The IPhO syllabus starts where A Level Physics stops. Almost every school topic appears, but you are expected to use calculus freely, to treat rotation, thermodynamics and electromagnetism in more depth, and to add special relativity and simple quantum ideas. Use this roadmap to see which topics you already have, which need stretching and which are new.
Go through the 38 topics below. Tick Got it when you are confident; topics marked New have a short bridge note to get you started.
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Study plan
Stage 1: Secure the school core4-6 weeks Work through the school topics until you can solve A Level questions quickly and without notes, then try the feeder problems.
Stage 2: Mechanics with calculus6-8 weeks Learn to differentiate and integrate as physics tools, then study orbits, rotation, accelerating frames and small oscillations. Mechanics is the backbone of every olympiad paper.
Stage 3: Thermal physics, waves and fields in depth8-10 weeks Take the stretch and new topics in thermodynamics, optics and electromagnetism one by one, doing the bridge note and then national-level problems for each.
Stage 4: Relativity, quantum physics and full papers6-8 weeks Add special relativity and the quantum topics, then sit full timed theory papers and practise data analysis for the experimental round.
Mechanics
Kinematics and vectorsIn your course
Constant-acceleration motion in one and two dimensions, vector resolution and projectiles, read from graphs or equations.
Beyond moments and pressure: equilibrium of extended bodies in 3D, tipping versus sliding, buoyancy on tilted bodies, surface tension and Bernoulli's equation.
Orbit problems test whether you can combine energy and angular momentum conservation instead of just setting gravity equal to the centripetal force.
A body of mass $m$ moving round a much heavier mass $M$ keeps two quantities fixed: its energy and its angular momentum about $M$. Gravity points at $M$, so it exerts no torque, and $L = mvr\sin\alpha$ stays constant. At the closest point (perigee) and furthest point (apogee) the velocity is perpendicular to the radius, so $v_p r_p = v_a r_a$.
Results to know:
Kepler 1: bound orbits are ellipses with $M$ at one focus. The semi-major axis is $a = (r_p + r_a)/2$.
Kepler 2: the radius sweeps equal areas in equal times. This is angular momentum conservation in disguise.
Kepler 3: $T^2 = \dfrac{4\pi^2}{GM}a^3$, the same as for a circle of radius $a$.
Orbital energy: $E = -\dfrac{GMm}{2a}$. It depends only on $a$. Setting $E = \tfrac12 mv^2 - GMm/r$ gives the vis-viva equation $v^2 = GM\left(\dfrac{2}{r} - \dfrac{1}{a}\right)$.
Escape needs $E \ge 0$, so the escape speed at distance $r$ is $\sqrt{2GM/r}$, which is $\sqrt2$ times the circular orbit speed there. When $m$ is not small compared with $M$, both bodies orbit their centre of mass; replace $M$ by $M + m$ in Kepler 3.
Worked example
A satellite orbits the Earth with perigee distance 7000 km and apogee distance 13 000 km (both from the Earth's centre). Take $GM_E = 3.99\times10^{14}\ \mathrm{m^3\,s^{-2}}$. Find the period and the speed at perigee.
Show the solution
$a = (7000 + 13\,000)/2 = 10\,000$ km $= 1.0\times10^7$ m.
Wheels, rods and spinning bodies appear in almost every olympiad mechanics paper, and school courses only treat point masses.
A rigid body rotating at angular speed $\omega$ about a fixed axis has every point moving in a circle. Adding $\tfrac12 m v^2$ over all its pieces gives rotational kinetic energy $\tfrac12 I\omega^2$, where the moment of inertia is $I = \sum m_i r_i^2$ ($r_i$ is the distance from the axis). $I$ plays the part of mass for rotation.
The rotational versions of Newton's laws are $\tau = I\alpha$ and $L = I\omega$; with no external torque, $L$ is conserved (the spinning skater pulling in her arms).
Values to know (mass $M$):
Thin rod, length $\ell$, about its centre: $\tfrac{1}{12}M\ell^2$; about one end: $\tfrac13 M\ell^2$.
Ring or thin hoop, radius $R$: $MR^2$. Solid disc or cylinder: $\tfrac12 MR^2$. Solid sphere: $\tfrac25 MR^2$.
Parallel axis theorem: $I = I_{\mathrm{cm}} + Md^2$ for an axis a distance $d$ from a parallel axis through the centre of mass.
Rolling without slipping means the contact point is momentarily at rest, so $v = \omega R$. The total kinetic energy is then $\tfrac12 Mv^2 + \tfrac12 I\omega^2$. Static friction does no work on a rolling body, so energy conservation still works.
Worked example
A solid cylinder rolls without slipping from rest down a slope, dropping a height of 0.60 m. Find its speed at the bottom ($g = 9.81\ \mathrm{m\,s^{-2}}$).
So $v = \sqrt{4gh/3} = \sqrt{4\times9.81\times0.60/3} = \sqrt{7.85} \approx 2.80\ \mathrm{m\,s^{-1}}$. A frictionless sliding block would reach $\sqrt{2gh} \approx 3.43\ \mathrm{m\,s^{-1}}$: part of the energy goes into spin.
Try it
A uniform rod 1.2 m long is pivoted at one end and released from rest horizontal. What is its angular speed as it passes the vertical?
