Physics as Magic?

There's a nice post over at Physics Buzz that I thought I might draw your attention to. The central quote for me is: "Speaking strictly about technology - which is often the knowledge attained by physicists put into practical use by engineers - physics has created some pretty amazing things. Cars, planes, iphones, medical treatments, lasers, 3-D movies, and the Large Hadron Collider. We are constantly WOWED by science. Unfortunately, the less someone understands how these things work, the more they begin to believe anything is possible. In other words, if you don't understand the parameters that allow for amazing things (like jets!) you also don't understand the parameters that would prevent other things (like energy generating heart replacements). If you don't understand anything about physics and technology, then it appears to be nothing short of magic, and magic has no bounds..." I think this gets at some of the heart of what we're interested in doing here at the virtuosi. By trying to strip away some of the mysticism around physics, we hope to bring people to a better understand of what we do. Sure, we work fun problems, and discuss interesting topics (at least, so I hope). More importantly though, while doing so we display both the tools of physics, and how physicists think. The reason I got started on this blog is because I feel there is a huge gap in understanding between what physicists do and how we do it, and what the general public perceives us as doing. I don't think this gap is good for anyone, and I think part of the reason for that is very well articulated in the above quote. Of course, part of my hope for this blog is that if more people are comfortable with physics, when I tell people at a dinner party that I'm a physicist the response won't be either "Oh, I hated physics in high school" or "Oh, that's nice." End of conversation.

Letting Air Out of Tires

Have you ever noticed how when you let air out of a bike tire (or, I suppose, a car tire) it feels rather cold? Today we're going to explore why that is, and just how cold it is. Many people consider the air escaping from a tire as a classic example of an adiabatic process. What is an adiabatic process? It is a process that happens so quickly there is no time for heat flow to occur. For our air in the bike tire this means we're letting it out of the tire so quickly that no energy can move into it from the surrounding air. This may not be exactly true, there may be a little energy flow, but there is little enough that we can ignore it. Given that, how do we talk about temperature change? Let's give a physical motivation first. Imagine a gas as a collection of hard spheres, like baseballs. Envision this bunch of baseballs in a box. Suppose you make the volume of the box smaller, you move the walls in. The baseballs will start to bounce around faster. Having trouble thinking of this? Think of a single baseball in a box. Imagine it hits a wall moving towards it. What happens? That's just like what happens when a baseball hits a baseball bat moving towards it, it goes flying. That is, it starts moving faster. The speed with which these gas particles are moving is what we measure as temperature. So shrinking our box increases the temperature. Likewise, expanding our box will decrease the temperature. The same principle holds here. Our gas is expanding from a small volume (the bike tire) into a larger volume (the surrounding world). Thus we expect the temperature to decrease. Got all that? Good. Now for some math. It turns out that using the ideal gas law, we can derive that for an (reversible) adiabatic process $$PV^\gamma=constant$$ where P is the pressure of the gas, V is the volume of the gas, and gamma is (for our purposes) just a number. The ideal gas law states that $$PV=NkT$$ Where N is the number of gas molecules we have, T is the temperature of the gas, and k is a constant. Since N doesn't change in the process we're considering, we can use this to rewrite the above equation, by substituting for V. This gives $$P^{1-\gamma}T^{\gamma}=constant$$ Where the constant is not the same as above. Because this is equal to a constant, we can say that our initial pressure and temperature are related to our final pressure and temperature by $$P_i^{1-\gamma}T_i^{\gamma}=P_f^{1-\gamma}T_f^{\gamma}$$ We can solve this for the final temperature giving $$T_f=T_i\left(\frac{P_i}{P_f}\right)^{\tfrac{1-\gamma}{\gamma}}$$ Finally, we can plug in some numbers. Gamma is 7/5 for diatomic gases (which is most of air). If we assume the air is about room temperature, 20 C, and the tire is at 60 psi, this gives (1 atmosphere of pressure is 15 psi): $$T_f=293K\left(\frac{60 psi}{15 psi}\right)^{\tfrac{1-7/5}{7/5}}=197K$$ Converting back from Kelvin to C, this is -76 C or -105 F. That's cold! For the Expert: There is actually a debate as to whether or not adiabatic cooling is responsible for the chill of air upon being let out of a tire. The argument for it is fairly straightforward. The released air does work on the surrounding atmosphere as it leaves, lowering the energy of the gas. If this is the primary effect, then the change in temperature is given above. However, it is possible that we can consider this a free adiabatic expansion. In a free adiabatic expansion (like a gas expanding into a vacuum), there is no work done, because gas is not acting 'against' anything. The other possibility is the Joule-Thomson effect. I don't claim to understand this effect very thoroughly, but it is another mechanism for cooling when air is let out of a well insulated valve. I've seen claims both ways as to which process is actually responsible for cooling. Fortunately, a simple experiment suggests itself. Helium heats up through the Joule-Thomson effect (when it starts abot \~50K). It will cool down through the above described adiabatic cooling. So, fill a bike tire with Helium gas, and let it out. See if the valve/gas feels hot or cold. This will determine the dominant effect. As an experimentalist, this appeals greatly to me. But if any theorists out there have ideas, please speak up.

