A Glimpse into the Special Theory of Relativity Siim Ainsaar January 2013 1 Why relativity?...... 1 4.6 Light­cones, simultaneity and 2 Postulates of special relativity...... 2 causality...... 6 3 Basic thought experiments...... 2 4.7 Lorentz transformations 3.1 Time dilation...... 2 algebraically...... 7 3.2 Length contraction...... 3 5 Dynamics...... 7 3.3 Proper time...... 3 5.1 Four­velocity and 4 Lorentz transformations...... 3 four­acceleration...... 7 4.1 Spacetime interval...... 3 5.2 Mass, momentum and energy...... 8 4.2 Minkowski spacetime, 5.3 Force...... 9 Poincaré transformations...... 4 6 Optical effects...... 9 4.3 Rapidity...... 4 7 Electromagnetism...... 9 4.4 Hyperbolic trigonometry...... 5 8 Basic problems...... 9 4.5 Length contraction, time dilation 9 Olympiad problems...... 9 and velocity addition...... 5 10 Further reading...... 9

1 Why relativity? frame and departing in another? Electromagnetism needs relativity Relativity is often seen as an intric­ for an explanation. ate theory that is necessary only when dealing with really high Photons (thus, much of optics) are speeds or ultra­precise measure­ always relativistic and other ments. However, there are some particles often are. Anything where quite often­encountered topics that the speed of light matters – for ex­ are paradoxical if treated non­ ample, the GPS measuring the time relativistically. These are also some for a radio signal to travel from of the main sources of Olympiad satellites – uses relativity. problems on relativity. Particle physics needs relativity in Think for a moment about two several aspects. Particles cannot be charged initially stationary controlled in a modern accelerator particles. They “feel” only the elec­ without taking into account their trostatic force from each another. relativistic dynamics. The only suc­ But in another, moving reference cessful quantum theory predicting frame there is also the magnetic the outcomes of particle collisions, force, in general, in a different dir­ quantum field theory, is relativistic. ection! How could force depend on Muons in cosmic rays would decay the choice of inertial reference long before reaching the ground, frame? What principles forbid the but we still detect them thanks to particles from colliding in one relativistic time dilation.

1 Relativistic theory of gravity – gen­ 2 Postulates of special eral relativity – allows to formulate relativity the physics independently of 1. The laws of all physics are whether the reference frame is iner­ the same in every inertial ref­ tial or not, thus unifying time and erence frame. space even more tightly. It is neces­ sary for astrophysics (precession of A reference frame is inertial if and planets' orbits, gravitational lens­ only if objects onto which no force ing, black holes), and cosmology acts move in a straight line with (history and future of large­scale constant velocity. structures). 2. The speed of light in vacu­ * * * um (c) is the same in every inertial reference frame. In the following, we shall derive the most important results of the spe­ In SI, after defining the second,* the cial theory of relativity, starting metre is defined through fixing (ex­ from the fundamental postulates. actly!)c= 299,792,458 m/s. Most steps of the derivation are giv­ en as problems, which are also 3 Basic thought good examples of what one can ask experiments in relativity and exercise the reader's ability to use the theory. 3.1 Time dilation

The most important general tech­ Problem 1. Consider a “light nique for problem­solving is rota­ clock” that works as follows. A tion of Minkowski spacetime in photon is emitted towards a complex coordinates, this is de­ mirror at a known distancel scribed in section 4. Section 3 shows and reflected back. It is detec­ the way from postulates to the use­ ted (almost) at the emitter ful techniques, its problems may be again. The time from the emis­ skipped if concentrating purely on sion to the detection (a “tick”) Olympiad preparation. Section 4 is is measured to bet. Now we mostly on kinematics, the following ones develop dynamics, optics and * One second is the duration of (briefly) electromagnetism from it. 9,192,631,770 periods of the Finally, some problems for prac­ radiation corresponding to the tising are given. transition between the two hyperfine levels of the ground state of the caesium-133 atom at rest at a temperature of 0 K.

