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Mathematical Review for Physical Chemistry Students
Thomas L. Beck
Department of Chemistry
University of Cincinnati
1
Introduction:
This set of notes reviews the basic mathematical techniques that are necessary
to understand the material in an upper level undergraduate physical chemistry
course. The notes are not intended to be a complete course in mathematical
methods, but rather they are designed to give you the essential tools that will
be required to solve the assigned problems. Physical chemistry is a fascinating
subject since it deals with a fundamental understanding of the world around us,
but it does require a modest level of skill in using mathematics. Mathematics is
the language that people use to describe and interpret the
physical
world, so just
as it would be necessary to learn a new language to function in another country,
one must learn the language of the physical sciences to be able to communicate
an understanding of physics and chemistry. So consider this a Berlitz Guide to
your first physical chemistry course, and use it to get functioning with the at
least partially new language. For some it will be just a review.
2
Calculus:
The two essential operations in calculus are differentiation and integration. We
will first consider functions of one variable such as
y
=
f
(x). We will assume
the functions are
well behaved
meaning the derivatives exist and the functions
are single valued for a given value of
x.
2.1
Derivatives:
The derivative of a function is the slope of a function at a point. It is the limit
of the ratio:
f
(x +
h)
−
f
(x)
df
(x)
= lim
.
h→0
dx
h
Some common derivatives we will need are:
(1)
1
de
x
=
e
x
dx
de
ax
=
ae
x
dx
d
cos
x
=
−
sin
x
dx
d
sin
x
= cos
x
dx
dx
n
=
nx
n−1
dx
d
log
x
= 1/x
dx
(2)
(3)
(4)
(5)
(6)
(7)
The chain rule says ‘take the derivative of the outside first and then work in-
ward. . . ’. Mathematically that means
df dg
df
(g(x))
=
.
dx
dg dx
(8)
Equation 3 above is an example of this rule. See the CRC Handbook for lots of
derivatives and integrals,
etc.
2.2
Partial derivatives:
Say we have a function of several variables:
f
(x,
y, z)
= 0. It may seem strange
to make it equal to zero, but let’s say there was a constant on the rhs, then
we could always just subtract it from both sides to get zero on the right and a
slightly more complicated expression on the left. The concept of partial deriva-
tive comes from taking the derivative with respect to one of the variables while
holding the others constant. We can write a small change in
f
(x,
y, z),
i.e.
df
as:
df
=
∂f
∂x
dx
+
y,z
∂f
∂y
dy
+
x,z
∂f
∂z
dz
x,y
(9)
where the subscripted variables mean they are both held constant while the
other one is varied.
For example, look at the equation for the surface of a sphere:
x
2
+
y
2
+
z
2
=
r
2
2
(10)
Then:
∂f
∂x
= 2x.
y,z
(11)
Or for the function
f
(x,
y, z)
=
x
2
y
3
z
4
we have:
∂f
∂z
= 4x
2
y
3
z
3
.
x,y
(12)
If we want to consider second derivatives, an example is (from the previous
function):
∂
2
f
∂z
2
= 12x
2
y
3
z
2
.
x,y
(13)
For a mixed derivative, we first take the derivative with respect to one variable,
holding the other two constant, and then do the same for the next variable.
∂
2
f
∂x∂y
= 6xy
2
z
4
.
(14)
The variables that are held constant are omitted since they change for the two
variables. For the cases that we will consider, the order of the differentiation
does not matter.
So it’s pretty simple.
2.3
Integral:
The integral is the limit of the sum of small areas under a curve:
N
I
= lim
h→0
f
(x
i
)h
i
(15)
where
f
(x
i
) is the height and
h
is the width of the small area centered at the
point
x
i
. In the limit of very small
h,
this equation goes over to the integral:
I
=
f
(x)dx
(16)
3
The integral above is an
indefinite
integral since there are no limits. The
definite
integral has limits, say
a
and
b:
b
I
=
a
f
(x)dx
(17)
Integration is the ‘inverse operation’ of differentiation, so if you know a deriva-
tive you can work backwards to get an integral. A couple examples of indefinite
integrals are:
I
=
e
x
dx
=
e
x
+
c
(18)
I
=
cos
xdx
= sin
x
+
c
(19)
If we change an indefinite integral to a definite integral then we take the differ-
ence of the function on the rhs at the two limits of integration, and since the
constant is the same on both sides, it drops out:
b
I
=
a
cos
xdx
= sin
x|
b
= sin
b
−
sin
a.
a
(20)
An important method for solving integrals is integration by parts. This comes
from the identity:
d(uv)
=
udv
+
vdu
If we move one term to the other side and then integrate, we get:
udv
=
uv
−
vdu.
(22)
(21)
2.4
Multiple integrals:
The concept of integral can be extended to multiple dimensions. In fact some
integrals in Quantum Mechanics are formally infinite dimensional! Don’t worry,
we won’t have to do those by hand. In words, you do one integral first, holding
the other variable constant, then do the second and so on. You write a two
dimensional integral like this:
4
b
d
I
=
a
c
f
(x,
y)dxdy.
(23)
If the function
f
(x,
y)
happens to be of the form
f
(x,
y)
=
g(x)h(y)
(separable)
then the integral breaks down into a product of two one dimensional integrals:
b
a
c
d
b
d
f
(x,
y)dxdy
=
a
g(x)dx
c
h(y)dy.
(24)
If you can make the integral separable, that’s the easiest way to solve it, but
most often that’s not so easy.
2.5
Integration change of variables:
Sometimes we need to change the variables for an integral, say from
x
and
y
to
r
and
θ,
polar coordinates. Remember
x
=
r
cos
θ
and
y
=
r
sin
θ.
The coordinate
ranges are: 0
≤
r
≤ ∞;
0
≤
θ
≤
2π. We need to express the function
f
(x,
y)
in
terms of
r
and
θ
and then change the ‘volume element’ (here area element) from
dxdy
to
Jdrdθ,
where
J
is called the Jacobian of the coordinate transformation.
The Jacobian is the
determinant
given below:
J
=
The 2-d determinant is given by:
a
c
b
d
=
ad
−
bc.
(26)
∂x
∂r
∂y
∂r
∂x
∂θ
∂y
∂θ
(25)
So if you do the four partial derivatives and plug them in, you get
J
=
r(cos
2
θ
+ sin
2
θ)
(27)
But we know that cos
2
θ
+ sin
2
θ
= 1 so
J
=
r
for the two dimensional change
of variables. Then the ‘volume element’ (here area element) change is
dxdy
→
rdrdθ.
Note the units stay the same (as they should), ı.e. area for both.
Remember angles don’t have units. For the 3d case, the variable change is:
x
=
r
sin
θ
cos
φ; y
=
r
sin
θ
sin
φ; z
=
r
cos
θ.
The coordinate ranges are: 0
≤
r
≤ ∞;
0
≤
θ
≤
π;
0
≤
φ
≤
2π. The determinant is a little more complicated
(you can look up the 3d determinant in a calculus text), and for the change of
volume elements we get
dxdydz
→
r
2
dr
sin
θdθdφ.
5

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