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Thank you for the help.

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- Thread starter betelgeuse91
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Thank you for the help.

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Oh the Lagrangian does not necessarily be T-U

- #3

wrobel

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

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wrobel

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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

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

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

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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

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- #9

wrobel

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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

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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.

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wrobel

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I apologize for the harshness

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

- #13

wrobel

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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##.

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

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

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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.

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wrobel

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wrobel

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

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- #20

wrobel

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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

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wrobel

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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).

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

- #24

wrobel

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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:

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