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Fields Test Example Questions Guide CAUTION: This guide covers 35 problems that will not appear on your Fields Test. Also under stand that the guide makes statements that are usually true without adding qualifying remarks or apologies. Advanced or special cases are not treated. In some cases, one or two steps are omitted. Complete the solutions are you read through this guide. STRATEGY: The best approach is to recognize the problem and solve it. If you are unsure, find physical reasons to eliminate as many of the five options as possible. Make an educated guess as you pick from among the remaining options. Mark the problem so that you can return to it later at which time you can attempt a more thorough analysis. As you work through the guide, pay special attention to the examples for which one or more of the options are found not to be viable. DO NOT MEMORIZE: Memorizing values and facts from this guide is not recommended. Do think about the concepts and values as you read this guide. Think and read just carefully enough to be left with a hazy recollection of the experience. THINK, think, think while you take the test. The sample questions appear boxed. Search for Ex. to find the next problem. MECHANICS AND RELATIVITY: Newton’s Laws: Use when asked about forces and acceleration at a single position and time. 2 acircle v r rˆ dv dt tˆ 2 r rˆ r tˆ Always account for the centripetal part. The tangential part applies if the speed is changing. Static friction acts with a magnitude from 0 to s N to keep the contacting surfaces from slipping relative to one another. One the surfaces are slipping, the friction force on each surface has magnitude equal to k N directed to oppose the relative motion. Factoid: k s. Work-Energy: Work is W = F dr suggesting that work-energy by used when a change is associated with a change in position. Mechanical energy: Ei = K1 i + K2 i + … + UA i + UB i + ….. K = ½ mv2 + ½ Icm2 = ½ McmVcm2 + ½ mvrel 2 + ... = .... Physics Handout Series – fields.tank page 1 Fields Test Example Questions Guide U = ½ k (stretch)2 + {mgh; - GMm/r} + choose near-earth or Universal form Ei + Wnon-conserve = Ef {q V; kQq/r} choose appropriate form Wnc-frintion = - k N (distance that surfaces slip relative to one another) A conservative is any force for which the work integral can be evaluated as the difference in the values of a scalar function at the integration endpoints. (Can do if the work integral around any closed path is zero; curl of the force is zero). Momentum and Collisions: Impulse is t2 t1 F dt suggesting that impulse-momentum methods be tried for a change that occurs as time changes – even for the time change from before to after. m k vk ,initial Fnet ,external mk vk , final tf Internal forces do not change the total momentum. ti (1) Collisions: Rule 1: Conserve momentum vectorially Rule 2: Kinetic energy is unchanged for an elastic collision. Some kinetic energy is converted to other forms if the collision is partially or totally inelastic. As the total momentum is conserved, only the kinetic energy associated with motion relative to the center of mass is available to be lost (converted). 2 Ktotal ½mi vi2 ½ M V 2 ½mi vrelative , i KCM K relative ; KCM i 2 Ptotal 2M i M: total mass; V: speed of the center of mass (CM); vrelative, i : speed of mass i relative to CM. In a totally inelastic collision, the particles stick together so Krelative is lost. KCM remains as required by conservation of momentum. Physics Handout Series – fields.tank page 2 Fields Test Example Questions Guide Ex. 1) The figure to the left shows two m, 2 vo particles with masses and velocities as indicated. The objects are moving on a flat, frictionless surface. When they collide, the 300 objects stick together. Their speed after the collection is most nearly: 300 (A) 0.67vo (B) 0.87vo (C) vo (D) 1.15 vo 2 m, vo (E) 1.73vo Rule 1 is to conserve momentum vectorially so a coordinate system is adopted. m, 2 vo y 300 x 300 2 m, vo The momentum of the particles before the collision are p1 m (2v0 )[ p2 (2 m) v0 [ 3 3 ˆ ½ ˆj ] and 2i ˆ ½ ˆj ] . The total mass is 3 m and the total momentum is Ptot 4 mv0 [ 2i Physics Handout Series – fields.tank 3 ˆ] . 2i page 3 Fields Test Example Questions Guide v final Ptotal 4 m v0 [ 3 2 iˆ ] 2 v0 ˆ i M total 3m 3 (inelastic; stick together) Was kinetic energy converted to other forms as a result of the collision? The collision was totally inelastic. Was all the kinetic energy lost? (2) Motion in a uniform gravitational field: Ex. 2) Ball 1 is dropped form a height h and ball 2 is dropped form a height ½ h. Which of the following gives the ratio of the speed of ball 1 to that of ball 2 just before they impact? (Assume that air resistance is negligible.) (B) 2 ½ (C) 1 (A) 2 (D) 2-½ (E) ½ First, the problem is a constant acceleration problem. Write down the master equations. x(t) = xo + vo t + ½ a t 2; v(t) = vo + a t ; v 2 = vo2 + 2 a (x – xo); vave = ½(v +vo) The balls fall from rest for a distance H. Using v 2 = vo2 + 2 a (x – xo), v 2 = 2 g H or v 2 g H . 2 g H1 v1 h v2 2 g H2 ½h Sanity Check. Three answers can be discarded immediately. Which three? Why? If the answer were (A), ball 1 would have twice the average speed of ball 2. What would be the ratio of the fall times in that case given that the first ball falls twice as far? (3) Law of Universal gravitation: Ex. 3) Two planets of mass m and M respectively have center-to-center spacing R. At what distance from the planet of mass M do the gravitational forces of the planets cancel each other. m R (B) 1 m R (A) M M (C) m R M m R (D) 1 M Physics Handout Series – fields.tank (E) R m 1 M page 4 Fields Test Example Questions Guide One need only consider points along the line joining the planets. Why? What must be true about the directions of two vectors that sum to zero? Prepare a sketch: A large well-drawn sketch provides the greatest benefit. Assume M > m. R-d m d M R Thought 1: Gravitational forces are attractive. Therefore, the ‘zero – force’ point must be between the planets and along the line joining them. Compare/contrast this situation with that in Ex. 10. Thought 2: The force is an inverse square (with distance) law force. Gm GM ( R d )2 m Rd m 2 2 2 M (R d ) d d M d Think about the result. At the point where their influences balance, the distance from the smaller planet is smaller than that from the large planet. (Include sanity checks as you proceed through your solution.) Rd m Since m M M d R d 1 m M d 1 Rm M is less than one, d > ½ R. The point is farther from the more massive planet. Reread the question. The distance d of the point from the more massive M planet is requested. The unknown label d is assigned to the value requested. Always review the question to ensure that you have answered the question that was asked. (4) Statics and Archimedes Principle In equilibrium, the sum of the forces acting on a rigid body is zero, and the sum of the torques (about any axis free to choose it where you wish) acting on the body is zero. Physics Handout Series – fields.tank page 5 Fields Test Example Questions Guide A body is buoyed up by the weight of the fluid that it displaces. FB = fluid Vdisplaced g. Ex. 4) A metal block is suspended in an empty tank from a scale that indicates a weight of W. The tank is then filled with water until the block is covered. If the density of the metal is three times the density of water, what apparent weight of the block does the scale now read? (A) ½ W (B) 2/3 W (C) W (D) 3/2 W (E) 3 W Examine your options. The new reading should be less than the original reading. Answer (B) could be chosen without much more, but one should solve the problem before choosing an answer. Prepare drawings T = Wdry T = Wwet wVg mVg Wdry = m V g = 3 w V g mVg Wwet =m V g - w V g = (m - w) V g * The ratio of the density of a material to that of water is called the specific gravity of the material. Wwet = Wdry ( - w/m) = Wdry ( - /) (5) Simple Harmonic Motion Equation of Motion: x(t) = A cos[ t + ] = C cos[ t ] + D sin[ t ] Take derivatives to find expressions for v(t) and a(t). v(t) = - A sin[ t + ]; a(t) = - 2 A cos[ t + ] Energy Conservation: ½ m v2 + ½ k x2 = ½ m (vmax)2 = ½ k A2 = Etotal mechanical = k m g I stiffness inertial property Physics Handout Series – fields.tank page 6 Fields Test Example Questions Guide A mass hanging from a linear spring executes simple harmonic motion about equilibrium. Potential energies are associated with systems of interacting entities. Potential energies are associated with pairs of things (at least) while a kinetic energy can be owned by a single entity. The potential energy can be shifted by an additive constant. For SHM, it is often set to zero for a particle positioned at equilibrium. Ex. 5) A 1-kilogram particle is attached to a spring and exhibits one-dimensional simple harmonic motion. The particle’s distance (!/* displacement) from the equilibrium position is given by the expression: y(t) = A sin[ t + /2], where A = 1 meter and = 0.5 rad/s. If the potential energy of the particle (system) at its equilibrium position is null (or zero), which of the following gives the total energy of the particle (system)? (A) 2 J (B) 1 J (C) ½ J (D) 1/8 J (E) 0 J v(t) = - A sin[ t + ]; a(t) = - 2 A cos[ t + ] ½ m v2 + ½ k x2 = ½ m (vmax)2 = ½ k A2 = Etotal mechanical The total energy is kinetic at equilibrium as the potential is set to zero at that point. ½ m (vmax)2 = ½ k A2 = Etotal mechanical The relation v(t) = - A sin[ t + ] shows that vmax = A = 0.5 m/s. Using m = 1 kg, Etotal = ½ m (vmax)2 = ½ (1 kg) (0.5 m/s)2 = 1/8 J. Write down the equation that is to be used. Substitute numerical values with units for each symbol writing the values in the same geometric pattern that was used for the symbols. Adopt procedures that reduce the chance of making a careless error. (6) Equilibrium with torques In equilibrium, the sum of the forces acting on a rigid body is zero, and the sum of the torques about any axis ( free to choose axis where you wish) acting on the body is zero. Physics Handout Series – fields.tank page 7 Fields Test Example Questions Guide r F ; r F r F r F sin The displacement from the axis to the point of application of the force is r , CCW is positive, and is the angle that takes you from the direction of r to the direction of F . Vector cross product: Extend the fingers of your right hand in along the direction of r . Orient your hand so that the fingers can curl toward the direction of F . The thumb of the right hand will be in the direction of r F . 