12—Tensors 314The velocity vector is
d~r
dt
=~e 1
dr
dt
+~e 2
dφ
dt
and Eq. (12.40) isdf
dt
=
∂f
∂xi
dxi
dt
= (∇f).~v
This sort of basis has many technical advantages that aren’t at all apparent here. At this point
this “coordinate basis” is simply a way to sweep some of the complexity out of sight, but with further
developments of the subject the sweeping becomes shoveling. When you go beyond the introduction
found in this chapter, you find that using any basis other than a coordinate basis leads to equations
that have complicated extra terms, which you want nothing to do with.
In spherical coordinatesx^1 =r, x^2 =θ, x^3 =φ
~e 1 =r, ~eˆ 2 =rθ, ~eˆ 3 =rsinθφˆ and ~e^1 =ˆr, ~e^2 =θ/r, ~eˆ^3 =φ/rˆ sinθ
The velocity components are now
dxi/dt={dr/dt, dθ/dt, dφ/dt}, and ~v=~eidxi/dt (12.44)
This last equation is central to figuring out the basis vectors in an unfamiliar coordinate system.
The use ofx^1 ,x^2 , andx^3 (xi) for the coordinate system makes the notation uniform. Despite
the use of superscripts, these coordinates arenotthe components of any vectors, though their time
derivatives are.
In ordinary rectangular coordinates, if a particle is moving along thex^1 -axis (thex-axis) then
dx^2 /dt= 0 =dx^3 /dt. Also, the velocity will be in thexdirection and of sizedx^1 /dt.
~e 1 =ˆx
as you would expect. Similarly
~e 2 =y, ~eˆ 3 =z.ˆ
In a variation of rectangular coordinates in the plane the axes are not orthogonal to each other,
but are rectilinear anyway.
0 1 2 3 4
0
1
2
3
α
~e 2
~e 1
Still keep the requirement of Eq. (12.44)