Transformation from spherical coordinates to rectangular coordinates

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[edit] Transformation of coordinates

The transformation from spherical coordinates $(r,θ,φ)$ to rectangular coordinates (Cartesian coordinates) $(x, y, z)$ is:

$x = r sinθcosφ$

$y = r sinθsinφ$

$z = r cosθ$

[edit] 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:

$d x = sinθcosφ d r + r cosθcosφ d θ - r sinθsinφ d φ$

$d y = sinθsinφ d r + r cosθsinφ d θ + r sinθcosφ d φ$

$d z = cosθ d r - r sinθ d θ$

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:

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

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:

$v x = v r sinθcosφ + v θ cosθcosφ + v φ ( - sinφ)$

$v y = v r sinθsinφ + v θ cosθsinφ + v φ cosφ$

$v z = v r cosθ + v θ ( - sinθ)$