Spherical Coordinates to Rectangular Coordinates

In summary, the conversation discusses the force on a particle of mass m moving in a central potential, V(r), where r represents the radial displacement from a fixed origin. The force can be calculated using spherical coordinates and is given by F = -\frac{dV}{dr} \cdot \frac{1}{\sin \theta \cos \phi}. The conversation also mentions the use of gradient and directional derivative to compute the force vector. It is concluded that the force only has a component in the radial direction since V is only a function of r.
  • #1
Domnu
178
0
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.

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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  • #2
Domnu said:
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.

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

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

Hi Domnu! :smile:

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

Try again! (hint: what "is" gradient?) :smile:
 
  • #3
Domnu said:
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]
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.
 
  • #4
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?
 
  • #5
Hmm... well I see that

[tex]F = - \nabla V = -\frac{\partial V}{\partial r} \hat{r}[/tex]

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.
 
  • #6
Domnu said:
Hmm... well I see that

[tex]F = - \nabla V = -\frac{\partial V}{\partial r} \hat{r}[/tex]

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! :smile:

Once you know that the gradient is in a particular direction, it's just the directional derivative in that direction. :wink:
 
  • #7
Yay :) I think I finally get the conversion between spherical/cylindrical coordinates. Awesome =)
 
  • #8
That is the correct answer. You probably need to show that this is the right answer.
 

1. What are spherical coordinates?

Spherical coordinates are a coordinate system used to locate points in three-dimensional space. They use three parameters: radius (r), inclination angle (θ), and azimuth angle (φ) to describe the position of a point in relation to a fixed origin.

2. How do you convert from spherical coordinates to rectangular coordinates?

To convert from spherical coordinates to rectangular coordinates, we use the following equations:

x = r*sin(θ)*cos(φ)

y = r*sin(θ)*sin(φ)

z = r*cos(θ)

3. How are spherical coordinates helpful in solving problems in physics and engineering?

Spherical coordinates are particularly useful in solving problems involving spherical objects or systems, such as celestial bodies, antennas, and satellite orbits. They also simplify calculations involving symmetry, such as spherical symmetry in electric and magnetic fields.

4. Can spherical coordinates be used in two-dimensional space?

No, spherical coordinates require three dimensions to fully describe a point. In two-dimensional space, polar coordinates (r, θ) are used instead.

5. What are some examples of real-world applications of spherical coordinates?

Spherical coordinates have many practical applications, including navigation and mapping systems, astronomy and astrophysics, and computer graphics for creating 3D objects. They are also used in geology to describe the location of mineral deposits and in fluid mechanics to solve problems involving fluid flow in spherical tanks or vessels.

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