Ask a Physicist: Volume I

We have received our first Ask a Physicist e-query! An entity known only to us as "Hungry" writes: "We had a dispute at a dinner party about whether blowing on hot food actually makes it cool down faster, or only gives you something to do while you wait for your food to cool." While it is questionable whether or not I am indeed a physicist (but people do pay me to do physics) and whether I will answer definitively (there'll be some hand-waving), I'll give it my darndest. Let's imagine a hot baked potato you just sliced open. The mass of potato is capable of transferring energy as heat to the surrounding air (and therefore cooling down) via a number of mechanisms. The most common we hear about are conduction, radiation, and convection. Conduction is the process by which two objects in intimate contact exchange heat. Radiation is a bit more mysterious, but in the same way that a poker in the fire can glow white hot, all things at non-zero temperature emit electromagnetic radiation, which carries away energy from your potato. Convection is even more complicated, but is one of the most important processes by which solids exchange heat with gases. Convection in our case occurs when the potato heats up the gas at its surface. The random motion of air molecules (now a little hotter) will "randomly" dissipate heat. However, the hotter gas is less dense than the room temperature gas, and buoyant forces (think Helium balloon) will create a current of hot gas upwards from the potato surface. Both of these processes constitute convection. Convection is REALLY hard to describe with physical models. You can do some computer simulations of the motion of the gas in convection. I found this cool picture on Wikipedia of someone who did just that: image Pretty cool. Now! To actually answer your question. To get a qualitative answer, we don't even need to consider models of convection, etc. Isaac Newton was kind enough to find an empirical relation for the process of cooling, aptly named "Newton's law of cooling". If we call the amount of heat transferred from the potato Q, the found that the rate of heat being lost by the potato is proportional to the temperature difference between the surrounding gas and the potato itself: $$ \text{Rate of Heat Loss}=\frac{\text{Amount of Heat Loss}}{\text{Amount of time}}=\frac{\Delta Q}{\Delta t} \propto T_{\text{Potato}}-T_{\text{Air }} $$ So the quantity which is of primary importance is the temperature difference between your potato and the surrounding air. Now we can answer the question. If you let your potato cool down naturally, then the potato heats the air which then starts to convect. This means that the air immediately above the potato is quite hot. If you blow on the potato, however, you get the air circulating faster than it would by convection alone, effectively replacing the hot air with slightly cooler air. And since the cooling rate is proportional to to temperature difference; this will result in a faster cooling rate. To first order, blowing on your food should help. So, Hungry, feel free to blow on your food, Miss Manners be damned.

Solar Sails II

[NOTE: In my hurry to make up for weeks of non-posts, I managed to almost immediately knock Nic's first post from the top of the page. It's got the LHC, black holes, and about 3 full cups of metric awesome, so make sure you check it out (after reading this one, of course).] Last time we did some calculations on how fast and far our solar sails can go, but those calculations were just for the sail itself. If you are going to do any science with it, you're going to need a payload. Let's take it a step further and make it an actual spaceship (with people and everything!). Comparing it with a typical people-carrying space hotel (the International Space Station), let's give our payload a mass of 300,000 kg. Remembering from the last post that a sigma of less that about 10^-4 g/cm^2 gave fairly nice results, we can make an effective sigma of our payload carrying sail as $$ \sigma_{eff} = \frac{m_{total}}{Area} = \frac{m_s + m_p}{Area}, $$ where m_s is the mass of the sail and m_p is the mass of our payload (the ship). Assuming the sail has some surface density of sigma and the sail is circular with some radius r, we have $$ \sigma_{eff} = \frac{\pi r^2 {\sigma}s + m_p}{\pi r^2} = {\sigma}_s + \frac{m_p}{\pi r^2} . $$ Now we can find the radius of our sail such that $$ \sigma \le 10^{-4} \frac{g}{{cm}^2}. $$ Rearranging our equation above and solving for radius, we find that $$ r \ge \left[\frac{m_p}{\pi \left( 10^{-4}\frac{g}{cm^2} - \sigma_s \right)} \right]^{1/2} cm. $$ Below is a plot of sail radius (in meters) against sail surface density (g/cm^2). image From this plot, we see that we will need our sail radius to be AT LEAST 10 km and the surface density of our sail must be less than 10^-4 g/cm^2. Now that's a big sail, but it's not obscenely big (depending, of course, on your definitions of obscenity). One could certainly imagine such a sail being built, but it would be an impressive engineering feat. So get to work engineers! I've already made a whole plot in Mathematica, I can't do everything.

