Spherical coordinates are three-dimensional set of coordinates where each point in space is determined by the distance r r r θ \theta θ ϕ \phi ϕ θ ∈ [ 0 , 2 π [ \theta\in[0,2\pi[ θ ∈ [ 0 , 2 π [ ϕ ∈ [ 0 , π [ \phi\in[0,\pi[ ϕ ∈ [ 0 , π [
Relation to Cartesian coordinates#
Spherical coordinates ( r , θ , ϕ ) (r,\theta,\phi) ( r , θ , ϕ ) Cartesian coordinates ( x , y , z ) (x,y,z) ( x , y , z ) Coordinate transformation
x = r cos θ sin ϕ , y = r sin θ sin ϕ , z = r cos ϕ x=r\cos\theta\sin\phi, \quad y=r\sin\theta\sin\phi, \quad z=r\cos\phi x = r cos θ sin ϕ , y = r sin θ sin ϕ , z = r cos ϕ and viceversa
r = x 2 + y 2 + z 2 , y = arctan ( x 2 + y 2 z ) , z = arctan ( y x ) r=\sqrt{ x^{2}+y^{2}+z^{2} },\quad y=\arctan\left( \frac{\sqrt{x^{2}+y^{2} }}{z} \right),\quad z=\arctan\left( \frac{y}{x} \right) r = x 2 + y 2 + z 2 , y = arctan ( z x 2 + y 2 ) , z = arctan ( x y ) To convert an integral from Cartesian to spherical, we find the determinant of the Jacobian of the conversion diffeomorphism σ ( x , y , z ) = ( r cos θ sin ϕ , r sin θ sin ϕ , r cos ϕ ) \sigma(x,y,z)=(r\cos\theta\sin\phi,r\sin\theta\sin\phi,r\cos\phi) σ ( x , y , z ) = ( r cos θ sin ϕ , r sin θ sin ϕ , r cos ϕ ) det J σ = r 2 sin ϕ \det J\sigma=r^{2}\sin \phi det J σ = r 2 sin ϕ
∭ d x d y d z = ∭ r 2 sin ϕ d r d θ d ϕ \iiint dxdydz=\iiint r^{2}\sin \phi\ drd\theta d\phi ∭ d x d y d z = ∭ r 2 sin ϕ d r d θ d ϕ Motion#
A unit vector in the unit sphere is R = ( cos θ sin ϕ , sin θ sin ϕ , cos ϕ ) \mathbf{R}=(\cos\theta\sin\phi,\sin\theta\sin\phi,\cos\phi) R = ( cos θ sin ϕ , sin θ sin ϕ , cos ϕ ) P = ( − sin θ , cos θ , 0 ) \mathbf{P}=(-\sin\theta,\cos\theta,0) P = ( − sin θ , cos θ , 0 ) cross product Q = R × P = ( − cos θ cos ϕ , − sin θ cos ϕ , sin ϕ ) \mathbf{Q}=\mathbf{R}\times\mathbf{P}=(-\cos\theta\cos\phi,-\sin\theta\cos\phi,\sin\phi) Q = R × P = ( − cos θ cos ϕ , − sin θ cos ϕ , sin ϕ ) Basis . The Moving frame is
{ r ( t ) ; R ( t ) , P ( t ) , Q ( t ) } \{\mathbf{r}(t);\mathbf{R}(t),\mathbf{P}(t),\mathbf{Q}(t)\} { r ( t ) ; R ( t ) , P ( t ) , Q ( t )} The position of a point is
r = r R \mathbf{r}=r\mathbf{R} r = r R The velocity is
v = r ˙ = ( r θ ˙ sin ϕ ) P + ( − r ϕ ˙ ) Q + r ˙ R \mathbf{v}=\dot{\mathbf{r}}=(r\dot{\theta}\sin\phi)\mathbf{P}+(-r\dot{\phi})\mathbf{Q}+\dot{r}\mathbf{R} v = r ˙ = ( r θ ˙ sin ϕ ) P + ( − r ϕ ˙ ) Q + r ˙ R and the acceleration
a = v ˙ = [ ( r θ ¨ + 2 r ˙ θ ˙ ) sin ϕ + 2 r θ ˙ ϕ ˙ cos ϕ ] P + [ r ( θ ˙ 2 sin ϕ cos ϕ − ϕ ¨ ) − 2 r ˙ ϕ ˙ ] Q + \mathbf{a}=\dot{\mathbf{v}}=[(r\ddot{\theta}+2\dot{r}\dot{\theta})\sin\phi+2r\dot{\theta}\dot{\phi}\cos\phi]\mathbf{P}+[r(\dot{\theta}^{2}\sin\phi\cos\phi-\ddot{\phi})-2\dot{r}\dot{\phi}]\mathbf{Q}+\ a = v ˙ = [( r θ ¨ + 2 r ˙ θ ˙ ) sin ϕ + 2 r θ ˙ ϕ ˙ cos ϕ ] P + [ r ( θ ˙ 2 sin ϕ cos ϕ − ϕ ¨ ) − 2 r ˙ ϕ ˙ ] Q + + [ r ¨ − r ( ϕ ˙ 2 + θ ˙ 2 sin 2 ϕ ) ] R +\ [\ddot{r}-r(\dot{\phi}^{2}+\dot{\theta}^{2}\sin^{2}\phi)
]\mathbf{R} + [ r ¨ − r ( ϕ ˙ 2 + θ ˙ 2 sin 2 ϕ )] R The antisymmetric derivative matrix is
( P ˙ Q ˙ R ˙ ) = ( 0 θ ˙ cos ϕ − θ ˙ sin ϕ − θ ˙ cos θ 0 ϕ ˙ θ ˙ sin ϕ − ϕ ˙ 0 ) ( P Q R ) \begin{pmatrix}\dot{\mathbf{P}} \\ \dot{\mathbf{Q}} \\ \dot{\mathbf{R}}\end{pmatrix}=\begin{pmatrix}0 & \dot{\theta}\cos\phi & -\dot{\theta}\sin\phi \\ -\dot{\theta}\cos\theta & 0 & \dot{\phi} \\ \dot{\theta}\sin\phi & -\dot{\phi} & 0\end{pmatrix}\begin{pmatrix}\mathbf{P} \\ \mathbf{Q} \\ \mathbf{R}\end{pmatrix} P ˙ Q ˙ R ˙ = 0 − θ ˙ cos θ θ ˙ sin ϕ θ ˙ cos ϕ 0 − ϕ ˙ − θ ˙ sin ϕ ϕ ˙ 0 P Q R 