Why do we get the same differential equations from both principles? Surely there is a fundamental connection between them? When written out, the two seem to have nothing in common.

$$\sum _i ( \mathbf F _i - \dot{\mathbf p}_i) \cdot \delta \mathbf r _i = 0$$

$$S[q(t)] = \int ^{t_2} _{t_1} \mathcal L (q,\dot{q},t)dt$$

After playing with d'Alembert's principle we find that we can rewrite the whole thing as $$\sum _i \left[ \frac{d}{dt} \left( \frac{\partial T}{\partial \dot{q}_i} \right) - \frac{\partial T}{\partial q_i}-Q_i\right] \delta q_i$$

This can further be rewritten under certain conditions so we get the exact form of the E-L equation.

It seems to me that both ways of arriving at the result are fundamentally different.

A function must obey the E-L equations in order to minimize the action over a path, but when we look at the virtual work, it appears that they come from the fact that (quoting Goldstein) "particles in the system will be in equilibrium under a force equal to the actual force plus a 'reversed effective force'."

I think I understand the principle of stationary action, I can see how it leads to the E-L equations, but d'Alembert's Principle seems so arbitrary, I can't see any motivation for it.

  • 1
    $\begingroup$ If I remember my classical mechanics correctly (what are the chances?) D'Alembert's principle works for nonholonomic conditions, while the framework of the variational principles that are based on generalized coordinates does not. So that leaves us with the question if D'Alembert guessed the correct mathematical condition for all nonholonomic cases? The answer to that can only come from experiments. If somebody can find a mechanical system that does not obey these equations, it will be obvious that they are either wrong or that they need to be expanded to something that works better. $\endgroup$
    – CuriousOne
    Commented Aug 17, 2014 at 6:25
  • $\begingroup$ Comment to the question (v1): The motivation for d'Alembert's principle is Newton's laws, cf. e.g. physics.stackexchange.com/q/82884/2451 . For how d'Alembert's principle and the principle of stationary action are related, see e.g. physics.stackexchange.com/q/78138/2451 . $\endgroup$
    – Qmechanic
    Commented Aug 17, 2014 at 9:05
  • $\begingroup$ I understand that D'Alembert's principle leads to the Lagrangian form ($L=T-V$), but what I'm really asking why. $\endgroup$
    – Astrum
    Commented Aug 17, 2014 at 22:00
  • $\begingroup$ Why as in how, or why as in why? $\endgroup$
    – Qmechanic
    Commented Aug 5, 2016 at 13:02

1 Answer 1


The principle of Least (Stationary) Action (aka Hamilton's Principle) is derived from Newton's axioms plus D'Alembert's principle of virtual displacements.

Because D'Alembert's principle allows to account for the (reactions of the) bonds between the components of a system in a transparent way, the Lagrangian and Hamiltonian formulations are possible.

Note1: Newton's axioms, as given, cannot derive neither the Lagrangian form nor the Hamiltonian as they would need the reactions of the bonds to be added literally inside the formalism, thus resulting in different dimensionality and equations for the same problem where the (reactions of the) constraints would appear as extra unknowns.

Note2: D'Alembert's principle is more general than the Lagrangian or Hamiltonian formalisms, as it can account also for non-holonomic bonds (in a slight generalisation).


When the forces are conservative, meaning derived from a potential $V(q_i)$ i.e $Q_i = -\frac{\partial V}{\partial q_i}$, and the potential is not depending on velocities $\dot{q_j}$ i.e $\partial V / \partial \dot{q_j} = 0$ (or the potential $V(q_i, \dot{q_i})$ can depend on velocities in a specific way i.e $Q_i = \frac{d}{dt} \left( \frac{\partial V}{\partial \dot{q_i}} \right) - \frac{\partial V}{\partial q_i}$, refered to as generalised potetial, like in the case of Electromagnetism), then the equations of motion become:

$$\frac{d}{dt} \left( \frac{\partial T}{\partial \dot{q_i}} \right) - \frac{\partial T}{\partial q_i}-Q_i = \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q_i}} \right) - \frac{\partial L}{\partial q_i}$$

where $L=T-V$ is the Lagrangian.

(ref: Theoretical Mechanics, Vol II, J. Hatzidimitriou, in Greek)


One can infact formulate D'Alembert's principle as an "action principle" but this "action" is in general very different from the known Hamiltonian/Lagrangian action.

  1. Variational principles of classical mechanics

  2. Variational Principles Cheat Sheet


  4. 1.2 Prehistory of the Lagrangian Approach


For a further generalisation of d'Alembert-Lagrange-Gauss principle to non-linear (non-ideal) constraints see the work of Udwadia Firdaus (for example New General Principle of Mechanics and Its Application to General Nonideal Nonholonomic Systems)

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    $\begingroup$ What do you mean by "bonds"? This doesn't explain why two seemingly mathematically and physically unrelated things can lead to equations that are almost identical: $\frac{d}{dt} \left( \frac{\partial T}{\partial \dot{q}_i} \right) - \frac{\partial T}{\partial q_i}-Q_i = 0 $ and $\frac{d}{dt} \left( \frac{\partial \mathcal L}{\partial \dot{q}_i} \right) - \frac{\partial \mathcal L}{\partial q_i} = 0$ $\endgroup$
    – Astrum
    Commented Aug 17, 2014 at 22:01
  • $\begingroup$ re-post comment: @Astrum, yeah. might not used best term (non-native speaker), "bonds" means "constraints". They are not mathematicaly nor physically unrelated, d'alembrt's principle is more general than E-L-H formulations, langrange's formulation, stems easily when forces are conservative (derived from a potential) $\endgroup$
    – Nikos M.
    Commented Aug 17, 2014 at 22:14
  • $\begingroup$ @Astrum, updated answer to reflect your comment $\endgroup$
    – Nikos M.
    Commented Aug 17, 2014 at 22:23
  • $\begingroup$ Quote from the wikipedia for virtual work " Among all of the possible displacements that a particle may follow, called virtual displacements, one will minimize the action, and, therefore, is the one followed by the particle by the principle of least action" Alembert's Principle suggests that the path that results in zero virtual work, is the path a particle will take. Is this correct? I feel this is the only conclusion to be drawn from the fact that the action and Alembert's principle leads us to the same equation. $\endgroup$
    – Astrum
    Commented Aug 18, 2014 at 0:23
  • $\begingroup$ @Astrum, no this is reversed, this of course is correct but in the reverse, meaning because of d'alemberts principle of virtual displacements the least action principle is derived. Action principle is based on d'alembert (which is more general as it holds in more cases). Presumably one can derive least action first and then a special form d'alemberts principle but you see the point here. $\endgroup$
    – Nikos M.
    Commented Aug 18, 2014 at 0:29

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