Solar Sails I

image Solar sails are in the news again, and this time not just for blowing up. The Japanese space agency is launching what they hope to be the first successful solar sail tomorrow. In honor of that, we will be discussing the physics of solar sails. First of all, what the heck are solar sails? Solar sails are a means of propulsion based on the simple observation that "Hey, sails work on boats. Therefore, they should work on interplanetary spacecraft (in space)." Boat sails work when air molecules hit into the sail and bounce back. By conservation of momentum, this gives the boat sail an itty bitty boost in momentum. Summing over the large number of air molecules moving as wind, the boat gets pushed along in the water. A similar process works with solar sails, but instead of air molecules doing the hitting, it's photons. Since each photon of a given wavelength has some momentum, by reflecting that photon the solar sail can gain a tiny bit of momentum. Summing over the large number of photons coming from the sun over a long time frame we can get a considerable boost. So let's see how good solar sails are. First we need to find the net force on our sail. We will certainly have to deal with gravitational forces (which will slow us down) : $$ F_{g} = \frac{-GM_{\odot}m}{r^2} $$ where big M is the mass of the sun and little m is the mass of the sail. Now we need to find the radiation force on the sail. Since force is just rate of change of momentum, we can find the change of momentum of one photon per unit time, then find how many photons are hitting our sail. So for one elastic collision of a photon with the sail, the change in momentum will be $$ \Delta p = 2 \frac{h\nu}{c} $$ and by conservation of momentum, this will also be the momentum gained by the sail. Now we want to find the number of photons incident on a given area in a given time. This will just be the energy flux output by the sun ( energy/ m^2 s ) divided by the energy per photon. In other words: $$ f_n = \frac{L_{\odot}}{4\pi r^2}\frac{1}{h\nu} .$$ So now we can get a force by $$ \text{Force} = \left(\frac{\Delta p}{\text{1 photon}} \right) \times \left(\frac{\text{number of photons}}{area \times time}\right) \times \left( Area\right) $$ which is just $$ F_{rad} = 2 \frac{h\nu}{c} \times \frac{L_\odot}{4\pi r^2 h\nu} \times \pi R^2 = \frac{L_{\odot} R^2}{2cr^2} .$$ So combining the radiation force with the gravitation force, we have a net force on the sail of $$ F = \left( \frac{L_{\odot} R^2}{2c} - GM_{\odot}m \right) \frac{1}{r^2} .$$ This can then be integrated over r to find an effective potential, giving: $$ U = \left( \frac{L_{\odot} R^2}{2c} - GM_{\odot}m\right)\frac{1}{r} .$$ For simplicity, let's just write that $$ \alpha = \frac{L_{\odot} R^2}{2c} - GM_{\odot}m $$ so $$ U = \frac{\alpha}{r} .$$ Now we can start saying some things about this sail. The most straightforward quantity to find would be the maximum velocity. By conservation of energy (and starting from some r_0 at rest), we have that $$ v_f = \left[\frac{2\alpha}{m} \left(\frac{1}{r_0} - \frac{1}{r_f} \right) \right]^{1/2} $$ So as r_f goes to very large values, the subtracted piece gets smaller and smaller. In the limit that r_f goes to infinity we have that $$ v_{max} = \left(\frac{2\alpha}{mr_0}\right)^{1/2} .$$ Plugging back in our long term for alpha and plugging in some numbers we get: $$ v_{max} = 42,000 m/s \left( \frac{1.5 \times 10^{-4}}{\sigma} - 1\right)^{1/2} $$ where sigma is just the surface mass density [g/cm^2] of the sail. Below is a plot of maximum velocity ( m/s) plotted against surface mass density (g/cm^2). For a sigma of 10^-4 g/cm^2, we get a max velocity of about 30,000 m/s. Not bad. image From this graph we see that there must be some maximum surface density, above which we don't get any (forward) motion at all. This makes sense since we want our radiation forces (which scale with area) to overcome our gravitational forces (which scale with mass). And below this maximal surface density we see a power law behavior. Cool. We can also find the distance traveled as a function of time. Taking the final velocity equation above and writing v as dr/dt, we see that $$ \frac{dr}{dt} = \left[ \frac{2\alpha}{m} \left( \frac{1}{r_0} - \frac{1}{r_f} \right)\right]^{1/2} $$ Rearranging and integrating, we can get time (in years) as a function of distance r (in AU): $$ t = \frac{0.11 \left(\sqrt{(-1+r) r}+\text{Log}\left[1+\sqrt{\frac{-1+r}{r}}\right]+\frac{\text{Log}[r]}{2}\right)}{\sqrt{-1+\frac{1.5 \times 10^{-4}}{\sigma}}}$$ A plot of t vs. r is shown below for typical solar system distances and a sigma of 10^-4 g/cm^2. We assume that we are launching from earth (1 AU). Since Pluto is at a distance of about 40 AU, we see that our sail could get there in less than 7 years. For comparison, the New Horizons probe will use conventional propulsion to get to Pluto in 9.5 years (and it is the fastest spacecraft ever made). image Zooming in to our starting point around 1 AU, we see that there is a period of acceleration and then the maximum velocity is reached after a few months. Just eyeballing it, it looks like it takes at least a month to reach appreciable speed. That it takes so long is a result of the very small forces involved due to radiation pressure. But even a small acceleration amounts to a considerable speed if applied for long enough! image Now Pluto is fine I guess (it's the second largest dwarf-planet!), but how about some interstellar flight? Well, the nearest star is Proxima Centauri which is about a parsec away. A parsec is 310^16 m, or about 200,000 AU. From, the plot below (or plugging in to the equation above), we see that such a trip would take of order 10,000 years. That's a long time, but its not too shabby considering this craft uses no fuel of its own. image So solar sails can do some fairly impressive things simply by harnessing the free energy of the sun. Though this only provides a very small acceleration, it can be taken over a long enough time to be useful. However, since the radiation pressure of the sun falls off as 1/r^2, we start to observe diminishing returns and the sail reaches a max velocity. But overall the numbers seem fairly impressive. All that remains now is whether they are feasible to construct. Right now my only data point for feasibility was that it was in Star Wars, but as I recall that was a long* time ago.

