Notes on “Differential geometry and relativitytheory” (Faber, 1983)

Robert B. Scott,1,2∗

1Institute for Geophysics, Jackson School of Geosciences, The University of Texas at Austin,

Austin, Texas, USA2Department of Physics, University of Brest,

Brest, France

∗To whom correspondence should be addressed; E-mail: rscott@ig.utexas.edu.

April 14, 2012

2

Chapter 1

Surfaces and the concept ofcurvature

1.1 Curves

Exercises I-1

1. Length of one turn of the helix.

β(t) = (a cos t, a sin t, bt).

ds =√

(x′dt)2 + (y′dt)2 + (z′dt)2

=√a2 + b2 dt (1.1)

Thus the length of the helix is 2π√a2 + b2.

2. Parabola α(t) = (t, t2/2).Velocity, α′(t) = (1, t).Acceleration, α′′(t) = (0, 1).

3. (a) Note the typo in this question. It should read “. . . and verify that

3

4

||α′(0)|| = V0.”

α′(t) = (V0 cos θ, V0 sin θ − 32t),

α′(0) = (V0 cos θ, V0 sin θ),

||α′(0)|| = V0. (1.2)

3. (b)The acceleration is:

α′′(t) = (0,−32). (1.3)

The direction is in the negative y−direction and the magnitude is 32 perhapsin [ft s−2].

4. Particle of matter moves on the branch of a hyperbole:

α(t) = (a cosh t, a sinh t).

with a > 0.Double check claim:

x2 − y2 = (a cosh t)2 + (a sinh t)2

= a2(

exp (2 t) + exp (−2 t) + 2

4− exp (2 t) + exp (−2 t)− 2

4

)= a2. (1.4)

Velocity:α′(t) = (a sinh t, a cosh t).

Acceleration:α′′(t) = (a cosh t, a sinh t) = α(t).

The force will be proportional to the acceleration by Newton’s secondlaw. The acceleration vector is identical to the position vector, and henceproportional to the distance from the origin.

5. The curve is the parabola, α(t) = (t, t2/2). It cannot be reparameter-ized in terms of arc length. Using the chain rule it was derived that

k(t) =1

(1 + t2)3/2.

Rob’s notes on Faber 5

(a) Find the equation of the osculating circle at α(0).The curvature k(t) is the inverse of the radius of the osculating circle, c.f.

p. 8. So the radius of the circle is

1

k(0)= (1)3/2 = 1.

By symmetry the centre of the circle lies on the y−axis. The curve bendsupward (see Problem 2), so the centre is a y = 1. So the equation is

x2 + (y − 1)2 = 1.

(b) Find the equation of the osculating circle at α(2).We must find the vector normal to the curve. The unit normal is N(s),

which is

N(s) = T′(s)/k(s) =1√

1 + t2(−t, 1).

We find the centre of the circle by moving a distance 1/k in the direction Nfrom α(2), see formula on p. 8 which, when expressed in terms of t is:

c(t) = α(t) +1

k(t)N(t)

= (t, t2/2) + (1 + t2)3/21√

1 + t2(−t, 1)

= (−t3, 1 + 3t2/2). (1.5)

So c(2) = (−8, 6) and

(x+ 8)2 + (y − 6)2 = 53.

6. (a) Show the curvature of a general curve α(t) = (x(t), y(t)) is

k(t) =

∣∣∣∣x′(t) y′′(t)− x′′(t) y′(t)(x′(t)2 + y′(t)2)3/2

∣∣∣∣Stuck! I confirmed that this expression is correct using Wikepedia, but I

can’t derive it.

6. (b) Show that the graph y = f(x) has the given curvature.

6

Hereα(x) = (x, f(x)).

