Kinetic Energy in Spherical Coordinates

Click For Summary

Homework Help Overview

The discussion revolves around deriving the expression for the kinetic energy of a classical particle in spherical coordinates, focusing on the appropriate mathematical formulation and the underlying physical principles.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive the kinetic energy expression using Cartesian coordinates and then transitioning to spherical coordinates. Participants question the necessity of taking time derivatives and the correct application of these derivatives in the context of kinetic energy.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on the need for time derivatives and the correct formulation of the kinetic energy expression. There is an ongoing exploration of the relationships between the variables in spherical coordinates, but no consensus has been reached on the final expression.

Contextual Notes

There are indications of confusion regarding the notation and the physical dimensions of the quantities involved, which may affect the clarity of the discussion. The original poster's approach has been critiqued for its lack of dimensional consistency.

mia705
Messages
3
Reaction score
0

Homework Statement


Derive the expression for kinetic energy of a classical particle in spherical coordinates.


Homework Equations


I believe the answer I am supposed to reach is:
T=\frac{1}{2} m (\dot{r}^2 + r^2\dot{\theta^2} + r^2\dot{\phi ^2}sin^2\theta)

The Attempt at a Solution


T=\frac{1}{2}mv^2
T=\frac{1}{2}m(\dot{x^2} + \dot{y^2} + \dot{z^2})
Knowing that:
x=rsin\theta cos\phi
y=rsin\theta sin\phi
z=rcos\theta

After plugging these in and working it out, I came up with:
T= \frac{1}{2}mr^2

My question now is, how do I get from the answer that I currently have to the solution I am trying to get to (that I listed above in part 2)?
 
Physics news on Phys.org
Did you remember to take the time derivatives?
 
I'm not quite sure what you mean by that?
 
mia705 said:
I'm not quite sure what you mean by that?
##\dot x = dx/dt##. Did you find an expression for ##\dot x##? Note that your answer does not have the correct units [dimensions] for energy.
 
I think I might understand now.

Since:
\dot{x}=\frac{dx}{dt} \ \ \ \dot{y}=\frac{dy}{dt} \ \ \ \dot{z}=\frac{dz}{dt}

Then:
\dot{x}=\frac{d(rsin \theta cos \phi)}{dt} \ \ \ \dot{y}=\frac{d(r sin\theta sin\phi)}{dt} \ \ \ \dot{z}=\frac{d(rcos \theta)}{dt}.

And since dr=dr \hat{r} + r d\theta \hat{\theta} + rsin \theta d\phi \hat{\phi}, this would then make \frac{dr}{dt} = \dot{r} + r \dot{\theta} + rsin\theta \dot{\phi}

Now, as I did before with squaring x, y, and z to get 1 (in my original post), the answer then becomes the solution of (\frac{dr}{dt})^2.

Which, when put back into my original answer is: T = \frac{1}{2} m (\dot{r}^2 + r^2 \dot{\theta}^2 + r^2 sin\theta^2 \dot{\phi}^2).

Is that more along the correct lines of thinking?
 
mia705 said:
I think I might understand now.

Since:
\dot{x}=\frac{dx}{dt} \ \ \ \dot{y}=\frac{dy}{dt} \ \ \ \dot{z}=\frac{dz}{dt}

Then:
\dot{x}=\frac{d(rsin \theta cos \phi)}{dt} \ \ \ \dot{y}=\frac{d(r sin\theta sin\phi)}{dt} \ \ \ \dot{z}=\frac{d(rcos \theta)}{dt}.
Did you work these out? You have ##x=x(r,\theta,\phi)##, so
$$\dot{x} = \frac{\partial x}{\partial r} \dot{r} + \frac{\partial x}{\partial \theta} \dot{\theta} + \frac{\partial x}{\partial \phi} \dot{\phi}.$$ Just calculate the partial derivatives and grind it out.

And since dr=dr \hat{r} + r d\theta \hat{\theta} + rsin \theta d\phi \hat{\phi}, this would then make \frac{dr}{dt} = \dot{r} + r \dot{\theta} + rsin\theta \dot{\phi}
You're being too sloppy with your notation, so what you wrote doesn't make sense. You have dr on both the lefthand side and righthand side of the first equation, but they don't represent the same quantity. You should have written
$$d\vec{r} = dr\,\hat{r} + r\,d\theta\,\hat{\theta} + r\sin\theta\,d\phi\,\hat{\phi}.$$ In the next equation, the unit vectors just mysteriously disappeared.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
4K
Replies
5
Views
2K
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
13
Views
3K