In "Classical Mechanics" by Goldstein and "A Students Guide to Lagrangians and Hamiltonians" by Hamill I noticed that both the virtual displacement derivatives and the normal displacement derivatives are used at different points of the proof, as shown below. My question is why can this mixing of real & virtual derivatives be done?

To simplify the equations it is assumed there is only one mass and one associated generalised variable with $x=x(q,t)$, with $\dot{x}$ meaning differential with respect to time.

The virtual displacement $\delta x$ is used to set up the virtual work equation via:

$$\delta x = \frac {\partial x}{\partial q} \delta q, \qquad \delta t=0, \tag{1}$$

being substituted into ($F$ is force, $a$ is acceleration):

$$(F/m) \delta x = a \delta x = a \frac {\partial x}{\partial q} \delta q.\tag{2}$$

The following equations (3) and (4) are used to transform the acceleration $a$ in the rhs of (2) into a form based on the kinetic energy $T$, using the usual velocity differential equation with possible explicit $t$-dependence:

$$v=\dot{x} = \frac {\partial x}{\partial q} \dot{q} + \frac {\partial x}{\partial t} \tag{3}$$

to derive:

$$ \frac {\partial v}{\partial \dot{q}} = \frac {\partial x}{\partial q}. \tag{4}$$

So it looks like virtual displacements are used in (1) & (2) and real displacements $$\delta x = \frac {\partial x}{\partial q} \delta q + \frac {\partial x}{\partial t} \delta t \tag{5}$$ are used in (3) & (4) parts of d'Alembert derivation of Lagrange equations.


1 Answer 1

  1. On one hand, holonomic constraints and the Lagrangian itself are certainly allowed to have explicit time dependence, cf. e.g. the last term in OP's eq. (3).

  2. On the other hand, it's a well-established fact that the relevant (infinitesimal) displacements in the d'Alembert's principle and the principle of stationary action -- the so-called (infinitesimal) virtual displacements -- are frozen in time $\delta t=0$. See e.g. this, this & this related Phys.SE posts.

  • $\begingroup$ Thanks. Is the following correct : It just 'looks' like (1) & (3) are used. It turns out that the differential calculus using (3) to get to (4) removes the term involving time ∂x/∂t using the fact that the constraint x(q,t) is holonomic. In effect this allows (1) to be used in the derivation of (4) instead of using (3) to get to (4) as the same result occurs (surprisingly). Hence there is no inconsistency in using real differential displacements as its consistent with just using virtual displacements. The question is whether this is accidental or very clever!? $\endgroup$
    – user248988
    Jun 2, 2020 at 12:42
  • $\begingroup$ Hm. That doesn't sound right. $\endgroup$
    – Qmechanic
    Jun 2, 2020 at 12:46
  • $\begingroup$ I checked where the virtual & actual differentials were used and as far as I can tell that is whats going on.The proof in Hamill in its Section 1.3 derives the equation and in deriving (4) it takes the derivative of the ∂x/∂t term with respect to q dot, and this term becomes zero. $\endgroup$
    – user248988
    Jun 2, 2020 at 12:51
  • $\begingroup$ That's merely a coincidence if that works. It is not the general reason. $\endgroup$
    – Qmechanic
    Jun 2, 2020 at 19:26

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