Simple Nature - Light and Matter

Page 658, problem 16:

By symmetry, the field is always directly toward or away from the center. We can therefore

calculate it along thexaxis, wherer=x, and the result will be valid for any location at that

distance from the center.

E=−

dV

dx

=−

d

dx

(

x−^1 e−x

)

=x−^2 e−x+x−^1 e−x

At smallx, near the proton, the first term dominates, and the exponential is essentially 1, so we
haveE∝x−^2 , as we expect from the Coulomb force law. At largex, the second term dominates,
and the field approaches zero faster than an exponential.
Page 666, problem 56:

sin(a+b) =

(

ei(a+b)−e−i(a+b)

)

/ 2 i

=

(

eiaeib−e−iae−ib

)

/ 2 i

= [(cosa+isina)(cosb+isinb)−(cosa−isina)(cosb−isinb)]/ 2 i

= cosasinb+ sinacosb

By a similar computation, we find cos(a+b) = cosacosb−sinasinb.

Page 666, problem 57:

Ifz^3 = 1, then we know that|z|= 1, since cubingzcubes its magnitude. Cubingztriples its

argument, so the argument ofzmust be a number that, when tripled, is equivalent to an angle

of zero. There are three possibilities: 0×3 = 0, (2π/3)×3 = 2π, and (4π/3)×3 = 4π. (Other

possibilities, such as (32π/3), are equivalent to one of these.) The solutions are:

z= 1, e^2 πi/^3 ,e^4 πi/^3

Page 666, problem 59:
We haveD=qandFx=qb=D∂Ex/∂x, which, as claimed, is consistent with the result of
example 7 on p. 589 and depends onqand`only viaD.

Solutions for chapter 11
Page 756, problem 51:
(a) For a material object,p=mv. The velocity vector reverses itself, but mass is still positive,
so the momentum vector is reversed.
(b) In the forward-time universe, conservation of momentum isp1,i+p2,i=p1,f+p2,f. In the
backward-time universe, all the momenta are reversed, but that just negates both sides of the
equation, so momentum is still conserved.
Page 756, problem 51:
(a) For a material object,p=mv. The velocity vector reverses itself, but mass is still positive,
so the momentum vector is reversed.
(b) In the forward-time universe, conservation of momentum isp1,i+p2,i=p1,f+p2,f. In the