500 F axis r r line of action of F Ex. 6) The figure shows a uniform rod of mass 4 kilograms that is pivoted at one end and supported by a string at the other end. If the rod is a rest (in equilibrium) the tension in the string is most nearly (A) 20 N (B) 26 N (C) 31 N (D) 40 N (E) 62 N Use the figure The rod does not have an assigned length. Assign a length of , 1 m or even 4 m. If the value is not given, the answer must be independent of its value. Choose an axis: The pivot point is a natural choice for the axis. Picture the situation. If the tension were changed, about what point would the rod rotate? The tension force tends to cause a CCW (+) ½ rotation so the associated torque about the 500 0 40 rT 40 pivot is T = + T sin(400). To compute torque, the weight of the boom can be assumed to be applied at its center of mass. W W T 0 The weight has a lever arm of ½ and tends to cause a CW rotation. W = - W (½ ). Physics Handout Series – fields.tank page 8 Fields Test Example Questions Guide The sum of the torques must be zero so: T ½ (4 kg ) (9.8 N kg ) ½W ½mg sin(400 ) sin(400 ) sin(400 ) ½ (4 kg ) (9.8 N kg ) .707 20 N 28.3 N .707 We expect an answer larger than 28 N so response (C) is chosen. The actual value is about 30.5 N. (7) Relativity Relative speed v Lengths are contracted along the direction of relative motion, but lengths running transverse to the motion are unchanged. 1 (v c ) 2 ; t A clock in motion relative to the observer runs slowly. t 1 (v c) 2 If a time t = 10 s elapses as measured by a moving clock, and it appears to be running slowly as viewed by the primed observer, then the corresponding interval t is longer than 10 s. Consider a primed observer moving at v iˆ relative to an unprimed observer with the initial conditions that their respective axes are parallel and the origins coincided at t = t = 0. Adopt the notations: v c and 1 (v c) 2 1 2 . x [ x ct ] c t [c t x ] y y z z t 1 t Ex. 7) A stick of length L lies in the x-y plane as shown. An observer moving at 0.8 c in the x direction measures the length of the stick. Which of the following gives the components of the length as measured by the moving observer? y Lx Ly (A) L cos 0.60 L sin (B) 0.6 L cos 0.60 L sin (C) 0.6 L cos L sin (D) 0.64 L cos 0.64 L sin (E) 0.78 L cos 0.78 L sin L O x Physics Handout Series – fields.tank page 9 Fields Test Example Questions Guide The component along the direction of motion should be contracted by a factor of = 0.6 while the transverse length should be unchanged. Lx = 0.6 (L cos) ; Ly = Ly = L sin Note that only answer (C) has an unchanged transverse component. ELECTROMAGNETISM AND CIRCUITS: (8) Basic Circuits and Circuit Elements The three basic passive elements are the capacitor, resistor and inductor. Each is characterized by its VC =( 1/C) Q ; voltage rule. Series Combos: Parallel Combos: 1 Cseries 1 1 C1 C2 Cparallel = C1 + C2 VR = I R ; VL = L dI/dt (Note 1/C in the VC relation.) Rseires= R1 + R2 1 Rparallel Lseries= L1 + L2 1 1 R1 R2 1 Lparallel 1 1 L1 L2 The voltage across a capacitor is a continuous function of time. Why? The current through an inductor is a continuous function of time. Why? Kirchhoff’s voltage rule: The sum of the voltages (potential changes) around a closed path is zero. Kirchhoff’s current rule: The algebraic sum of the currents into a node is zero. Elements are in series if the wiring of the circuit requires that the current through the elements is the same. There can be no branch points between them. Elements are in parallel if the wiring of the circuit requires that the voltage across the elements is the same. One lead from each goes to a common point and the other two leads from the elements join at another common point (connection path to join of high conductivity). Physics Handout Series – fields.tank page 10 Fields Test Example Questions Guide Ex. 8) A capacitor of capacitance 125 microfarads is initially charged such that the voltage across the plates is found to be 100 volts. If the capacitor is the connected in series to (with) a pure inductor of inductance 0.2 Henry, what is the maximum value of the current that is observed in the inductor? (A) 1.0 A (B) 2.5 A (C) 4.5 A (D) 5.0 A (E) 10.0 A Prepare a sketch: When the switch is closed, the two elements will be connected in series and parallel. The capacitor initially has a charge Q = C VC = (125 F)(100 V) = 12500 F. C 100 V Q KVR: /C + L dI/dt = 0. Using I = - dQ/dt, L 125 F d 2Q 1 d 2x Q 0 compare 2 x 0 2 2 dt LC dt 0.2 H A simple harmonic oscillation at = (LC)- ½. Q(t) = Qi cos[t + ]; I(t) = - Qi sin[t + ] The current through an inductor is a continuous function of time. It is zero just before the switch is closed so it is zero just after it is closed. I(t) = - Qi sin[] = 0 = 0 and Qi is just Q(t = 0) or 12500 F. The frequency is (LC)- ½ = (125 x 0.2 x 10-6 x 102)- ½ = (25 x 10-4 )- ½ = 20 rad/s. I(t) = - Qi sin[t + ] Imax = Qi = (12500 F) (20 rad/s) = 250000 A. The potential drops as one moves down in both Switch closed at t = 0. elements. Note the pattern in which the static charges accumulate on the inductor to cause the dI +Q C +++ dt d 2Q --- dI /dt . Starting at the lower left and proceeding CW, the sum of the potential changes is Q/C – (L dI/dt) = 100 V -Q dt 2 0.2 H 0.When the current is in the direction chosen as positive for dI/dt, the charge Q decreases so I = dQ /dt. Better: Recall that an LC circuit is an oscillator. d 2Q 1 Q0 dt 2 LC Physics Handout Series – fields.tank page 11 Fields Test Example Questions Guide Ex. 9) If the V is the potential difference between points I R R and II in the diagram above and all three resistance have the same resistance R, what is the total current between I and II? I (A) V/3R II R (B) 3 VR (C) 2V/3R (D) 3VR/2 (E) 3V/2R What is not said? The lower conductor continues out of the field of view. The components illustrated are embedded as a unit in some larger circuit. Solution: Start with the voltage versus current relation: VR = I R. The voltage across a resistor is equal to the current through it times its resistance. The first conclusion is that I = V/R for an individual resistor with each symbol representing its value for that resistor. 1st Conclusion: Responses (B) and (D) have incorrect dimensions and so are eliminated. 2nd Conclusion: There is a current V/R in the lowest resistor. A smaller current exists in the upper branch of two resistors. The currents are to be summed. V/R < I < 2 V/R. They make it rather easy as only response (E) corresponds to a current greater than V/R. Related Techniques: When you encounter a network of passive elements, you should make series and parallel combinations in sequence to reduce the network. The upper branch resistors are combined in series (What is required for two elements to be in series?) to yield an effective resistance of 2R. That 2R and the R in the lower branch are then combined in parallel to yield a net effective resistance of 2/3 R. If one is asked about the current, voltages, … for individual elements, one should start with the fully collapsed network and find all the values. Next back out one step (in Physics Handout Series – fields.tank page 12 Fields Test Example Questions Guide our case to R and 2R in parallel. Analyze to find all the values. Next, back out one more step and analyze …. . (10) Coulomb’s Law and Electrostatics A physics major should know Coulomb’s law, the Biot-Savart law and Maxwell’s equations. equation integral form Gauss’s Law V Faraday’s Law A A Charge Conservation (Continuity) r 4 0 rp rs p rp rS all rs rs rp rs 2 J (rs ) d 3rs rp rS E Bt B nˆ dA 0 A J Et nˆ dA 0 0 0 dQinside dt J nˆ dA (rs ) d 3rs all rs B(r ) 0 4 0 3 all r / B 0 B o J o o Et V t dV Fq q Eothers vq Bothers Lorentz Force Law E (rp ) V B d V E V 0 dV 0 E d Bt nˆ dA A Gauss’s Law - Magnetism Ampere (Maxwell's 4th) Qinside E nˆ dA differential form J t Fq q Eothers vq Bothers (rs ) d 3rs qs rˆsp rˆsp 2 4 0 rsp 4 0 rsp2 Coulomb 0 I d s rp rS qs vs rp rS 0 3 3 4 4 rp rS rp rS Biot-Savart Ex. 10) A charge of +Q is placed on the x-axis at x = - 1 meters, and a charge of -2Q is placed at x = + 1 meters, y as shown to the left. At what position on the x-axis will a test charge of +q experience zero net force? (A) (3 8) m -2Q Q 1 m 1 m x (D) 1 /3 m (B) – 1/3 m (E) (C) 0m (3 8) m Physics Handout Series – fields.tank page 13 Fields Test Example Questions Guide One should think about a problem before launching into a solution. The net force on the test charge due to the two existing charges is to be zero. Two vectors can sum to zero only if they are antiparallel. A little reflection makes it clear that the solution points must lie on the (extended) line joining the two charges. That is: The solution point is somewhere on the x axis. y y q I -2Q Q 1 m 1 m II Q III -2Q x x In order to sum to zero, the forces must have equal magnitude and opposite direction. For points on the x-axis, the test charge must lie to the right of both charges or to the left of both charge to meet the opposite direction requirement. To meet the equal magnitude