12—Tensors 315
points,∆x^1 =one cm, sodx^1 /dt= 1cm/sec. The speed however, iscscαcm/sec because the
distance moved is greater by that factor. This means that
|~e 1 |= cscα
and this is greater than one; it is not a unit vector. The magnitudes of~e 2 is the same. The dot product
of these two vectors is~e 1 .~e 2 = cosα/sin^2 α.
Reciprocal Coordinate Basis
Construct the reciprocal basis vectors from the direct basis by the equation
~ei.~ej=δij
In rectangular coordinates the direct and reciprocal bases coincide because the basis is orthonormal.
For the tilted basis of Eq. (12.45),
~e 2 .~e^2 = 1 =|~e 2 ||~e^2 |cos
(
90 ◦−α
)
= (cscα)|~e^2 |sinα=|~e^2 |
The reciprocal basis vectors in this case are unit vectors.
The direct basis is defined so that the components of velocity are as simple as possible. In
contrast, the components of the gradient of a scalar field are equally simple provided that they are
expressed in the reciprocal basis. If you try to use the same basis for both you can, but the resulting
equations are a mess.
In order to compute the components ofgradf(wherefis a scalar field) start with its definition,
and an appropriate definition should not depend on the coordinate system. It ought to be some sort
of geometric statement that you can translate into any particular coordinate system that you want.
One way to definegradfis that it is that vector pointing in the direction of maximum increase off
and equal in magnitude todf/dwhereis the distance measured in that direction. This is the first
statement in section8.5. While correct, this definition does not easily lend itself to computations.
Instead, think of the particle moving through the manifold. As a function of time it sees changing
values off. The time rate of change offas felt by this particle is given by a scalar product of the
particle’s velocity and the gradient off. This is essentially the same as Eq. (8.16), though phrased in
a rather different way. Write this statement in terms of coordinates
d
dt
f
(
x^1 (t),x^2 (t),x^3 (t)
)
=~v.gradf
The left hand side is (by the chain rule)
∂f
∂x^1
∣∣
∣∣
x^2 ,x^3
dx^1
dt
+
∂f
∂x^2
∣∣
∣∣
x^1 ,x^3
dx^2
dt
+
∂f
∂x^3
∣∣
∣∣
x^1 ,x^2
dx^3
dt
=
∂f
∂x^1
dx^1
dt
+
∂f
∂x^2
dx^2
dt
+
∂f
∂x^3
dx^3
dt
(12.46)
~vis expressed in terms of the direct basis by
~v=~ei
dxi
dt
,
now expressgradfin the reciprocal basis
gradf=~ei
(
gradf
)
i (12.47)