Show the answer
The centre falls $\ell/2$: $Mg\ell/2 = \tfrac12\cdot\tfrac13 M\ell^2\omega^2$, so $\omega = \sqrt{3g/\ell} = \sqrt{3\times9.81/1.2} \approx 4.95\ \mathrm{rad\,s^{-1}}$.
Going deeper: The rotation chapters of a university first-year mechanics text, then the rigid-body chapter of a classical mechanics text.
Accelerating and rotating framesNew
Solving problems from inside an accelerating or rotating frame by adding inertial forces: the uniform frame force, centrifugal and Coriolis forces.
Many olympiad set-ups (a bead on a spinning hoop, a block on a sliding wedge) are far easier in the frame that moves with the apparatus.
Newton's laws hold only in inertial frames. You can still work in an accelerating frame if you add an inertial force to every body.
Frame with acceleration $\vec a_0$: add $-m\vec a_0$ to each mass. It acts like extra gravity, so the effective gravity is $\vec g_{\mathrm{eff}} = \vec g - \vec a_0$.
Frame rotating at steady $\omega$: add the centrifugal force $m\omega^2 r$ outwards from the axis, and, for bodies moving in that frame, the Coriolis force $-2m\,\vec\omega\times\vec v$.
The method: pick the frame in which the body you care about is at rest or moves simply, draw the free-body diagram including the inertial forces, then use ordinary statics or $F = ma$. The centrifugal force can be given a potential energy $-\tfrac12 m\omega^2 r^2$, so a bead on a spinning wire sits at a minimum of $U_{\mathrm{grav}} - \tfrac12 m\omega^2 r^2$. The Coriolis force does no work because it is always perpendicular to the velocity.
Two checks: inertial forces have no reaction partner, and you must never add them when working in the ground frame as well.
Worked example
A pendulum 0.80 m long hangs in a car that accelerates steadily at $3.0\ \mathrm{m\,s^{-2}}$ on a level road. Find the angle the string makes with the vertical and the period of small swings ($g = 9.81\ \mathrm{m\,s^{-2}}$).
Show the solution
In the car frame, the bob feels $mg$ down and $ma_0$ backwards. It hangs along $\vec g_{\mathrm{eff}}$: $\tan\theta = 3.0/9.81$, so $\theta \approx 17^\circ$, tilted backwards.
$g_{\mathrm{eff}} = \sqrt{9.81^2 + 3.0^2} \approx 10.26\ \mathrm{m\,s^{-2}}$, so $T = 2\pi\sqrt{0.80/10.26} \approx 1.75$ s, a little shorter than the 1.79 s at rest.
Try it
A bucket of water is in a lift that accelerates upwards at $2.0\ \mathrm{m\,s^{-2}}$. What is the pressure due to the water 0.30 m below the surface? (Water density $1000\ \mathrm{kg\,m^{-3}}$, $g = 9.81\ \mathrm{m\,s^{-2}}$.)
Show the answer
$g_{\mathrm{eff}} = 11.81\ \mathrm{m\,s^{-2}}$, so $p = \rho g_{\mathrm{eff}} h = 1000\times11.81\times0.30 \approx 3.5$ kPa.
Going deeper: The non-inertial frames chapter of a university classical mechanics text.
Thermodynamics and statistical physics
Heat, temperature and calorimetryIn your course
Specific and latent heat, mixing problems, thermal equilibrium and energy balance with losses.
First law, gas processes and heat enginesGoes further
School gives the first law; olympiads add molar heat capacities at constant volume and pressure, adiabatic processes, work from p-V areas and the efficiency of cycles.
Gas cycles and adiabatic changes (rising air, engines, pumps) are a favourite way to test the first law with real numbers.
For $n$ moles of ideal gas, the internal energy depends only on temperature: $U = nC_VT$, with $C_V = \tfrac32R$ for a monatomic gas and $\tfrac52R$ for a diatomic gas near room temperature. At constant pressure the gas also does work, so $C_p = C_V + R$. The ratio $\gamma = C_p/C_V$ is $5/3$ (monatomic) or $7/5$ (diatomic).
The first law: $Q = \Delta U + W$, where $W$ is the work done by the gas, the area under the p-V curve.
Isothermal: $\Delta U = 0$, $W = nRT\ln(V_2/V_1)$.
Isobaric: $W = p\,\Delta V$, $Q = nC_p\Delta T$.
Adiabatic (no heat, for example a fast change): $pV^\gamma$ = constant, so $TV^{\gamma-1}$ = constant, and $W = -\Delta U = nC_V(T_1 - T_2)$.
An adiabat is steeper than an isotherm on a p-V diagram. For a cycle, $\Delta U = 0$ over one loop, the net work is the enclosed area, and the efficiency is $\eta = W_{\mathrm{net}}/Q_{\mathrm{in}}$, counting only heat taken in.
Worked example
Air ($\gamma = 1.4$) at 300 K is compressed adiabatically to one tenth of its volume, as in a diesel engine. Find its final temperature.
Show the solution
$T_2 = T_1\left(\dfrac{V_1}{V_2}\right)^{\gamma-1} = 300\times10^{0.4} = 300\times2.512 \approx 754$ K, hot enough to ignite the fuel without a spark.