The Beer Diet

I know it's been pretty quiet over here this week. The semester is winding down (a week of classes left), and that means that things have been kicked up into another gear. We've got four or five ideas bouncing around at the moment, so hopefully we'll get some up soon. Today I'd like to talk about the beer diet. A while back, there was a rumor going around that if you drank ice cold beer your body would burn more calories heating the beer than the beer contained. It turns out that this false, and I think the claim relied on a lack of knowledge that the American food Calorie is actually one thousand calories (note the difference in capitalization). Let's prove this to ourselves. Let us consider 12 fluid oz of beer. To any good approximation we can treat this as 12 oz of water. Let us assume that the beer starts at 0C and our body raises it to body temperature, 37C. This takes energy Q given by $$Q=mc\Delta T$$ Where m is the mass of the beer, c the specific heat of water, and the last term the change in temperature. We convert out of the archaic units (thanks google!) to get 12 oz = .35 liters. The density of water is 1000kg/m^3, which is 1kg/liter. This gives us the mass of the beer as .35 kg. The specific heat of water is 1 kcal/kgC (note: kcal = kilocalorie = 1 Calorie, i.e. 1000 calories), so the energy it takes to change the temperature of our beer is $$Q=(.35kg)(1kcal/kgC)(37C-0C)$$ $$Q=13kcal$$ If memory serves, some beer company recently had ads toting how their light beer was 'under 100 Calories'. So, I imagine 100 kcal is a fairly good lower bound on the calories in a 12 oz beer. So you can't lose weight just by drinking beer. Sad, isn't it? Finally, we can see that if we didn't understand the distinction between Calorie and calorie, we might think that, since it took 1300 calories to heat up the beer, and beer only contains 100 Calories, this would be a great way to lose weight! I leave it as a question for you, dear readers, as to how much ice water you'd have to drink to have an effective ice water diet. If it does become the next fad diet, you heard about it here first!

Cell Phone Brain Damage... or not.

