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125

5.3 Tensor calculus in polar coordinates

5.3 Te n s o r c a l c u l u s i n p o l a r co o rd i n a t e s

The fact that the basis vectors of polar coordinates are not constant everywhere, leads to some problems when we try to differentiate vectors. For instance, consider the simple vector ex, which is a constant vector field, the same at any point. In polar coordinates it has components ex → ( rx, θ x) = (cos θ , −r−1 sin θ ). These are clearly not constant, even though ex is. The reason is that they are components on a nonconstant basis. If we were to differentiate them with respect to, say, θ , we would most certainly not get ∂ex/∂θ , which must be identically zero. So, from this example, we see that differentiating the components of a vector does not necessarily give the derivative of the vector. We must also differentiate the nonconstant basis vectors. This is the key to the understanding of curved coordinates and, indeed, of curved spaces. We shall now make these ideas systematic.

Figure 5.6

Derivatives of basis vectors

Since ex and ey are constant vector fields, we easily find that

 

 

 

 

∂

 

∂

 

 

 

 

 

 

er =

 

(cos θ ex + sin θ ey) = 0,

(5.37a)

 

∂r

∂r

 

∂

 

∂

 

 

 

 

 

er =

 

(cos θ ex + sin θ ey)

 

 

 

∂θ

∂θ

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

= − sin θ ex + cos θ ey =

 

 

eθ .

(5.37b)

 

 

 

 

r

These have a simple geometrical picture, shown in Fig. 5.6. At two nearby points, A and B, er must point directly away from the origin, and so in slightly different directions. The derivative of er with respect to θ is just the difference between er at A and B divided by θ .

y

B

A

θ

x

Change in er, when θ changes by θ .

126

Preface to curvature

The difference in this case is clearly a vector parallel to eθ , which then makes Eq. (5.37b) reasonable.

Similarly,

 

∂

 

∂

 

 

 

 

eθ =

 

(−r sin θ ex + r cos θ

ey)

 

 

∂r

∂r

 

 

 

1

 

 

 

 

= − sin θ ex + cos θ ey =

 

eθ ,

(5.38a)

 

 

r

 

∂

eθ = −r cos θ ex − r sin θ ey = −r er.

(5.38b)

 

∂θ

The student is encouraged to draw a picture similar to Fig. 5.6 to explain these formulas.

Derivatives of general vectors

Let us go back to the derivative of ex. Since

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ex = cos θ er −

 

sin θ

eθ ,

 

(5.39)

 

 

 

r

 

we have

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

∂

ex =

 

∂

 

(cos θ ) er + cos θ

∂

(er)

 

 

 

 

 

 

 

 

 

 

 

 

∂θ

∂θ

∂θ

(eθ )

(5.40)

 

 

 

− ∂θ

r sin θ

eθ − r sin θ ∂∂θ

 

 

 

 

 

∂

1

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

= − sin θ er + cos θ r

eθ

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

−

 

cos θ eθ −

 

sin θ (−r er).

 

(5.41)

 

 

 

r

r

 

To get this we used Eqs. (5.37) and (5.38). Simplifying gives

 

 

 

 

 

 

 

 

 

 

 

∂

= 0,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ex

 

 

 

 

 

 

 

(5.42)

 

 

 

 

 

 

 

 

∂θ

 

 

 

 

 

 

 

just as we should have. Now, in Eq. (5.40) the first and third terms come from differentiating the components of ex on the polar coordinate basis; the other two terms are the derivatives of the polar basis vectors themselves, and are necessary for cancelling out the derivatives of the components.

r θ

A general vector V has components (V , V ) on the polar basis. Its derivative, by analogy

with Eq. (5.40), is

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

∂V

=

∂

 

r

er + V

θ

eθ )

 

 

 

 

 

 

 

 

(V

 

 

 

 

 

 

 

 

∂r

∂r

 

 

 

 

 

 

 

 

 

=

∂Vr

e

 

Vr

∂er

+

∂Vθ

e

Vθ

∂eθ

,

 

 

∂r

 

 

∂r

 

 

 

r +

 

 

∂r

θ +

 

∂r

and similarly for ∂V/∂θ . Written in index notation, this becomes

 

∂V

=

 

∂

(Vα e )

=

∂Vα

e

Vα

∂eα

.

∂r

∂r

∂r

 

 

α

α +

 

∂r

(Here α runs of course over r and θ .)

127

5.3 Tensor calculus in polar coordinates

This shows explicitly that the derivative of V is more than just the derivative of its components Vα . Now, since r is just one coordinate, we can generalize the above equation to

∂V

=

∂Vα

e

Vα

∂eα

,

(5.43)

∂xβ

∂xβ

 

α +

∂xβ

 

 

where, now, xβ can be either r or θ for β = 1 or 2.

