Lec31 - The second term is the Newtonian kinetic energy The...

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Unformatted text preview: The second term is the Newtonian kinetic energy. The first term has dimensions of energy, but it is not zero even when the speed goes down to zero. We therefore interpret mc2 as an energy associated with an object at rest. We therefore define E := γ mc2 to be the total relativistic energy, so that E 2 = p2 c2 + m 2 c4 . We justify this definition below. In the rest frame of the object, and only in that rest frame, p = 0, so that ERF = mc2 . Because E 2 − p2 c2 = m2 c4 is an equation describing an invariant, i.e., E 2 − p 2 c 2 = E ￿2 − p ￿2 c 2 , different observers will disagree on measurements of E, p, but will always agree on this particular combination thereof, in complete analogy with the spacetime interval, and hence all observers agree on the mass m of a particle. Relativistic dynamics: kinetic energy To show that our definition of E makes sense and is not arbitrary, we next show that E is the sum of the rest energy, and what we naturally call the relativistic kinetic energy. We define the force to be the rate of change of momentum, fi := dpi / dt, where the time t is in some reference frame. Then the kinetic energy of an object accelerated by force f is ￿u ￿u ￿u ￿u dp dp dx K= f dx = dx = dt = u dp u=0 u=0 dt u=0 dt dt u=0 Next, calculate dp: dp = d(γ mu) = d Therefore, K= ￿ u ￿ ￿ 1 − u2 /c2 1 1− u2 c2 mu ￿ =￿ m 1− u2 c2 ￿ 3 du 2 u=0 as indeed one would expect, or E = mc2 + K . ￿ mu 1− 2 ￿ ￿ 3 du = mc u2 2 c2 − 1 = mc2 (γ − 1) ≡ E − mc2 The law of conservation of energy applies to the total energy of a closed system, so that E is the conserved quantity. Therefore, part of the total energy can be transferred from K to the rest energy and vice versa. 106 Example: Inelastic collision of two balls. Let m1 , m2 be two balls with initial relativistic kinetic energies K1 , K2 , respectively. Then, before the collision, E1 = m1 c2 + K1 and E2 = m2 c2 + K2 . The total energy of the system before the collision is therefore Ei = E1 + E2 = (m1 + m2 )c2 + (K1 + K2 ). After the collision the two balls stick together (perfectly inelastic collision) and are in practice a single ball of mass M and kinetic energy K , so that it’s energy is Ef = M c2 + K . Conservation of energy implies that Ei = Ef , or M c2 + K = (m1 + m2 )c2 + (K1 + K2 ). The change in kinetic energy is ∆K = K − (K1 + K2 ) = (m1 + m2 )c2 − M c2 = (m1 + m2 − M )c2 = ∆M c2 . Notice that if ∆K < 0, then M > m1 + m2 . That is, the kinetic energy lost is manifested by an increase in the mass of the object. This is an example for how mass and energy can convert into one another. Recall now that the momentum p = γ mv . Therefore, v v pc = (γ mv )c = (γ mc2 ) = E . c c We therefore find that v/c = pc/E . For a photon, v = c, so that Ephoton = pc. As E 2 = p2 c2 + m2 c4 , we find that mphoton = 0. In fact, this conclusion holds for any particle moving at the speed c: only zero mass particles move at the speed of light, and they cannot move at any other speed. Relativistic Doppler Effect Suppose one wavefront arrives at the observer, and the next wavefront is a distance λ from it. By the time t it gets to the observer, the observer has moved a distance vt. Therefore (for the case the observer is moving away from the source), the total distance traveled by the second wavefront is ct = λ + vt where c is speed of the wave and v is speed of the observer. Note, that in the reference frame of the source, t is the time coordinate in the source’s frame. Therefore, t= c Substituting λ = f , we obtain λ = c−v c λ 1 − v λ = c λ 1 −v c 1 . f c λ . t= 1 1 = v f − cf 1− v c · The frequency is measured not in the source’s frame, but rather in the observer’s frame, and there is relative motion with speed v between the two frames. As the measurement is done 107 at two different times (for the two wavefronts) in the same place in the observer’s reference frame —two colocal events— the time difference between