Try it
A monatomic gas at 400 K expands adiabatically to twice its volume. What is its new temperature?
Show the answer
$T = 400\times2^{-2/3} = 400/1.587 \approx 252$ K.
Going deeper: The thermodynamics chapters of a university first-year physics text.
Entropy and the second lawNew
Entropy as heat divided by temperature for reversible heat flow, the Carnot limit on engines and refrigerators, and entropy changes in irreversible processes.
The second law sets hard limits on engines and decides which way processes go, and IPhO expects you to calculate with it.
The first law says energy is conserved; the second law says which way heat flows. The tool is entropy $S$. When a body at temperature $T$ takes in a small amount of heat $\mathrm{d}Q$ reversibly, its entropy rises by $\mathrm{d}S = \mathrm{d}Q/T$. Entropy is a state variable, so to find $\Delta S$ for any change you may invent a reversible path with the same start and end.
The second law: the total entropy of an isolated system never decreases. It stays constant only for reversible processes.
Heat $Q$ at constant $T$ (melting, a large reservoir): $\Delta S = Q/T$.
Heating a body of heat capacity $C$: $\Delta S = C\ln(T_2/T_1)$.
Ideal gas: $\Delta S = nC_V\ln(T_2/T_1) + nR\ln(V_2/V_1)$.
Carnot limit: an engine between $T_h$ and $T_c$ has efficiency at most $1 - T_c/T_h$. A refrigerator has coefficient of performance at most $T_c/(T_h - T_c)$.
The Carnot limit follows directly: in one cycle the engine's entropy is unchanged, the hot reservoir loses $Q_h/T_h$ and the cold one gains $Q_c/T_c$, and the total cannot fall, so $Q_c/Q_h \ge T_c/T_h$.
Worked example
1000 J of heat flows by conduction from a reservoir at 400 K to one at 300 K. Find the total entropy change.
Show the solution
Hot reservoir: $-1000/400 = -2.50\ \mathrm{J\,K^{-1}}$. Cold reservoir: $+1000/300 = +3.33\ \mathrm{J\,K^{-1}}$. Total $+0.83\ \mathrm{J\,K^{-1}}$: positive, so the process is irreversible, as expected.
Try it
An ideal engine works between 500 K and 300 K and takes 2000 J from the hot reservoir each cycle. What is the most work it can do, and how much heat does it then reject?
Show the answer
$\eta_{\max} = 1 - 300/500 = 0.40$, so at most 800 J of work, and it rejects 1200 J.
Going deeper: A university text on thermal physics, the chapters on entropy and heat engines.
Boltzmann factor and equipartitionNew
How likely a state is at a given energy and temperature, equipartition of energy, and the exponential atmosphere.
Statistical physics links the microscopic and the measured world, and the Boltzmann factor turns up in atmospheres, semiconductors and reaction rates.
In a system at temperature $T$, the chance of finding a particle in a particular state of energy $E$ is proportional to the Boltzmann factor $e^{-E/kT}$, where $k = 1.38\times10^{-23}\ \mathrm{J\,K^{-1}}$. States much higher than $kT$ above the ground state are rarely occupied. At room temperature $kT \approx 0.025$ eV.
Results to know:
Two levels with energy gap $\Delta E$ and equal degeneracy: $N_2/N_1 = e^{-\Delta E/kT}$.
Isothermal atmosphere: the number density falls as $n(h) = n_0e^{-mgh/kT}$, with scale height $H = kT/mg$.
Equipartition: each quadratic term in the energy (such as $\tfrac12mv_x^2$ or $\tfrac12kx^2$) holds on average $\tfrac12kT$. So a monatomic gas has $\tfrac32kT$ per molecule, a diatomic gas $\tfrac52kT$ (two rotations added) and a solid about $3kT$ per atom.
Equipartition explains $C_V = \tfrac32R$ and $\tfrac52R$ from the gas laws, and the root-mean-square speed $\sqrt{3kT/m}$. It fails when $kT$ is small compared with the spacing of quantum levels: vibrations are frozen out at room temperature.
Worked example
Estimate the scale height of nitrogen ($m = 28\times1.66\times10^{-27}$ kg) in an isothermal atmosphere at 290 K.
Show the solution
$m = 4.65\times10^{-26}$ kg. $H = \dfrac{kT}{mg} = \dfrac{1.38\times10^{-23}\times290}{4.65\times10^{-26}\times9.81} \approx 8.8\times10^3$ m. The density falls by a factor $e$ about every 9 km.
Try it
Two energy levels with equal degeneracy are 0.10 eV apart. At 300 K, what is the ratio of the populations of the upper and lower levels?
Show the answer
$kT = 0.0259$ eV, so the ratio is $e^{-0.10/0.0259} = e^{-3.87} \approx 0.021$.
Going deeper: A university introduction to statistical physics or thermal physics.
Black-body radiation and heat conductionGoes further
School gives Wien and Stefan-Boltzmann for stars; olympiads use them for planets and heated bodies in energy balance, and add Fourier's law of conduction through layers.
School treats springs and simple pendulums; olympiads ask you to find the frequency of any system near equilibrium, from its energy or from the curvature of its potential.
Almost every equilibrium in physics oscillates when disturbed, and finding the period of an unusual system is a standard olympiad part.