Solar Sails Addendum I

As requested, below is an explicit evaluation of the silly looking integral in Solar Sails I. If you just want some hints to do the integral, see Solar Sails Addendum II. Here we present a step-by-step solution of the differential equation: $$ \frac{dr}{dt} = \left[ \frac{2\alpha}{m} \left( \frac{1}{r_0} - \frac{1}{r} \right)\right]^{1/2} $$ This is just a separable equation, so we rearrange to get an integral equation: $$ \int_{r_0}^{r_f} \frac{dr}{\left[ \frac{2\alpha}{m} \left( \frac{1}{r_0} - \frac{1}{r} \right)\right]^{1/2}} =\int_{0}^{t}dt$$ We see that the right hand side here will just evaluate to t. Let's rearrange the left hand side to get the integration variable to be dimensionless. This is important because it allows the integral to just be a number, with all the unit-dependent terms pulled outside. So we have $$ \int_{r_0}^{r_f} \frac{dr}{\left[ \frac{2\alpha}{mr_0} \left( 1 - \frac{r_0}{r} \right)\right]^{1/2}} =t$$ \noindent Now we can do a change of variables to get the dimensionless variable u = r/r0. This is just giving us our distance in terms of our initial distance. In the problem I took r_0 to be 1 AU. So u just gives our distance now in terms of AU: $$ \int_{1}^{u_f} \frac{du}{\left[ \frac{2\alpha}{m{r_0}^3} \left( 1 - \frac{1}{u} \right)\right]^{1/2}} =t$$ So lets set $$ k = (m{r_0}^3}/{2 \alpha)^{1/2} ,$$ which just gives $$ k\int_{1}^{u_f} \frac{du}{\left[ \left( 1 - \frac{1}{u} \right)\right]^{1/2}} =t$$ Now we are ready to get started! Typically, when I see something with a square root in the denominator that's giving me trouble, I just blindly try trig substitutions. Let's try u = [csc(x)]^2, so 1/u = [sin(x)]^2 and du = -(csc x)^2 * cot x, and $$ \int_{1}^{u_f} \frac{du}{\left[ \left( 1 - \frac{1}{u} \right)\right]^{1/2}} =\int_{x_0}^{x_f} \frac{-2\csc^2x \cot x dx}{\left(1-\sin^2x \right)^{1/2}} = \int_{x_0}^{x_f} -2\csc^3x dx ,$$ where x_0 = pi/2 and x_f = arcsin u_f . The last equality above just comes from simplifying the trig expressions. So now how do we solve this "easier" problem? As a wise man once said, "When in doubt, integrate by parts." So let's try that. Expanding out a bit we see that: $$ -\int_{x_0}^{x_f} \csc^3x dx = \int_{x_0}^{x_f}\csc x \left(-\csc^2x \right)dx $$ Remembering that integration by parts goes like $$ \int u dv = uv - \int v du $$ we can set u = csc x and v = \cot x$ to get $$ \int_{x_0}^{x_f}\csc x \left(-\csc^2x \right)dx = \csc x \cot x \Big |{x_0}^{x_f} - \int^{x_f} \cot x \left(-\csc x \cot x \right) dx $$ which is just $$ -\int_{x_0}^{x_f}\csc^3x dx = \csc x \cot x \Big |{x_0}^{x_f} + \int^{x_f} \cot^2 x \csc x dx $$ Remembering that $$ \cot^2 x = \csc^2 x -1 ,$$ we have $$ -\int_{x_0}^{x_f}\csc^3x dx = \csc x \cot x \Big |{x_0}^{x_f} + \int^{x_f} \left(\csc^2 x - 1 \right) \csc x dx $$ which we can expand to $$ -\int_{x_0}^{x_f}\csc^3x dx = \csc x \cot x \Big |{x_0}^{x_f} - \int^{x_f} \csc x dx + \int_{x_0}^{x_f} \csc^3 x dx$$ But this is just what we want! Rearranging we now have that $$ -2\int_{x_0}^{x_f}\csc^3x dx = \csc x \cot x \Big |{x_0}^{x_f} - \int^{x_f} \csc x dx$$ Remembering that the integral for csc is $$ \int \csc x dx =-\ln | \csc x + \cot x| + C $$ and evaluating our limits we have that $$ -2\int_{x_0}^{x_f}\csc^3x dx = \csc x_f \cot x_f - \csc x_0 \cot x_0 +\ln \left( \frac{| \csc x_f + \cot x_f|}{| \csc x_0 + \cot x_0|} \right) $$ Now we just need to evaluate at x_0 = pi/2 and x_f = arcsin u_f. We can draw a right triangle with far angle x_f, hypoteneuse of length sqrt{u}, and legs of length 1 and sqrt{u-1} to see that csc x_f = sqrt{u} and cot x_f = sqrt{u-1}, so $$ -2\int_{x_0}^{x_f}\csc^3x dx = \sqrt{u}\sqrt{u-1} + \ln{ | \sqrt{u} + \sqrt{u-1}|} $$ And this is what we have sought from the beginning. Using the last two equations on the first page we see that $$ t = k\int_{1}^{u_f} \frac{du}{\left[ \left( 1 - \frac{1}{u} \right)\right]^{1/2}} = k\left[-2\int_{x_0}^{x_f}\csc^3x dx \right] = k \left[\sqrt{u}\sqrt{u-1} + \ln{ | \sqrt{u} + \sqrt{u-1}|} \right] $$ Plugging back in our value of k, we have $$ t = \left(\frac{m{r_0}^3}{2\alpha} \right)^{1/2}\left[\sqrt{u}\sqrt{u-1} + \ln{ | \sqrt{u} + \sqrt{u-1}|} \right] $$ Simplifying a bit and dropping the absolute value bars since u will always be bigger than u-1, we have our final answer: $$ t = \left(\frac{m{r_0}^3}{2\alpha} \right)^{1/2}\left[\sqrt{u(u-1)} + \ln{ \left( \sqrt{u} + \sqrt{u-1}\right)} \right] $$ where u = r / r_0 is our non-dimensional distance measurement. And this is (up to some rearranging) exactly what we get in the initial post.

