Lecture notes on modern physics

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Modern Physics Notes © J Kiefer 2006 Table of Contents TABLE OF CONTENTS..............................................................................................................1 I. RELATIVITY ....................................................................................................................... 2 A. Frames of Reference.......................................................................................................................................... 2 B. Special Relativity ............................................................................................................................................... 5 C. Consequences of the Principle of Special Relativity ....................................................................................... 8 D. Energy and Momentum .................................................................................................................................. 14 E. A Hint of General Relativity........................................................................................................................... 19 II. QUANTUM THEORY ................................................................................................... 21 A. Black Body Radiation...................................................................................................................................... 21 B. Photons ............................................................................................................................................................. 27 C. Matter Waves................................................................................................................................................... 30 D. Atoms................................................................................................................................................................ 37 III. QUANTUM MECHANICS & ATOMIC STRUCTURE (ABBREVIATED)........... 45 A. Schrödinger Wave Equation—One Dimensional ......................................................................................... 45 B. One-Dimensional Potentials............................................................................................................................ 47 D. The Hydrogen Atom........................................................................................................................................ 52 E. Multi-electron Atoms ...................................................................................................................................... 59 1I. Relativity A. Frames of Reference Physical systems are always observed from some point of view. That is, the displacement, velocity, and acceleration of a particle are measured relative to some selected origin and coordinate axes. If a different origin and/or set of axes is used, then different numerical values    are obtained for r , v , and a , even though the physical event is the same. An event is a physical phenomenon which occurs at a specified point in space and time. 1. Inertial Frames of Reference a. Definition An inertial frame is one in which Newton’s “Laws” of Motion are valid. Moreover, any frame moving with constant velocity with respect to an inertial frame is also an inertial frame of   reference. While r and v would have different numerical values as measured in the two   frames, F = ma in both frames. b. Newtonian relativity Quote: The Laws of Mechanics are the same in all inertial reference frames. What does “the same” mean? It means that the equations and formulae have identical forms, while the numerical values of the variables may differ between two inertial frames. c. Fundamental frame It follows that there is no preferred frame of reference—none is more fundamental than another. 2. Transformations Between Inertial Frames a. Two inertial frames Consider two reference frames—one attached to a cart which rolls along the ground. Observers on the ground and on the cart observe the motion of an object of mass m.  