What is required to be able to grasp the ideas of physics and solve problems so well that you could get a gold medal at IPhO? Do you need just to be gifted? Of course not, there are other students, who have solved a lot of problems – while you are thinking hard trying to "invent a bicycle", they are already writing the solution, because they had solved a similar problem earlier. Is it enough to solve a lot of problems and read a lot of problem solutions? Most often, no. Just solving or reading solutions, of course, will increase your technical skills, but you also need to think over, what were the main ideas which made it possible to solve the problem, and take these ideas into your permanent arsenal; if you solve too many problems, you don't have time to think over. Is it possible to learn "the art of problem solving" and if yes then how? Well, 99% of the Olympiad problems are solved using a rather limited set of ideas (for mathematics, that set is somewhat larger). So, if you acquire those ideas well enough – so that you can recognize them even if they are carefully hidden – then the IPhO gold will be yours! Do not worry, no-one expects you to discover a solving technique which has been never seen before, because that would be an achievement worth of a Nobel Prize!

Since we started the topic of Nobel Prize – is it enough to be the absolute winner of the IPhO to get, at a later stage, a Nobel Prize? (Each year, there is one Nobel Prize in physics – similarly to the absolute winner of IPhO.) Of course, it is not; however, you'll have better chances than anyone else. Becoming a great physicist requires several components, one of which is having brilliant problem solving skills (tested at IPhO). Another one is ability to make solvable models - formulate problems which can be solved and which reflect important aspects of reality. Third component is ability to distinguish, which problems are important and which are not. You can be very skillful and smart, but if you study problems of marginal interest, no-one will pay attention to your research results. Finally, you need a considerable amount of luck. Indeed, that particular field of physics in which you start your studies, e.g. start making your PhD thesis, depends on somewhat random decisions – it is almost impossible to foresee, where are the biggest scientific challenges after five or ten years. Also, in order to perfect yourself in regard of the above-mentioned three components, you need excellent supervisors and excellent lab; although you have some freedom in choosing your supervisor and lab, you still need to be very lucky to find outstanding ones!

Now follow my other posts below to know more about a set of solving techniques, fragments of the whole arsenal needed for a perfect problem solver. With a large number of pieces, the picture would become recognizable, but we need to start making it piece by piece. While some "tiles" will be useful for solving a spectrum of problems, other tiles are aimed to give more insight into certain physical concepts.

__Physics Olympiad Problem Solving Technique: Maximum or Minimum?__

It is well-known that a
system is stable at the minimum of its potential energy. But why? Why is a
minimum different from a maximum? For Fermat' principle it is clear: there is
no longest optical path between two
points – the ray could just go "zig-zag" -, but there is definitely
one which is the shortest!

The reason is simple – at an
equilibrium state, the kinetic energy has always minimum (as long as masses are
positive). What we actually do need for a stability is a conditional extremum of
one conserved quantity (such as the net energy), under the assumption that the other
conserved quantities are kept constant (unconditional
extremum is OK, too). Consider the motion of a body along x-axis and let us
describe it on the phase plane, with coordinates x and p (the momentum).
The overall energy is E =U(x)+p

^{2}/2m. Now, if we depict this energy as a surface in 3-dimensional space with coordinates x, p and E, the point describing the state of the system will move along the intersection line of that surface with a horizontal plane E=Const. At the minimum of U(x), with p=0, this intersection line would be just a single point, because this is the lowest point of that surface. The near-by trajectories will be obtained if we ascend the horizontal plane a little, E =E_{min}+e, so that it no longer just touches the surface, but cuts a tiny ellipse from it. All the points of that trajectory (the ellipse) are close to the equilibrium point, so the state is, indeed, stable.
It appears that a system can
be stable also because of a conditional maximum of the net energy: while an
unconditional extremum of the kinetic energy can only be a minimum, things are
different for conditional extrema. Perhaps the simplest example is the
rotation of a rigid body. Let us consider a rectangular brick with length a, width b, and thickness c (a>b>c). Let I

