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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 = rsinθcosφ

y = rsinθsinφ

z = rcosθ

[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:

dx = sinθcosφdr + rcosθcosφdθ − rsinθsinφdφ

dy = sinθsinφdr + rcosθsinφdθ + rsinθcosφdφ

dz = cosθdrrsinθdθ

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:

vx = vrsinθcosφ + vθcosθcosφ + vφ( − sinφ)

vy = vrsinθsinφ + vθcosθsinφ + vφcosφ

vz = vrcosθ + vθ( − sinθ)

[edit] Transformation of unit vectors

Let us consider the unit velocity:

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

In other words:

vr = 1

vθ = 0

vφ = 0

Using the velocity transformation, we obtain:

vx = sinθcosφ

vy = sinθsinφ

vz = cosθ

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:

vr = 0

vθ = 1

vφ = 0

Using the velocity transformation, we obtain:

vx = cosθcosφ

vy = cosθsinφ

vz = − sinθ

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:

vr = 0

vθ = 0

vφ = 1

Using the velocity transformation, we obtain:

vx = − sinφ

vy = cosφ

vz = 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}

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