The unit tangent vector is found from the chain rule:

T =dα(s)

ds

=dα(x)/dx

ds/dx

(1.6)

To find the relation between arc length and x we note that (Boas, 1983,Chapter 9):

ds =√

(dx)2 + (f ′ dx)2,

=√

1 + f ′2 dx.

ds/dx =√

1 + f ′2. (1.7)

So the unit tangent vector is given by:

T(x) =dα(x)/dx

ds/dx,

=(1, f ′)√1 + f ′2

. (1.8)

And we use the chain rule again to find the curvature:

k =

∥∥∥∥dT(s)

ds

∥∥∥∥ ,=

∥∥∥∥dT(x)/dx

ds/dx

∥∥∥∥ .(1.9)

This gets a bit tedious, but is straightforward. Write T = (T x, T y). Thex component is

dT x

dx=

(1 + f ′2)−1/2

dx,

=−f ′ f ′′

(1 + f ′2)3/2. (1.10)

Rob’s notes on Faber 7

And a little algebra gives:

dT y

dx=f ′ (1 + f ′2)−1/2

dx,

=f ′′

(1 + f ′2)3/2. (1.11)

So

dT(s)

ds=

1

ds/dx

(dT x

dx,dT y

dx

)=

1√1 + f ′2

(−f ′ f ′′

(1 + f ′2)3/2,

f ′′

(1 + f ′2)3/2

),

=

(−f ′ f ′′

(1 + f ′2)2,

f ′′

(1 + f ′2)2

). (1.12)

And finally the curvature is the magnitude of this vector, which simplifiesto

k =

∥∥∥∥dT(x)/dx

ds/dx

∥∥∥∥ ,=

∣∣∣∣ f ′′

(1 + f ′2)3/2

∣∣∣∣ . (1.13)

6. (c) Compute the curvature of a semicircle

y =√a2 − x2.

From the part (b) above,

k =

∣∣∣∣ f ′′

(1 + f ′2)3/2

∣∣∣∣ . (1.14)

Here f(x) = y =√a2 − x2 so

f ′ =−x√a2 − x2

, (1.15)

and

f ′′′ =−a2

(a2 − x2)3/2, (1.16)

8

Fortunately this reduces to

k =

∣∣∣∣ f ′′

(1 + f ′2)3/2

∣∣∣∣ ,=

∣∣∣∣∣ −a2

(a2 − x2)3/21

( a2

a2−x2 )3/2

∣∣∣∣∣ ,=

1

a, (1.17)

as it should since a was the radius of the semicircle.

7. Letα(t) = (a cos t, b sin t),

for 0 ≤ t ≤ 2π, which defines an ellipse

x2

a2+y2

b2= 1.

Compute the curvature k(t) at t = 0 and t = π/2 using the formula fromExercise 6(a). We’ll need:

x(t) = a cos t,

x′ = −a sin t,

x′′ = −a cos t.

and

y(t) = b sin t,

y′ = b cos t,

y′′ = −b sin t.

So

k(0) =a

b2, (1.18)

k(π/2) =b

a2. (1.19)

We’ll need these later for solving Exercise I-2 problem 3.

8. This is clearly an important problem.

Rob’s notes on Faber 9

1.2 Gauss curvature (informal treatment)

The sign of the curvature is determined by the directions of the unit normalto the curve U and the “principal” normal vector of the curve αv known asthe “normal section”. But what is the “principal” normal vector.

Exercises I-2

1.3 Surfaces in E3

Definition of surface involved a vector-valued function of two variables, X :D → E3, defined on an open subset D of <2. Note that D is an open subsetof <2. (D will be referred to in later sections without definition.)

Definition of regular on p. 22. In my words, amounts to being able to movein 2 independent directions on the surface by changing the two parameters(u, v). In Faber’s words, p. 24, the regularity ensures that for each point ofD ∈ <2 there is an open neighbourhood Ω about this point X is one-to-oneand there is a continuous inverse function X(Ω) → Ω. So each point onX(Ω) ⊂M has a unique set of coordinates (u, v).

Defines a tangent vector and tangent plane, denoted TpM , on p. 26.

Exercises I-3

1 (a). Show X is regular if f(u) 6= 0, α′(u) 6= 0.

