Modern Physics Fall/Winter 1900, Max Planck’s paper « Ueber das Gesetz der Energieverteilung im Normalspectrum », Annalen der Physik IV, 553 (1901) – peak In 1920s/30s Two major parts: modern relativity, first 4 – 6 lectures Quantum mechanics and its applicatlons, rest of the course -also main content of Phys 312 to follow next quarter What is Physics all about? concepts and their connection, i. e. mathematically formulated equations/laws, concepts and laws are derived from interplay between theory and experlment, this theories get better o some « fundamental » older than Physics an (Kanes « a priori » conc orles survive and org nc•2F. ipa and time are much nse knowledge nherited genetically but: Werner Heisenberg (in Physics and Philosophy) « Any concepts or words which we have formed in the past through the interplay between the world and ourselves are not really sharply defined With respect to their meanlng; that is to Say, we do not know exactly how far they will help us in finding our way ln the world. This is true even of the simplest and most general concepts like « existence » and « space and time »… The concepts may, however, be sharply defined With regard to th Swipe to nex: page their connections.
This is actually the fact when the concepts become a part of a system of axioms and definition which can be expressed consistently by a mathematical scheme. Such a group of connected concepts may be applicable to a Wide fjeld of experience and Will help us to find our way in this field. So modern physics will be in large parts contrary to our intuition because it deals With the very fast, i. e. speeds comparable to the speed of light and the very Small, atoms, molecules, elemental particles.
Our (possibly inherited) lack of appreciation that the world of the very fast and the world of the very Small may well be very ifferent from the world we are used to makes modern physics difficult to comprehend, but Heisenberg showed the way, see above, we have to stick to the mathematical schemes that connect concepts and have to redefine well known concepts, such as space, time, causality, to fit these schemes Relativistic Mechanics – Special Relativity Galileo, Newton: Inertial reference frame: Newton’s (first) law: a body continues to be at rest or continues moving With constant velocity f there is no net force acting on it v = 0 or constant if Ea = = O becauseEF = ma and v = ds m = inertia, that is where the reference frame gets its name from
Galileo, Newton: all laws of mechanics are the same in inertial reference frames, all reference frames are equally valid, there is no prefer same in inertial reference frames, all reference frames are equally valid, there is no preferred inertial reference frame – classical concept of relatlvity Classic relations between an event as observed in two different (v = O, v’ > O) inertial reference frames are related by Galilean transformation dt Galilean transformations two different sets of coordinates: space coordinates (x,y,z), time coordinate t at rest; in motion x = r + vty=y’ vent has one set of coordinates in one system and another set of coordinates in another system (back transformation are the same except for sign of v) leads to vector addition law of velocities, if event moves in unprimed frame With velocity u, v and u add up ux=x2 -xl v ü -tl t’2 ux =ux’+v uy = uy’ uz = uz Nota Bene: space and time coordinates do not mix, 4 importance of these equation is that they ensure the physical laws that are invariant With respect to these equations are valid everpnhere and at all times (if we use our common sense ideas of space and time) Result of Galileian relativity: there is no mechanical experiment hat can detect absolute motion ou can eat your dinner in an air plane (when it is not accel is movine rather fast With PAGF3CFq accelerating) which is moving rather fast With respect to the earth – ust as well as on your dinner table at home – which is moving even faster With respect to the sun ln 18705 -1904 some new idea of how to measure absolute motion czi- 1 PE -7 -1 -12 2-1 -2 0 0 -rmA C N m prediction of Maxwell’s 1 860 set of equations, permeability, EO permittivity of free space (i. e. vacuum) c 2. 9792458 108 ms constant (and now exact per definition) according to what/whom has c this value ??? Maxwell’s own answer: luminiferous ether (sornething quite strange, present everywhere even in the nearly absolute vacuum of free space, but allows planets and other objects to move through it freely, , and which is in absolute rest) Not only c = constant in vacuum but the other laws by Maxwell’s do not Obey a Galilean transformation, so at last there seemed to be a way of detecting motion, if you do an electromagnetic experiment such as measuring the speed of light in an airplane or on earth, you should get the relative speed With respect to the ether which is supposed to be at absolute rest.