2 look at the clock from a refer­ ror in our reference frame, if ence frame where the whole the distance in the stationary apparatus is moving with velo­ frame isl? city v perpendicularly to the light The answer is in the following fact. beam. Assume that the lengths perpendicular to the motion do Fact 2. If the length of a station­ not change. How long is the ary rod isl, then its length in a tick for us? (Hint: the light reference frame moving in beam follows a zig­zag path.) parallel to the rod with speed The answer is given by the follow­ v is l/γ . ing fact. Lengths are contracted (compressed) in the direction of motion. Fact 1. If the time interval between to events happening 3.3 Proper time at a stationary point ist, then A spaceship flies in a reference frame where the Problem 3. speed of the point isv the freely from (t 1 ,x1 , y1 , z1) time interval is γt , where (event 1) to (t 2 , x2 , y2 , z2) the Lorentz factor (event 2). What is the proper 1 time τ – time measured by a γ= . passenger on the spaceship – v2 1− between these events? [Answer: 2 c2 2 c 2 t t 2 x x 2 c τ = ( 2− 1) −( 2− 1) ] √ 2 2 Another useful quantity in relativ­ −(y 2−y 1) −(z 2−z 1) istic calculations is β=v/c . As γ>1 , we see everything in a mov­ 4 Lorentz transformations ing vehicle take longer than in a sta­ tionary one – time is dilated 4.1 Spacetime interval (stretched) in a moving reference In ordinary, Galilean relativity, frame. lengths and time intervals are abso­ lute. As we have now seen, the pos­ 3.2 Length contraction tulates of Einsteinian relativity im­ Problem 2. Now consider the ply that neither is so, once speeds same “light clock” as in Prob­ become comparable toc. However, lem 1., but moving in parallel to proper time – time in a comoving the light beam, with velocityv. frame – must clearly be independ­ What is the distance to the mir­ ent of our reference frame. There­ fore we can define a new invariant

3 quantity with the dimension of numbers. Namely, the invariant length. quantity is ic t 2 x 2 y 2 z 2 Fact 3. The spacetime interval = √( Δ ) +(Δ ) +(Δ ) +(Δ ) 2 2 2 2 2 is now expressed just like Py­ s= c (Δt) −(Δx) −(Δy) −(Δz) √ thagorean theorem. So, the Euc­ is independent of the choice lidean distance between two events of reference frame. in the spacetime of (ict,x,y,z) is in­ Ifs is a real number, the interval is dependent of reference frame. called time­like; ifs is imaginary, the interval is space­like. Ifs is zero, the What transformations of Euclidean interval is light­like. space leave lengths invariant? Rota­ tions and translations and combina­ Fact 4. The interval between two tions thereof! events on the same light­ray (in vacuum) is zero – thus, Fact 5. Changes of inertial refer­ light­like. ence frames correspond to ro­ tations and shifts in the space­ 4.2 Minkowski spacetime, time coordinates ict, x, y andz. Poincaré transformations We can say that spacetime points In general, such transformations are * are represented by position four­vec­ called the Poincaré transformations * μ and, if we only rotate and do not tors x =(ct,x,y,z) and the in­ shift the coordinates, the Lorentz terval calculates the length of the μ transformations. displacement four­vector Δx . However, this law of calculating the 4.3 Rapidity length has important minus signs in By what angle should we rotate the it, so these four­vectors form a axes? Clearly, as one axis has ima­ Minkowski spacetime, not the usual ginary numbers on it, the angle Euclidean space, where lengths must also be complex. would be calculated using the usual Pythagoras' law. Luckily this poses no problems in drawing the angle, as long as we We can reuse our familiar laws of consider only one­dimensional mo­ geometry if we introduce complex tion: it turns out to be a purely ima­ ginary angle, so its cosine

* Four-vectors are customarily labelled by Greek indices written * The Poincaré transformations are as superscripts; subscripts have a also known as the inhomogeneous meaning in more advanced theory. Lorentz transformations.