Near a stable equilibrium at $x_0$ any potential looks like a parabola: $U(x) \approx U(x_0) + \tfrac12U''(x_0)(x - x_0)^2$. So the motion is simple harmonic with $\omega = \sqrt{U''(x_0)/m}$.
The energy method handles systems with several moving parts. Write the total energy in terms of one small coordinate $x$ and its rate $\dot x$:
Use small-angle forms such as $1 - \cos\theta \approx \tfrac12\theta^2$ and drop terms beyond second order.
Physical pendulum (pivot a distance $d$ from the centre of mass, moment of inertia $I$ about the pivot): $T = 2\pi\sqrt{I/(mgd)}$.
Floating body of cross-section $A$, mass $m$: $\omega = \sqrt{\rho gA/m}$.
Liquid in a U-tube, total column length $\ell$: $\omega = \sqrt{2g/\ell}$.
If $U''(x_0) \lt 0$ the equilibrium is unstable and there is no oscillation. A good habit: once you have $\omega$, check that it has units of $\mathrm{s^{-1}}$ and that it rises when the system gets stiffer and falls when it gets heavier.
Worked example
A uniform rod 1.00 m long swings about one end. Find the period of small oscillations ($g = 9.81\ \mathrm{m\,s^{-2}}$).
Show the solution
$I = \tfrac13m\ell^2$ and $d = \ell/2$, so $T = 2\pi\sqrt{\dfrac{\ell^2/3}{g\ell/2}} = 2\pi\sqrt{\dfrac{2\ell}{3g}} = 2\pi\sqrt{0.0680} \approx 1.64$ s. A simple pendulum of the same length has 2.01 s.
Try it
Water fills a U-tube with a total column length of 0.50 m. What is the period of its sloshing?
Show the answer
$T = 2\pi\sqrt{\ell/2g} = 2\pi\sqrt{0.50/19.62} \approx 1.0$ s.
Going deeper: The oscillations chapter of a university mechanics text, and a book on vibrations and waves.
Superposition, interference and standing wavesIn your course
Two-source interference, path and phase difference, standing waves on strings and in pipes.
IGCSE draws ray diagrams; olympiads use the lens and mirror equations, lens combinations, apparent depth and Fermat's principle, and trace rays through drops and prisms.
Optics questions reward quick, accurate use of a few equations, and they appear at every level from the first national round upwards.
Use the real-is-positive sign rule: distances to real objects and images are positive, to virtual ones negative; a converging lens or concave mirror has positive $f$.
Thin lens and mirror equation: $\dfrac1u + \dfrac1v = \dfrac1f$, with magnification $m = -v/u$ (negative means inverted). For a spherical mirror $f = R/2$.
Lenses in contact: powers add, $\dfrac1f = \dfrac1{f_1} + \dfrac1{f_2}$. For separated lenses, the image from the first is the object for the second.
Refraction: $n_1\sin\theta_1 = n_2\sin\theta_2$; critical angle $\sin c = 1/n$; looking straight down into a liquid, apparent depth = real depth / $n$.
Fermat's principle: light takes a path of stationary travel time. Snell's law and the law of reflection both follow from it, and it is quick for curved surfaces.
Because $n$ depends on wavelength, different colours bend by different amounts. A ray inside a sphere (a raindrop) reflects once and is turned through a total angle that has a minimum; the light piles up at that angle.
Worked example
An object is 30 cm from a converging lens of focal length 20 cm. Find the image position and magnification.
Show the solution
$\dfrac1v = \dfrac1{20} - \dfrac1{30} = \dfrac1{60}$, so $v = +60$ cm: a real image on the far side. $m = -60/30 = -2$: inverted and twice the size.
Try it
Two thin converging lenses of focal lengths 10 cm and 15 cm are placed in contact. What is the focal length of the pair?
Show the answer
$1/f = 1/10 + 1/15 = 1/6$, so $f = 6.0$ cm.
Going deeper: The geometrical optics chapters of a university first-year physics text.
Thin films and other interference devicesNew
Interference by reflection from thin films and wedges, including phase changes on reflection, Newton's rings and anti-reflection coatings.
Thin films test whether you can track both path difference and phase changes, the two ideas behind every interference problem.
Light reflected from the top and the bottom of a thin transparent film interferes. Two things set the phase difference between the two reflected rays.
Extra path: at near-normal incidence the lower ray travels an extra $2t$ inside the film, which counts as an optical path of $2nt$ ($n$ is the film's refractive index).
Phase change on reflection: reflection off a medium of higher refractive index adds a phase change of $\pi$ (half a wavelength). Reflection off a lower index adds nothing.
Count the $\pi$ changes. If exactly one ray has one (soap film in air), bright reflection needs $2nt = (m + \tfrac12)\lambda$ and dark needs $2nt = m\lambda$. If both or neither have one (a coating on glass), the conditions swap.
A wedge of angle $\alpha$ gives straight fringes spaced $\lambda/(2n\alpha)$ apart. A lens on a flat plate gives rings (Newton's rings) whose radii grow as $\sqrt m$. A gap that is very thin compared with $\lambda$ looks dark in reflection when only one reflection has the phase change.
Worked example
A soap film ($n = 1.33$) in air is 300 nm thick. Which visible wavelength is reflected most strongly at normal incidence?
