Fragments: Posts tagged 'physics'urn:http-www-tfeb-org:-fragments-tags-physics-html2015-10-10T10:31:28ZMelting the Antarctic ice sheeturn:http-www-tfeb-org:-fragments-2015-10-10-melting-the-antarctic-ice-sheet2015-10-10T10:31:28Z2015-10-10T10:31:28ZTim Bradshaw
<p>How long might this take, in the worst case?</p>
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<p>The Antarctic ice sheet has a volume of about
<script type="math/tex">26.5\times 10^6\text{km}^3</script>, according to <a href="https://www.bas.ac.uk/project/bedmap-2/">Bedmap2</a><sup><a href="#2015-10-10-melting-the-antarctic-ice-sheet-footnote-1-definition" name="2015-10-10-melting-the-antarctic-ice-sheet-footnote-1-return">1</a></sup>. This is
<script type="math/tex">2.7\times 10^{16}\text{m}^3</script> of ice. The density of ice is about
<script type="math/tex">10^3 \text{kg}/\text{m}^3</script> (about a tonne per cubic metre, which is approximately the same as water of course), so this is about
<script type="math/tex">2.7\times 10^{19}\text{kg}</script> of ice. The <a href="https://en.wikipedia.org/wiki/Enthalpy_of_fusion">enthalpy of fusion</a> of water is about
<script type="math/tex">3.3\times10^5\text{J}/\text{kg}</script> so, if we assume that the ice is all at freezing point<sup><a href="#2015-10-10-melting-the-antarctic-ice-sheet-footnote-2-definition" name="2015-10-10-melting-the-antarctic-ice-sheet-footnote-2-return">2</a></sup>, then we require
<script type="math/tex">8.9\times 10^{24}\text{J}</script> to melt it all.</p>
<p>Let’s assume we use the Sun to do this. The <a href="https://en.wikipedia.org/wiki/Solar_constant">solar constant</a> is about
<script type="math/tex">1.4\times 10^3 \text{W}/\text{m}^2</script>: this is the amount of power per square meter that the Sun provides at the top of the atmosphere. So, imagine we use <em>all</em> of the power that the Earth intercepts from the Sun to do this<sup><a href="#2015-10-10-melting-the-antarctic-ice-sheet-footnote-3-definition" name="2015-10-10-melting-the-antarctic-ice-sheet-footnote-3-return">3</a></sup>. Well, the Earth’s radius is about
<script type="math/tex">6.4\times 10^6\text{m}</script> so the total power available is about
<script type="math/tex">1.4\times 10^3 \times \pi \times (6.4\times 10^6)^2\text{W}\\ \approx 1.8\times 10^{17}\text{W}</script>, or about
<script type="math/tex">1.8\times 10^{17}\text{J}/\text{s}</script>.</p>
<p>So to melt the Antarctic ice cap, using all of the power from the Sun that reaches the top of the atmosphere would take</p>
<p>
<script type="math/tex; mode=display">
\frac{8.9\times 10^{24}}{1.8\times 10^{17}}\text{s}
= 4.9\times 10^7\text{s}</script></p>
<p>Well, there are about
<script type="math/tex">32\times 10^6</script> seconds in a year, so this is about 1.5 years.</p>
<p>Of course we can’t use all the Sun’s power: even if we had the technology to do this (which we are <em>not anywhere near</em> doing!), doing this would cause an inconceivable catastophe for the rest of the planet: this would be a winter night which lasted for a year and a half. Everyone would die.</p>
<p>A plausible figure might be a tenth of one percent<sup><a href="#2015-10-10-melting-the-antarctic-ice-sheet-footnote-4-definition" name="2015-10-10-melting-the-antarctic-ice-sheet-footnote-4-return">4</a></sup>: in this case the Antarctic ice sheet would melt in about
<script type="math/tex">1500</script> years.</p>
<hr />
<div class="footnotes">
<ol>
<li id="2015-10-10-melting-the-antarctic-ice-sheet-footnote-1-definition" class="footnote-definition">
<p><a href="http://www.bbc.co.uk/news/science-environment-21692423">BBC news article on Bedmap2</a>, <a href="http://www.the-cryosphere.net/7/375/2013/tc-7-375-2013.pdf">paper (PDF)</a>. <a href="#2015-10-10-melting-the-antarctic-ice-sheet-footnote-1-return">↩</a></p></li>
<li id="2015-10-10-melting-the-antarctic-ice-sheet-footnote-2-definition" class="footnote-definition">
<p>The ice is, of course, far below freezing so the actual energy required will be much greater. <a href="#2015-10-10-melting-the-antarctic-ice-sheet-footnote-2-return">↩</a></p></li>
<li id="2015-10-10-melting-the-antarctic-ice-sheet-footnote-3-definition" class="footnote-definition">
<p>This is enormously more than the amount of power that we could plausibly use: see later. <a href="#2015-10-10-melting-the-antarctic-ice-sheet-footnote-3-return">↩</a></p></li>
<li id="2015-10-10-melting-the-antarctic-ice-sheet-footnote-4-definition" class="footnote-definition">
