Transformation from spherical coordinates to rectangular coordinates

A transformation from spherical coordinates to rectangular coordinates

Transformation of coordinates

The transformation from spherical coordinates $(r,\theta ,\phi )$ to rectangular coordinates (Cartesian coordinates) $(x,y,z)$ is:

$x=r\sin \theta \cos \phi$

$y=r\sin \theta \sin \phi$

$z=r\cos \theta$

Transformation of velocities

If we take the total derivatives of these equations, we obtain:

$dx={\frac {\partial x}{\partial r}}dr+{\frac {\partial x}{\partial \theta }}d\theta +{\frac {\partial x}{\partial \phi }}d\phi$

$dy={\frac {\partial y}{\partial r}}dr+{\frac {\partial y}{\partial \theta }}d\theta +{\frac {\partial y}{\partial \phi }}d\phi$

$dz={\frac {\partial z}{\partial r}}dr+{\frac {\partial z}{\partial \theta }}d\theta +{\frac {\partial z}{\partial \phi }}d\phi$

The partial derivatives are easily obtained:

${\frac {\partial x}{\partial r}}=\sin \theta \cos \phi ,{\frac {\partial x}{\partial \theta }}=r\cos \theta \cos \phi ,{\frac {\partial x}{\partial \phi }}=-r\sin \theta \sin \phi$

${\frac {\partial y}{\partial r}}=\sin \theta \sin \phi ,{\frac {\partial y}{\partial \theta }}=r\cos \theta \sin \phi ,{\frac {\partial y}{\partial \phi }}=r\sin \theta \cos \phi$

${\frac {\partial z}{\partial r}}=\cos \theta ,{\frac {\partial z}{\partial \theta }}=-r\sin \theta ,{\frac {\partial z}{\partial \phi }}=0$

The total derivatives are therefore:

Failed to parse (unknown function "\dx"): {\displaystyle \dx = \sin \theta \cos \phi dr + r \cos \theta \cos \phi d \theta - r \sin \theta \sin \phi d \phi}

Failed to parse (unknown function "\dy"): {\displaystyle \dy = \sin \theta \sin \phi dr + r \cos \theta \sin \phi d \theta + r \sin \theta \cos \phi d \phi}

$dz=\cos \theta dr-r\sin \theta d\theta$

The total derivatives are easily converted to derivatives wrt time:

${\dot {x}}=\sin \theta \cos \phi {\dot {r}}+\cos \theta \cos \phi (r{\dot {\theta }})-\sin \phi (r\sin \theta {\dot {\phi }})$

${\dot {y}}=\sin \theta \sin \phi {\dot {r}}+\cos \theta \sin \phi (r{\dot {\theta }})+\cos \phi (r\sin \theta {\dot {\phi }})\;\;(*)$

${\dot {z}}=\cos \theta {\dot {r}}-\sin \theta (r{\dot {\theta }})$

Now the velocity of a point particle in 3-space (3D space) may be expressed in either rectangular or spherical coordinates.

In rectangular coordinates, the infinitesimal displacement vector is:

$d{\vec {s}}=(dx{\hat {x}}+dy{\hat {y}}+dz{\hat {dz}})$ $=({\frac {dx}{dt}}{\hat {x}}+{\frac {dy}{dt}}{\hat {y}}+{\frac {dz}{dt}}{\hat {z}})dt$ $=({\dot {x}}{\hat {x}}+{\dot {y}}{\hat {y}}+{\dot {z}}{\hat {z}})dt$

But the infinitesimal displacement vector may also be expressed as:

$d{\vec {s}}={\vec {v}}dt$ $=(v_{x}{\hat {x}}+v_{y}{\hat {y}}+v_{z}{\hat {z}})dt$

When these two expressions are compared, it becomes obvious that:

$v_{x}={\dot {x}}$

$v_{y}={\dot {y}}$

$v_{z}={\dot {z}}$

In spherical coordinates, the infinitesimal displacement vector is:

$d{\vec {s}}=(dr{\hat {r}}+rd\theta {\hat {\theta }}+r\sin \theta d\phi {\hat {\phi }})$ $=(dr({\hat {r}})+d\theta (r{\hat {\theta }})+d\phi (r\sin \theta {\hat {\phi }}))$ $=({\frac {dr}{dt}}({\hat {r}})+{\frac {d\theta }{dt}}(r{\hat {\theta }})+{\frac {d\phi }{dt}}(r\sin \theta {\hat {\phi }}))dt$ $=({\dot {r}}({\hat {r}})+{\dot {\theta }}(r{\hat {\theta }})+{\dot {\phi }}(r\sin \theta {\hat {\phi }}))dt$