requirement, the test charge must be closer to the smaller magnitude charge. That places the solution point in region I, x < - 1 m. The particular value of x is found using the inverse square law nature of the Coulomb force. Look for equal magnitudes for points in region three. The distance from Q is | x – (-1)| and the distance from -2Q is | x – 1|. kQ k (2 Q) 2( x 1) 2 ( x 1) 2 or x2 + 6 x + 1 = 0 2 2 ( x 1) ( x 1) This equation has a root (3 8) m in region I and a root (3 8) m in region II. Our previous arguments have identified region I (x < - 1 m) as only region in which both the equal magnitude and opposite direction requirements can be met. Recall that quadratic equations often return one physical result and one not-so-physical result. Review your options. Which of the proposed answers corresponds to a point to the left of both charges? Test them, simplest value first, to see if the equal magnitude condition is met. (11) Gauss’s Law, Ampere’s Law and E&M Knowledge Points: Physics Handout Series – fields.tank page 14 Fields Test Example Questions Guide Gauss: V E nˆ dA ( 0 )1 dV Qenc 0 V V E nˆ dA EA Qenc 0 Qenc directed away o A E A : The enclosing surface through the point where the field is to be determined is to be made of portions perpendicular to the field and parallel to it. A : net area that is perpendicular to the field. A cyl = 2 r L ; A plane = 2 A ; A sph = 4 r2 For applications to conductors, make a sketch. (For equilibrium,) The net charge is on the surface(s). The field at points in the conducting material is zero. A small patch on the surface is essentially planar, and the field only pokes outside. Qenc = local A; A = A; E local away. o Ampere: A B d I enc J nˆ dA I enc A B d Bi A i i B o I enc 0 i I enc i directed to circulate (RHR) The encircling path through the point where the field is to be determined is to be made of portions perpendicular to the field and parallel to it. constant non-zero magnitude. long st. wire : the net parallel length along which the field has a =2r; solenoid = ; toroid =2r Knowledge points: (a) Field and potential of a point charge: V (rsp ) q 4 0 rsp ; E (rsp ) q rsp 4 r 3 0 sp q rˆsp 4 0 rsp2 rsp : displacement from source point to field point (b) The field of a uniform shell of charge is zero inside and the same as a point charge with qtotal located at the center outside. Physics Handout Series – fields.tank page 15 Fields Test Example Questions Guide (c) The field of a spherical ball of charge increases linearly with r inside and is the same as a point charge with qtotal located at the center outside. (d) The field of a long uniform line of charge is the charge per length divided by 2 times the distance from the wire. The field is directed away from a uniform line of positive charge. (e) much, much more … Ex. 11) A uniform insulating sphere of radius r has a total charge Q and a uniform charge density. The electric field at a point R/3 from the center of the sphere is given by which of the following? (A) Q 12 0 R 2 (B) Q 8 0 R (C) 2 Q 6 0 R (D) 2 Q 4 0 R 2 (E) 3Q 4 0 R2 By knowledge: The field grows linearly inside a uniformly charge sphere so the field has 1/3 of the 3 4 Q R strength at r = R or 1 2 . The direction is radially away by symmetry. 0 By Gauss: E Qenc directed away. o A For the point R/3 from the center, the Gaussian surface is a sphere surface of radius R/3 concentric with the sphere. (The point at which the field is to be found must lie on the Gaussian surface. The surface has the symmetry of the problem.) Qenc = (1/3)3 Q; A = (1/3)2 (4 R2) (12) Faraday’s Law A B nˆ dA A t E d Faraday’s Flux Rule = Emf = - dB dt B B nˆ dA B A A Ex. 12) Four meters of wire form a square that is placed perpendicular to a uniform magnetic field of strength 0.1 Tesla. The wire is reduced in length by 4.0 centimeters per second while still maintaining its square shape. Which one of the following gives the initial induced emf across the ends of the wire? (A) 1 mV (B) 2 mV (C) 4 mV (D) 8 mV Physics Handout Series – fields.tank (E) 16 mV page 16 Fields Test Example Questions Guide The magnetic field is not varying in time, so we will use the flux rule form: Emf = - dB . dt The easy way: The side s is the perimeter divided by 4. The side at time zero is s(0) = 1 m. At one second, s(1 sec) = 0.99 m. Using B B A , B(0 s) = (0.1 T) (1 m2) and B(1 s) = (0.1 T) (.992 m2) Emf = - dB dt B (1 s) B (0 s) (0.1T )(1 m2 ) (0.1T )((.99)(.99) m2 ) (0.1T )[1 0.98]V = 2 mV 1s 1s As a function of time. 2 B B A = B side2 = B (perimeter/4)2 = (0.1 ) (4.0 – 0.04 t ) /16 Tm2 d/ dt B =(0.1 ) ( [2 (4.0 – 0.04 t) (- 0.04)]/16) Tm2s-1 At time zero, the magnitude of the emf is |(0.1) ( [8 (- 0.04)]/16)| Tm2s-1 = 2 mV. OPTIC, WAVES, THERMO AND STAT: (13) Properties of an E&M plane wave and the Poynting Vector EM waves are transverse which means that the electric and magnetic fields are perpendicular to the direction of propagation of the wave. The electric and magnetic fields are mutually perpendicular. The wave propagates in the direction of the Poynting vector, S ( E H 01 E B ) . The E and B fields oscillate in phase with one another and B/E = v, the wave speed (= c in vacuum). The Poynting vector has units W/m2 and represents the power flow of the wave. Energy Densities: E = ½ o E2 ; B = ½ 2/o Physics Handout Series – fields.tank page 17 Fields Test Example Questions Guide Ex. 13) A linearly polarized electromagnetic plane z wave carries energy in the positive z direction. At y some positions r and time t, the magnetic field points along the positive x-axis, as shown in the figure to the right. At that position and time , the electric field points x B along the (A) positive y-axis (B) negative y-axis (C) positive z-axis (D) negative z-axis (E) negative x-axis EM waves are transverse which means that the electric and magnetic fields are perpendicular to the direction of propagation of the wave. The electric and magnetic fields are mutually perpendicular. As the wave propagates in the z direction and at that point and time has the magnetic field in the x direction, the electric field must be in the y direction. The Poynting vector S represents the intensity of the wave times its propagation direction. Find the direction of S . S E H 01 ( E B) S 01 ( E B) and ˆj iˆ kˆ and ( ˆj ) iˆ kˆ The electric field is in the negative y direction at that point and time. (14) Diffraction and Interference Standing waves As a first guess, diffraction limits angular resolution to min = /d where d is the width of the aperture. (If the aperture is circular, use min = 1.22 /D where D is the diameter of the aperture.) (Ex. 14) An observer looks through a slit of width 5 x 10-4 meters at two lanterns a distance of 1 kilometer from the slit. The lantern emits light of wavelength 5 x 10-7 meters. The minimum separation of the lanterns at which the observer can resolve the lantern lights is most nearly (A) 0.01 m Physics Handout Series – fields.tank page 18 Fields Test Example Questions Guide (B) 0.1 m (C) 1m (D) 10 m (E) 100 m In this case, = x 10-7 meters and d = 5 x 10-4 meters so min = 10-3 radian or one milli-radian. One should prepare a sketch, but one can guess that the answer is 10-3 of one kilometer or one meter. Polazation, polarizers: The interaction of E-M waves with matter is described by the Lorentz force law, Fq q Eothers vq Bothers . For a wave, |B| c-1 E so Fmagnetic (vq/c) Felectric. The electric interaction dominates so the polarization of the wave is associated with the direction of the electric field. That field is transverse to the direction of propagation and there are two independent transverse directions so there are two independent polarizations for a given propagation directions. These polarizations might be chosen to be vertical and horizontal linear polarizations or left- and righthanded circular polarizations. Circular polarizations carry angular momentum along the direction of propagation. Linear polarizations carry zero net angular momentum. Natural (unpolarized light) can be assumed to be an equal mix of the two independent polarizations. A linear polarizer absorbs the component of the polarization perpendicular to its pass axis. It does this by adding a wave polarized perpendicular to the pass axis that is out of phase with the incident wave and of equal amplitude. The incident wave has an E field that makes an Incident angle w.r.t. the pass axis. The charges in the components polarizer move perpendicular to the pass axis and added by polarizer absorb energy effectively adding a field perpendicular to the pass axis that is out of transmitted phase. Superposing, the perpendicular part is pass axis eliminated on only the part along the axis passes. The wave immediately after a polarizer has a polarization parallel to the pass axis and an average electric field strength of Eincident cos. As the intensity of a wave is proportional to the square of its Physics Handout Series – fields.tank page 19 Fields Test Example Questions Guide amplitude, Itransmitted = Iincident cos2. (The Law of Malus.) Natural light is an equal mix of the two polarizations. If natural light is incident on a polarizer, light of one-half the intensity that is 100% polarized along the pass axis is transmitted. Io natural ½ Io cos2 along last pass axis ½ Io vertical ½ Io cos2cos2 along last pass axis One half the intensity of the incident intensity of natural light passes through the first polarizer. All the light that passes is polarized along the pass axis direction. Incident polarized light obeys the Law of Malus and the transmitted light is 100% along the pass axis of the last polarizer. (15) Lens Problems (Ex. 15) Two thin converging lenses A and B, each having a focal length of 6 centimeters. are placed 10 centimeters apart, as shown in the figure above. If an object is