# Time dependence of generalized coordinates and virtual displacement

The Cartesian coordinates of particles are related to the generalized coordinates via a transformation (for the $$x$$ component of the $$j$$-th particle) as:

$$x_j = x_j(q_1, q_2, \ldots, q_N, t)$$

What I can't understand is why in the virtual displacement which occurs in constant time i.e. $$\delta t=0$$ isn't zero? We can write the virtual displacement as:

$$\delta x_j = \sum_{i=1}^N \frac{\partial{x_j}}{\partial{q_i}}\cdot \delta q_i$$

but because the generalized coordinates can also be considered functions of time then:

$$\delta x_j = \sum_{i=1}^N \frac{\partial{x_j}}{\partial{q_i}}\cdot \dot{q_i} \cdot \delta t$$

If time is frozen isn't virtual displacement also $$0$$?

• More on virtual displacement. Commented Feb 21, 2022 at 19:45
• I guess that $\dot{q}^j = \frac{dq^j}{dt}$. To use this identity, there must be a curve $q^j=q^j(t)$. Here it is not given. Commented Feb 21, 2022 at 19:46
• @ValterMoretti Yes that is what I meant by the dot. Isn't though generalized coordinates functions of time? Commented Feb 21, 2022 at 19:51
• No, generalized coordinates and time are independent variables when you describe constraints. Look at concrete examples to grasp an idea... Commented Feb 21, 2022 at 19:53
• @ValterMoretti Wikipedia refers that each generalized coordinate is function of time. I am completely lost...I will try to find some examples. Commented Feb 21, 2022 at 20:00

Consider a system of $$N$$ material points described by position vectors $${\bf x}_1, \ldots, {\bf x}_N$$ in a reference frame $${\cal R}$$. These position vectors are not free to assume any configuration in the physical space, but they are constrained to satisfy some constraints which, possibly, may depend on time, $$f_j({\bf x}_1, \ldots, {\bf x}_N, t) =0\quad j=1,\ldots, c < 3N\:.\tag{1}$$ If these functions are smooth and satisfy a condition of functional independence (I do not want to enter into the details), we can choose $$n:= 3N-c$$ abstract coordinates $$q^1,\ldots, q^n$$ which can be used the embody the constraints into the formalism. This result holds locally around every admitted configuration and around a given time $$t$$.

As a matter of fact, we can locally (in space and time) represent the position vectors $${\bf x}_1, \ldots, {\bf x}_N\quad i=1,2,\ldots, N$$ as known functions of the said free coordinates.

$${\bf x}_i = {\bf x}_i(q^1,\ldots, q^n,t)\tag{2}$$

When $$q^1,\ldots, q^n, t$$ varies in their domain (an open set in $$\mathbb{R}^{n+1}$$), the vectors $${\bf x}_i(q^1,\ldots, q^n,t)$$ automatically satisfy the constraints (1). The admissible configuration are therefore determined by the free coordinates $$q^1,\ldots, q^n$$ at each time $$t$$.

Now we pass to the notion of virtual displacement compatible with the set of constraints (1). It is defined by fixing $$t$$ and computing the differential of the functions $${\bf x}_i$$ as functions of the remaining variables. The virtual displacement of the system at time $$t_0$$ around a permitted configuration determined by $$q_0^1,\ldots, q_0^n$$ is the set of $$N$$ vectors in the real space

$$\delta {\bf x}_i = \sum_{k=1}^n \left.\frac{\partial {\bf x}_i}{\partial q^k}\right|_{(q^1_0,\ldots, q_0^n,t_0)}\delta q^k\:, \quad i=1,\ldots, N\:.$$

Above the numbers $$\delta q^k\in {\mathbb R}$$ are arbitrary, not necessarily "infinitesimal" (which does mean anything!).

Example

Consider a point of position vector $${\bf x}$$, constrained to live on a circle of radius $$r=\sqrt{1+ct^2}$$, where $$c>0$$ is a known constant. This circle is centered on the origin and stays in the plane $$z=0$$.

Here we have just two constraint $$f_1({\bf x}, t)=0, \quad f_2({\bf x}, t)=0$$ where $$f_1({\bf x}, t) := x^2+ y^2+ z^2 - (1+ct^2)\:, \quad f_1({\bf x}, t) := z$$ if $${\bf x}= x{\bf e}_x+ y{\bf e}_y+ z{\bf e}_z$$

Locally we can use, for instance, the coordinate $$q^1=x$$ to describe a portion of circle and we have

$$x = q^1\:, \quad y = \sqrt{(1+ct^2) - x^2}\:, \quad z=0\:.$$ The relations (2) here read $${\bf x}(q^1,t)= q^1 {\bf e}_1+ \sqrt{(1+ct^2) - (q^1)^2}{\bf e}_2$$

The virtual displacements at time $$t_0$$ are the vectors of the form

$$\delta {\bf x} = \delta q^1 {\bf e}_1+ \frac{\delta q^1}{\sqrt{(1+ct_0^2) - (q^1)^2}}{\bf e}_2$$ for every choice of $$\delta q^1$$.

