7/29/2019 Delsphspherical Coordinates

1/5

Spherical Coordinates

Transforms

The forward and reverse coordinate transformations are

r= x2+ y

2+ z

2

!= arctan x2 + y2 ,z"# $

%

&= arctan y,x( )

x = rsin cos"

y = rsin!sin"

z = rcos!

where weformally take advantage of the two argumentarctan

function to eliminate quadrant confusion.

Unit Vectors

The unit vectors in the spherical coordinate system are functions of position. It is convenient to express them in terms of

the spherical coordinates and the unit vectors of the rectangularcoordinate system which are notthemselves functions of

position.

r=

!r

r=xx +yy+z z

r= x

sin!cos

"+ ysin

!sin

"+ z

cos!

"= z # rsin!

= $ xsin"+ y cos"

!= "# r= x cos!cos"+ y cos!sin"$ zsin!

Variations of unit vectors with the coordinates

Using the expressions obtained above it is easy to derive the following handy relationships:

!r

!r= 0

!r

!"= x cos" cos#+ y cos"sin#$ zsin"= "

!r

!#= $ x sin"sin#+ ysin"cos# = $ x sin#+ ycos#( )sin" = #sin"

!"!r

= 0

!"!#

= 0

!"!"

= $ x cos"$ y sin"= $ rsin#+ #cos#( )

!"!r

= 0

!"!"

= # xsin"cos$# y sin"sin$# zcos"= #r

!"!$

= # x cos"sin$+ y cos"cos$= $cos"

x

y

z

r^

!^

"^

!

"

r

r

7/29/2019 Delsphspherical Coordinates

2/5

Path increment

We will have many uses for the path increment d!r expressed in spherical coordinates:

d!r= d rr( ) = r dr+ rdr= rdr+ r

!r

!rdr+

!r

!"d"+

!r

!#d#

$

%&

'

()

= r dr+

"rd

"+

#rsin

"d#

Time derivatives of the unit vectors

We will also have many uses for the time derivatives of the unit vectors expressed in spherical coordinates:

r=!r

!rr+

!r

!"" + !r

!##= ""+ ##sin"

"= !"!r

r+

!"!"

"+ !"!#

# = $r"+ ##cos"#= !#

!rr+

!#!"

"+ !#!#

#= $ rsin"+ "cos"( )#

Velocity and Acceleration

The velocity and acceleration of a particle may be expressed in spherical coordinates by taking into account the associated

rates of change in the unit vectors:

!v =

!r=

r r+ rr

!v = rr+

!r!+ "r"sin!

!a =

!v = rr+ rr+ !r!+ !r!+ !r! + "r"sin!+ "r"sin!+ "r"sin!+ "r"!cos!

=! !+ ""sin!( )r+ rr+ #r!+ ""cos!( )r!+ !r!+ !r!+ # rsin!+ !cos!( )"[ ]r"sin!+ "r"sin!+ "r"sin!+ "r"!cos !

!a = r r! r

"2 ! r#2 sin"( ) + " r"+ 2 r"! r#2 sin"cos"( ) + # r#sin"+2r"#cos" + 2r#sin"( )

The del operator from the definition of the gradient

Any (static) scalar field u may be considered to be a function of the spherical coordinates r, , and . The value ofu

changes by an infinitesimal amount du when the point of observation is changed by d!r. That change may be determined

from the partial derivatives as

du =!u

!rdr+

!u

!"d"+

!u

!#d# .

But we also define the gradient in such a way as to obtain the result

du =

!!u " d

!r

Therefore,

!u

!rdr+

!u

!"d"+

!u

!#d#=

!$u %d

!r

or, in spherical coordinates,

7/29/2019 Delsphspherical Coordinates

3/5

!u

!rdr+

!u

!"d"+

!u

!#d#=

!$u( )

rdr+

!$u( )

"rd" +

!$u( )

#rsin"d#

and we demand that this hold for any choice ofdr, d, and d. Thus,

!

!u( )r

=

"u

"r,

!

!u( )#=

1

r

"u

"#,

!

!u( )$=

1

rsin#

"u

"$,

from which we find

!!= r

"

"r+

#r

"

"#+

$rsin#

"

"$

Divergence

The divergence!!" !A is carried out taking into account, once again, that the unit vectors themselves are functions of the

coordinates. Thus, we have

!!"

!A = r

#

#r+

$r

#

#$+

%rsin$

#

#%

&

'(

)

*+ " Arr+A$ $ +A%%( )

where the derivatives must be taken before the dot product so that

!!"

!A = r

#

#r+

$r

#

#$+

%rsin$

#

#%

&

'(

)

*+ "

!A

= r"#!A

#r+$r" #

!A

#$+

%rsin$

" #!A

#%

= r"#A

r

#rr+

#A$#r

$ + #A%#r

%+Ar#r#r

+A$

#$#r

+A%

#%#r

&

'(

)

*+

+

$r" #Ar

#$r+

#A$#$

$ + #A%#$

%+ Ar

#r

#$+ A$

#$#$

+ A%

#%#$

&

'(

)

*+

+

%

rsin$ "#A

r

#%r+ #

A$

#%$+ #A%#% %+Ar #

r

#%+A$ #

$

#%+A% #

%

#%&

'( )

*+

With the help of the partial derivatives previously obtained, we find

!!"

