Dork Scratchings

I got into a debate with a flat-earther the other day.  After a while I grew to respect the position.  After all, not taking anything on trust is a cornerstone of science.  I pointed out that I've verified the Round Earth theory personally, by creating a sundial that tells both the time and the date .  My sundial was created using a mathematical model that assumes Earth is spherical, spinning about it's axis, and orbiting the sun.  And I've tested it works.  (I also had a job writing code for transceivers that talk to satellites, so you could say I'm a fully signed up round-earther!) Unfortunately my argument did not convince because it required either That the flat-earther trusted me, or That he was willing and able to follow the maths behind the model This led to to wonder what's the simplest way to demonstrate the Earth is round?  It has to involve no maths, and be easy to reproduce by skeptics anywhere.  The video is my answer to that question. PS

UBI stands for: Universal Basic Income The idea is pretty simple - it's a benefit$^*$ that is given to everyone (hence Universal) and provides enough to live on (hence Basic Income).  It's very much en vogue with economists at the moment but it tends to be received badly by voters .  "Why should millionaires get benefits?" they ask.  In this post I'll explain why I think they should! Imagine a country in which benefits are means tested, so that a citizen with no income receives £10K, but loses 50p of it for each extra £1 of income up to £20K.  After that they receive no benefits and start to pay tax.  The tax rate is a smooth curve that starts at zero, grows to 25% of a £40K income, 30% of a 50K income, and 40% of a 100K income.  That sounds fair. In the country next door there's a flat 50% tax rate on all income, and a £10K universal basic income.  Taxing people with barely any income at the same rate as higher earners and giving millionaires bene

Any orbit can be represented given enough epicycles. The Geocentric Model The geocentric model was wonderful for human self aggrandizement.  We belonged at the centre of the universe and everything revolved around us.  That early humans believed this isn't that surprising.  After all the Sun, moon, and stars do seem to move in circles around us (albeit different ones).  What's more interesting is how we attempted to cling on to this theory in the light of a) conflicting evidence, and b) a far better explanation. David Deutsch in The Fabric of Reality puts forward the view that the point of a scientific theory is to explain .  The more a theory explains - whilst still remaining consistent with observable facts - the better it is.  This is a philosophical justification for Occam's Razor.  Why is an explanation with fewer postulates better than one with more postulates? because it leaves less unexplained! The earliest theory for why the planets did not move in sim

I discovered the card game Dobble a few days ago whilst visiting relatives in Germany.  There are 57 cards with 8 symbols on each.  To play the most basic version of the game you deal the cards face down between the players such that one card remains (if there are an odd number of players you may need to hold back more than 1 card).  Then you turn face up the remaining card (or one of the remaining cards) and each person turns their top card face up. The first person to spot a symbol on their own card that matches one on the left over card shouts out the name of that symbol (e.g. "car!") and gets to move their card to the top of the shared stack; the others move their top card to the bottom of their own stack.  The winner is the person who gets rid of their cards first. At first sight this doesn't seem very mathematical.  However, after a while I noticed something odd about the cards: every card shares exactly one symbol with each other card in the pack.  I thou

I've been wondering: is it possible to visualize the rules of calculus? 1. Integration by parts To simplify things, in the above visualization $v(x_1) = 0$, so that $\int^{x'}_{x_1}\frac{dv}{dx}dx = v(x')$.  This means that $v(x_2)$ is simply the area of the cross-section facing us on the LHS, and the volume  on the LHS is $u(x_2)v(x_2)$. The RHS shows the same volume split into two parts.  The rectangle embedded in the first is $u\frac{dv}{dx}$, and so its volume is $\int^{x_2}_{x_1}u\frac{dv}{dx}dx$.  The cross section of the 2nd part is $v(x)$, so its volume is $\int^{u_2}_{u_1}vdu$, which becomes $\int^{x_2}_{x_1}v\frac{du}{dx}dx$ when rewritten as an integral over $x$. So the picture is visual proof that the equation in pink holds, provided that $v(x_1) = 0$.  To prove it in general we just need to check that when we replace $v$ with $v+v_1$ in the equation, both sides change by the same amount. 2. The chain rule This picture demonstrates the cha