Solar Sails Addendum II

This is the schematic version, if you just wanted hints. The full solution is given in Solar Sails Addendum I. Here we present a schematic solution of the differential equation: $$ \frac{dr}{dt} = \left[ \frac{2\alpha}{m} \left( \frac{1}{r_0} - \frac{1}{r} \right)\right]^{1/2} $$ This is just a separable equation, so we rearrange to get an integral equation: $$ \int_{r_0}^{r_f} \frac{dr}{\left[ \frac{2\alpha}{m} \left( \frac{1}{r_0} - \frac{1}{r} \right)\right]^{1/2}} = \int_{0}^{t}dt$$ From here it's nice to non-dimensionalize, so our integration variable is just a number (with no units attached). This allows us to get the integral into a form like $$ k\int_{1}^{u_f} \frac{du}{\left[ \left( 1 - \frac{1}{u} \right)\right]^{1/2}} = t$$ for appropriate values of k and u. Now we are ready to get started! Typically, when I see something with a square root in the denominator that's giving me trouble, I just blindly try trig substitutions. After an appropriate trig substitution, we get something of the form $$ \int_{1}^{u_f} \frac{du}{\left[ \left( 1 - \frac{1}{u} \right)\right]^{1/2}} = \int_{x_0}^{x_f} -2\csc^3x dx$$, So now how do we solve this "easier" problem? As a wise man once said, "When in doubt, integrate by parts." So let's try that. $$ \left[ \mbox{HINT:} -\int_{x_0}^{x_f} \csc^3x dx = \int_{x_0}^{x_f}\csc x \left(-\csc^2x \right)dx \right]$$ Remembering that integration by parts goes like $$ \int u dv = uv - \int v du $$ we can pick appropriate values of u and v to get something nice, which eventually leads to $$ -\int_{x_0}^{x_f}\csc^3x dx = \csc x \cot x \Big |{x_0}^{x_f} - \int^{x_f} \csc x dx + \int_{x_0}^{x_f} \csc^3 x dx$$ But this is just what we want! Rearranging we now have that $$ -2\int_{x_0}^{x_f}\csc^3x dx = \csc x \cot x \Big |{x_0}^{x_f} - \int^{x_f} \csc x dx$$ Evaluating our integrals, we see that $$ -2\int_{x_0}^{x_f}\csc^3x dx = \sqrt{u}\sqrt{u-1} + \ln{ | \sqrt{u} + \sqrt{u-1}|} $$ And this is what we have sought from the beginning. Plugging back in to our earlier equations and rearranging gives $$ t = \left(\frac{m{r_0}^3}{2\alpha} \right)^{1/2}\left[\sqrt{u(u-1)} + \ln{\left( \sqrt{u} + \sqrt{u-1}\right)} \right] $$ where u = r / r_0 is our non-dimensional distance measurement. And this is (up to some algebra) exactly what we get in the initial post.

Why Black Holes from the Large Hadron Collider Won't Destroy the World

Hi everyone. As this is my first post, I thought I'd introduce myself. Like the rest of the Virtuosi, I'm a graduate student in physics at Cornell University. I work in experimental particle physics, in particular on the Compact Muon Solenoid, one of the detectors at the Large Hadron Collider. I'll post more on what I actually do at some point in the future, but I thought I'd start with a post in the spirit of some of the other fun calculations that we've done. My goal is to convince you that black holes created by the LHC cannot possibly destroy the world. To start with, the main reason no one working on the LHC is too concerned about black holes is because of Hawking radiation. While we usually think of black holes as objects that nothing can escape from, Stephen Hawking predicted that black holes actually do emit some light, losing energy (and mass) in the process. In the case of the little bitty black holes that the LHC could produce, they should just evaporate in a shower of Hawking radiation. That's great you say, but Hawking radiation has never actually been observed. What if Hawking is wrong and the black holes won't evaporate? Well, the usual next argument is that cosmic rays from space bombard the earth all the time, producing collisions many times more energetic than what we'll be able to produce at the LHC. To me, this is a fairly