Laplacian from Rectangular to Spherical Coordinates

Laplacian from Rectangular to Spherical Coordinates

The Laplacian from Rectangular to Spherical Coordinates
Swapnil Sunil Jain
June 4, 2006

Laplacian from Rectangular to Spherical Coordinates

We begin by recognizing the familiar conversion from rectangular to

spherical coordinates

1

1

Note that convention in use here:

ϕ

is the azimuthal angle, whereas

θ

is the

polar angle

.

x

=

r

⁢

sin

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

,

y

=

r

⁢

sin

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

)

,

z

=

r

⁢

cos

⁡

(

θ

)

,

(1)

and conversely from spherical to

rectangular coordinates

r

=

x

2

+

y

2

+

z

2

,

θ

=

arccos

⁡

(

z

r

)

,

ϕ

=

arctan

⁡

(

y

x

)

.

(2)

Now, we know that the Laplacian in rectangular coordinates is defined in the following way

2

2

Readers should note that we do not have to define the Laplacian this way. A more rigorous approach would be define the Laplacian using a coordinate-free expression.

∇

2

⁡

f

=

∂

2

⁡

f

∂

⁡

x

2

+

∂

2

⁡

f

∂

⁡

y

2

+

∂

2

⁡

f

∂

⁡

z

2

.

(3)

We also know that the

partial derivatives

in rectangular coordinates can be

expanded

in the following way by using the

chain rule

∂

⁡

f

∂

⁡

x

=

∂

⁡

f

∂

⁡

r

⁢

∂

⁡

r

∂

⁡

x

+

∂

⁡

f

∂

⁡

θ

⁢

∂

⁡

θ

∂

⁡

x

+

∂

⁡

f

∂

⁡

ϕ

⁢

∂

⁡

ϕ

∂

⁡

x

,

(4)

∂

⁡

f

∂

⁡

y

=

∂

⁡

f

∂

⁡

r

⁢

∂

⁡

r

∂

⁡

y

+

∂

⁡

f

∂

⁡

θ

⁢

∂

⁡

θ

∂

⁡

y

+

∂

⁡

f

∂

⁡

ϕ

⁢

∂

⁡

ϕ

∂

⁡

y

,

(5)

∂

⁡

f

∂

⁡

z

=

∂

⁡

f

∂

⁡

r

⁢

∂

⁡

r

∂

⁡

z

+

∂

⁡

f

∂

⁡

θ

⁢

∂

⁡

θ

∂

⁡

z

+

∂

⁡

f

∂

⁡

ϕ

⁢

∂

⁡

ϕ

∂

⁡

z

.

(6)

The next step is to manipulate the right-hand side of each of the above three equations so that it is only a function of

r

,

θ

and

ϕ

. We can do this by substituting the following expressions in their respective places in the above three equations

3

3

These expression can be derived simply by taking the proper

derivative

of the expressions in (2) and then substituting the expressions from (1) in order to eliminate any dependence on

x

,

y

, and

z

.

∂

⁡

r

∂

⁡

x

=

sin

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

,

∂

⁡

θ

∂

⁡

x

=

1

r

⁢

cos

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

,

∂

⁡

ϕ

∂

⁡

x

=

-

1

r

⁢

sin

⁡

(

ϕ

)

sin

⁡

(

θ

)

,

∂

⁡

r

∂

⁡

y

=

sin

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

)

,

∂

⁡

θ

∂

⁡

y

=

1

r

⁢

cos

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

)

,

∂

⁡

ϕ

∂

⁡

y

=

1

r

⁢

cos

⁡

(

ϕ

)

sin

⁡

(

θ

)

,

∂

⁡

r

∂

⁡

z

=

cos

⁡

(

θ

)

,

∂

⁡

θ

∂

⁡

z

=

-

1

r

⁢

sin

⁡

(

θ

)

,

∂

⁡

ϕ

∂

⁡

z

=

0

.

(7)

After the substitution, equation (4) looks like the following

∂

⁡

f

∂

⁡

x

=

sin

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

r

+

1

r

⁢

cos

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

θ

-

1

r

⁢

sin

⁡

(

ϕ

)

sin

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

ϕ

.