X(u, v) = (f(u) cos v, f(u) sin v, g(u))

The tangent vectors are

X1 =∂X

∂u= (f ′(u) cos v, f ′(u) sin v, g′(u))

X2 =∂X

∂v= (−f(u) sin v, f(u) cos v, 0) (1.20)

10

As long as their cross-product

X1 ×X2 = (−g′f cos v,−g′f sin v, f ′f)

(1.21)

is nonzero then have nonzero orthogonal components,

|X1 ×X2|2 = (g′2 + f ′2)f 2 = α′2 f 2

whereα′(u) = (f ′, 0, g′)

1 (b). (i) u−parameter curves.

Hold v = v0 fixed.

X(u, v0) = (f(u) cos v0, f(u) sin v0, g(u))

In the x− z plane we see the original curve,

α(u) = (f(u), 0, g(u))

but squished in the x−direction by cos v0. In the y − z plane it’s the samecurve but squished by sin v0. In short, it’s like the curve α(u) has beenrotated by v0 from the x − z plane. The coordinates in the x − z plane arethose of the shadow one would see if a light were shown through the curvetoward the x − z plane. To obtain the y coordinates one shines the lightthrough the curve toward the y − z plane.

(ii) v−parameter curves. Hold u = u0 fixed.

X(u0, v) = (f(u0) cos v, f(u0) sin v, g(u0))

This is a circle at height x = g(u0) with radius f(u0).

(iii) u−parameter and v−parameter curves intersect orthogonally.The tangent vectors

X1 =∂X

∂u= (f ′(u) cos v, f ′(u) sin v, g′(u))

X2 =∂X

∂v= (−f(u) sin v, f(u) cos v, 0) (1.22)

Rob’s notes on Faber 11

point in the directions of the u−parameter and v−parameter curves. Theyare orthogonal because

X1 ·X2 = 0

2 (a). Sketch profile curve in x−z plane and surface in 3D, show X, andgive equation for the surface in the form g(x, y, z) = 0.

X(u, v) = (u cos v, u sin v, a u)

In the x − z plane the curve is a straight line with slope one. The curvesform a cone with apex at the origin.

The tangent vectors are

X1 =∂X

∂u= (cos v, sin v, a)

X2 =∂X

∂v= (−u sin v, u cos v, 0) (1.23)

Their cross-product

X1 ×X2 = (−au cos v,−au sin v, u)

(1.24)

has magnitude|X1 ×X2|2 = u2(1 + a2)

so for u 6= 0, X is regular.We can express the surface in the form g(x, y, z) = 0 by

a2(x2 + y2)− z2 = 0

Substituting the components of X into this equation confirms it applies.

3(a) Show that the helicoid (example 9) is a ruled surface.

HelicoidX = (u cos v, u sin v, bv)

12

By inspection,

β(v) = (cos v, sin v, 0)

so then

uβ(v) = (u cos v, u sin v, 0)

and we require

α(v) = (0, 0, bv)

And thus we have expressed the helicoid surface in the form of a ruled surface:

X = α(v) + uβ(v)

3(b) Show that the hyperbolic parabola (example 5) is a doubly ruledsurface.

Hyperbolic parabola:

X(u, v) = (u− v, u+ v, 2uv)

By inspection,

β(u) = (−1,+1, 2u)

so then

vβ(u) = (−v,+v, 2uv)

and we require

α(u) = (u, u, 0)

And thus we have expressed the helicoid surface in the form of a ruled surface:

X = α(u) + vβ(u)

Interchanging u and v in X(u, v) = (u− v, u+ v, 2uv) we see that we canrepeat the above argument to obtain:

X = α(v) + uβ(v)

so the hyperbolic parabola is doubly ruled.

Rob’s notes on Faber 13

To show that it is regular we find the tangent vectors:

X1 = (1, 1, 2v)

X2 = (−1, 1, 2u) (1.25)

These are never parallel, as is clear from

||X1 ×X2||2 = ||2(u− v,−(u+ v), 1)||2 = 4[(u− v)2(u+ v)2 + 1] 6= 0

We should also confirm that these equations are indeed the hyperbolic parabolaof Example 5, which was expressed as

z =1

2(y2 − x2)

Substituting from X we find z = 2uv and

1

2(y2 − x2) =

1

2[(u+ v)2 − (u− v)2] = 2uv = z,

confirming it is the hyperbolic parabola.

