Problem Set 1: Solutions

Transcription

1 PH 253 / LeClair Fall 213 Proble Set 1: Solutions 1. Daily proble due 23 Aug 213: How fast ust a rocket trael relatie to the earth so that tie in the rocket slows down to half its rate as easured by earth-based obserers? Do present-day jet planes approach such speeds? Solution: The question wants to know how uch tie is slowed down copared to the earthbased obserers, which eans the earth-based obserers hae the proper tie. The rocket ust experience dilated tie by coparison. Thus, the elapsed tie ust be related by t rocket = γ t earth = t earth 1 2 /c 2 1) Tie slowing down by a factor two iplies t rocket t earth = 2 = = 1 4 = 1 2 = = /c 2 2) c c.866c /s 4) 3) 2. Daily proble due 26 Aug 213: A cube of etal with sides of length a sits at rest in frae S with one edge parallel to the x-axis. Therefore, in S the cube has olue a 3. Frae S oes along the x-axis with speed u. As easured by an obserer in frae S, what is the olue of the etal cube? Solution: Since relatie otion is inoled, there ust be length contraction for the oing obserer - the person in S since we are obsering the block. Since there is relatie otion only along the x axis, there is length contraction only along that axis. The cube therefore appears shortened by a factor γ along the x axis, but its diensions along y and z are the sae. The olue in the two fraes is thus: V = a a a = a 3 in S 5) V = a γ a a = a3 γ in S 6) The probles below are due by the end of the day on 28 Aug 213.

2 3. One of the waelengths of light eitted by hydrogen atos under noral laboratory conditions is λ = n, in the red portion of the electroagnetic spectru. In the light eitted fro a distant galaxy this sae spectral line is obsered to be Doppler-shifted to λ = n, in the infrared portion of the spectru. How fast are the eitting electrons oing relatie to the earth? Are they approaching the earth or receding fro it? Solution: The relatiistic Doppler shift is gien by λ o = λ s c + c 7) where λ o is the waelength obsered in relatie otion at elocity with respect to the source, and λ s its he waelength obsered in the source s frae. Positie elocities correspond to obserers approaching the source. We are gien both waelengths: λ o = n and λ s = n. Since λ o > λ s, fro the equation aboe it is clear that we ust hae < for this to be true, which already tells the source is receding. All we need to do is sole the aboe for to find the speed, which gies c = λo λ s 1 λo λ s ) 4. Two particles in a high-energy accelerator experient approach each other head-on with a relatie speed of.89c. Both particles trael at the sae speed as easured in the laboratory. What is the agnitude of the elocity of one particle relatie to the other? Solution: What we are gien is the speed of the two particles relatie to each other. That is, if we were in the reference frae of one of the particles, we would say the other approaches with u=.89c. In the obserer s frae call it S ), we see the particles oing toward each other, each with the sae speed. In the S frae of reference, we would hae to say that adding the two elocities together gies us u=.89c. u = /c 2 =.89c 9) Soling this for gies us elocity of one particle relatie to the other in the lab frae. The result is = 2 ± 4 4u 2 /c 2 2u/c 2 = c2 u 1 ± ) 1 u 2 /c 2 {.611c, 1.64c} 1) Clearly, the second root, while atheatically allowed, does not ake physical sense - we e

3 already established elocities can t be greater than c. Therefore we reject it as unphysical, and the reaining alid solution is.611c. 5. a) Through what potential difference does an electron hae to be accelerated, starting fro rest, to achiee a speed of.98c? b) What is the kinetic energy of the electron at this speed? Express your answer in both joules and electron olts. Solution: The key is to reeber that a charge q oing through a potential difference V changes its potential energy by U = q V. If we are not worrying about resistie forces, this change in potential energy is equal to the charge s change in kinetic energy. Starting fro rest, we know that ust be K=γ 1) c 2, with γ=1/ 1 2 /c 2. For an electron q = e, and c 2 = 511 kev, and with =.98c we hae γ Putting it all together, K = U = e V = γ 1) c 2 = ) ev ) ev 11) Since e V is the particle s change in both potential and kinetic energy, this is already the answer to the second part of the question: the particle s kinetic energy is about 2.6 MeV, or about.33 pj p=1 12 ). The corresponding potential difference is by definition 2.6 MV, illustrating how handy a unit the electron olt is. 6. Use the following two equations: p = E = 1 2 /c 2 c /c 2 12) 13) to derie the following relationship: E = pc + c 2 Solution: No big trick, just grind through it. Since you know the result you want to get to, start by finding p 2 c 2. p 2 c 2 = 2 2 c 2 ) 1 2 /c 2 = 2 c 4 2 c ) Now add 2 c 4 and rearrange, and you e got it.

