our observation point ought to be 180 - our latitude +/- 23.5 degrees. So on either side of the sun, the difference in this angle from summer to winter is just twice 23.5 degrees of 47 degrees. So, regardless of where you are on the earth, between when the sun is highest in the sky, and when it is lowest in the sky, there ought to be a 47 degree difference. (Assuming of course you can see the sun year round, which you can't if you live too far north). Great. So we know that the difference between the height of the traces in the picture and the bottom needs to be 47 degrees. But that still doesn't help us calculate the latitude the picture was taken at.

Finding Latitude in theory

Looking at our diagram again, we notice that when the sun is lowest in the sky, we can calculate directly what angle the sun should make with respect to the horizon (psi) (We already obtained the upper angle in our triangle, the last part is realizing that the angle to the horizon from the radial line is just 90 degrees). We obtain $$\psi = \theta - 90^\circ + 23.5^\circ = \theta + 66.5^\circ$$ So, turning it on its head, we can calculate the latitude (theta) we are at if we know the lowest angle the sun makes in the sky $$\theta = 66.5^\circ - \psi$$ Great.

Finding Latitude in Practice

So, now I know two things. The total angle difference between the upper most the sun gets and the lowest point in the sky is 47 degrees. And I also know that if I know the angle between the horizon and the lowest point the sun gets in the sky, I can calculate the latitude. So how am I going to do this, well, I just took the picture and opened it up in Gimp, and used the measure tool to get the difference between the highest and lowest points in the sky in pixels, and the distance to the horizon in pixels. Here's one example of my guesses:

Getting the differences in height for the sun was relatively straight forward. In order to get the distance to the horizon, I had to guess a little bit. I just lined it up by eye to where I thought the horizon should be in the picture since its not visible. Why does this do me any good? Well assuming the picture doesn't have too much distortion, since I know the one angle is 47 degrees, I can make a comparison between the pixel lengths. Lets say in the picture above I call the length of the green line B, and the length of the yellow line S, then I can estimate $$\frac{\psi}{S} = \frac{47^\circ}{B}$$ which gives me a measure on psi. Since I was doing this by eye, I repeated the measurement ten times independently for each, obtaining in the end $$B = 356 \pm 2 \qquad S = 66 \pm 6$$ And I have a complete formula for my estimation of the latitude $$\theta = 66.5^\circ - 47^\circ \frac{S}{B}$$

Results

This allowed me to estimate the latitude at 58 +/- 1 degree latitude. Naturally, I can't tell the difference between 58 degrees north or 58 degrees south. To get a better idea of what this guess looks like, here I've overlayed the possible locations in green

You'll notice that there isn't much life down at 58 degrees south of the equator. So, I reckon the picture was taken in the northern hemisphere. Surfing around on the webpage where the picture was found, you'll first notice that its taken from a co.uk site (Great Britain), and if you go to the homepage, Mr. Mallon has a weather widget which says Bellshill. Google Maps tells me that Bellshill, UK is at about 56 degrees North. I've tried to put a green dot on the map above near where Bellshill actually is (its hard to see). Not bad for a night's work.

The malleability of theory

There were some undergrads in my office for office hours the other day, asking questions for the midterm. I'm teaching the engineering course on optics and waves (also known as Things that go Sine) and the students were looking at a problem dealing with quarter-wave plates. A quarter-wave plate, the question goes, is a kind of optical instrument that turns light with linear polarization into circular polarization. Show, it continues, that it also turns light with circular polarization linear. After a minute or two of discussing how a quarter-wave plate work ("what, does it act differently on different axes?" giggled one, and everybody gasped when I said yes) we went to the math. I wrote down the equations for light with linear and circular polarization, $$E_0 \hat{x} \cos\left(kz-\omega t\right) \to \frac{E_0}{\sqrt{2}}\hat{x}\cos\left(kz-\omega t\right) + \frac{E_0}{\sqrt{2}}\hat{y}\sin\left(kz-\omega t\right)$$ with a little arrow in between them to signify that the quarter-wave plate turns one into the other. This was acceptable, so we continued. What happens when light with circular polarization enters the plate? Well, we start out the other way: $$\frac{E_0}{\sqrt{2}}\hat{x}\cos\left(kz-\omega t\right) + \frac{E_0}{\sqrt{2}}\hat{y}\sin\left(kz-\omega t\right)$$ and then break it up into components. The first term obviously behaves exactly the same as in the previous case, $$\frac{E_0}{\sqrt{2}}\hat{x}\cos\left(kz-\omega t\right) \to \frac{E_0}{2}\hat{x}\cos\left(kz-\omega t\right) + \frac{E_0}{2}\hat{y}\sin\left(kz-\omega t\right)$$ while the second term is easy to extrapolate from there, $$\frac{E_0}{\sqrt{2}}\hat{y}\sin\left(kz-\omega t\right) \to -\frac{E_0}{2}\hat{x}\cos\left(kz-\omega t\right) + \frac{E_0}{2}\hat{y}\sin\left(kz-\omega t\right)$$ and adding them both up, we get on the other side of the quarter-wave plate $$E_0\hat{y}\sin\left(kz-\omega t\right)$$ a linearly polarized wave. Simple, right? Almost simple. There's one trick in there, which you may have noticed if you were following. My student noticed it - but she's a clever one - and she asked me why I put that minus sign in the second transformation equation. And my honest answer was that I put it there to get the result I wanted. Sure, I can back it up with more math (handedness, or maybe I can figure it out by redrawing the axes) but when I was writing those equations on the board, I put that minus sign there because I was expecting a certain result. And really, we do this all the time. The math isn't neutral; it's an ally (or an enemy, sometimes) that we're bending to get the result we want - because of intuition, because of a previous results, because this is a homework problem with a known solution - or because we have an experimental result that we really want to be in accord with. This isn't a problem, necessarily. Sometimes it leads us to wrong results, sometimes it helps us reach the correct ones. It usually makes our life easier, and sometimes makes them a lot harder. Keeping this bias in mind when we do our math should hopefully lead to more of the first and less of the second.

Bubbles!

Ever wonder why don't you see a standard rainbow when looking at a thin film such as soap stretched across a membrane ready for bubble making? Well, I encountered this problem when I presented my intro physics section with a quiz question today. Properly stated, the question was... "Suppose white light is incident on a thin film (a soap bubble of n=1.33) hanging vertically inside of a square loop. The minimum thickness of the film at the top of the loop is 900nm and it increases linearly (due to gravity) to 1300nm by the bottom of the loop which is 10cm away. This means that the thickness as a function of distance from the top of the loop is $$d(x) = \text{900nm} + \text{400nm} * \left( \frac{x}{\text{10cm}} \right)$$ What wavelengths will be most strongly reflected as a function of distance along our bubble film?" So I got to thinking - don't the partially interfering wavelengths also contribute to the image that our eyes see? Isn't there a better mass profile to use such as an exponential? Linear is just silly. As for the first question, if you consider a single ray entering the thin film and reflecting off both the first interface as well as the second, then there is a phase difference between the two reflected waves, $$\Delta \phi = \pi + 2\pi * \frac{2d}{\lambda}$$ where d is the thickness of the film and lambda is the wavelength. If we consider these two waves as two standing waves added together with a phase then we see that the superposition of their electric fields, for example, is $$E_{tot} = E_0 \cos(\omega t) + E_0 \cos(\omega t + \Delta \phi)$$ $$E_{tot} = E_0 \sin(\omega t) \cos(\Delta \phi)$$ The intensity that our eyes see then goes like the square of this giving an effective damping to certain wavelengths as given by $$\delta = \cos^2(\Delta \phi)$$ Using a more realistic exponential mass profile and this damping factor for wavelengths in the visible spectrum, I created the top image using OpenIL (maybe it's called DevIL, hard to say).