placed 10 centimeters to the left of the lens A, the final image is (A) 30 cm to the right of lens B (B) 30 (C) 30 (D) 30 /11 cm to the right of lens B /10 cm to the right of lens B /11 cm to the left of lens B Physics Handout Series – fields.tank page 20 Fields Test Example Questions Guide (E) 30 /10 cm to the left of lens B A positive lens converges the light that passes through it. Light diverges for a real object. Real objects have positive object distances. Light converges to a real image with a positive image distance. The focal length of the lenses in this problem is f = 6 cm. The object is 10 cm before the left lens and so it has an object distance + 10 cm. Using the lens equation, 1 1 1 dObj dImg f 1 1 1 3 1 5 1 2 10 dImg 6 30 dImg 30 dImg 30 Lens A would form an image 15 cm to its right if there were no second lens. That would be 5 cm to the right of lens B. Lens B is converging so the rays should converge closer to lens B than 5 cm. (Lens B is converging so we are focusing on responses (B) and (C).) The appearance of 5 cm and 6 cm makes the 30/11 cm value look attractive. The light arriving at lens B is converging. That case is called a virtual object and corresponds to a negative object distance. (Light diverges from a real object and converges to form real image. These cases correspond to positive object and image distances. Converging light incident on a lens and diverging light leaving the lens corresponds to virtual objects and virtual images negative distances for the lens formula.) Applying the formula to lens B, 1 1 1 1 1 1 6 1 5 1 11 dObj dImg f 5 dImg 6 30 dImg 30 dImg 30 The image distance is positive so the light converges to a point on the side of the lens opposite to that from which the light was incident. Ray Trace of the action of the first lens only. NOT drawn to scale!!!!! 15 cm Real image: the rays actually converge to and pass through a real image point. Physics Handout Series – fields.tank page 21 Fields Test Example Questions Guide Rays are converging as they reach the second optic lens two has a virtual object negative object distance. add construction ray through lens center toward lens 1 image; undeflected 15 cm action of the second lens in parallel; out through focal point Physics Handout Series – fields.tank page 22 Fields Test Example Questions Guide Virtual image: The rays appear to diverge from an image point that the rays did not converge to and pass through. Rays are diverging immediately after the last optic. (16) Basic Wave Properties Basic Wave Plug*: v=f The wave-speed is the frequency of the wave times its wavelength. THE BASIC WAVE PLUG (BWP) is the most important wave property relation. Wave Speed: v = Tension stiffness mass / length inertial density Traveling Wave: y(x,t) = A sin[ k x - t + ] Standing Wave: y(x,t) = A cos[t + ] sin[ k x + ] k = 2; T = 2 node to node spacing: ½ node to anti-node spacing: ¼ Wave types: longitudinal and transverse There are two transverse directions so transverse waves have polarizations Waves reflect with a sign change when they reflect from a region with lower wave speed. Waves reflect with a sign change when they reflect from a fixed end. (Ex. 16) Two identical sinusoidal waves travel in opposite directions in a wire 15 meters long and produce a standing wave in the wire. The traveling waves have a speed of 12 meters per second and the standing wave has 6 nodes, including those at the two ends. Which of the following gives the wavelength and frequency of the standing wave? Wavelength Frequency (A) 3 m 2 Hz (B) 3m 4 Hz (C) 6m 2 Hz Physics Handout Series – fields.tank page 23 Fields Test Example Questions Guide (D) 6m 3 Hz (E) 12 m 2 Hz Prepare a sketch! The four interior nodes divide the wire into five equal length segments, node-tonode spacings. Thus, 5 (/2) = 15 m or = 30 m/5 = 6 m. From the BWP, f = v/ = 12/6 s -1 = 2 Hz. (17) Kinetic Theory of Gases Gas molecules have an average translational kinetic energy of 3/2 k T (three-halves times Boltzmann’s constant times the absolute temperature). The pressure arises due to the many molecular collisions per second with the walls so it increases for higher speeds v (v increases if temperature increases) and for higher density. The molecules have a cross sectional area for interaction and so sweep out a volume v t in time t. The mean time between collisions is found as the time for a molecule to sweep out the average volume per molecule. free v tcollision = V/N so tfree = V/Nv The mean free path free is the distance traveled to sweep out V/N. free = V/N. Ex. 17) Cubical tanks X and Y have the same volumes and share a common wall. There is 1 gram of helium in tank X and 2 grams of helium in tank Y, and both samples are held at the same temperature. Which of the following is the same for both samples? (A) the number of molecular collisions per second on the common wall (B) the average speed of the molecules Physics Handout Series – fields.tank page 24 Fields Test Example Questions Guide (C) the pressure exerted by the helium (D) the density of the helium (E) the mean free path of the molecules The number of collisions per second, pressure and density for sample Y are double the X values. The mean free path for sample Y is half that for sample X because there are twice as many things to hit in the same volume. The average speed depends on the temperature and molecular mass. It is identical for samples X and Y. (18) Ideal Gas Law isothermal: process at constant temperature isobaric: process at constant pressure iso-entropic: process at constant entropy adiabatic: process in which there is no heat transfer; insulated from the environment isochoric, isometric, iso-volumetric -------- process at constant volume The Ideal Gas Law: PV=nRT PV=NkT number of moles: n ideal gas constant R = 8.31 (J/ mol-K) number of molecules: N Boltzmann’s constant: k = 1.38 x 10-23 J/K LaChatlier’s Principle: A stable system acts to counter any change in its parameters. If a the volume of a sample of gas is reduced, its pressure increases to fight additional decreases in volume. (Ex. 18) If one mole of an ideal gas doubles its volume as it undergoes an isothermal expansion, it pressure is (A) quadrupled (B) doubled (C) unchanged (D) halved (E) quartered Physics Handout Series – fields.tank page 25 Fields Test Example Questions Guide The process is isothermal (same temperature). P1 V1 = n R T = P2 V2 V P2 P1 1 ½ P1 V2 Use ratios whenever possible. (19) Thermodynamics and efficiency Entropy Change: S Qrev T Use absolute temperatures. Ideal Heat Engine Efficiency: e Thigh Tlow Thigh 0 0C = 273 K Ex 19.) A power plant takes in steam at 527 0C to power turbines and then exhausts the steam at 127 0C. In any given time, it consumes 100 megawatts of heat energy from the steam. The maximum output power of the plane is (A) 10 MW (B) 20 MW (C) 50 MW (D) 75 MW (E) 100 MW The problem statement is a little fuzzy with regard to its use of the terms power and energy! Using 0 0C = 273 K, Thigh = 800 K and Tlow = 400 K. The ideal thermal efficiency for a reversible engine operating between these temperatures is 50% so the maximum possible mechanical/electrical power out is 50 MW. QUANTUM MECHANICS AND ATOMIC PHYSICS: Particles have associated wave properties. As a crude characterization, particles propagate like waves and interact as (point) particles. It may be helpful to assume that each particle has a guide wave that feels out the space to generate the probability distribution for the particle’s location. Waves such as optical radiation also display particle-like character. When they interact, energy transferred into and out of the wave in quanta (photons) each with energy hv (and momentum hv/c.) h p k; E hv ; h 6.626 1034 Js; 1.0546 1034 Js The information that can be known about a system is encoded in a wavefunction (x,t) that satisfies Physics Handout Series – fields.tank page 26 Fields Test Example Questions Guide the Schrödinger equation. The probability to find the particle between x and x + dx is (x,t)(x,t)dx (Born). Additional information can be extracted from the wavefunction using the operator for the item of interest. The operator for x is x. The operator for px is - i x. Dynamic quantities are functions of position, momentum and time: Q(x,y,z, px,py,pz,t) Q(x,y,z, –i x, –i y, –i z, t); K 2 2m 2 ; V (r ) V (r ) Shrödinger’s equation identifies two operators for energy, the Hamiltonian and i t. In other words, the Hamiltonian represents the total energy (kinetic + potential) and is the time-development operator for wavefunction. 2 2 Hˆ (r , t ) V (r ) (r , t ) i (r , t ) t 2m (r , t ) u(r ) T (t ); (Use separation.) 