2
$\begingroup$

Goldstein's book of Classical Mechanics derive the Euler-Lagrange equations from two different principles:

  1. Hamilton's principle states that

$$\delta S = \delta\int_{t_1}^{t_2}L(q^{i},\dot{q}^{i},t)dt=0,$$

where no conditions on $L(q^{i},\dot{q}^{i},t)$ seem to be required, and from here one can directly obtain Euler-Lagrange equations.

  1. D'Alembert's principle (in terms of generalized coordinates) states that

$$\sum_{j}\left\lbrace\bigg[\frac{d}{dt}\bigg(\frac{\partial T}{\partial \dot{q}^{j}}\bigg)-\frac{\partial T}{\partial q^{j}}\bigg]-Q_{j}\right\rbrace\delta q^{j}=0,$$

where one needs to consider that the kinetic energy $T$ is a quadratic function of the velocities $\dot{q}^{j}$.

In order to obtain the Euler-Lagrange equations from this principle, one also needs to consider that the generalized force $Q_j$ is derivable from the potential energy function $V$, which must be a function dependent only on the coordinates $q^{j}$. The form of the Lagrangian must be $L=T-V$ in this case.

  1. Is the Hamilton's principle more powerful in the sense that puts no restrictions on the form of the Lagrangian? or there is some missing restriction?

  2. Is there another way of deriving Euler-Lagrange equations from D'Alembert's principle without restricting the forms of $T$, $V$ and $L$?

$\endgroup$
  • $\begingroup$ 1st question: Yes, 2nd question: S'Alembert's principle just provides a justification for Hamilton's principle $\endgroup$ – Boltzee Nov 30 '16 at 1:21
  • $\begingroup$ So, the Hamilton's principle does not require that the forces of constraint do no virtual work? $\endgroup$ – Saavestro Nov 30 '16 at 2:00
  • $\begingroup$ @bgr95 1) No. The Hamilton principle in that form holds under the conditions of 2) (which is specified in the Goldstein itself). $\endgroup$ – gented Nov 30 '16 at 9:26
3
$\begingroup$

I) Actually, it's the other way around. Within the context of Newtonian mechanics, the hierarchy is the following from most to least applicable:

  1. Newton's laws are always applicable.

  2. D'Alembert's principle or Lagrange equations. E.g. sliding friction typically violates D'Alembert's principle.

  3. The stationary action principle $S=\int\! dt~L$, with Lagrangian $L=T-U$, and its Euler-Lagrange equations. E.g. a generalized force might not have a generalized potential $U$.

II) In point 3 we have tacitly assumed that the Lagrangian is of the form $$L~=~T-U,\tag{1}$$ as is customary. $T$ and $U$ in eq. (1) may be viewed as representing the kinematic and the dynamical side of Newton's 2nd law, cf. e.g. this Phys.SE post. The linear structure of eq. (1) also reflects a categorical-like composition rules for how to build physical models out of physical subsystems.

There exist strictly speaking exceptions to the form (1), cf. e.g. this Phys.SE post, but these exceptions often lacks categorical-like composition rules, which make them unsuitable for useful model building.

III) For further details and discussions, see e.g. my related Phys.SE answers here, here, and links therein.

$\endgroup$
  • $\begingroup$ I don't see clearly that, on the derivation of Euler-Lagrange equations done by Goldstein, Hamilton's principle requires that the generalized force must be derivable from a potential function. He states that the system must be monogenic ($\boldsymbol{F}=-\nabla V$), but I don't see how this actually enters the formulation... $\endgroup$ – Saavestro Nov 30 '16 at 15:45
  • $\begingroup$ Sure... the Lagrangian must always be written in terms of the energies (although not necessarily $L=T-V$). If you can't write the force as a potential function it is impossible to have the covariant property of the Lagrangian formulation (something in terms of scalars only). That is the point, right? $\endgroup$ – Saavestro Nov 30 '16 at 16:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.