Final Review. A Puzzle... Special Relativity. Direction of the Force. Moving at the Speed of Light

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1 Final Review A Puzzle... Diretion of the Fore A point harge q is loated a fixed height h above an infinite horizontal onduting plane. Another point harge q is loated a height z (with z > h) above the plane. The two harges lies on the same vertial line. If z is only slightly larger than h, then the fore on the top harge is learly upward. But for larger values of z, is the fore still always upward? Hint: Try to solve this without doing any alulations. Think dipole. There are two negative image harges on the other side of the plane, at the mirror-image loations. For very large z values of the top harge q, the lower q and its image harge -q look like a dipole from afar, whih has a repulsive (upward) field that falls off like z 3. But the attrative (downward) field from the other image harge -q behaves like (2 z) 2. This has a smaller power of z in the denominator, so it dominates for large z. The fore on the top harge q is therefore downward for large z. So the answer to the stated question is "No." Moving at the Speed of Light One of the interesting quirks about the veloity addition formula is that if you start off moving at in one frame, then you move in in another frame. This begs some interesting questions, suh as what happens if you aelerate a ar to the speed of light, and you turn on your headlights. Would the light move at speed relative to you, would it all pool inside of the headlight, or would something altogether different happen? Mihael Stevens has an amazing YouTube video analyzing this very question. Speial Relativity The following problems all deal with the setup shown below. Two trains, A and B, eah have proper length L and move in the same diretion. A s speed is 4, and B s speed is 3. A starts behind B. Printed by Wolfram Mathematia Student Edition

2 2 Leture 8 - Final Review.nb We define the following two events, Event E : "The front of A passing the bak ofb" Event E 2 : "The bak of A passing the front ofb" () Length Contration/Time Dilation What is the differene in time Δt C and spae Δx C between E and E 2, as viewed by person C on the ground? Relative to C on the ground, the γ fators assoiated with A and B are Therefore, their lengths in the ground frame are γ A (C frame) = γ B (C frame) = /2 = 3 (2) /2 = 4 (3) L (C L frame) A = γ(c frame) = 3 A L (4) (C L frame) L B = γ(c frame) = 4 B L () While overtaking B, A must travel farther than B, by an exess distane equal to the sum of the lengths of the (C trains, whih is L frame) (C A + L frame) B = 3 L + 4 L = 7 L. The relative speed of the two trains (as viewed by C on the ground) is the differene of the speeds, whih is. The total time is therefore Δt C = 7 L = 7 L (6) The distane between both events equals the distane that A travels minus the length of A (sine E happens at the front of A and E 2 happens at the rear of A). Therefore for the two events. Veloity Addition Δx C = Δt C 4-3 L = L (7). Find Δt B, Δx B between E and E 2, as viewed by a stationary observer in B s frame? Repeat for Δt A, Δx A for a stationary observer in A s frame. 2. Person D walks at onstant speed from the bak of train B to its front, suh that he oinides with both events E and E 2. Compute Δt D, Δx D.. In B s referene frame, A moves at a slower speed found by the veloity addition formula Printed by Wolfram Mathematia Student Edition

3 Leture 8 - Final Review.nb 3 with an assoiated γ fator whih implies that A has a length while B has its rest length u A (B frame) = 4-3 γ A (B frame) = L A (B frame) = Therefore the time between events E and E 2 equals The distane between the two events is simply - 4 Δt B = L A (B frame) +L B (B frame) 3 = /2 = 3 (8) 2 (9) L γ(b frame) = 2 A 3 L (0) L B (B frame) = L () u(b frame) = 2 3 L = L A 3 (2) Δx B = L (3) It is straightforward to repeat all of these alulations in A s frame. Everything will be symmetri (exept with everything now moving in the opposite diretion), so that Δt A = L (4) Δx A = -L () 2. In B s frame, D must walk with speed Δx B = in order to oinides with both events. In D s referene frame, Δt B A will have veloity and B will have veloity u A (D frame) = u B (D frame) = = (6) = - (7) (D In hindsight, it makes sense that u frame) (D A = -u frame) B beause of the symmetry of the problem (i.e. that D oinides with both events and that A and B both have proper length L). The γ fator and length of A will be (D γ frame) A = = - 2 /2 2 6 (8) L A (D frame) = and, by symmetry, this must equal the length of B, Therefore, Lastly, sine D oinides with both events, Δt D = L A (D frame) +L B (D frame) L γ(d frame) = 2 6 L A (9) L B (D frame) = 2 6 L (20) u(d frame) A -u(d frame) = 2 6 L = 2 L 6 B (2) Δx B = 0 (22) Note: There are several double heks we an perform. For example, the speed of D with respet to the ground frame (i.e. C frame) an be obtained by relativistially adding 3 and or subtrating from 4. These both give the same answer, namely, as they must. The γ fator between D and the ground is therefore 7 (C γ frame) D = = /2 2 6 (23) We an now use time dilation to say that someone on the ground sees the overtaking take a time of Printed by Wolfram Mathematia Student Edition

