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Indefinite Quadratic Forms

and the Invariance of the Interval

in Special Relativity

John H. Elton

Abstract. In this note, a simple theorem on proportionality of indefinite real quadratic forms

is proved, and is used to clarify the proof of the invariance of the interval in special relativity

from Einstein’s postulate on the universality of the speed of light; students are often rightfully

confused by the incomplete or incorrect proofs given in many texts. The result is illuminated

and generalized using Hilbert’s Nullstellensatz, allowing one form to be a homogeneous poly-

nomial which is not necessarily quadratic. Also a condition for simultaneous diagonalizability

of semi-definite real quadratic forms is given.

1. INTRODUCTION. In the special theory of relativity, an event is a point in space-

time whose coordinates with respect to an inertial reference frame correspond to some

point (t, x, y, z) in R4. Coordinates of events in different inertial reference frames are

assumed to be connected by linear transformations, based on the assumption of homo-

geneity and isotropy of space-time. A famous postulate of Einstein is the universality

of the speed of light: the speed of light in a vacuum is the same in all inertial reference

frames, independent of the motion of the source. One can use the postulate of the uni-

versality of the speed of light, together with the assumption that changes of coordinates

are linear, to determine what changes of coordinates are possible. The idea is to use

this postulate to directly show the invariance of a certain quadratic function of the co-

ordinates, which can in turn be used to determine the linear transformations connecting

the coordinates (called Lorentz transformations). Defining the Lorentz transformations

as the group of linear transformations which leave this quadratic function invariant is

geometrically very appealing. To be most satisfying, and not circular, the invariance of

the quadratic function should be shown to be a simple and immediate consequence of

the postulates; the Lorentz transformations should only then be developed after that.

Suppose points in space-time are specified by (t, x, y, z) in one inertial reference

frame K , and by (t ′, x ′, y′, z′) in a second inertial reference frame K ′ whose origin

coincides with the first (that is, t = 0, x = 0, y = 0, z = 0 in K corresponds to the

same event as t ′ = 0, x ′ = 0, y′ = 0, z′ = 0 in K ′). Let a pulse of light be emit-

ted at this common event. Then events on the wave front have coordinates satisfying

x2 + y2 + z2 − c2t2 = 0 in system K , and also x ′2 + y′2 + z′2 − c2t ′2 = 0 in system

K ′, where c, the speed of light, is the same in both systems. This is from Einstein’s

postulate.

In 1966 the author was taking a course in “modern” physics, and remembers being

puzzled by the next step taken in the text [8, p. 58]. The text simply assumed without

further ado that x2 + y2 + z2 − c2t2 = x ′2 + y′2 + z′2 − c2t ′2 for all events (not just

doi:10.4169/000298910X492826

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c© THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 117

those on the wave front of the pulse, when both expressions are zero) and proceeded to

use that for a derivation of the form of the Lorentz transformations. Looking in some

other texts, we found the same “unconscious” assumption of the invariance of the

interval x2 + y2 + z2 − c2t2. In [6, p. 90], it is even stated that “(3.27) x2 + y2 + z2 −

c2t2 = 0”; “(3.28) x ′2 + y′2 + z′2 − c2t ′2 = 0”; and then the amazing statement “...

equating lines 3.27 and 3.28, we conclude x2 + y2 + z2 − c2t2 = x ′2 + y′2 + z′2 −

c2t ′2.” So our confusion remained unresolved for the moment, puzzled by the logic of

“things that are equal when zero are always equal” that seemed to be used in these

books.

Next semester the author took a course in classical mechanics using the text by

J. B. Marion [2]. Appendix G of that book has a demonstration of the invariance of

the interval arguing directly from Einstein’s postulates, acknowledging the issue that

concerned us. (This text is still popular today.) Here is the beginning of the proof given

in Appendix G, p. 558, of that book:

(∗) The wave front is described by x2 + y2 + z2 − c2t2 = s2 = 0 in K , and x ′2 +

y′2 + z′2 − c2t ′2 = s ′2 = 0 in K ′. “The equations of the transformation that con-

nect the coordinates (t, x, y, z) in K and (t ′, x ′, y′, z′) in K must themselves be

linear. In such a case the quadratic forms s2 and s ′2 can be connected by, at

most, a proportionality factor: s ′2 = κs2.” (It is then shown by further argu-

ments using homogeneity, isotropy, and continuity that in fact κ = 1.)

We are of the opinion that the statement above about the reason for the proportion-

ality of the quadratic forms would be misleading to many readers. It is not generally

true that if one quadratic form is the result of making a linear change of variables

in another quadratic form, and the two quadratic forms have the same zero set, then

they must be proportional (even when this zero set has infinitely many points). Here

is a somewhat arbitrary example with three variables: Let s2 = 2x2 + 2y2 + z2 −

2xz − 2yz, and let s ′2 = 2x ′2 + 2y′2 + z′2 − 2x ′z′ − 2y′z′, where x ′ = −2x − 2y + z,

y′ = 2y − 2z, and z′ = −2z, so the coordinates are connected by a linear transforma-

tion. Algebra shows that s ′2 = 8x2 + 16y2 + 10z2 + 16xy − 16xz − 24yz, which is

clearly not proportional to s2. Yet both quadratic forms are zero on the same infinite

set {(x, y, z) ∈ R3 : z = x + y, x = y}, which is apparent after we reveal that actually

s2 = (x + y − z)2 + (x − y)2 and s ′2 = 8(x + y − z)2 + 2(2y − z)2, noting that if

x + y = z, then x = y if and only if 2y = z. For another sort of example (not really

related to the statement in Marion but relevant later in this paper), in two variables, let

s2 = x2 + y2 − 2xy and s ′2 = x2 − y2. Then s2 = 0 ⇒ s ′2 = 0, yet these quadratic

forms are not even simultaneously diagonalizable.

So it would seem the statement about proportionality of the quadratic forms could

use further explanation. The author fashioned a proof for himself, but remained puz-

zled why the books seemed unconcerned about the logical gap.

Fast-forwarding 43 years, we recently had occasion, after not thinking about physics

since being an undergraduate, to come upon this topic again. The 1985 text on general

relativity by Schutz [7, p. 32] gives a logically correct argument for the proportionality

of the quadratic forms in (∗). But this does not seem to have been propagated to the

community of physics students and textbook writers. From the 2006 relativity text

[3], we find on page 10 essentially the same puzzling statements that occurred in the

1964 text [6] mentioned above: “c2t2 − x2 − y2 − z2 = 0”; “c2t ′2 − x ′2 − y′2 + z′2 =

0”; “These are equal, so c2t2 − x2 − y2 − z2 = c2t ′2 − x ′2 − y′2 + z′2.” And we have

evidence, from the Physics Forums [5], that indeed other physics students are still

finding themselves confused by exactly the same thing that we found unexplained

so long ago! The answers we saw given by other students there were unfortunately

June–July 2010]

NOTES

541

not correct and were essentially on the level of the “unconscious” proofs of some of

those texts, along with some rather arrogant statements about the students who didn’t

understand the “proofs” they saw in their books.

So we decided this time to fill in the gap, for the benefit of others who might be

confused, by stating and proving a more general but very simple result about indefinite

quadratic functions that settles the matter. This result about containment of zero sets

suggests a more general result, proved using Hilbert’s Nullstellensatz. Also we prove a

simple result about simultaneous diagonalization of semidefinite quadratic forms and

containment of zero sets.

2. A THEOREM ON INDEFINITE QUADRATIC FORMS. A function q :

R

n → R is a real quadratic form if there is a symmetric bilinear functionq̃ :

R

n × Rn → R such that q(x) =q̃(x, x). In matrix language, this means there is a

symmetric n × n matrix Q = [Qij ] of real numbers such that q(x1,... , xn) = q(x) =

∑n

i=1

∑n

j=1 Qij xi x j , i.e., q(x) = xt Qx for x ∈ Rn . The elements of the matrix Q are

the components ofq̃ in the standard basis.

A real quadratic form q is indefinite if it takes both positive and negative values;

this is equivalent to the matrix Q having at least one positive eigenvalue and at least

one negative eigenvalue. See [4] for example, or any book on linear algebra.

For a real quadratic form q, define Zq = {x ∈ Rn : q(x) = 0}; this is the zero set

of q.

Theorem 1. Let q be an indefinite real quadratic form on Rn, and let r be a real

quadratic form on Rn such that Zq ⊂ Zr ; that is, q(x) = 0 ⇒ r(x) = 0. Then r is

proportional to q; that is, there exists a real number α such that r(x) = αq(x) for all

x. If α is not zero, then r is also indefinite and has the same zero set as q.