(8)

Assuming that

f

is a sufficiently

differentiable function

, we can replace

f

by

∂

⁡

f

∂

⁡

x

and arrive at the following

∂

2

⁡

f

∂

⁡

x

2

=

sin

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

⁢

∂

∂

⁡

r

⁢

[

∂

⁡

f

∂

⁡

x

]

+

1

r

⁢

cos

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

⁢

∂

∂

⁡

θ

⁢

[

∂

⁡

f

∂

⁡

x

]

-

1

r

⁢

sin

⁡

(

ϕ

)

sin

⁡

(

θ

)

⁢

∂

∂

⁡

ϕ

⁢

[

∂

⁡

f

∂

⁡

x

]

.

(9)

Now the trick is to substitute equation (8)

into

equation (9) in order to eliminate any partial derivatives with respect to

x

. The result is the following equation

∂

2

⁡

f

∂

⁡

x

2

=

sin

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

⁢

∂

∂

⁡

r

⁢

[

sin

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

r

+

1

r

⁢

cos

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

θ

-

1

r

⁢

sin

⁡

(

ϕ

)

sin

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

ϕ

]

+

1

r

⁢

cos

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

⁢

∂

∂

⁡

θ

⁢

[

sin

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

r

+

1

r

⁢

cos

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

θ

-

1

r

⁢

sin

⁡

(

ϕ

)

sin

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

ϕ

]

-

1

r

⁢

sin

⁡

(

ϕ

)

sin

⁡

(

θ

)

⁢

∂

∂

⁡

ϕ

⁢

[

sin

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

r

+

1

r

⁢

cos

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

θ

-

1

r

⁢

sin

⁡

(

ϕ

)

sin

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

ϕ

]

.

In the hopes of simplifying the above equation, we operate the derivatives on the operands and get

∂

2

⁡

f

∂

⁡

x

2

=

sin

(

θ

)

cos

(

ϕ

)

[

sin

(

θ

)

cos

(

ϕ

)

∂

2

⁡

f

∂

⁡

r

2

-

1

r

2

cos

(

θ

)

cos

(

ϕ

)

∂

⁡

f

∂

⁡

θ

+

1

r

cos

(

θ

)

cos

(

ϕ

)

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

θ

+

1

r

2

sin

⁡

(

ϕ

)

sin

⁡

(

θ

)

∂

⁡

f

∂

⁡

ϕ

-

1

r

sin

⁡

(

ϕ

)

sin

⁡

(

θ

)

∂

2

⁡

f

∂

⁡

ϕ

⁢

∂

⁡

r

]

+

1

r

cos

(

θ

)

cos

(

ϕ

)

[

cos

(

θ

)

cos

(

ϕ

)

∂

⁡

f

∂

⁡

r

+

sin

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

⁢

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

θ

-

1

r

⁢

sin

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

θ

+

1

r

⁢

cos

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

⁢

∂

2

⁡

f

∂

⁡

θ

2

+

1

r

⁢

sin

⁡

(

ϕ

)

⁢

cos

⁡

(

θ

)

sin

2

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

ϕ

-

1

r

sin

⁡

(

ϕ

)

sin

⁡

(

θ

)

∂

2

⁡

f

∂

⁡

θ

⁢

∂

⁡

ϕ

]

-

1

r

sin

⁡

(

ϕ

)

sin

⁡

(

θ

)

[

-

sin

(

θ

)

sin

(

ϕ

)

∂

⁡

f

∂

⁡

r

+

sin

(

θ

)

cos

(

ϕ

)

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

ϕ

-

1

r

cos

(

θ

)

sin

(

ϕ

)

∂

⁡

f

∂

⁡

θ

+

1

r

cos

(

θ

)

cos

(

ϕ

)

∂

2

⁡

f

∂

⁡

θ

⁢

∂

⁡

ϕ

-

1

r

cos

⁡

(

ϕ

)

sin

⁡

(

θ

)

∂

⁡

f

∂

⁡

ϕ

-

1

r

sin

⁡

(

ϕ

)

sin

⁡

(

θ

)

∂

2

⁡

f

∂

⁡

ϕ

2

]

.

