A particle of mass [tex]m[/tex] moves in a "central potential," [tex]V(r),[/tex] where [tex]r[/tex] denotes teh radial displacement of the particle from a fixed origin.(adsbygoogle = window.adsbygoogle || []).push({});

a) What is the (vector) force on the particle? Use spherical coordinates.

We have

[tex]F = -\nabla V = -\frac{\partial V}{\partial x} \hat{i} - \frac{\partial V}{\partial y} \hat{j} - \frac{\partial V}{\partial z} \hat{k}[/tex]

Now, note that

[tex]\frac{\partial V}{\partial x} = \frac{\partial V}{\partial r} \frac{\partial r} \partial x},[/tex]

since [tex]V is only dependent on [tex]r[/tex] (and not [tex]\theta[/tex] or [tex]\phi[/tex]). Since [tex]x = r \sin \theta \cos \phi[/tex], we have that

[tex]\frac{\partial r}{\partial x} = \frac{1}{\sin \theta \cos \phi},[/tex]

so finally,

[tex]F = -\frac{dV}{dr} \cdot \frac{1}{\sin \theta \cos \phi}[/tex]

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# Homework Help: Spherical Coordinates to Rectangular Coordinates

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