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Notes: i) the inverse transformation is obtained by replacing v with –v.ii) for v c, these reduce to the Galilean transformation.c.4-vectorsSuppose that when O O’, a flash of light is emitted from the origin O. In the S frame, thedistance the light wave front travels in time t is r 2 x 2 y 2 z 2 c 2 t 2 . Measured in the S’frame, it’s r ′ 2 x ′ 2 y ′ 2 z ′ 2 c 2 t ′ 2 . Subtract the second expression from the first andcollect the S frame on one side of the equal sign, the S’ frame on the other side.r 2 r ′ 2 c 2t 2 c 2t ′2r 2 c 2t 2 r ′ 2 c 2t ′2There is this quantity, a generalized displacement (call it s) which is the same in the two inertialreference frames.s 2 s′2We see that the quantity (ict) “acts like” a component of displacement along a fourth axis. Theinterval between any two events in space-time is s 2 x 2 y 2 z 2 c 2 t 2 . The intervalis invariant under the Lorentz Transformation. That is, as measured in any two inertial frames, s 2 s′2 . This is an extension of the invariance of lengths under a rotation of the coordinateaxes.d.Transformation of velocitiesSince displacements and time intervals are transformed, obviously relative velocities won’t addsimply, either.In the S’ frame an object moves with constant velocity along the x axis; u ′x dx ′. Transform todt ′dx vu vγ (dx vdt )and similarly for the y and zthe S frame; u ′x dt xv 1 v dx 1 v uγ dt 2 dx xc 2 dtc2c components. While dy & dz are not contracted, dt is still dilated.example:u A 0.5c and u B 0.8c , both as measured in the S frame. The S’ frame rides along with spaceship B. Therefore, v u B .12

uA v 0.5c ( 0.8c) c 0.8cv21 2 uA1 2 0.5cccBe careful with the directions of the velocities.u ′A Note that when u c and v c , thenvu 0 and u ′ u v . On the other hand, if u c ,c2 v c 1 c vc c.then u ′ cvv1 21 cc13

D.Energy and MomentumWe require that all the “Laws” of Physics be the same in all inertial reference frames. Werequire further that when v c, we recover the familiar Newtonian forms of the “Laws.” Thislatter requirement is called a Correspondence Principle. What are those “Laws”?1.Conservation of MomentumWe define a relativistic momentum so that the two conditions above are satisfied. p γmuThis m is the rest mass—the mass measured by an observer at rest with respect to the object.This quantity should be the same in all inertial reference frames. With this definition, pinitial p final in all inertial reference frames.2.Relativistic Energya.Work-energy theorem (one dimensional)The work done by a force on an object changes its kinetic energy, thusx2 K W12 Fdx .x1x2dp dt dx K x1t2 K dp dxdtdt dtt1 K udpIntegrate by parts.u2u2 pdu K up u 1 K upRecall that udu u2muu2u1 u2u1m 2 u1duu21 2cdu 2.2 K upLook up the formu1 dxa bxin a math tables book.14du 21 u2c2

uu K up u212 u2 2 1 2 m c 1 2 2 c u1u 2 mu 2u2 2 K mc 1 2 2c u 1 2 uc1u 2 222 2 mu mc mu mc K 2uu2 1 21 2 ucc 1Now, if we started from rest, then u1 0 and u2 u and K mc 21 2uc2 mc 2 . Therefore, wedefine the relativistic kinetic energy to beK mc 22 mc 2 .uc2The quantity mc2 is called the rest energy, because it’s independent of u. The total relativisticenergy is E K mc2 V, where V is the potential energy, if any. If V 0, thenE K mc 2 γmc 2 .1 b.Energy-momentum relationTake a look at the quantity (V 0) 2 2m cm c u22 42 42 m u m c cE m c u2u2u2 1 21 21 2ccc 22 42 2E m c c p2 42 2 2E 2 c 2 p 2 m2c 4For photons, m 0 and E pc.15 .

