Laplacian in Spherical Coordinates

The Laplacian operator in spherical coordinates is

( ) ( ) 2 1--∂- 2-∂- ---1----∂- -∂- ---1----∂2- ∇ sph = r2∂r r ∂r + r2sin 𝜃∂𝜃 sin𝜃∂ 𝜃 + r2sin2 𝜃∂ ϕ2

(1)

The derivation is fairly straight forward and begins with locating a vector r in spherical coordinates as shown in the figure.

Figure 1. Spherical Coordinates

The z component of the unit vector in direction of r is given from the simple right triangle

-z- cos𝜃 = |ˆr|

Since a unit vector has a length of 1, the z component is

z = cos𝜃

To get the x component, we need to get the r projected onto the xy-plane

|ˆrxy| cos(90 − 𝜃) = ----- |ˆr|

using the trig identity

cos (90 − 𝜃 ) = sin 𝜃

the projected unit vector is

|ˆrxy| = sin𝜃

Finally, the x component is reached through the right triangle

x cosϕ = ----- |ˆrxy|

giving

x = cosϕ sin𝜃

the y component follows the x component through

--y-- sin ϕ = |ˆrxy|

which yields

y = sinϕ sin𝜃

The vector in spherical coordinates is then

r = rˆr = r[sin𝜃cosϕ ˆi + sin 𝜃sinϕ ˆj + cos𝜃 ˆk]

(2)

Now describing the unit vectors of a moving particle r,𝜃,ϕ shown in the spherical coordinates figure is a little more tricky. If a particle moves in the r direction, it only moves in or out along r so

∂r ---= k ˆr ∂r

For 𝜃 and ϕ, think of uniform circular motion like a record, so they can be calculated from

ˆ -∂∂r𝜃- 𝜃 = |∂r| ∂𝜃

∂r ˆ𝜃 = -∂ϕ- |∂∂rϕ|

Next, take the partial derivatives of (2) and then calculate their magnitudes to get the above unit vectors.

∂r --- = r[cos𝜃cosϕˆi + cos𝜃 sinϕˆj − sin𝜃ˆk ] ∂ 𝜃

∂r --- = r[− sin𝜃sin ϕˆi + sin 𝜃cosϕˆj] ∂ ϕ

∂r √ ----- ∘ ------------------------------------ |---| = x ⋅ x = r2[cos2𝜃cos2ϕ + cos2𝜃 sin2ϕ + sin2𝜃] ∂ 𝜃

∂r- ∘ ----2-----2-------2--------2- |∂𝜃| = r cos 𝜃 (cos ϕ + sin ϕ) + sin 𝜃]

Using the basic trig identity

∂r |--| = r ∂ 𝜃

∂r- ∘ ----2-----2-------2-- |∂ϕ| = r sin 𝜃(sin ϕ + cos ϕ)]

∂r- |∂ϕ | = rsin𝜃

Then plug in these values to get

ˆ ˆ ˆ ϕ = − sinϕi + cosϕj

ˆ ˆ ˆ ˆ 𝜃 = cos𝜃 cosϕi + cos𝜃sinϕ j − sin 𝜃k

and from before we had

ˆr = sin𝜃 cosϕˆi + sin𝜃sin ϕˆj + cos𝜃ˆk

The generalized differential for curvilinear coordinates is

∂r ∂r ∂r dr = ---du1 + ----du2 + ----du3 ∂u1 ∂u2 ∂u3

For spherical coordinates we have

u1 = r

u2 = 𝜃

u3 = ϕ

and from earlier we learned

∂r- ∂r- |∂r |ˆr = ∂r

|∂r-|ˆ𝜃 = ∂r- ∂𝜃 ∂𝜃

|∂r-|ˆ𝜃 = ∂r- ∂ϕ ∂ϕ

Plugging these values into the generalized differential yields

dr = drˆr + rd𝜃𝜃ˆ+ rsin 𝜃dϕˆϕ

The next major step is to see how the gradient fits into the definition of the differential for a function f

∂f ∂f ∂f df = ---dr + --d 𝜃 + ---dϕ ∂r ∂𝜃 ∂ϕ

df = ∇f ⋅ dr

So we see that the following must be equal and we need to solve for the gradient’s components.