After further simplifying the above equation, we arrive at the following form

∂

2

⁡

f

∂

⁡

x

2

=

[

sin

2

(

θ

)

cos

2

(

ϕ

)

∂

2

⁡

f

∂

⁡

r

2

-

1

r

2

cos

(

θ

)

sin

(

θ

)

cos

2

(

ϕ

)

∂

⁡

f

∂

⁡

θ

+

1

r

cos

(

θ

)

sin

(

θ

)

cos

2

(

ϕ

)

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

θ

+

1

r

2

sin

(

ϕ

)

cos

(

ϕ

)

∂

⁡

f

∂

⁡

ϕ

-

1

r

sin

(

ϕ

)

cos

(

ϕ

)

∂

2

⁡

f

∂

⁡

ϕ

⁢

∂

⁡

r

]

+

[

1

r

cos

2

(

θ

)

cos

2

(

ϕ

)

∂

⁡

f

∂

⁡

r

+

1

r

⁢

sin

⁡

(

θ

)

⁢

cos

⁡

(

θ

)

⁢

cos

2

⁡

(

ϕ

)

⁢

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

θ

-

1

r

2

⁢

sin

⁡

(

θ

)

⁢

cos

⁡

(

θ

)

⁢

cos

2

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

θ

+

1

r

2

⁢

cos

2

⁡

(

θ

)

⁢

cos

2

⁡

(

ϕ

)

⁢

∂

2

⁡

f

∂

⁡

θ

2

+

1

r

2

sin

⁡

(

ϕ

)

⁢

cos

⁡

(

ϕ

)

⁢

cos

2

⁡

(

θ

)

sin

2

⁡

(

θ

)

∂

⁡

f

∂

⁡

ϕ

-

1

r

2

cos

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

)

⁢

cos

⁡

(

ϕ

)

sin

⁡

(

θ

)

∂

2

⁡

f

∂

⁡

θ

⁢

∂

⁡

ϕ

]

+

[

1

r

sin

2

(

ϕ

)

∂

⁡

f

∂

⁡

r

-

1

r

⁢

sin

⁡

(

ϕ

)

⁢

cos

⁡

(

ϕ

)

⁢

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

ϕ

+

1

r

2

⁢

cos

⁡

(

θ

)

⁢

sin

2

⁡

(

ϕ

)

sin

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

θ

-

1

r

2

⁢

cos

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

⁢

sin

⁡

(

ϕ

)

sin

⁡

(

θ

)

⁢

∂

2

⁡

f

∂

⁡

θ

⁢

∂

⁡

ϕ

+

1

r

2

sin

⁡

(

ϕ

)

⁢

cos

⁡

(

ϕ

)

sin

2

⁡

(

θ

)

∂

⁡

f

∂

⁡

ϕ

+

1

r

2

sin

2

⁡

(

ϕ

)

sin

2

⁡

(

θ

)

∂

2

⁡

f

∂

⁡

ϕ

2

]

.

Notice that we have derived the first term of the right-hand side of equation (3) (i.e.

∂

2

⁡

f

∂

⁡

x

2

) in terms of spherical coordinates. We now have to do a

similar

arduous derivation for the rest of the two terms (i.e.

∂

2

⁡

f

∂

⁡

y

2

and

∂

2

⁡

f

∂

⁡

z

2

). Lets do it!

After we substitute the values of (7) into equation (5) we get

∂

⁡

f

∂

⁡

y

=

sin

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

r

+

1

r

⁢

cos

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

θ

+

1

r

⁢

cos

⁡

(

ϕ

)

sin

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

ϕ

.

(10)

We then

differentiate

both sides with respect to

y

and arrive at the following

∂

2

⁡

f

∂

⁡

y

2

=

sin

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

)

⁢

∂

∂

⁡

r

⁢

[

∂

⁡

f

∂

⁡

y

]

+

1

r

⁢

cos

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

)

⁢

∂

∂

⁡

θ

⁢

[

∂

⁡

f

∂

⁡

y

]

+

1

r

⁢

cos

⁡

(

ϕ

)

sin

⁡

(

θ

)

⁢

∂

∂

⁡

ϕ

⁢

[

∂

⁡

f

∂

⁡

y

]

.