convincing argument. However, let's pretend we don't know about these cosmic rays and that there's no Hawking radiation. We can calculate what effect black holes produced by the LHC would have on the earth if they do stick around. To start out with, the most massive black hole the LHC could produce would be around 10 Tera-electron-volts, or TeV. We're probably overestimating here. The eventual goal is for the LHC collisions to be 14 TeV, but producing a particle with the entire collision energy is incredibly unlikely (see Tomasso Dorigo's post for more details on why, along with more details than you probably wanted to know about hadron colliders). However, we want to think about the worst case scenario here, and we're just going to do an order of magnitude calculation, so 10 TeV is a good number. Note that I'm using a particle physics convention here of giving masses in terms of energies using E=mc^2. For reference, 10 TeV is about 1000 times smaller than a small virus. Now from the mass of our black hole, we can get its size by calculating something called the Schwarzschild radius. The Schwarzchild radius for a black hole of mass m is given by $$r = \frac{2Gm}{c^2}\text{.}$$ Here G is Newton's gravitational constant and c is the speed of light. Plugging our mass in gives us $$r = 10^{-50} \text{meters.}$$ This is incredibly small! In fact as I was writing this, I realized that it's actually smaller than the Planck length, which means our equation for the Schwarzschild radius may be somewhat suspect. Nonetheless, let's hope that if we ever figure out quantum gravity, it gives us a correction of order one and proceed with our calculation, which is just an order-of-magnitude affair anyway. Now, anything that enters the Schwarzschild radius of the black hole is absorbed by it. The lightest thing that we could imagine the black hole swallowing is an electron. Let's figure out how long on average a black hole would have to travel through material with the density of the earth before it absorbs an electron. In the spirit of considering the worst case scenario, we'll have the black hole travel at the speed of light, and consider the earth to be the density of lead. We could do a complicated cross-section calculation to find the rate at which the black hole accumulates mass, but we can also get it right up to factors of pi through unit analysis. We know that the answer should involve the area of the black hole, the density of the earth, and the speed of the black hole. We want our answer to have units of mass per time to represent the mass accumulation rate of the black hole. The only combination that gives the right units is $$a=\frac{\text{mass}}{\text{time}} = \rho c r^2=\frac{10,000\text{kg}}{\text{m}^3}\frac{3\times 10^8 \text{m}}{\text{s}}(10^{-50}\text{m})^2} = 10^{-88}\text{kg/s}\text{.}$$ Alright, now that we know how fast our black hole accumulates mass, let's figure out how long it takes it to accumulate an electron. The electron mass is $$m_e = 10^{-30}\text{kg,}$$ so the time to accumulate an electron is $$t = \frac{m_e}{a} = 10^{50}\text{s.}$$ Now, the current age of the universe is 10^17 seconds. The time it takes our black hole to accumulate an electron is longer than the age of the universe by many orders of magnitude! So, if the LHC produces black holes, and if Hawking is wrong, the black holes will just fly straight through the earth without interacting with anything. Even if we take the size of the black hole to be the Planck length, our black hole accumulates an electron in 10^25 seconds, which is still much longer than the age of the universe. So the moral of the story is that you should be excited about the new discoveries that the LHC might produce, and you don't need to worry about black holes.