The S’-frame is moving with velocity v relative to the S-frame. As observed in the two frames: 2′ ∆ x ′ In S’ we’d measure ∆ t’, ∆ x’, and u = . x ′ ∆ t ∆ x In S we’d measure ∆ t, ∆ x, and u = . x ∆ t b. Galilean transformation ′ Implicitly, we assume that ∆ t = ∆ t . Also, we assume that the origins coincide at t = 0. Then ′ ′ x = x + v ∆ t x ′ ′ y = y + v ∆ t y ′ ′ z = z + v ∆ t z ′ ∆ t =∆ t The corresponding velocity transformations are ′ dx dx ′ u = = + v = u + v x x x x dt dt ′ dy dy ′ u = = + v = u + v y y y y dt dt ′ dz dz ′ u = = + v = u + v z z z z dt dt For acceleration du dv ′ x x a = = a + x x dt dt du dv y ′ y a = = a + y y dt dt du dv ′ z z a = = a + z z dt dt ′ ′ ′ Note that for two inertial frames, the a = a , a = a , and a = a . x x y y z z 3Example S-frame     du dp F = ma = m = , if m is constant. dt dt S’-frame         ′ dp      du dv du   ′ ′ ′ F = ma = , where p′= mu′ . But u = u− v , so F′= m − = m = F . That is,   dt dt dt dt     a = a′ , as they must for 2 inertial reference frames. nd Notice the technique. Write the 2 “Law” in the S’-frame, then transform the position and velocity vectors to the S-frame. 4B. Special Relativity 1. Michelson-Morley a. Wave speeds th Midway through the 19 century, it was established that light is an electromagnetic (E-M) wave. 8 Maxwell showed that these waves propagate through the vacuum with a speed c≈ 3x10 m/sec. Now, wave motion was well understood, so it was expected that light waves would behave exactly as sound waves do. Particularly the measured wave speed was expected to depend on the frame of reference.   ′ In the S-frame, the speed of sound is u ; in the S’-frame the speed is u . The source and the    ′ medium are at rest in the S-frame. We find (measure) that u = u + v , in conformity with Newtonian or Galilean relativity. We may identify a “preferred” reference frame, the frame in which the medium is at rest. b. Michelson-Morley th Throughout the latter portion of the 19 century, experiments were performed to identify that preferred reference frame for light waves. The questions were, what is the medium in which light waves travel and in what reference frame is that medium at rest? That hypothetical medium was given the name luminiferous ether (æther). As a medium for wave propagation, the ether must be very stiff, yet offer no apparent resistance to motion of material objects through it. The classic experiment to detect the ether is the Michelson-Morley experiment. It uses interference to show a phase shift between light waves propagating the same distance but in different directions. The whole apparatus (and the Earth) is presumed to be traveling through the  ether with velocity, v . A light beam from the source is split into two beams which reflect from the mirrors and are recombined at the beam splitter— forming an interference pattern which is projected on the screen. Take a look at 5the two light rays as observed in the ether rest frame. The sideward ray: The time required for the light ray to travel from the splitter to the mirror is obtained from 1 − 2 2    v 2 2 2   (ct) =  + (vt) ⇒ t = 1− . 2   c c   Now c v, so use the binomial theorem to simplify 2    1 v   n(n+1) t ≈ 1+ . −n 2  2  () 1− x =1+ nx+ x + c 2 c   2 2   2 1 v   The total time to return to the splitter is twice this: . t = 2t ≈ 1+ 1 2   c 2 c   For the forward light ray, the elapsed time from splitter to mirror to splitter is 2 2       2 v 2 v t = + = 1−  ≈ 1+ . 2 2 2     c− v c+ v c c c c     λ The two light rays recombine at the beam splitter with a phase difference letτ = .: c 2 2 ∆ t c 2 1 v c  v =() t − t = = . 2 1 2 2 τ λ c 2 λ λ c c ∆ t Since ≠ 0 , the two light rays are out of phase even though they have traveled the same τ distance. By measuring ∆ t one could evaluate v .  