Problem Set 1: Solutions

Transcription

1 Uniersity of Alabama Department of Physics and Astronomy PH 253 / LeClair Fall 2010 Problem Set 1: Solutions 1. A classic paradox inoling length contraction and the relatiity of simultaneity is as follows: Suppose a runner moing at 0.75c carries a horizontal pole 15 m long toward a barn that is 10 m long. The barn has front and rear doors. An obserer on the ground can instantly and simultaneously open and close the two doors by remote control. When the runner and the pole are inside the barn, the ground obserer closes and then opens both doors so that the runner and pole are momentarily captured inside the barn and then proceed to exit the barn from the back door. Do both the runner and the ground obserer agree that the runner makes it safely through the barn? Solution: Gien the relatie elocity between the reference frames = 0.75c, we will hae to account for length contraction and the relatiity of simultaneity. One point to make clear: in the barn s reference frame, the doors close and then open immediately, and at the same time. They do not stay closed. From the point of iew of the person in the barn, the pole is moing toward them at elocity =0.75c, and its length is contracted by a factor γ = / Thus, instead of 15 m the person in the barn sees the pole as haing a length 15/γ 9.9 m, and obseres the runner to make it safely through. From the point of iew of the runner, the barn is in motion, and appears shortened by a factor γ to 10/γ 6.6 m. When do the doors close from the runner s point of iew? The runner will not see the doors close at the same time, since the runner is in motion relatie to the doors and they are spatially separated. If the person in the barn sees the doors closing with time delay t = 0 i.e., simultaneously, the runner sees a time delay t goerned by the Lorentz transformation: t = γ t + x 2 Here x is the separation between eents the front door closing and the back door closing in the barn s reference frame, i.e., the length of the barn L B. If we call the front door opening the first eent and the back door closing the second eent, then we are implying the runner s heading

2 is positie, and that makes the distance x negatie. This makes x negatie, which means that the runner sees the front door close first and then the back: t = γl B 37.8 ns 3 From the person in the barn s point of iew, he will see the runner reach the front door, then close it immediately. At that point, the runner also thinks the pole is at the front door, and that part of its 15 m length according to him is still sticking out of the back of the barn which is 6.6 m long according to the runner. The runner then has time t to get that much of the pole in through the back door before it closes. The runner doesn t need to get out of the barn completely remember, the doors will close and then open again immediately, the runner just can t be caught in the middle of either door while that happens. So, the runner must go the length of the pole minus the length of the barn before t passes. That length is: L p L B = 15 m 10 m γ 8.38 m 4 At elocity, going this length takes 8.38 m/0.75c 37.2 ns, leaing 0.6 ns to spare before the rear door closes. Thus, both the runner and the obserer in the barn agree that the runner makes it through. Thus, both obserers agree that the pole does not smash into the doors, but makes it through the barn safely. We hae only gien a quick discussion here, one can perform a much more rigorous analysis. For further discussion see, for example: And an applet for good measure: The links should be clickable. 2. A pilot is supposed to fly due east from A to B and then back again to A due west. The elocity of the plane in air is u and the elocity of the air with respect to the ground is. The distance between A and B is l and the plane s air speed u is constant.

3 a If =0 still air show that the time for the round trip is t o =2l/u. b Suppose that the air elocity is due east or west. Show that the time for a round trip is then t E = t o 1 2 / u 2 c Suppose the air elocity is due north or south. Show that the time for a round trip is then t N = t o 1 2 / u 2 d In parts b and c we must assume that <u. Why? Solution: Though this problem does not inole relatiity at all, classical problems inoling relatie elocities look strikingly similar. Find / Gien: We are to find the round-trip transit time for the plane with wind speed in arious directions. We are gien the plane s speed and the wind s speed and direction. Sketch: In part a we simply hae a trip of total distance 2l at speed u. In part b we hae the wind s elocity with and against that of the plane, and must add and subtract the two elocities to get the net elocity. In part c we hae the net elocity at a right angle to the wind, and must find it by ector decomposition. a u out u in b wind u + = u u + = u + out in c wind u 2 2 u u 2 2 u Releant equations: Since the situation inoles motion at constant elocity, we need only the relationship between distance l, net elocity, and time t: t = l/. Additionally, we will need the rules for addition of ectors.

