## Archive for the Cute Problems Category

Posted in Bad Statistics, Cute Problems with tags , , , , on November 25, 2016 by telescoper

I just came across this interesting little problem recently and thought I’d share it here. It’s usually called the ‘Neyman-Scott’ paradox. Before going on it’s worth mentioning that Elizabeth Scott (the second half of Neyman-Scott) was an astronomer by background. Her co-author was Jerzy Neyman. As has been the case for many astronomers, she contributed greatly to the development of the field of statistics. Anyway, I think this example provides another good illustration of the superiority of Bayesian methods for estimating parameters, but I’ll let you make your own mind up about what’s going on.

The problem is fairly technical so I’ve done done a quick version in latex that you can download

here, but I’ve also copied into this post so you can read it below:

I look forward to receiving Frequentist Flak or Bayesian Benevolence through the comments box below!

## Reflections on Quantum Backflow

Posted in Cute Problems, The Universe and Stuff with tags , , , , on November 10, 2016 by telescoper

Yesterday afternoon I attended a very interesting physics seminar by the splendidly-named Gandalf Lechner of the School of Mathematics here at Cardiff University. The topic was one I’d never thought about before, called quantum backflow. I went to the talk because I was intrigued by the abstract which had been circulated previously by email, the first part of which reads:

Suppose you are standing at a bus stop in the hope of catching a bus, but are unsure if the bus has passed the stop already. In that situation, common sense tells you that the longer you have to wait, the more likely it is that the bus has not passed the stop already. While this common sense intuition is perfectly accurate if you are waiting for a classical bus, waiting for a quantum bus is quite different: For a quantum bus, the probability of finding it to your left on measuring its position may increase with time, although the bus is moving from left to right with certainty. This peculiar quantum effect is known as backflow.

To be a little more precise about this, imagine you are standing at the origin (x=0). In the classical version of the situation you know that the bus is moving with some constant definite (but unknown) positive velocity v. In other words you know that it is moving from left to right, but you don’t know with what speed v or at what time t0 or from what position (x0<0) it set out. A little thought, (perhaps with the aid of some toy examples where you assign a probability distribution to v, t0 and x0) will convince you that the resulting probability distribution for moves from left to right with time in such a way that the probability of the bus still being to the left of the observer, L(t), represented by the proportion of the overall distribution that lies at x<0 generally decreases with time. Note that this is not what it says in the second sentence of the abstract; no doubt a deliberate mistake was put in to test the reader!

If we then stretch our imagination and suppose that the bus is not described by classical mechanics but by quantum mechanics then things change a bit.  If we insist that it is travelling from left to right then that means that the momentum-space representation of the wave function must be cut off for p<0 (corresponding to negative velocities). Assume that the bus is  a “free particle” described by the relevant Schrödinger equation.One can then calculate the evolution of the position-space wave function. Remember that these two representations of the wave function are just related by a Fourier transform. Solving the Schrödinger equation for the time evolution of the spatial wave function (with appropriately-chosen initial conditions) allows one to calculate how the probability of finding the particle at a given value of evolves with time. In contrast to the classical case, it is possible for the corresponding L(t) does not always decrease with time.

To put all this another way, the probability current in the classical case is always directed from left to right, but in the quantum case that isn’t necessarily true. One can see how this happens by thinking about what the wave function actually looks like: an imposed cutoff in momentum can imply a spatial wave function that is rather wiggly which means the probability distribution is wiggly too, but the detailed shape changes with time. As these wiggles pass the origin the area under the probability distribution to the left of the observer can go up as well as down. The particle may be going from left to right, but the associated probability flux can behave in a more complicated fashion, sometimes going in the opposite direction.

Another other way of thinking about it is that the particle velocity corresponds to the phase velocity of the wave function but the probability flux is controlled by the group velocity

For a more technical discussion of this phenomenon see this review article. The exact nature of the effect is dependent on the precise form of the initial conditions chosen and there are some quantum systems for which no backflow happens at all. The effect has never been detected experimentally, but a recent paper has suggested that it might be measured. Here is the abstract:

Quantum backflow is a classically forbidden effect consisting in a negative flux for states with negligible negative momentum components. It has never been observed experimentally so far. We derive a general relation that connects backflow with a critical value of the particle density, paving the way for the detection of backflow by a density measurement. To this end, we propose an explicit scheme with Bose-Einstein condensates, at reach with current experimental technologies. Remarkably, the application of a positive momentum kick, via a Bragg pulse, to a condensate with a positive velocity may cause a current flow in the negative direction.

Fascinating!

## A Cosmic Microwave Background Dipole Puzzle

Posted in Cute Problems, The Universe and Stuff with tags , , , , , on October 31, 2016 by telescoper

The following is tangentially related to a discussion I had during a PhD examination last week, and I thought it might be worth sharing here to stimulate some thought among people interested in cosmology.

First here’s a picture of the temperature fluctuations in the cosmic microwave background from Planck (just because it’s so pretty).