For better or worse, cell phones are a part of our lives. I say this because they can be convenient when needed, but there's nothing more annoying than a dropped call or an inconveniently timed ring tone. Since many people carry their phones with them daily, there has been a number of studies which ask what are the long-term health effects of cell phone usage. While the controversy rages among medical researchers, I decided to find my own answers by doing a calculation based on the power output of a simple hand set. The key physical idea in this calculation is Faraday's law. You can check Wikipedia for a lengthy article describing it, but I'm more interested in getting down to business. Now, from the antenna of every cell phone, an electromagnetic wave is emitted. This EM wave, has magnetic and electric components which oscillate back and forth across space and through time. To quantify the effects of this EM wave on a human being, I'm going to focus on just the brain. In the simplest case, I would imagine it as a homogeneous blob. Unfortunately, this model is too simplistic and looses the all the interesting physiology. Instead, I'm going to treat the brain as a collection of loops made by neurons. In this model, I imagine that every neuron is connected to many other neurons by electrically conducting dendrites and axons. Some neurons will be connected to each other in two places, thus creating a small continuous loop. This loop has some area $$ A $$ and as the EM wave from the cell phone passes by, it induces an EMF a la Faraday's law: $$ \epsilon = \frac{ d\phi}{dt}. $$ Well, since the flux $$ \phi = \textbf{B} \cdot \textbf{A} \approx \pi r^2 E / c,$$ and the electric field is an oscillatory function $$ E = E_0 \cos \omega t,$$ we have $$ \epsilon = \frac{\pi r^2 \omega E_0}{c} \cos \omega t, $$ where $$r$$ is the radius of the neuron loop, $$\omega = 2 \pi f$$ is the frequency of cell phone carrier waves and $$c $$ is the speed of light. Taking the root mean square gets rid of the cosine factor and gives $$ \epsilon_{rms} = \frac{ \pi r^2 \omega}{c} E_{rms}. $$ So, if I know the RMS amplitude of the EM waves coming from my cell phone, then I will be able to calculate the induced voltage in a given neuron loop. Fortunately, there's a convenient set of formulae which relates power output to the RMS amplitude: $$ \frac{ \langle P \rangle}{A} = I = \langle u_E \rangle c = \epsilon_0 c E^2_{RMS} \quad \rightarrow \quad E_{RMS} = \left( \frac{ \langle P \rangle}{\epsilon_0 c A} \right)^{1/2}. $$ This line of gibberish is relating the intensity $$ I $$ to the power per unit area as well as the average energy density, which in turn is expressed in terms of $$E_{RMS}. $$ The final expression can now be substituted to find $$ \epsilon_{rms} = \frac{2 l^2 f}{c} \left( \frac{ \langle P \rangle}{\epsilon_0 c 4 \pi R^2} \right)^{1/2}. $$ Here, I've made a few additional substitutions to get the formula in terms of real numbers. In particular, I've converted from $$r, $$ the radius of a neuron loop to $$l, $$ the length of an individual neuron when two of them form a loop. Also, I'm using $$R$$ to denote the distance from the cell phone antenna to the neuron loop in your head. I think a reasonable number for this should be about 10 cm, yeah? Okay, now here comes the fun part. I asked around to look at peoples cell phones, and I found that roughly, they all run at 1 W of power. Furthermore, according to an introductory physics text, cell phones operate at around $$ 10^9 Hz $$ and from a little web browsing, $$ l = 10^{-3} \sim 10^1 $$ meters. Since the length of neurons seems to vary quite a bit, I've written the expression as $$ \epsilon = l^2 \times (360 V/m^2). $$ So for two neurons about a mm long in the same vicinity, there's a pretty decent chance they'll make contact with each other at two points. Thus, the induced voltage is about 0.36 mV. On the off chance two longer neurons form a loop (taking $$l \sim 10^{-2} $$ meters), the induced voltage is about 36 mV. Finally if two REALLY long neurons ($$ l \sim 1$$ meter) make a loop (mind you, the chances of this are zero since the circular area they would form is bigger than your body!), then the induced voltage is 360 V. Electrifying! To compare these numbers to something useful, consider that the brain is constantly sending electrical signals at around 100 mV (this is called the Action Potential). Since the most probbable induced voltage (0.36 mV) in significantly smaller than the action potential (less than 1%), we can conclude that the cell phonne EM waves are essentially negligable effects. Like I said earlier, this is an extremely simple model for the brain, but it does illustrate a particular mechanism by which common technology can interfere with our biology. And yes, I trust that the calculation is correct, but while the experiments are showing higher incidence of tumors coinciding with cell phone use, well, I'm just limit my talk time.

Some of the Best Advice You'll Ever Receive

I came across what might be the best advice any student (nay any human being) could possibly ever receive reading a book today...

