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B Proof / derivation of d'Alembert principle?

  1. Oct 25, 2016 #1
    I can't find a derivation of d'Alembert principle. Wikipédia says there is no general proof of it. Same with stackexchange. I find it surprising so I thought I'd come here to check with you guys. D'Alembert principle has indeed no proof ?
     
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  3. Oct 25, 2016 #2

    Simon Bridge

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    I always thought of it as a general expression of the definition of force in Newton's second law.
    ie. it is a definition of a concept - or a consequence of a definition. There is no proof - outside of demonstrating that the concept is empirically useful.
     
  4. Oct 26, 2016 #3
    Thank you !
     
  5. Oct 26, 2016 #4

    vanhees71

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    Well, I'd say the d'Alembert principle can be derived from the Hamilton principle, which is the fundamental principle of all contemporary physics. You could, of course, as well argue the other way around and take the d'Alembert principle as fundamental and derive Hamilton's from it (as is the tradition in many textbooks on analytical mechanics).
     
  6. Oct 26, 2016 #5
    Any chance you'd have this derivation available somewhere ?
     
  7. Oct 26, 2016 #6
    After going through the stackexchange link it seems to me that D'Alembert's principle is just a restatement of Newton's 2nd law.
     
  8. Oct 26, 2016 #7

    dextercioby

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    Can one succintly describe the relevance of d'Alembert's principle ? I've taken two courses in classical mechanics (Newtonian and "analytical") and never heard of it. Since this bears no relevance to real theories of physics (electrodynamics, relativity, QM and QFT), I don[t understand why bring it in the first place...
     
  9. Oct 26, 2016 #8
    What I understood is that this is true only in under certain conditions.

    I find it peculiar as well. It seems it's useful to solve mechanic problems.
     
  10. Oct 26, 2016 #9
    It states that constraint and internal forces do not produce a change in momentum of that body. I guess because it is so obvious many have a hard time understanding why it's so important.
     
  11. Oct 27, 2016 #10

    vanhees71

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    I also never understood the relevance of the d'Alembert principle from a modern point of view. It's just historically earlier than the Hamilton action principle, which nowadays is the fundamental starting point of all dynamical theories. It's an extension of naive Newtonian mechanics in so far as it allows to incorporate holonomous and anholonomous constraints as does of course the action principle. Given the fundamental lemma of variational calculus the principles are pretty much equivalent:

    The Hamilton principle defines the dynamics by the stationarity of the action
    $$A[q]=\int_{t_1}^{t_2} \mathrm{d} t L(q,\dot{q},t),$$
    where ##q## are ##f## generalized coordinates. If there are constraints, they have to be incorporated as restrictions on the allowed variations, i.e.,
    $$\delta q^j f_j^{(\alpha)}(q,t)=0,$$
    where ##j## runs from ##1,\ldots,f## (Einstein summation implied), and ##\alpha## labels the constraints, ##\alpha \in \{1,\ldots,k \}##, ##k<f##.

    Then the constraints can be worked in with help of Lagrange parameters, leading to
    $$\delta A[q]=\int_{t_1}^{t_2} \mathrm{d} t \left (\frac{\partial L}{\partial q^j} -\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}^j} + \sum_{\alpha} \lambda^{\alpha} f_j^{(\alpha)} \right ) \delta q^j \stackrel{!}{=}0,$$
    and since this should be true for all ##\delta q^j##, you can leave out the integral, and what you get is d'Alembert's principle.
     
  12. Oct 27, 2016 #11

    dextercioby

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    The time independent constraints restrict the number of degrees of freedom one naïvely computes (number of particles times the dimension of space), hence the Lagrangian variables (generalized coordinates) are the ones encoding the true dynamics. Their variations are free.
    For time dependent constraints, things are not easily separated.
     
  13. Oct 27, 2016 #12
    I once asked a similar question. The Hamilton's principle can be derived as a classical limit of the QM path integral formulation.

    However I still don't now if this is more fundamental, since the path integrals also make use of Hamiltonian or Lagrange formulations.
     
  14. Nov 1, 2016 #13
    Thank you for the answers !
     
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