4 i i e α +e − α Problem 5. Calculate cosα and cosα= (a projection of the 2 sinα . unit direction vector) is real (can be drawn on the realx­axis) and its Fact 7. cosα=γ , sinα=βγ/i . eiα−e−iα The quantity ϕ=α/i is a real di­ sine sinα= is purely ima­ 2i mensionless number and is called ginary (can be drawn on the ima­ the rapidity. ginary ict­axis). 4.4 Hyperbolic trigonometry Problem 4. Take two coordinate Some imaginary unitsi and some systems,O and O', with the minuses can be eliminated by using spatial axes parallel and the hyperbolic trigonometry. Employ­ α −α * e −e (spatial) origin ofO' moving in ing the formulae sinhα= , thex­direction with velocityv. 2 eα+e−α sinhα Calculate the angle α between coshα= , tanhα= 2 coshα thex­ and x'­axis. (Hint: make and cosh 2α−sinh 2 α=1 you can a diagram with ict on one axis prove the following. andx on another. Add the ict'­ and x'­axes. Calculate thex­ Problem 6. Prove for the rapidity and ­coordinate of one arbit­ ict ϕ that tanh ϕ=β , coshϕ=γ rary point the spatial origin of and sinhϕ=βγ . O' passes through. The ratio of Consequently, using the inverse these coordinates is tan .) α function of hyperbolic tangent, Such a Lorentz transformation in­ α=iϕ=i artanhβ . volving only time and one spatial coordinate is called the Lorentz boost 4.5 Length contraction, time in thex­direction. The answer to the dilation and velocity addition problem is the following useful fact. Problem 7. Prove again the length contraction formula of Fact 2. Fact 6. A Lorentz boost in thex­ Here use rotation of Minkowski direction from standstill to ve­ spacetime. locityv corresponds to rota­ tion ofx­ and ict­axis by an Problem 8. Prove similarly the angle of time dilation formula of Fact 1. v β α=arctan =arctan . ic i Fact 8. If an object moves with re­ spect to reference frame O' * In the spatial origin, x=y=z=0, with velocityu and O' moves but ict changes.

5 with respect to frameO with The world­line of a photon cuts a velocityv in the same direc­ very special wedge from the dia­ tion, then the velocity of the gram: the inside of the wedge can object inO is be influenced event at the tip of the u+v cone; the outside cannot. The region w= . uv where an event can have influence 1+ c2 in is called the light­cone of the event. Problem 9. Prove the velocity ad­ dition formula in the last fact. Fact 10. If the spacetime diagram tanα+tanβ (Hint: tan(α+β)= is scaled so thati metres (on 1−tanα tanβ the ict­axis) is at the same dis­ and tance from the origin as 1 tanhα+tanhβ tanh(α+β)= .) metre (on thex­axis), then the 1+tanhαtanhβ world­line of a photon is at Problem 10. Show that the velocity 45º from either axis. addition formula implies the Fact 11. Simultaneity is relative. postulate that the speed of light is universal. (Hint: u=±c .) Problem 12. In reference frameO, two events take place at the Problem 11. Prove that ifu andv in same timet = 0, but with spa­ the velocity addition formula tial separation Δx . What is the are both between −c and c , time Δt' between them in ref­ then so isw. (Hint: show that dw erence frameO', which is mov­ >0 – hencew is monoton­ du ing in thex­direction with velo­ ous – and use the result of the cityv? [Answer: 2 last problem that u=±c cor­ Δt'=−γvΔx/c ] responds to w=±c .) Fact 12. The order of two events Fact 9. If there exists a reference with time­like or light­like frame where an object moves separation is absolute. For slower than light, then it does space­like separation, the or­ so in every reference frame. der depends on the reference frame. 4.6 Light­cones, simultaneity and causality This means that only time­like or The trajectory of a particle in the light­like separation allows one event to be the cause of another. De­ space­time is called its world­line. manding that the causality should

