Lecture 8

Metrics

Objectives:

• More on the metric and how it transforms.

Reading: Hobson, 2.

8.1 Riemannian Geometry

The intervalds2 = gαβ dxα dxβ,

is a quadratic function of the coordinate differentials.

This is the definition of Riemannian geometry, or more correctly, pseudo-Riemanniangeometry to allow for ds2 < 0.

Example 8.1 What are the coefficients of the metric tensor in 3D Euclidean

space for Cartesian, cylindrical polar and spherical polar coordinates?

Answer 8.1 The “interval” in Euclidean geometry can be written in Carte-

sian coordinates as Introducing anobvious notationwith x standingfor the xcoordinate index,etc.

ds2 = dx2 + dy2 + dz2.

The metric tensor’s coefficients are therefore given by

gxx = gyy = gzz = 1,

with all others = 0.

In cylindrical polars:

ds2 = dr2 + r2dφ2 + dz2,

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LECTURE 8. METRICS 31

so grr = 1, gφφ = r2, gzz = 1 and all others = 0.

Finally spherical polars:

ds2 = dr2 + r2dθ2 + r2 sin2 θdφ2,

gives grr = 1, gθθ = r2 and gφφ = r2 sin2 θ.

Example 8.2 Calculate the metric tensor in 3D Euclidean space for the

coordinates u = x + 2y, v = x − y, w = z.

Answer 8.2 The inverse transform is easily shown to be x = (u + 2v)/3,y = (u − v)/3, z = w, so

dx =1

3du +

2

3dv,

dy =1

3du −

1

3dv,

dz = dw,

so

ds2 =

(

1

3du +

2

3dv

)

2

+

(

1

3du −

1

3dv

)

2

+ dw2,

=2

9du2 +

5

9dv2 +

2

9dudv + dw2.

We can immediately write guu = 2/9, gvv = 5/9, gww = 1, and guv = gvu =1/9 since the metric is symmetric. This metric still describes 3D Euclidean

flat geometry, although not obviously.

8.2 Metric transforms

The method of the example is often the easiest way to transform metrics,however using tensor transformations, we can write more compactly:

gα′β′ =∂xγ

∂xα′

∂xδ

∂xβ′gγδ.

This shows how the components of the metric tensor transform under coor-dinate transformations but the underlying geometry does not change.

Example 8.3 Use the transformation of g to derive the metric components

in cylindrical polars, starting from Cartesian coordinates.

LECTURE 8. METRICS 32

Answer 8.3 We must compute terms like ∂x/∂r, so we need x, y and z in

terms of r, φ, z:

x = r cos φ,

y = r sin φ,

z = z.

Find ∂x/∂r = cos φ, ∂y/∂r = sin φ, ∂z/∂r = 0. Consider the grr component:

grr =∂xi

∂r

∂xj

∂rgij,

where i and j represent x, y or z. Since gij = 1 for i = j and 0 otherwise,

and since ∂z/∂r = 0, we are left with:

grr =

(

∂x

∂r

)

2

+

(

∂y

∂r

)

2

= cos2 φ + sin2 φ = 1.

Similarly

gθθ =

(

∂x

∂φ

)

2

+

(

∂y

∂φ

)

2

= (−r sin φ)2 + (r cos φ)2 = r2,

and gzz = 1, as expected.

This may seem a very difficult way to deduce a familiar result, but the point is

that it transforms a problem for which one otherwise needs to apply intuition

and 3D visualisation into a mechanical procedure that is not difficult – at

least in principle – and can even be programmed into a computer.

8.3 First curved-space metric

We can now start to look at curved spaces. A very helpful one is the surfaceof a sphere.

LECTURE 8. METRICS 33

Figure: Surface of a sphere parameterised by distance r froma point and azimuthal angle φ

The sketch showsthe surface“embedded” in3D. This is apriviledged viewthat is not alwayspossible. Youneed to try toimagine that youare actually stuckin the surfacewith no “height”dimension.

Two coordinates are needed to label the surface. e.g. the distance from apoint along the surface, r, and the azimuthal angle φ, similar to Euclideanpolar coords.

The distance AP is given by R sin θ, so a change dφ corresponds to distanceR sin θ dφ. Thus the metric is

ds2 = dr2 + R2 sin2 θ dφ2.

or since r = Rθ,

ds2 = dr2 + R2 sin2

( r

R

)

dφ2.

This is the metric of a 2D space of constant curvature.

Circumference of circle in this geometry: set dr = 0, integrate over φ

C = 2πR sinr

R< 2πr.

e.g. On Earth (R = 6370 km), circle with r = 10 km shorter by 2.6 cm thanif Earth was flat.

Exactly the same is possible in 3D. i.e we could find that a circle radius rhas a circumference < 2πr owing to gravitationally induced curvature.

8.4 2D spaces of constant curvature

Can construct metric of the surface of a sphere as follows. First write theequation of a sphere in Euclidean 3D

x2 + y2 + z2 = R2.

LECTURE 8. METRICS 34

If we switch to polars (r, θ) in the x–y plane, this becomes

r2 + z2 = R2.

In the same terms the Euclidean metric is

dl2 = dr2 + r2dθ2 + dz2.

But we can use the restriction to a sphere to eliminate dz which implies

2r dr + 2z dz = 0,

and so

dl2 = dr2 + r2dθ2 +r2dr2

z2,

which reduces to

dl2 =dr2

1 − r2/R2+ r2dθ2.

Defining curvature k = 1/R2, we have

dl2 =dr2

1 − kr2+ r2dθ2,

the metric of a 2D space of constant curvature. k > 0 can be “embedded”in 3D as the surface of a sphere; k < 0 cannot, but it is still a perfectly validgeometry. [A saddle shape has negative curvature over a limited region.]

A very similar procedure can be used to construct the spatial part of themetric describing the Universe.