PHY 2061
Steven Detweiler
Fall 2008
The lack of mathematical sophistication is a leading cause of difficulty for students in
Electricity & Magnetism.
An official pre-requisite of phy2061 is phy2048, or the equivalent, and the math requirements include MAC 2311 and 2312 with a co-requisite of 2313, Vector Calculus. These math courses together cover derivatives and integrals of trig and log functions, series and sequences, analytic geometry, vectors and partial derivatives and multiple integrals. We will casually be using math from all of these subjects. None of these should be completely unfamiliar to you.
The following discussions and questions are grouped by subject and in approximate order of difficulty—easiest first. These are representative of the level of mathematics which I expect. You should feel comfortable with mathematics at this level, at least through section
C, and section D is probably new, but it can help make trigonometry easy. The answers to the questions are not always given. If you do not know that your answer is correct then you are not comfortable with mathematics at this level.
Sections D, E, F, G and H are more advanced than is necessary for this course, although some of the ideas will be used in lectures. If you understand the Taylor series in section
F, then you are likely to find section G, on calculators, amusing. Don’t be surprised if my discussions seem confusing at first: To understand math and physics often requires multiple readings while working out algebraic details with paper and pencil in hand.
Section H involves an ordinary differential equation that has an interesting application to radioactivity.
1
A.
Algebra
Solve for x : f ( x ) = ax
2
+ bx + c = 0 .
For what value of x is f ( x ) a maximum or a minimum?
Make a sketch of the function y ( x ) where y = mx + b and where m and b are constants.
What are the meanings of the constants m and b in terms of your sketch?
Factor:
( a
2
+ 4 ab + 4 b
2
) and ( a
2
− 9 b
2
)
B.
Vector Algebra
Let
~
= 1ˆ + 2ˆ  and
~
= 4ˆ + 5ˆ  .
What is | ~ | ?
What is
~ · B ?

What is
~
×
~
?
What is the cosine of the angle between
~ and
~
?
If
~
·
~
= 10 and the angle between
~ and
~ is 30
◦
, then what is the magnitude of
~
?
2
C.
Calculus
If x
0
, v
0 and a are constants and x ( t ) = x
0
+ v
0 t +
1
2 at
2 then what is dx/dt ? What is d
2 x/dt
2
? If a < 0, does the function x ( t ) curve up or down?
If x is negative when t = 0 and x is positive when t is very large: then for precisely which values of t is x positive?
Evaluate the derivative d dt
A cos( ωt + φ ) where A , ω and φ are constant.
Evaluate the following integrals (the second is not easy, but it is used in the first chapter about the electric force in the textbook):
Z
π sin θ dθ,
0
Z
L/ 2 dz
− L/ 2
( z 2 + a 2 ) 3 / 2 where L and a are constant,
Z k x 2 dx where k is a constant, and
Z x k
1 x dx.
D.
Trigonometry
Euler’s identity, e iθ
= cos θ + i sin θ, is probably new to you. But it provides a convenient and easy way to derive some of the basic trig identities such as e i ( α + β )
= e iα e iβ cos( α + β ) + i sin( α + β ) = (cos α + i sin α ) × (cos β + i sin β )

3 or, after multiplying out the right hand side, cos( α + β ) + i sin( α + β ) = cos α cos β − sin α sin β + i (cos α sin β + sin α cos β )
The real and the imaginary parts of this equation give the well known trig identities: cos( α ± β ) = cos α cos β ∓ sin α sin β and sin( α ± β ) = sin α cos β ± cos α sin β.
Use the Euler identity to show that sin
2
θ + cos
2
θ = 1 .
Hint: start with e iα e
− iα = 1 and use the Euler Identity.
E.
Sums
Question: Evaluate the sum
S ( x ) =
∞
X x n n =0 for | x | < 1 .
Ans: Note that
S ( x ) =
∞
X x n n =0
= 1 +
∞
X x n n =1
∞
= 1 + x
X x n n =0
= 1 + xS ( x )
So we have and, finally
S = 1 + xS
(1 − x ) S = 1
S ( x ) =
1
1 − x
.
F.
Taylor expansions of a function
Any differentiable function f ( x ) may be approximated in the neighborhood of a point x
0 by the Taylor expansion df f ( x ) = f ( x
0
) + ( x − x
0
) dx x = x
0
+ · · · +
+
1
2
( x − x
0
1 n !
( x − x
0
) n d n f dx n x = x
0
+
)
2
· · · d 2 f dx 2 x = x
0
+
1
6
( x − x
0
)
3 d 3 f dx 3 x = x
0