Show the solution
One reflection (top surface) has a phase change, so bright needs $2nt = (m + \tfrac12)\lambda$. $2nt = 798$ nm, so $\lambda = 798/(m + \tfrac12)$: 1596 nm, 532 nm, 319 nm, ... Only 532 nm (green) is visible.
Try it
A magnesium fluoride coating ($n = 1.38$) on glass ($n = 1.50$) is to cancel reflection of 550 nm light. What is the minimum thickness?
Show the answer
Both reflections have a phase change, so cancellation needs $2nt = \lambda/2$: $t = 550/(4\times1.38) \approx 100$ nm.
Going deeper: The interference chapter of a university optics text.
Diffraction and resolving powerGoes further
School covers the grating; olympiads add single-slit minima, the Rayleigh criterion for telescopes and the eye, and Bragg reflection from crystals.
Gauss's law gives in one line fields that would otherwise need integration, and conductors appear in many olympiad electrostatics problems.
The electric flux through a surface counts field lines crossing it: for a uniform field $E$ at right angles to an area $A$, the flux is $EA$. Gauss's law says the total flux out of any closed surface equals the charge inside divided by $\varepsilon_0$:
$$\oint \vec E\cdot\mathrm{d}\vec A = \frac{Q_{\mathrm{inside}}}{\varepsilon_0}.$$
It is always true, but it only gives the field when symmetry tells you $E$ is constant over a well-chosen surface.
Sphere or point charge: outside, $E = \dfrac{Q}{4\pi\varepsilon_0r^2}$ (as if all charge were at the centre). Inside a hollow charged shell, $E = 0$. Inside a uniformly charged solid sphere, $E$ grows in proportion to $r$.
Long line with charge per length $\lambda$: $E = \dfrac{\lambda}{2\pi\varepsilon_0r}$.
Large flat sheet with charge per area $\sigma$: $E = \dfrac{\sigma}{2\varepsilon_0}$ on each side, independent of distance.
Conductors in equilibrium: $E = 0$ inside the metal, all excess charge sits on the surface, the whole conductor is at one potential, and just outside the surface $E = \sigma/\varepsilon_0$, perpendicular to it. A charge inside a hollow conductor induces an equal and opposite charge on the inner wall.
Worked example
A long straight wire carries $2.0\ \mathrm{nC}$ per metre. Find the field 5.0 cm from it ($\varepsilon_0 = 8.85\times10^{-12}\ \mathrm{F\,m^{-1}}$).
Show the solution
Use a cylinder of radius $r$ and length $\ell$ round the wire: $E\times2\pi r\ell = \lambda\ell/\varepsilon_0$. So $E = \dfrac{2.0\times10^{-9}}{2\pi\times8.85\times10^{-12}\times0.050} \approx 720\ \mathrm{V\,m^{-1}}$, pointing away from the wire.
Try it
A uniformly charged solid sphere has a field of $4000\ \mathrm{V\,m^{-1}}$ at its surface. What is the field at half the radius, and at twice the radius?
Show the answer
Inside, $E \propto r$: $2000\ \mathrm{V\,m^{-1}}$ at $R/2$. Outside, $E \propto 1/r^2$: $1000\ \mathrm{V\,m^{-1}}$ at $2R$.
Going deeper: The electrostatics chapters of a university electricity and magnetism text.
D.C. circuits and networksGoes further
School covers Kirchhoff's laws; olympiads add symmetric and infinite networks, non-ideal meters, maximum power transfer and non-linear components.
Charged particles in magnetic and crossed fieldsGoes further
School covers circles and the velocity selector; olympiads add helical paths, the cyclotron frequency, drift in crossed fields and the Hall effect in detail.
To solve induction and force problems you first need the field, and these two laws give it for every standard shape.
Every short piece of wire $\mathrm{d}\vec\ell$ carrying current $I$ makes a field at a point a distance $r$ away given by the Biot-Savart law:
$$\mathrm{d}\vec B = \frac{\mu_0}{4\pi}\,\frac{I\,\mathrm{d}\vec\ell\times\hat r}{r^2}.$$
Add (integrate) over the whole circuit. Ampère's law is the shortcut when there is symmetry: round any closed loop, $\oint\vec B\cdot\mathrm{d}\vec\ell = \mu_0I_{\mathrm{enclosed}}$.
Results to know ($\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}$):
Long straight wire: $B = \dfrac{\mu_0I}{2\pi r}$, circling the wire (right-hand grip).
Centre of a circular loop of radius $R$: $B = \dfrac{\mu_0I}{2R}$; for $N$ turns multiply by $N$.
Long solenoid with $n$ turns per metre: $B = \mu_0nI$ inside, almost zero outside.
Parallel wires a distance $d$ apart: force per metre $\dfrac{\mu_0I_1I_2}{2\pi d}$, attractive for currents in the same direction.
A small current loop of area $A$ acts like a magnetic dipole of moment $m = IA$; in a field $B$ it feels a torque $mB\sin\theta$. Fields from several wires add as vectors, so draw each one's direction before you add magnitudes.
Worked example
Two long parallel wires 10 cm apart carry 5.0 A and 8.0 A in the same direction. Find the force per metre between them.