The C hristoffel symbols

The final term in Eq. (5.43) is obviously of great importance. Since ∂eα /∂xβ is itself a vector, it can be written as a linear combination of the basis vectors; we introduce the symbol μαβ to denote the coefficients in this combination:

∂eα

=

μ e .

(5.44)

∂xβ

αβ μ

 

The interpretation of μαβ is that it is the μth component of ∂eα /∂xβ . It needs three indices: one (α) gives the basis vector being differentiated; the second (β) gives the coordinate with respect to which it is being differentiated; and the third (μ) denotes the component of the resulting derivative vector. These things, μαβ , are so useful that they have been given a name: the Christoffel symbols. The question of whether or not they are components of tensors we postpone until much later.

We have of course already calculated them for polar coordinates. From Eqs. (5.37) and (5.38) we find

(1)

∂er/∂r = 0 μrr = 0

for all μ,

 

 

 

 

 

(2)

∂er/∂θ

=

1r

eθ

rrθ

=

0,

θ rθ

=

1r ,

(5.45)

 

 

1

 

r

 

θ

1

 

 

(3)

∂eθ /∂r

 

 

eθ

 

0,

 

 

,

 

 

 

r

θ r

 

θ r

 

r

 

 

 

 

=

 

 

=

 

 

=

 

 

 

(4)

 

 

 

 

 

 

 

 

∂eθ /∂θ = −r er rθ θ = −r, θ θ θ = 0.

 

In the definition, Eq. (5.44), all indices must refer to the same coordinate system. Thus, although we computed the derivatives of er and eθ by using the constancy of ex and ey, the Cartesian bases do not in the end make any appearance in Eq. (5.45). The Christoffel symbols’ importance is that they enable us to express these derivatives without using any other coordinates than polar.

The covariant derivative

Using the definition of the Christoffel symbols, Eq. (5.44), the derivative in Eq. (5.43) becomes

∂V

 

∂Vα

α

 

μ

 

 

=

 

 

eα + V

 

 

αβ eμ.

(5.46)

∂xβ

∂xβ

 

In the last term there are two sums, on α and μ. Relabeling the dummy indices will help here: we change μ to α and α to μ and get

128

 

 

 

Preface to curvature

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

∂V

 

∂Vα

μ

 

α

 

 

 

 

 

 

 

 

 

 

 

=

 

 

eα + V

 

 

μβ eα .

(5.47)

 

 

 

∂xβ

∂xβ

 

The reason for the relabeling was that, now, eα can be factored out of both terms:

∂xβ

=

∂xβ + V

 

 

μβ

∂V

 

 

∂Vα

μ

 

α

β

So the vector field ∂V/∂x has components

∂Vα /∂xβ + Vμ α μβ .

Recall our original notation for the partial derivative, notation and define a new symbol:

eα . (5.48)

(5.49)

∂Vα /∂xβ = Vα,β . We keep this

Vα ;β := Vα ,β + Vμ α μβ .

(5.50)

Then, with this shorthand semicolon notation, we have

∂V

/∂xβ

=

Vα

 

(5.51)

 

;β eα ,

a very compact way of writing Eq. (5.48).

β

Now ∂V/∂x is a vector field if we regard β as a given fixed number. But there are two

values that β can have, and so we can also regard ∂V/∂xβ as being associated with a

 

11

 

 

β

 

 

 

 

tensor field which maps the vector eβ into the vector ∂V

/∂x , as in Exer. 17, § 3.10.

This

 

 

 

 

tensorfield is called the covariant derivative of V, denoted, naturally enough, as V. Then

its components are

( V)α β = ( β V)α = Vα ;β .

(5.52)

On a Cartesian basis the components are just Vα ,β . On the curvilinear basis, however, the derivatives of the basis vectors must be taken into account, and we get that Vα ;β are the

components of V in whatever coordinate system the Christoffel symbols in Eq. (5.50) refer to. The significance of this statement should not be underrated, as it is the foundation

of all our later work. There is a single

11

 

tensor called V. In Cartesian coordinates its

components are ∂V

α

/∂x

β

. In general

coordinates

{

xμ

}

its components are called Vα

 

 

 

 

 

 

 

;β

and can be obtained in either of two equivalent ways: (i) compute them directly in {xμ } using Eq. (5.50) and a knowledge of what the α μ β coefficients are in these coordinates; or (ii) obtain them by the usual tensor transformation laws from Cartesian to {xμ }.

What is the covariant derivative of a scalar? The covariant derivative differs from the partial derivative with respect to the coordinates only because the basis vectors change. But a scalar does not depend on the basis vectors, so its covariant derivative is the same as

its partial derivative, which is its gradient:

 

 

= ˜

 

α f =

∂f /∂xα ;

f

(5.53)

 

 

df .
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