Freezing in Space II - Turn On The Sun!

Yesterday I considered how long it would take a human to freeze in space. However, I considered only what would happen if you were not absorbing any radiation from nearby sources. Today we consider what happens if you do have hot objects nearby. Namely, the sun. The sun provides a lot of energy, even as far away from it as we are. It keeps the earth at a comfortable \~20 C, good for us humans, and provides the energy for life on earth, also good for us humans. That's a lot of energy. So maybe the sun can keep you alive when you're adrift in space. Or at least keep you warm. I still think you'll asphyxiate. From here on out we're going to assume that we are adrift in space near earth. You were out for a joyride in that new spaceship of yours and something went horribly wrong. We could go through a whole song and dance of calculating how much power the sun delivers to the earth, but we won't (if you're interested, let me know, an I can do that later). Instead, we'll quote the known result, that the sun delivers \~1370W/m^2 in the vicinity of the earth. To find out what temperature we would cool to we set the power we absorb from the sun equal to the power we radiate $$P_{sun}=P_{rad}$$ $$1370W/m^2A_{ab}e_{ab}=e_{rad}A_{rad}\sigmaT^4$$ Where A_ab is the surface area absorbing the suns power, e_ab is a factor between 0 and 1 that indicates how much of the incident power we actually absorb, and e_rad is the emissivity of us, while A_rad is our radiating area. Note that the emitting and absorbing areas are not the same! Take a simple example. If you put a sheet in space, and face the flat side towards the sun, it will only absorb energy from the sun on one side, but it will radiate energy from both sides. Likewise e_ab and e_rad are not necessarily equal because we are radiating and absorbing at different wavelengths. We can solve the above equation for T, giving $$T=\left(\frac{A_{ab}}{A_{rad}}\right)^{1/4}\left(\frac{e_{ab}}{e_{rad}}\right)^{1/4}\left(\frac{1370W/m^2}{\sigma}\right)^{1/4}$$ For a first pass, we'll make the simplifying assumption that e_ab=e_rad. Given this, $$T=394K\left(\frac{A_{ab}}{A_{rad}}\right)^{1/4}$$ Now, the absorption area of an object is just the shape of the object flattened into the plane the incident radiation is perpendicular to. That is, the absorption area of a sphere is a circle (a sphere projected to 2D is always a circle), while the absorption area of a cylinder could be a sheet or a circle, or something stranger. The best area ratio we can ever have is that of a flat sheet, which gives 1/2. For a sphere, like the earth, the ratio is 1/4. As an aside, this gives an equilibrium temperature of the earth as \~5C, which is too cold. It turns out that we shouldn't neglect either the emissivity ratio or the natural greenhouse effect in the case of the earth. Now, we need to figure out the area ratio for us. In a previous post I modeled myself as a cylinder with height 1.8 m and radius .14 m. Let us assume we are facing the sun dead on, beating down on our chests. This gives the cross sectional area of a sheet with width 2.14 m =.28 m and height 1.8 m. This is an area of .5 m^2, while my total surface area is 1.7 m^2. This gives an area ratio of \~.3, or an equilibrium temperature of $$T=394K(.3)^{1/4}\approx292K$$ That is an equilibrium temperature of 19 C. Not too cold, but certainly not body temperature! So the sun will not save us. We also have to factor in the fact that we reflect better in the visual that we do in the infrared, so the emissivity ratio we set to 1 probably is less than that, reducing our equilibrium temperature even more. It is interesting to note, though, that if we model a human as a two sided sheet instead of a cylinder, we can bring our equilibrium temperature up to 331 K. That's \~58 C! So in our model our geometrical assumptions change whether or not we freeze or die of heat stroke. Finally, since it looks like the sun may not save us, lets see how much it might slow down our temperature loss. Instead of a net loss of 860W at body temperature, as we calculated yesterday, sticking with our cylindrical human we'll have a net loss of $$860W-1370W/m^2(.5m^2)=175W$$ Similarly at our lowered body temperature of \~30 C, we'll be losing a net of \~105 W. Once again taking a geometric average gives an average power loss of \~135 W. Using the energy to cool we found yesterday it would take 16300 s, or 4.5 hours to freeze! Also note that if you're getting too hot or cold, given how much the geometry plays into things, by changing your orientation to the sun you'll be able to have a certain amount of control over how much you heat up or cool down. Also, make sure you rotate yourself so that you end up evenly heated, and not roasted on one side and frozen on the other!