Induction links mechanics, energy and circuits, so it is ideal for long multi-part olympiad problems.
Faraday's law: the induced e.m.f. is $\mathcal{E} = -\mathrm{d}\Phi/\mathrm{d}t$, where $\Phi$ is the flux linkage. The minus sign is Lenz's law: the induced current opposes the change.
Motional e.m.f.: a rod of length $\ell$ moving at $v$ across a field $B$ has $\mathcal{E} = B\ell v$. If a current $I$ flows, the field pushes back on the rod with $F = BI\ell$. The mechanical power you supply, $Fv$, equals the electrical power $I\mathcal{E}$; this energy check catches sign errors.
Self-inductance: a coil's own changing current induces $\mathcal{E} = -L\,\mathrm{d}I/\mathrm{d}t$. A solenoid of $N$ turns, length $\ell$, area $A$ has $L = \mu_0N^2A/\ell$. The energy stored is $\tfrac12LI^2$, the magnetic partner of $\tfrac12CV^2$.
LR circuit: the current grows or decays exponentially with time constant $L/R$.
LC circuit: charge oscillates like a mass on a spring with $\omega = 1/\sqrt{LC}$; energy swaps between $\tfrac12q^2/C$ and $\tfrac12LI^2$.
An inductor resists sudden changes of current, just as a capacitor resists sudden changes of voltage.
Worked example
A rod 0.40 m long slides at a steady 3.0 m/s along rails in a field of 0.50 T. The circuit resistance is $2.0\ \Omega$. Find the current, the force needed to keep the rod moving and the power supplied.
Show the solution
$\mathcal{E} = B\ell v = 0.50\times0.40\times3.0 = 0.60$ V, so $I = 0.30$ A. $F = BI\ell = 0.50\times0.30\times0.40 = 0.060$ N. Power $Fv = 0.18$ W, which matches $I^2R = 0.09\times2.0 = 0.18$ W.
Try it
What is the natural frequency of an LC circuit with $L = 20$ mH and $C = 5.0\ \mu$F?
Real circuits contain capacitors and coils, and IPhO problems use impedance and resonance freely.
With a sinusoidal supply of angular frequency $\omega$, a capacitor and an inductor each limit the current, like resistors, but they also shift its phase. Their reactances are
with $I_{\mathrm{rms}} = V_{\mathrm{rms}}/Z$. Complex numbers do the same job: $Z_R = R$, $Z_L = i\omega L$, $Z_C = 1/(i\omega C)$, combined with the ordinary series and parallel rules.
Resonance: $X_L = X_C$ at $\omega_0 = 1/\sqrt{LC}$, where $Z = R$ and the current is largest.
Power: only the resistor dissipates energy on average, $P = I_{\mathrm{rms}}^2R = V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\phi$.
Worked example
A series circuit has $R = 40\ \Omega$, $L = 0.10$ H and $C = 50\ \mu$F on a 230 V rms, 50 Hz supply. Find the current and the average power.
$P = I^2R \approx 4.48^2\times40 \approx 800$ W. Since $X_C \gt X_L$, the current leads the voltage, by about $39^\circ$.
Try it
At what frequency does the circuit above resonate?
Show the answer
$f_0 = \dfrac{1}{2\pi\sqrt{0.10\times5.0\times10^{-5}}} \approx 71$ Hz.
Going deeper: The a.c. circuits chapter of a university electricity and magnetism or electronics text.
Special relativity
Time dilation and length contractionNew
Einstein's two postulates and their consequences for clocks, lengths and simultaneity.
Read the bridge note
Special relativity is on the IPhO syllabus, and fast particles in many problems need it.
Special relativity rests on two postulates: the laws of physics are the same in every inertial frame, and light travels at $c$ in every inertial frame. Together they force time and space to depend on the observer.
Define the Lorentz factor $\gamma = \dfrac{1}{\sqrt{1 - v^2/c^2}}$. It is 1 at rest, 1.25 at $0.6c$ and grows without limit as $v \to c$.
Time dilation: a clock moving at $v$ ticks slowly. If it records a proper time $\Delta\tau$ (time between two events at the same place in its own frame), you measure $\Delta t = \gamma\Delta\tau$.
Length contraction: a rod with proper length $L_0$ (measured at rest) moving along its length has length $L_0/\gamma$ for you. Lengths at right angles to the motion do not change.
Relativity of simultaneity: events that happen at the same time in one frame, at different places, need not be simultaneous in another.
The commonest mistake is applying the formulas the wrong way round. Always ask: in which frame do the two events happen at the same place? That frame measures the proper time. In which frame is the object at rest? That frame measures the proper length.
Worked example
Muons made high in the atmosphere have a lifetime of $2.2\ \mu$s in their own frame and travel at $0.995c$. How far does a typical muon travel in the Earth's frame?
Show the solution
$\gamma = 1/\sqrt{1 - 0.995^2} = 1/\sqrt{0.00998} \approx 10.0$, so the lifetime in the Earth frame is about $22\ \mu$s. Distance $\approx 0.995\times3.0\times10^8\times22\times10^{-6} \approx 6.6$ km. Without relativity you would get only 0.66 km, and few muons would reach the ground.
Try it
A rod has a proper length of 2.0 m and moves along its length at $0.60c$. How long is it in your frame?