Now there are two ways in which you can increase your understanding of these issues. One way is to remember the general ideas and then go home and try to figure out what commands you need and make sure you don't leave one out. Make the set shorter or longer for convidence and try to understand the tradeoffs by trying to do problems with your choice. This is the way I would do it because I have that kind of personality! It's the way I study -- to understand somethingby trying to work it out or, in other words, to understand something by creating it. Not creating it one hundred percent, of course; but taking a hint as to which direction to go but not remembering the details. These you work out for yourself. The other way, which is also valuable, is to read carefully how someone else did it. I find the first method best for me, once I have understood the basic idea. If I get stuck I look at a book that tells me how someone else did it. I turn the pages and then I say "Oh, I forgot that bit", then close the book and carry on. Finally, after you've figured out how to do it you read how they did it and find out how dum your solution is and how much more clever and efficient theirs is! But this way you can understand the cleverness of their ideas and have a framework in which to think about the problem. When I start straight off to read someone else's solution I find it boring and uninteresting, with no way of putting the whole picture together. AT least, thats the way it works for me! Throughout the book, I will suggest some problems for you to play with. You might feel tempted to skip them. If they're too hard, fine. Some of them are pretty difficult! But you might skip them thinking that, well, they've probably been done by somebody else; so what's the point? Well, of course they've been done! But so what? Do them for the fun of it. That's how to learn the knack of doing things when you have to do them. Let me give you an example. Suppose I wanted to add up a series of numbers, 1 + 2 + 3 + 4 + 5 + 6 + 7 ... up to say, 62. No doubt you know how to do it; but when you play with this sort of problem as a kid, and you haven't been shown the answer ... it's fun trying to figure out how to do it. Then, as you go into adulthood, you develop a certain confidence that you can discover things' but if they've already been discovered, that shouldn't bother you at all. What one fool can do, so can another, and the fact that some other fool beat you to it shouldn't disturb you: you should get a kick out of having discovered something. Most of the problems I give you in this book have been worked over many times, and many ingenious solutions have been devised for them. But if you keep proving stuff that other s have done, getting confidence, increasing the complexities of your solution for the fun of it -- then one day you'll turn around and discover that nobody actually did that one! And that's the way to become a scientist.

The quote is from none other than Richard Feynman in his Lectures on Computation (a great read by the way). [reproduced without permission] Honestly I think that's one of the biggest problems with science and math education in this country. They have taken something which is fun and challenging, a path of discovery, and reduced it to memorizing a list of dogmatic equations handed down from high. Students are all but taught that the men and women that discovered these laws are somehow above the rest of us. One of the best kept secrets of science is that scientists are in fact mere mortals. They have their faults, and deficiencies just like the rest of us. You don't need anything special to do science, all you need is a sense of curiosity and wonder, and the resolve to spend some time working at it. The more time you work at it, the better you get. Science is not a list of facts. It is not plugging numbers into equations. It is neither history nor accounting. It is, in its purest form, the recognition that as human beings we are each endowed with the power to come to understand the world around us. I firmly believe that anyone of a sane mind is capable of becoming a physicist. I'm not saying that everyone should be; people should follow their passions, but I worry that too many students are turned away thinking they aren't cut out to be scientists, and that's rubbish. And that's my two cents.

End of the Earth Physics III- Asteroids!