6 hold and, thus, no information may over around two different axes, re­ be sent to the past, we get the fol­ member the result and then repeat, lowing fact. switching the axes. The result of two successive boosts in different Fact 13. Information cannot directions is actually not just a boost propagate faster than light in in a third direction, but adds some vacuum. rotation that depends on the order This means, among many other im­ of the boosts. plications, that everything must be somewhat deformable: if we push 5 Dynamics one end of a long rod, then the push will propagate to the other end 5.1 Four­velocity and slower thanc (probably much four­acceleration slower). Generalising from the position four­ vector xμ=(ct,x,y,z) intro­ 4.7 Lorentz transformations duced in section 4.2, we now in gen­ algebraically erality define a four­vector as a col­ Fact 14. When going to a reference lection of four numbers t x y z frame moving in thex­direc­ qμ=(q ,q ,q ,q ) that trans­ tion with velocityv, the time forms under Lorentz transforma­ and space coordinates of an tions. The spatial components x y z event transform under Lorentz (q ,q , q )≡⃗q rotate just like a transformations as follows. usual vector. The time­ and space­ vx components are mixed by Lorentz t '=γ(t− 2 ) c boosts that act as rotations in the x ' x vt t =γ( − ) four­space of (iq ,⃗q). A boost in y '=y thex­direction is given just as in z'=z Fact 14. Problem 13. Prove the last fact. qt '=γ(q t −βq x ) Problem 14. Show algebraically qx '=γ(q x−βq t ) that if boosting in bothx­ and The Lorentz­invariant length of the y­directions, the order of boosts four­vector is matters. μ t 2 x 2 y 2 z 2 Intuitively, as boosts are rotations, ∣q ∣=√(q ) −(q ) −(q ) −(q ) . their order should matter just like We already know that periods of the order of ordinary spatial rota­ proper time dτ are Lorentz­ tions matters: try turning a book

7 invariant. Thus, the following deriv­ discussing motion in several dimen­ atives can be formed. sions. Therefore, in this studying material, we refer tom as just the μ μ d x mass. This mass is an intrinsic prop­ Fact 15. Four­velocity v = dτ erty of any object and does not de­ and four­acceleration pend on the reference frame. μ μ d v a = of a particle are dτ Fact 17. The four­momentum of a four­vectors. particle with massm is the four­vector pμ=mv μ .

Problem 15. Show that the four­ μ velocity of a particle moving Fact 18. p =(E/c,⃗p) where the 2 with speedv in thex­direction total energy E=γmc and is (γc,γv,0,0) . the relativistic momentum ⃗p=γm⃗v. Problem 16. What Lorentz­invari­ ant quantity is the length of the Note that here ⃗vis the usual three­ four­velocity from the last velocity and not the spatial part of problem? [Answer:c] the four­velocity that has an addi­ tional γ in it. As thex­direction was arbitrary, we can generalize the answer as fol­ Fact 19. The length of the four­ lows. momentum is mc, whatever the velocity is. Therefore, Fact 16. The length of any four­ 2 2 2 2 velocity isc. E =(pc) +(mc ) . For massless particles (such as Problem 17. Show that the four­ photons), E=pc . acceleration of a particle mov­ ing and accelerating in thex­ Fact 20. In interactions, four­ direction with a three­accelera­ momentum is conserved. tion of magnitude a=dv/ dt is This encompasses both the conser­ 4 4 (β γ a ,γ a ,0,0) with in­ vation of energy and the conserva­ variant lengtha. tion of momentum.

5.2 Mass, momentum and energy Fact 21. The total energy can be Some texts about relativity distin­ separated into the rest energy guish the rest mass or invariant mass 2 E rest=mc and the kinetic m from the relativistic mass γm , 2 but this would be misleading for energy E k=(γ−1)mc .