4. If α = α(u) is a smooth curve in E3 and β is a non-zero vector withconstant coefficients, then the ruled surface X(u,v) = α(u) + vβ is calleda cylinder. (The special case of a right circular cylinder results when α is acircle and β is perpendicular to the plane of α.)

(a) Show that X is regular except where α′ × β = 0. Describe the u−and v− parameter curves.

Clearly X1 = α′ and X2 = β so,

X1 ×X2 = α′ × β

The surface will be regular when the tangent vectors are not parallel, whichmeans that X1 ×X2 6= 0.

The u−parameter curves are given by α(u), a general smooth function ofu producing a curve in E3.

The v−parameter curves are given by α(u0)+vβ. These are straight linesin the direction of the vector β that pass through α(u0).

14

(b) Sketch the cylinder for which

α(u) = (u, u2, 0), β = (0,−1, 2)

The u−parameter curves are parabolae parallel to the x − y plane. Theapex of the parabolae falls on the y − z plane along the line z = −2y. Thev−parameter curves are straight lines passing through the parabolae and areparallel to z = −2y in the y − z plane. The surface looks a bit like the hullof a ship but it extends forever downward and upward.

7. Let α(u) = (cos(u), sin(u), 0). Through each point of α(u), pass a unitline segment with midpoint α(u) and direction vector

β(u) =(

sinu

2

)α(u) +

(cos

u

2

)(0, 0, 1)

The resulting surface,

X(u, v) = α(u) + vβ(u), −1

2≤ v ≤ 1

2

is called a Mobis strip!

(7a) The coordinate functions:

x(u, v) = cos u+ v sinu

2cosu

y(u, v) = sinu+ v sinu

2sinu

z(u, v) = v cosu

2(1.26)

1.4 The first fundamental form

p. 34. Middle of page, unfortunate notation: “Any open subset of R2 mayserve as the domain of X”. Here, R is not the larger radius of the torus, itmeans R from Rn, the product space of ordered n−tuplets of real numbers.In fact, R2 was used on p. 22.

p. 35 Middle of page, typo: “In the new rotation” should be “notation”not “rotation”.

Rob’s notes on Faber 15

Exercises I-4

1. Derive the formula ds2 + dr2 + r2dθ2 for the differential of arc lengthin polar coordinates by substituting

x = r cos θ y = r sin θ

into the formula ds2 = dx2+dy2 for the differential of arc length in Cartesiancoordinates.

dx =∂x(r, θ)

∂rdr +

∂x(r, θ)

∂θdθ = cos θdr − r sin θdθ

dy =∂y(r, θ)

∂rdr +

∂y(r, θ)

∂θdθ = sin θdr + r cos θdθ (1.27)

Squaring and adding we find the cross terms cancel, leaving the desired resultds2 + dr2 + r2dθ2 .

2(a). L =√

2(exp(β)− exp(α))

2(b). L = 8a

3(a). Find (gij), (gij), det (gij),U, where U is the unit normal, for the

sphere of Example 7, see also Fig. I-16.

X(u, v) = (R cosu cos v,R sinu cos v,R sin v)

The tangent vectors are:

X1 = (−R sinu cos v,R cosu cos v, 0)

X2 = (−R cosu sin v,−R sinu sin v,R cos v) (1.28)

16

The metric tensor components are:

g11 = X1 ·X1 = R2 cos2 v

g12 = X1 ·X2 = 0

g21 = g12 = 0

g22 = X2 ·X2 = R2 (1.29)

Because the metric tensor is diagonal, the components of the inversemetric tensor are simply the inverse of the diagonal coefficients of the metrictensor (and zero in the off diagonal components):

gij = 1/gij

The determinant isdet(gij) = R4 cos2 v

The unit normal vector U is

U =X1 ×X2

||X1 ×X2||

=R2(cosu cos2 v, sinu cos2 v, cos v sin v)

||R2(cosu cos2 v, sinu cos2 v, cos v sin v)||= (cosu cos v, sinu cos v, sin v) (1.30)

5. Compute the metric form (or first fundamental form) and the unitnormal vector U for the general surface of revolution (Exercise 1, Section 3).