The analysis of these fluctuations yields a huge amount of information about the universe, including its matter content and spatial geometry as well as the form of primordial fluctuations that gave rise to galaxies and large-scale structure. The variations in temperature that you see in this image are small – about one-part in a hundred thousand – and they show that the universe appears to be close to isotropic (at least around us).

I’ll blog later on (assuming I find time) on the latest constraints on this subject, but for the moment I’ll just point out something that has to be removed from the above map to make it look isotropic, and that is the Cosmic Microwave Background Dipole. Here is a picture (which I got from here):

This signal – called a dipole because it corresponds to a simple 180 degree variation across the sky – is about a hundred times larger than the “intrinsic” fluctuations which occur on smaller angular scales and are seen in the first map. According to the standard cosmological framework this dipole is caused by our peculiar motion through the frame in which microwave background photons are distributed homogeneously and isotropically. Had we no peculiar motion then we would be “at rest” with respect to this CMB reference frame so there would be no such dipole. In the standard cosmological framework this “peculiar motion” of ours is generated by the gravitational effect of local structures and is thus a manifestation of the fact that our universe is not homogeneous on small scales; by “small” I mean on the scales of a hundred Megaparsecs or so. Anyway, if you’re interested in goings-on in the very early universe or its properties on extremely large scales the dipole is thus of no interest and, being so large, it is quite easy to subtract. That’s why it isn’t there in maps such as the Planck map shown above. If it had been left in it would swamp the other variations.

Anyway, the interpretation of the CMB dipole in terms of our peculiar motion through the CMB frame leads to a simple connection between the pattern shown in the second figure and the velocity of the observational frame: it’s a Doppler Effect. We are moving towards the upper right of the figure (in which direction photons are blueshifted, so the CMB looks a bit hotter in that direction) and away from the bottom left (whence the CMB photons are redshifted so the CMB appears a bit cooler). The amplitude of the dipole implies that the Solar System is moving with a velocity of around 370 km/s with respect to the CMB frame.

Now 370 km/s is quite fast, but it’s much smaller than the speed of light – it’s only about 0.12%, in fact – which means that one can treat this is basically a non-relativistic Doppler Effect. That means that it’s all quite straightforward to understand with elementary physics. In the limit that v/c<<1 the Doppler Effect only produces a dipole pattern of the type we see in the Figure above, and the amplitude of the dipole is ΔT/T~v/c because all terms of higher order in v/c are negligibly smallFurthermore in this case the dipole is simply superimposed on the primordial fluctuations but otherwise does not affect them.

My question to the reader, i.e. you,  is the following. Suppose we weren’t travelling at a sedate 370 km/s through the CMB frame but instead enter the world of science fiction and take a trip on a spacecraft that can travel close to the speed of light. What would this do to the CMB? Would we still just see a dipole, or would we see additional (relativistic) effects? If there are other effects, what would they do to the pattern of “intrinsic” fluctuations?

## A Problem in Lagrangian Mechanics

Posted in Cute Problems with tags , , on April 28, 2016 by telescoper

Today, as well as saying goodbye to Sally Church, I managed to finish my lecture course on Theoretical Physics. There’s still another week of teaching to go, but I have covered all the syllabus now and can use the remaining sessions for revision. The last bit of the course module concerned the calculus of variations and a brief introduction to Lagrangian mechanics so for a bit of fun I included this example.

Professor Percy Poindexter of the University of Neasden has invented a new theory of mechanics in which the one-dimensional motion of a particle in a potential $V(x)$ is governed by a Lagrangian of the form

$L=mx\ddot{x} +2V(x)$.

Use Hamilton’s Principle and an appropriate form of the Euler equation to derive the equation of motion for such a particle and comment on your answer.

UPDATE: Since nobody has commented I’ll just reveal the point of this question, which is that if you follow the instructions the equation of motion you should obtain is

$m\ddot{x}= -\frac{\partial V}{\partial x},$

which is exactly the same as you would have got using the usual Lagrangian

$L= \frac{1}{2}m\dot{x}^{2} - V(x).$

Anyone care to comment on that?

## A Potential Problem with a Sphere

Posted in Cute Problems with tags , , on March 9, 2016 by telescoper

Busy busy busy again today so I thought I’d post a quick entry to the cute problems folder. I set this as a problem to my second-year Theoretical Physics students recently, which is appropriate because I encountered it when I was a second-year student at Cambridge many moons ago!

HINT: You can solve this by finding the general solution for the potential at any point inside the sphere, but that isn’t the smart way to do it!

FURTHER HINT: The question asks for the Electric Field at the origin. What terms in the solution for the potential can contribute to this?

OUTLINE SOLUTION: A numerically correct answer has now been posted so I’ll give an outline solution. The potential V inside the sphere is governed by Laplace’s equation, the general solution of which is a series expansion in powers of r and Legendre polynomials, i.e. rn Pn(θ). The coefficients of this expansion can be determined for the given boundary conditions (V=V0 at r=a for θ = +1, V=0 for cos θ = -1). However this is a lot more work than necessary. The question asks for the electric field, i.e. the gradient of the potential, and if you look at the form of the potential there is only one term that can possibly contribute to the field at r=0, namely the one involving rP1(cosθ) =rcos θ (which is actually z). Any higher power of r would give a derivative that vanishes at the origin. Hence we just have to determine the coefficient of one term. Using the orthogonality properties of the Legendre polynomials this can easily be seen to be 3V0/4a. The electric field is thus -3V0/4a in the z-direction, i.e. vertically downwards from the top of the sphere.