2 2 2m V (r ) un (r ) Enun (r ); En Tn (t ) i Tn (t ) t The values En are the energy eigenvalues, the total energies, a set of real values that are characteristic of the problem and the associated temporal dependence is: Tn(t) = e in t e A general solution has the form, (r , t ) i En t . an un (r ) ein t . all states n Eigenvalue equation: px (x,t) = –i x(x,t) = (momentum eigenvalue) (x,t) H n(x,t) = En n (x,t) Operators for physical observables are Hermitian (or self-adjoint) an, as such, they have real eigenvalues and expectation values. Wavefunction Behavior: 2 (r , t ) 2m 2 En V (r ) (r , t ) In a classically allowed region, the kinetic energy K, E – V, is greater than zero so the net curvature of has a sign opposite to that of . The function value oscillates back and forth across zero. In the case that V > E or K < 0 which can not happen classically, the region is classically forbidden, and the curvature and function have the same sign behavior like growing and decaying exponentials. Physics Handout Series – fields.tank page 27 Fields Test Example Questions Guide All the functions oscillate in the classically allowed region and decay in the forbidden regions. The decay is more rapid for larger energy deficits. Each higher state has an additional node. Free Particle States for piecewise constant potentials are of the form ei ( kx t ) where k 2m E V ( x) . At any abrupt change in V(x), the wave is partially transmitted and partially reflected. The wave function ca tunnel through classically forbidden regions (V > E k = i and the wave is like ex). Fails to show the reflected wave! Energy eigenstates are stationary the associated probability density * is time-independent. n (r , t ) un (r ) ein t un (r ) e i En t ; n (r , t ) n (r , t ) un (r ) 2 22 Energy Level Spacing: Energy levels in an infinite well with width a have the form n 2 so 2 2ma 22 the n to n + 1 spacing is (2 n 1) ; the spread grows with n and is larger for smaller a 2 2ma (tighter confinement). The finite-well states penetrate into the forbidden region and so are less tightly confined leading to lower energies as compared to the infinite well with the same inside Physics Handout Series – fields.tank page 28 Fields Test Example Questions Guide width. For the hydrogen atom problem, the electron’s range approaches infinite as the energy approaches zero from below. The level spacing approaches zero in this limit. For positive energies, the particle is not confined and the allowed energies run continuously (zero spacing). Commutation: If the operators for two quantities do not commute, then those quantities have a minimum uncertainty product. [x,px] = i x px ½ . If operators commute, they can have simultaneous eigenvalues. For the hydrogen problem, H, L2 and Lz (plus the operator for electron spin) commute. The hydrogen atom states are labeled by nm representing the eigenvalues of energy ( - 13.6 eV/n2), orbital angular momentum squared ([+1]2) and z component of orbital angular momentum (m) (plus spin – up or down). Generalized Uncertainty Principle. Experimental measurements are performed on systems in order to extract average values and the corresponding uncertainties. In quantum systems, constraints can arise between two uncertainties due to the nature of the corresponding operators. Given two operators A and B, and knowledge of the commutator [A,B] = iC, then there exists a hard lower limit on the product of uncertainties A B >= ½ <C>. Example: [x,px] = i x px ½ . If two operators commute, then there is no lower-level uncertainty constraint on knowledge of the eigenvalues, and the expectation values of these operators serve as a very useful label to identify and characterize a system. For the bare Coulomb potential (simple H-atom problem), the operators H, L2, and Lz commute, therefore we identify individual states by the expectation values (energy, a.m. magnitude, a.m. z-comp) or their corresponding labels n,,m (quantum numbers). Expectation value: Q everywherethat *(r , t ) Qˆ (r , t ) d 3 r is non zero Mixed State: (r , t ) an un (r ) ein t more than one an 0. Here, an is the amplitude to be all states n found in state n and | an |2 is the probability for the system to be found in state n. As energy eigenstates have been used above, E 2 an En . A precise measurement of the energy will all states n yield one of the energy eigenvalues. For a set of identically prepared systems, the eigenvalue En will Physics Handout Series – fields.tank page 29 Fields Test Example Questions Guide be found with probability | an |2. Making a precise measurement and finding the value En places the system in a state that has that energy. If a measurement on the state: hydrogen a100 100 a210 200 a211 210 a211 211 a211 211 a310 310 returns the energy, - 3.40 eV, then the atom must be in a state of the form after b210 200 b211 210 b211 211 b211 211 immediately after the measurement. It will time develop as directed by the Hamiltonian and is not necessarily a state consistent with the measured eigenvalue after some time has passed. If the operator for the eigenvalue commutes with Hamiltonian, the expectation value for the quantity will be time independent. i Q Hˆ , Qˆ t i Hˆ , Qˆ d Q dt d Q dt time development of expectation values for operators Q without explicit time dependence Example: p i d p dt i The time rate of change is momentum is the ‘force’. Hˆ , p V (r ) Ehrenfest’s Theorem Every Hermitian operator has a complete set of eigenstates (Q m = qm m) which can be used as an basis set for representing a general solution. (r , t ) cm m (r , t ) . It follows that: eigenstates m Q cm 2 qm . A precise measurement of Q will return an eigenvalue and place the system eigenstates m in a state with that eigenvalue immediately after the measurement. (20) Ionization and the Periodic Chart Physics Handout Series – fields.tank page 30 Fields Test Example Questions Guide The inner electrons screen the nuclear charge seen by the outer electrons. The apparent unscreened charge increases as you move across the table filling an electron shell. It is a maximum for group VII and a minimum for the closed shell group VIII’s. The effective unscreened charge increases as you move to down the table to the higher Z atoms. Group I has one electron outside a closed shell and are the easiest to ionize (easiest at top after H; lowest screened charge). In group VII, the effective charge gets higher as you move down the table most electron hungry (highest electron affinity). Electron affinity increases as you move down and across from I to VII. VIII is the closed shell nobles. Outer valence shell has two S plus 6 P states. Physics Handout Series – fields.tank page 31 Fields Test Example Questions Guide Electron shell filling sequence. Except for (H, He), the first electron in an S shell leads to an alkali metal (strong electron donor). Two s electrons yield a metal. The nth valence shell consists of the two ns and six np states. Beginning with period n = 4, inner shell (d and f electrons of lower n) fill after the nS, but before the nP. Shells fill first with spins parallel (say up) and add spins anti-parallel only after the sub-shell is half filled. Elements with approximately half-filled d and f sub-shells exhibit magnetic properties associated with the unpaired spins. Ex 20.) For which of the following elements is the ionization energy of a neutral atom the lowest? (Z) is the atomic number. (A) Oxygen (Z = 8) (B) Fluorine (Z = 9) (C) Neon (Z = 10) (D) Sodium (Z = 11) (E) Magnesium (Z = 12) The first period of elements includes hydrogen and helium with one and two electrons. In the next period, there are eight electrons in the valance shell. Oxygen and fluorine need electrons to close the shell and so are hard to ionize. Neon is the very stable closed shell configuration. Sodium has just one electron outside a closed shell and so is easy to ionize. For Magnesium, the nucleus plus inner electron shells combined have a net charge of plus 2 and so hold the outer electrons more tightly bound than does the sodium nucleus plus inner shell electrons (net charge +1). Sodium is a light alkali metal and so is an electron donor. In compounds such as NaCl, it is strongly ionic with the sodium being positive which means its electron has transferred to the chlorine. After hydrogen, the Group I atoms are willing electron donors and easy to ionize. Group 7 is the group of the strongest electron acceptors. Top Left strong donor; Bottom Right (grp.7) strong acceptor. (21) Hydrogen Atom spectrum levels, … Physics Handout Series – fields.tank page 32 Fields Test Example Questions Guide En E 13.6 eV n2 E h h c upper 1 1 Elower 13.6 eV 2 2 nlower nlower Ex 21.) The Paschen series for hydrogen corresponds to transitions that end in the state of quantum number n = 3. The shortest wavelength line in the Paschen series is closest to which of the following? (The ionization energy of hydrogen is 13.6 eV and hc = 1,200 eV nm) (A) 125 nm (B) 250 nm (C) 400 nm (D) 800 nm (E) 1800 nm The shortest wavelength would correspond to the transition with the largest E from the highest bound level down to the n = 3 level. 13.6 eV 31 1 13.69eV ; E h c 2 2 9 hc 9 1200 eV nm 794 nm 13.6 eV 13.6 eV (22) Blackbody radiation Power per area = Stephan Boltzmann constant times T to the fourth power = T 4 Wien’s Displacement Law: peak T = constant = 2.898 x 106 nm K Planck’s Law: I ( ) dv 2hv 3 1 dv hv 2 c e kT 1 I ( ) d 2c 2 h 5 1 e hc kT 1 d Wien: I() is a function of T so the peak occurs for a set value of T peakT = constant. Stephan: 0 I ( ) dv ? change of variable 3 3 2hv3 1 kT 2hu 1 kT dv h c 2 eu 1 h du c 2 ehv kT 1 0 I ( ) dv T 4 2k 4 h3c 2 0 u du T 4 u e 1 Ex 22.) Which of the following describes the effect of doubling the absolute temperature of a blackbody on its power output per square meter and on the wavelength where the radiation Physics Handout Series – fields.tank page 33 Fields Test Example Questions Guide distribution is a maximum? (A) The output power is increased by a factor of 16 and the maximum of the distribution shifts to twice its original wavelength. (B) The output power is increased by a factor of 16 and the maximum of the distribution shifts to half its original wavelength. C) The output power is increased by a factor of 8 and the maximum of the distribution shifts to twice its original wavelength. (D) The output power is increased by a factor of 16 and the maximum of the distribution shifts to half its original wavelength. (E) The output power is increased by a factor of 2 and the maximum of the distribution shifts to four its original wavelength. Higher temperature suggests higher energy and hence shorter wavelength. Answers (B) and (D) are viable! The power increase factors of 24 or 21 for (B) and (D). The fourth power should be familiar. E hv hc 1 eV 1240 nm hc = 197 eV·nm (atomic physics) = 197 MeV·fm (nuclear physics) color (nm) v (1012 Hz) E(eV) IR > 700 < 428 < 1.77 red 625-700 405-480 1.77-2.0 orange 585-620 