# D'Alembert's Principle: Necessity of virtual displacements

1. Why is the d'Alembert's Principle $$\sum_{i} ( {F}_{i} - m_i \bf{a}_i )\cdot \delta \bf r_i = 0$$ stated in terms of "virtual" displacements instead of actual displacements?

2. Why is it so necessary to "freeze" time in displacements?

3. Also, what would $\sum_{i} ( {F}_{i} - m_i \bf{a}_i )\cdot d \bf r_i$ correspond to if anything at all? In other words, what will be the value of the expression with real displacements instead of virtual ones?

• I would like to make a prediction about the future. You will soon be getting an answer from a user whose name begins with a "Q" and ends with a "c". Oct 22, 2013 at 6:15
• The framework is Lagrangian Mechanics. To find the real movement (path), you have to extremize an action, and the variation of the action calculated for the real path and a virtual infinitely close path must be zero. At fixed time $t$, the infinitesimal variation of coordinates between these 2 infinitely close paths is a virtual displacement $\delta \vec r(t)$. Oct 22, 2013 at 8:48
• .......The D'Alembert principle, is a related philosophy, with constraints, and the notions of virtual work , generalized coordinates, generalized forces, and generalized equations of motion. The latter, with a conservative force $F_i = - \frac{\partial U(\vec r)}{\partial x^i}$, are equivalent to Euler-Lagrange equations. Oct 22, 2013 at 8:48
• @joshphysics You see! Physics is superdeterministic after all! Oct 24, 2013 at 23:54
• Or at least one person is superpredictable :) May 25, 2019 at 16:39

Let us consider a non-relativistic Newtonian problem of $$N$$ point particles with positions

$${\bf r}_i(q,t), \qquad i\in\{1, \ldots, N\},\tag{1}$$

with generalized coordinates $$q^1, \ldots, q^n$$, and $$m=3N-n$$ holonomic constraints.

Let us for simplicity assume that the applied force of the system has generalized (possibly velocity-dependent) potential $$U$$. (This e.g. rules out friction forces proportional to the velocity.)

It is then possible to derive the following key identity

\begin{align}\sum_{i=1}^N \left({\bf F}_i-\dot{\bf p}_i\right)\cdot \left(\delta {\bf r}_i - \frac{\partial {\bf r}_i}{\partial t}\delta t\right) ~=~& \sum_{i=1}^N \left({\bf F}_i-\dot{\bf p}_i\right)\cdot \sum_{j=1}^n\frac{\partial {\bf r}_i}{\partial q^j}\delta q^j\cr ~=~& \sum_{j=1}^n \left(\frac{\partial L}{\partial q^j}-\frac{\mathrm d}{\mathrm dt} \frac{\partial L}{\partial \dot{q}^j} \right) \delta q^j,\end{align} \tag{2}

where

$${\bf p}_i~=~m{\bf v}_i, \qquad {\bf v}_i~=~\dot{\bf r}_i, \qquad L~=~T-U,\qquad T~=~\frac{1}{2}\sum_{i=1}^Nm_i {\bf v}_i^2. \tag{3}$$

Here $$\delta$$ denotes an arbitrary infinitesimal$$^1$$ displacement in $$q$$s and $$t$$, which is consistent with the constraints. There are infinitely many such displacements $$\delta$$.

The actual displacement (i.e the one which is actually being realized) is just one of those with $$\delta t >0$$.

In contrast, a virtual displacement $$\delta$$ has by definition

$$\delta t~=~0. \tag{4}$$

It is customary to refer to the time axis as horizontal, and the $$q^j$$ directions as vertical. Then we may say that a virtual displacement is vertical (4), while an actual displacement never is.

Note that both the lhs. and the rhs. of eq. (2) do effectively not depend on $$\delta t$$.

We can chose between the following first principles:

$$\text{D'Alembert's principle } \Leftrightarrow \text{ Lagrange equations }\Leftrightarrow\text{ Stationary action principle}. \tag{5}$$

I) On one hand, d'Alembert's principle says that

$$\sum_{i=1}^N \left({\bf F}_i-\dot{\bf p}_i\right)\cdot \delta {\bf r}_i~=~0 \tag{6}$$

for all virtual displacements $$\delta$$ satisfying eq. (4). This is equivalent to saying that the lhs. of eq. (2) vanishes for arbitrary (not necessarily vertical) displacements. Then Lagrange equations

$$\frac{\partial L}{\partial q^j}-\frac{\mathrm d}{\mathrm dt} \frac{\partial L}{\partial \dot{q}^j}~=~0\tag{7}$$

follows via eq. (2) from the fact that the virtual displacements $$\delta q^j$$ in the generalized coordinates are un-constrained and arbitrary.

Conversely, when the Lagrange eqs. (7) are satisfied, then the lhs. of eq. (2) vanishes. This leads to d'Alembert's principle (6) for vertical displacements. It does not lead to d'Alembert's principle (6) for non-vertical displacements.

II) On the other hand, if we integrate the rhs. of eq. (2) over time $$t$$, we get (after discarding boundary terms) the infinitesimal virtual/vertical variation

$$\delta S ~=~ \int \! \mathrm dt \sum_{j=1}^n \left(\frac{\partial L}{\partial q^j}-\frac{\mathrm d}{\mathrm dt} \frac{\partial L}{\partial \dot{q}^j} \right) \delta q^j\tag{8}$$

of the action $$S= \int \!\mathrm dt~L$$. The principle of stationary action then yields Euler-Lagrange equations (7).

III) Finally let us stress the following points:

1. Note in both case (I) and (II) that the freedom to perform arbitrary virtual displacements or virtual variations is what allows us to deduce the Lagrange eqs. (7).

2. Note in both case (I) and (II) that the displacements are vertical (4), i.e. no horizontal variation $$\delta t$$.

References:

1. H. Goldstein, Classical Mechanics, Chapter 1 and 2.

--

$$^1$$ All displacements and variations in this answer are implicitly assumed to be infinitesimal.

• ”This e.g. rules out friction forces proportional to the velocity“ - why? If we are considering velocity dependent potential, we really want to include those forces, no? Oct 24, 2013 at 20:55
• @Kazz8: Yes, we allow velocity dependent potentials $U$. And No, the friction force ${\bf F}_f= -k{\bf v}$ does not have a velocity dependent potential $U$. See e.g. this Phys.SE answer and the text around eq. (1.67) in Goldstein. Oct 24, 2013 at 21:53

Terms virtual displacements, as well as corresponding virtual works, are used to ensure that during these displacement all acting forces remain the same. Real displacements are, usually, supplemented by changes in forces.

The virtual displacement is collinear with the resulting force and acceleration of a particle. Now imaging, what if the REAL displacement is perpendicular to the force (it is possible if forces are changing). In this case the direction of the acceleration can not be defined. The virtual displacement is a vector value and it is not arbitrary.

• So why do we "need" the forces to be same? Oct 22, 2013 at 11:17
• Short answer is to find the real accelerations at the instant moment of time.The virtual displacement is collinear with the resulting force and acceleration of the particle. Now imaging, what if the REAL displacement is perpendicular to the force (it is possible). In this case the direction of the acceleration can not be defined. Oct 22, 2013 at 12:44

1. Virtual displacements are used because without them, the theorem would be useless in deriving useful equations of motion. With them, we can derive $N-m$ independent differential equations of motion where $N$ are the number of unconstrained degrees of freedom, $m$ the number of constraints.