# Variation of Lagrangian w/r to canonical momenta

• A
• muscaria
In summary, there is a passage in the book "The Variational Principles of Mechanics" by Cornelius Lanczos that introduces the Legendre transformation and motivates a variational principle for Hamilton's equations of motion. The variation of the Lagrangian with respect to the generalized momenta seems to have an extra term, but it can be shown that this term vanishes due to the duality of the Legendre transformation. This leads to the conclusion that variations of the Lagrangian with respect to the generalized momenta only enter as a second order effect and can be put to 0 when looking for stationary values of the action. Additionally, the Lagrangian and Hamiltonian are defined on all possible potential paths, making the generalized coordinates and
muscaria
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!

#### Attachments

• Canonical Integral_Lanczos.pdf
145.4 KB · Views: 241
To compute the coefficient of the variation of p you use the partial derivative NOT the total derivative. The qdot's have no explicit dependence on p, so they're just constants. True, in the physical space there is a dependence - which would have to be taken into account if, for some reason, you were taking a total derivative here (which actually wouldn't make much sense). But for the partial, qdots are constants multiplying p so you get simply qdot for that first term in 64.2, coming from the first term in 64.1.

As for what you tried to do in your "attempt at understanding what is going on" - it may work out somehow; I didn't examine it. Certainly not the right way to go, though you get points for trying!

Last edited:
When I first learned about Lagrangian mechanics and that q and qdot were to be treated as formally independent variables, I was similarly confused. How can q and qdot be independent if qdot is fully determined by q? It finally made sense to me when I realized that the Lagrangian (and Hamiltonian) are defined not just along the path of the motion, but on all possible potential paths. In the case of Hamiltonian, H is a function of p and q, and p and q are just coordinates, not the actual position and momentum. You can ask, what is the energy at this instant if the position is q and the momentum is p, regardless of if the system will actually reach that point in phase space. Therefore, p and q are independent. The advantage of this formulism is that you can use it for quantum mechanics, where there is no precise trajectory.

Hi guys, thank you for your responses and apologies for the rather late reply!
Khashishi said:
When I first learned about Lagrangian mechanics and that q and qdot were to be treated as formally independent variables, I was similarly confused. How can q and qdot be independent if qdot is fully determined by q?
This was not what I was confused about, ##\dot{q}## is only determined by ##q(t)##, i.e once you have solved the equations of motion.. they're independent dynamical variables which need to be specified independently of each other in order for a given solution trajectory to be taken - I get that. Without meaning to be presumptuous and I appreciate the response, from your reply I think you may have misunderstood/overlooked what I was asking. My question was basically, if the "new" dynamical space is the set of independent dynamical variables ##(q,p)##, then any function ##f## of the motion (be it actual or tentative) is a function of these dynamical variables -> ##f(q,p)##. The generalised velocities are no different in this respect, i.e. for each point in phase space there is a definite set of associated ##\dot{q}_i=\dot{q}_i(q_1,\cdots,q_n,p_1,\cdots,p_n)##; so my line of thinking was that OK well, a variation of ##p_i## will induce a variation of ##\dot{q}_i##. And so in considering ##\delta L## resulting from ##\delta p_i##, why does ##\delta\dot{q}_i## not play any part?
secur said:
To compute the coefficient of the variation of p you use the partial derivative NOT the total derivative.
I don't see how I was computing a total derivative..? I computed a partial just in the same way as you would for the following situation say: Given some ##f(x,y)=xg(x,y)## (meant to represent ##p_i\dot{q}_i##), the variation ##\delta f## which results from varying ##x## is simply $$\delta f =\frac{\partial\left(xg(x,y)\right)}{\partial x}\delta x$$
secur said:
But for the partial, qdots are constants multiplying p so you get simply qdot for that first term in 64.2, coming from the first term in 64.1.
But not because
secur said:
The qdot's have no explicit dependence on p, so they're just constants
I think I've just finally got it actually, or just about.. the point seems to be that the ##\dot{q}_i## are slopes of a given function in ##(q,p)## space, namely ##\frac{\partial H(q,p)}{\partial p_i}##, so although ##\dot{q}_i=\dot{q}_i(q_1,\cdots,q_n,p_1,\cdots,p_n)## is true, there's a deeper structure which follows from the Legendre transform... It's not simply a point transformation we're carrying out on our dynamical space in going from Lagrangian to Hamiltonian, we're substituting a set of variables by the slope of a given function in this dynamical space. But given that the "new" set of variables correspond to (1st order) slopes in the "old" set of variables,a variation of these will only manifest as a second order effect in the reciprocal space, through the duality of the Legendre Transform and the E-L equations of motion. And the reciprocity is due to the structure of how the E-L e.o.m : ##\dot{p}_i =\frac{\partial L}{\partial q_i}## and the Legendre transform combine. Does this line of reasoning sound reasonable? I guess that's essentially what the calculation that came out in the original post is saying also, because I don't see how it is wrong..

## 1. What is the Lagrangian in classical mechanics?

The Lagrangian is a function that describes the dynamics of a system in classical mechanics. It is defined as the difference between the kinetic energy and potential energy of the system.

## 2. What are canonical momenta in classical mechanics?

Canonical momenta are a set of variables that describe the motion of a system in classical mechanics. They are defined as the derivatives of the Lagrangian with respect to the velocities of the particles in the system.

## 3. How is the Lagrangian related to the canonical momenta?

The Lagrangian is related to the canonical momenta through the Euler-Lagrange equations, which state that the time derivative of the canonical momenta is equal to the negative partial derivative of the Lagrangian with respect to the corresponding coordinate.

## 4. What is the variation of Lagrangian with respect to canonical momenta?

The variation of Lagrangian with respect to canonical momenta is a technique used in classical mechanics to find the equations of motion for a system. It involves taking the derivative of the Lagrangian with respect to the canonical momenta and setting it equal to zero.

## 5. Why is the variation of Lagrangian with respect to canonical momenta important?

The variation of Lagrangian with respect to canonical momenta allows us to derive the equations of motion for a system in a more efficient way than using Newton's laws of motion. It also allows us to easily incorporate constraints into the system, making it a more versatile tool for studying classical mechanics problems.

Replies
19
Views
1K
• Mechanics
Replies
7
Views
1K
• Electromagnetism
Replies
2
Views
919
Replies
9
Views
1K
• Classical Physics
Replies
5
Views
1K
• Other Physics Topics
Replies
1
Views
1K
Replies
7
Views
5K
• Mechanics
Replies
3
Views
622
• Classical Physics
Replies
2
Views
3K
• Classical Physics
Replies
19
Views
2K