(11)

Now we substitute equation (10) into equation (11) in order to eliminate any partial derivatives with respect to

y

. The result is the following

∂

2

⁡

f

∂

⁡

y

2

=

sin

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

)

⁢

∂

∂

⁡

r

⁢

[

sin

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

r

+

1

r

⁢

cos

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

θ

+

1

r

⁢

cos

⁡

(

ϕ

)

sin

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

ϕ

]

+

1

r

⁢

cos

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

)

⁢

∂

∂

⁡

θ

⁢

[

sin

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

r

+

1

r

⁢

cos

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

θ

+

1

r

⁢

cos

⁡

(

ϕ

)

sin

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

ϕ

]

+

1

r

⁢

cos

⁡

(

ϕ

)

sin

⁡

(

θ

)

⁢

∂

∂

⁡

ϕ

⁢

[

sin

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

r

+

1

r

⁢

cos

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

θ

+

1

r

⁢

cos

⁡

(

ϕ

)

sin

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

ϕ

]

.

Now we operate the operators and get

∂

2

⁡

f

∂

⁡

y

2

=

sin

(

θ

)

sin

(

ϕ

)

[

sin

(

θ

)

sin

(

ϕ

)

∂

2

⁡

f

∂

⁡

r

2

-

1

r

2

cos

(

θ

)

sin

(

ϕ

)

∂

⁡

f

∂

⁡

θ

+

1

r

cos

(

θ

)

sin

(

ϕ

)

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

θ

-

1

r

2

cos

⁡

(

ϕ

)

sin

⁡

(

θ

)

∂

⁡

f

∂

⁡

ϕ

+

1

r

cos

⁡

(

ϕ

)

sin

⁡

(

θ

)

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

ϕ

]

+

1

r

cos

(

θ

)

sin

(

ϕ

)

[

sin

(

θ

)

sin

(

ϕ

)

∂

⁡

f

∂

⁡

r

⁢

∂

⁡

θ

+

cos

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

r

-

1

r

⁢

sin

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

θ

+

1

r

⁢

cos

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

)

⁢

∂

2

⁡

f

∂

⁡

θ

2

-

1

r

⁢

cos

⁡

(

ϕ

)

⁢

c

⁢

o

⁢

s

⁢

(

θ

)

sin

2

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

ϕ

+

1

r

cos

⁡

(

ϕ

)

sin

⁡

(

θ

)

∂

2

⁡

f

∂

⁡

θ

⁢

∂

⁡

ϕ

]

+

1

r

cos

⁡

(

ϕ

)

sin

⁡

(

θ

)

[

sin

(

θ

)

cos

(

ϕ

)

∂

⁡

f

∂

⁡

r

+

sin

(

θ

)

sin

(

ϕ

)

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

ϕ

+

1

r

cos

(

θ

)

cos

(

ϕ

)

∂

⁡

f

∂

⁡

θ

+

1

r

cos

(

θ

)

sin

(

ϕ

)

∂

2

⁡

f

∂

⁡

θ

⁢

∂

⁡

ϕ

-

1

r

sin

⁡

(

ϕ

)

sin

⁡

(

θ

)

∂

⁡

f

∂

⁡

ϕ

+

1

r

cos

⁡

(

ϕ

)

sin

⁡

(

θ

)

∂

2

⁡

f

∂

⁡

ϕ

2

]

,

and after some simplifications

∂

2

⁡

f

∂

⁡

y

2

=

[

sin

2

(

θ

)

sin

2

(

ϕ

)

∂

2

⁡

f

∂

⁡

r

2

-

1

r

2

sin

(

θ

)

cos

(

θ

)

sin

2

(

ϕ

)

∂

⁡

f

∂

⁡

θ

+

1

r

sin

(

θ

)

cos

(

θ

)

sin

2

(

ϕ

)

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

θ

-

1

r

2

sin

(

ϕ

)

cos

(

ϕ

)

∂

⁡

f

∂

⁡

ϕ

+

1

r

sin

(

ϕ

)

cos

(

ϕ

)

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

ϕ

]

+

[

1

r

cos

(

θ

)

sin

(

θ

)

sin

2

(

ϕ

)

∂

⁡

f

∂

⁡

r

⁢

∂

⁡

θ

+

1

r

⁢

cos

2

⁡

(

θ

)

⁢

sin

2

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

r

-

1

r

2

⁢

sin

⁡

(

θ

)

⁢

cos

⁡

(

θ

)

⁢

sin

2

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

θ

+

1

r

2

⁢

cos

2

⁡

(

θ

)

⁢

sin

2

⁡

(

ϕ

)

⁢

∂

2

⁡

f

∂

⁡

θ

2

-

1

r

2

sin

⁡

(

ϕ

)

⁢

cos

⁡

(

ϕ

)

⁢

cos

2

⁡

(

θ

)

sin

2

⁡

(

θ

)

∂

⁡

f

∂

⁡

ϕ

+

1

r

2

cos

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

)

⁢

cos

⁡

(

ϕ

)

sin

⁡

(

θ

)

∂

2

⁡

f

∂

⁡

θ

⁢

∂

⁡

ϕ

]

+

[

1

r

cos

2

(

ϕ

)

∂

⁡

f

∂

⁡

r

+

1

r

⁢

cos

⁡

(

ϕ

)

⁢

sin

⁡

(

ϕ

)

⁢

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

ϕ

+

1

r

2

⁢

cos

⁡

(

θ

)

⁢

cos

2

⁡

(

ϕ

)

sin

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

θ

+

1

r

2

⁢

cos

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

⁢

sin

⁡

(

ϕ

)

sin

⁡

(

θ

)

⁢

∂

2

⁡

f

∂

⁡

θ

⁢

∂

⁡

ϕ

-

1

r

2

sin

⁡

(

ϕ

)

⁢

cos

⁡

(

ϕ

)

sin

2

⁡

(

θ

)

∂

⁡

f

∂

⁡

ϕ

+

1

r

2

cos

2

⁡

(

ϕ

)

sin

2

⁡

(

θ

)

∂

2

⁡

f

∂

⁡

ϕ

2

]

.

Now its time to derive

∂

2

⁡

f

∂

⁡

z

2

. After our substitution of value in (7) into equation (6) we get

∂

⁡

f

∂

⁡

z

=

cos

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

r

-

1

r

⁢

sin

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

θ

.

(12)

We then differentiate both sides of the above equation with respect to

z

resulting in the following

∂

2

⁡

f

∂

⁡

z

2

=

cos

⁡

(

θ

)

⁢

∂

∂

⁡

r

⁢

[

∂

⁡

f

∂

⁡

z

]

-

1

r

⁢

sin

⁡

(

θ

)

⁢

∂

∂

⁡

θ

⁢

[

∂

⁡

f

∂

⁡

z

]

.

(13)

Now we substitute equation (12) into equation (13) in order to eliminate any partial derivatives with respect to

z

and we arrive at

∂

2

⁡

f

∂

⁡

z

2

=

cos

⁡

(

θ

)

⁢

∂

∂

⁡

r

⁢

[

cos

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

r

-

1

r

⁢

sin

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

θ

]

-

1

r

⁢

sin

⁡

(

θ

)

⁢

∂

∂

⁡

θ

⁢

[

cos

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

r

-

1

r

⁢

sin

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

θ

]

.

After operating the operators we get

∂

2

⁡

f

∂

⁡

z

2

=

cos

⁡

(

θ

)

⁢

[

cos

⁡

(

θ

)

⁢

∂

2

⁡

f

∂

⁡

r

2

+

1

r

2

⁢

sin

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

θ

-

1

r

⁢

sin

⁡

(

θ

)

⁢

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

θ

]

-

1

r

sin

(

θ

)

[

-

sin

(

θ

)

∂

⁡

f

∂

⁡

r

+

cos

(

θ

)

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

θ

-

1

r

cos

(

θ

)

∂

⁡

f

∂

⁡

θ

-

1

r

sin

(

θ

)

∂

2

⁡

f

∂

⁡

θ

2

]

,

and then simplifying

∂

2

⁡

f

∂

⁡

z

2

=

[

cos

2

(

θ

)

∂

2

⁡

f

∂

⁡

r

2

+

1

r

2

cos

(

θ

)

sin

(

θ

)

∂

⁡

f

∂

⁡

θ

-

1

r

cos

(

θ

)

sin

(

θ

)

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

θ

]

+

[

1

r

sin

2

(

θ

)

∂

⁡

f

∂

⁡

r

-

1

r

sin

(

θ

)

cos

(

θ

)

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

θ

+

1

r

2

sin

(

θ

)

cos

(

θ

)

∂

⁡

f

∂

⁡

θ

+

1

r

2

sin

2

(

θ

)

∂

2

⁡

f

∂

⁡

θ

2

]

.