c.Units of mass-energyIt is convenient to express energy in units of electron-volts (eV). An electron-volt is the energygained by an electron upon being accelerated through a one Volt potential difference. Thus 1 eV 1.60x10-19Joules. The rest energy of an electron is()2mc 2 9.11x10 31 kg 3 x10 8 m / sec 8.20 x10 14 J 0.511x10 6 eV 0.511MeV .Often, mass is expressed in terms of MeV/c2 so that the electron mass is 0.511MeV/c2.Sometimes, the c2 is dropped, but it’s understood to still be there. Similarly, momentum isexpressed in terms of MeV/c, since pc units of MeV.3.Relativistic Mechanicsa.ForceWe want the “Laws” of Mechanics to be invariant under the Lorentz Transformation. Also, wewant to recover the classical result when u c. So, we define the relativistic force componentmu xdp.to be Fx x , where p x dtu21 2cLet’s say the motion and force are entirely along the x-direction. dumd 1 d mu F mu2 2 dt2 dt dt uu 1 u 1 1 c2c 2 c 2 F m1 u2c22du 1 u mu 1 2 dt 2 c 32 2u du 2 c dt 1 3 du u 2 2 u 2 u 2 2 1 F m 2 1 2 dt c 2 c c Solve for the acceleration. 1F m u2 1 2c 32du F u 2 1 dt m c 2 16dudt32

du 0 , no matter how large the applied force. At the otherdtdu F .extreme, when u c,dt mThe result is, that as u c ,b.Collisions—conservation of momentumConsider the collision of two billiard balls. They have equal masses, m. Let’s say that one ballis initially at rest while the second ball has momentum po and energy Eo before the collision.After the collision, both balls have the same energy, E, and mass, m. It’s an elastic collision.Momentum and energy are conserved.In the x direction, po 2 p cosθ . Substitute for po and p using E2 p2c2 m2c4.12E o2 m 2 c 4 E 2 m 2 c 4 cosθccConservation of energy allows us to eliminate E, since it was given that E o mc 2 2 E . Keepin mind that Eo is the relativistic total energy of the second ball, while mc2 is the rest energy ofthe first (target) ball. At the same time, we solve for cosθ , the cosine of the scattering angle.cosθ Eo2 m 2 c 4(Eo mc 2)2 4m 2 c 4cosθ (E mc )(E mc )(E 3mc )(E mc )ooEo mc 2Eo 3mc 2In the classical limit, Eo mc 2 and therefore2mc 21cos θ θ 45 o . But, as Eo 224mc2mc , cosθ 1 θ 0 o !c.Decay of a high-energy particleAn unidentified high-energy particle is observed todecay into two pions ( π mesons), as shown.1722oo22

Knowing the momenta and masses of the decay products, we determine the mass of the incidentparticle, hoping to identify it.MeVMeV, p 2 323, m1c 2 m2 c 2 mc 2 139.6Mev .p1 910ccThe energy and momenta are conserved. The total energy isE o E1 E 2 p12 c 2 m 2 c 4 p 22 c 2 m 2 c 4E o 921MeV 352MeV 1273MeVThe quickest way to obtain the magnitude of the incident momentum is to use the law of cosines:p o2 c 2 p12 c 2 p 22 c 2 2 p1 p 2 c 2 cos θ 1034229 MeV 2po c 1017 MeVNow that we have the total energy and the kinetic energy, the mass is obtained fromE o2 po2 c 2 mo2 c 4mo c 2 E o2 po2 c 2 765MeVEvidently, the incident particle was a ρ meson. What was its speed before it decayed? Well,the total energy is also Eo2 mo2c 4, so solve that for u.u21 2cmo2 c 4u 1 0.8cEo2d.Mass-energy equivalenceWhen we speak of the total energy being conserved that includes the total rest energy. Forinstance, consider the decay of a neutron that is initially at rest.n p e νThe neutron decays into a proton, an electron and an anti-neutrino. The three product particlesare observed to have total kinetic energy of K 0.781 MeV. The initial energy is just the restenergy of the neutron, Ei 939.57 MeV. The total final energy isE f m p c 2 me c 2 K 938.28MeV 0.511MeV 0.781MeV 939.57 MeVNotes: i) The rest energy of the anti-neutrino is too small to bother with.ii) Keep in mind the rounding of numbers and significant digits when substitutingnumerical values into the formulae.iii) Notice that mn m p me . A portion of the neutron’s rest energy has beenconverted into kinetic energy.18

