- #1

T-7

- 64

- 0

Hi,

I'm stuck on what I suspect is a really easy proof. I'd appreciate some rapid help!

For a trajectory with given initial position and velocity, we can always rotate the system of coordinates such that initially [tex]d\theta / dt = 0 [/tex] and [tex]\theta= \pi/2[/tex] say, t = 0.

Using the Euler-Lagrange equations, show that [tex]\theta[/tex] remains at [tex]\pi/2[/tex] for all times t.

Having obtained the Lagrangian (L=mv^2/2 + U, rewriting v as [tex]\dot{r}^{2}+r^{2}(\dot{\phi}^{2}sin^{2}\theta + \dot{\theta}^{2})[/tex] in spherical coordinates, and taken partial derivatives with respect to [tex] \dot{\theta} [/tex] and [tex] \theta [/tex], and [tex]\phi[/tex] and [tex]\dot{\phi}[/tex], I have found the following relations (I think they're correct?)

[tex]d/dt [mr^2\dot{\theta}] = mr^2\dot{\phi^2}cos(\theta)sin(\theta)[/tex]

[tex]d/dt [mr^{2} \dot{\phi} sin^{2} (\theta)] = 0[/tex]

(There is a third expression, from the partial derivatives wrt. [tex]r[/tex] and [tex]\dot{r}[/tex], but I doubt it's relevant here.)

I'd be tempted to integrate both sides of the first to get an expression for [tex]\dot{\theta}[/tex], but that's where I become unstuck just now... I'm not sure how to integrate the RHS with respect to time, given what I know about the components. But presumably integration is going to be involved here, or the initial conditions would never come into play (?). I can deduce from the second that [tex]mr^{2} \dot{\phi} sin^{2} (\theta) = const = c[/tex], and so at t = 0 [tex]mr^{2} \dot{\phi} = c[/tex], but nothing more of interest seems to be following at the moment...

Incidentally, I take the geometrical meaning to be that the motion is confined to a plane orthogonal to L, ang. momentum is conserved.

Advice gratefully received (as soon as possible!) Many thanks :-)

I'm stuck on what I suspect is a really easy proof. I'd appreciate some rapid help!

## Homework Statement

For a trajectory with given initial position and velocity, we can always rotate the system of coordinates such that initially [tex]d\theta / dt = 0 [/tex] and [tex]\theta= \pi/2[/tex] say, t = 0.

Using the Euler-Lagrange equations, show that [tex]\theta[/tex] remains at [tex]\pi/2[/tex] for all times t.

## The Attempt at a Solution

Having obtained the Lagrangian (L=mv^2/2 + U, rewriting v as [tex]\dot{r}^{2}+r^{2}(\dot{\phi}^{2}sin^{2}\theta + \dot{\theta}^{2})[/tex] in spherical coordinates, and taken partial derivatives with respect to [tex] \dot{\theta} [/tex] and [tex] \theta [/tex], and [tex]\phi[/tex] and [tex]\dot{\phi}[/tex], I have found the following relations (I think they're correct?)

[tex]d/dt [mr^2\dot{\theta}] = mr^2\dot{\phi^2}cos(\theta)sin(\theta)[/tex]

[tex]d/dt [mr^{2} \dot{\phi} sin^{2} (\theta)] = 0[/tex]

(There is a third expression, from the partial derivatives wrt. [tex]r[/tex] and [tex]\dot{r}[/tex], but I doubt it's relevant here.)

I'd be tempted to integrate both sides of the first to get an expression for [tex]\dot{\theta}[/tex], but that's where I become unstuck just now... I'm not sure how to integrate the RHS with respect to time, given what I know about the components. But presumably integration is going to be involved here, or the initial conditions would never come into play (?). I can deduce from the second that [tex]mr^{2} \dot{\phi} sin^{2} (\theta) = const = c[/tex], and so at t = 0 [tex]mr^{2} \dot{\phi} = c[/tex], but nothing more of interest seems to be following at the moment...

Incidentally, I take the geometrical meaning to be that the motion is confined to a plane orthogonal to L, ang. momentum is conserved.

Advice gratefully received (as soon as possible!) Many thanks :-)

Last edited: