- #1

- 32

- 0

Thank you for the help.

- I
- Thread starter betelgeuse91
- Start date

- #1

- 32

- 0

Thank you for the help.

- #2

- 32

- 0

Oh the Lagrangian does not necessarily be T-U

- #3

wrobel

Science Advisor

- 742

- 454

Hamilton's principle is derived from Newton's laws + hypothesis on constrains they must be ideal and holonomic

Last edited:

- #4

- 17,059

- 8,164

- #5

wrobel

Science Advisor

- 742

- 454

Let's clarify it by a simplest example.

Assume we have a Lagrangian

$$L=\frac{1}{2}g_{ij}\dot x^i\dot x^j,\quad i,j=1,\ldots, m.$$ The coefficients ##g_{ij}## form a Riemann metric; for simplicity we assume that all ##g_{ij}## are constants.

Consider** a variational problem **of finding extremums of a functional

$$ x(\cdot)\mapsto \int_{t_1}^{t_2}Ldt$$ in the class of smooth functions ##x(t)=(x^1,\ldots,x^m)(t)## such that ##x(t_i),\quad i=1,2## are fixed and $$a_i(x)\dot x^i=0,\quad t\in[t_1,t_2].\qquad (*)$$

From Calculus of Variations [see for example Bruce van Brunt: The Calculus of Variations. Springer 2004] it is known that the extremum to this problem is a solution to the Lagrange equations with the following Lagrangian ##\mathcal L(x,\dot x)=L-\lambda a_k\dot x^k##. Simultaneously the multiplier ##\lambda=\lambda(t)## is chosen to fulfill equations (*).

The system with Lagrangian ##\mathcal L## is written as follows

$$g_{si}\ddot x^i-\dot\lambda a_s-\lambda\dot x^p\Big(\frac{\partial a_s}{\partial x^p}-\frac{\partial a_p}{\partial x^s}\Big)=0. \qquad (**)$$

To express ##\dot\lambda## let us differentiate formula (*) in ##t##:

$$\frac{\partial a_k}{\partial x^n}\dot x^k\dot x^n+a_k\ddot x^k=0.$$

Substituting here ##\ddot x## from (**) we finally yield

$$\dot\lambda=-\frac{1}{g^{ij}a_ia_j}\Big(\frac{\partial a_i}{\partial x^n}\dot x^n\dot x^i+\lambda a_i\dot x^sg^{ik}\Big(\frac{\partial a_k}{\partial x^s}-\frac{\partial a_s}{\partial x^k}\Big)\Big) .\qquad (***)$$

**Generally equations (**)-(***) describe the vaconomic mechanics of system with Lagrangian ##L##. To obtain a solution to this system we must impose the initial conditions: ##x(0),\dot x(0),\lambda(0).## For the same ##x(0),\dot x(0)## we can take different ##\lambda(0)## and obtain different solutions ##x(t)##. This means that from the "classical" standpoint when we trace only the function ##x(t)##, system (**)-(***) does not possesses the uniqueness property. Thus in general it can not be a system of classical mechanics.**

We point just a simplest case when system (**)-(***) turns to be a system of classical mechanics. Assume that $$\frac{\partial a_s}{\partial x^p}-\frac{\partial a_p}{\partial x^s}=0.\qquad (!)$$ Then substituting ##\dot\lambda## from (***) to (**) we get

$$g_{si}\ddot x^i+\frac{a_s}{g^{ij}a_ia_j}\frac{\partial a_i}{\partial x^n}\dot x^n\dot x^i=0.$$

It is not hard to show that these equations describe the classical Lagrangian system with Lagrangian ##L## and ideal constraints (*).

**Introduce a differential form ##\omega=a_idx^i##. Then the condition (!) implies that locally there exists a function ##f=f(x)## such that ##df=\omega## i.e. the constraint (*) is integrable and our system is holonomic. In this case the described above variational problem tuns to be the Hamilton principle. Generally, necessary and sufficient condition of integrability of constraint (*) is as follows ##\omega\wedge d\omega=0##. In case of nonholonomic constraint the formulated above variational problem does not give equations of classical mechanics.**

Thus it important not to mix up Hamilton's variational principle and the Lagrange-d'Alembert principle

Assume we have a Lagrangian

$$L=\frac{1}{2}g_{ij}\dot x^i\dot x^j,\quad i,j=1,\ldots, m.$$ The coefficients ##g_{ij}## form a Riemann metric; for simplicity we assume that all ##g_{ij}## are constants.

Consider

$$ x(\cdot)\mapsto \int_{t_1}^{t_2}Ldt$$ in the class of smooth functions ##x(t)=(x^1,\ldots,x^m)(t)## such that ##x(t_i),\quad i=1,2## are fixed and $$a_i(x)\dot x^i=0,\quad t\in[t_1,t_2].\qquad (*)$$

From Calculus of Variations [see for example Bruce van Brunt: The Calculus of Variations. Springer 2004] it is known that the extremum to this problem is a solution to the Lagrange equations with the following Lagrangian ##\mathcal L(x,\dot x)=L-\lambda a_k\dot x^k##. Simultaneously the multiplier ##\lambda=\lambda(t)## is chosen to fulfill equations (*).

The system with Lagrangian ##\mathcal L## is written as follows

$$g_{si}\ddot x^i-\dot\lambda a_s-\lambda\dot x^p\Big(\frac{\partial a_s}{\partial x^p}-\frac{\partial a_p}{\partial x^s}\Big)=0. \qquad (**)$$

To express ##\dot\lambda## let us differentiate formula (*) in ##t##:

$$\frac{\partial a_k}{\partial x^n}\dot x^k\dot x^n+a_k\ddot x^k=0.$$

Substituting here ##\ddot x## from (**) we finally yield

$$\dot\lambda=-\frac{1}{g^{ij}a_ia_j}\Big(\frac{\partial a_i}{\partial x^n}\dot x^n\dot x^i+\lambda a_i\dot x^sg^{ik}\Big(\frac{\partial a_k}{\partial x^s}-\frac{\partial a_s}{\partial x^k}\Big)\Big) .\qquad (***)$$

We point just a simplest case when system (**)-(***) turns to be a system of classical mechanics. Assume that $$\frac{\partial a_s}{\partial x^p}-\frac{\partial a_p}{\partial x^s}=0.\qquad (!)$$ Then substituting ##\dot\lambda## from (***) to (**) we get

$$g_{si}\ddot x^i+\frac{a_s}{g^{ij}a_ia_j}\frac{\partial a_i}{\partial x^n}\dot x^n\dot x^i=0.$$

It is not hard to show that these equations describe the classical Lagrangian system with Lagrangian ##L## and ideal constraints (*).

Thus it important not to mix up Hamilton's variational principle and the Lagrange-d'Alembert principle

Last edited:

- #6

- 17,059

- 8,164

$$A[q]=\int_{t_1}^{t_2} \mathrm{d} t L(q,\dot{q},t)$$

for all variations around the stationary point ##q(t)## which are subject to the constraints

$$a_{ik}(q,t) \delta q^{k}=0.$$

To implement them you take the variation of ##A## and add the usual Lagrange-multiplier terms:

$$\delta A[q] = \int_{t_1}^{t_2} \mathrm{d} t \delta q^{k} \left [\frac{\partial L}{\partial q^k}-\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}^k} - \lambda^i a_{ik} \right] \stackrel{!}{=}0,$$

which leads to the equations of motion

$$\frac{\partial L}{\partial q^k}-\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}^k}=\lambda^i a_{ik}.$$

Now, for the "true trajectory" you can write the non-holonomic contraints as

$$a_{ik} \dot{q}^k=0.$$

Note that the vaskonomic dynamics and the correct d'Alembert dynamics coincide if and only if the an-holonomic constraints are integrable, i.e., can be implemented as holonomic constraints.