However, no such phase difference was/is observed So, there is no ether, no v with respect to such an ether. This null result is obtained no matter which way the apparatus is turned. The conclusion must be that either the “Laws” of electromagnetism do not obey a Newtonian relativity principle or that there is no universal, preferred, rest frame for the propagation of light waves. c. Expedients to explain the null result length contraction—movement through the ether causes the lengths of objects to be shortened in the direction of motion. ether-drag theory—ether is dragged along with the Earth, so that near the Earth’s surface the ether is at rest relative to the Earth. 6Ultimately, the expedients were rejected as being too ad hoc; it’s simpler to say there is no ether. This still implies that the “Laws” of electromagnetism behave differently under a transformation from one reference frame to another than do the “Laws” of mechanics. 2. Postulates of Special Relativity a. Principle of Special Relativity It doesn’t seem sensible that one “part” of Physics should be different from another “part” of 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 first. ‘The speed of light in a vacuum is (measured to be) the same in all inertial reference frames.’ When the speed of light is measured in the two reference frames, it is found that c ≠ c′+ v , rather c= c′. Evidently, the Galilean Transformation is not correct, or anyway not exact. In any case, we assume the postulates are true, and work out the consequences. 7C. Consequences of the Principle of Special Relativity 1. Time Dilation a. Events An event may be regarded as a single observation made at a specific location and time. One might say that an event is a point in space-time (x,y,z,t). Two events may be separated by intervals in either space or in time or in both. b. Time intervals Consider a kind of clock: We observe two events: i) the emission of a flash at O’ and ii) the reception of the flash at O’. In this case, ∆ x′=∆ y′=∆ z′= 0. The ′ 2d time interval between the two events is ∆ t′= . c Now let’s view the same two events from the point of view of another frame, S. As shown below, the S’-frame is moving to the right with speed v relative to the S-frame. In the S-frame, ∆ x ≠ 0 . 2d 2 2 2 ′ The elapsed time is ∆ t = , where d = d + . Substitute for c ′ ′ d , d , and  in terms of ∆ t , ∆ t , c, and v. 2 2 2 2 2 2 c ∆ t c ∆ t′ v ∆ t = + 4 4 4 1 2 2   ′ c ∆ t ′  Solve for ∆ t =∆ t = . 2 2   2 c − v   v 1− 2 c example (prob. 1-11 in the text) −8 The lifetime of a pion in its own rest frame is ∆ t′= 2.6x10 sec. Consider a pion moving with speed v= 0.95c in a lab—what will be measured as its lifetime in the lab? 8 −8 −8 ′ ∆ t 2.6x10 sec 2.6x10 sec −8 ∆ t = = = = 8.33x10 sec . 2 2 0.312 v 1− 0.95 1− 2 c The lifetime of a fast-moving particle is measured by noting how far it travels before decaying. −8 In this example = v∆ t = 0.95c⋅8.33x10 sec= 23.7 m. In practice, we measure  and compute ∆ t . c. Proper time The proper time is the time interval measured by an observer for whom the two events occur at ′ ′ ′ the same place, so that ∆ x =∆ y =∆ z = 0. 2. Length Contraction a. “Contraction” Consider an object, such as a meter stick, of length L in its own rest frame, S. A second frame, S’, moves to the right with a speed v relative to S. We observe two events: i) the point A passes the left end of the stick ii) the point A passes the right end of the stick. ′ ′ As measured in the S’ frame, L = v∆ t ′ and ∆ x = 0 . 92 2 ′ ∆ t v v In the S frame, ∆ x = L and ∆ t = L′= v∆ t 1− = L 1− . Therefore, . 2 2 2 c c v 1− 2 c An observer in the S’ frame observes the stick to be shorter (contracted) than does the observer in the S frame. Notice particularly that the stick is at rest in the S frame. The contraction takes place in the direction of the relative motion. Lengths perpendicular to v are not affected. So for instance in the situation discussed above the width and thickness of the meter stick are still measured the same in both reference frames. b. Proper length The proper length of an object is that length measured in the rest frame of the object. 