Recall from Exercise 1, Section 3:

X(u, v) = (f(u) cos v, f(u) sin v, g(u))

The tangent vectors were

X1 =∂X

∂u= (f ′(u) cos v, f ′(u) sin v, g′(u))

X2 =∂X

∂v= (−f(u) sin v, f(u) cos v, 0) (1.31)

Rob’s notes on Faber 17

The metric form (or first fundamental form) (see Eqs. 6 & 4 and p. 35) canbe written:

ds2 = g11du2 + 2g12 du dv + g22dv

2

with

g11 = X1 ·X1 = f ′2 + g′2

g12 = X1 ·X2 = 0

g22 = X2 ·X2 = f 2 (1.32)

For the unit normal vector,

U =X1 ×X2

||X1 ×X2||=

X1 ×X2√det(gij)

=(−g′f cos v,−g′f sin v, f ′f)

f√f ′2 + g′2

=(−g′ cos v,−g′ sin v, f ′)√

f ′2 + g′2(1.33)

6(a). Show that the area A of the general surface of revolution (Exer-cise 1, Section 3) between a ≤ u ≤ b and 0 ≤ v ≤ 2π is

A = 2π

∫ b

a

f(u)√f ′(u)2 + g′(u)2du

Recall from Eq. (13),

dA =√g du dv

A =

∫ 2π

0

∫ b

a

√g du dv

=

∫ 2π

0

∫ b

a

f√f ′2 + g′2 du dv

= 2π

∫ b

a

f√f ′2 + g′2 du (1.34)

18

Although X(u, v) : D → E3 where D was an open subset of <2, (c.f. p. 22),Eq. (13) was claimed to be valid for X(u, v) : Ω→ E3 with Ω a closed subsetsof <2, (c.f. top of p. 38), with X(u, v) being regular and one-to-one only onthe interior of Ω.

6(b). Show that the area A of the surface obtained by revolving thegraph y = f(x), a ≤ x ≤ b about the x−axis is

A = 2π

∫ b

a

f(x)√

1 + f ′(x)2 dx

Consider a small element of this surface consisting of a circular ring aboutthe x−axis with radius f(x) and centred at x. Its area will be 2πf(x)dl, where

dl2 = dx2 + dy2 = dx2 + f ′2 dx2

thus dl =√

1 + f ′2 dx and

A = 2π

∫ b

a

f(x)dl = 2π

∫ b

a

f(x)√

1 + f ′(x)2 dx

1.5 Second Fundamental Form

A strange and subtle thing happens after Faber introduced The Second Fun-damental Form

Lij ui′ uj

′

on p. 44. He first tells us how to computer the coefficients, Lij in Eq. (20),but then says nothing about ui

′uj

′. In principles, these should be familiar

ui′ ≡ dui

ds

But their computation deserves at least a sentence or two. Then he worksthrough two examples, Example 13 and 14 on pp. 44–46 and doesn’t sayanything about ui

′uj

′. Here’s my two cents on their computation. Let’s go

back to the first fundamental form:

ds2 = gijduiduj

Rob’s notes on Faber 19

So if we suppress summation over repeated indices, we can write:

dui

ds=

1√gii

In the next section (I-6 on Gauss curvature) Faber will speak of the secondfundamental form as being just Lij. It will be explained in that section thatone can compute the Gauss curvature K using just the gij and Lij, seeTheorem I-5.

Exercises I-5Most of these are straightforward computations. Exceptions are 1, 6, and

8. The first is quite easy. Here are my solutions for 6 and 8.

6. Prove Meusnier’s Theorem.

If one jumps in and starts computing the curvature of the surface etc.without thinking through what’s going on, one can quickly get lost in com-plicated computations. (I know from personal experience!) On the otherhand if one understands Eqs. (16, 17, 18), then the answer if very obvious.

