Mathematical Tools for Physics

12—Tensors 385

The way that the scalar product looks in terms of these bases, Eq. ( 32 ) is

~v.gradφ=~ei

dxi

dt

.~ej

(

gradφ

)

j=v

i(gradφ)

i (42)

Compare the two equations ( 41 ) and ( 42 ) and you see

gradφ=~ei

∂φ

∂xi

(43)

For a formal proof of this statement consider three cases. When the particle is moving along thex^1 direction
(x^2 &x^3 constant) only one term appears on each side of ( 41 ) and ( 42 ) and you can divide byv^1 =dx^1 /dt.
Similarly forx^2 andx^3. As usual with partial derivatives, the symbol∂φ

/

∂xiassumes that the other coordinates
x^2 andx^3 are constant.
In the case of polar coordinates this equation for the gradient reads, using Eq. ( 40 ),

gradφ=~e^1

∂φ

∂x^1

+~e^2

∂φ

∂x^2

=

(

ˆr

)∂φ

∂r

+

( 1

r

θˆ

)∂φ

∂θ

which is the standard result, Eq. (8.18). Notice again that the basis vectors are not dimensionless. They can’t
be because∂φ/∂rdoesn’t have the same dimensions as∂φ/∂θ.

2

x

x

1

2

e^1

e^2

0 1 2

0

1

Example
I want an example to show that all this formalism actually gives the correct answer in a
special case for which you can also compute all the results in the traditional way. Draw
parallel lines a distance 1 cm apart and another set of parallel lines also a distance 1 cm
apart intersecting at an angleαbetween them. These will be the constant values of the
functions defining the coordinates, and will form a coordinate system labeledx^1 andx^2.
The horizontal lines are the equationsx^2 = 0,x^2 = 1cm,etc.
Take the case of the non-orthogonal rectilinear coordinates again. The components ofgradφin the~e^1
direction is∂φ/∂x^1 , which is the derivative ofφwith respect tox^1 holdingx 2 constant, and this derivative is
notin the direction along~e^1 , but in the direction wherex^2 =a constant and that is along thex^1 -axis, along~e 1.
As a specific example to show that this makes sense, take a particularφdefined by

φ(x^1 ,x^2 ) =x^1

For this function gradφ=~e^1

∂φ

∂x^1

+~e^2

∂φ

∂x^2

=~e^1