E.A Hint of General Relativity1.EquivalenceIn Special Relativity it is asserted that all inertial reference frames are equivalent—the “laws” ofphysics are the same in all inertial reference frames. No experiment done in one frame candetect its uniform motion relative to another frame. Can the same be said for reference framesthat have a relative acceleration?a.ElevatorRecall the past discussion of a person standing in anelevator. If the elevator moves perfectly smoothly andthere are no floor indicator lights, then the person insidewill have no perception of the elevator’s motion, exceptfor feeling perhaps the elevator floor pressing upwardon his or her feet. [Keep in mind: the person gets noinformation from any source outside the reference frameof the elevator.] Contrast this situation with that ofanother person standing in a similar elevator, but thiselevator is simply resting level on the Earth’s surface.The person in this elevator also feels the floor pressingupward on his or her feet, also has no perception of theelevator’s motion. We, as omniscient externalobservers, know that this second elevator is resting on the surface of a planet, and that what theperson inside is experiencing is the gravitational force exerted by that planet. The point is thatthere is no experiment that either of the persons inside the elevators could perform that woulddistinguish between the two situations. Pendula would swing back and forth just the same;projectiles would follow the same kinds of arcs, etc.b.Light and gravityImagine ourselves as observers far from any source of gravitational force. Nearby, we observe a closed “elevator” which is accelerating, relative to us, at a constant rate, ao . A person standinginside the “elevator” sends a series of light pulsestoward one wall—he or she and we see the light pulsesdropping toward the floor as they approach the wall.The light follows a curved path inside the elevator.The Postulate of General Relativity asserts that the“laws” of physics have the same form for observers inany frame of reference, regardless of its accelerationrelative to another frame. We have seen that anaccelerated frame is equivalent to one in a gravitationalfield. It follows that the force of gravity must affect abeam of light just as it affects the motion of a massiveprojectile. Indeed, experiment has shown that it does.But, light has no mass.19

2.CurvatureClassically, we would say that a mass, such as a planet, exerts a gravitational force on anothermass, such as a moon or a person. However, a person in an “elevator” cannot determine whetherhis or her “elevator” is in the gravitational field of a planet or is being accelerated at a constantrate by, say rocket motors. If the “elevator” is in a gravitational field, we can nonethelessmathematically transform the “laws” of physics into versions of the same mathematical form thatdo not include gravity yet which make equally accurate predictions of the motions of particlesand of light beams.What Einstein did was to formulate such a version of the “laws” of motion. Objects and lightbeams move always in straight lines, but in a curved space-time. Empty space-time is flat, butthe presence of mass at any location curves space-time to a degree proportional to the amount ofmass that is present.Predictions of General Relativity:Precession of orbits—MercuryGravitational redshift—time runs slower in intense gravitational fieldGravitational lensing—light paths curvedGravitational waves—ripples in space time(?); slowing binary neutron stars20

