- #1

muscaria

- 125

- 29

Hi,

I've been working through Cornelius Lanczos book "The Variational Principles of Mechanics" and there's something I'm having difficulty understanding on page 168 of the Dover edition (which is attached). After introducing the Legendre transformation and transforming the Lagrangian equations of motion into their canonical form, he is now trying to motivate a ("extended") variational principle which gives rise directly to Hamilton's equations of motion, in which the action integral is varied w/r to the 2n positional phase space variables ##(q,p)##. Basically, I don't understand the passage from equation (64.1) to (64.2), where he considers the variation of $$L=\sum_{i=1}^n p_i\dot{q}_i - H,$$ with respect to the ##p_i##, which he writes as $$\delta L = \sum_{i=1}^n\left(\dot{q}_i - \frac{\partial H}{\partial p_i}\right)\delta p_i.$$ My problem is aren't the ##\dot{q}_i## a function of ##(q,p)##, so that there should be an extra term?

My attempt at understanding what is going on:

Because of the duality of the Legendre transform, we can write $$L= \sum_{i=1}^n p_i\frac{\partial H}{\partial p_i} - H,$$ so that the variation of ##L## w/r to the ##p_i## in this form, is what we had above but with an extra piece: $$\delta L = \sum_{i=1}^n\left(\dot{q}_i - \frac{\partial H}{\partial p_i}\right)\delta p_i + \sum_{i,j}p_i\frac{\partial^2H}{\partial p_j\partial p_i}\delta p_j.$$ The first term vanishes on account of the duality of the Legendre transformation but what's the meaning of the 2nd term, if there is any? If we follow the assumption that there is no coupling between different canonical momenta (which seems reasonable otherwise conservation of momentum would be violated, right?), then we will only have the diagonal terms of the double sum:##\sum_{i,j} p_i\frac{\partial^2H}{\partial p_j\partial p_i}\delta p_j = \sum_i p_i \frac{\partial^2 H}{\partial p_i^2}\delta p_i##, i.e $$\delta L = \sum_{i=1}^n\left(\dot{q}_i - \frac{\partial H}{\partial p_i}\right)\delta p_i + \frac{1}{2}\sum_i \frac{\partial^2 H}{\partial p_i^2}\delta (p_i)^2.$$ Is this suggesting in some way that a variation of the Lagrangian w/r to the ##p_i## only enters as a second order effect and can be put to 0 given that we are looking for stationary values of the action? Or is there something I'm missing or misinterpreting which will make all of this obvious?

As a side note, you can get to the result if you consider the variation of the action integral w/r to the ##p_i## (as opposed to variation of the bare Lagrangian) then integration by parts shows that there is no variation of the action from arbitrary variations of the ##p_i##, so all fine, you can still form the ##2n## differential equations in the way he does on page 169, but still, I'd like to understand how he can just go between those 2 equations.. To me it seems like what he is getting at only works if the variation is integrated out w/r to time, we need that integration by parts.

Thanks for your help and I hope this message reaches you in good spirits!

I've been working through Cornelius Lanczos book "The Variational Principles of Mechanics" and there's something I'm having difficulty understanding on page 168 of the Dover edition (which is attached). After introducing the Legendre transformation and transforming the Lagrangian equations of motion into their canonical form, he is now trying to motivate a ("extended") variational principle which gives rise directly to Hamilton's equations of motion, in which the action integral is varied w/r to the 2n positional phase space variables ##(q,p)##. Basically, I don't understand the passage from equation (64.1) to (64.2), where he considers the variation of $$L=\sum_{i=1}^n p_i\dot{q}_i - H,$$ with respect to the ##p_i##, which he writes as $$\delta L = \sum_{i=1}^n\left(\dot{q}_i - \frac{\partial H}{\partial p_i}\right)\delta p_i.$$ My problem is aren't the ##\dot{q}_i## a function of ##(q,p)##, so that there should be an extra term?

My attempt at understanding what is going on:

Because of the duality of the Legendre transform, we can write $$L= \sum_{i=1}^n p_i\frac{\partial H}{\partial p_i} - H,$$ so that the variation of ##L## w/r to the ##p_i## in this form, is what we had above but with an extra piece: $$\delta L = \sum_{i=1}^n\left(\dot{q}_i - \frac{\partial H}{\partial p_i}\right)\delta p_i + \sum_{i,j}p_i\frac{\partial^2H}{\partial p_j\partial p_i}\delta p_j.$$ The first term vanishes on account of the duality of the Legendre transformation but what's the meaning of the 2nd term, if there is any? If we follow the assumption that there is no coupling between different canonical momenta (which seems reasonable otherwise conservation of momentum would be violated, right?), then we will only have the diagonal terms of the double sum:##\sum_{i,j} p_i\frac{\partial^2H}{\partial p_j\partial p_i}\delta p_j = \sum_i p_i \frac{\partial^2 H}{\partial p_i^2}\delta p_i##, i.e $$\delta L = \sum_{i=1}^n\left(\dot{q}_i - \frac{\partial H}{\partial p_i}\right)\delta p_i + \frac{1}{2}\sum_i \frac{\partial^2 H}{\partial p_i^2}\delta (p_i)^2.$$ Is this suggesting in some way that a variation of the Lagrangian w/r to the ##p_i## only enters as a second order effect and can be put to 0 given that we are looking for stationary values of the action? Or is there something I'm missing or misinterpreting which will make all of this obvious?

As a side note, you can get to the result if you consider the variation of the action integral w/r to the ##p_i## (as opposed to variation of the bare Lagrangian) then integration by parts shows that there is no variation of the action from arbitrary variations of the ##p_i##, so all fine, you can still form the ##2n## differential equations in the way he does on page 169, but still, I'd like to understand how he can just go between those 2 equations.. To me it seems like what he is getting at only works if the variation is integrated out w/r to time, we need that integration by parts.

Thanks for your help and I hope this message reaches you in good spirits!