Show the answer
$\gamma = 1.25$, so $L = 2.0/1.25 = 1.6$ m.
Going deeper: An introductory special relativity text, or the relativity chapters of a university modern physics text.
Lorentz transformation and velocity additionNew
Transforming event coordinates between frames, adding velocities relativistically and the relativistic Doppler effect.
For differences between two events, replace $x$ and $t$ by $\Delta x$ and $\Delta t$. The quantity $c^2\Delta t^2 - \Delta x^2$ (the interval) is the same in every frame.
Velocity addition: an object moving at $u$ in S moves at $u' = \dfrac{u - v}{1 - uv/c^2}$ in S'. Equivalently, if it moves at $u'$ in S', then in S it moves at $\dfrac{u' + v}{1 + u'v/c^2}$, which never exceeds $c$.
Relativistic Doppler effect: for a source approaching at speed $\beta c$, $f = f_0\sqrt{\dfrac{1+\beta}{1-\beta}}$; for a receding source swap the signs. For small $\beta$ this gives the school result $\Delta f/f \approx \beta$.
The $vx/c^2$ term in $t'$ is the source of the relativity of simultaneity: a time difference appears from a space difference alone.
Worked example
A rocket moving at $0.60c$ fires a probe forwards at $0.80c$ relative to the rocket. How fast is the probe moving relative to the ground?
Two lamps 600 m apart flash at the same time in the ground frame. For an observer moving at $0.60c$ along the line joining them, how far apart in time are the flashes?
Show the answer
$\Delta t' = \gamma\left(0 - \dfrac{v\,\Delta x}{c^2}\right) = 1.25\times\dfrac{0.60\times600}{3.0\times10^8} = 1.5\ \mu$s. The lamp further ahead in the direction of motion flashes first.
Going deeper: The Lorentz transformation chapter of an introductory special relativity text.
Relativistic energy and momentumGoes further
A Level uses mass-energy equivalence for binding energy; olympiads add relativistic momentum, total energy and the energy-momentum relation for collisions, decays and photons.
Particle collisions and decays need exact relativistic energy and momentum conservation, and they appear on most IPhO papers.
For a particle of mass $m$ moving at $v$, the relativistic momentum and total energy are
$$p = \gamma mv, \qquad E = \gamma mc^2.$$
At rest $E = mc^2$ (the rest energy); the kinetic energy is $E - mc^2 = (\gamma - 1)mc^2$, which reduces to $\tfrac12mv^2$ when $v \ll c$.
Energy-momentum relation: $E^2 = (pc)^2 + (mc^2)^2$. It holds in every frame and avoids computing $v$.
Speed from energy: $v/c = pc/E$.
Massless particles (photons): $E = pc$.
In collisions and decays, total energy and total momentum are both conserved; kinetic energy and mass separately are not. Work in units of MeV for energy, MeV/$c$ for momentum and MeV/$c^2$ for mass; then the factors of $c$ disappear from the arithmetic. The electron's rest energy is 0.511 MeV and the proton's 938 MeV.
A useful trick: the combination $E_{\mathrm{tot}}^2 - (p_{\mathrm{tot}}c)^2$ for a group of particles is the same in every frame, so you can evaluate it in whichever frame is easiest, often the centre-of-momentum frame.
Worked example
An electron is accelerated from rest through 1.0 MV. Find its momentum (in MeV/$c$) and its speed.
Show the solution
Kinetic energy 1.0 MeV, so $E = 1.0 + 0.511 = 1.511$ MeV and $\gamma = 2.96$.
Light carrying momentum explains Compton scattering, solar sails and laser cooling, all recurring olympiad themes.
A photon of wavelength $\lambda$ carries energy $E = hc/\lambda$ and momentum $p = h/\lambda = E/c$. So light pushes on whatever absorbs or reflects it.
Radiation force: a beam of power $P$ fully absorbed gives a force $P/c$; fully reflected at normal incidence, $2P/c$. Divide by area for radiation pressure, $I/c$ or $2I/c$.
Compton scattering: a photon bouncing off a free electron at rest loses energy. Conserving relativistic energy and momentum gives $\lambda' - \lambda = \dfrac{h}{m_ec}(1 - \cos\theta)$, where $\theta$ is the photon's scattering angle and $h/m_ec = 2.43$ pm. The shift is only noticeable for X-rays and gamma rays.
Atom recoil: an atom absorbing a photon picks up a speed $h/(m\lambda)$. Many absorptions from a beam facing the atom slow it down; this is the idea behind laser cooling.
A useful number: $hc = 1240$ eV nm, so a 500 nm photon has 2.48 eV. In every one of these problems, write momentum conservation as a vector equation first, then resolve it along sensible axes.
Worked example
X-rays of wavelength 50.0 pm scatter from free electrons through $90^\circ$. Find the scattered wavelength and the kinetic energy given to the electron.
Show the solution
$\lambda' = 50.0 + 2.43(1 - 0) = 52.43$ pm.
Photon energies: $1240/0.0500 = 24\,800$ eV before and $1240/0.05243 \approx 23\,650$ eV after. The electron gains about 1.15 keV.
Try it
A 2.0 W laser beam falls on a black plate that absorbs it all. What force does it exert? What if the plate were a perfect mirror?