But the infinitesimal displacement vector may also be expressed as:

$d{\vec {s}}={\vec {v}}dt$ $=(v_{r}{\hat {r}}+v_{\theta }{\hat {\theta }}+v_{\phi }{\hat {\phi }})dt$

When these two expressions are compared, it becomes obvious that:

$v_{r}={\dot {r}}$

$v_{\theta }=r{\dot {\theta }}$

$v_{\phi }=r\sin {\theta }{\dot {\phi }}$

If we now take these expressions for the velocity components, in both rectangular and spherical coordinates, and plug them into the set of equations labelled (*), then we obtain the following velocity transformation:

$v_{x}=v_{r}\sin \theta \cos \phi +v_{\theta }\cos \theta \cos \phi +v_{\phi }(-\sin \phi )$

$v_{y}=v_{r}\sin \theta \sin \phi +v_{\theta }\cos \theta \sin \phi +v_{\phi }\cos \phi$

$v_{z}=v_{r}\cos \theta +v_{\theta }(-\sin \theta )$

Transformation of unit vectors

Let us consider the unit velocity:

${\vec {v}}={\hat {r}}$ $=(1,0,0)$ in the spherical system

In other words:

$v_{r}=1$

$v_{\theta }=0$

$v_{\phi }=0$

Using the velocity transformation, we obtain:

$v_{x}=\sin \theta \cos \phi$

$v_{y}=\sin \theta \sin \phi$

$v_{z}=\cos \theta$

Therefore the unit velocity in the rectangular system is:

${\vec {v}}=\sin \theta \cos \phi {\hat {x}}+\sin \theta \sin \phi {\hat {y}}+\cos \theta {\hat {z}}$

We therefore have a transformation for the unit vector ${\hat {r}}$

${\hat {r}}=\sin \theta \cos \phi {\hat {x}}+\sin \theta \sin \phi {\hat {y}}+\cos \theta {\hat {z}}$

Let us consider the unit velocity:

${\vec {v}}={\hat {\theta }}$ $=(0,1,0)$ in the spherical system

In other words:

$v_{r}=0$

$v_{\theta }=1$

$v_{\phi }=0$

Using the velocity transformation, we obtain:

$v_{x}=\cos \theta \cos \phi$

$v_{y}=\cos \theta \sin \phi$

$v_{z}=-\sin \theta$

Therefore the unit velocity in the rectangular system is:

${\vec {v}}=\cos \theta \cos \phi {\hat {x}}+\cos \theta \sin \phi {\hat {y}}-\sin \theta {\hat {z}}$

We therefore have a transformation for the unit vector ${\hat {\theta }}$

${\hat {\theta }}=\cos \theta \cos \phi {\hat {x}}+\cos \theta \sin \phi {\hat {y}}-\sin \theta {\hat {z}}$

Let us consider the unit velocity:

${\vec {v}}={\hat {\phi }}$ $=(0,0,1)$ in the spherical system

In other words:

$v_{r}=0$

$v_{\theta }=0$

$v_{\phi }=1$

Using the velocity transformation, we obtain:

$v_{x}=-\sin \phi$

$v_{y}=\cos \phi$

$v_{z}=0$

Therefore the unit velocity in the rectangular system is:

${\vec {v}}=-\sin \phi {\hat {x}}+\cos \phi {\hat {y}}+0{\hat {z}}$

We therefore have a transformation for the unit vector ${\hat {\phi }}$

${\hat {\phi }}=-\sin \phi {\hat {x}}+\cos \phi {\hat {y}}+0{\hat {z}}$

Summarizing, the transformation equations for the unit vectors are:

${\hat {r}}=\sin \theta \cos \phi {\hat {x}}+\sin \theta \sin \phi {\hat {y}}+\cos \theta {\hat {z}}$

${\hat {\theta }}=\cos \theta \cos \phi {\hat {x}}+\cos \theta \sin \phi {\hat {y}}-\sin \theta {\hat {z}}$

${\hat {\phi }}=-\sin \phi {\hat {x}}+\cos \phi {\hat {y}}+0{\hat {z}}$