Connecting Geometry and Algebra Matrix determinants have the following strange definition that seems to have been pulled out of thin air: $$ det(A) = \sum_{\sigma}{sign(\sigma)a_{1\sigma(1)}...a_{n\sigma(n)}} $$ where $A = (a_{ij})$ is a real $n\times n$ matrix $\sigma$ ranges over all permutations of $\{1,...,n\}$ $sign(\sigma)$ is $+1$ if $\sigma$ is a product of an even number of transpositions and $-1$ otherwise$^\dagger$ However, in the geometric world the definition is far more intuitive: The determinant of A is the volume of the parallelepiped formed by its columns, multiplied by minus one if these have the opposite handedness to the unit vectors. Why are these two definitions the same?  To begin to answer this we need to first define elementary matrices and then show that every square matrix can be written as a product of these. Definition The elementary matrices are $E_{i, j}$ for $1 \le i,j \le n$ $E_{i,\lambda}$ for every real $\lambda$ and $1 \l

In this post I will attempt to show that sitting down and having a nice cup of tea is the best approach to solving any new mathematical problem.  So, here's a problem: A plane has 100 seats and 100 passengers each with an allocated seat number.  The first passenger to get on the plane is blind and chooses a seat at random.  Each subsequent passenger to board chooses their own seat if still available, or a seat at random if not.  What is the probability that the 100th passenger gets their allocated seat? This is more subtle than it seems at first.  The last passenger could get their own seat because the blind passenger chooses the correct seat, or because the 2nd passenger chooses the blind person's seat, or because the first 87 passengers occupy the first 87 seats.  In fact there's a huge number of ways in which it could happen. Knuckleheaded Compsci solution Suppose we've forgotten to have a cup of tea.  Then we might just dive in and start modelling

Queqiao and Chang'e-4 Exciting news from the BBC Website: China Moon mission lands Chang'e-4 spacecraft on far side .  The page includes a video with sinister background music as if to suggest they're the baddies (a la Drax in Moonraker).  However, the part that really intrigued me was the mention of the L2 Lagrange point - the first I've ever seen in a news story! As I described in my post Lagrange Points   there are 5 locations in the Earth-moon-Sun plane in which - in the rotating frame of reference and taking centrifugal forces into account - there is no overall force and an object can be parked indefinitely.  One of them is just beyond the moon and is called L2 . Now the problem with landing a probe on the far side of the moon is that you can't talk directly to it: there's a big rock in the way!  So, according to the BBC article the Chinese Space Agency has parked a satellite Queqiao at L2 to relay messages.  This left me a bit confused, as

If the sky were covered in moons it'd be almost a bright as day! Walking to the pub through the Yorkshire Dales on a particularly brilliant moonlit evening, when everything was clearly visible, I wondered just how much less light there was than during daytime?  It turns out to be quite easy to estimate an upper bound - all you need to do is measure the angle subtended by the moon! Let $A_e$ and $A_m$ be the cross sectional areas of the Earth and moon, $d$ be the distance to the moon, and $r$ be the distance to the Sun.  Now, suppose the Sun releases some energy $E_s$ then the amounts $E_{se}$ and $E_{sm}$ which land on the Earth and the moon are given by: $$ \begin{align} E_{se} &= \frac{A_e E_s}{4\pi r^2}\\ E_{sm} &\approx \frac{A_m E_s}{4\pi r^2} \end{align} $$ On a full moon, let's assume for the sake of calculating an upper bound that the moon reflects all of $E_{sm}$ equally in all hemispheric directions.  Then the energy reflected to the Earth is giv

Theorem of the week This week's theorem of the week is the fundamental theorem of algebra , and the picture is the proof! Theorem Every non degree-zero polynomial $p(z) = a_nz^n + ...+ a_1z +  a_0$ has a root in $\mathbb{C}$. Picture proof To see how the picture proves this, write $z$ as $Re^{i\theta}$, then for all $k$ $$ z^k = R^ke^{k i \theta} $$ So for sufficiently large $R$ the $a_nz^n$ term dwarfs all the others and so the image of $\{z\in\mathbb{C}: \lvert z \rvert = R\}$ must go around the origin $n$ times, like the rubber band in the photo.  But when $R= 0$ the image is just $\{a_0\}$ which goes around the origin zero times.  So, for some $0 < r < R$ the image of $\{z\in\mathbb{C}: \lvert z \rvert = r\}$ must cross the origin.  QED. Less handwavy proof In order to obtain a contradiction assume $p(z)$ has no zeros.  Then $\frac{z^{n-1}}{p(z)}$ is everywhere differentiable, which in turn means that its closed loop integrals are zero$^{(\dagger)}$.