imageNo day of earth destroying celebration would be complete without that apocalyptic all-time favorite: the asteroid. And to be fair, it deserves to be the favorite. Of all the doomsday predictions out there (nuclear holocaust for the cynics, death by snoo-snoo for the optimists) it is the only one that is just about certain to occur at some point in the geological near future. On top of that, it's one that we could potentially avoid with enough time and some neat ideas . Let's model an asteroid impact on the earth. We will assume an asteroid starting from rest at infinity and falling to the surface of the earth due to earth's gravity only. Now obviously these assumptions are not exactly correct. We can imagine our asteroid not starting from rest or feeling some force due to the sun. Each of these would certainly change our answer, but should still be within an order of magnitude of our result. By conservation of energy, we have that $$ \text{KE}{i} + \text{PE} = \text{KE}{f} + \text{PE} $$ Plugging in the formulas for kinetic and potential energy, we have $$ \frac{1}{2}m{v_i}^{2} + \frac{-GM_{\oplus}m_a}{{r_{i}}} = \text{KE}f + \frac{-GMm_a}{{r_{f}}}$$ Now, plugging in our initial conditions of starting at rest at infinity and rearranging, we have that $$ \text{KE}f = \frac{GMm_a}{{R_{\oplus}}} $$ Assuming that all of this kinetic energy is deposited into the earth as heat, we have that our total energy added to the earth is $$ \text{E} = \frac{GM_{\oplus}m_a}{{R_{\oplus}}} = \frac{(7 \times 10^{-11} J m kg^{-2})(6 \times 10^{24} kg)}{6 \times 10^{6} m}m_a \approx (10^{8} J)(\frac{m_a}{1kg}) $$ Now let's consider an asteroid made out of iron. Iron is about 10 times denser that water so it has a density of 10^4 kg/m^3. If our asteroid is a sphere, we can find its radius given $$ m_a = \frac{4}{3} \pi {R_a}^{3} \rho $$ Plugging this into our energy equation gives energy as a function of asteroid radius: $$ \text{E} \approx (4 \times 10^{8} J) \times ( \frac{R_a}{1m} )^{3} $$. Now we have the energy released during an asteroid impact in terms of either the size or the mass. So let's do some destruction. We have heard in the last decade or so that even relatively small changes in average global temperatures can result in catastrophe. So let's see how big of an asteroid we would need to deposit enough energy to raise global temperatures by 1 degree (NOTE: this is very much a toy model. For a more realistic and sadistically addicting model, check out this site). To do this, let's pretend increasing global temperatures is the same as increasing ocean temperatures. Since water has a much much higher specific heat than air, this seems like a reasonable assumption. Plus, we have already made this calculation before. In the first End of the Earth post a bit ago, we found that the amount of energy required to raise the temperature of the world's oceans by 100 degrees is $$ E_{\text{100}} \approx 10^{26} J $$ Since the scaling is linear, we can see that the energy to heat up the oceans by 1 degree is just one hundredth of this value, or $$ E_{\text{heat}} \approx 10^{24} J $$ So how big of an asteroid is this? Using our equation from before, setting E = Eheat, we have that $$ R_a = \frac{E_\text{heat}}{4 \times 10^{8} J m^{-1}} = (\frac{10^{24}}{4 \times 10^8})^{1/3}m = (2.5)^{1/3} \times{10^5}m \approx 100 km $$ For comparison, the asteroid that killed the dinosaurs was thought to be about 30 km or so. This is a big rock. FUN FACT FINALE: We all know that an asteroid killed the dinosaurs. But what would happen if instead of an asteroid, we dropped a dinosaur from infinity? According to wikipedia, a brontosaurus was about 30 tons which is about 30,000 kg. Using the equation for energy as a function of mass derived earlier, we have that $$\text{E} = (10^{8} J)(\frac{m_a}{1kg}) = 3 \times 10^{12} J \approx 1 \text{kiloton} \text{TNT} $$ For comparison, the Hiroshima bomb was about 15 kilotons. So to reproduce the Hiroshima energy yield we need about 15 brotosauruses falling from infinity.

Earth Day - Earth Units

In honor of Earth day, I thought I would take a look at what it would mean to do physics in 'Earth' units. What do I mean by that? Well lets be anti-Copernican here, in fact lets assume the opposite of the Copernican principle, and state that the Earth is privileged in the universe and define all of our units around the Earth.


So, I will put a little subscript earth on all of the 'earth' units. They are to be read as 'earth meters' or 'earth amps', etc. We will take as our starting point the mass, radius and day of the earth, normalizing all of our standards to that. This gives us our initial conversion factors $$ 1 g_{\oplus} = M_{\oplus} = 5.9742 \times 10^{34} \text{ kg} $$ $$ 1 m_{\oplus} = R_{\oplus} = 6378.1 \text{ km} $$ $$ 1 s_{\oplus} = T_{\oplus} = 86,400 \text{ s} $$ From this, we can figure out what all of the other 'earth' units would be.


Some things we are used to talking about start to look a little simpler in these units. The surface area of the earth would be $$ 4 \pi\ m_{\oplus}^2 \sim 12.6\ m_{\oplus}^2 $$ and the volume of the earth would be $$ \frac{4\pi}{3}\ m_{\oplus}^3 \sim 4.2\ m_{\oplus}^3 $$ One earth velocity would be $$ 1\ \frac{ m_{\oplus} }{ s_{\oplus} } = 73.8\ \frac{ m }{ s} $$ so that the velocity of a person standing at the equator would be about $$ 2 \pi\ \frac{ m_{\oplus }}{s_{\oplus} } \sim 6.3\ \frac{ m_{\oplus } }{ s_{\oplus } } $$ and one earth acceleration would be $$ 1\ \frac{ m_{\oplus} }{ s_{\oplus}^2 } = 8.5 \times 10^{-4} \frac{ m}{s^2} $$ so that the gravitational acceleration we feel on the earth in earth accelerations would be $$ g \sim 1.15 \times 10^4\ \frac{m_{\oplus}}{s_{\oplus}^2 } $$ which is a little more awkward than the 10 that it is in SI. After this all of the numbers start to get pretty silly.