4 4 Leture 8 - Final Review.nb Δt C = Δt D γ D (C frame) = 2 L 6 Likewise, the γ fator between D and either train equals say ground overtaking 7 = 7 L, in agreement with equation (6) from the previous problem. 2 6 γ D (A frame) = γ D (B frame) = = - 2 /2 2 6 (24) Therefore, the time between events as viewed in A or B equals Δt A = Δt B = Δt D γ D (A frame) = 2 L 6 agreement with equations (2) and (4) above. = L in 2 6 Note that we annot use simple time dilation to relate the ground to A or B, beause the two events don t happen at the same plae in the train frames! But sine both events happen at the same plae in D s frame, namely right at D, we an indeed use time dilation to go from D s frame to any other frame. Lorentz Transformations Verify that the values of Δx and Δt found in the above problems satisfy the Lorentz transformations between the six pairs of frames, namely AB, AC, AD, BC, BD, CD. Gathering together all of the values found above, Δt A B C D L L 7 L 2 L 6 Δx -L L L 0 We also list the various veloities and γ fators from the above examples between eah pair of frames The Lorentz transformations are AB AC AD BC BD CD v 3 3 γ Δx = γ (Δx' + v Δt) (2) Δt = γ Δt' + v Δx' 2 (26) For eah of the six pairs, we ll transform from the faster frame to the slower frame. This means that the oordinates of the faster frame will be on the right-hand side of the Lorentz transformations. The sign on the right-hand side of the Lorentz transformations will therefore always be a +. In the AB ase, for example, we will write, Frames B and A, in that order, to signify that the B oordinates are on the left-hand side, and the A oordinates are on Printed by Wolfram Mathematia Student Edition

5 Leture 8 - Final Review.nb the right-hand side. We ll simply list the Lorentz transformations for the six ases, and you an hek that they do indeed all work out. Frames B and A Frames C and A Frames D and A Frames C and B Frames B and D Frames C and D Eletrostatis L = 3 2 -L + 3 L L = 3 2 L + 3 (-L) 2 L = 3 -L + 4 L 7 L = 3 L + 4 (-L) 2 0 = 2 6 -L + L 2 L 6 = L ( ) (-L) 2 L = 4 L + 3 L 7 L = 4 L + 3 (L) 2 L = L 6 L = L 6 + ( ) (0) 2 L = L 6 7 L = 7 2 L (0) 2 (27) (28) (29) (30) (3) (32) Sphere and Cones (a) Consider a fixed hollow spherial shell with radius R and surfae harge density σ. A partile with mass m and harge -q that is initially at rest falls in from infinity. What is its speed when it reahes the enter of the shell? (Assume that a tiny hole has been ut in the shell, to let the harge through.) (b) Consider two fixed hollow onial shells (that is, ie ream ones without the ie ream) with base radius R, slant height L, and surfae harge density σ, arranged as shown in figure (b) below. A partile with mass m and harge -q that is initially at rest falls in from infinity, along the perpendiular bisetor line, as shown. What is its speed when it reahes the tip of the ones? (The answers to both parts of this problem should relate very niely!) Printed by Wolfram Mathematia Student Edition

6 6 Leture 8 - Final Review.nb s (a) This problem is terrifially simple if we use the eletri potential. Reall that for a spherial shell, the potential equals ϕ[r] = k Q where Q = 4 π R 2 σ for r R. Inside the sphere, the eletri field is zero so that the potential must r be a onstant; by ontinuity, the eletri potential inside the shell equals ϕ[r] = k Q for r R. R Therefore the work required to bring a partile from to the enter of the spherial shell equals -q ϕ[0], so that the partiles energy due to free fall would be 2 m v2 = q ϕ[0] = k q Q or equivalently R v = 2 k q Q /2 = 2 q R σ R ϵ 0 /2 (33) We ould also do this (more painfully) by integrating the eletri field and solving the orresponding differential equation. The eletri field for r R equals E = k Q r so that the fore aelerating the partile radially inward r 2 satisfies m v dv dr = m r = - k Q q r (34) 2 where we have used the relation d2 r = r = dr dr dt 2 dr dv = dv v. Separating the variables and integrating t differential dr equation above, v 0 m v dv = r - k Q q dr r (3) 2 2 m v2 = k Q q r (36) This formula is valid from r = until r = R, at whih point the eletri field beomes zero and the veloity remains onstant until the partile hits the enter, moving at speed 2 m v2 = k Q q R (37) The solution proeeds as above. (b) Sine ones may seem like foreign objets, let start by first integrating over the surfae area of the one to make sure that we orretly find its surfae area (the surfae area of a one is π R L, as per Wikipedia). We will integrate in rings from z = 0 to z = H = L 2 - R 2, as shown below. Printed by Wolfram Mathematia Student Edition