4 Symbolic solution: In all three cases, we may find the time for the outward trip t out and inward trip t in separately and add them together to get the total round-trip time. For each leg, we diide the distance coered l by the net forward elocity. a For each leg of the trip we coer distance l at elocity : t out = l u 5 t in = l u t net = l u + l u = 2l u t o 7 b Suppose the wind elocity is due east. On the outward trip, we are going with the wind, so the net forward elocity is that of the plane plus that of the wind. On the inward trip, going westward, we are going against the wind, and the net forward elocity is the plane s elocity minus that of the wind. i Thus, l t out = u + l t in = u t net = l u + + l u = l u + l u u + u = 2lu u 2 2 = 2l 1 u 1 2 /u 2 = t o 1 2 /u c If the plane is to go forward with a side-to-side wind, the pilot must steer into the wind as shown in the sketch. This will reduce the net forward progress, and the net forward elocity is gien by the ector diagram in the sketch, u 2 2. The elocity is the same in both outward and inward trips, so we may just calculate one of the trips and double the result: t net = 2t out = 2l u 2 2 = 2l u /u 2 = 2l u 1 2 /u 2 = t o 1 2 /u 2 11 d In part b, we must assume the wind s elocity is smaller than that of the plane for two reasons. First, mathematically the inward trip time would be negatie, which is unphysical. Second, if the wind speed were higher than that of the plane, it would not be able to make any forward progress to eer complete the outward trip! In part c, the ector diagram makes it clear that if the wind speed were larger than the plane s speed, no forward progress could be made. Mathematically, the net forward elocity would be negatie in this case... which also means that the plane simply can t i If we chose the wind going westward, the result would be the same.

5 go forward. Numeric solution: n/a. Double check: First, we can check that in all cases we hae the correct units: distance diided by elocity gies time. Second, if we set the wind speed to zero for parts b and c we should recoer the original result from part a. 3. The length of a spaceship is measured to be exactly half its proper length. a What is the speed of the spaceship relatie to the obserer s frame? b What is the dilation of the spaceship s unit time? Solution: This problem inoles only time dilation and length contraction. Find / Gien: We are to find the speed of the spaceship and the dilation factor of the ship s time gien the ratio of the length of the ship in its own reference frame to that seen by a moing obserer. Sketch: n/a. Releant equations: For the first portion, we can relate the ratio of the ship s length in its own frame L p, its proper length to that measured by the moing obserer L through the length contraction formula: L = L p /γ. With the definition of γ below, we can find the ship s elocity. For the second portion, the time dilation formula relates the elapsed time according to the moing obserer t to that of the ship t: t =γ t. L = L p /γ t = γ t γ = / Symbolic solution: Since the moing obserer sees the ship at half its rest length, the proper length is twice as big as that measured by the obserer, and γ=l p /L =2.

6 γ = γ 2 = / / 1 2 = 1 γ 2 1 c = 1γ 2 = = The time dilation factor is the ratio of the obserer s time to that of the ship, which is γ Numeric solution: For the first part, c = 3 2 factor For the second part, γ = 2 is the dilation Double check: If we let γ 1, we should recoer the classical non-relatiistic result. In the first part, that implies that L p =L, as expected, and in the second it implies that the dilation factor is zero, as expected. We also note that the formula for /c is dimensionless in the first part, as is the formula for the dilation factor in the second. Finally, our elocities are less than c, so there is no physical impossibility here. 4. Derie the relatiistic acceleration transformation a x = a x u x 3 3/2 in which a x =du x /dt and a x =du x/dt. Hint: du x/dt =du x/dt dt/dt Solution: This problem teaches us a nice lesson: what is conenient for eeryday physics is hideous and ugly relatiistically. That isn t because relatiity is ugly it is because our eeryday definitions of things like acceleration are faulty in the domain of relatiity. One can define proper elocities and accelerations which transform nicely, but unfortunately those quantities are a bit annoying for eeryday situations. Find / Gien: We are to derie the relatiistic acceleration transformation, armed with little more than the Lorentz transformations and calculus. Sketch: n/a. Releant equations: We need the Lorentz transformation for time, the elocity addition rule in

7 one dimension, and the chain rule: u x = u x 1 u x c 2 t = γ t x dy dx = dy du du dx Symbolic solution: The acceleration in the primed frame is a x =du x/dt, while in the unprimed frame it is du x /dt. The difficulty here is that we must transform both the elocity and time parts, and this is where the chain rule comes in handy: a x = du x dt = du x/dt dt /dt 22 The numerator and denominator can now be found from the elocity addition rule and the time transformation, respectiely. We must take care that u x aries in time but does not. Noting that du x /dt=a x, and handling the numerator first, ii du x dt = d ux dt 1 u = a x x 1 u + u x 1 a x c x 2 1 u x 2 23 = a x 1 u x 1 u + u x ax u a x a x u x + a x c x 1 u x 2 = 2 x a 2 x c 2 c 1 u x c 2 a x 1 2 c = 2 1 u x 2 25 Whew! Now for the denominator noting that since is constant so is γ: dt dt = d [ γ t x ] dt = γ 1 a x 26 Putting it all together, and recalling γ=1/ 1 2 / : a x = du x dt = du x/dt dt /dt = a x ux 2 γ 1 a x = a x 1 2 γ 1 u x 3 = a x 1 2 3/2 1 u x 3 27 Numeric solution: n/a. ii Also remember that d dx f g = f g fg g 2.