Recall sound: travels in air and any kind of body, speeds: 243 ms in air at 293 K, 249 ms in air at 303 K and normal pressure, 3800 ms in concrete at 293 K, needs actually a medium to propagate, if you have a potential source of sound in vacuum – you can’t hear it as the wave can’t propagate So upwind so source of sound in vacuum So upwind sound travels faster – as it is carried along With the Wlnd itself, downwind sound travel slower slnce the medium (alr) travels in the opposite direction — Galilean transformations seem to apply Conundrum: Since ether seemed to be so special – it should deflne a very articular frame of reference, i. e. the only one in which Maxwell’s equatians are correct, in all ather frames of reference, i. e. our earth, there should be deviations from Maxwells laws, on the other hand, these lavvs work quite well, how can this be? 6 7 Michelson-Morley Experiment, 1887-1904 Designed to detect the ether and earth’s relative motion With respect to the ether by detecting Small changes in the speed of light, i. e. deviations from Maxwell’s « c constant laW’ by interferometry Light source A, semitransparent mirror beam splitter B, two mirrors C and E all mounted on a rigid base
Mirrors C and E are placed at equal distances L from beam splitter, so that the two resultlng beams have (apparently the) same path Iength (2L) to go in perpendicular directions, reach the mirrors C and D and get reflected back to the beam splitter where they are joined together again If time taken for the light to go from Bto E and back is the same as the time from B to C and back, emerging beams D and F will be in phase and reinforce each other It these two and back, emerging beams D and F will be in phase and reinforce each other It these two times differ slightly, beams D and will produce interference pattern If apparatus is at rest With respect to the ether, times should be exactly equal because the Iengths the light must travel are exactly equal – if it is moving towards the right With a velocity u, there should be a difference in the times, resulting in an interference pattern. Why should that be? Time to go from B to E and back = tl return time Eto B t2 (tl t2 because of movement of apparatus to right) it the apparatus moves, while light is on its way to from B to E, the mirror together With the whole apparatus moves away, this distance is u tl, i. e. he light must travel With speed c the Iength L + utl in order to reach the mirror tl =L+ut1 tl which means that velocity of the Ilgh With respect to apparatus is for return travel velocity of light With respect to apparatus must therefore be c u, because the beam splitter B and the light beam are moving in opposite directions t2 — L/ (c u) and L ut2 total time for Bto E and back is tl +t2 = 2Lc/(c -u c(1-u2 22 2L now the other path: B – C and back, again assumption is apparatus is moving to the right (because we want to measure this movement by an anticipated shift in the interference pattern) during time t3 mirror C Will move to the right an anticipated shift in the interference pattern) uring time t3 mirror C Will move to the right by a distance ut3 light has therefore to travel along the hypotenuse of right triangle BCIQB’ (Ct3)2 – Q + (Lit3)2 L2 – c2t32 – u2t32 = (c2 u2)t32 LS0t3=UC2 1-u2 c2 triangle is symmetric, so time it takes for the ight to return to B is 2 t3 2L2t3ZC1-u2 tl+t2ZC(1-u2) = y (Lorentz factor) >1 difference is just factor 1 1- u2 denominators represent modifications in time caused by motion of the apparatus, they are not the same so we should see an interference pattern and from this we could calculate u veloclty of earth With respect to ether- the whole point of the experiment minor technical point, we can’t make the Iengths L exactly equal, we can compensate for this in the interference pattern, then we can turn apparatus around by degrees and should see a shift of interference pattern between two sets of settings 1) arbitrary orientation and 2) 900 rotated With respect to 1) But no shift in interference pattern was ever observed, we do know u 21 12L2t3 -c 1- +t2- 1- u2 9 c2 c2c2 so it seems as if Iength of the path is B to E and back is contracted by a factor y (Lorentz and Fitzgerald) 2. Result, the speed of ligh th) is in all directions light (in air at earth) is in all directions equal regardless of any elative movement of the earth With respect to the ether that should result in an « ether wind » analogous to the wind that affects the speed of sound apparent 0 velocity of earth and constant velocity of light results can both be explained by Lorentz Transformations (1904) , moving frame ‘ at t = O both frames coincide x = 1 (x’+vt’) [m] = [m + ms] 1—v2 s 1-v2 regardless of the movement of the observer ! reverse transformations (t’+ vx’) c2 m2 s2 srn2 10 in which Maxwell’s law are invariant, i. e. ave the same form x’- 1 (x-vt) [m]=[m+ ms ] 1-v2 s e- 1 1-v2 (t — vx) c2 m2s2 sm2 consequences (some of which not fully realized by Lorentz): . space and time coordinates mix, i. e. they are the same sort of thing, for v 01 v2 1- , (lets call c2 c2 h+l min+25 sec or 1 h+3 min+5 s ? say vo 100 km h-l 3600 s At’=1h+5sor1 h +15 us or 1h+18 ps ? if al (lets call 4. relativistic addition of velocities Lorentzfactor / :At’ 1 1-a y(At’+vAx’ ) = 1+(v ) = ux’•v Ax y(Lx’+vAt’) (Ax’At’)+v ux’+v 14 c2 c2 At’1+C2 (AY’ uy=Ay = Ay’ = Ati) = y y 2} y •{I+(V 2)ŒX’ » Y c u – AzAz’ c At (Oz’Ot’) c2 uz’ z At = y ‘{Ot’+vx’ reverse transformations ux’•v Ax (AxAt)-v ux -v ux -v ux’ = At uy=Ay = ay = at) = y Aty •Ut-vx 2} Y •{I-(V 2)Œx y •(l-ux (Ay u
