Michael Corral: Vector Calculus

4.6 Gradient, Divergence, Curl and Laplacian 183
Goal: Show that the gradient of a real-valued function F(ρ,θ,φ) in spherical coordinates
is:
∇F= ∂F
∂ρ e 1 ∂F
ρ+
ρsinφ ∂θ e θ+ 1 ∂F
ρ∂φ e φ
Idea: In the Cartesian gradient formula∇F(x,y,z)= ∂F
∂x i+∂F ∂y j+∂F ∂z
k, put the Cartesian
basis vectors i, j, k in terms of the spherical coordinate basis vectors e ρ , e θ , e φ
and functions ofρ,θandφ. Then put the partial derivatives ∂F
∂x , ∂F
∂y , ∂F ∂F
∂z
in terms of
∂ρ ,
∂F
∂θ , ∂F
∂φ
and functions ofρ,θandφ.
Step 1: Get formulas for e ρ , e θ , e φ in terms of i, j, k.
We can see from Figure 4.6.2 that the unit vector e ρ in theρdirection at a general
point (ρ,θ,φ) is e ρ = r
‖r‖, where r= xi+yj+zk is the position vector of the point in
Cartesian coordinates. Thus,
e ρ = r
‖r‖ = xi+yj+zk √
x 2 +y 2 +z 2,
so using x=ρsinφcosθ, y=ρsinφsinθ, z=ρcosφ, andρ= √ x 2 +y 2 +z 2 , we get:
e ρ = sinφ cosθi+sinφ sinθj+cosφk
Now, since the angleθis measured in the xy-plane, then the unit vector e θ in theθ
direction must be parallel to the xy-plane. That is, e θ is of the form ai+bj+0k. To
figure out what a and b are, note that since e θ ⊥ e ρ , then in particular e θ ⊥ e ρ when
e ρ is in the xy-plane. That occurs when the angleφisπ/2. Puttingφ=π/2 into the
formula for e ρ gives e ρ = cosθi+sinθj+0k, and we see that a vector perpendicular
to that is−sinθi+cosθj+0k. Since this vector is also a unit vector and points in the
(positive)θdirection, it must be e θ :
Lastly, since e φ = e θ ×e ρ , we get:
e θ =−sinθi+cosθj+0k
e φ = cosφ cosθi+cosφ sinθj−sinφk
Step 2: Use the three formulas from Step 1 to solve for i, j, k in terms of e ρ , e θ , e φ .
This comes down to solving a system of three equations in three unknowns. There
are many ways of doing this, but we will do it by combining the formulas for e ρ and
e φ to eliminate k, which will give us an equation involving just i and j. This, with the
formula for e θ , will then leave us with a system of two equations in two unknowns (i
and j), which we will use to solve first for j then for i. Lastly, we will solve for k.
First, note that
sinφe ρ +cosφe φ = cosθi+sinθj