- #7

wrobel

Science Advisor

- 742

- 454

ok, then cite please a book which contains such a statement:We should not repeat this superfluous discussion again

and this statement is called the Hamilton variational principle (for nonholonomic constraints, you know).This means you have to minimize

A[q]=∫t2t1dtL(q,˙q,t)A[q]=\int_{t_1}^{t_2} \mathrm{d} t L(q,\dot{q},t)

for all variations around the stationary point q(t)q(t) which are subject to the constraints

aik(q,t)δqk=0.

It would be great if you cut the corresponding text from that book and post the scan here

- #8

- 17,059

- 8,164

- #9

wrobel

Science Advisor

- 742

- 454

Because this is not a variational problem. Variational problem is a problem about extremums of a functional which is defined on some manifold with constraints. When you solve a variational problem you can not choose variations as you want. The constraints determine the space of variations.But, I have derived the nonholonomic equations of motion from the variational principle in the way shown in my previous posting. So why are you saying that's wrong?

So the phrase:

is meaningless in the mathematical sense. The constraints ##a_i\dot x^i=0## imply the following space of variations ##\{\delta x^j\mid a_i\frac{d}{dt}(\delta x^i)+\frac{\partial a_i}{\partial x^k}\dot x^i\delta x^k=0\}##. This corresponds to the vaconomic mechanics. But the Largange-D'Alembert equations do not correspond to conditional extremum of any functional in nonholonomic case.This means you have to minimize

A[q]=∫t2t1dtL(q,˙q,t)A[q]=\int_{t_1}^{t_2} \mathrm{d} t L(q,\dot{q},t)

for all variations around the stationary point q(t)q(t) which are subject to the constraints

aik(q,t)δqk=0.

And please never tell to specialists in dynamical systems that Landau&Lifshitz vol. 1 is a good book they will die laughing

- #10

- 17,059

- 8,164

I don't bother, how you call what I did. I find this very straight forward, and I never understood, why non-holnomic constraints are more complicated than holonomic ones. Again the logic is the following. One defines the action functional as usualBecause this is not a variational problem. Variational problem is a problem about extremums of a functional which is defined on some manifold with constraints. When you solve a variational problem you can not choose variations as you want. The constraints determine the space of variations.

So the phrase:

is meaningless in the mathematical sense. The constraints ##a_i\dot x^i=0## imply the following space of variations ##\{\delta x^j\mid a_i\frac{d}{dt}(\delta x^i)+\frac{\partial a_i}{\partial x^k}\dot x^k\delta x^i=0\}##. This corresponds to the vaconomic mechanics. But the Largange-D'Alembert equations do not correspond to conditional extremum of any functional in nonholonomic case.

And please never tell to specialists in dynamical systems that Landau&Lifshitz vol. 1 is a good book they will die laughing

$$A[q]=\int_{t_1}^{t_2} \mathrm{d} t L(q,\dot{q},t).$$

Then one defines the (in general nonholonomic) constraints as constraints on the allowed variations

$$a_{ik}(q,t) \delta q^k=0.$$

These have to be included by the introduction of Lagrange multipliers in the usual way, leading to

$$\delta A[q] = \int_{t_1}^{t_2} \mathrm{d} t [\delta L-\lambda^{i} a_{ik} \delta q^k] \stackrel{!}{=}0.$$

Thanks to the Lagrange multipliers now we can vary the ##\delta q^k## independently and get the usual equations of motion as from d'Alembert's principle,

$$\frac{\partial L}{\partial q^k}-\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\dot{q}^k}=\lambda^{i} a_{ik}.$$

That's, how it's derived in Landau&Lifshitz vol. I, Section 38 too, as I've just checked.

- #11

wrobel

Science Advisor

- 742

- 454

I apologize for the harshness

- #12

- 17,059

- 8,164

No problem; physics discussions sometimes get excited :-)).

- #13

wrobel

Science Advisor

- 742

- 454

O, it seems I have found proper words.

