Recitation 8
Chapter 9
Problem 9.23. Calculate the momentum of an electron moving with a speed of (a) 0.0100c,(b) 0.500c,(c) 0.900c.
The momentum (using relativistic mass) is given by
p=mv =γm0v(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=1
2mv2to 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−m0c2=mc2−m0c2= (γ−1)m0c2=
1
q1−v2
c2
−1
m0c2="1−v2
c2−0.5
−1#m0c2(8)
Using the Taylor series expansion around x= 0,
(1 + x)b≈1 + bx +O(x2) (9)
we see that for small x=−v2/c2, the kinetic energy looks like
K=1+0.5v2
c2+O(v4/c4)−1m0c2= 0.5m0v2+O(m0v4/c2)≈1
2m0v2(10)
with the relative error on the order of v2/c2<0.01 = 1% for v < 0.1c, which is what we set out to show.
Alternatively, we can compare the exact kinetic energy with the nonrelativistic form for v= 0.1c. From the above analysis,
we see that the relative error will be less for slower v.
γ=1
√1−0.12= 1.00503782 (11)
Krel = (γ−1)m0c2= 0.00503782γm0c2(12)
Knon =1
2m0v2= 0.5·m0·0.01c2= 0.005m0c2(13)
Knon
Krel
=0.005
0.00503782 = 0.992494 (14)
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.