## The Three-Card Puzzle

Posted in Books, Talks and Reviews, Cute Problems on February 29, 2016 by telescoper

As promised I did my turn at the Brighton Science Festival yesterday. The Sallis Benney Theatre wasn’t quite full but there was a decent crowd, which was mildly surprising because the event I was involved in hadn’t really been advertised very well. If you want to know how my talk went then you should ask someone who was in the audience because I wasn’t really paying attention. However, I was preceded by John Haigh (seen below in mid-talk) whose presentation included a nice puzzle for the “Cute Problems” file:

Imagine you have three cards. One is blue on both sides. One is pink om both sides. One is pink on one side and blue on the other. Other than the colours the cards are identical. For the demonstration John glued playing cards together, but they don’t have to be playing cards. Anyway, you put the three cards into a bag (seen on the stage in the picture), pull out one card “at random” and look at the colour of one side but not the other. If the colour you see is blue, what is the probability that the other side is also blue?

Try to answer this without googling. I’ll post the solution when there have been enough responses to the poll:

OK. Over a hundred people responded so I have now closed the voting.

As always seems to be the case with this sort of problem, the majority went for the “obvious” answer, which turns out to be wrong!

SOLUTION: If the card is blue on one side then it must be either the blue-blue or blue-pink one. I think most people voted for 1/2 because there are two possible cards. But the relevant consideration is that there are three possible sides: side 1 of the blue-blue card; side 2 of the blue-blue card; and the blue side of the blue-pink card. Each of these is equally likely and two of them result in the other side being blue. The correct answer is therefore 2/3; it is twice as likely for the other side to be blue as it is to be pink.

## A Question of Magnitude

Posted in Cute Problems, Education, The Universe and Stuff with tags , , , on January 30, 2016 by telescoper

A frequent complaint raised by students of Astronomy is that astronomers insist on using funny units. Chief among them is the use of magnitudes to quanitify the brightness of an object. Why not use the observed intensity (or brightness or flux) of the light from the star, which can be expressed straightforwardly in SI units, instead of faffing around with a clunky logarithmic measure? The reason we use the magnitude scale is primarily historical and based on the fact that the eye’s response to light is more-or-less logarithmic and that in the days before calculators it was easier to deal with very large and very small numbers using logarithms.Most relevant calculations involve divisions and multiplications which become subtractions and additions when you use logarithmic quantities.

It was Norman Pogson who first suggested that a magnitude scale be defined such that a difference of five magnitudes should correspond to a factor of 100 in actual brightess. This was because the brightest naked-eye stars – those of first magnitude – are about 100 times brighter than the faintest naked-eye stars, which are of sixth magnitude. That was in 1856 and we’ve been stuck with it ever since!

Although the magnitude system may appear strange, it’s not really that hard to use when you get used to it. A beginner really just needs to know a few key things:

1.  Bright things have lower magnitudes (e.g. first magnitude stars are brighter than second magnitude stars);
2.  If two stars have apparent magnitudes $m_1$ and $m_2$ respectively then $m_2-M_1=2.5\log_{10} (I_1/I_2)$ where $I_1$ and $I_2$ are respectively the fluxes received from the two stars;
3. The intensity of light falls off with the square of the distance from the source;
4.  The absolute magnitude is the apparent magnitude a star would have if it were 10 parsecs from the observer;
5. Most stars have roughly black-body spectra so their total intrinsic luminosity depends on the product of their surface area (i.e. on the square of the radius) and the fourth power of the surface temperature.

Got it?

To test your understanding you could try these little problems. To warm up you might look at I posted the first of them a while ago. Anyway, here we go:

1. A binary system at a distance of 100 pc has such a small separation between its component stars that it is unresolved by a telescope. If the apparent visual magnitude of the combined image of the system is 10.5, and one star is known to have an absolute visual magnitude of 9.0, what is the absolute visual magnitude of the other star?
2. Two stars are observed to have the same surface temperature, but their apparent visual magnitudes differ by 5. If the fainter star is known to be twice as far away as the brighter one, what is the ratio of the radii of the two stars?
3. A binary system consists of a red giant star and a main-sequence star of the same intrinsic luminosity. The red giant has a radius 50 times that of the main-sequence star. (i) If the main-sequence star has a surface temperature of 10,000 K, what is the surface tempature of the red giant star? (ii) If the two stars can’t be resolved the combined system has an apparent magnitude of 12, what are the apparent magnitudes the two component stars would have if they could be observed separately?

Answers through the comments box please! The first correct entry wins a year’s free subscription to the Open Journal of Astrophysics…