Logic of variational problem is as follows. (little bit informally) What is ##\delta A##? To calculate it we do the following procedure:

##\delta A=\frac{d}{ds}\Big|_{s=0}\int_{t_1}^{t_2}L\Big(x+s\delta x,\frac{d}{dt}\Big( x+s\delta x\Big)\Big)dt.##That is we differentiate the functional ##A## on the set ##\{ x+s\delta x\}## with ##x(t)## fixed. To determine the space of variations ##\delta x## we must do the same for the equation of constraint:

$$a_i(x)\dot x^i=0\Longrightarrow\frac{d}{ds}\Big|_{s=0}\Big[a_i(x+s\delta x)\frac{d}{dt}\Big(x^i+s\delta x^i\Big)\Big]=0.$$

It is exactly the same as when we look for conditional minimum for example for the function ##f(u,v)=u-v## under constraint ##u^2+v^2=1##.

Logic of variational problem is as follows. (little bit informally) What is ##\delta A##? To calculate it we do the following procedure:

##\delta A=\frac{d}{ds}\Big|_{s=0}\int_{t_1}^{t_2}L\Big(x+s\delta x,\frac{d}{dt}\Big( x+s\delta x\Big)\Big)dt.##That is we differentiate the functional ##A## on the set ##\{ x+s\delta x\}## with ##x(t)## fixed. To determine the space of variations ##\delta x## we must do the same for the equation of constraint:

$$a_i(x)\dot x^i=0\Longrightarrow\frac{d}{ds}\Big|_{s=0}\Big[a_i(x+s\delta x)\frac{d}{dt}\Big(x^i+s\delta x^i\Big)\Big]=0.$$

It is exactly the same as when we look for conditional minimum for example for the function ##f(u,v)=u-v## under constraint ##u^2+v^2=1##.

Last edited:

- #14

- 17,059

- 8,164

$$a_i \delta x^i=0.$$

In your description the contraint rather reads

$$\delta x^k \partial_k a_i(x) \dot{x}^i+a_i \frac{\mathrm{d}}{\mathrm{d} t} \delta x^i=0.$$

This is something different than the correct d'Alembertian virtual displacement constraint given above.

- #15

wrobel

Science Advisor

- 742

- 454

Sure! it is vaconomic againHm, but aren't you then lead back to the vakonomic dynamics again

Sure, but this equation must be imposed if we want to minimize the functional ##A## on the space of functions ##\{x(t)\mid a_i(x(t))\dot x^i=0\}##. It is obviously not the same as ##a_i\delta x^i=0##. It is the main thing I am trying to say: the Lagrange-d'Alembert equations are not generated by the problem of minimization.In your description the contraint rather reads

δxk∂kai(x)˙xi+aiddtδxi=0.

- #16

wrobel

Science Advisor

- 742

- 454

- #17

- 17,059

- 8,164

- #18

wrobel

Science Advisor

- 742

- 454

what exactly is wrong? Just cite my phraseYes, and IT IS WRONG!

- #19

- 17,059

- 8,164

- #20

wrobel

Science Advisor

- 742

- 454

I have never assumed that. Please read cearfully what I have writtenTo assume that you can derive the correct equations by treating it as you say

- #21

- 17,059

- 8,164

- #22

wrobel

Science Advisor

- 742

- 454

1) The problem of minimization of the functional ##A## on the space ##\{x(t)\mid a_i(x(t))\dot x^i=0\}## leads to the vaconomic equations. It is a fact from variational calculus, I have already cited the textbook.

2) In general the Lagrange-D'Alembert equations do not follow from the variational principle formulated in 1).

- #23

- 17,059

- 8,164

So I guess, we have to leave it at that and agree to disagree.

- #24

wrobel

Science Advisor

- 742

- 454

it is an interesting storyI don't understand, how one came to the idea to use (1)

and from the Preface of Bloch's book:

- #25

- 17,059

- 8,164

- Last Post

- Replies
- 2

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 1K

- Replies
- 1

- Views
- 3K

- Replies
- 3

- Views
- 6K

- Last Post

- Replies
- 0

- Views
- 958

- Last Post

- Replies
- 1

- Views
- 2K

- Replies
- 7

- Views
- 9K

- Last Post

- Replies
- 4

- Views
- 944

- Replies
- 18

- Views
- 2K

- Replies
- 4

- Views
- 1K