Now that we have all three terms of the right hand side of equation (3)(i.e.

∂

2

⁡

f

∂

⁡

x

2

,

∂

2

⁡

f

∂

⁡

y

2

and

∂

2

⁡

f

∂

⁡

z

2

), we add them all together (because of equation (3)) to get the laplacian in terms of

r

,

θ

and

ϕ

∇

2

⁡

f

=

[

sin

2

(

θ

)

cos

2

(

ϕ

)

∂

2

⁡

f

∂

⁡

r

2

-

1

r

2

cos

(

θ

)

sin

(

θ

)

cos

2

(

ϕ

)

∂

⁡

f

∂

⁡

θ

+

1

r

cos

(

θ

)

sin

(

θ

)

cos

2

(

ϕ

)

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

θ

+

1

r

2

sin

(

ϕ

)

cos

(

ϕ

)

∂

⁡

f

∂

⁡

ϕ

-

1

r

sin

(

ϕ

)

cos

(

ϕ

)

∂

2

⁡

f

∂

⁡

ϕ

⁢

∂

⁡

r

]

+

[

1

r

cos

2

(

θ

)

cos

2

(

ϕ

)

∂

⁡

f

∂

⁡

r

+

1

r

⁢

sin

⁡

(

θ

)

⁢

cos

⁡

(

θ

)

⁢

cos

2

⁡

(

ϕ

)

⁢

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

θ

-

1

r

2

⁢

sin

⁡

(

θ

)

⁢

cos

⁡

(

θ

)

⁢

cos

2

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

θ

+

1

r

2

⁢

cos

2

⁡

(

θ

)

⁢

cos

2

⁡

(

ϕ

)

⁢

∂

2

⁡

f

∂

⁡

θ

2

+

1

r

2

sin

⁡

(

ϕ

)

⁢

cos

⁡

(

ϕ

)

⁢

cos

2

⁡

(

θ

)

sin

2

⁡

(

θ

)

∂

⁡

f

∂

⁡

ϕ

-

1

r

2

cos

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

)

⁢

cos

⁡

(

ϕ

)

sin

⁡

(

θ

)

∂

2

⁡

f

∂

⁡

θ

⁢

∂

⁡

ϕ

]

+

[

1

r

sin

2

(

ϕ

)

∂

⁡

f

∂

⁡

r

-

1

r

⁢

sin

⁡

(

ϕ

)

⁢

cos

⁡

(

ϕ

)

⁢

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

ϕ

+

1

r

2

⁢

cos

⁡

(

θ

)

⁢

sin

2

⁡

(

ϕ

)

sin

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

θ

-

1

r

2

⁢

cos

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

⁢

sin

⁡

(

ϕ

)

sin

⁡

(

θ

)

⁢

∂

2

⁡

f

∂

⁡

θ

⁢

∂

⁡

ϕ

+

1

r

2

sin

⁡

(

ϕ

)

⁢

cos

⁡

(

ϕ

)

sin

2

⁡

(

θ

)

∂

⁡

f

∂

⁡

ϕ

+

1

r

2

sin

2

⁡

(

ϕ

)

sin

2

⁡

(

θ

)

∂

2

⁡

f

∂

⁡

ϕ

2

]

+

[

sin

2

(

θ

)

sin

2

(

ϕ

)

∂

2

⁡

f

∂

⁡

r

2

-

1

r

2

⁢

sin

⁡

(

θ

)

⁢

cos

⁡

(

θ

)

⁢

sin

2

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

θ

+

1

r

⁢

sin

⁡

(

θ

)

⁢

cos

⁡

(

θ

)

⁢

sin

2

⁡

(

ϕ

)

⁢

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

θ

-

1

r

2

⁢

sin

⁡

(

ϕ

)

⁢

cos

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

ϕ

+

1

r

sin

(

ϕ

)

cos

(

ϕ

)

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

ϕ

]

+

[

1

r

cos

(

θ

)

sin

(

θ

)

sin

2

(

ϕ

)

∂

⁡

f

∂

⁡

r

⁢

∂

⁡

θ

+

1

r

cos

2

(

θ

)

sin

2

(

ϕ

)

∂

⁡

f

∂

⁡

r

-

1

r

2

⁢

sin

⁡

(

θ

)