Show the answer
$F = P/c = 2.0/3.0\times10^8 \approx 6.7\times10^{-9}$ N; a mirror doubles it to about $1.3\times10^{-8}$ N.
Going deeper: The early quantum physics chapters of a university modern physics text.
Energy levels and the Bohr modelGoes further
A Level uses given energy levels; olympiads expect you to derive the Bohr energies for hydrogen-like ions and apply them to spectra.
The Bohr model is a short derivation that combines circular motion, electrostatics and quantisation, and olympiads ask you to adapt it to new systems.
Bohr's model treats an electron as orbiting a nucleus of charge $Ze$ in a circle, with one extra rule: the angular momentum is quantised, $m_evr = n\hbar$ with $n = 1, 2, 3, \ldots$ and $\hbar = h/2\pi$.
Combine it with the circular-motion equation $\dfrac{m_ev^2}{r} = \dfrac{Ze^2}{4\pi\varepsilon_0r^2}$ and you get
radii $r_n = \dfrac{n^2}{Z}a_0$, with the Bohr radius $a_0 = 0.0529$ nm;
photon energy for a jump: $hf = 13.6\,Z^2\left(\dfrac1{n_1^2} - \dfrac1{n_2^2}\right)$ eV.
The total energy is negative and equals minus the kinetic energy, a result of the inverse-square force. The 13.6 eV is proportional to the orbiting particle's mass, so for a different particle (a muon, or a nucleus that moves too) use the reduced mass $\mu = \dfrac{m_1m_2}{m_1+m_2}$ instead of $m_e$.
Quantum mechanics replaces orbits by standing waves, and the de Broglie picture gives the same rule: a whole number of wavelengths $h/(m_ev)$ must fit around the circle.
Worked example
Find the wavelength of the photon emitted when the electron in He$^+$ ($Z = 2$) drops from $n = 3$ to $n = 2$.
Uncertainty and box estimates give quick order-of-magnitude answers for atoms, nuclei and nanostructures.
Every particle with momentum $p$ has a de Broglie wavelength $\lambda = h/p$. Two consequences follow.
Uncertainty principle: a particle confined to a region of size $\Delta x$ has a momentum spread of at least $\Delta p \ge \dfrac{\hbar}{2\Delta x}$. Confinement costs energy: roughly $E \sim \dfrac{(\Delta p)^2}{2m}$. This is why electrons do not fall into the nucleus, and it gives the size of atoms and the energies of nuclear particles to within a factor of a few.
Particle in a box: a particle trapped in a one-dimensional box of length $L$ must fit a whole number of half-wavelengths, $L = n\lambda/2$. So $p = nh/(2L)$ and
$$E_n = \frac{n^2h^2}{8mL^2}, \qquad n = 1, 2, 3, \ldots$$
The lowest energy is not zero (zero-point energy). The levels spread out as $n^2$, and they get further apart for lighter particles and smaller boxes.
For estimates, olympiad answers accept order-of-magnitude results from uncertainty arguments, but say clearly which form ($\hbar$, $\hbar/2$ or $h$) you used.
Worked example
An electron is trapped in a one-dimensional box 0.50 nm long. Find its ground-state energy and the photon energy emitted when it drops from $n = 2$ to $n = 1$.
$E_2 = 4E_1$, so the photon carries $3E_1 \approx 4.5$ eV.
Try it
Estimate the minimum kinetic energy of a proton ($1.67\times10^{-27}$ kg) confined within a nucleus, taking $\Delta x = 1.0\times10^{-14}$ m.
Show the answer
$\Delta p \ge \hbar/(2\Delta x) = 1.055\times10^{-34}/(2.0\times10^{-14}) \approx 5.3\times10^{-21}\ \mathrm{kg\,m\,s^{-1}}$. $E \sim p^2/2m \approx 8\times10^{-15}$ J, about 0.05 MeV. A cruder estimate with $\Delta p \sim \hbar/\Delta x$ gives about 0.2 MeV; both show nuclear energies are far above atomic ones.
Going deeper: An introductory quantum physics text, the chapters on matter waves and the Schrodinger equation in one dimension.
Nuclear physics and radioactivityIn your course
Decay equations, half-life and exponential decay, binding energy per nucleon, fission and fusion.
Dimensional analysisCheck the units of every answer before you move on. When you are stuck, a quick dimensional argument often gives the form of the answer up to a number.
Limiting cases and symmetryTest your formula by sending a mass to zero, an angle to 90 degrees or a speed to the speed of light. Symmetry tells you where fields vanish and which way forces point before you calculate anything.
ApproximationsOlympiad problems expect small-angle forms, the first terms of the binomial expansion and the neglect of small terms. Say which approximation you make and why it is justified.
EstimationMany questions ask for an order of magnitude. Pick sensible round values, keep one significant figure and finish in a minute.
Experimental skillsThe experimental exam is almost half the IPhO marks. Practise planning measurements, estimating uncertainties, linearising a relationship to get a straight-line graph and finding a gradient with its uncertainty.
Reading long problemsIPhO theory problems run to several pages and build a model step by step. Read the whole problem first, note every given quantity, and carry earlier answers forward symbolically so a slip does not spoil later parts.
Ready to practise?Problems with staged hints and a timed drill.IPhO practice →