α′′(s)

is the vector proportional to the curvature of the curve at point α(s). Itcan be projected onto the tangent plane of the surface giving α′′tan and theremainder projects onto the normal to the surface U giving α′′nor, Eq. (14).The curvature of the surface in the direction of the curve T(s) = α′(s) is themagnitude

||α′′nor|| = Lij ui′ uj

′

So Meusnier’s Theorem is essentially contained in Eq. (22)

kn(v) = Lij ui′ uj

′= α′′ ·U = ||α′′|| cosφ

where v = α′(s) is the direction of the curve at point s and φ is the anglebetween α′′(s) and U, noting of course that U is a unit vector.

20

8. First try writing the unit normal vector U, see p. 33, as

U =1√gX1 ×X2 =

1√g

det

x y zx1 y1 z1x2 y2 z2

where we have used Eq. (10) on p. 35 relating the determinant of the matrixg to the magnitude squared of X1 × X2. Here x is the unit vector in thex−direction, and similarly for the other 2 directions in Euclidean space. Nowsince Lij = Xij ·U (see Eq. (20) p. 44), it’s the x−component of Xij thatmultiplies the x component of U. So we can simply replace the unit vectorsabove with the corresponding components of Xij, giving the desired relation

Lij =1√g

det

xij yij yijx1 y1 z1x2 y2 z2

Chapter 6

Extended Indices

6.1 Index of symbols

•α

A smooth curve, see p. 1.

•α′(t) ≡ dα(t)

dt

The derivative vector or “velocity” of the smooth curve α parameterizedby t, see p. 2.

•α′′(t) ≡ d2α(t)

dt2

The acceleration vector of the smooth curve α parameterized by t, seep. 4.

•D

Open subset of R2, see p. 22.

•E3

Three dimensional Euclidean space, see p. 1.

21

22

•k(s) = ||T′(s)||

Curvature of a smooth curve α at point α(s).

•N(s) =

T′(s)

||T′(s)||Principal Normal vector, see p. 6.

•s

Arc length of a smooth curve, see pp. 2, 3. Use as a parameter of thecurve, see p. 5.

•t

Often used as a real parameter for the curve α(t), not necessarily timebut clearly meant to imply that in some contexts.

•T(s) = α′(s)

Unit tangent vector, see p. 5.

•T′(s) = α′′(s)

Curvature vector, see p. 5.

•TpM

Tangent plane at point P of surface M. See pp. 26, 47.

•U

Unit normal vector, p. 33,

Rob’s notes on Faber 23

6.2 Index of terms

• Catenoid: p. 48, see also Exercise 2(f) of section I-3 on p. 28, andExercise 9(e) of Section I-6.

• Principal Normal Vector, p. 6. See also symbol N(s).

• Unit normal vector, p. 33, U.

24

6.3 List of Definitions, Theorems, and Lem-

mas

1. Definition I-1: Curvature (of a curve in E3), pp. 5-6.

2. Definition I-2: Tangent vector (to a surface in E3), p. 26.

3. Definition I-3: Normal Curvature (of a surface in E3 in the direction ofa vector), p. 46.

4. Definition I-4: Principal Curves (of a surface in E3), principal directions(of the corresponding curves), and Gauss curvature (of correspondingsurface) p. 49.

5. Theorem I-5: The Gauss curvature at any point P of M is given byK(P) = L/g. p.50

6. Theorem I-6: p.54U1 ×U2 = K(X1 ×X2)

7. Definition I-7: Geodesic Curvature (of a curve on a surface in E3),p. 58.

8. Definition I-8: Geodesic (of a surface in E3), p. 59.

9. Theorem I-9: p. 65. Let α = α(s), a ≤ s ≤ b, be a curve on thesurface M, where s is arc length. If α is the shortest possible curve onM connecting its two end points, then α is a geodesic.

10. Theorem I-10: p. 68. Given a point P of M and a unit tangent vectorv at P, there exists a unique geodesic α(s) such that α(0) = P andα′(0) = v.

Bibliography

Boas, M. L., 1983: Mathematical methods in the physical sciences . JohnWiley and Sons, 793 pp.

Faber, R. L., 1983: Differential geometry and relativity theory: An Introduc-tion. Marcel Dekker. 255 + X pp.

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