<p>This is just a number I have pulled out of thin air: one percent seems too high, so perhaps a tenth of one percent is plausible. <a href="#2015-10-10-melting-the-antarctic-ice-sheet-footnote-4-return">↩</a></p></li></ol></div>Black body planeturn:http-www-tfeb-org:-fragments-2015-09-30-black-body-planet2015-09-30T14:51:09Z2015-09-30T14:51:09ZTim Bradshaw
<p>A model of the planets as black bodies is surprisingly accurate, except in one interesting case<sup><a href="#2015-09-30-black-body-planet-footnote-1-definition" name="2015-09-30-black-body-planet-footnote-1-return">1</a></sup>.</p>
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<h2 id="in-theory">In theory</h2>
<p>A <a href="https://en.wikipedia.org/wiki/Black_body">black body</a> is an ideal object which absorbs all radiation which falls on it, and then reemits it as thermal radiation. Real objects are, of course, not black bodies, but they are often surprisingly close. One nice thing about black bodies is that there are is a nice equation which relates the amount of power they radiate to their temperature:</p>
<p>
<script type="math/tex; mode=display"> \frac{P}{A} = \sigma T^4</script></p>
<p>Where
<script type="math/tex">P</script> is power,
<script type="math/tex">A</script> is area and
<script type="math/tex">\sigma</script> is the <a href="https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_constant">Stefan-Boltzmann constant</a> which is
<script type="math/tex">\sigma \approx 5.67\times10^{−8}\text{W}\text{m}^{-2}\text{K}^{−4}</script>: this formula tells you the power per unit area that a black body radiates, at a given temperature.</p>
<p>Well, obviously you can derive this formula:</p>
<p>
<script type="math/tex; mode=display"> T = \left(\frac{P}{A\sigma}\right)^{1/4}</script></p>
<p>In other words temperature goes as the fourth root of power.</p>
<p>If you consider a ball of radius
<script type="math/tex">r</script> then its surface area is
<script type="math/tex">A = 4\pi r^2</script>, so a perfectly spherical black body of radius
<script type="math/tex">r</script> at a uniform temperature
<script type="math/tex">T</script> radiates a total output power
<script type="math/tex">P_O</script></p>
<p>
<script type="math/tex; mode=display"> P_O = 4\pi r^2 \sigma T^4</script></p>
<p>or equivalently</p>
<p>
<script type="math/tex; mode=display"> T = \left(\frac{P_O}{4\pi r^2 \sigma}\right)^{1/4}</script></p>
<p>OK, so consider a planet, which is a perfect black-body, orbiting at a radius
<script type="math/tex">R</script> from the Sun<sup><a href="#2015-09-30-black-body-planet-footnote-2-definition" name="2015-09-30-black-body-planet-footnote-2-return">2</a></sup>. Let the input power flux from the Sun, at the point directly facing the Sun, be
<script type="math/tex">S</script>. For Earth,
<script type="math/tex">S \approx 1360\text{W}\text{m}^{-2}</script>. We can calculate two things from this.</p>
<p>The total output power of the Sun,
<script type="math/tex">P_S</script> is given by the integral of
<script type="math/tex">S</script> over a sphere of radius
<script type="math/tex">R</script>:</p>
<p>
<script type="math/tex; mode=display"> P_S = 4\pi R^2 S</script></p>
<p>The total power falling on the planet,
<script type="math/tex">P_I</script>, is given by the integral of
<script type="math/tex">S</script> over the surface of the disk of the planet which faces the Sun, and this is</p>
<p>
<script type="math/tex; mode=display"> P_I = \pi r^2 S</script></p>
<p>So the first of these equations can be used to work out
<script type="math/tex">S</script> in terms of
<script type="math/tex">P_S</script> and then substituted into the second one:</p>
<p>
<script type="math/tex; mode=display"> P_I = P_S\frac{r^2}{4 R^2}</script></p>
<p>If the planet is at equilibrium, then
<script type="math/tex">P_O = P_I</script> or, in other words, output and input power is the same<sup><a href="#2015-09-30-black-body-planet-footnote-3-definition" name="2015-09-30-black-body-planet-footnote-3-return">3</a></sup>. So</p>
<p>
<script type="math/tex; mode=display"> T = \left(\frac{P_S}{16\pi \sigma R^2}\right)^{1/4}</script></p>
<p>Finally, given
<script type="math/tex">S</script> and
<script type="math/tex">R</script> we can work out
<script type="math/tex">P_S</script>, and for our Sun, based on
<script type="math/tex">R \approx 1.50\times 10^{11}\text{m}</script> for Earth we get
<script type="math/tex">P_S \approx 3.85\times 10^{26}\text{W}</script>.</p>
<h2 id="in-practice">In practice</h2>
<p>So, we can compute the surface temperatures for the rocky planets, assuming they were black bodies.</p>
<p><strong>Earth.</strong>