Freezing in Space I - Blackest Night

In the last post I made, I discussed the fact that humans radiate energy. In that post I calculated that we actually radiate quite a lot of power. This immediately raises a few questions, the most obvious one being: How long would it take you to freeze in space? This question is multifaceted, and I'm going to split it between two parts. This first part, 'Blackest Night' is how quickly we'd freeze if we were completely lost in space, nothing anywhere near. The second part, 'Turn On The Sun!' will address what would happen in near earth orbit. We need to clarify what we mean when we say freezing in space. The fatal temperature change for a human is (according to the all knowing internet), roughly a drop of 7C. If you remember my previous post, we calculated that we would radiate about 860 W. Now, we have to ask how much energy it takes to change our temperature by 7 C. Well, as I've discussed a few times, the energy it takes to change the temperature is given by $$Q=mc\Delta T$$ The mass of a human is \~75 kg. We're mostly water, so let's just assume that we are all water. This gives us a specific heat of 4.2 kJ/kgK. The energy needed for our 7 C temperature change is then $$Q=(75kg)(4.2kJ/kgK)(7K)=2.2 MJ$$ By the time we have dropped to 30 C, we are only radiating a power of $$P=eA\sigma T^4\=(.97)(1.7m^2)(5.67\cdot10^{-8}W/m^2K^4)(303K)^4=790W$$ Let us assume that the average power radiated is the geometric average of these two powers, $$P_{avg}=\sqrt{(860W)(790W)}\approx825W$$ This gives us a time to freeze of $$t=\frac{2.2MJ}{825W}\approx2700s$$ This is \~45 minutes. So you've got 45 minutes until a deadly freeze in deep space. Seems a rather long time, does it not? I'm fairly certain that you'd freeze much faster in antarctica than deep space. Why? Because in Antarctica you have more cooling mechanisms that just radiation, you have conduction in the air around you and convection of that warmer air away from your skin. If you're adrift in space, for all that it is rather cold, the good news is that you'll asphyxiate before you freeze! All of this was done assuming that there's no energy gain from anywhere. That is, that we're stuck somewhere in the deepest space, the blackest night. Tomorrow we'll consider what would happen if you were in near earth orbit, with all of these lovely energy sources around, particularly the sun.