484-513 yellow 585-570 513-526 green 570-505 526-594 blue 500-440 600-682 violet 440-400 682-750 UV < 400 > 750 > 3.10 ranges are approximate and arbitrary no universal agreement on the ranges Ex 23.) Let H denote an Hermitian operator and suppose that H |= a |, where | is an eigenvector of H. Which of the following is true of the eigenvalue a? (The symbols Re and Im Physics Handout Series – fields.tank page 34 Fields Test Example Questions Guide denote the real and imaginary parts, respectively.) (A) Re(a) = Im(a) (B) Re(a) = - Im(a) C) Re(a) = 0 (D) Im(a) = 0 (E) H = a The eigenvalues of a Hermitian operator are real. The eigenvectors associated with distinct eigenvalues are orthogonal. Ex 24.) The quantum numbers used to label the radial wave function solutions to the Shroedinger equation for hydrogen atom are the principal quantum number n and the angular momentum quantum number . If the principal quantum number is n = 2, which of the following gives the possible values for the angular momentum quantum number ? (A) 1, 0 (B) (C) 2, 1, 0 (D) (E) 3/2, 1/2 The states of the hydrogen atom are labeled by the quantum numbers n, and m plus the intrinsic spin. The allowed values are n = 1, 2, 3, ….; = 0, 1, … , (n -1); m = -, - +1, … , 0, 1, …, ). That is: m assumes the 2+ 1 values spaced by 1 running from - to . For n = 2, can be 0 or 1 and m can be -1, 0, +1. Each combination can be combined with spin up or down. The value is the orbital angular momentum which assumes integer values. The total angular momentum is the orbital plus spin, and it assumes half-integer values for an electron in hydrogen. Ex 25.) The three operators (Lx, Ly, Lz) for the components of the angular momentum commute with the Hamiltonian of a particular particle. Therefore, the angular momentum of the particle is (A) equal to zero (B) equal to the energy in magnitude (C) always equal to Lz Physics Handout Series – fields.tank page 35 Fields Test Example Questions Guide (D) a unit vector (E) a constant of the motion The Hamiltonian is the energy and time-development operator for a quantum system. It results that d Q dt i [ H , Q] for time-independent operators. Quantities with time-independent operators that commute with the Hamiltonian are constants of the motion. A quantum state can be labeled with good quantum numbers corresponding to it eigen-energy quantum number and quantum numbers each additional member of the largest set of mutually commuting operators that includes the Hamiltonian. For example, the members of the set (Lx, Ly, Lz) do not commute with one another. They do commute with L2 = Lx2 + Ly2 + Lz2. Considering that it is a given that (Lx, Ly, Lz) commute with the Hamiltonian, then L2 will also commute with the Hamiltonian. In such a case, there would be good quantum numbers corresponding to H, L2 and one of the components of angular momentum. Lz has the simplest representation in spherical coordinates so it is usually chosen. For the hydrogen atom problem, the quantum numbers n, and m are associated with H, L2 and Lz. Ex 26.) A particle of energy E is in an eigenstate of the square well potential shown above, with wave function (x). Which of the following is a correct expression for the expectation value of x2 for this particle? a (A) x 2 ( x ) dx a (B) (C) (D) x 2 ( x) dx *( x) x ( x) dx a a *( x) x 2 ( x) dx Physics Handout Series – fields.tank page 36 Fields Test Example Questions Guide (E) *( x) x 2 ( x) dx The correct procedure to compute an expectation value is to slap the operator for the quantity between *(x) and (x) and then to integrate of the full range of x for which x) is non-zero. This square well has a finite depth so the particle penetrates into the classically forbidden region |x| > a. In this case the integrals must run over the full range (-,). SPECIAL TOPICS: (27) The Pauli Matrices: x 0 1 = ; 1 0 y 0 i = ; i 0 z 1 0 = 0 1 Ex 27.) Given the Pauli spin matrices shown above and the operators defined by - = x (A) -i y, = x +i y and which of the following is NOT correct. 1 0 (B) 0 0 + + 0 1 2 (C) 1 0 + 1 0 2 (D) 0 1 - z 1 1 0 0 (E) 0 0 1 1 z Method: Realize that you have no idea what the problem is about, but also realize that it just requires some matrix multiplication. Find the explicit forms for + and -, and brute force compute each matrix product until you find the one that is NOT correct. Background: ** Skip to the next problem if you are in a hurry. (28) Counting Statistics Suppose that a set of 2 x 2 matrices is sought such that any 2 x 2 matrix can be represented as a sum of the members of that set. Clearly the set must have at least four members as 2 x 2 matrices have four independent elements. One choice that could be made is: 1 0 0 1 0 0 0 0 , , , 0 0 0 0 1 0 0 1 This set of matrices is not very exciting. Spice it up; add the requirement that each member of the set Physics Handout Series – fields.tank page 37 Fields Test Example Questions Guide be Hermitian (equal to the complex conjugate of its transpose). The simplest matrix that has imaginary elements that meets this requirement is: 0 i i 0 An independent off-diagonal matrix needs to have elements that are equal rather than the negative of one another. Equal off-diagonal elements and Hermitian restricts the elements to be real. 0 1 1 0 Following the equal and negative scheme for the on diagonal matrices, the remaining members are: 1 0 1 0 0 1 and 0 1 This particular set of 2 X 2 matrices is the set of Pauli matrices. They have assigned labels: 1 0 = ; 0 1 x = 1 0 1 = ; 1 0 y = 2 0 i = ; i 0 z = 3 1 0 = 0 1 The first matrix is the identity, and the final three are the Pauli matrices. The set of the four matrices is a basis for the collection of all 2 X 2 matrices. a b a d 1 0 b c 0 1 c b 0 i a d 1 0 c d 2 0 1 2 1 0 2 i i 0 2 0 1 An arbitrary 2 x 2 matrix can be represented as a linear combination of the four matrices, but not as linear combination of any set with fewer than four members. (A basis must be complete in the sense that all elements (matrices) of interest can be represented as linear combinations of the members of the set and economical in the sense that every member is necessary. If even one member of the set is removed then there will be at least one 2 x 2 matrix that cannot be represented as a linear combination of the remaining members.) Why are the Pauli matrices introduced? The Pauli matrices are introduced to act on two-row column matrices. Matrices of the forms: a 1 0 b , 0 and 1 . The column vectors 1 0 0 and 1 are a basis set for all the a b . The column 1 0 vectors and are to be called UP () and DOWN (). 0 1 Physics Handout Series – fields.tank page 38 Fields Test Example Questions Guide If you are not familiar with modern physics, skip this paragraph. In quantum mechanics, the Pauli matrices add the flexibility that allows a wavefunction to represent the spin character of an electron. A two row column vector is appended to a function of position and time. a (r , t ) b 1 The column vector represents spin up while 0 0 1 represents spin down. The spin part is to be a a a * b * a normalized independently so: b a a * bb* 1 . b b † Special properties of the Pauli matrices: 1 1 = 2 2 = 3 3 = The determinants are each of the three Pauli matrices is – 1. The determinant of the identity is, of course, equal to one. | 1| = | 2| = | 3|= 1 Actions of the Pauli Set: x The operator x y z 0 0 1 0 1 1 = 1 0 1 0 1 0 = 0 i 1 0 i 0 0 i 1 ; y 0 1 = 0 i 0 1 i 0 1 i 0 lowers the UP state to i times the DOWN and raises the DOWN state to –i times UP. z The operator x lowers the UP state to DOWN and raises the DOWN state to UP. y The operator 1 0 1 1 0 0 = 1 0 0 1 ; 1 0 = 1 0 1 1 0 1 0 (1) 0 ; z 1 0 = 1 0 0 0 0 1 1 (1) 1 on the UP state to 1 times the UP and, on the DOWN state, returns – 1 times the DOWN state. The operator z measures the up or down character of the state. Physics Handout Series – fields.tank page 39 Fields Test Example Questions Guide The combination of operators + The operator + + = x +i y annihilates 1 0 2 1 0 0 = 0 0 0 0 ; + UP and raises DOWN to 2 times UP. 0 0 2 0 1 1 = 0 0 1 2 0 is called the raising operator. Exercise: Find the action of - = x +i y on the states UP and DOWN. Propose a name for -. ** The Pauli matrices are multiplied by ½ when they are to be used in anger. They become the operators for the components of the electron’s spin angular momentum. z =½ z 1 0 = ½ ; 0 1 The eigenvalues of z 2 =(½ )2 [ 2 x + y 2 2 z ] + are ½ , and the eigenvalue of 2 1 0 = 3/4 2 0 1 is 3/4 2 = s(s + 1)2 (28) Counting Statistics We will assume that the quick and dirty answers are enough to fake our way through the problem. When a counting experiment yields N counts, the uncertainty in that number is N ½ (or a fractional uncertainty of N -½ large count numbers are relatively more precise.). The value N ½ is the expected standard deviation. If the counting experiment is repeated many times, the expected results are assumed to follow a standard distribution. 