3. Simultaneity a. Space-time Each event has associated with it four numbers: x, y, z coordinates and a “value of time” which we read off a clock located at that spatial location. There is no central universal clock, rather there is a clock at every point in space. b. Synchronization We would like all clocks in a reference frame to display exactly the same reading simultaneously, but can this be arranged? Only by the exchange of signals, which is another way of saying only in terms of intervals. However, as we have seen, intervals are not the same for observers in different inertial reference frames. Therefore, the concept of two events being simultaneous has no absolute meaning. c. Non-simultaneity Two events viewed as simultaneous in one frame will not be seen as occurring simultaneously in another frame. example: a train moving with constant velocity on a straight, smooth track. One observer rides on the train, the other observer stands beside the track. Flashes of light are emitted at the points C and C when the origins (O & O’) of the two frames 1 2 coincide. To the trackside observer at O, the flashes are simultaneous. To the observer on the train, however, the flash emitted at C’ is received before the flash emitted at C’ . Yet both 2 1 observers measure the same speed of light, c. 104. Lorentz Transformation Now we wish to derive the transformation equations for the displacement and velocity of an object—the relativistic version of the Galilean transformation equations. In what follows, we’ll 1 be setting γ = . 2 v 1− 2 c a. Two frames Consider two inertial reference frames, S & S’ and assume that O = O’ at t’ = 0. What is the x-distance from O to the point P, as measured in the S’ frame? In effect, then, we’ll have ∆ t = t and ∆ t′= t′ . ′= x′+ vt′ x In the S frame, = x , so ′= also. Set ‘em equal. γ x = x′+ vt′ γ x =γ() x′+ vt′ ′ x On the other hand, as measured in the S frame, x = vt+ . Set them equal. γ ′ x γ() x′+ vt′ = vt+ γ Solve for t. ′ x vt =γ() x′+ vt′ − γ  v  ′ ′ t =γ t + x   2  c  b. Transformation equations We have, then, for relative motion along the x-axis: v   ′ ′ ′ ′ ′ ′ x =γ() x + vt ; y = y ; z = z ; t =γ t + x   2  c  11Notes: i) the inverse transformation is obtained by replacing v with –v. ii) for v c, these reduce to the Galilean transformation. c. 4-vectors Suppose that when O = O’, a flash of light is emitted from the origin O. In the S frame, the 2 2 2 2 2 2 distance the light wave front travels in time t is r = x + y + z = c t . Measured in the S’ 2 2 2 2 2 2 frame, it’s r′ = x′ + y′ + z′ = c t′ . Subtract the second expression from the first and collect the S frame on one side of the equal sign, the S’ frame on the other side. 2 2 2 2 2 2 r − r′ = c t − c t′ 2 2 2 2 2 2 r − c t = r′ − c t′ There is this quantity, a generalized displacement (call it s) which is the same in the two inertial reference frames. 2 2 ′ s = s We see that the quantity (ict) “acts like” a component of displacement along a fourth axis. The 2 2 2 2 2 2 interval between any two events in space-time is ∆ s =∆ x +∆ y +∆ z − c ∆ t . The interval is invariant under the Lorentz Transformation. That is, as measured in any two inertial frames, 2 2 ′ ∆ s = ∆ s . This is an extension of the invariance of lengths under a rotation of the coordinate axes. d. Transformation of velocities Since displacements and time intervals are transformed, obviously relative velocities won’t add simply, either. ′ dx In the S’ frame an object moves with constant velocity along the x axis; u′ = . Transform to x ′ dt dx − v u − v γ() dx− vdt dt x the S frame; u′ = = = and similarly for the y and z x v dx v v   1− 1− u γ dt− dx   x 2 2 2 dt c c c   components. While dy & dz are not contracted, dt is still dilated. example: u =−0.5c and u =−0.8c , both as measured in the S frame. The S’ frame rides along with A B   spaceship B. Therefore, v = u . B 12u − v − 0.5c− (−0.8c) c A ′ u = = = A v 0.8c 2 1− u 1− 0.5c A 2 2 c c Be careful with the directions of the velocities. vu ′ Note that when u c and v c , then → 0 and u = u− v . On the other hand, if u = c , 2 c v   c − 1  c v − c   ′ then u = = = c . cv v 1− 1− 2 c c 13D. Energy and Momentum We require that all the “Laws” of Physics be the same in all inertial reference frames. We require further that when v c, we recover the familiar Newtonian forms of the “Laws.” This latter requirement is called a Correspondence Principle. What are those “Laws”? 