Human Radiation

Things are still busy here at the Virutosi. Hopefully in a week or so we'll be back to normal, and much more active than we've been recently Anyways, today I'd like to consider human radiation. It is well known that any object will radiate energy based on its temperature. Even more interesting, we radiate at all wavelengths, though at the human body temperature our radiation is sharply peaked in the infrared. Even so, we still put out some x-ray radiation. As a professor of mine once said, consider that next time you sleep with someone! Given all this, the question on my mind today is: how does the energy we radiate daily compare to the energy we consume? That is, why don't I lose weight sitting here typing on the computer? We physicists call perfect radiators black bodies (something that radiates perfectly also must absorb perfectly). For perfect radiators, the power radiated is given by $$P=\sigma A T^4$$ where sigma is the stephan-boltzman constant, A is the surface area of the radiator, and T is the temperature of the radiator. For objects that are not perfect emitters or perfect absorbers, we through in a fudge factor, e the emissivity, which is between 0 and 1. This makes the power emitted $$P=e \sigma A T^4$$ To figure out the power radiated by a human, we need to know three things. The first is the emissivity of human skin. It turns out this is .97. The second is the temperature of a human body, \~37 C. The third is the surface area of a human. This requires a little estimation. I'm about 180 cm tall, and I wear 35" waist pants, so my radius is \~14 cm. Modeling myself as a cylinder, I have a surface area of \~1.7 m^2. Now we can estimate my power output: $$P=.975.67\cdot 10^{-8}W/m^2K^4 1.7 m^2 * (310K)^4$$ $$P \approx 860W$$ I'm powerful! That's about 14 (60W) lightbulbs! We'd like to compare that to our daily energy intake, so we need to turn this power into an energy. Well, there are 86400 s in a day. So we radiate 7410^6 J per day. If you read my beer diet post, you'll know the conversion between J and Calories (note the capitol C). If not, suffice it to say that 1 Calorie = 4.210^3 J. If I consume about 2000 Calories a day (typical, right?), then I take in about 8.410^6 J per day. So, dear reader, why haven't I lost weight while typing this post? There are a few answers. I'm wearing insulating layers, clothing which keeps in some of my radiated heat. Also, our skin temperature is lower than our internal temperature of 37 C. But more than that, I'm absorbing energy from the surroundings. The earth is a fairly good blackbody radiator, with an average temperature of \~20 C. This means that my net power loss, with no clothing, would be about $$\Delta P \approx (5.67\cdot 10^{-8} W/m^2K^4)(1.7m^2)((310K)^4-(293K)^4)$$ $$\Delta P = 180W$$ This is \~1510^6 J per day. While still more than my intake, this is much closer, and you can imagine that the rest of the difference can be made up by our lower skin temperature, and clothing and such instruments of men. Of course, these numbers are rough, so I don't recommend the 'naked diet', where you try to lose weight by walking around naked. Or if you do try it, don't say I told you to when you're taken in for indecent exposure!

Letting Air Out of Tires II

In a recent post I calculated how cold air coming out of bike tires should feel. However, at the end of the post, I did note that there are competing explanations for why the air cools. There's the approach I took, which is adiabatic cooling, but there's also something called the Joule-Thomson effect. The Joule-Thomson effect has the interesting property that helium being let out of a bike tire would actually be warmer, which suggests an immediate way to test which effect is dominant. We pressurize a bike tire with helium, and see if the valve gets cold or hot. This is exactly what I did. With the help of Mark and Vince, our local equipment gurus, I was able to pressurize a bike tire to 26 psi with helium. Using one of those little thermal measurers you can buy at radio shack, we measured the initial temperature of the valve as 80 F. We then released the helium, and measured the temperature of the valve as 73 F. The adiabatic approach is the winner! Our experiment confirms that the dominant effect of the cooling is the adiabatic cooling I talked about yesterday. The Joule-Thomson effect may be at play, but if so it takes a secondary role to the adiabatic cooling. Now, some of you may be saying: wait a second, you predicted the air would be -100 F! It doesn't feel that cold! Nor did your valve cool down to -100 F! To which I reply: Yes, I did predict very cold air. But you have to remember that it is mixing with a lot of room temperature air, so it won't feel as cold as I predicted. Nor will it transfer much heat to the valve (recall, we predicted this would be an adiabatic process, with absolutely no heat transfer, something that is obviously false). Also, I didn't have 60 psi of pressure. If we do the calculation, 26 psi only gives a temperature of 250K = - 10 F. Hopefully that answers your question. And now, dear reader, as I've wanted to say for a while: Myth Busted!