# Virtual displacement

Currently I am reading Classical Dynamics written by Donald Greenwood. I have a doubt in the section discussing about virtual displacement. As far as I understood, Virtual displacement (virtual or imaginary) is an infinitesimal change in the configuration of the system which conforms to any instantaneous constraints and forces of the system.

On the left hand side of the page, it is given that the condition for $\delta{x}$ be replaced by $dx$ is that the constraint equation has to be holonomic, but my actual question is from my understanding, virtual displacement is not actual displacement then how can one replace $\delta{x}$ by $dx$ when the contraint is holonomic. It doesn't mean like virtual displacement is same as actual displacement when constraint is holonomic? Or did I understood the concept wrongly?

• Related: physics.stackexchange.com/q/129786/2451 , physics.stackexchange.com/q/456771/2451 and links therein. – Qmechanic Aug 31 '16 at 16:06
• I hope my question is definitely not a duplicate one. My question is related to a case where virtual displacement is replaced by actual displacement. In the link you are provided, the questioner has the difficulty in understanding about the term virtual itself. – Muthu manimaran Aug 31 '16 at 16:56
• Virtual displacement is when you "freeze time" and check what infinitesimal displacements you can make. Actual displacements are the infinitesimal displacements you can make when you don't "freeze time". If a constraint is time-dependent, then there is a difference between the two, for example. Since if you freeze time, the constraint's time evolution also freezes, so then the possible displacements are different than if you also took the constraint's evolution into account. – Bence Racskó Aug 31 '16 at 17:09

Consider a system of particles with position vectors we denoted as $\vec{r_i}$. The system may be subjected to a constraint $f(\vec{r_i},\dot{\vec{r_i}},t)=0$. Let us consider a sub class of constraints $f(\vec{r_i},t)=0$. These constraints are said to be holonomic constraints. Now by differentiating this once, we obtain $$\sum_i\frac{\partial f}{\partial\vec{r_i}}\cdot\vec{v_i}+\frac{\partial f}{\partial t}=0$$
This equation is satisfied by the velocities of the particles. The velocities that obey this equation are said to be allowed velocities. Let us define allowed displacements by $\mathrm d\vec{r_i}=\vec{v_i}~\mathrm dt$. This satisfies, $$\sum_i\frac{\partial f}{\partial\vec{r_i}}\cdot{\mathrm d\vec{r_i}}+\frac{\partial f}{\partial t}~\mathrm dt=0$$We can now define virtual displacements as the difference between two allowed displacements. i.e. $\delta\vec{r_i}=(\vec{v}^\prime_i-\vec{v}_i)~\mathrm dt$. Therefore this satisfies, $$\sum_i\frac{\partial f}{\partial\vec{r_i}}\cdot\delta{\vec{r_i}}=0$$Let us do an example. Consider a simple pendulum with the string length $l$ attached to an oscillating support. Let the coordinates of the bob be $(x,y)$ and let the coordinates of the point of support be given by $(L \cos(\Omega t),0)$. Then the constraint here is $f(x,y,t)=(x-L\cos(\Omega t))^2+y^2-l^2=0$. Then the allowed displacements satisfy $2(x-L\cos(\Omega t)~\mathrm dx+2y~\mathrm dy+2(x-L\cos(\Omega t))L\Omega\sin(\Omega t)~\mathrm dt=0$ whereas the virtual displacements satisfy $2(x-L\cos(\Omega t))~\delta x+2y~\delta y=0$.
• Actually I don't understand the part where we defined virtual displacement as the difference between two allowed displacements. I can not understand this part. Why we are doing this? Usually the virtual displacement happens instantaneously right? then dt=0 and R.H.S also becomes zero, which makes $$\delta{\vec{r_i}}=0$$ right? (Sorry If I am wrong) Actually I could understood the math, but I am unable to understand the situation physically. – Muthu manimaran Aug 31 '16 at 17:05