Proof. There exists a basis {v1,... , vn} for Rn such that the matrix Qij =q̃(vi , v j )

representing q in this basis is diagonal, with only 1’s, −1’s, and 0’s on the diagonal, and

Qii = 1 for 1 ≤ i ≤ k; Qii = −1 for k + 1 ≤ i ≤ k + m; Qii = 0 for k + m + 1 ≤

i ≤ n; and Qij = 0 for i

= j . The numbers k and m here are unique: k is the number

of positive eigenvalues and m is the number of negative eigenvalues of any matrix

representing q (Sylvester’s law of inertia; see [4, p. 202]). Since q is indefinite, k > 0

and m > 0. So without loss of generality, in the proof which follows we will just

assume that Q is a diagonal matrix with k ones and m negative ones and the rest

(if any) zeroes on the diagonal, in order, as described above. (In the application to

invariance of the interval which motivated this discussion, Q is already of this form,

but we wanted to treat the general case.) Let R be the symmetric matrix representing

r in this basis.

The idea is to make judicious choices of points where q is zero, and to conclude

that R must also be diagonal and that the on-diagonal elements of R are a common

multiple of those of Q.

To that end, let j be an integer such that k + 1 ≤ j ≤ k + m. Let x have compo-

nents x1 = 1, x j = 1, and all other components zero. Then q(x) = Q11x21 + Q jj x2j =

1 − 1 = 0, so r(x) = R11x21 + R jj x2j + 2R1 j x1x j = R11 + R jj + 2R1 j = 0, by hy-

pothesis. Now change the sign of the j th component of x so that x j = −1 but leave

the other components of x unchanged; then q(x) = 0 still, so r(x) = R11x21 + R jj x2j +

2R1 j x1x j = R11 + R jj − 2R1 j = 0 also. These two equations together imply R1 j = 0

and then R jj = −R11. Then for 1 < i ≤ k, using i in place of 1 in the argument above

shows Rii = −R jj = R11, and Rij = 0.

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c© THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 117

Next let j be an integer (if any) such that m + k + 1 ≤ j ≤ n. First let x be

the vector with x j = 1 and all other components zero. Then q(x) = Q jj = 0, so

r(x) = R jj = 0. Next let 1 < i ≤ k, k + 1 ≤ l ≤ k + m, and let x be the vector

with components xi = 1, xl = 1, x j = 1, and all other components zero. Then q(x) =

Qii + Qll + Q jj = 1 − 1 + 0 = 0, so r(x) = Rii + Rll + R jj + 2Ril + 2Rij + 2Rlj =

2Rij + 2Rlj = 0 also. Changing x so that xi = −1 and x is otherwise unchanged leads

to −2Rij + 2Rlj = 0. This implies that Rij = 0, and then Rlj = 0.

Suppose that k ≥ 2. Let 1 ≤ i < j ≤ k, k + 1 ≤ l ≤ k + m, and let x be the vec-

tor with components xi = 3, x j = 4, xl = 5, and all other components zero. Then

q(x) = Qii x2

i + Q jj x2j + Qll x2

l = 9 + 16 − 25 = 0, so r(x) = Rii x2

i + R jj x2j +

Rll x2

l + 2Rij xi x j + 2Ril xi xl + 2R jl x j xl = R11(9 + 16 − 25) + 2Rij (12) = 0 also

(note we have already shown that Ril = R jl = 0 and the proportionality of the di-

agonal elements). This proves Rij = 0. Similarly, if m ≥ 2 or n − (k + m) ≥ 2, the

corresponding off-diagonal terms of R are zero.

This completes the proof that R = R11 Q, and the proof of the theorem.

3. AN ALTERNATE PROOF USING HILBERT’S NULLSTELLENSATZ,

AND A STRONGER RESULT. The containment of zero sets in the hypothesis

of Theorem 1 suggests Hilbert’s Nullstellensatz [1, p. 254], of importance in alge-

braic geometry. We can also prove Theorem 1 using this theorem rather than using

diagonalization and bases as we did above; and although the proof above is certainly

simple enough, there is some insight to be gained from this alternate proof, and a more

general result can be proved this way as well. The Nullstellensatz concerns zero sets

of ideals in the ring of polynomials in several variables over an algebraically closed

field. For our application the ideal in question will be simply the principal ideal gen-

erated by a single polynomial q. If the reader is not familiar with ideal theory and the

Nullstellensatz, it will not matter because we shall use only the following immediate

consequence of Hilbert’s theorem:

If q(x) and r(x) are complex polynomials in n variables such that x ∈ Cn and

q(x) = 0 implies r(x) = 0, then r p(x) = q(x)s(x) for some polynomial s(x) and pos-

itive integer p. If q is square-free (that is, the irreducible factors of q occur only to the

first power), p can be taken to be one.

