Sometimes there is a need to deal with spherical coordinates in more than three dimensions.
In this section, this concept is defined and formulas are derived for these coordinate systems.
Recall polar coordinates are of the form
y1 = ρcosθ
y2 = ρsin θ
where ρ > 0 and θ ∈ ℝ. Thus these transformation equations are not one to one
but they are one to one on
(0,∞ )
× [0,2π). Here I am writing ρ in place of r to
emphasize a pattern which is about to emerge. I will consider polar coordinates as
spherical coordinates in two dimensions. I will also simply refer to such coordinate
systems as polar coordinates regardless of the dimension. This is also the reason
I am writing y_{1} and y_{2} instead of the more usual x and y. Now consider what
happens when you go to three dimensions. The situation is depicted in the following
picture.
PICT
From this picture, you see that y_{3} = ρcosϕ_{1}. Also the distance between
(y1,y2)
and
(0,0)
is ρsin
(ϕ1)
. Therefore, using polar coordinates to write
(y1,y2)
in terms of θ and this
distance,
y1 = ρsin ϕ1cosθ,
y2 = ρsin ϕ1sin θ,
y3 = ρcosϕ1.
where ϕ_{1}∈ ℝ and the transformations are one to one if ϕ_{1} is restricted to be in
[0,π ]
. What
was done is to replace ρ with ρsinϕ_{1} and then to add in y_{3} = ρcosϕ_{1}. Having
done this, there is no reason to stop with three dimensions. Consider the following
picture:
PICT
From this picture, you see that y_{4} = ρcosϕ_{2}. Also the distance between
where ϕ_{2}∈ ℝ and the transformations will be one to one if
ϕ2,ϕ1 ∈ (0,π),θ ∈ (0,2π ),ρ ∈ (0,∞ ).
Continuing this way, given spherical coordinates in ℝ^{p}, to get the spherical coordinates in
ℝ^{p+1}, you let y_{p+1} = ρcosϕ_{p−1} and then replace every occurance of ρ with ρsinϕ_{p−1} to
obtain y_{1}
⋅⋅⋅
y_{p} in terms of ϕ_{1},ϕ_{2},
⋅⋅⋅
,ϕ_{p−1},θ, and ρ.
It is always the case that ρ measures the distance from the point in ℝ^{p} to the origin in ℝ^{p},
0. Each ϕ_{i}∈ ℝ and the transformations will be one to one if each ϕ_{i}∈
(0,π)
, and
θ ∈
(0,2π)
. Denote by h_{p}
( )
ρ,⃗ϕ,θ
the above transformation.
It can be shown using math induction and geometric reasoning that these coordinates map
∏_{i=1}^{p−2}
(0,π)
×
(0,2π)
×
(0,∞)
one to one onto an open subset of ℝ^{p} which is everything
except for the set of measure zero Ψ_{p}
(N )
where N results from having some ϕ_{i} equal to 0 or
π or for ρ = 0 or for θ equal to either 2π or 0. Each of these are sets of Lebesgue
measure zero and so their union is also a set of measure zero. You can see that
h_{p}
(∏ )
p−i=12(0,π) × (0,2π)× (0,∞ )
omits the union of the coordinate axes except for maybe
one of them. This is not important to the integral because it is just a set of measure
zero.
Theorem 9.10.1Let y = h_{p}
(⃗ϕ,θ,ρ)
be the spherical coordinate transformations inℝ^{p}. Then letting A = ∏_{i=1}^{p−2}
(0,π)
×
(0,2π)
, it follows h maps A×
(0,∞)
one to one ontoall of ℝ^{p}except a set of measure zero given by h_{p}
andθ.^{1}Then if f is nonnegative and Lebesgue measurable,
∫ ∫ ∫ ( ( )) ( )
f (y)dmp = f (y)dmp = f hp ⃗ϕ,θ,ρ ρp−1Φ ⃗ϕ,θ dmp (9.20)
ℝp hp(A ) A
(9.20)
Furthermore whenever f is Borel measurable and nonnegative, one can apply Fubini’stheorem and write
∫ ∫ ∞ ∫ ( ( ) ) ( )
p f (y)dy = ρp− 1 f h ⃗ϕ,θ,ρ Φ ⃗ϕ,θ d⃗ϕdθdρ (9.21)
ℝ 0 A
(9.21)
where here d
⃗ϕ
dθ denotes dm_{p−1}on A. The same formulas hold if f ∈ L^{1}
(ℝp)
.