8 Problem 18. Show that for low 6 Optical effects m v2 speeds, E ≈ . k 2 Problem 20. What is the apparent length of a rod with rest length Note that if an object has any intern­ l moving with velocityv in par­ al structure and, thus, internal en­ allel to the rod, if you take into ergy, then it must be taken into ac­ account the finite travel times count in its rest energy and, thus, its of photons from its ends to our (rest) mass. eyes? On the other hand, for any ul­ Fact 25. Doppler shift of the fre­ trarelativistic object moving almost c−v with a speed ofc, the rest energy quency of light: ν'=ν . 0 c v and the rest mass can be neglected; √ + thus, E≈pc . Problem 21. Prove the formula, Since the speed of light,c, corres­ considering the world­lines of ponds to γ=∞ , we can deduce the two wave­crests. following. Problem 22. Reprove the for­ mula using E h . Fact 22. It takes infinite energy to = ν accelerate a massive object At least two important relativistic toc. Massless particles move optical effects have been left out of only with a speed ofc. this studying material, but are still worthwhile to think about: 5.3 Force • Measuring the Astronomic­ d⃗p d(γmv) al Unit through aberration Fact 23. F⃗= = . d t d t • Compton scattering Fact 24. Four­force μ μ μ d p 7 Electromagnetism F =m a = . dτ The Lorentz force acting on a particle Problem 19. Show that if all the with chargeq moving in an electro­ F⃗ q E⃗ q v B⃗ motion is inx­direction, then magnetic field is = + ⃗ × . If μ we separate the fields into compon­ F =(β γF,γF,0,0) . ents parallel and perpendicular to μ In general, F =(γ ⃗vF⃗⋅/c,γ F⃗) ⃗v, it can be shown that the electric where ⃗v⋅F⃗=dE/ dt is the power. and magnetic fields transform into

9 each other upon Lorentz transform­ Problem 25. The characteristic life­ ations: time of a muon at rest is τ=2.2⋅10−6 s. How long a path ⃗ ⃗ ⃗ ⃗ Fact 26. E∥ '=E∥ , B∥'= B∥ , s can it travel since its creation, if its speed isv = 0.999c? [An­ E⃗ '=γ( E⃗ +⃗v×B⃗ ) , ⊥ ⊥ ⊥ swer: γvτ=14.7 km] (PK202) ⃗ ⃗ ⃗ 2 B ⊥ '=γ( B⊥−⃗v×E ⊥ /c ) . Problem 26. A pion at rest decays 8 Additional problems into a muon and a neutrino. Find the total energyE and the The following problems have been * kinetic energyT of the muon, if translated from an Estonian book. the rest masses of the pion and the muon are, respectively, m Problem 23. A rod with rest length π and m ; the rest mass of the l is moving translationally μ 0 neutrino is zero. [Answers: with speedv in such a way that (m 2+m 2)c 2 the line connecting its end­ E= π μ , 2m π points at an instant forms an 2 2 (m −m ) c angle ϕ with the direction of T= π μ ] (PK234) 2m motion. Find its length. [An­ π l swer: 0 ] (PK200) Problem 27. A muon at rest decays γ √1−β2sin 2 ϕ into an electron and two neutri­ nos. The rest mass of the muon Problem 24. A body is moving uni­ is μ , the mass of the electron formly in a circle, an orbit takes ism, the mass of the neutrino is t = 3 h. A clock inside the body zero. Find the maximum pos­ sees it to take τ=30min . Find sible energy Emax of the elec­ the radiusR of the orbit. [An­ 2 2 1+(μ/m) c t2 2 2 (PK201) tron. [Answer: mc ] swer: √ −τ /( π) ] 2(μ/m) (PK235)

Problem 28. At least how big must * Paul Kard, “Elektrodünaamika ja be the energyE of a pion, if its spetsiaalse relatiivsusteooria collision with a nucleon at rest ülesannete kogu” (“A collection of produces a nucleon­antinucle­ problems on electrodynamics and on pair and the pion is ab­ special relativity”), Tartu State University 1961. Here sorbed? The rest masses of the we cite it as “PK”, followed by the nucleon and the pion are, re­ problem number.