8 Double check: We hae mathematically proen the desired result, no additional checks are necessary. 5. A charge q at x = 0 accelerates from rest in a uniform electric field E which is directed along the positie x axis. a Show that the acceleration of the charge is gien by a = qe m 1 2 3/2 b Show that the elocity of the charge at any time t is gien by = qet/m 1 + qet/mc 2 c Find the distance the charge moes in a time t. Hint: integrals.wolfram.com Solution: I changed the notation in the problem slightly since I found it annoying after the fact just for elocity and a for acceleration, since it is a 1D problem anyway. Find/gien: We are to find the acceleration, elocity, and position as a function of time for a particle in a uniform electric field. We are gien the electric force and the boundary conditions x=0, =0 at t=0. Sketch: n/a. Releant equations: We will need only F = dp/dt, p = γm, the definition of γ gien in preious problems, and a good knowledge of calculus including the chain rule gien in the last problem. Symbolic solution: First, we must relate force and acceleration relatiistically. Since elocity is explicitly a function of time here, so is γ, and we must take care.

9 F = dp dt = d γm = γmd dt dt + mdγ 28 dt dγ dt = d 1 1 dt 1 2 c = 2 2 d dt / 3/2 = / 3/2 29 F = γm d dt + m2 1 d 1 2 / 3/2 dt = ma /c / 3/2 30 ma F = 1 2 / 3/2 31 That accomplished, we can set the net force equal to the electric force qe and sole for acceleration: ma F = qe = 1 2 / 3/2 32 a = qe 1 m 1 2 / 3/2 33 We can find elocity by writing a as d/dt and noticing that the resulting equation is separable. a = d dt = qe 1 m 1 2 / 3/2 34 qe m dt = du 1 2 / 3/2 35 We can now integrate both sides, noting from the boundary conditions that if time runs from 0 to t, the elocity runs from 0 to. 0 d 1 2 / 3/2 = 1 2 / 0 t 0 qe m = qet m 1 2 / = qet m dt 36 t Soling for, we first square both sides...

10 2 1 2 / = q2 E 2 t 2 m = 1 2 q 2 E 2 t 2 m q2 E 2 t 2 m 2 = q2 E 2 t 2 m 2 41 qet/m = qet/mc 2 We can find position by integrating through time from 0 to t. Since is only a function of time aboe, this is straightforward. x = t qet/m dt = mc qet/mc 2 qe 0 qet mc 2 t 0 = mc2 qe 1 + qet mc Classically, we would expect a parabolic path, but in relatiity we find the path is a hyperbola. Also note that the position, elocity, and acceleration depend oerall on the ratio between the particle s rest energy m to the electric force qe note energy/force is distance. Double check: If we let the ratio /c tend to zero, we should recoer the classical result for acceleration: a = qe m / 3/2 lim a = qe γ 0 m 44 As a separate check on the elocity expression, we note that if the classical acceleration is qe/m, and =at in the classical limit. Plugging that in our expression for elocity, = qet/m 1 + qet/mc 2 = at 1 + at/c 2 45 If c, then the denominator tends toward 1, and we recoer =at. We can do the same for the expression for x to show that it reduces to the correct classical limit. Finally, dimensional analysis will show that the units work out correctly for all three expressions - the ratio qe/mc has units of force/momentum which gies inerse seconds, so qet/mc is dimensionless. The ratio qe/m is force/mass which has units of acceleration, so qet/m has units of elocity. The ratio m /qe has units of energy/force which gies distance. 6. Show that the angular frequency of a charge moing in a uniform magnetic field B is gien by

11 ω = qb m 1 u 2 / Solution: If we hae a charge moing with angular frequency in a uniform magnetic field, we should immediately recognize that we hae uniform circular motion, which implies the charge s elocity is perpendicular to the magnetic field. Find/gien: We wish to find the angular frequency of a charge q of mass m moing at elocity u gien a magnetic field B. Sketch: Releant equations: We need the relatiistic equation for force, the constraint for circular motion, the magnetic force, the definition of angular elocity, and the definition of γ gien in preious problems. Note that the constraint on acceleration for circular motion is a purely geometric result, and holds true with or without relatiity. Since and B are at a right angle, the magnetic force may be written as simply qb. F = dp dt = d γm = γmd dt dt + mdγ 46 dt 2 d r = circ 47 dt F B = qb 48 ω = r 49 Symbolic solution: We set the total force equal to the magnetic force. In uniform circular motion, speed is constant, but since the direction is constantly changing elocity is not constant. Howeer, γ depends only on 2, i.e., on the speed, so dγ/dt is zero. Thus, F = dp dt = d γm = γmd dt dt + mdγ dt = γmd dt 50 Using the circular motion constraint, F = qb = γm d dt = γm2 = γmω r 51 ω = qb mγ = qb mγ = qb 1 2 m 52

