Physics 103 Hour Exam #5 Solution

Point values are given for each problem. Within a problem, the points are not necessarily evenly divided between the parts. The total is 50 points.

1. [10 points] A certain transition between energy levels in hydrogen atoms emits radio waves with frequency 1.4204 GHz. Radio waves emitted by hydrogen near the center of the Andromeda galaxy received at Earth are measured to have a frequency of 1.4218 GHz.

(a) Is the Andromeda galaxy moving toward Earth or away from Earth? The received frequency is higher than the emitted frequency, so it is moving toward Earth . One way to think about this is that each successive wave crest emitted in the Andromeda galaxy has to travel a less far to get to Earth, so each crest arrives a little early. Hence the period is shorter and the frequency higher than in the galaxy’s rest frame. Another way to think about this is that the velocity v calculated in part (b) is negative. According to the convention used to derive the Doppler formulas, that means the source is moving toward the receiver.

(b) At what speed is it moving toward Earth or away from Earth?

 v  f ' 1 − f r c s f v r ' 1 − fs c v f ' 1 − r c fs  f   1421.8 GHz  v ' 1 − r c = 1 − c fs 1420.4 GHz m = −0.00099c = −3.0 × 105 s

1 2. [10 points] In the following, you must do calculations by hand and explicitly show how you made your estimates. If you do not show your work, you will not get credit. You may use a calculator to check your answers.

(a) Estimate p1 without using a calculator. 1.1

1 1 p = 1.1 (1 + 0.1)1/2 = (1 + 0.1)−1/2 1 = 1 + ( -2) (0.1) = 0.95

p (b) Estimate 110 without using a calculator.

p 110 = p(100)(1.1) p = 10 1.1 = 10(1 + 0.1)1/2  1  = 10 1 + 2 (0.1) = 10(1.05) = 10.5

2 3. [10 points] In a particle accelerator, protons (p+) and antiprotons (p−) are accelerated to equal but opposite velocities. They collide with each other and turn into an positron (e+), a electron (e−), and two photons:

p+ + p − ! e + + e − + 2 photons

If the electron and positron created in a collision each have speed 0.9999995c, and if the two photons each have wavelength 1.0 × 10−6 nm, what were the speeds of the proton and antiproton prior to the collision?

Notes: Mass values are given on the equation sheet. The mass of an antiproton is the same as the mass of a proton. The mass of a positron is the same as the mass of electron.

Use conservation of energy. First find the energy after the collision by adding the electron and positron energies plus the photon energies. This must be the total energy of the proton and antiproton, so use it to find their speeds.

Electron and positron. The electron and positron have Lorentz factors

1 1 = = = 1000, q 2 q 1 − V
1 − (0.9999995)2 ( c so each has total energy E = mc2 = (1000)(0.511 MeV) = 511 MeV, and the combined energy of both is (2)(511 MeV)=1020 MeV.

Photons. Each of the two photons has energy 1240 eV nm 1240 eV nm E = = = 1240 × 106 eV = 1240 MeV,  1 × 10−6 nm so the combined energy of both is (2)(1240 MeV)=2480 MeV.

Total energy. The total final energy is 1020 MeV+2480 MeV=3500 MeV.

Proton and anti-proton. The total final energy must equal the total initial energy, and it must have been equally shared by the proton and the antiproton. Thus each particle (proton and antiproton) had energy (3500 MeV)/2=1750 MeV. Then

E 1750 MeV E = mc2 ) = = = 1.86, mc2 938 MeV and the speed of each is

r 1 r 1 m V = c 1 − = c 1 − = 0.843 c = 2.52 × 108 2 1.862 s

3 4. [20 points] Imagine that the United States has converted to the metric system, and football fields are now marked o in metric. This means that there are lines indicating distances of 0 m, 10 m, 20 m, and so on, up to 100 m drawn on the ground. Three things happen on a football field (events A, B, and C). Their positions and times are given in the reference frame of the ground:

• Firecracker A goes o at x=30.0 m • Firecracker B goes o at x=50.0 m at the same time as firecracker A • Firecracker C goes o at x=50.0 m, 77.0 ns after firecracker A

These events are observed by people in a rocket ship which is passing by, moving in the +x direction at V = 0.866c. m Hint: use c = 0.3 in this problem. ns (a) What is the distance between events A and B according to observers in the rocket? 1 1 The Lorentz factor is = q p = 2. V 2 1−0.8662 1−( c ) In the football field frame,
x = 50.0 m−30.0 m = 20.0 m and
t = 0. The Lorentz transform equation gives

x0 = (
x − V
t) = 2 (20 m − 0) = 40 m.

(b) What is the distance between events A and C according to observers in the rocket?

As in part (a), the Lorentz factor is = 2. m m The rocket speed is V = 0.866c =(0.866) (0.3 ) = 0.26 . ns ns In the football field frame,
x = 50.0 m − 30.0 m = 20.0 m and
t = 77.0 ns−0 = 77.0 ns The Lorentz transform equation gives

x0 = (
x − V
t)  m   = 2 20 m − 0.26 (77 ns) ns = 2 (20 m − 20 m) = 0.

This problem is continued on the next page....

4 Problem 4, continued....

(c) Hopefully your answer to either part (a) or part (b) of this problem was zero or, if not exactly zero, a number close to zero. Give a physical explanation of what this means. Use words and/or diagrams (not equations). The answer to part (b) was zero. This means that events A and C happened at the same place in the rocket reference frame. Why did this happen? The speed of the rocket is such that it goes (0.26 m/ns)(77 ns)=20 m in 77 ns. Hence if it was right over the 30 m mark at time t = 0, it would be right over the 50 m mark at time t = 77 ns. If the coordinate system is centered on the rocket, both events happen at x0 = 0 (see figure).

V t=0ns x'=0

x=0x=10 x=20 x=30x=40 x=50 x=60

t=77 ns x'=0

x=0x=10 x=20 x=30x=40 x=50 x=60

(d) Is there some reference frame in which event C happens before event A? Justify your answer. No. The spacetime separation between the events is

s2 = (c
t)2 − (
x)2 h m  i2 = 0.3 (77 ns) − (20 m)2 ns = 534 m2 − 400 m2 = 134 m2 .

Positive values of s2 correspond to time-like events, which can be causally connected and which cannot change order.

Alternatively, for C to happen before A in some frame, we must have
t0 < 0 in that frame. What would the rocket velocity, V , be for that frame? The Lorentz transformation of this is (
t − (V/c2)
x) < 0. With a little algebra, this leads to
t < (V/c2)
x and finally to V > c2
t/
x. Plugging in numbers for
t and
x gives V > 0.35 m/ns. But this is faster than the speed of light, c = 0.30 m/ns. So there cannot be such a reference frame.

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