the two wavefronts is the observer’s proper time t0 . The time t in the source’s frame is therefore dilated according to t = γ t0 . Therefore, ￿ 1− 1− 1 1 1 t0 = t = · γ γ 1− v c 1 ·= f v2 c2 v c 1 ·= f ￿ ￿ (1 − v )(1 + v ) 1 1+ c c ·= v 1− c f 1− v c v c · 1 f and as t0 is the time measurement by the observer between two successive wavefronts, it is also the period of the waves as measured by the observer. Therefore, the frequency measured by the observer is 1 f0 = = t0 ￿ 1− 1+ v c v c ·f ￿ ￿ ￿ v 1 v2 1 − + · 2 − ... × f c 2c where the last step is the Taylor series expansion about v = 0 to second order in v/c. Compare this to the non-relativistic formula )with speed of wave v taken to be the speed of light c) ￿ fNR D 1 ± vv = ·f S 1 ∓ vv which in our case (observer moving away from static source) reduces to ￿ vD ￿ ￿ fNR = 1 − ·f c which is identical with the relativistic result to first order in v/c, but is different to second order. Recalling that λ = c f we find ￿ 1+ 1− v c v c c λ0 = = f0 ·λ￿ ￿ ￿ v 1 v2 1 + + · 2 + ... × λ c 2c ￿v 1+ c We next define the redshift as z + 1 := λ0 so that z + 1 = λ0 = . Note that from λ λ 1− v c the Principle of Relativity it doesn’t matter which frame moves; only relative motion is important. 108 If, in the observer’s frame, the source is moving away with angle θ, we find in complete analogy with the previous discussion t= λ 1 1 = · v c − v cos θ 1 − c cos θ f 1 1 1 1 t= · · v γ γ 1 − c cos θ f c after dividing top and bottom by c and using λ = f . Therefore, t0 = and 1 v = γ (1 − cos θ)f . t0 c which is the relativistic Doppler shift when there is an angle θ between the line of sight and the velocity of the source. Note, that if θ = 90◦ then cos θ = 0 and then we find the transverse Doppler effect, specifically f0 = f0⊥ 1 = γ f =⇒ f = f0⊥ = γ ￿ 1− v2 · f0⊥ . c2 Notice, that f0⊥ ￿= f unless v = 0 for transverse motion. The transverse Doppler effect is always blueshift. In the classical model the transverse Doppler effect doesn’t happen, whereas in the relativistic model there will still be a transverse Doppler effect, resulting from the time dilation effect. 109 Question: What is the relative size of the transverse Doppler effect compared with the longitudinal effect? Answer: In the transverse effect, the change in frequency is ∆f⊥ = (γ − 1)f , and for the longitudinal effect ∆f￿ = [γ (1 ∓ v/c) − 1]f . Their ratio is ￿ ￿￿ ￿￿ ￿ ￿ ∆f￿ ￿ ￿ γ (1 ∓ v/c) − 1 ￿ ￿ ￿ ￿ ￿=￿ ￿ = ￿1 ∓ γ v/c ￿ ≈ 2 c ￿ 1 for v ￿ c ￿ ∆f⊥ ￿ ￿ ￿￿ γ−1 γ − 1￿ v which means that measuring the transverse effect is not easy from the experimental point of view, because of the smallness of the effect: any small deviation from 90◦ introduces a confounding factor of the longitudinal effect which is larger than the transverse effect. The relativistic transverse Doppler effect was first confirmed in the Ives–Stilwell experiment (1938). Because it is hard to measure the effect directly, the experimental method was to detect changes in the results of the combined longitudinal and transverse effects, recalling that the transverse effect is always blueshift whereas the much larger longitudinal effect is either blueshift or redshift. Without the extra Lorentz factor γ , the longitudinal shift is f0 /f = 1 ∓ v/c (velocity parallel or anti–parallel to the line of sight), and the important thing here is that the two directions are equally separated from the unshifted frequency. The extra Lorentz factor makes the separation of the two lines from the unshifted one asymmetrical, and this asymmetry is easier to detect experimentally than the transverse effect itself. The experiment measures in fact the Lorentz pre-factor, and in the transverse case the only factor is this Lorentz factor, so that it in fact measures the transverse effect. 110 ...
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This note was uploaded on 07/19/2011 for the course PH 113 taught by Professor Liorburko during the Spring '09 term at University of Alabama - Huntsville.

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