Proof: Formula 9.19 is obvious from the definition of the spherical coordinates because in
the matrix of the derivative, there will be a ρ in p − 1 columns. The first claim is also
clear from the definition and math induction or from the geometry of the above
description. It remains to verify 9.20 and 9.21. It is clear h_{p} maps A× [0,∞) onto ℝ^{p}.
Since h_{p} is differentiable, it maps sets of measure zero to sets of measure zero.
Then
ℝp = hp(N ∪ A × (0,∞ )) = hp(N )∪ hp(A × (0,∞ )),
the union of a set of measure zero with h_{p}
(A ×(0,∞ ))
. Therefore, from the change of
variables formula,
∫ ∫ ∫ ( ( )) ( )
p f (y)dmp = f (y)dmp = f hp ⃗ϕ,θ,ρ ρp− 1Φ ⃗ϕ,θ dmp
ℝ hp(A×(0,∞)) A×(0,∞)
which proves 9.20. This formula continues to hold if f is in L^{1}
(ℝp)
. Finally, if f ≥ 0 or in
L^{1}
(ℝn )
and is Borel measurable, then it is ℱ^{p} measurable as well. Recall that ℱ^{p} includes the
smallest σ algebra which contains products of open intervals. Hence ℱ^{p} includes the Borel
sets ℬ
(ℝp )
. Thus from the definition of m_{p}
∫ ( ( )) ( )
f hp ⃗ϕ,θ,ρ ρp−1Φ ⃗ϕ,θ dmp
A×(0,∞ )
∫ ∫ ( ( )) ( )
= f hp ⃗ϕ,θ,ρ ρp− 1Φ ⃗ϕ,θ dmp−1dm
∫(0,∞) A ∫
p−1 ( (⃗ )) (⃗ )
= (0,∞)ρ Af hp ϕ,θ,ρ Φ ϕ,θ dmp− 1dm
Now the claim about f ∈ L^{1} follows routinely from considering the positive and negative
parts of the real and imaginary parts of f in the usual way. ■
Note that the above equals
∫ ( ( )) ( )
f hp ⃗ϕ,θ,ρ ρp−1Φ ⃗ϕ,θ dmp
A¯×[0,∞ )
and the iterated integral is also equal to
∫ ∫ ( ( ) ) ( )
ρp− 1 f hp ⃗ϕ,θ,ρ Φ ⃗ϕ,θ dmp −1dm
[0,∞ ) ¯A
because the difference is just a set of measure zero.
Notation 9.10.2Often this is written differently. Note that from the spherical coordinateformulas, f
( ( ))
h ⃗ϕ,θ,ρ
= f
(ρω)
where
|ω|
= 1. Letting S^{p−1}denote the unit sphere,
p
{ω ∈ ℝ : |ω| = 1}
, the inside integral in the above formula is sometimes writtenas
∫
f (ρω)dσ
Sp−1
where σ is a measure on S^{p−1}. See [20]for another description of this measure. It isn’t animportant issue here. Either 9.21or the formula
∫ ∞ (∫ )
ρp−1 f (ρω)dσ dρ
0 Sp−1
will be referred to as polar coordinates and is very useful in establishing estimates. Hereσ
( )
Sp−1
≡∫_{A}Φ
( )
⃗ϕ,θ
dm_{p−1}.
Example 9.10.3For what values of s is the integral∫_{B}
(0,R)
( )
1+ |x |2
^{s}dy boundedindependent of R? Here B
(0,R )
is the ball,
p
{x ∈ ℝ : |x | ≤ R}
.
I think you can see immediately that s must be negative but exactly how negative? It
turns out it depends on p and using polar coordinates, you can find just exactly what is
needed. From the polar coordinates formula above,
∫ ( 2)s ∫ R ∫ ( 2)s p−1
1+ |x| dy = p−1 1+ ρ ρ dσdρ
B(0,R) 0 ∫ SR
= Cp (1 + ρ2)sρp−1dρ
0
Now the very hard problem has been reduced to considering an easy one variable problem of
finding when
∫ R
ρp− 1(1+ ρ2)sdρ
0
is bounded independent of R. You need 2s +
(p − 1)
< −1 so you need s < −p∕2.
class=”left” align=”middle”(Y )9.11. BROUWER FIXED POINT THEOREM