10 spectively,M andm. [Answer: c(γ+3γ 2 /4)(2+3γ+5γ 2/4) −1 ] (8M 2−m2)c 2/(2M) ] (PK244) (Part of PK249)

Problem 29. At least how big should be the energyE of a nuc­ 9 Olympiad problems leon, if its collision with a nucle­ See the following pages for the ori­ on at rest produces a nuc­ ginal texts of the problems. leon­antinucleon pair and the original nucleons are both • Cuba 1991 (Relativistic intact? The rest mass of the nuc­ Square) 2 leon isM. [Answer: 7Mc ] (PK245) • Iceland 1998 (Faster than Problem 30. An atom with Light?) rest massm, at rest, radiates a • Taiwan 2003 (Neutrino De­ photon with frequency ν . What cay) is the rest mass m0 of the atom after the process? [Answer: • China 1994 (Relativistic 2 Particle) m√1−2hν/(mc ) ] (PK248) • Australia 1995 (Gravitation­ Problem 31. The difference al Red­shift) between an excited energy level and the ground level of an atom • Physics Cup 2012 (Electron­ is ΔE . What should the speedv Positron annihilation) of the excited atom be, if we want a photon, that is radiated 10 Further reading in the direction of motion, have a frequency of ΔE/h ? The rest • Ta­Pei Cheng, “Relativity, mass of the atom in its ground gravitation, and cosmology: state ism. [Answer: a basic introduction”, Ox­ ford University Press, 2005, 2006

11 THEORETICALPROBLEMS

Problem 1 The figure 1.1 shows a solid, homogeneous ball radiusR. Before falling to the floor its center of mass is at rest, but the ball is spinning with angular velocityω 0 about a horizontal axis through its center. The lowest point of the ball is at a heighth above the floor.

When released, the ball falls under gravity, and rebounds to a new height such that its lowest point is now ah above the floor. The deformation of the ball and the floor on impact may be considered negligible. Ignore the presence of the air. The impact time, although, is finite.

The mass of the ball ism, the acceleration due the gravity isg, the dynamic coefficient of friction between the ball and the floor isµ k, and the moment of inertia of the ball about the given axis is: 2mR 2 I = 5 You are required to consider two situations, in the first, the ball slips during the entire impact time, and in the second the slipping stops before the end of the impact time.

Situation I: slipping throughout the impact. Find: a)tan θ , whereθ is the rebound angle indicated in the diagram; b)the horizontal distance traveled in flight between the first and second impacts; c)the minimum value ofω 0 for this situations.

Situation II: slipping for part of the impacts. Find, again: a)tan θ; b)the horizontal distance traveled in flight between the first and second impacts. Taking both of the above situations into account, sketch the variation of tanθ withω 0.

Problem 2 In a square loop with a side length L, a large number of balls of negligible radius and each with a charge q are moving at a speed u with a constant separationa between them, as seen from a frame of reference that is fixed with respect to the loop. The balls are arranged on the loop like the beads on a necklace, L being much greater than a, as indicated in the figure 2.1. The no conducting wire forming the loop has a homogeneous charge density per unit length in the in the frame of the loop. Its total charge is equal and opposite to the total charge of the balls in that frame. Consider the situation in which the loop moves with velocityv parallel to its side AB (fig. 2.1) through a homogeneous electric field of strength E which is perpendicular to the loop velocity and makes an angleθ with the plane of the loop.

Taking into account relativistic effects, calculate the following magnitudes in the frame of reference of an observer who sees the loop moving with velocityv: a)The spacing between the balls on each of the side of the loop,a AB ,a BC ,a CD , ya DA. b)The value of the net charge of the loop plus balls on each of the side of the loop:Q AB, QBC, QCD y,Q DA c)The modulusM of the electrically produced torque tending to rotate the system of the loop and the balls. d)The energyW due to the interaction of the system, consisting of the loop and the balls with the electric field. All the answers should be given in terms of quantities specified in the problem. Note. The electric charge of an isolated object is independent of the frame of reference in which the measurements takes place. Any electromagnetic radiation effects should be ignored.

Some formulae of special relativity

Consider a reference frameS ’ moving with velocityV with reference to another reference frameS. The axes of the frames are parallel, and their origins coincide at = 0.V is directed along the positive direction of thex axis.

