# 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". – joshphysics Oct 22 '13 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 $\delta \vec r(t)$. – Trimok Oct 22 '13 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. – Trimok Oct 22 '13 at 8:48
• Euler-Langrange equations can be derived in their complete form just by using D'Almebert's principle and the fact that the constraints are holonomic.So we can't invoke the action principle or the Euler-Langrange equations as that just leads to circular reasoning. – Sandesh Kalantre Oct 22 '13 at 9:57
• @joshphysics You see! Physics is superdeterministic after all! – WetSavannaAnimal Oct 24 '13 at 23:54

## 4 Answers

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

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

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:

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

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? – pppqqq Oct 24 '13 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. – Qmechanic Oct 24 '13 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? – Sandesh Kalantre Oct 22 '13 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. – freude Oct 22 '13 at 12:44

Answering your points:

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.

2. A virtual displacement is a displacement that doesn't necessarily take place in the problem, but is imagined to take place, while remaining compatible with the constraints. Even in a statics problem with a marble at the bottom of a spherical well, any imagined displacement is virtual with time frozen because in reality, it's fixed at the bottom.

3. The value of the expression with displacements that really happen will be zero, since virtual displacements can take place in any direction compatible with the constraints of the problem

ad 3.) If your constraints are not depending on time this corresponds to the work that the constraint forces perform for the actual time evolution of you system. If you want this expression to vanish for all possible displacements you demand that the constraint forces can not perform work for any possible displacement.