Spherical Coordinates to Rectangular Coordinates

[Domnu]

A particle of mass m moves in a "central potential," V(r), where r denotes teh radial displacement of the particle from a fixed origin.

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

We have

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

Now, note that

\frac{\partial V}{\partial x} = \frac{\partial V}{\partial r} \frac{\partial r} \partial x},

since V is only dependent on r (and not \theta or \phi). Since x = r \sin \theta \cos \phi, we have that<br /> <br /> \frac{\partial r}{\partial x} = \frac{1}{\sin \theta \cos \phi},<br /> <br /> so finally,<br /> <br /> F = -\frac{dV}{dr} \cdot \frac{1}{\sin \theta \cos \phi}

 

[tiny-tim]

A particle of mass m moves in a "central potential," V(r), where r denotes teh radial displacement of the particle from a fixed origin.

a) What is the (vector) force on the particle? Use spherical coordinates.
…
F = -\frac{dV}{dr} \cdot \frac{1}{\sin \theta \cos \phi}

Hi Domnu!

Nooo … the force has to depend only on r, doesn't it?

Try again! (hint: what "is" gradient?)

 

[D H]

Use spherical coordinates.

We have

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

You were asked to use spherical coordinates. So why did you use cartesian coordinates?

One reasonable answer is that you have not been taught how to compute the gradient in spherical coordinates. If that is the case, you have only computed one of the three cartesian elements of the force vector. The force will have y and z components as well as an x component.

 

[Domnu]

Hmm... I haven't really had a whole lot of experience with gradients, etc. Let's say we have a function F, and we take the gradient of it. Would the resulting gradient just be the vector (after substituting x,y,z points) which lies on the plane which is tangent to F?

 

[Domnu]

Hmm... well I see that

F = - \nabla V = -\frac{\partial V}{\partial r} \hat{r}

only... the dv/dtheta and dv/dphi terms are both zero, causing the theta-hat and phi-hat terms to be zero. So, would this just be the answer? It is only in terms of r, since V is only a function of r.

 

[tiny-tim]

Hmm... well I see that

F = - \nabla V = -\frac{\partial V}{\partial r} \hat{r}

only... the dv/dtheta and dv/dphi terms are both zero, causing the theta-hat and phi-hat terms to be zero. So, would this just be the answer? It is only in terms of r, since V is only a function of r.

That's it!

Once you know that the gradient is in a particular direction, it's just the directional derivative in that direction.

 

[Domnu]

Yay :) I think I finally get the conversion between spherical/cylindrical coordinates. Awesome =)

 

[D H]

That is the correct answer. You probably need to show that this is the right answer.