II.Quantum TheoryA.Black Body Radiation1.Equilibrium Between Matter and Radiationa.Thermal equilibriumImagine a closed oven, maintained at a constant temperature, To. Inside, E-M waves bouncefrom wall to wall, being absorbed and re-emitted over and over. Ultimately, the radiation isrendered homogeneous, isotropic and unpolarized. A thermometer placed in the center of theoven will stabilize at a temperature T To . The radiation is said to be in thermal equilibriumwith the walls of the oven.b.EmissivityFrom thermodynamics (Kirchhoff) the power radiated by a body in thermal equilibrium withradiation is expressedEfJ ( f ,T ) ,Afwhere J is the power radiated per unit area per unit frequency, Ef is the emissivity or intensity perunit frequency of the radiation emitted by the body and , Af is the fractional absorption of thebody for radiation of frequency f. Notice that the substance of which the body is made is notimportant. For a black body, A f 1.0 for all f. The observed spectrum of radiation emitted by ablack body looks qualitatively like this:c.Model for a black bodyOne physical model for an ideal black body is a small opening in thewall of a heated cavity. Because the opening is small, a light rayentering through the opening is very unlikely to bounce back outagain. Conversely, any light ray that exits through the opening willhave reached equilibrium with the interior walls, having bounced offthe walls many times. The black body is not the oven as a whole,but the opening in the oven wall.21

2.Stefan-Boltzmann “Law”a.EmissivityThe total emissivity is obtained by integrating Efdf. E E f df0This quantity was found experimentally by J. Stefan to be proportional to T4.E σT 4Subsequently, Boltzmann derived this result from Maxwell’s equations. The proportionalityWconstant is called the Stefan-Boltzmann constant, σ 5.67 x10 8 2 4 or Wm-2K-4. For nonm K4black bodies, E aσT , where a 1.b.Wien’s displacement “law”Experimentally, λmax T 2.898 x10 3 m K , or3.Rayleigh-Jeans “Law”a.Wien’s exponential “law”cf max T 2.898 x10 3 m K .4J ( f ,T ) ,cwhere u is the energy per unit volume per unit frequency. As derived from thermodynamics andMaxwell’s equations,u( f , T ) β fTu ( f , T ) Af 3 ewhere A and β are constants. When tested by experiment, this expression fails at longwavelengths. But, Wien’s exponential “law” fits well near the peak, at f max .b.Rayleigh’s approachRayliegh proposed that the energy density be expressed as the product of the number of standingwave modes in the cavity and the average energy of each mode. Let N(f)df be the number ofmodes between f and f df.udf N ( f ) E df22

We need to obtain expressions for N(f) and for E . Picture standing waves in a cavity. In onedimension:We imagine the whole volume of the cavity occupied withstanding E-M waves, of many different frequencies. The radiationis in equilibrium with the walls of the cavity, at temperature T.Classically, the probability that there will be a mode of energy, E,in the cavity is given by the Boltzmann distribution. EP ( E ) e k BTThe average energy per mode is therefore E Ee0 e E Ek BTk BTdE(k B T )2 dEk BT k BT0As for N(f), the density of states, consider a cubical box of side L.Inside the box, the x-component of the E-field satisfies the wave equation 2 E x k 2 E x 0 . Ifwe assume that E x u ( x)v( y ) w( z ) , then we get three separated equations. The one for u(x) isd 2u k x2 u 0 ,2dxwhich has the solution u ( x) u o sin k x x . Similarly, v( y ) vo sin k y y and w( z ) wo sin k z z .The quantity k2 is like the square of a radius in k-space: k 2 k x2 k y2 k z2 . Since we must havestanding waves in the box, with the electric field vanishing at the walls, k x n yπn xπ, ky LLn zπ, where the nx, ny and nz are positive integers. In other words, only discrete pointsLin k-space designate the allowed energy modes in the cubical cavity. So, we count the number ofk-points lying in a spherical shell of radius k and thickness dk.and k z 23

3 π We see that each point occupies a volume so the number of points in the shell is L 14πk 2 dkVk 2 dk ,N (k )dk 832π 2 π L 3where V L . Finally, there are two perpendicular polarizations for each mode, so the number ofmodes per unit volume (the density of states) isN (k )k 2 dkdk 2 .Vπ2N( f )8πf dfkc, since f .In terms of frequency,df 3V2πcAt last the energy density is8πf 2k B Tdf .c3Alter

Physics. Let’s assume that they are not different, and work out the consequences. This is what Einstein did. He postulated that ‘All the “Laws” of Physics are the same in all inertial reference frames.’ b. Second Postulate The second postulate follows from the f