4 1. PROLOGUE

Vectors x such that

x2

0,

x2

= 0, or

x2

0 are called timelike, lightlike, or

spacelike, respectively.

In this framework, the notation for derivatives on

R4

is

∂μ =

∂

∂xμ

,

so that, with respect to traditional space-time coordinates x and t,

(1.3) (∂0,...,∂3) =

(c−1∂t,

∇x),

(∂0,...,∂3)

=

(c−1∂t,

−∇x),

and

∂2

is the wave operator or d’Alembertian:

∂2

= ∂0

2

− ∂1

2

− ∂2

2

− ∂3

2

=

c−2∂t 2

−

∇x.2

Integrals. Physicists like to indicate the dimensionality of their integrals ex-

plicitly by writing the volume element on

Rn

as

dnx.

This convention is occasionally

an aid to clarity, and we shall generally follow it. Another convention for integrals

more commonly used in physics than in mathematics is to put the differential next

to the integral sign: dx f(x) rather than f(x) dx. This can also be an aid to

clarity, particularly in multiple integrals where writing

b

a

dx

d

c

dy f(x, y) in pref-

erence to

b

a

d

c

f(x, y) dy dx makes it easier to see which variables go with which

limits of integration. However, for reasons of inertia more than anything else, we

shall not adopt this convention.

Fourier transforms. If we are considering functions on the space Rn equipped

with the Euclidean inner product (x, y) → x · y, the Fourier transform f → f and

its inverse f → f

∨

will be defined by

f(y) =

e−ix·yf(x) dnx,

f

∨(x)

=

eix·yf(y)

dny

(2π)n

.

The Parseval identity is then

(1.4)

|f(x)|2 dnx

=

|f(y)|2

dny

(2π)n

.

However, most of the time the Fourier transform will pertain to functions on

R4

equipped with the Lorentz metric, in which case we define

f(p) =

eipμxμ

f(x)

d4x,

f

∨

(x) =

e−ipμxμ

f(p)

d4p

(2π)4

.

(The Parseval identity is again (1.4), with n = 4.) Thus the Lorentz sign conven-

tion in the exponent agrees with the Euclidean one as far as the space variables

are concerned, but it is reversed for the time variable. In this situation, generally

x is in “position space” and p is in “momentum space,” in which case the

pμxμ

in the exponent really needs to be divided by a normalization factor so that the

argument of the exponent is a pure number, independent of the units of measure-

ment. In quantum mechanics this factor is Planck’s constant , but it will usually

be suppressed since we will be using units in which = 1.

Lord Kelvin once quipped that a mathematician is someone to whom it is

obvious that

∞

−∞

e−x2

dx =

√

π. In the same spirit, I submit that a mathematical

physicist is someone to whom it is obvious that

(1.5) δ(x) =

eix·y

dny

(2π)n

,