<script type="math/tex">R \approx 1.50\times 10^{11}\text{m}</script>, giving
<script type="math/tex">T \approx 278\text{K}</script> or
<script type="math/tex">5^\circ\text{C}</script>. Actual mean surface temperature is
<script type="math/tex">287\text{K}</script> or
<script type="math/tex">14^\circ\text{C}</script>: this is reasonably accurate.</p>
<p><strong>Mercury.</strong>
<script type="math/tex">R \approx 5.79\times 10^{10}\text{m}</script> giving
<script type="math/tex">T \approx 448\text{K}</script> or
<script type="math/tex">175^\circ\text{C}</script>. Actual mean surface temperature is
<script type="math/tex">452\text{K}</script> or
<script type="math/tex">179^\circ\text{C}</script>. This is also OK.</p>
<p><strong>Mars.</strong>
<script type="math/tex">R \approx 2.28\times 10^{11}\text{m}</script>, giving
<script type="math/tex">T \approx 226\text{K}</script> or
<script type="math/tex">-47^\circ\text{C}</script>. Actual mean surface temperature is
<script type="math/tex">226\text{K}</script> or
<script type="math/tex">-47^\circ\text{C}</script>. This is pretty spooky: it has no right to be this good, and is probably only this good by chance.</p>
<p><strong>Pluto.</strong>
<script type="math/tex">R \approx 5.91\times 10^{12}\text{m}</script>, giving
<script type="math/tex">T \approx 44\text{K}</script> or
<script type="math/tex">-229^\circ{C}</script>. Actual mean surface temperature is
<script type="math/tex">44\text{K}</script> or
<script type="math/tex">-229^\circ\text{C}</script>. The same is true for this: it is probably only this good by chance.</p>
<p><strong>Venus.</strong>
<script type="math/tex">R \approx 1.08\times 10^{11}\text{m}</script>, giving
<script type="math/tex">T \approx 328\text{K}</script> or
<script type="math/tex">55^\circ\text{C}</script>. Actual mean surface temperature is
<script type="math/tex">730\text{K}</script> or
<script type="math/tex">460^\circ\text{C}</script>. This is hopeless, as you would expect it to be.</p>
<p>These are astonishingly accurate, except for Venus, which is out by a factor of nearly two: Venus is the hottest planet in the Solar system because of a runaway greenhouse effect.</p>
<h2 id="notes">Notes</h2>
<p>Data came from <a href="http://www.wolframalpha.com/">Alpha</a> which, while I have considerable qualms about Wolfram products, is a reasonably good source of this sort of thing. I have been very casual about rounding, generally rounding to integers for both
<script type="math/tex">\text{K}</script> and
<script type="math/tex">{}^\circ\text{C}</script>. The temperatures don’t come with an uncertainty but I suspect that for Venus is less accurate than the others. I didn’t deal with the gas giants because they don’t have well-defined surfaces and I was just too lazy.</p>
<hr />
<div class="footnotes">
<ol>
<li id="2015-09-30-black-body-planet-footnote-1-definition" class="footnote-definition">
<p>This is mostly an experiment with maths in frog. The conclusion is that it’s possible, but does not look great: I will stick to LaTeX. <a href="#2015-09-30-black-body-planet-footnote-1-return">↩</a></p></li>
<li id="2015-09-30-black-body-planet-footnote-2-definition" class="footnote-definition">
<p>Assume also the planet is distant from the Sun so we don’t need to worry about whether light leaks around the side of the planet and so on. <a href="#2015-09-30-black-body-planet-footnote-2-return">↩</a></p></li>
<li id="2015-09-30-black-body-planet-footnote-3-definition" class="footnote-definition">
<p>This assumes that the planet has a constant surface temperature, which will be true only if it conducts heat perfectly, but turns out to be a good enough approximation. <a href="#2015-09-30-black-body-planet-footnote-3-return">↩</a></p></li></ol></div>Covariance and contravarianceurn:http-www-tfeb-org:-fragments-2015-07-21-coco2015-07-21T16:22:16Z2015-07-21T16:22:16ZTim Bradshaw
<p>Physicists seem still to be taught about tensors as being, essentially, multidimensional arrays with special transformation rules which must be learned by rote. So I wrote a document which tries to present a more useful approach.</p>
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<p>The aim of this document was to present a more modern, ‘geometrical’ approach, while not requiring too much background in differential geometry. I don’t suppose anyone read it when I posted pointers to it on reddit, and certainly no-one will read it now, but <a href="/texts/2015/coco.pdf">here it is</a><sup><a href="#2015-07-21-coco-footnote-1-definition" name="2015-07-21-coco-footnote-1-return">1</a></sup>.</p>
<hr />
<div class="footnotes">
