Euler Lagrange equation of motion

[bobred]

Homework Statement

Find the equations of motion for both r and [itex]\theta[/itex] of

Homework Equations

My problem is taking the derivative wrt time of

and
[itex]\dfrac{\partial\mathcal{L}}{\partial\dot{r}}=m \dot{r} \left( 1 + \left( \dfrac{\partial H}{\partial r}\right)^2 \right)[/itex]

The Attempt at a Solution

So

I think this is correct. I have the solution to the second expression but I am unsure how it was found
[itex]
m \ddot{r} \left( 1 + \left( \dfrac{\partial H}{\partial r}\right)^2 \right) + 2 m \dot{r}^2 \dfrac{\partial H}{\partial r}\dfrac{\partial^2 H}{\partial r \partial t}
[/itex]
The first part of this expression is fine it is the second part I am unsure of.

 

[Orodruin]

Please specify what ##H## is a function of.

 

[bobred]

HI, sorry H(r).

 

[Orodruin]

So if H is only a function of r, the only possible derivative of H is the total derivative wrt r. Since r depends on t, the derivative wrt t is given by the chain rule. This also goes for dH/dr, which is also a function of r. The partial derivative of H wrt t does not make sense if H does not depend explicitly on t.

 

[Ray Vickson]

HI, sorry H(r).

So, if we write ##K(r) = (\partial H(r) /\partial r)^2## (which is just some computable function of ##r##) you have
[tex] {\cal{L}} = \frac{1}{2} m \left((1+ K(r)) \dot{r}^2 + r^2 \dot{\theta} \right) - mgH(r) [/tex]
Thus,
[tex] {\cal{L}}_r \equiv \partial {\cal{L}} /\partial r = \frac{1}{2}m \left( K'(r) \dot{r}^2 + 2 r \dot{\theta} \right) - m g H'(r) [/tex]
so
[tex] \frac{d}{dt} {\cal{L}}_r = \frac{1}{2} m \left( K''(r) \dot{r}^3 + K'(r) 2 \dot{r} \ddot{r} + 2 \dot{\theta} \right) - mg H''(r) \dot{r} [/tex]

If ##H## does not depend explicitly on ##t## there should be no ##\partial H/\partial t## anywhere.

 

[Orodruin]

So, if we write ##K(r) = (\partial H(r) /\partial r)^2## (which is just some computable function of ##r##) you have
[tex] {\cal{L}} = \frac{1}{2} m \left((1+ K(r)) \dot{r}^2 + r^2 \dot{\theta} \right) - mgH(r) [/tex]
Thus,
[tex] {\cal{L}}_r \equiv \partial {\cal{L}} /\partial r = \frac{1}{2}m \left( K'(r) \dot{r}^2 + 2 r \dot{\theta} \right) - m g H'(r) [/tex]
so
[tex] \frac{d}{dt} {\cal{L}}_r = \frac{1}{2} m \left( K''(r) \dot{r}^3 + K'(r) 2 \dot{r} \ddot{r} + 2 \dot{\theta} \right) - mg H''(r) \dot{r} [/tex]

This, although correct, is not relevant for the Euler-Lagrange equations which contain what you would call ##d\mathcal L_{\dot r}/dt##. Also, please allow the OP to try to do the derivatives for himself.

 

[Ray Vickson]

This, although correct, is not relevant for the Euler-Lagrange equations which contain what you would call ##d\mathcal L_{\dot r}/dt##. Also, please allow the OP to try to do the derivatives for himself.

You are right. Anyway, the OP did try to obtain the derivatives for himself, but made some errors. I admit maybe I should have not written so much detail (which does not matter anyway, since I did the wrong calculation!).