I'm reading Goldstein's Classical Mechanics and he says the following:

A virtual (infinitesimal) displacement of a system refers to a change in the configuration of the system as the result of any arbitrary infinitesimal change of the coordinates $\delta \mathbf{r}_i$, consistent with the forces and constraints imposed on the system at the given instant $t$. The displacement is called virtual to distinguish it from an actual displacement of the system ocurring in a time interval $dt$, during which the forces and constraints may be changing.

Then he discusses virtual work and so on. Now, I can't grasp what this thing of virtual really is. By this text there's a diference between one infinitesimal change and one virtual change and I really don't get what this virtual really is.

Also, this is based on infinitesimals. How can this be expressed rigourously without refering to infinitesimals? I tried looking on Spivak's Physics for Mathematicians where he considers these virtual displacements as tangent vectors to a certain manifold, but I'm not sure this is the most "standard" way to do it rigorously.


4 Answers 4


Let $Q$ denote the set of all possible configurations of the system (the configuration manifold). Consider a point $q_0\in Q$. For the sake of conceptual clarity, and to make contact with physics notation, let's work in some local coordinate patch around $q_0$.

Suppose that $q_0$ represents the position of the system under consideration at time $t_0$. At a given time $t$ later, the system will be at some position say $q(t)$ that is determined by the evolution equations (the Euler-Lagrange equations if we are doing Lagrangian mechanics), and the quantity \begin{align} q(t) - q(t_0) = q(t) - q_0 \end{align} would be the displacement of the system after a time $t-t_0$. Suppose, instead we consider some other curve $\gamma(s)$ in the configuration space which starts at the point $s_0$; \begin{align} \gamma(s_0) = q_0, \end{align} and suppose that we compute the displacement \begin{align} \gamma(s) - \gamma(s_0) = \gamma(s) - q_0 \end{align} that would result from moving along this other curve of our choosing. We call this displacement the virtual displacement after a "time" $s-s_0$ corresponding to moving along the curve $\gamma$. It's called virtual because it is the displacement in the position of the system that would occur if the system were to move along the curve $\gamma$ of our choosing -- a "virtual" curve as opposed to the "real" curve along which the system travels according to the Lagrangian evolution of the system.

Note. As Qmechanic suggested in the comments, I used the parameter $s$ for the virtual curve $\gamma$ instead of $t$ to emphasize that moving along that curve does not correspond to time-evolution, but rather any curve of our choosing.

Now what about virtual "infinitesimal" displacements? Well, recall that the term "infinitesimal" in physics essentially always refers to "first order" approximations, see, e.g. this SE post:

Rigorous underpinnings of infinitesimals in physics

So when we are discussing a virtual infinitesimal displacement, what we have in mind is taking the virtual displacement $\gamma(s) - q_0$, Taylor expanding it to first order in $s$, and extracting only the first order term. Let's do this: \begin{align} \gamma(s) - q_0 = \gamma(s_0) + \dot\gamma(s_0) (s-s_0) + O((s-s_0)^2) - q_0 \end{align} Using the fact that $\gamma(s_0) = q_0$, we see that the Taylor expansion of the virtual displacement is \begin{align} \gamma(s) - q_0 = \dot\gamma (s_0) (s-s_0) + O((s-s_0)^2), \end{align} and now we notice that to first order in $s$, the size of the virtual displacement is controlled by the coefficient of $s-s_0$, namely $\dot\gamma(s_0)$. In other words, virtual infinitesimal displacements (meaning we just keep the first order contribution in $s-s_0$), are determined by the velocity vector of the chosen "virtual curve" at $s_0$. But if you've taken a differential geometry course, then you know that velocities of curves on a manifold are simply tangent vectors to that manifold!

So virtual infinitesimal displacements can be associated with tangent vectors to the configuration manifold. The intuition to keep in mind here as that a virtual displacement just tells us how far we would get away from a certain point on the manifold if we were to travel on a certain curve of our choosing that may not coincide with the actual motion of the system determined by time evolution. The "infinitesimal" part and identifying this part with tangent vectors comes simply from considering what happens only to first order.

  • 4
    $\begingroup$ Thanks for your answer, it is much clearer now where the "virtual" name comes from, but I have one doubt. Goldstein says that these virtual displacements should be consistent with the forces and constraints. That wouldn't make the curve $\gamma$ be exactly the solution of the evolution equations? What he really means by that then? $\endgroup$
    – Gold
    Commented Aug 6, 2014 at 22:37
  • 6
    $\begingroup$ @user1620696 Imagine that you have a particle constrained to move on the surface of a sphere but such that the particle is otherwise free. If the particle is sitting at some point and you give it some initial velocity, then it will travel along a particle great circle (the one whose tangent is in the same direction as the initial velocity). However, even though the initial conditions tell us that the particle will move in a particular direction, we could have considered sending it in any direction along some curve that lies on the sphere; this would still be consistent with the constraints. $\endgroup$ Commented Aug 6, 2014 at 22:42
  • 3
    $\begingroup$ Suggestion to the answer (v1): When discussion virtual displacements, call the curve parameter something else than $t$, e.g. $s$ or $u$ (as the reader may confuse $t$ with time). Recall that a virtual displacement takes place at a frozen instant of time. $\endgroup$
    – Qmechanic
    Commented Aug 6, 2014 at 22:42
  • $\begingroup$ @Qmechanic Yeah I was on the fence as to whether to do that or not, but I think you're right; it might be confusing as written for that reason. I'll change the notation. Thanks for the suggestion. $\endgroup$ Commented Aug 6, 2014 at 22:43
  • $\begingroup$ (+1) Really nice explanation. But that would be much nicer if you could provide some simple example which may help to understand the abstract reasoning. :) $\endgroup$ Commented Sep 30, 2016 at 18:54

In short: virtual displacement is "pretend you are moving, but don't really move". In other words - you move by such a small amount that you don't change the state of the system - but it gives you insight (through work done etc) in what would happen if you did move.

In other words - if the system is really moving, you can look at an interval $dt$ to see how little it moved in that time. That's an "infinitesimal" motion. With virtual motion, you pretend you moved by $dx$ - but not because the system is in motion, but just imagining that you made the tiniest motion (in finite time - so there is no velocity, $\frac{dx}{dt}=0$)

  • $\begingroup$ "pretend you are moving, but don't really move". In other words - you move by such a small amount " - if we pretend to move a larger distance , would it not be virtual displacement then. Because Wikipedia article says " all possible virtual paths"... $\endgroup$
    – Shashaank
    Commented May 12, 2017 at 13:51

You may look into this. It has elaborate works on understanding what virtual displacement is. https://www.researchgate.net/publication/2174249_On_Virtual_Displacement_and_Virtual_Work_in_Lagrangian_Dynamics


As I understand, it must be a displacement in generalized coordinates. If they are orthogonal space coordinates they are not virtual.


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