In Theorem 1, q and r are quadratic forms with real coefficients, q is indefinite, and

the real zeroes of q are assumed to be zeros of r by hypothesis. If q were not square-

free, it would be the square of a linear polynomial or the negative of such a square,

contrary to the indefiniteness of q, so we can take p to be one in our application (it is

easy to see that q is actually irreducible when its rank exceeds two, but we don’t need

that). Thus all we need to do is to show that, as a consequence of the indefiniteness of

q, the complex zeroes of q are also zeroes of r , and the conclusion of Theorem 1 will

follow from the Nullstellensatz, since the degrees of q and r being two requires s to

be constant.

To that end, suppose q(x + iy) = 0, for some x, y ∈ Rn , so q(x) − q(y) = 0 and

q̃(x, y) = 0. If q(x) = 0 (hence q(y) = 0) then q(x + y) = 0, so r(x) = r(y) = r(x +

y) = 0, which impliesr̃(x, y) = 0 and so r(x + iy) = 0.

Suppose then that q(x) > 0 (the opposite case would be handled similarly); by

rescaling assume q(x) = 1. Since q is indefinite, there is u ∈ Rn such that q(u) <

0. Let w = u −q̃(u, x)x −q̃(u, y)y, soq̃(w, x) = 0 andq̃(w, y) = 0 (a “Gram-

Schmidt” construction). Now q(w) =q̃(w, u) = q(u)−q̃(u, x)2 −q̃(u, y)2 < 0. By

rescaling we may assume that q(w) = −1,q̃(w, x) = 0, andq̃(w, y) = 0. Thus q(w +

αx + βy) = −1 + α2 + β2 = 0 whenever α2 + β2 = 1, so by hypothesis r(w +

αx + βy) = r(w) + α2r(x) + β2r(y) + 2αr̃(w, x) + 2βr̃(w, y) + 2αβr̃(x, y) = 0

June–July 2010]

NOTES

543

also for α2 + β2 = 1. Taking α = ±1, β = 0, we conclude thatr̃(w, x) = 0, and sim-

ilarlyr̃(w, y) = 0. Then choosing α = 2−1/2, β = ±α, we conclude thatr̃(x, y) = 0.

Then choosing α = 1, β = 0 and α = 0, β = 1, we see that r(x) = r(y), and thus

r(x + iy) = 0, concluding the proof.

This proof from the Nullstellensatz is perhaps slightly cleaner than the first proof

of Theorem 1. But also one can prove more this way, with a little more work. The

quadratic form r is a polynomial in n variables in which each term has degree 2. In

general, a polynomial in n variables for which each term has the same degree d is

called a homogeneous polynomial of degree d.

Theorem 2. Suppose r is a homogeneous real polynomial in n variables, not neces-

sarily a quadratic form, with the other hypotheses of Theorem 1 unchanged. Then q is

a factor of r; that is, r(x) = q(x)s(x) for some polynomial s(x).

Proof. We only need to show that any complex zeroes of q are zeroes of r . Suppose

q(x + iy) = 0, and suppose that q(x) > 0. As above, we can assume that q(x) =

q(y) = 1,q̃(x, y) = 0, and there is w such that q(w) = −1, w is q-orthogonal to x

and y, and q(w + αx + βy) = 0, so r(w + αx + βy) = 0 also, whenever α2 + β2 = 1.

Suppose that r has even degree 2m. Now r(γw + αx + βy) is a homogeneous polyno-

mial of degree 2m in the variables α, β, and γ that is zero when γ = 1 and α2 + β2 =

1. We may write r(γw + αx + βy) = ∑ j+k≤2m α jβkγ 2m− j−kc( j, k) where the indices

j and k are nonnegative. By changing the signs of α and β separately, and then to-

gether, we see that for γ = 1 and α2 + β2 = 1,

∑

j+k≤2m, j odd, k even

α jβkc( j, k) = 0,

∑

j+k≤2m, j even, k odd

α jβkc( j, k) = 0,

∑

j+k≤2m, j and k odd

α jβkc( j, k) = 0,

and

∑

j+k≤2m, j and k even

α jβkc( j, k) = 0.