Relativistic sum of velocities

If a particle is moving with velocityu ’ in thex ’ direction , as measured inS’, the velocity of the particle measured in S is given by: ′ + Vu u = ′Vu 1+ c 2 Relativistic Contraction

’ If an object at rest in frameS has lengthL 0 in thex-direction, an observer in frameS (moving at velocity V in thex-direction} will measure its length to be:

v 2 L = L− 1 0 c 2

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½� Theoretical Question 3 Part A Neutrino Mass and Neutron Decay

A free neutron of mass mn decays at rest in the laboratory frame of reference into three non-interacting particles: a proton, an electron, and an anti-neutrino. The rest mass of the proton is mp, while the rest mass of the anti-neutrino mv is assumed to be nonzero and much smaller than the rest mass of the electron me. Denote the speed of light in vacuum byc. The measured values of mass are as follows: 2 2 2 mn＝939.56563 MeV/c , mp＝ 938.27231 MeV/c , me＝0.5109907 MeV/c In the following, all energies and velocities are referred to the laboratory frame. LetE be the total energy of the electron coming out of the decay.

(a) Find the maximum possible valueE max ofE and the speed v m of the anti-neutrino

whenE =E max. Both answers must be expressed in terms of the rest masses of the 2 particles and the speed of light. Given thatm v < 7.3 eV/c , computeE max and the

ratio vm /c to 3 significant digits. [4.0 points] Theoretical Problem 1 RELATIVISTIC PARTICLE In the theory of special relativity the relation between energyE and momentumP or a free particle with rest massm 0 is

22 242 E 0  mccmcp

When such a particle is subject to a conservative force, the total energy of the

22 42 particle, which is the sum of  0 cmcp and the potential energy, is conserved. If the energy of the particle is very high, the rest energy of the particle can be ignored (such a particle is called an ultra relativistic particle). 1) consider the one dimensional motion of a very high energy particle (in which rest energy can be neglected) subject to an attractive central force of constant magnitudef. Suppose the particle is located at the centre of force with initial

momentump 0 at timet=0. Describe the motion of the particle by separately plotting, for at least one period of the motion:x against timet, and momentum p against space coordinatex. Specify the coordinates of the “turning points” in

terms of given parametersp 0 andf. Indicate, with arrows, the direction of the progress of the mothon in the (p,x) diagram. There may be short intervals of time during which the particle is not ultrarelativistic. However, these should be neglected. Use Answer Sheet 1. 2) A meson is a particle made up of two quarks. The rest massM of the meson is equal to the total energy of the two-quark system divided byc 2. Consider a one--dimensional model for a meson at rest, in which the two quarks are assumed to move along thex-axis and attract each other with a force of constant magnitude f It is assumed they can pass through each other freely. For analysis of the high energy motion of the quarks the rest mass of the quarks can be neglected. At time t=0 the two quarks are both atx=0. Show separately the motion of the two quarks graphically by a (x,t) diagram and a (p,x) diagram, specify the coordinates of the “turning points” in terms ofM andf, indicate the direction of the process in your (p,x) diagram, and determine the maximum distance between the two quarks. Use Answer Sheet 2. 3) The reference frame used in part 2 will be referred to as frameS, the Lab frame, referred to asS, moves in the negativex-direction with a constant velocity v=0.6c. the coordinates in the two reference frames are so chosen that the point

2 x=0 inS coincides with the point x 0in S  at time tt  0. Plot the motion of the two quarks graphically in a ( x , t) diagram. Specify the coordinates of the turning points in terms ofM,f andc, and determine the maximum distance between the two quarks observed in Lab frame S. Use Answer Sheet 3. The coordinates of particle observed in reference framesS and S  are related by the Lorentz transformation

    ctxx )(   x  tt   )(  c

where  /cv , 1/1   2 andv is the velocity of frameS moving relative

to the frame S  . 4) For a meson with rest energy Mc2=140 MeV and velocity 0.60c relative to the Lab frame S  , determine its energy E in the Lab Frame S  .