7 Leture 8 - Final Review.nb 7 The surfae area of a ring between z and z + d z is given by (surfae area of ring between z and z + dz) = 2 π (radius of ring) (width of ring) (38) Noting the similar triangles, the radius of a ring at height z equals z R. The vertial height of the ring is dz, but H we want to know the slant width (i.e. the diagonal length) of the ribbon. Sine the ring goes vertially up by d z and outward by R H dz, the slant width is + R H 2 /2 d z. Thus (surfae area of ring between z and z + dz) = 2 π z R H + R H 2 /2 dz (39) and the surfae area of the entire one equals surfae area = 0 H 2 π z R H + R H 2 /2 dz = π R (H 2 + R 2 ) /2 = π R L as desired. Having onfirmed this neat fat, let us return to the problem. The area element of a one is given by Equation (40), sine we just showed that this integral overs the full surfae of the one and yields the orret surfae area formula. The potential at the base of the two ones, ϕ[0], equals twie the ontribution from a single one, written in the familiar form k dq, equals r H ϕ[0] = 2 k 2 π z R H + R /2 H 2 dz σ 0 z + R H 2 /2 = 4 π k σ R H 0 H dz = σ R ϵ 0 Sine this has the exat same form as the potential ϕ[0] = k Q R = σ R ϵ 0 we found in Part (a), the solution proeeds as above, and we one again find the same veloity v = 2 q R σ /2. What a remarkable oinidene! ϵ 0 Condutors (40) (4) Image Charges for Two Planes This setion investigates the following problem very deeply! A point harge q is loated between two parallel infinite onduting planes, a distane b from one and l - b from the other. Where should image harges be loated so that the eletri field is everywhere perpendiular to the Printed by Wolfram Mathematia Student Edition

8 8 Leture 8 - Final Review.nb planes? image harges everywhere perpendiular Set one plate at z = 0 and one at z = l, and let the positron be at z = b. The problem is learly one dimensional, so we fix x = y = 0 for all points we disuss. If we just onsider the bottom plate, we an plae a negative harge at -b, but then we need to take are of both of these harges with images beyond the top plate with a negative harge at z = 2 l - b and a positive harge at z = 2 l + b. In other words, there will be a asade of image harges (of alternating signs!) In the figure below, the two given planes are indiated by the bold lines, and the given real harge is labeled R. It turns out that we will need an infinite number of image harges, as shown. Solid dots are positive, hollow dots are negative (assuming the given real harge is positive). The following Manipulate shows the real harge q (in blak) together with the positive image harges (blue) and negative image harges (orange). The real onduting planes are in the middle and are darker than the other planes at z = -l, ±2 l, ±3 l... b Show E Out[3]= Real harge Q Image harge Q Image harge -Q The pattern learly emerges: Printed by Wolfram Mathematia Student Edition

9 Leture 8 - Final Review.nb 9 Positive Charges Negative Charges l + b 4 l - b 2 l + b 2 l - b 0 l + b 0 l - b -2 l + b -2 l - b -4 l + b -4 l - b If you want, you an group the harges into two sets the odds and evens, as indiated by the onneting lines in the figure above. Eah odd harge orrets the effet of the previous odd harge, with respet to alternating planes. Likewise for the evens. In the speial ase where the given real harge is loated midway between the two planes, all the image harges are similarly loated midway between the (imaginary) planes in the figure above. So the net fore on the given harge is zero, as it should be. The Limit b l In the limit b l when the harge q is very lose to one of the plates, find an approximate expression for the fore on the harge. To zeroth order, the fore on the harge q omes solely from the image harge nearby at z = -b, whih is a fore F k q2 (2 b) 2 (43) downwards. But we an do muh better than that! All the remaining pairs of harges an be approximated as dipoles; two of these dipoles are at a distane 2 l, another two are at 4 l, and so on. Reall the formula 2 k q d for the r 3 eletri field of a dipole along its axis (where d is the distane between the two harges and r is the distane from the dipole s enter). Note that every dipole pushes the harge q in the +z-diretion, the total downwards fore on the dipole equals F = k q2-2 {2 k q 2 (2 b)} (2 b) 2 = k q2 4 b 2 - k q2 b l (2 l) 3 (4 l) 3 (6 l) 3 + where the first fator of 2 in the seond term omes from the fat that there are two pairs of dipoles at distane 2 l, 4 l... (on the left and right side of the harge). The fator in parenthesis equals Zeta[3].202 (44) Sum, {n,, } 3 n Zeta[3] You an also alulate the total fore by looking at the fores from the positive and negative image harges separately. From the figure above, the fore on the real harge q from the other positive harges will always be 0 by symmetry, but the downwards fore from the negative harges equals F = k q2 - k q 2 (2 b) 2 n= = k q2 - k q2 4 b 2 4 l 2 n= n 2 - (2 n l-2 b) 2 (2 n l+2 b) b n l 2 + b n l 2 k q2 - k q2 4 b 2 4 l 2 n= + 2 b n 2 n l b = k q2 - k q2 b 4 b 2 l 3 n= n 3 n l (4) Printed by Wolfram Mathematia Student Edition