68% of the time the result is within 95% of the time the result is within 2 99.7% of the time the result is within 3 The average count number is N , and we assume the standard deviation is N ½. ** Know: 68% with and 95% within p ( n) Physics Handout Series – fields.tank ( n N ) ]2 1 e½[ 2 page 40 Fields Test Example Questions Guide Ex. 28) An experimenter measures the counting rate from a radioactive source as 10,150 counts in 100 minutes. Without changing any conditions, the experimenter counts for one minute. There is a probability of about 15% that the number of counts recorded will be less than (A) 50 (B) 70 (C) 90 (D) 100 (E) 110 The first line established the average count rate as 101.5 per minute. We expect 101.5 counts in one minute on average. The standard deviation of the count number in one minute is assumed to be the square root of that value or 10.075. There is a 68% probability that the counts falls between 101.5 – 10.075 and 101.5 + 10.075 or a 32% chance that the number of counts is outside this range. The counts can be high or low, so there is a 16% chance that the number of counts in a minute is less than 101.5 – 10.075 = 91.42. about 15% likelihood to be less than 90. We expect 101.5 counts means that if the one minute count were repeated a large number of times that various integer results would be recorded each time. The results 101 and 102 would occur about equally, and larger and smaller values would occur at a lower frequency. The values around N occurring about 0.6times as often as those around N . p ( n) ( n N ) ]2 1 , e-½ 0.6. e½[ 2 (29) Graphs and Basic Lab Ex.29) A motion sensor is used to measure the position x versus time t for a cart traveling down a ramp. A spreadsheet is then used to make a linear fit to the plot of x vs. t, as shown in the graph to the left. The equation for the best fit line 2 appears on the graph. Which of the following gives the acceleration of the cart? x(t) = 0.3190 t + 0.0013 Physics Handout Series – fields.tank page 41 Fields Test Example Questions Guide This one is a gift! The acceleration is the second derivative of x(t) with respect to t or 0.6380. Adding dimensions, x(t) = 0.3190 m/s 2 t 2 + 0.0013 m yielding a = 0.6380 m/s 2. Finally, everyone knows that we could not have gotten data that was accurate to even two significant figures. Whenever possible, work from the definition. You are always correct to do so. (30) Lagrangian and Hamiltonian Mechanics: We restrict our treatment to the simple cases for which the constraints are good, the potentials are time independent and for which there is no dissipation. The Lagrangian L is T – U, the kinetic energy minus the potential energy. The Lagrangian is a natural function of the coordinates qi, the coordinate velocities qi and time t. The Lagrange equations of motion are: d L dt qi L qi The signs and details can be recovered by considering: L( x, x, t ) ½ m x 2 V ( x) . d L L V Fx m x dt x x x (recovered Newton’s 2nd Law) It is crucial to note that the time derivative is a total derivative while the others are partial derivatives. The momentum conjugate (associated with) a coordinate is found as: pi L . Using the example qi L( x, x, t ) ½ m x 2 V ( x) , px m x . Another familiar result is p m r 2 which appears for the planar orbit problem for which L(r , , r , ) ½ m[r 2 r 2 2 ] V (r ) . Conservation or constants of the motion: pi L qi d L dt qi L qi L d pi dt qi Physics Handout Series – fields.tank page 42 Fields Test Example Questions Guide The momentum pi is a constant of the motion if the Lagrangian does not depend on the coordinate conjugate to that momentum. Given L(r , , r , ) ½ m[r 2 r 2 2 ] V (r ) , there is no dependence of the angular coordinate so the conjugate momentum p m r 2 is a constant of the motion. In Hamiltonian mechanics, the coordinates and the momenta are the natural variables. The Hamiltonian is defined to be H = pi qi - L. Here pi is the momentum conjugate to qi as defined i based on the Lagrangian above. For the simple case, L( x, x, t ) ½ m x 2 V ( x) , we found px m x . H(x, px) = px x (½ m x 2 V ( x)) px x ½ m x 2 V ( x ) . We find: 2 H ( x, px ) ½ m x 2 V ( x) px 2 m V ( x) The Hamiltonian is to be expressed as a function of the coordinates and momenta, so the second form is the one that must be used. For time-independent problems of the type usually encountered, the Hamiltonian represents the total energy of the system. The equations of motion are: qi qi H pi 2 H H and pi . Using the example: H ( x, px ) px 2 m V ( x) , pi qi x px H and pi qi m Constants of the Motion: The relation pi mx V Fx x H shows that pi is a constant of the motion of the qi Hamiltonian is independent of qi. (Recall that we are restricting our attention to Hamiltonians that do not depend explicitly on time.) Ex. 30) Two masses m1 and m2 on a horizontal straight frictionless track are connected by a spring of spring constant k, as shown in the figure to the left. The spring is initially at its equilibrium length. If x1 and x2 give the displacements of the masses from their equilibrium positions, the Lagrangian L for the system is given by which of the following? The dot denotes differentiation with respect to time. Physics Handout Series – fields.tank page 43 Fields Test Example Questions Guide The Lagrangian is to be the kinetic energy minus the potential energy. The kinetic energy is the sum of the kinetic energies of the two particles, T (or K) = ½ m1 x12 ½ m2 x22 , and the potential energy is ½ times the spring constant times the stretch of the spring squared, U = ½ k (x2 – x1)2. Note that it helps to roll the words around in your mind as you attempt to decipher a problem. Conclusion: L( x1 , x2 , x1 , x2 ) ½ m1 x12 ½ m2 x22 ½ k ( x2 x1 ) 2 Note that four of the candidates identify the kinetic energy as: ½ m1 x12 ½ m2 x22 , and three identify the potential as ½ k (x2 – x1)2. Only two have the negative sign for L = T – U. Possibility (E) includes the awe-inspiring reduced mass; it has the wrong sign for the potential. In addition, (E) depends only on the relative position and relative velocity. That is: the center of mass motion has been suppressed. For example, if q = x2 – x1, then (E) is ½ q 2 + ½ k q2. Forgetting that it should be ½ q 2 - ½ k q2, there is the issue of the kinetic energy. ½ m1 x12 ½ m2 x22 = ½ (m1 m2 ) X 2 + ½ q 2 where X is the center of mass velocity. Kinetic energy in terms of the center of mass and relative coordinates. Let V be the velocity of the center of mass of a system of particles, vi be the velocity of particle i and ui be the velocity of particle i relative to the center of mass. Ktotal KCM K relative or ½ m v i 2 i ½ M V 2 ½ mi ui2 i i For a system with only two particles, the reduced mass can be used to get the special form: 2 ½ m1 v12 ½ m2 v22 ½ M V 2 ½ | v2 v1 |2 ½ M V 2 ½ vrel where M = m1 + m2, m1 m2 and vrel v2 v1 m1 m2 (31) Nuclear Decays Physics Handout Series – fields.tank page 44 Fields Test Example Questions Guide Each nucleus has Z protons (aka atomic number) and N neutrons with a mass number of A =Z+N. A particular nucleus is denoted A Z X N ; such as 146C8 , however the Z and N are usually omitted because they are redundant and are known from A and the symbol used for X. particle nucleus AX photon alpha beta minus - or ebeta plus+ or e+ Electron-neutrino ve electron-antineutrino ve charge Z 0 2 -1 +1 0 nucleon # A 0 4 0 0 0 lepton # 0 0 0 +1 -1 +1 0 0 -1 Alpha particle is a tightly bound system of 2 protons and 2 neutrons (A=4). The very high binding energy (~28 MeV) of the alpha and its very high first internal excited state make it favorable to be treated as a unit object. (The same could be said for the proton and neutron which in reality are quark composites.) Nuclei will spontaneously decay into another nucleus if the final system has lower rest-masss energy than the initial. There are four decay modes; gamma decay, beta decay, alpha decay, and fission. The conserved quantities are total energy, linear and angular momentum, along with nucleon number and lepton number. The rest-mass energy difference between the initial and final systems is called the Q-value and appears as kinetic energy and internal excitation energy of the products. In gamma decay, an excited nucleus decreases its level of excitation by emitting a gamma ray (photon)The process may be denoted: A(X*) AX + 0 or just AX + The number of neutrons and protons is unchanged. No leptons are involved so lepton number need not be considered. In alpha decay, a nucleus emits an alpha particle (2 protons and 2 nucleons for a total of 4 nucleons). The final state nucleus, the daughter, has Z–2 protons, N-2 neutrons and A-4 nucleons. This can be denoted A Z X A 4 Z 2 Y + 42 or just A-4Y + . No leptons are involved so lepton number need not be considered. Physics Handout Series – fields.tank page 45 Fields Test Example Questions Guide There are three fundamental processes in beta decay. For convenience they can be considered to be (with Q-values given). p e ve 0.78 MeV beta minus decay n beta plus decay p n e ve x.xx MeV electron capture p e n ve x.xx MeV Beta minus decay can occur to an isolated neutron, but because the Q-values for the remaining two processes are negative, the reactions can only occur in regions where the proton is ‘disturbed’; that is, in a nucleus. Electron capture occurs when a ‘proton grabs an electron out of its s-state orbit’. Note that the reactions above conserve charge, nucleon number, and lepton number. For nuclei undergoing beta decay, the reactions would be written, beta minus decay A Z X A Z 1 Y + 01 e or just beta plus decay A Z X A Z 1 Y + ee electron capture A Z X A Z 1 A Z 1 Y + e. + ve Y + e Note that the reactions above conserve charge, nucleon number, and lepton number. The existence of three-bodies in the final state means that the available energy or Q value of the reaction is not apportioned to the particles in a set ratio, but rather the particles of each type emerge with a distribution of energies. In fission, the nucleus breaks into two large pieces (fission fragments) and a few neutrons with Qvalues on the order of 200 MeV. A commonly used radioactive fission source is 252Cf. The nuclei 235 U and 239Pu do not themselves spontaneously fission, but fission can be induced by striking them with a neutron. The conserved quantities are charge, total energy, linear and angular momentum, along with nucleon number and lepton number *************** WORDS NOT USED: Alpha particles have a nuclear binding energy of 28.3 MeV or about 7.1 MeV per nucleon. The relatively high binding energy of the alpha makes it favorable to be ejected as a unit. The binding Physics