12 Numeric solution: n/a Double check: Dimensionally, our answer is correct: 2 / is dimensionless, and qb/m has units of inerse seconds as required. In the limit c 1, the expression under the radical is 1, and we recoer the familiar classical result. 7. The effectie mass of a photon bundle of electromagnetic radiation of zero rest mass and energy hf can be determined from the relationship m=e/. Compute the effectie mass for a photon of waelength 500 nm isible and for a photon of waelength 0.1 nm X-ray. Solution: This problem is looking ahead to our discussion of radiation and energy quanta coming up shortly. Find/gien: Gien the waelength of a photon, and thus its energy, we are to find the rest mass that has the same energy. Releant equations: We need the photon energy, the waelength-frequency relationship, and the rest mass-energy relationship: E = hf λf = c m = E Symbolic solution: Putting the photon energy in terms of waelength, we hae c E = hf = h = λ hc λ 56 We now wish to equate this energy to the rest energy of a particle with mass: hc λ = mc2 m = h λc Numeric solution: Using h J s, c m/s, and the gien waelengths, one finds m = kg for λ = 500 nm 59 m = kg for λ = 0.1 nm 60

13 Double check: We can check that the units come out correctly: Planck s constant has units of J s, so m = [J s] [m][m/s] = [kg m2 /s] [m 2 = [kg] 61 /s] 8. A meterstick makes an angle of 30 with respect to the x -axis of O. What must be the alue of if the meterstick makes an angle of 45 with respect to the x-axis of O? Solution: This problem inoles length contraction along the x direction, but not along the y direction, so we hae to handle the projections of the meter stick along each axis separately. Find/gien: Gien the angle a meter stick makes with the x in one coordinate system a second coordinate system, we are to find the relatie elocity of the second coordinate system. Releant equations: We need only the formula for length contraction along the direction of motion, L x =L x /γ and trigonometry. Sketch: Symbolic solution: In the rest frame of the meterstick unprimed, we hae tan θ = L y L x 62 In the frame moing with respect to the meter stick primed, we hae tan θ = L y L x 63 The meter stick s length along the x direction in the primed frame will be contracted, L x = L x /γ, whereas the length along the y direction will remain the same. Thus, tan θ = L y L = L y x L x /γ = γl y = γ tan θ L x 64 tan θ γ = tan θ = / 65 Numeric solution:

14 Plugging in θ=30 and θ =45, we find γ= 3, which gies = 2/ c. Double check: The classical limit corresponds to γ = 1, which would imply that the angle is the same in both frames. 9. A particle appears to moe with speed u at an angle θ with respect to the x axis in a certain system. At what speed and angle will this particle appear to moe in a second system moing with speed with respect to the first? Why does the answer differ from that of the preious problem? Solution: It is most straightforward to assume that the two systems hae their horizontal x axes aligned. This is still quite general, since we are still letting the particle moe at an arbitrary angle θ, we may consider it to be a choice of axes and nothing more. Let the first frame, in which the particle moes with speed u at an angle θ be the unprimed frame x, y, and the second the primed frame x, y. Along the x direction in the primed frame, both perceied time and distance will be altered. Taking only the x component of the elocity, we consider the particle s motion purely along the direction of relatie motion of the two frames, and we may simply use our elocity addition formula. The x component of the particle s elocity will in the primed frame become u x = u x 1 u x / 66 Along the y direction in the primed frame, since we consider motion of the particle orthogonal to the direction of relatie motion of the frames, there is no length contraction. We need only consider time dilation. We deried this case in class, and the proper elocity addition for directions orthogonal to the relatie motion leads to u y = u y γ 1 u x / where γ = / 67 The particle s speed in the primed frame is then easily calculated: u = = ux u xu x + u yu y = 1 u x / u x 2 + u 2 y/γ 2 1 u x / 2 = 2 + u x 2 + u 2 y/γ 2 u y γ 1 u x / u x / 69 As a double-check, we can set θ = 0, such that u y = 0, which corresponds to the particle moing along the x axis. Our expression then reduces to the usual one-dimensional elocity addition for-

15 mula. The direction of motion in the primed frame is also found readily: tan θ = u y u x = u y 1 u x / γ 1 u x / u x = uy 1 u x 2 / Consider a charged parallel-plate capacitor, whose electric field in its rest frame is uniform neglecting edge effects between the plates and zero outside. Find the electric field according to an obserer in motion at constant elocity a along a line running through the center of the capacitor between the plates, and b along a line perpendicular to the plates. Solution: See the lecture notes on radiation, Notes/blackbody.pdf, where I e worked this problem out in detail.