Mathematical Tools for Physics - Department of Physics - University

12—Tensors 313

x^1

x^2

x^1 =constant

x^3 =constant

x^2 =constant

x^3 =constant

Specify the equationsx^2 = 0andx^3 = 0for thex^1 coordinate axis. For example in rectangular

coordinatesx^1 =x,x^2 =y,x^3 =z, and thex-axis is the liney= 0, andz= 0. In plan polar

coordinatesx^1 =r=a constant is a circle andx^2 =φ=a constant is a straight line starting from the

origin.

Start with two dimensions and polar coordinatesrandφ. As a basis in this system we routinely

use the unit vectorsˆrandφˆ, but is this the only choice? Is it the best choice? Not necessarily. Look

at two vectors, the differentiald~rand the gradient∇f.

d~r=rdrˆ +φrdφˆ and ∇f=rˆ

∂f

∂r

+φˆ

1

r

∂f

∂φ

(12.39)

And remember the chain rule too.

df

dt

=

∂f

∂r

dr

dt

+

∂f

∂φ

dφ

dt

(12.40)

In this basis the scalar product is simple, but you pay for it in that the components of the vectors have

extra factors in them — therand the 1 /r. An alternate approach is to place the complexity in the

basis vectors and not in the components.

~e 1 =r, ~eˆ 2 =rφˆ and ~e^1 =r, ~eˆ^2 =φ/rˆ (12.41)

These are reciprocal bases, as indicated by the notation. Of course the originalˆr-φˆis self-reciprocal.

The preceding equations (12.39) are now

d~r=~e 1 dr+~e 2 dφ and ∇f=~e^1

∂f

∂r

+~e^2

∂f

∂φ

(12.42)

Make another change in notation to make this appear more uniform. Instead ofr-φfor the coordinates,

call themx^1 andx^2 respectively, then

d~r=~e 1 dx^1 +~e 2 dx^2 and ∇f=~e^1

∂f

∂x^1

+~e^2

∂f

∂x^2

(12.43)

rˆ–φˆ ~e 1 – ~e 2 ~e^1 – ~e^2