Handout Series – fields.tank page 46 Fields Test Example Questions Guide energy per nucleon increases form 7.1 MeV for the alpha (He4) to a maximum of 8.8 MeV for Fe56. Nuclei lighter than iron-56 are candidates for fusion; those heavier are candidates for fission. Heavier nuclei have a larger ratio of neutrons to protons than light nuclei so after a sequence of a few alpha decays, the resulting daughter is neutron rich compared to the stable nuclei of that atomic number. At this point the nucleus needs to reduce the number of neutrons and increase the number of protons. Beta decay does it. The net change in the nucleus is that one neutron has changed into a proton. The appearance of an electron means that the lepton number has grown by one. STOP: Lepton number is conserved. The leptons to consider are the electron and the electron neutrino e plus the positron e+ and the electron anti-neutrino ve . The first two are particles and have lepton number +1 while the final two are antiparticles with lepton number -1. (The e- and e+ are also called the beta minus and beta plus.) After conserving lepton number, the decay becomes: XA Z+1YA + e ve which for lepton number is: 0 0 + (+1)(-1) Z The existence of three-bodies in the final state mean that the available energy or Q value of the reaction is not apportioned to the particles in a set ratio, but rather the particles of each type emerge with a distribution of energies. For electron capture by a nucleus: XA + e- Z-1YA + e Z For a beta plus decay: XA Z-1YA + ee Z The basic beta decay process is: n 1p1 + e- + ve 0 1 The neutron has a rest energy of 939.573 MeV; the proton has 938.280 MeV; the electron has 0.511 MeV; the neutrino has zero rest mass and therefore rest energy. The energetics of the reaction are: 939.573 MeV 938.280 MeV + 0.511 MeV + Q The Q value is 0.782 MeV so the electrons are emitted with a distribution of kinetic energies ranging up to almost 0.782 MeV. Similarly, the neutrinos have an energy distribution ranging up to almost Physics Handout Series – fields.tank page 47 Fields Test Example Questions Guide 0.782 MeV. The lifetime of a free neutron is about 886 seconds. Neutrons inside a nucleus have a binding energy of about 8 MeV that stabilizes them against decay. ******** Ex. 31) Thorium with atomic mass 228, decays by alpha emission to a daughter nucleus which also decays by alpha decay to radon. Which of the following is true of the decay product, radon? (The atomic number of thorium is 90. Atomic Mass 220 220 224 224 228 (A) (B) (C) (D) (E) Atomic Number 82 86 82 88 91 The process is a sequence of two alpha decays: A Z X A 4 Z 2 Y + 42 A 8 Z 4 Y + 42 + 42 Atomic Mass: A = 228 Afinal – 4– 4 Atomic Number: Z = 90 Zfinal = 90 – 4 = 86 228 Th 224Ra + 220Rn + + Thorium Radium + alpha Radon + alpha + alpha (31) Spontaneous Elementary Particle Decays: As with other decays, the electric charge, energy, linear and angular momentum must be conserved. Because there is only one initial particle, the Q-value for the reaction must be positive. The excess energy or Q appears as kinetic energy and excitation energy of the products. Anti-particles have the same rest energy as their corresponding particles. Charge is conserved. mpc2 = 939.573 MeV * mec2 = 0.511 MeV * mpc2 = 939.6 MeV * mc2 = 106 MeV * mc2 = 1775 MeV Physics Handout Series – fields.tank mc2 = 0 * page 48 Fields Test Example Questions Guide mc2 = 140 MeV charged; 135 MeV neutral mc2 = 0 * Be aware of the values marked * and that a pion is a little more massive than a muon. Ex. 32) Which of the following decays is possible in a vacuum? (A) + - + (- + + e- + e+ (C)0 e- + p (D)p n + e+ + e (E)n p + e- + e (A) Fails. Does not conserve charge. Q-value > 0 is OK. (B) Fails for two reasons. Does not conserve charge. Does not conserve energy because Q<0, that is, the products have more mass than the initial particle. (C) Fails. Conserves charge, however the products have much more massive than the pion. (D) Fails. Conserves charge, however, the neutron is slightly more massive than the proton. (E) Possible. Conserves charge and the Q>0. An isolated neutron is unstable and decays as shown with a lifetime of 886 seconds. Neutrons are stabilized in the nucleus by their binding energy which is greater than the Q of decay (E). Because the fundamental building blocks are the 6 quarks (u,d,s,c,b,t) and 6 leptons (x,x,x,x,x,x) there are actually additional conservation rules. These are not as trivial to apply because one must be told the quark content of the composite particles. **** Not used below There are three families of leptons (electron, muon, tau) each with its anti-particle and neutrino/antineutrino pair). The muonic family group is {-, +, the anti-particle, the and the anti-neutrino ) with muon-lepton numbers {1,-1,1,-1}. Lepton number is conserved separately for each family. For the decay to proceed spontaneously, the total rest energy of the products must be less than that of the initial state particles. *** Physics Handout Series – fields.tank page 49 Fields Test Example Questions Guide Particles and Intrinsic Spin Intrinsic spin S of most objects (electron, proton, neutron, quarks, leptons) is described by the label (quantum number) s = ½ . The length is the intrinsic spin vector is calculated as s(s 1) with possible z-components of ms , where ms can have the values + ½ or – ½ . These z-components are often referred to as spin-up and spin-down, respectively. Particles with half-integer spins are fermions, particle that obey the Pauli exclusion principle, only one particle per each distinct quantum state. The pattern of electrons filling atomic states is an example of state filling by fermions. The building blocks of matter are typically fermions. Particles with integer-valued intrinsic spin are bosons which can, and in fact prefer, to multiply occupy states. Examples of bosons include pions, photons, gluons and gravitons. Many bosons are associated with forces or interactions. Those with zero rest mass give rise to forces with infinite range (1/r2) forces such as the Coulomb and gravitational forces. Quantum states with many bosons in the same or nearby states gives the classical behavior of electromagnetic waves, etc. Interactions exchanging massive bosons decrease exponentially for large separations. Mediator pion gluon photon graviton Spin 0 1 1 2 Interacting Pair nucleon-nucleon quark-quark charge-charge mass-mass Composite Particles: A group of particles that is bound together acts like a particle of the net spin as long as it is probed only weakly compared to its binding energy. Pairs of quarks can bind to form mesons (bosons, ½ and ½ integer value). Triplets of quarks for fermions such as protons and neutrons (fermion, ½ and ½ and ½ ½ integer value). Effective bosons such as helium atoms (2 protons + 2 neutrons + 2 electrons integer) can collect in large numbers in a single (ground) state, the condensate, leading to the superfluid state of liquid helium below 2.2 K. Electrons bound in pairs (bosons) form similar condensates in super-conductors. A nucleus can be modeled as a composite particle if it is probed weakly using say electric and magnetic fields. If you shoot a 40 MeV proton at it, the probe will see the inner structure invalidating the composite particle model. Physics Handout Series – fields.tank page 50 Fields Test Example Questions Guide Ex. 33) For atoms in a simple cubic structure, the maximum percentage of the total available volume that can be occupied by the atoms is approximately (A) ( (C) (D)60% (E)70% The idea is to stack solid spheres in the prescribed pattern and to compute the fraction of the total volume that is in the spheres. A cube has eight corners and a side length s. A sphere of radius ½ s can be centered on each corner. One eighth of each sphere will be in the cube times eight corner or one sphere of radius ½ s per cube of side s. (Prepare a sketch!) 4 (½ s)3 4 fraction ½ 0.52 3 3 s 3 8 3 The closest packing of hard spheres (74%) is achieved in the hexagonal close-packed and facecentered cubic configurations. Body-centered cubic is about 68 %. simple cublic body-centered cubic face-centered cubic hexangonal c.p. MISCELLANEOUS TOPICS: (34) Astro-Knowledge Physics Handout Series – fields.tank page 51 Fields Test Example Questions Guide Ex. 34) Which of the following gives the distance to Andromeda (M31), the spiral galaxy nearest to the Milky Way? (A) 2 x 100 (2 x 102 (C)2 x 104 (D)2 x 106 (E)2 x 108 The earth is 8 light minutes from the sun. The sun is 4 light years from the nearest star. Galaxies have lots of stars so Galaxies should be thousands of light years in dimension. During a 1965 episode, science officer Spock remarked that it was 40,000 light years across the Milky Way (our galaxy). Google says that the main disk of the Milky Way is about 105 light years in diameter. Galaxies should be separated by 10 to 100 galaxy diameters. Look for 10-100 x 105 (D) The universe is 13-15 x 109 years of age and so is about 13-15 x 109 light years in dimension. The separation of galaxies should be small compared to the extent of the universe. More ASTRO facts: (35) Math Methods Ex. 35) Which of the following is the derivative with respect to x of the function x2 cos[3 x4 + 1]? The function is the product of x2 and cos[3 x4 + 1] so by the product rule, df /dx = 2x cos[3 x4 + 1] + x2 d/dx(cos[3 x4 + 1]) Using the chain rule, d/dx(cos[3 x4 + 1]) = - sin[3 x4 + 1] (12 x3) Combining terms: df/dx = 2x cos[3 x4 + 1] – 12 x5 sin[3 x4 + 1] All students should be intimately familiar with every concept and detail presented in SP351/2. Physics Handout Series – fields.tank page 52 Fields Test Example Questions Guide TO ADD Kepler’s Laws Astro star cycle sizes and times create heavy elements Physics Handout Series – fields.tank page 53