Consider the last expression above (with both indices even), which can be rewritten

with a change of indices as

∑

j+k≤m

(α2) j (1 − α2)kc(2 j, 2k) = 0

for α2 ≤ 1. This is a polynomial of degree 2m in α; the coefficient of the highest power

term α2m must be zero because of the constancy of the polynomial on an infinite set,

so

∑

j≤m

(−1)m− j c(2 j, 2m − 2 j) = 0.

Next consider the next-to-last expression (with both indices odd) which can be rewrit-

ten

αβ

∑

j+k≤m−1

(α2) j (1 − α2)kc(2 j + 1, 2k + 1) = 0,

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c© THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 117

so for 0 < α2 < 1,

∑

j+k≤m−1

(α2) j (1 − α2)kc(2 j + 1, 2k + 1) = 0.

Setting the coefficient of the highest power term in the polynomial in α (which occurs

when k = m − j − 1) to zero, we get

∑

j≤m−1

(−1)m− j−1c(2 j + 1, 2m − 2 j − 1) = 0.

Since r is homogeneous of degree 2m,

r(x + iy) = r(0w + 1x + iy)

=

∑

j≤m

(−1)m− j c(2 j, 2m − 2 j)

+ i

∑

j≤m−1

(−1)m− j−1c(2 j + 1, 2m − 2 j − 1),

because i2m−2 j = (−1)m− j and i2m−2 j−1 = (−1)m− j−1i . The results just proved show

this is zero, completing the proof of the theorem when the degree of r is even and

q(x) > 0.

Now suppose r has odd degree 2m − 1. Then

r(γw + αx + βy) =

∑

j+k≤2m−1

α jβkγ 2m−1− j−kc( j, k),

and this breaks into four sums equaling zero when γ = 1 and α2 + β2 = 1 as before,

depending on the parities of the indices. Consider first the sum corresponding to j odd

and k even; this can be rewritten

α

∑

j+k≤m−1

(α2) j (1 − α2)kc(2 j + 1, 2k) = 0

for α2 ≤ 1. Setting the coefficient of the highest power to zero gives

∑

j≤m−1

(−1)m− j−1c(2 j + 1, 2m − 2 j − 2) = 0.

Now consider the sum corresponding to j even and k odd, which can be rewritten

β

∑

j+k≤m−1

(α2) j (1 − α2)kc(2 j, 2k + 1) = 0,

so

∑

j+k≤m−1

(α2) j (1 − α2)kc(2 j, 2k + 1) = 0

for α2 < 1, which implies

∑

j≤m−1

(−1)m− j−1c(2 j, 2m − 2 j − 1) = 0.

June–July 2010]

NOTES

545

But

r(x + iy) = r(0w + 1x + iy)

=

∑

j≤m−1

(−1)m− j−1c(2 j + 1, 2m − 2 j − 2)

+ i

∑

j≤m−1

(−1)m− j−1c(2 j, 2m − 2 j − 1),

so this is zero, and the proof is concluded when r is of odd degree and q(x) > 0.

The case q(x) < 0 is handled in a similar way. Finally, if q(x + iy) = 0 and q(x) =

0, then since q(y) = 0 andq̃(x, y) = 0, q(x + αy) = 0 and thus r(x + αy) = 0 for all

real numbers α. Now r(x + αy) is a polynomial in α which is identically zero, so all

its coefficients are zero, and this clearly implies that r(x + iy) = 0, which concludes

the proof of Theorem 2.

4. SIMULTANEOUS DIAGONALIZATION OF QUADRATIC FORMS. Theo-

rem 1 implies a result on simultaneous diagonalizability: if q is an indefinite real

quadratic form on Rn and r is a real quadratic form on Rn such that Zq ⊂ Zr , then

q and r are simultaneously diagonalizable (meaning there is a basis in which the ma-

trices representing q and r are both diagonal).

However, if q is a semi-definite real quadratic form on Rn (semi-definite means

q(x) ≥ 0 for all x or q(x) ≤ 0 for all x), and r is a real quadratic form on Rn such

that Zq ⊂ Zr , then q and r are not necessarily simultaneously diagonalizable. For

an example in R2, let q(x) = (x − y)2 and let r(x) = x2 − y2; this example (already

mentioned in the introduction) satisfies the conditions and it is easy to see these are

not simultaneously diagonalizable.

But if q and r are both assumed semi-definite, there is a similar (and similarly easy)

result on containment of zero sets implying simultaneous diagonalizability.

Theorem 3. Let q and r be semi-definite real quadratic forms on Rn such that Zq ⊂

Zr . Then r and q are simultaneously diagonalizable.

