- #1

H_man

- 145

- 0

If we take the K.E. of a particle in spherical polar coords.

[tex]

T = \frac{1}{2}m( \dot r^2 + r^2 \dot\theta^2 + r^2 sin^2 \theta \dot \phi^2)

[/tex]

And

[tex]

T' = \frac{1}{2}( \dot r^2 + r^2 \dot\theta^2 + r^2 sin^2 \theta \dot \phi^2)

[/tex]

Now plugging this in Lagrange's equation:

[tex]

\frac {d} {dt} \frac {\partial T'} {\partial \dot \theta} - \frac {\partial T'} {\partial \theta} = r^2 \ddot \theta + 2r \dot r \dot \theta - r^2 \dot \phi^2 sin \theta cos \theta

[/tex] --------- Line 3

Now, as far as I understand the above equation is [tex]a_\theta[/tex]. That is, the [tex]\theta[/tex] component of acceleration.

*However*, it seems I am wrong.

My book tells me I have to divide the expression by [tex]h_\theta[/tex]

where

[tex]

h_\theta = \left ( \left ( \frac {\partial x} {\partial \theta} \right )^2

+ \left ( \frac {\partial y} {\partial \theta} \right )^2

+ \left ( \frac {\partial z} {\partial \theta} \right )^2 \right )^\frac {1}{2} = ( (r cos \theta cos \phi )^2 + ( (r cos \theta sin \phi )^2 + r^2 sin^2 \theta )^\frac {1}{2} = r

[/tex]

Producing [tex]a_\theta = r \ddot \theta + 2 \dot r \dot \theta - r \dot \phi^2 sin \theta cos \theta [/tex]

So making the reasonable assumption that the book is correct and I am not. What does the expression (Line 3) that I thought was the acceleration represent?

This is especially confusing as I know that if we do not divide T by m then line 3 should produce the force [tex]F_\theta[/tex]??

:uhh: