




The gradient of a scalar
and divergence and curl of a vector have a simple form in rectilinear coordinates.
This form is maintained if you move the origin or rotate the coordinates, but
still use rectilinear coordinates.
Their forms change, however, under if you choose arbitrary coordinates,
and they look quite different even in polar coordinates in the plane, and in
cylindric and spherical coordinates in three dimensions.
We will derive the form of the divergence and curl in an arbitrary orthogonal
coordinate systems by application of Stokes' theorem and the divergence theorem.
The character of a coordinate system that affects the form
of these objects is encapsulated in the expression for distance in the coordinate
system.
In rectilinear coordinates distance is described by the Pythagorian Theorem:
ds^{2} = dx^{2}
+ dy^{2} + dz^{2}.
In a general orthogonal coordinate system, with variables w_{1
}, w_{2}
and w_{3}, distance
can take on the more general form
ds^{2} = u_{1}^{2}dw_{1}^{2}
+ u_{2}^{2}dw_{2}^{2}
+ u_{3}^{2}dw_{3}^{2},
where the u's are each functions of the variables.
Thus,
In polar coordinates with w_{1 }=
r and w_{2 }=
q, we have
u_{1 }=1 and u_{2 }= r: ds^{2} = dr^{2} + r^{2}d^{2}
^{ }In spherical coordinates with w_{1 }= , w_{3 }= and w_{2 }=, and r = rsin, we have
u_{1 }=1, u_{3 }= r =sin and u_{2}=: ds^{2} = d^{2} + r^{2}d^{2} + ^{2}d
In cylindric coordinates, we have
ds^{2} = dr^{2} + r^{2}d^{2} + dz^{2}
We now apply the divergence theorem to a little
rectangular box of space around the point r oriented with sides parallel
to the directions of w_{1 },
w_{2}, and w_{3},
so that the the coordinates change by dw_{1}, dw_{2} and dw_{3}
in it. We make these quantities so small that div v is essentially constant
inside it, and v is essentially constant on it boundaries.
We get:
(div v(r))V = d(A_{1}v_{1}(r)) + d(A_{2}v_{2}(r)) + d(A_{3}v_{3}(r))
where A_{i }is the area of the part of the boundary
of our box on which w_{i} is constant, and V is the volume of the box.
We can differentiate to get
and observe that
and V = u_{1}u_{2}u_{3}dw_{1}dw_{2}dw_{3}
Substituting these facts into the last equation and dividing by V gives us
The key observation here is that both v_{i }and A_{i} can change across the box, so that one must differentiate the product v_{i }A_{i }and not just v_{i }.
In two dimensions, there is no w_{3} and the same equation holds without any last term and with u_{3 }=1.
Thus, in polar coordinates, we get:
In spherical coordinates we find