1. Conservation of Momentum We define a relativistic momentum so that the two conditions above are satisfied.   p =γmu This 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,   p = p in all inertial reference frames. initial final 2. Relativistic Energy a. Work-energy theorem (one dimensional) The work done by a force on an object changes its kinetic energy, thus x 2 ∆ K = W = Fdx . 12 ∫ x 1 x 2 dp ∆ K = dx ∫ dt x 1 t 2 dp dx ∆ K = dt ∫ dt dt t 1 ∆ K = udp ∫ Integrate by parts. u 2 u 2 ∆ K = up − pdu ∫ u 1 u 1 u 2 u mu 2 ∆ K = up − du ∫ u 1 2 u u 1 1− 2 c 2 du Recall that udu = . 2 2 m du u 2 ∆ K = up − ∫ u 1 2 2 u 1− 2 c dx Look up the form in a math tables book. ∫ a+ bx 14u 2 2   u 2 1−  2 u m c   2 ∆ K = up − u   1 1 2 −   2 c     u 1 u 2     2 2 mu u  2  ∆ K = + mc 1− 2   2 c u   1− 2   c   u 1 u 2         2 2 2 2 mu + mc − mu mc     ∆ K = =∆     2 2 u u     1− 1− 2 2     c c     u 1 2 mc 2 ∆ K = − mc Now, if we started from rest, then u = 0 and u = u and . Therefore, we 1 2 2 u 1− 2 c define the relativistic kinetic energy to be 2 mc 2 K = − mc . 2 u 1− 2 c 2 The quantity mc is called the rest energy, because it’s independent of u. The total relativistic 2 energy is E = K + mc + V, where V is the potential energy, if any. If V = 0, then 2 2 E = K + mc =γmc . b. Energy-momentum relation Take a look at the quantity (V = 0)     2 4 2 2 2 2 2 m c m c u m u 2 2 4 2 4 2  E − m c = − m c = = c . 2 2 2   u u u 1− 1− 1−  2 2 2 c c  c  2 2 4 2 2 E − m c = c p 2 2 2 2 4 E = c p + m c For photons, m = 0 and E = pc. 15c. Units of mass-energy It is convenient to express energy in units of electron-volts (eV). An electron-volt is the energy gained by an electron upon being accelerated through a one Volt potential difference. Thus 1 eV -19 = 1.60x10 Joules. The rest energy of an electron is 2 2 −31 8 −14 6 mc = 9.11x10 kg() 3x10 m / sec = 8.20x10 J = 0.511x10 eV = 0.511MeV . 2 2 Often, mass is expressed in terms of MeV/c so that the electron mass is 0.511MeV/c . 2 Sometimes, the c is dropped, but it’s understood to still be there. Similarly, momentum is expressed in terms of MeV/c, since pc = units of MeV. 3. Relativistic Mechanics a. Force We want the “Laws” of Mechanics to be invariant under the Lorentz Transformation. Also, we want to recover the classical result when u c. So, we define the relativistic force component dp mu x x to be F = , where p = . x x 2 dt u 1− 2 c Let’s say the motion and force are entirely along the x-direction.         d mu m du d 1     F = = + mu     2 2 2 dt dt dt u u u     1− 1− 1− 2 2 2     c c c     3 − 2 2   m du 1 u 2u du       F = + mu − 1− −      2  2 2 dt 2 dt c c     u   1− 2 c 1 3 − −   2 2 2 2 2     du u u u       F = m 1− + 1−  2  2  2    dt c c c       3 2     1 du   F = m 2   u dt 1−   2 c   Solve for the acceleration. 3 2 2   du F u   = 1− 2   dt m c   16du The result is, that as u → c , → 0 , no matter how large the applied force. At the other dt du F extreme, when u c, = . dt m b. Collisions—conservation of momentum Consider the collision of two billiard balls. They have equal masses, m. Let’s say that one ball is initially at rest while the second ball has momentum p and energy E before the collision. o o After the collision, both balls have the same energy, E, and mass, m. It’s an elastic collision. Momentum and energy are conserved. 2 2 2 2 4 In the x direction, p = 2 p cosθ . Substitute for p and p using E = p c + m c . o o 1 2 2 2 4 2 2 4 E − m c = E − m c cosθ o c c 2 Conservation of energy allows us to eliminate E, since it was given that E + mc = 2E . Keep o 2 in mind that E is the relativistic total energy of the second ball, while mc is the rest energy of o the first (target) ball. At the same time, we solve for cosθ , the cosine of the scattering angle. 