Can you always reorient a four legged chair to make it stable on an uneven surface? If we can assume the feet form a perfect square, then: Yes! Without loss of generality let's suppose that the chair starts off in the position shown top left, with its front right foot in the air and the others on the ground.  If we keep the two left feet on the ground while pushing down on the other two, the above ground foot will eventually touch the floor and the remaining foot will sink below it. Now imagine instead rotating the chair by 90$^\circ$, keeping the same 3 feet touching the ground at all times, and ending with the two back feet in the positions previously occupied by the two left feet.  The two back feet and the front left foot occupy three corners of the square described by the feet at the end of the previous paragraph, so the fourth must be below ground.  Since the front right foot starts off in the air and ends up below ground we can infer from the interme

 Déjà vu Reality Checkpoint Cambridge ... the feeling of having been somewhere or experienced something before when you know you haven't.  Déjà vu demonstrates that knowing we've been somewhere before and feeling that we have are different things. What makes it so difficult to convince skeptics of the reality of Everett's parallel worlds is that it feels as if time flows through the present moment like water flowing through a hosepipe.  It doesn't feel like the present moment just exists as a droplet in a sea of moments, some of which happen to be a bit future- like , some a bit past- like (but most neither).  Past-like moments invoke a feeling of recognition in us, and we line them up like dominoes and call them the past.  But if you've ever experienced déjà vu you know that such feelings of recognition can't be trusted. So, if we could determine what neural activity or hormone , is associated with déjà vu, then maybe we could develop a drug th

A few Christmases ago I was playing with a toy xylophone meant for my nephew when I noticed something odd.  I expected going up an octave to halve the bar length, but it didn't.  A quick measurement confirmed my suspicion: to go up one octave you divide by $\sqrt{2}$ instead of by $2$.  I remembered this earlier this week while at my son's school concert, and decided to see if I could work out why! Strings First a bit of background.  Why did I expect halving the length to result in a note one octave higher (i.e. double the frequency)?  The answer is because I'd learned that this was the case with stringed instruments.  Assume that the tension is a constant $T$, and let $z(y,t)$ be the vertical displacement of the string at position $y$ and time $t$.  Then the upward force on a small element of  size $\delta y$ is approximately $$ T \left(\left. \frac{\partial z}{\partial y}\right|_{y+\delta y} - \left.\frac{\partial z}{\partial y}\right|_y\right) = T\delta y \frac{\

Image by Goodtiming8871 CC-BY-SA 4.0 Most people learn how AM and FM radio works at some point or another.  These are methods for transmitting and receiving an analogue signal, but nowadays most radio signals are digital.  How is it possible to send bytes from one place to another using analogue electromagnetic radiation?  I thought I'd write a post to try to demystify it.

Imagine you were a hamster in a hamster ball living on a hilly surface.  Base camp is surrounded on all sides by high summits, but you have a powerful catapult there that can fire you to the top of any of them.  Once fired you can influence your trajectory, but it's hard work and you don't have much energy in your little legs.  Suppose you know where you want to end up beyond the hills.  What's the best strategy for getting there?

ESA/Rosetta/NavCam – CC BY-SA IGO 3.0 In November 2014 the Philae lander touched down on 67P.  As I tried to imagine what was happening there now, I realized that "there now" doesn't really have much meaning when "there" is 30 light minutes away.  And maybe "what's happening" doesn't have much meaning either, given that the region of spacetime outside of one's light cone provides the sort of causal isolation quantum computing engineers would kill for. In a lapse of character brought on by mental fug I penned a poem On Everett's Peak  Rosetta, Philae, half an hour away if you're travelling light Packed with meters, big and small And a single transistor failure could ruin it all Cosmic ray, beta decay, a single gamma misplaced State changed, plan deranged, non-redundant memory defaced It hasn't happened yet, at mission control, as far as it is known At mission control, it hasn't happened yet, anyti

The following equation can be viewed in a couple of ways $$ action =  \int_{\Omega} \mathcal{L} (\phi_a, \partial_{\mu}{\phi_a}) d^4x $$ 1. Classically Every physical law can be written in terms of an Action Principle .  An action principle states that measurable values $\phi_a(\mathbb{x},t)$ over a region of spacetime $\Omega$ will be such that the action is stationary.  Or to put it another way, if you infinitesimally deform the $\phi_a$ then either  however you do it the action will increase, or  however you do it the action will decrease. There is an important caveat though: the action is stationary because we only consider deformations of the $\phi_a$ that preserve its values on the boundary $\partial \Omega$. This is an alternative to the differential way of describing the universe, in which only a single point of spacetime is considered.  In the differential formulation, instead of being told the values of $\phi_a$ on a boundary of a region of spacetime, and asked