4
For example, consider f ( x ) = 1 / (1 − x ), expanded about x
0
= 0. Then df f ( x ) = 1 / (1 − x )
= [1 / (1 − x )
2
] x =0
= 1 dx x =0 d
2 f dx 2 x =0 d 3 f dx 3 x =0 d n f dx n x =0
= 2[1 / (1 − x )
3
] x =0
= 2
= 6[1 / (1 − x )
4
] x =0
= 6
= n ![1 / (1 − x ) n +1
] x =0
= n !
The Taylor expansion for 1 / (1 − x ) with x
0
= 0 is now
1
1 − x
= 1 + x +
1
2 x
2 × 2 +
1
6 x
3 × 6 + · · · +
1 n !
x n × n ! + · · ·
And this is easily seen to be
1
1 − x
=
∞
X x n
, n =0 the same as our example for doing sums above!
Taylor expansions of this sort are extremely useful in physics. For example in special relativity when we are interested to see how close special relativity is to Newtonian physics for small speeds v , we usually make the assumption that v/c 1. Then we make Taylor expansions of the relevant formulae, and include only the terms proportional to v/c or maybe also v 2 /c 2 .
Common Taylor expansions give approximations such as
1
1 −
(1 + ) n
= 1 + + O (
2
)
= 1 + n + O (
2
)
√
1 − = 1 −
1
√
1 −
= 1 +
1
2
1
2
+ O (
2
)
+ O (
2
) e i
1 e = 1 + +
= 1 + i −
2
1
2
2
2
1
+
6
− i
1
6
3
3
1
+
24
1
+
24
4
+ O (
4
+ O (
5
)
5
) .
The O ( n ) term here is standard mathematical notation to mean a function which is less than some constant times n bigger than something times in the limit that n .
→ 0. In other words for small , O ( n ) is no

5
We can use the Euler identity e i = cos + i sin to easily pick off the purely real terms from this last expansion which give the expansion of cos for a small angle , and the purely imaginary terms, which give the expansion of sin for small : and
1 cos = 1 −
2
1 sin = −
6
2
+
1
24
4
3
+ O (
5
) .
+ O (
6
)
G.
Calculators
When solving a physics problem, think with your brain not with your calculator! Before touching your calculator, check to see that your algebraic answer has the correct units and that it has the expected behavior for various limits. It is nearly impossible to check the correctness of an answer once you touch your calculator.
You might find it amusing that the number 10 100 has been given the name googol, and
10 googol is called googolplex—and these names were coined well before the internet was invented. But, note the difference in spelling between googol and the name of the internet search engine. As an aside: The internet was invented by physicists who wanted to exchange easily experimental data between the United States and Europe.
Here are a couple examples which are relevant to one of the homework problems for this course. Let f ( n ) = n 2 , where n is an integer. First evaluate f (10 2 ) − f (10 2 − 1) on your calculator. You should get 199.
Now try to evaluate f (10 100 ) − f (10 100 − 1). Your calculator will choke on this problem, but your brain can easily find the answer to 100 significant digits. Note that f ( n ) − f ( n − 1) = n
2 − ( n − 1)
2
= n
2 − ( n
2 − 2 n + 1) = 2 n − 1 .
With n = 10 100 , it is easy to see that f ( n ) − f ( n − 1) = 2 × 10 100 − 1 ≈ 2 × 10 100 with 100 significant digits.
Here is a second, more challenging, problem. Let f ( n ) = n
− 2
Evaluate f (10
100 − 1) − f (10
100 where n is an integer.
). Your calculator will also choke on this problem, but again you can easily find the answer to about 100 significant digits. Use the Taylor expansion df f ( n + δn ) = f ( n ) + δn dn df
+ . . .
⇒ f ( n + δn ) − f ( n ) = δn dn
2
+ . . .
= − δn n 3
+ . . .
With n = 10 where the . . .
100 and δn = − 1, we easily have f (10 100 − 1) − f (10 100 ) = 2 × 10
− 300 represents terms which are comparable to 1 /n 4 = 10
− 400
+ . . .
or smaller. Thus, the
, answer is correct for the first 100 digits.
For a final example which reveals the limitations of your calculator, evaluate
1 −
√
1 − 3 × 10 − 30
The answer is not zero. Analytically, find an approximation to the answer. In this context, the word “analytically” means that you should use algebra and calculus to find the answer.
And you shouldn’t touch a calculator or computer.
Hint: use a Taylor expansion.