10 0 Leture 8 - Final Review.nb where in the third step we used the Taylor series - 2 ϵ + O[ϵ] 2. This result mathes equation (44), as (+ϵ) 2 expeted. Ciruits Two Light Bulbs Certain light bulbs an be treated as resistors, with the brightness of the bulb proportional to the power dissipated in the bulb s resistor. (a) Two light bulbs are onneted in parallel, and then onneted to a battery, as shown in figure (a). You observe that bulb is twie as bright as bulb 2. Whih bulb s resistor is larger, and by what fator? (b) The bulbs are now onneted in series, as shown in figure (b). Whih bulb is brighter, and by what fator? How bright is eah bulb ompared with bulb in Part (a)? (a) The power dissipated takes the form V2. Both bulbs have the same voltage drop V, so if bulb is twie as bring R as bulb 2, it must have half the resistane, R 2 = 2 R. In parallel, the larger resistor is dimmer. (b) The power dissipated also takes the form I 2 R. Both bulbs now have the same urrent I, so if Bulb 2 has twie the resistane, as we found in Part (a), then it is twie as bright - the opposite of the ase in Part (a). In series, the larger resistor is brighter. We an also ompare the total power dissipated in eah ase. If the resistanes are R and 2 R, then in Part (a) the total power dissipated is V2 R + V2 2 R = 3 V2 2 R. In Part (b) the total power is I2 R + I 2 (2 R) = 3 I 2 R, where I = V. So the 3 R power is V2 3 R. This is 2 V2 of the power in Part (a). In units of 9 R, the power in Part (a) are and, while in Part (b) 2 they are 9 and 2 9. Attenuator Chain (a) Find the equivalent resistane between terminals A and B in the infinite ladder of resistors shown below. Hint: Call the input resistane R, and note that it will not be hanged by adding a new set of resistors to the front end of the hain to make it one unit longer. (b) Show that, if voltage V 0 is applied at the input to suh a hain, the voltage at suessive nodes dereases in a geometri series. What should the ratio of the resistors be so that the ladder is an attenuator that halves the voltage Printed by Wolfram Mathematia Student Edition

11 Leture 8 - Final Review.nb geometri voltage at every step? () Obviously a truly infinite ladder would not be pratial. Can you suggest a way to terminate it after a few setions without introduing any error in its attenuation? (a) If R is the effetive resistane of the infinite hain, then the hain is equivalent to the iruit shown below. Thus, R = R + R2 R R2 +R (46) whih implies R= R + R2 +4 R R2 2 /2 (47) where we have hosen the positive root. (b) To demonstrate the stated geometri series result, onsider four points A, A', B, B' that form a square somewhere within the iruit. Given the voltage V ' between A' and B', what is the voltage V between A and B? Let the urrent flowing towards A ' be I, so that (using the effetive resistane R of the infinite hain) the urrent splits into I = I2 = R2 R+R2 R R+R2 I (48) I (49) What will be the urrent between A and B? By symmetry, it must be V V' = I I = R2 R2 +R I I2, I so that (0) (As a quik hek, you an make sure that the voltage around the square loop going from A ' to A to B to B' is zero, as it must be.) This result is independent of where along the hain we pik the adjaent nodes, so the voltages Printed by Wolfram Mathematia Student Edition

12 2 Leture 8 - Final Review.nb be.) independent along pik adjaent voltages aross suessive nodes derease in a geometri series. If we want V = V' 2, then we must have R = R 2. Equation (47) then yields 2 R 2 - R = (R R R 2 ) /2 () (2 R 2 - R ) 2 = R R R 2 (2) R 2 = 2 R (3) If we instead wanted V (that is, the voltage hardly dereases), then we need R R V' 2, whih implies R R 2. On the other hand, if we want V (that is, the voltage dereases quikly), then we need R R V' 2, whih implies R R 2. These results make intuitive sense. () To terminate the ladder after any setion, without hanging its resistane from that of the infinite hain, we an simply onnet a single resistor R given by Equation (47) in parallel with the last R 2, beause this R mimis the rest of the infinite hain. Mathematia Initialization Printed by Wolfram Mathematia Student Edition