One earth energy is $$ 1\ J_{\oplus} = 3.3 \times 10^{28}\ J $$ and earth force $$ 1\ N_{\oplus} = 5.1\times 10^{21}\ N $$ the gravitational constant becomes $$ G = 3.88 \times 10^{-25} \ \frac{m_{\oplus}^3 }{ g_{\oplus} s_{\oplus}^2} $$ earth power $$ 1\ W_{\oplus} = 3.8 \times 10^{23} \ W $$ earth pressure $$ 1\ Pa_{\oplus} = 1.2 \times 10^8 \ Pa $$

Electrical and Thermal

Additionally if I take the Boltzmann constant and electrical constant as fundamental dimensionfull quantities, and set them equal to 1 (i.e. CGS-type Earth units), I can use them to discover earth units dealing with electrical or thermal phenomenon. An earth kelvin is $$ 1\ K_{\oplus} = 3.2 \times 10^{28} \ K $$ earth coulomb $$ 1\ C_{\oplus} = 4.6 \times 10^{17} \ C $$ earth amp $$ 1\ A_{\oplus} = 5.3 \times 10^{12} \ A $$ an earth volt $$ 1\ V_{\oplus} = 7.1 \times 10^{10} \ V $$ an earth farad $$ 1\ F_{\oplus} = 6.4\times 10^6 \ F $$ an earth ohm $$ 1\ \Omega_{\oplus} = 14 m\Omega$$ a earth henry $$ 1\ H_{\oplus} = 1170 \ H $$ an earth electric field $$ 1\ \frac{V_{\oplus}}{m_{\oplus}} = 11200\ \frac{V}{m} $$ an earth tesla $$ 1\ T_{\oplus} = 152 \ T $$ etc....


So, it looks like if we really decide to fly in the face of the Copernican principle and look to the earth as something fundamental in the universe, these considerations can suggest a bunch of other relevant values for other kinds of dimensionfull quantities in the world. If the earth really was something super special in the universe, and if somehow its design was intimately connected with the properties of physics at large, then all of these different values ought to have some kind of deep meaning. Unfortunately, as far as I can tell, they are just a bunch of random numbers. Looks like the Copernican principle wins again. Nobody should tell the earth. Its feelings might get hurt.

Nobody Really Gets Quantum

Nobody Really Gets Quantum / Etgar Keret On Yom Kippur eve Quantum walked over to Einstein's house to seek forgiveness. "I'm not home," shouted Einstein from behind a closed door. On the way home people taunted him and somebody even hit him with an empty can of coke. Quantum pretended not to care, but deep inside he was really hurt. Nobody really gets Quantum, everybody hates him. "You parasite" people cry out when he's walking down the street, "why are you dodging the draft?" - "I wanted to enlist," Quantum tries to say, "but they wouldn't take me, because I'm so small." Not that anybody listens to Quantum. Nobody listens to Quantum when he tries to speak up for himself, but when he says something that can be misconstrued, oh, then suddenly everybody's paying attention. Quantum can say something innocent like "wow, what a cat!" and right away the news says he's making provocations and run off to talk to Schrodinger. And anyway, the media hates Quantum most, because once when he was interviewed in Scientific American Quantum said that the observer affects the observed event, and all the journalists thought he was talking about the coverage of the Intifada and claimed he was deliberately inciting the masses. And Quantum can keep talking until tomorrow about how he didn't mean it and he has no political affiliation, nobody believes him anyway. Everybody knows he's friends with Yuval Ne'eman. A lot of people think Quantum is heartless, that he has no feelings, but that's not true at all. On Friday, after a documentary on Hiroshima, he was on the expert panel. And he couldn't even speak. Just sat in front of the open mic and cried, and all of the viewers at home, that don't really know Quantum, couldn't understand that Quantum was crying, they just thought he was avoiding the question. And the sad thing about it is, even if Quantum writes dozens of letters to the editors of all the scientific journals in the world and proves beyond any doubt that for the whole atomic bomb thing he was just being used and he never thought it would end this way, it wouldn't help him, because nobody really gets quantum. Least of all the physicists. from the original Hebrew by me; posted without permission. Originally from The Girl on the Fridgewhere you can find somebody else's translation of this and other surreal short stories by the very talented Etgar Keret. Incidentally, the original story is written in the plural because in Hebrew we call quantum mechanics, roughly, "theory of the quantas;" I switched it to singular here because I think it works better this way in English. I'm not sure I've done it justice - I'm not sure you can actually do Keret's writing justice reading it out of the Israeli cultural context (for instance, many physicists will know Ne'eman for his work on QCD but only Israelis know he was politically active in a far-right party) - but I thought it was worth a shot.