_{c}be its moment of inertia for the axis passing its center of mass and perpendicular to the (a,b)-plane; I_{b}and I_{a}are defined in a similar way. For a generic case, the moment of inertia I will depend on the orientation of the rotation axis, but it is quite clear that I_{c}>= I_{ }>= I_{a}(it can be shown easily once you learn how to use tensor calculations). Now, let us throw the brick rotating into air and study the motion in a frame which moves together with the centre of mass of the brick (in that frame, we can ignore gravity). There are two conserved quantities: angular momentum L, and rotation energy K=L^{2}/2I . We see that for a fixed L, the system has minimal energy for I_{ }= I_{c}(axis is parallel to the shortest edge of the brick), and maximal energy for I_{ }= I_{a}(axis is parallel to the longest edge of the brick). You can easily check experimentally that both ways of rotation are, indeed, stable! Not so for the axis parallel to the third edge… This phenomenon is demonstrated in a video made by NASA on the International Space Station.
Well, actually the rotation
with the minimal energy is still a little bit more stable than that of with the
maximal energy; the reason is in dissipation. If we try to represent the motion
of the system in the phase space (as described above), we would start with
touching a top of an hill with a horizontal plane E =E

_{max }(so that the intersection is just a point), but due to dissipation, the energy will decrease, E =E_{min}– e, and the phase trajectory would be a slowly winding-out spiral. So, while you are probably used to know that dissipation draws a system towards a stable state, here it is vice versa, it draws the system away from the stable state!__Physics Olympiad Problem Solving Technique: Fast or Slow?__

What is an adiabatic
process? Most of the readers would probably answer that this is a process with
a gas which is so fast that there is no heat exchange with the surroundings.
However, this is only a half of the truth, and actually the less important
half. In fact, it is quite easy to understand that this is not entirely
correct: consider a cylinder, which is divided by a thin wall into two halves;
one half is filled with a gas at a pressure p, and the other one is
empty. Now, let us remove momentarily the wall: the gas from one half fills the
entire cylinder. Since no external work is done (the wall can be removed
without performing a work), the energy of the gas is preserved, hence, the
temperature remains the same as it was at the beginning. Meanwhile, for an
adiabatic process we would expect a decrease of temperature by a factor
of 2

^{γ-}^{1}: part of the internal energy is supposed to be spent on a mechanical work performed by the expanding gas. However, if the piston moves faster than the speed of sound, the gas will be unable to catch up and push the piston. So, the adiabatic law was not followed because the process was too fast!
It appears that the adiabatic law for thermodynamics
has also a counterpart in classical mechanics – the conservation of the
adiabatic invariant. For mechanical systems (oscillators) performing
periodic motion, the adiabatic invariant is defined as the area of the
closed curve drawn by the system in phase space (which is a graph
where the momentum p is plotted as a function of the
respective coordinate x), and is (approximately) conserved when the
parameters of the system are changed adiabatically, i.e., slowly
as compared with the oscillation frequency. For typical applications, the
accuracy of the conservation of the adiabatic invariant is exponentially good
and can be estimated as e

^{-fτ}, where f is the Eigen frequency of the oscillator, and τ is the characteristic period of the variation of the system parameters.
How are these parameters related to each other
(a) adiabatic invariant and (b) adiabatic process with a gas? The easiest way to understand this is
to consider a one-dimensional motion of a molecule between two walls, which
depart slowly from each other (Figure 1). Let us use the system of reference
where one of the walls is at rest, and the other moves with a velocity u << v,
where v is the velocity of the molecule (the interaction of
the molecule with the walls is assumed to be absolutely elastic). One can say
that such a molecule represents an oscillator with a slowly changing potential: the
potential energy U(x) = 0 for 0<x< X (where X = a +ut)
and otherwise, U(x) = ∞. The trajectory of the molecule
in the phase space is a rectangle of side lengths X and 2mv.
So, the adiabatic invariant is 2mvX; hence, vX = Const.
For a one-dimensional gas, the distance Xbetween the walls plays
the role of the “volume” V, and mv

^{2}/2=kT/2, hence v ~ T^{1/2}(here "~" means “is proportional to”). So, the adiabatic invariant can be written as V^{2}T = Const. On the other hand, from the adiabatic law for an ideal gas, we would expect TV^{γ}^{‑1}= Const. For the one-dimensional gas, the number of the degrees of freedom i = 1, hence γ = c_{p}/c_{V }= (i+2)/i =3, and TV^{2}= Const, i.e., we can conclude that the adiabatic invariant and the adiabatic gas law give us exactly the same result!
How to prove that for an adiabatic forcing of an
oscillator, the adiabatic invariant is conserved? Well, this is not a too simple mathematical task
and thus we skip the proof here (it can be found in good textbooks of
theoretical mechanics). However, for a simple particular case of an elastic
ball between two walls (see above), it can be done more easily. Indeed, with
each impact with the departing wall, the speed of the ball is decreased by 2u,
and this happens once per time interval t = 2X/v.
So, the ball decelerates with the rate of dv/dt = 2u/t =uv/X,
hence dv/v= udt/X = –dX/X.
Integrating this differential equation gives us directly Xv = Const.

