Why are there 2s -1 independent integrals of motion?

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  • #1
yucheng
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I was reading Mechanics by Landau and Lifshitz and I am confused when it is stated in chapter 2 section 6 that one of the integrals of motion is not independent and it can be considered an additive constant of time. Hence I tried searching it up online.

https://physics.stackexchange.com/q...f-motion-vs-first-integrals?noredirect=1&lq=1

According to the OP in the link above (first paragraph second sentence), we need to specify 2N initial conditions, one of them is the initial time, the others the initial positions and velocity.

However, shouldn't it be N position and N velocities? Can it be shown to be equivalent? Plus aren't we working with an autonomous system of equations? Is this why we need ##t - t_0## so that it is independent of time?

https://physics.stackexchange.com/questions/13832/integrals-of-motion

The answer provided above seems interesting. However, how correct is it? There are several points that I would like to verify...

Questions:
  1. Because the ##\mathcal{L} (q, \dot q)##, the Lagrangian is independent of the acceleration. Hence $$\frac{d}{dt} \frac{\partial \mathcal{L} }{\partial \dot q} - \frac{\partial \mathcal{L}}{\partial q}$$
    which only involved one time derivative, only introduces terms linear in ##\ddot q##.
  2. According to the author (see a comment below the post as well), $$\ddot q_i =\frac{\text{d}\dot q_i}{\text{d} q_1} \frac{\text{d}q_1}{\text{d}t} =\dot q_1\frac{\text{d}\dot q_i}{\text{d}q_1}$$. How does a total time derivative become a total derivative in ##q_1##? Are we performing a change of variables by inverting ##q_1(t)## to get time as a function of ##q_1## then all coordinates become ##q_i(t(q_1))##?
  3. Are there other proofs of this?
 
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Answers and Replies

  • #2
anuttarasammyak
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However, shouldn't it be N position and N velocities? Can it be shown to be equivalent? Plus aren't we working with an autonomous system of equations? Is this why we need t−t0 so that it is independent of time?
Hi. Position and velocity change in time in general so they cannot be integral constants e.g. energy.
 
  • #3
yucheng
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Hi. Position and velocity change in time in general so they cannot be integral constants e.g. energy.
Actually, for that part of the question, I am referring to the initial conditions (which of course determines the ##2s## integrals of motions). I asked it here because it is very relevant! ;)
 
  • #4
yucheng
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Ooops by the way, ##N=s## (degrees of freedom)
 
  • #5
yucheng
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Because the L(q,q˙), the Lagrangian is independent of the acceleration. Hence $$\frac{d}{dt} \frac{\partial \mathcal{L} }{\partial \dot q} - \frac{\partial \mathcal{L}}{\partial q}$$
which only involved one time derivative, only introduces terms linear in q¨.
Is this correct?
 

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