End of the Earth II - Blaze of Glory

image In honor of earth day today, many bloggers are posting things about how to save the earth, or retrospectives on earth days past. We here at The Virtuosi decided, what better way to celebrate the earth than to figure out how to destroy it? So that is exactly what we intend to do. This post will focus on the destruction of the earth by a laser beam. This is a familiar concept. Whether it is Marvin the Martian or the Death Star, destroying planets with lasers (or threatening to) is a common theme. We will be considering two questions today. The first is fairly obvious: how much power would the death star need to destroy the earth? The second relates to a topic of continuing interest of mine: how much would the death star recoil upon firing? First, the power we would need. Most estimates of the death star's power seem to rely on simply overcoming the gravitational binding energy of the earth. We're going to assume the earth is a uniform density. We can easily calculate the gravitational binding energy to be $$U_G=\frac{3GM_{earth}^2}{5R_{earth}}$$ $$U=2.3\cdot 10^{32} J$$ However, looking at the video of the death star blast, it looks like the earth was more than just gravitationally unbound. It looks like we've actually managed to atomize some of the constituent molecules. Let us assume that about half of the earth gets atomized. We're also going to assume that the earth is made entirely of iron (not too bad an assumption). In a lattice, iron will have \~6 bonds, and the energy of each bond will be on the order of \~2 eV. We can calculate the number of iron atoms, N, needed to give the mass of the earth by: $$N=\frac{M_{earth}}{m_{iron}}$$ $$N=3\cdot 10^{50} atoms$$ We assume half of these have all of their bonds broken, this gives an energy required to break the bonds of $$E \approx 3N(2eV)=2.88\cdot 10^{32} J$$ This gives us a total destruction energy of $$E_{tot}=5.2\cdot 10^{32} J$$ Analyzing the video of the firing of the first death star tells us that the laser fired for about 4s, so this is a power of 1.310^32 W! As an aside, that much energy is about the total energy output of the sun over a day. That means the impulse this laser delivered (momentum per second, or force) to the death star must be the power over c, the speed of light (see my earlier post for a discussion of laser gun recoil). This is a force of 4.310^23 N. We now need to estimate the death star mass. According to confidential sources the death star had a diameter of 150 km. The death star is made of metal, but it also has a lot of empty space inside of it. We'll go ahead and assume it has the density of water. This gives is a mass of $$M_{ds}=\tfrac{4}{3}\pi r^3\rho=2.4\cdot 10^{13} kg$$ A quick glance at our numbers reveals that we're going to need special relativity here! Otherwise we'd be accelerating this thing well past the speed of light. We have a total momentum change of 1.710^{24}kgm/s. This is a relativistic momentum, which we can solve for v. The algebra is a tiny bit ugly, but it turns out that velocity in terms of relativistic momentum is $$v=\sqrt{\frac{p^2}{m^2+p^2/c^2}} \approx c $$ It turns out that the death star would be moving so close to speed of light as to not matter. That's fast! I think we can safely say that this laser recoil would be noticed!