Proof. Without loss of generality assume they are both positive semi-definite. First ob-

serve that the zero sets are subspaces: Let q(x) = 0 and q(y) = 0; then q(ax + by) =

a2q(x) + b2q(y) + 2abq̃(x, y) = 2abq̃(x, y) ≥ 0 for all real numbers a, b implies

q̃(x, y) = 0, so q(ax + by) = 0. This is quite different from the indefinite case where

the zero sets are cones and not subspaces.

There is a subspace M such that Rn = M ⊕ Zq (choose any basis for Zq and ex-

tend it to a basis for Rn , and M is the span of those added-on basis vectors). Let

x = y + z with y ∈ M and z ∈ Zq . Then q(y + αz) = q(y)+ 2αq̃(y, z)+ α2q(z) =

q(y)+ 2αq̃(y, z) ≥ 0 for all real α impliesq̃(x, y) = 0, so q(y + z) = q(y). Similarly,

r(y + z) = r(y) for y ∈ M and z ∈ Zq , since Zq ⊂ Zr .

Thus q and r may be considered as positive semi-definite quadratic forms on M ,

and in fact q is positive definite on M , because if y ∈ M and q(y) = 0, then y ∈ Zq

by definition, so y = 0. By a well-known theorem [4, p. 218], this implies q and r are

simultaneously diagonalizable on M , and they are then simultaneously diagonalizable

on Rn = M ⊕ Zq , with zeroes on the diagonal corresponding to the basis vectors for

Zq .

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5. APPLICATION TO THE PROOF OF INVARIANCE OF THE INTERVAL.

Suppose the coordinates x = (t, x, y, z) in K and x′ = (t ′, x ′, y′, z′) in K ′ are con-

nected by a linear transformation, so x′ = Lx for some 4 × 4 matrix L . Let q(x) =

−c2t2 + x2 + y2 + z2 = xt Qx, where Q is the diagonal matrix with diagonal en-

tries (−c2, 1, 1, 1). Let r(x) = −c2t ′2 + x ′2 + y′2 + z′2 = (Lx)t QLx = xt(Lt QL)x,

so r(x) = xt Rx, where R = Lt QL . Now q is indefinite, and r(x) = 0 precisely when

q(x) = 0, from (∗) above. So the conditions of Theorem 1 are in force, and we may

conclude that r is proportional to q, which is equivalent to the statement from (∗) that

we wanted to prove, namely that s ′2 is proportional to s2.

ACKNOWLEDGMENT. We would like to thank Michael Loss for suggesting looking at Hilbert’s Nullstel-

lensatz for a connection with the topic in this note.

REFERENCES

1. N. Jacobson, Lectures in Abstract Algebra, vol. III, Van Nostrand, New York, 1964.

2.

J. B. Marion, Classical Dynamics of Particles and Systems, Academic Press, New York, 1965.

3. D. McMahon, Relativity Demystified, McGraw-Hill, New York, 2006.

4. G. D. Mostow and J. H. Sampson, Linear Algebra, McGraw-Hill, New York, 1969.

5. Physics Forums, Proving invariance of spacetime interval, available at http://www.physicsforums.

com/showthread.php?t=115451.

6. W. G. V. Rosser, An Introduction to the Theory of Relativity, Butterworths, London, 1964.

7. B. F. Schutz, A First Course in General Relativity, Cambridge University Press, Cambridge, 1985.

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School of Mathematics, Georgia Institute of Technology, Atlanta, GA 30332

and Accusoft-Pegasus Imaging, 4001 N. Riverside Drive, Tampa, FL 33603

elton@math.gatech.edu

Monotone Convergence Theorem

for the Riemann Integral

Brian S. Thomson

Abstract. The monotone convergence theorem holds for the Riemann integral, provided (of

course) it is assumed that the limit function is Riemann integrable. It might be thought, though,

that this would be difficult to prove and inappropriate for an undergraduate course. In fact the

identity is elementary: in the Lebesgue theory it is only the integrability of the limit function

that is deep. This article shows how to prove the monotone convergence theorem for Riemann

integrals using a simple compactness argument (i.e., invoking Cousin’s lemma). This material

could reasonably and appropriately be used in classroom presentations where the students are

indoctrinated on this antiquated, but still popular, integration theory.

The monotone convergence theorem is usually stated and proved for the Lebesgue in-

tegral, but there is little difficulty in formulating and proving a version for the Riemann

integral.

doi:10.4169/000298910X492835

June–July 2010]

NOTES

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