Conservation laws play a central role both for the
physical processes, and for the physics as a science (cf “Minimum of Maximum”), and adiabatic invariant is not an
exception. Perhaps the most important role of it is related to the
quantum mechanics. Namely, during adiabatic processes, the system will not
leave the stationary quantum state it has taken (as long as the state itself
does not disappear). To motivate this claim, let us consider a biatomic
molecule, which can be modelled as an oscillator. When treating the process
classically, the trajectory of a harmonic oscillator in the phase space is an
ellipse of surface area J = πp

_{0}x_{0}, where p_{0}and x_{0}are the amplitudes of the momentum and coordinate. Note that p_{0}= mx_{0}ω_{0}, where ω_{0}is the circular Eigen frequency of the oscillator; therefore, the full energy of the oscillator (calculated as the maximal kinetic energy) is E = p_{0}^{2}/2m = p_{0}x_{0}ω_{0}/2 = J ω_{0}/2π = J f_{0}. Hence, the adiabatic invariant J = E/f_{0}: during adiabatic processes, the oscillation energy is proportional to the frequency. According to the quantum mechanics, the stationary energy levels of the oscillator are given byE_{n}=hf_{0}(n +1/2), where n is an integer representing the order number of the energy level. Comparing the classical and quantum-mechanical results leads us to the conclusion that during adiabatic processes = Const: the system will remain at the stationary state of the same order number where it was (Figure 2) . (While it is not always completely correct to combine classical and quantum-mechanical results, classical mechanics is a macroscopic limit of the quantum mechanics and hence, the conservation laws of both theories need to be compatible.)
Now, suppose our bi-atomic molecule is forced by an
electromagnetic field in the form of an adiabatic pulse. In terms of classical
mechanics we say that such a forcing is unable to pump energy into oscillations
of the molecule, because the adiabatic invariant is conserved and hence, the
energy of oscillations depends only on the current Eigen frequency. In terms of
quantum mechanics we’ll say exactly the same, but the motivation will be
different: the adiabatic pulse contains no photons which are resonant with the
oscillator.

Another important role of the adiabatic invariant is
protecting us from the cosmic radiation (in “collaboration” with the magnetic
field of the Earth). It appears that the motion of a charged particle in a
magnetic field can be represented as an Hamiltonian motion (we skip here the
definition of the Hamiltonian motion as it would go too deeply into the subject
of theoretical mechanics), with a re-defined momentum. It appears also that
with this new momentum (the so-called generalized momentum), the adiabatic
invariant of a gyrating (helicoidally moving) charged particle is its magnetic
dipole moment (which is proportional to the magnetic flux embraced by the
trajectory, hence this flux is also conserved). So, if a charged particle moves
helicoidally along magnetic field lines towards a stronger magnetic field, due
to the conservation of its magnetic moment, the perpendicular (to the magnetic
field) component of its velocity will increase. Owing to the conservation of
its kinetic energy, the parallel component of the velocity will decrease, and
at a certain point, it becomes equal to zero: the particle is reflected back
(Figure 3). This is exactly what happens with a majority of the charged
particles approaching Earth along the field lines of its magnetic field.

Adiabatic invariant has simple every-day applications,
too. Suppose you try to carry a cup of coffee – this will be quite simple even
if the cup is completely full. Now try the same with a plate of soup – at least
with full plate, this will be quite difficult! Finally, with a large full
photographic tray, this will be nearly impossible! The reason is that when you
try to keep your hands motionless, they still move slightly, but the feedback
from your vision allows you to correct the mistakes. The characteristic
time-scale of such a motion of hands is of the same order of magnitude as your
reaction time, in the range of 0.2 – 0.4 s. This is to be compared with the
reciprocal of the circular Eigen frequency ω