Laser Launching

image Lasers seem to be on my mind recently. Just yesterday, the class I TA for (E&M for engineers) talked about the momentum carried by E&M waves. This called to mind a discussion I had with a housemate a few weeks back. He had heard somewhere that 'they' were thinking of launching satellites with lasers. No way, I thought to myself. Satellites are too heavy. However, his question has been hovering around in my mind, so I've decided to try and answer it: can we use a laser to launch a satellite into orbit? Let us begin with a few simplifying assumptions. I'm going to assume that we want our launched satellite to reach the escape velocity of the earth. Of course, we don't want a satellite to escape orbit, but escape velocity is calculated without considering any kind of drag forces on our launched object. To first order then, I expect achieving escape velocity will get our satellite into a relatively high orbit without actually escaping earth's gravity well. Second, I'm going to assume our laser is on the ground (reusable launching device!), and that our satellite is perfectly reflective, so we're not going to be melting it. Finally, I'm going to assume that the laser will remain effective at targeting our object up to 15 km above the surface, around the effective range for lasers used as guides for adaptive optics in telescopes. Now, to the meat of the problem. First, we need to find the escape velocity from the earth. This is defined as the velocity we would need to give an object to overcome the gravitational energy of the earth upon the object. This is found by setting the kinetic energy of the object equal to the change in gravitational potential energy from the earth's surface to infinity: $$ KE=\Delta PE $$ so $$\tfrac{1}{2}mv_{esc}^2 = \frac{GM_{earth}m}{R_{earth}}$$ (for those interested, the gravitational potential energy at infinity is defined to be zero). We can easily solve this for v, $$ v_{esc}=\sqrt{\frac{2GM_{earth}}{R_{earth}}$$ To me, the most remarkable feature of the escape velocity is that it is independent of the mass of the object! We can now plug in numbers, and find an escape velocity if \~11.2 km/s. We now make use of our constraint, that we have to achieve this escape velocity over 15 km. We assume the laser outputs a constant number of photons per second, n. The force the laser provides the satellite is the change in momentum with time, $$F=\frac{dp}{dt}$$ How is our laser imparting momentum to our system? Well, light is composed of photons, tiny packets of light. Each photon has momentum. Assume each photon reflects perfectly, that is, straight backwards and with no energy loss. Then the photon has reversed its momentum. Since momentum is conserved, the satellite has gained a momentum of 2p. The momentum of a photon is $$p=\frac{h}{\lambda}$$ where h is Planck's constant. This gives $$F=\frac{2nh}{\lambda}$$ Given a constant force, the time taken to reach some velocity is just $$t=mv/F$$ In our case the v is the escape velocity. Now, given a constant force, and no initial velocity, an object of mass m travels a distance d in a time t given by $$d=\frac{F}{2m}t^2$$ We can substitute our expression for t into this expression, giving $$d=\left(\frac{F}{2m}\right)\left(\frac{mv}{F}\right)^2$$ $$d=\frac{mv^2}{2F}$$ What we are really interested in is n, the number of photons per second this process takes, so we substitute our expression for the force, and solve for n: $$d=\frac{mv^2\lambda}{4nh}$$ s0 $$n=\frac{mv^2\lambda}{4hd}$$ Now that we have the number of photons per second our laser supplies, all that is left is to find the power of laser this would take. The power of a laser will be given by the number of photons per second times the energy per photon. The energy per photon is given by E=hf where f is the frequency of our light. So, the power, P, of our laser is $$P=nE$$ or $$P=\frac{mv_{esc}^2\lambda f}{4d}$$ so $$P=\frac{mv_{esc}^2 c}{4d}$$ Where we have recognized that the frequency times the wavelength of a photon is just the speed of light, c. All that remains is to plug in numbers. A medium sized satellite might be about 750kg, giving a laser power of 4.710^14W. This is .5 PW. According to wikipedia, the greatest power output of a continuous operation laser is on the order of 1 kW, or \~10^12 times less than our necessary power! There's no way we're getting a normal sized satellite into orbit with a laser. What about a smaller satellite? In recent years picosatellites have proposed, with masses of \~.1kg. This gives a necessary power of 6.210^10, or 620 TW. This is still \~10^8 times greater than our most powerful continuous laser. In fact, reversing the calculation, our laser could launch a satellite with a mass of \~1.6 mg. A little googling reveals this is the same order of magnitude as a grain of rice! Simply put, we're not going to be launching satellites with lasers anytime soon.