12—Tensors 316
The way that the scalar product looks in terms of these bases, Eq. (12.33) is
~v.gradf=~ei
dxi
dt
.~ej(gradf)
j=v
i(gradf)
i (12.48)
Compare the two equations (12.46) and (12.48) and you see
gradf=~ei
∂f
∂xi
(12.49)
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 (12.46) and (12.48) and you can
divide byv^1 =dx^1 /dt. Similarly forx^2 andx^3. As usual with partial derivatives, the symbol∂f
/
∂xi
assumes that the other coordinatesx^2 andx^3 are constant.
For polar coordinates this equation for the gradient reads, using Eq. (12.41),
gradf=~e^1
∂f
∂x^1
+~e^2
∂f
∂x^2
=
(
rˆ
)∂f
∂r
+
( 1
r
φˆ
)∂f
∂φ
which is the standard result, Eq. (8.27). Notice again that the basis vectors are not dimensionless.
They can’t be, because∂f/∂rdoesn’t have the same dimensions as∂f/∂φ.
0
1
2
0 1
x^1
x^2
~e^1
~e^2
α
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 ofgradfin
the~e^1 direction is∂f/∂x^1 , which is the derivative offwith respect tox^1 holdingx 2 constant, and
this derivative isnotin 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 particularf
defined by
f(x^1 ,x^2 ) =x^1
For this function gradf=~e^1
∂f
∂x^1
+~e^2
∂f
∂x^2
=~e^1
~e^1 is perpendicular to thex^2 -axis, the line x 1 =constant, (as it should be). Its magnitude is the
magnitude of~e^1 , which is one.
To verify that this magnitude is correct, calculate it directly from the definition. The magnitude
of the gradient is the magnitude ofdf/dwhereis measured in the direction of the gradient, that is,
in the direction~e^1.
df
d`
=
∂f
∂x^1
dx^1
d`
= 1.1 = 1
Why 1 fordx^1 /d`? The coordinate lines in the picture arex^1 = 0, 1 , 2 ,.... When you move on the
straight line perpendicular tox^2 =constant (~e^1 ), and go fromx^1 = 1tox^2 = 2, then both∆x^1 and
∆sare one.