⁢

cos

⁡

(

θ

)

⁢

sin

2

⁡

(

ϕ

)

⁢

∂

⁡

f

∂

⁡

θ

+

1

r

2

⁢

cos

2

⁡

(

θ

)

⁢

sin

2

⁡

(

ϕ

)

⁢

∂

2

⁡

f

∂

⁡

θ

2

-

1

r

2

⁢

sin

⁡

(

ϕ

)

⁢

cos

⁡

(

ϕ

)

⁢

cos

2

⁡

(

θ

)

sin

2

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

ϕ

+

1

r

2

cos

⁡

(

θ

)

⁢

sin

⁡

(

ϕ

)

⁢

cos

⁡

(

ϕ

)

sin

⁡

(

θ

)

∂

2

⁡

f

∂

⁡

θ

⁢

∂

⁡

ϕ

]

+

[

1

r

cos

2

(

ϕ

)

∂

⁡

f

∂

⁡

r

+

1

r

cos

(

ϕ

)

sin

(

ϕ

)

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

ϕ

+

1

r

2

⁢

cos

⁡

(

θ

)

⁢

cos

2

⁡

(

ϕ

)

sin

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

θ

+

1

r

2

⁢

cos

⁡

(

θ

)

⁢

cos

⁡

(

ϕ

)

⁢

sin

⁡

(

ϕ

)

sin

⁡

(

θ

)

⁢

∂

2

⁡

f

∂

⁡

θ

⁢

∂

⁡

ϕ

-

1

r

2

⁢

sin

⁡

(

ϕ

)

⁢

cos

⁡

(

ϕ

)

sin

2

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

ϕ

+

1

r

2

cos

2

⁡

(

ϕ

)

sin

2

⁡

(

θ

)

∂

2

⁡

f

∂

⁡

ϕ

2

]

+

[

cos

2

(

θ

)

∂

2

⁡

f

∂

⁡

r

2

+

1

r

2

cos

(

θ

)

sin

(

θ

)

∂

⁡

f

∂

⁡

θ

-

1

r

cos

(

θ

)

sin

(

θ

)

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

θ

]

+

[

1

r

⁢

sin

2

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

r

-

1

r

⁢

sin

⁡

(

θ

)

⁢

cos

⁡

(

θ

)

⁢

∂

2

⁡

f

∂

⁡

r

⁢

∂

⁡

θ

+

1

r

2

⁢

sin

⁡

(

θ

)

⁢

cos

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

θ

+

1

r

2

⁢

sin

2

⁡

(

θ

)

⁢

∂

2

⁡

f

∂

⁡

θ

2

]

.

It may be hard to believe but the truth is that the above expression, after some miraculous simplifications of course, reduces to the following succinct expression and we finally arrive at the Laplacian in spherical coordinates!

∇

2

⁡

f

=

∂

2

⁡

f

∂

⁡

r

2

+

1

r

2

⁢

∂

2

⁡

f

∂

⁡

θ

2

+

1

r

2

⁢

1

sin

2

⁡

(

θ

)

⁢

∂

2

⁡

f

∂

⁡

ϕ

2

+

2

r

⁢

∂

⁡

f

∂

⁡

r

+

1

r

2

⁢

cos

⁡

(

θ

)

sin

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

θ

.

(14)

By further manipulating the above expression, we can write the Laplacian in a more compact form as

∇

2

⁡

f

=

1

r

2

⁢

∂

∂

⁡

r

⁢

[

r

2

⁢

∂

⁡

f

∂

⁡

r

]

+

1

r

2

⁢

sin

⁡

(

θ

)

⁢

∂

∂

⁡

θ

⁢

[

sin

⁡

(

θ

)

⁢

∂

⁡

f

∂

⁡

θ

]

+

1

r

2

⁢

sin

2

⁡

(

θ

)

⁢

∂

2

⁡

f

∂

⁡

ϕ

2

.

(15)

Title

Laplacian from Rectangular to Spherical Coordinates

Canonical name

LaplacianFromRectangularToSphericalCoordinates1

Date of creation

2013-03-11 19:24:44

Last modified on

2013-03-11 19:24:44

Owner

swapnizzle (13346)

Last modified by

(0)

Numerical id

1

Author

swapnizzle (0)

Entry type

Definition

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