!A = r"

#Ar

#rr+

#A$#r

$ + #A%#r

%+ 0 + 0 +0&'(

)

*+

+

$r" #Ar

#$r+

#A$#$

$ + #A%#$

%+Ar$ +A$ ,r( ) + 0&'()

*+

+

%rsin$

"#A

r

#%r+

#A$#%

$+ #A%#%

%+Arsin$ %+ A$ cos$%+A% , rsin$+ $ cos$( )[ ]&'()

*+

= #Ar#r

&'( )

*+ + 1

r#A$#$

+ Ar

r&'( )

*+ + 1

rsin$#A

%

#%+ Ar

r+ A$ cos$

rsin$&'( )

*+

=

#Ar

#r+

2Ar

r

&'(

)*++

1

r

#A$#$

+

A$cos$

rsin$

&'(

)*++

1

rsin$

#A%

#%

!!"

!A =

1

r2

#

#rr2Ar( ) +

1

rsin$

#

#$A$ sin$( ) +

1

rsin$

#A%

#%

7/29/2019 Delsphspherical Coordinates

4/5

Curl

The curl!!" !A is also carried out taking into account that the unit vectors themselves are functions of the coordinates.

Thus, we have

!!"

!A = r

#

#r+

$r

#

#$+

%rsin$

#

#%

&

'

()

*

+" Arr+A$$ +A%%( ) where the derivatives must be taken before the cross product so that

!!"

!A = r

#

#r+

$r

#

#$+

%rsin$

#

#%

&

'(

)

*+"

!A

= r"#!A

#r+$r" #

!A

#$+

%rsin$

" #!

A

#%

= r"#A

r

#rr+

#A$#r

$+ #A%#r

%+ Ar #r#r

+A$

#$#r

+A%

#%#r

&

'(

)

*+

+

$r

" #Ar

#$

r+#A$

#$

$+ #A%#$

%+ Ar

#r

#$

+A$

# $#$

+A%

#%#$

&

'(

)

*+

+

%rsin$

"#A

r

#%r+

#A$#%

$ + #A%#%

%+ Ar #r#%

+A$

#$#%

+A%

#%#%

&

'(

)

*+

With the help of the partial derivatives previously obtained, we find

!!"

!A = r"

#Ar

#rr+

#A$#r

$+ #A%#r

%+ 0 + 0 + 0&'(

)

*+

+$r" #Ar

#$r+

#A$#$

$+ #A%#$

%+ Ar$ + A$ ,r( ) + 0&

'(

)

*+

+

%rsin$ "

#Ar

#%r+

#A$

#%$ + #A%

#%%+ Arsin$% +A

$cos$%+A

% ,rsin$+ $ cos$( )[ ]

&

'(

)

*+

=#A$#r

%, #A%#r

$&'(

)

*+ + ,

1

r

#Ar

#$%+ 1

r

#A%

#$r+

A$

r%&

'(

)

*+

+

1

rsin$

#Ar

#%$, 1

rsin$

#A$#%

r,A%

r

$+ A% cos$rsin$

r&

'(

)

*+

= r1

r

#A%

#$, 1

rsin$

#A$#%

+

A% cos$

rsin$

&

'(

)

*+

+ $ , #A%#r

+1

rsin$

#Ar

#%,A%

r

&

'(

)

*+

+% #A$

#r, 1r

#Ar#$

+A$

r

&'( )

*+

!!"

!A =

r

rsin#

$

$#A% sin#( ) &

$A#$%

'

()

*

+,+

#rsin#

$Ar

$%& sin#

$

$rrA%( )

'

()

*

+,+

%r

$

$rrA#( ) &

$Ar

$#

'

()*

+,

7/29/2019 Delsphspherical Coordinates

5/5

Laplacian

The Laplacian is a scalar operator that can be determined from its definition as

!2u =!!"

!!u( ) = r #

#r+

$r

#

#$+

%rsin$

#

#%

&

'(

)

*+ " r

#u

#r+

$r

#u

#$+

%rsin$

#u

#%

&

'(

)

*+

= r" ##r

r#u

#r+

$r

#u

#$+

%

rsin$#u

#%&'( )

*+

+$r

" ##$

r

#u

#r+$r

#u

#$+

%rsin$

#u

#%

&

'(

)

*+

+

%rsin$

" ##%

r

#u

#r+

$r

#u

#$+

%rsin$

#u

#%

&

'(

)

*+

With the help of the partial derivatives previously obtained, we find

!2u = r" r#2u

#r2$

%r2

#u

#%+

%r

#2u

#%#r$ &

r2sin%

#u

#&+

&rsin%

#2u

#r

'

()

*

+,

+

%r

" % #u#r

+ r

#2u

#r#%$

r

r

#u

#%+

%r

#2u

#%2$

&cos%rsin

2%

#u

#&+

&rsin%

#2u

#%

'

()

*

+,

+&

rsin%" &sin%#u

#r+ r

#2u

#r#&+&cos%

r

#u

#%+%r

#2u

#%#&$ rsin%+ %cos%

rsin%

#u

#&+

&rsin%

#2u

#&2'

()

*

+,

=#2u

#r2

'

()

*

+, +

1

r

#u

#r+

1

r2

#2u

#%2

'

()

*

+, +

1

r

#u

#r+

cos%

r2sin%

#u

#%+

1

r2sin

2%

#2u

#&2

'

()

*

+,

=#2u

#r2+

2

r

#u

#r

'

()

*

+, +

1

r2

#2u

#%2+

cos%

r2sin%

#u

#%

'

()

*

+, +

1

r2sin

2%

#2u

#&2

'

()

*

+,

=

1

r2

#

#rr

2 #u

#r

'()

*+, +

1

r2sin%

#

#% sin%#u

#%

'()

*+, +

1

r2sin

2%

#2u

#&2

Thus, the Laplacian operator can be written as

!2 =1

r2

"

"rr2 "

"r

#$%

&'(+

1

r2sin)

"

")sin)

"

")

#$%

&'(+

1

r2sin

2 )

"2

"*2