∂f ---dr = ∇rf dr ∂r

∂f ---d𝜃 = ∇ 𝜃f d𝜃 ∂𝜃

∂f ---dϕ = ∇ϕf dϕ ∂ϕ

Therfore our scale factors are

∇r = 1

1 ∇ 𝜃 = -- r

1 ∇ ϕ = ------ rsin𝜃

which finally gives us the gradient in spherical coordinates

∂ ∂ ˆ𝜃 ∂ ˆϕ ∇sph = ---ˆr + -----+ --------- ∂r ∂𝜃 r ∂ϕ rsin𝜃

The last tedious calculation is then the Laplacian, which is our goal

( ) ( ) 2 ∂-- -∂-ˆ𝜃- -∂- --ˆϕ--- ∂-- -∂-ˆ𝜃- -∂- --ˆϕ--- ∇ = ∇ ⋅ ∇ = ∂rˆr + ∂ 𝜃r + ∂ ϕ rsin 𝜃 ⋅ ∂rˆr + ∂ 𝜃r + ∂ ϕ rsin 𝜃

Let us break it up by components to make it easy to view, so carrying out only part of the dot product our 1st term is

( ) ∂ˆr ∂ ∂2 ∂ˆ𝜃1 ∂ ∂ 1 ∂ ˆ𝜃 ∂2 ∂ ˆϕ 1 ∂ ∂ ( 1 ) ∂ ˆϕ ∂2 ˆr⋅ ----- + ˆr--2-+ -------+ 𝜃ˆ-------+ ------ + ------------ + ˆϕ--- ------ --- + ------------ ∂r∂r ∂r ∂rr ∂𝜃 ∂r r∂𝜃 r∂r ∂𝜃 ∂rrsin 𝜃∂ ϕ ∂r rsin 𝜃 ∂ϕ rsin𝜃 ∂r∂ϕ

While this looks a little scary, all but the 2nd term is zero. The first term is zero because there is no r in r

ˆr = sin𝜃 cosϕˆi + sin𝜃sin ϕˆj + cos𝜃ˆk

so taking the derivative with respect to r yeilds zero. Similarly, the 3rd and 6th term are zero when taking the partial derivatives. The 4th,5th,7th and 8th terms are zero because the dot product of two orthogonal vectors (90^o) is zero so

ˆ ˆr ⋅𝜃 = 0

ˆr ⋅ϕˆ= 0

This leaves only the second term

∂2 ∂2 rˆ⋅ ˆr∂r2 = ∂r2-

(3)

Now onto the 𝜃 term

ˆ ( 2 ˆ ˆ 2 ˆ ( ) ˆ 2 ) 𝜃⋅ ∂rˆ-∂-+ ˆr -∂---+ ∂𝜃-1-∂-+ 𝜃ˆ∂--1-∂-+ 𝜃-∂--+ ∂ϕ---1----∂-+ ϕˆ-∂- --1--- ∂--+ --ϕ----∂--- r ∂𝜃 ∂r ∂𝜃∂r ∂𝜃 r∂𝜃 ∂𝜃 r∂ 𝜃 r∂ 𝜃2 ∂𝜃 rsin𝜃 ∂ϕ ∂𝜃 rsin𝜃 ∂ϕ rsin𝜃∂ 𝜃∂ϕ

The dot product gets rid of the 2nd, 3rd, 7th and 8th terms, while the 4th and 6th terms are zero when taking the derivatives

∂-1-= 0 ∂𝜃r

∂-ˆϕ ∂ 𝜃 = 0

Two terms remain, for the first term we need to calculate

∂-ˆr ˆ ˆ ˆ ˆ ∂ 𝜃 = cos𝜃 cosϕi + cos𝜃sinϕ j − sin 𝜃k = 𝜃

This yields

ˆ𝜃 ∂ 1 ∂ -⋅ ˆ𝜃---= ---- r ∂r r∂r

(4)