4 ) p 2 c c 4 = 2 c c c 2 2 = 2 c ) c ) = 2 c 4 c 2 ) c 2 2 = 2 c /c 2 = γ 2 2 c 4 16) p 2 c c 4 = γc 2 = E 17) 7. A charge q at x = accelerates fro rest in a unifor electric field E which is directed along the positie x axis. a) Show that the acceleration of the charge is gien by a = qe 1 2 c 2 ) 3/2 b) Show that the elocity of the charge at any tie t is gien by = qet/ 1 + qet/c c) Find the distance the charge oes in a tie t. Hint: http: // integrals. wolfra. co Solution: We are to find the acceleration, elocity, and position as a function of tie for a particle in a unifor electric field. We are gien the electric force and the boundary conditions x=, = at t =. We will need only F = dp/dt, p = γ, the definition of γ gien in preious probles), and a good knowledge of calculus including the chain rule once again). First, we ust relate force and acceleration relatiistically. Since elocity is explicitly a function of tie here, so is γ, and we ust take care. F = dp dt = d γ) = γd dt dt + dγ 18) dt dγ dt = d ) ) 1 1 dt 1 2 c = 2 2 d c 2 dt /c 2 ) 3/2 = 1 c /c 2 ) 3/2 19) F = γ d ) ) dt d d c /c 2 ) 3/2 dt = 1 2 dt 1 2 /c + c /c 2 ) 3/2 2) ) c F = a /c 2 ) 3/2 + c 2 a 1 2 /c 2 ) 3/2 = 1 2 /c 2 ) 3/2 21) If you decided not to use γ and wrote eerything explicitly in ters of 1/ 1 2 /c 2, that is fine.

5 The end result is the sae. That accoplished, we can set the net force equal to the electric force qe and sole for acceleration: a F = qe = 1 2 /c 2 ) 3/2 22) a = qe 1 2 /c 2) 3/2 23) We can find elocity by writing a as d/dt as it was aboe) and noticing that the resulting equation is separable. a = d dt = qe 1 2 /c 2) 3/2 qe dt = d 1 2 /c 2 ) 3/2 24) 25) We can now integrate both sides, noting fro the boundary conditions that if tie runs fro to t, the elocity runs fro to. d 1 2 /c 2 ) 3/2 = 1 2 /c 2 t qe = qet 1 2 /c = qet 2 dt 26) t 27) 28) Soling for, we first square both sides /c 2 = q2 E 2 t ) ) 2 = 1 2 q 2 E 2 t 2 c 2 2 3) q2 E 2 t 2 ) 2 c 2 = q2 E 2 t ) qet/ = 32) 1 + qet/c

6 We can find position by integrating through tie fro to t, which is straightforward. x = t qet/ dt = c qet/c qe qet c t = c2 qe 1 + qet c 1 33) Classically, we would expect a parabolic path, but in relatiity we find the path is a hyperbola. i Also note that the position, elocity, and acceleration depend oerall on the ratio between the particle s rest energy c 2 to the electric force qe note energy/force is distance). 8. Show that for the preceding question the particle s speed approaches c as t. Solution: This is basically a way of double-checking that our preious result akes sense. It also reinforces the idea that een with a constant, steady acceleration for infinite tie nothing is going to reach the speed of light. All we need to do is take the t liit of our elocity expression. Diide eerything by t and it is straightforward. li = li qet/ qe/ = li = t t 1 qe/ + qet/c t 1/t 2 + qe/c qe/c = c 34) 9. At what speed is the oentu of a particle twice as great as the result obtained fro the non-relatiistic expression? Express your answer in ters of the speed of light. Solution: Relatiistic oentu is p=γ, classically we would write p=. The latter is off by a factor of two when γ = 2 γ = 2 36) 3 = 2 c 37) 35) 1. Light traels with respect to earth at s. A rocket traels at s with respect to earth in opposite direction of the light. What is the speed of light as iewed fro the rocket? Solution: It is light, the speed is always c in acuu. i If you square both sides, the equation for xt) can be put in the for x 2 /a 2 t 2 /b 2 = 1, the standard for of the equation for a hyperbola.