2 2 4 2 2 E − m c () E + mc(E − mc) o o o cosθ = = 2 2 2 () E + 3mc(E − mc) 2 2 4 o o () E + mc − 4m c o 2 E + mc o cosθ = 2 E + 3mc o 2 In the classical limit, E ≈ mc and therefore o 2 2mc 1 o cosθ ≈ = ⇒θ = 45 . But, as E o 2 2 4mc 2 o mc , cosθ → 1⇒θ → 0 c. Decay of a high-energy particle An unidentified high-energy particle is observed to decay into two pions (π mesons), as shown. 17Knowing the momenta and masses of the decay products, we determine the mass of the incident particle, hoping to identify it. MeV MeV 2 2 2 p = 910 , p = 323 , m c = m c = mc = 139.6Mev . 1 2 1 2 c c The energy and momenta are conserved. The total energy is 2 2 2 4 2 2 2 4 E = E + E = p c + m c + p c + m c o 1 2 1 2 E = 921MeV + 352MeV = 1273MeV o The quickest way to obtain the magnitude of the incident momentum is to use the law of cosines: 2 2 2 2 2 2 2 2 p c = p c + p c − 2 p p c cosθ = 1034229MeV o 1 2 1 2 p c = 1017MeV o Now that we have the total energy and the kinetic energy, the mass is obtained from 2 2 2 2 4 E = p c + m c o o o 2 2 2 2 m c = E − p c = 765MeV o o o Evidently, the incident particle was a ρ meson. What was its speed before it decayed? Well, 2 4 m c 2 o the total energy is also E = , so solve that for u. o 2 u 1− 2 c 2 4 m c u o = 1− = 0.8 2 c E o d. Mass-energy equivalence When we speak of the total energy being conserved that includes the total rest energy. For instance, consider the decay of a neutron that is initially at rest. n→ p+ e+ν The neutron decays into a proton, an electron and an anti-neutrino. The three product particles are observed to have total kinetic energy of K = 0.781 MeV. The initial energy is just the rest energy of the neutron, E = 939.57 MeV. The total final energy is i 2 2 E = m c + m c + K = 938.28MeV + 0.511MeV + 0.781MeV = 939.57MeV f p e Notes: 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 substituting numerical values into the formulae. m ≠ m + m . A portion of the neutron’s rest energy has been iii) Notice that n p e converted into kinetic energy. 18E. A Hint of General Relativity 1. Equivalence In Special Relativity it is asserted that all inertial reference frames are equivalent—the “laws” of physics are the same in all inertial reference frames. No experiment done in one frame can detect its uniform motion relative to another frame. Can the same be said for reference frames that have a relative acceleration? a. Elevator Recall the past discussion of a person standing in an elevator. If the elevator moves perfectly smoothly and there are no floor indicator lights, then the person inside will have no perception of the elevator’s motion, except for feeling perhaps the elevator floor pressing upward on his or her feet. Keep in mind: the person gets no information from any source outside the reference frame of the elevator. Contrast this situation with that of another person standing in a similar elevator, but this elevator is simply resting level on the Earth’s surface. The person in this elevator also feels the floor pressing upward on his or her feet, also has no perception of the elevator’s motion. We, as omniscient external observers, know that this second elevator is resting on the surface of a planet, and that what the person inside is experiencing is the gravitational force exerted by that planet. The point is that there is no experiment that either of the persons inside the elevators could perform that would distinguish 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 gravity Imagine 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, a . A person standing o inside the “elevator” sends a series of light pulses toward one wall—he or she and we see the light pulses dropping 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 in any frame of reference, regardless of its acceleration relative to another frame. We have seen that an accelerated frame is equivalent to one in a gravitational field. It follows that the force of gravity must affect a beam of light just as it affects the motion of a massive projectile. Indeed, experiment has shown that it does. But, light has no mass. 19