_{0}^{-1}of the water level oscillations. (ω_{0}^{-1}differs from the full period T by 2π; ω_{0}^{-1 }serves as a better reference here, because the corrective motion of hands represents no more than a quarter of a full period of an oscillatory motion.) For a plate of depth h and length L, the smallest Eigen frequency can be estimated as the frequency of standing waves of wavelength 2L (see also problem No 2 of IPhO-1984). The speed of shallow water waves is (gh)^{1/2}, so that the Eigen frequency will be f_{0 }= (gh)^{1/2}/2L. For a cup of coffee, the diameter and depth can be estimated as 7cm, hence the characteristic time scale of oscillations will be ω_{0}^{-1 }≈ 0.03s; with respect to such oscillations, the hand motion is adiabatic – even if we apply our smallest estimate of 0.2s (note that counter-intuitively, here a slow reaction is better than a fast one). For a plate of H = 3cm and L = 25cm we get ω_{0}^{-1 }≈ 0.15s – the hand motion is already not very adiabatic. Finally, for a photographic tray of H = 3cm and L = 60cm, we obtain ω_{0}^{-1 }≈ 0.35s, which is really difficult to handle.
Finally, in the context of adiabaticity, it is
interesting to analyse the IPhO problem about tides,
which was posed in 1996 in Oslo (as Problem No 3). The problem is, indeed, very
interesting: you are given a simplified model of a complex and important
phenomenon, which, regardless of simplicity, gives you reasonable estimate and
teaches valuable physical concepts. Let us read its text and comment the model
assumptions.

In this problem we
consider some gross features of the magnitude of mid-ocean tides on earth. We
simplify the problem by making the following assumptions:

(i) The earth and the moon
are considered to be an isolated system,

/a very reasonable
assumption: even the effect of the Sun is small in the reference frame of
Moon-Earth centre of mass, where the inertial force and Sun gravity cancel each
other out/

(ii) the distance between
the moon and the earth is assumed to be constant,

/also reasonable: there
are small variations, but nothing to worry about/

(iii) the earth is
assumed to be completely covered by an ocean,

/this is definitely not
the case, but at least the Pacific Ocean is very large; as a model, why not …/

(iv) the dynamic effects
of the rotation of the earth around its axis are neglected, and

/Did you understand what
they wanted to say? If not, you need to learn reading the problem texts! Well,
it means that the forcing of the water by the Moon is to be assumed to be
adiabatic (slow), so that the water level will take a quasi-equilibrium position
(ie. equilibrium, where the equilibrium state changes slowly in time). The
validity of this assumption will be discussed below./

(v) the gravitational
attraction of the earth can be determined as if all mass were concentrated at
the centre of the earth.

Again, a perfectly
reasonable assumption: the gravitational field of a sphere (assuming that the
mass density depends only on the distance from the centre) is outside the
sphere the same as that of a point mass. The departure of the Earth's shape
from a sphere is small, indeed.

And so, is the tide
forcing really adiabatic? We need to compare the period of forcing with the
Eigen frequency, or, the speed of the "piston" with the speed of
waves. The speed of the "piston" is the Earth perimeter divided by 24
h, ie. v = 460 m/s. The relevant wave is, in effect, a tsunami with the
estimated speed of (gH)

^{1/2}= 200 m/s (here, H = 4000 m is an estimate for the average ocean depth). So, the forcing is far from being adiabatic, we could say that the assumption (iv) is horribly wrong. On the other hand, if we solve the problem according to these assumptions, we obtain for the tide amplitude h = 27 cm, which has at least a correct order of magnitude; why? Well, because for a typical resonance response curve, the response amplitude at a double Eigen frequency (which we would need as the "piston" speed is ca twice the wave speed) is of the same order of magnitude as that of a zero frequency (which is obtained in this Problem). Further, since the tidal motion of the water is by no means quasi-stationary, the ocean boundaries will play an important role. What will happen is very similar to the motion of tea in a cup, when you push the tea by a spoon: basin boundaries reflect the moving water, creating vortices and complex pattern of tidal heights? To conclude, we learned that the above tide model fails for water tides (providing a very rough estimate of the tidal height); perhaps it can be used somewhere else with a better accuracy? The answer is "yes, for the tides of the Earth crust "! Indeed, the mantle thickness is of the order of few thousands km, which corresponds to almost ten-fold tsunami speed and makes the Moon as a "piston" reasonably adiabatic. The relative crust deformation due to tidal movements is so small that the elastic response of the crust is also negligible: the result h = 27 cm is indeed very close to reality.**This article is the original work of Jaan Kalda, Academic Committee, IPhO 2012.**

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