The 5th term is just the dot product

ˆ𝜃- ˆ𝜃-∂2-- 1--∂2- r ⋅r ∂𝜃2 = r2∂𝜃2

(5)

Finally, the ϕ part of the dot product is

( 2 2 ( ) 2 ) --ˆϕ---⋅ ∂-ˆr-∂- + ˆr-∂----+ ∂-ˆ𝜃1--∂-+ 𝜃ˆ∂--1-∂- + ˆ𝜃--∂---+ ∂ˆϕ---1---∂--+ ϕˆ-∂- --1--- -∂-+ --ˆϕ---∂-- rsin𝜃 ∂ ϕ∂r ∂ϕ ∂r ∂ ϕr ∂𝜃 ∂ϕ r∂ 𝜃 r ∂ϕ∂𝜃 ∂ϕ rsin𝜃 ∂ϕ ∂ϕ rsin𝜃 ∂ϕ rsin𝜃∂ ϕ2

Once again the dot product makes the 2nd, 4th and 5th terms zero. The first term derivative

∂ˆr- ˆ ˆ ˆ-∂- ∂ϕ = − sin 𝜃sinϕ i + sin𝜃 cosϕj = sin𝜃ϕ ∂r

this leads to

--ˆϕ--- ˆ-∂- 1-∂-- rsin𝜃 ⋅ sin 𝜃ϕ∂r = r ∂r

(6)

The 3rd term derivative is

∂ ˆ𝜃 --- = − cos𝜃 sin ϕˆi + cos𝜃cosϕ ˆj = cos𝜃ˆϕ ∂ ϕ

this leads to

--ˆϕ--- ˆ1-∂- -cos𝜃---∂- rsin𝜃 ⋅ cos𝜃ϕ r∂ 𝜃 = r2sin𝜃 ∂𝜃

(7)

The 6th term can be seen to be zero because the derivative of ϕ with respect to ϕ is a vector perpendicular to ϕ (feel free to carry out this calculation), so the dot product will be zero.

The 7th term derivative is zero

( ) -∂- --1--- ∂ ϕ rsin𝜃 = 0

Then we keep the 8th term

ˆϕ ϕˆ ∂2 1 ∂2 ------⋅---------2 = -2---2----2- rsin𝜃 rsin 𝜃∂ϕ r sin 𝜃 ∂ϕ

(8)

Putting the results from (3),(4),(5),(6),(7) and (8) we get

2 2 2 ∇2 = ∂---+ 1-∂--+ -1 ∂---+ 1-∂--+ -cos𝜃--∂-+ ---1----∂-- sph ∂r2 r ∂r r2 ∂𝜃2 r ∂r r2sin 𝜃∂ 𝜃 r2sin2𝜃∂ ϕ2

Certainly, we could finish here, but let us combine some terms and notice the following relationships (check these for yourself)

[ ] ( ) ∂2 2 ∂ 1 ∂ 2 ∂ ∂r2-+ r-∂r- = r2 ∂r- r ∂r-

[ ] ( ) 1 ∂2 cos𝜃 ∂ 1 ∂ ∂ r2∂-𝜃2 + r2sin𝜃-∂𝜃- = r2sin-𝜃∂-𝜃 sin𝜃∂-𝜃

This gives us the equation given in (1), the Laplacian in spherical coordinates

1 ∂ ( ∂ ) 1 ∂ ( ∂ ) 1 ∂2 ∇2sph = -2--- r2--- + --2------- sin𝜃--- + -2---2----2 r ∂r ∂r r sin 𝜃∂𝜃 ∂ 𝜃 r sin 𝜃∂ ϕ

References

[1] Marsden, J., Tromba, A. ”Vector Calculus” Fourth Edition. W.H. Freeman Company, 1996.

[2] Ellis, R., Gulick, D. ”Calculus” Harcourt Brace Jovanovich, Inc., Orlando, FL, 1991.

[3] Benbrook, J. ”Intermediate Electromagnetic Theory”, lecture notes, University of Houston, Fall 2002.

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