6
H.
Radioactivity and a simple differential equation
The radioactive nucleus
14
C spontaneously decays into
14
N a β
−
ν e
. That is to say, carbon–14 decays into nitrogen–14, a beta particle (also known as an electron), and an anti-electron-neutrino, which is generically described as just a neutrino. If you start with a glass full of 14 C today, then in 5730 years you will only have half a glass of 14 C. After a total of 11460 years only a quarter of a glass will remain. And so forth. We say that the half-life of 14 C is 5730 years.
In general for any radioactive particle, if we start at t = 0 with N
0 a time t the number remaining is particles, then after
N ( t ) = N
0
1
2 t/t
1 / 2 where t
1 / 2 is the half-life.
Radioactive decay gives one example of a number N ( t ) whose rate-of-change in time dN ( t ) /dt is proportional to the number N ( t ) itself. In other words, dN ( t )
∝ N ( t ) .
dt
For definiteness assume that dN ( t )
= − λN ( t ) , dt where λ is a constant. We solve this differential equation by rewriting it as dN ( t )
= − λ dt
N and integrating both sides
Z dN ( t )
N
= −
Z
λ dt or ln N = − λt + constant
N = N
0 e
− λt is a general solution to the differential equation where ln( N
0
) = constant is a constant of integration determined by the initial conditions .
The last line follows by taking the logarithm of both side of the previous equation.
With radioactivity, we often define the “ e -folding time” τ ≡ 1 /λ , which also happens to be the “mean-lifetime” of the particle, so that
N ( t ) = N
0 e
− t/τ
.
τ is called the e -folding time because the number of particles decreases by a factor of e after a time τ .
It is easy to see the relationship between t
1 / 2 and τ by starting with
N
0 e
− t/τ
= N
0
1
2 t/t
1 / 2
.

7
Now divide out the N
0
, and take the natural logarithm of both sides t
−
τ
= t t
1 / 2 ln
1
2
.
(We are using ln( A
B
) = B ln( A ) and ln e = 1)
Finally, cancel the t , invert each side of the equation, and use the fact that ln(1 / 2) = − ln 2.
The result is t
1 / 2
= τ ln 2 .
Note that
τ (
14
C) = 5370 yr / ln 2 = 7750 yr is the e -folding time of
14
C.
The mean-lifetime ( e -folding time) of a muon is about 2 µ s. So, the half-life of a muon is about t
1 / 2
(muon) = ln 2 × 2 µ s ≈ 1 .
4 µ s. .
1.
Carbon dating
Carbon–14 14 C and carbon–13 13 C interact chemically in nearly identical ways.
14 C is created naturally by nuclear reactions from cosmic ray neutrons colliding with atmospheric nitrogen 14 N; the nuclear event also gives off a free proton (hydrogen nucleus). Living things, like you or like a tree, absorb chemically both stable. But once the tree dies,
13 C and 14 C while they are growing.
13 C is
14 C is no longer absorbed by the tree and the amount of 14 C decays with a half-life of 5730 yr. By measuring the ratio of 14 C to 13 C in the atmosphere, and comparing it to the ratio of what fraction of the original
14
C to
13
C in a wooden spear, archeologists can calculate
14
C in the spear has decayed, and from that fraction they can determine how long ago the spear was a growing tree branch.
This same basic technique is used to date accurately many archeological relics over time scales of hundreds of years to a few ten-thousands of years. For shorter times, not enough of the 14 C has decayed to measure accurately, and for longer time not enough of the 14 C remains to measure accurately.
Similar techniques, using different radioactive isotopes are used to determine the “lifetime” of rocks from the moon, rocks on earth and meteorites. In this context the “lifetime” of a rock is the time since the rock solidified.
2.
A note on nuclear power
Today, we have the technology to manufacture nuclear power. By and large, this can be done safely if it is done carefully. The typical daily radioactive waste emission from a nuclear power plant, directly into the atmosphere, is actually less than daily radioactive waste emission into the atmosphere from a coal-fired power plant. However, a significant roadblock to “going nuclear” is the disposal of nuclear waste which is collected inside the reactor and stored “off-site.”
Every nuclear reactor (or explosion of a nuclear weapon) creates nuclear waste with a mixture of many different radioactive isotopes. Some of these have reasonably short halflifes such as iodine-126 with a half-life of 13 days—the iodine-126 from a nuclear test is essentially gone after a year. (A year is about 24 half-lifes and (1 / 2)
− 24 ≈ 6 × 10
− 8 .
Only
1 part in 10 million remains after a year, and only 1 part in 10 14 after two years.) Some of

8 the radioactive isotopes have very long half-lifes, such as Uranium-238 with a half-life of 4.5
billion years. Essentially none of the 238 U decays over a time span comparable to the time that Man has been walking on this planet.
The disposal of nuclear waste remains a significant technological problem.