ANSWER SHEET 1 ANSWER SHEET 2 1) 2) x x , x 1 2

t t O

p p p 1 2

x x x 1 2 O O O

Quark1 Quark2 The maximum distance between the two quarks isd=

3 Theoretical Question 1 Gravitational Red Shift and the Measurement of Stellar Mass

(a) (3 marks) A photon of frequencyf possesses an effective inertial massm determined by its energy. Assume that it has a gravitational mass equal to this inertial mass. Accordingly, a photon emitted at the surface of a star will lose energy when it escapes from the star’s gravitational field. Show that the frequency shift Δf of the photon when it escapes from the surface of the star to infinity is given by

Δf � −GM f Rc2 for Δf�f where: •G = gravitational constant •R = radius of the star •c = velocity of light •M = mass of the star. Thus, the red-shift of a known spectral line measured a long way from the star can be used to measure the ratio M/R. Knowledge ofR will allow the mass of the star to be determined. (b) (12 marks) An unmanned spacecraft is launched in an experiment to measure both the massM and radius R of a star in our galaxy. Photons are emitted from He+ ions on the surface of the star. These photons can be monitored through resonant absorption by He+ ions contained in a test chamber in the spacecraft. Resonant absorption accors only if the He+ ions are given a velocity towards the star to allow exactly for the red shifts. As the spacecraft approaches the star radially, the velocity relative to the star (v=βc) of the He + ions in the test chamber at absorption resonance is measured as a function of the distanced from the (nearest) surface of the star. The experimental data are displayed in the accompanying table. Fully utilize the data to determine graphically the massM and radiusR of the star. There is no need to estimate the uncertainties in your answer. Data for Resonance Condition Velocity parameter β= v/c(×10 −5) 3.352 3.279 3.195 3.077 2.955 Distance from surface of star d(×10 8m) 38.90 19.98 13.32 8.99 6.67

(c) (5 marks) In order to determineR andM in such an experiment, it is usual to consider the frequency correction due to the recoil of the emitting atom. [Thermal motion causes emission lines to be broadened without displacing emission maxima, and we may therefore assume that all thermal effects have been taken into account.] (i) (4 marks) Assume that the atom decays at rest, producing a photon and a recoiling atom. Obtain the relativistic expression for the energy hf of a photon emitted in terms of ΔE (the difference in rest energy between the two atomic levels) and the initial rest massm 0 of the atom. (ii) (1 mark) Δf Hence make a numerical estimate of the relativistic frequency shift for the case of � f �recoil He+ ions. Your answer should turn out to be much smaller than the gravitational red shift obtained in part (b).

Data: Velocity of lightc =3.0× 10 8ms−1 2 Rest energy of Hem 0c = 4× 938(MeV) 13.6Z2 Bohr energyE n =− (eV) n2 Gravitational constantG =6.7× 10 −11Nm2kg−2 Problem No 9 | IPhO Estonia 2012 http://www.ipho2012.ee/physicscup/problem-no-9/

Home First Circular Second Circular Third Circular Registration Problem No 9 Program Leaders and observers Electron, initially at rest, is accelerated with a voltage , where is the electron's rest mass, Students – the elementary charge, – the speed of light, and – a dimensionless number. The electron hits a Opening Ceremony motionless positron and annihilates creating two photons. The direction of one emitted photon defines the direction of the other one. Find the smallest possible value of the angle between the directions of the Lecture: Sir Harold Kroto two emitted photons (express it in terms of and provide a numrical value for ). Closing Ceremony Problems Solutions Experimental apparatus Results Gold medalists Silver medalists Bronze medalists Honorable Mentioned Special prizes Statistics Steering Committee Academic Committee Organizing Committee IPhO Homepage History Statutes Syllabus Newsletter Press Releases Invitation movie Short version Long version Sponsors Travel and accommodation Estonia Tallinn Tartu IPhO 2012 Eestis Feedback Competition “Physics Cup – IPhO2012” Formula sheet Frequently asked questions Physics solver’s mosaic 1. Minimum or maximum? 2. Fast or slow? 3. Force diagrams or generalized coordinates? 4. Are Trojans stable? 5. Images or roulette? Problem 0 Solution Problem No 1 Results after Problem 1 Solution Problem No 10 Intermediate conclusion Solution Problem No 2 Results after Problem 2 Solution Problem No 3 Results after Problem 3 Solution

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