<ol>
<li id="2015-07-21-coco-footnote-1-definition" class="footnote-definition">
<p>This URL might change (and has already changed: it used to be on Dropbox). <a href="http://www.tfeb.org/fragments/2015/07/21/coco/">This post itself</a> is a better link to remember as I will update the pointer if I move the document. <a href="#2015-07-21-coco-footnote-1-return">↩</a></p></li></ol></div>The electron at the edge of the universeurn:http-www-tfeb-org:-fragments-2015-02-22-the-electron-at-the-edge-of-the-universe2015-02-22T22:05:03Z2015-02-22T22:05:03ZTim Bradshaw
<p>How easy are physical systems to predict?</p>
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<p>Well, here are a couple of rather lovely examples. Both of these are due to Michael Berry and are mentioned in a book called ‘A Passion for Science’ which is in fact a set of collected transcripts of BBC radio programmes from sometime in the mid 1980s: I heard them on the radio originally, and they have stayed with me – I didn’t find the paper versions until quite recently. I’m giving these from memory: they might differ slightly from the versions described in the book.</p>
<p>For both of them imagine a universe where everything is completely Newtonian, so no quantum mechanics in particular. There is Newtonian gravity.</p>
<h2 id="billiards">Billiards</h2>
<p>The first case is billiards, and we’ll consider a completely idealised billiard table: completely smooth, flat and rigid, completely round balls with completely known properties (so how elastic they are etc), and the same for the cushions. Now someone makes a shot, and we either know the direction and force exactly or are allowed to measure the cue ball’s position and velocity exactly shortly after the shot. We don’t know one thing: there are some people standing around the table, and we don’t know where they are, so we don’t know what their gravitational fields look like. Now we want to predict where the balls go, and we’ll say that the prediction fails when a ball leaves a collision 90 degrees from where we predicted – it’s obvious that after that point we can’t usefully predict anything. How many collisions can we predict ahead?</p>
<h2 id="the-electron-at-the-edge-of-the-universe">The electron at the edge of the universe</h2>
<p>The second case is an ideal gas: a lot of little ideal particles in an ideal box. Again we know everything: the starting conditions are known completely, the box is completely understood &c &c. The box insulates everything but (Newtonian) gravity to make things simpler. And this time we also know everything about the rest of the universe as well: we don’t need to predict it forward, we’re just given all the data about how it evolves (in fact I think that without loss of generality we can assume an empty universe outside the box, which reduces the data volume considerably). Except that there’s an electron at the edge of the universe and we don’t know where it is (apart from how far away it is), and so again we don’t know its gravitiational field. Now we want to predict this system forwards and we’ll use the same criterion for failure: when some particle leaves a collision 90 degrees out from where we predict. How many collisions before that happens?</p>
<hr />
<p>The answers are seven or eight for the first case, and about fifty for the second.</p>Playing cards with the Devilurn:http-www-tfeb-org:-fragments-2014-12-30-playing-cards-with-the-devil2014-12-30T13:05:55Z2014-12-30T13:05:55ZTim Bradshaw
<p>You are playing cards with the Devil, the prize being your soul.</p>
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<p>The game is very simple:</p>
<ol>
<li>the Devil deals two cards face down;</li>
<li>you both turn over the cards – if they are the same colour then you win, if they are different colours the Devil wins.</li></ol>
<p>You play six times, and the Devil wins every time. He has obviously stacked the deck.</p>
<p>You suggest to the Devil that you should play the opposite game: if the cards are <em>different</em> colours you win. He agrees, and you play six times: he wins every time. Obviously He either saw this coming or He has changed the deck while you were not looking.</p>
<p>You suggest a third game:</p>
<ol>
<li>the Devil will deal two cards, face down;</li>
<li>after the cards are dealt you will choose which game to play – the one where you win if they are the same, or the one where you win if they are different;</li>
<li>the cards are turned over.</li></ol>
<p>Much to your surprise he agrees to play once more, and you play six times. The Devil wins every time.</p>