

Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
Solutions to various problems in relativistic physics, focusing on the calculation of momentum and energy of moving particles. It includes problems involving electrons and protons, and demonstrates the use of relativistic equations to find their velocities and total energies.
Typology: Exercises
1 / 3
This page cannot be seen from the preview
Don't miss anything!
Chapter 9
Problem 9.23. Calculate the momentum of an electron moving with a speed of (a) 0. 0100 c, (b) 0. 500 c, (c) 0. 900 c.
The momentum (using relativistic mass) is given by
p = mv = γm 0 v (1)
(a)
γ = 1. 0000500 (2) p = 5.11 keV/c (3)
(b)
γ = 1. 155 (4) p = 295 keV/c (5)
(c)
γ = 2. 294 (6) p = 1.06 MeV/c (7)
Problem 9.30. Show that, for any object moving at less than one-tenth the speed of light, the relativistic kinetic energy agrees with the result of the classical equation K = 12 mv^2 to within less than 1%. Therefore, for most purposes the classical equation is good enough to describe these objects, whose motion we call nonrelativistic.
The kinetic energy is the energy that is due to the objects motion. In other words, the increase in the total energy over the rest mass (in the absence of potential energies etc.). In math
K = E − m 0 c^2 = mc^2 − m 0 c^2 = (γ − 1)m 0 c^2 =
1 − v 2 c^2
(^) m 0 c^2 =
v^2 c^2
m 0 c^2 (8)
Using the Taylor series expansion around x = 0,
(1 + x)b^ ≈ 1 + bx + O(x^2 ) (9)
we see that for small x = −v^2 /c^2 , the kinetic energy looks like
v^2 c^2
m 0 c^2 = 0. 5 m 0 v^2 + O(m 0 v^4 /c^2 ) ≈
m 0 v^2 (10)
with the relative error on the order of v^2 /c^2 < 0 .01 = 1% for v < 0. 1 c, which is what we set out to show. Alternatively, we can compare the exact kinetic energy with the nonrelativistic form for v = 0. 1 c. From the above analysis, we see that the relative error will be less for slower v.
γ =
Krel = (γ − 1)m 0 c^2 = 0. 00503782 γm 0 c^2 (12)
Knon =
m 0 v^2 = 0. 5 · m 0 · 0. 01 c^2 = 0. 005 m 0 c^2 (13) Knon Krel
for a 7.51% relative underestimate.
Problem 9.31. An electron has a kinetic energy five times greater than its rest energy. Find (a) its total energy and (b) its speed.
The total energy is (a) E = K + m 0 c^2 = (5 + 1)m 0 c^2 = 6 · 511 keV = 3.07 MeV (15)
(b)
E = mc^2 = γm 0 c^2 = 6m 0 c^2 (16)
γ =
1 − v 2 c^2
v^2 c^2
v c
v = 0. 986 c (20)
Problem 9.35. The rest energy of an electron is 0. 511 MeV. The rest energy of a proton is 938 MeV. Assume that both particles have kinetic energies of 2. 00 MeV. Find the speed of (a) the electron and (b) the proton. (c) By how much does the speed of the electron exceed that of the proton? (d) Repeat the calculations assuming that both particles have kinetic energies of 2 , 000 MeV.
First we’ll work out the solution symbolically, since we’ll need it twice. The total energy yields γ, which in turn yields v.
E = K + m 0 c^2 = γm 0 c^2 (21)
γ =
m 0 c^2
γ =
1 − v 2 c^2
γ^2
v^2 c^2
v c
γ^2
v =
γ^2
· c (26)
Applying these formula to our various situations, we get (a)
γa =
2 .00 MeV 0 .511 MeV
va = 0. 979 c (28)
(b)
γb =
2 .00 MeV 938 MeV
vb = 0. 0652 c (30)
(c) va − vb = 0. 914 c (31)
(d)
γe =
2 .00 GeV 0 .511 MeV
ve = (1 − 3. 26 · 10 −^8 )c (33)
γp =
2 .00 GeV 938 MeV
vp = 0. 948 c (35) ve − vp = 0. 0523 c (36)