The geometric meaning of $$\delta {\bf x}$$ should be evident: it is nothing but a vector (of arbitrary length) tangent to the circle at time $$t_0$$ emitted by a configuration determined by the value $$q^1$$.

REMARK It is worth stressing that varying $$t$$, the circle changes! Virtual displacement are defined at given time $$t$$.

The discussed example is actually quite general. The virtual displacement are always vectors which are tangent to the manifold of admitted configurations at given time $$t_0$$.

The prosecution of Lagrangian approach (once assumed the validity of the postulate of ideal constraints and introducing interactions for instance defined by a Lagrangian) consists of finding the evolution of the system not in terms of curves $${\bf x}_i = {\bf x}_i(t)\:, \quad i=1\,\ldots, N$$ in the physical space. The motion is described directly in terms of free coordinates, i.e., curves $$q^k=q^k(t)\:, \quad k=1,\ldots, n$$

Just at this level it makes sense to introduce the notation $$\dot{q}^k = \frac{dq^k}{dt}$$, because here we have a curve $$t$$-parametrized describing the evolution of the system. Before this step $$q^1,\ldots, q^n$$ and $$t$$ are independent variables.

A posterori, if we have a description of the motion of the system in terms of free coordinates, we also have the representation in the physical space by composing these curves with the universal (= independent of any possible motion) relations (2), $${\bf x}_i(t) = {\bf x}_i(q^1(t), \ldots, q^n(t), t)\:, \quad i=1\,\ldots, N\tag{3}$$

It is finally interesting to compare virtual displacements with real displacements when we have a motion $$q^k=q^k(t)\:, \quad k=1,\ldots, n\:.$$ In the physical space, the velocities with respect to the reference frame $${\cal R}$$ are given by taking the derivative with respect to $$t$$ of (3), $${\bf v}_i(t) = \sum_{k=1}^n \frac{\partial {\bf x}_i}{\partial q^k}\frac{dq^k}{dt} + \frac{\partial {\bf x}_i}{\partial t}\:.$$ An approximate displacement ascribed to an interval of time $$\Delta t$$ is $$\Delta {\bf x}_i = {\bf v}_i(t)\Delta t = \sum_{k=1}^n \frac{\partial {\bf x}_i}{\partial q^k}\frac{dq^k}{dt}\Delta t + \frac{\partial {\bf x}_i}{\partial t}\Delta t\:,$$ which can be rephrased to $$\Delta {\bf x}_i = \sum_{k=1}^n \frac{\partial {\bf x}_i}{\partial q^k}\Delta q^k + \frac{\partial {\bf x}_i}{\partial t}\Delta t\:. \tag{4}$$ This identity has to be compared with the definition of virtual desplacement $$\delta {\bf x}_i = \sum_{k=1}^n \frac{\partial {\bf x}_i}{\partial q^k}\delta q^k\:.$$ Even if we choose $$\delta q^k = \Delta q^k$$, the right-hand sides are different in view of the term $$\frac{\partial {\bf x}_i}{\partial t}\Delta t$$ which accounts for a part of the displacement, in real motion, due to the fact that constraints may depend on $$t$$ explicitly, as in the example above.

• The independence of $q$'s and $t$ cleared up some of my confusion. If I get it right, without even knowing the time dependence of $q$'s we can describe the position of a particle at each time by taking a snapshot in the values of $(q_1, q_2, \ldots, q_n, t)$ and mapping to the vector $\mathbf{r}$. Now what I can't get is why $\delta q$ can be arbitrary large. If it is too large then the new position (e.g. in the circle case) doesn't satisfy the constraints. Also, in the comparison of virtual and real displacement even if constraints don't depend on time, isn't necessary that are equal, right? Commented Feb 22, 2022 at 19:18
• Indeed, the fact that virtual displacements connect admissible configurations is just a formal intuition. For small $\delta q^k$ it is approximately true. If the contraints do not depend on $t$, then it is true that virtual and "real" displacements coincide. Commented Feb 22, 2022 at 19:32
• Consider the motion defined by a curve satisfying the equation of motion $q^k=q^k(t)$. Fix a configuration along that motion determined by $q^1(t_0), \ldots, q^n(t_0)$. The configuration after a short period of time $\Delta t$ is approximately $\delta {\bf x}_i = \sum_k \frac{\partial {\bf x}_i}{\partial q^k}|_{(q^1(t_0), \ldots, q^n(t_0))} \dot{q}^k(t_0) \Delta t$. This coincides to the virtual displacement $\delta {\bf x}_i = \sum_k \frac{\partial {\bf x}_i}{\partial q^k}|_{(q^1(t_0), \ldots, q^n(t_0))} \delta q^k$ if $\delta q^k = \dot{q}^k(t_0) \Delta t$. Commented Feb 22, 2022 at 20:48
• The procedure is reversible. Starting from a virtual displacement $\delta {\bf x}_i = \sum_k \frac{\partial {\bf x}_i}{\partial q^k}|_{(q^1(t_0), \ldots, q^n(t_0))} \delta q^k$, we can consider the unique motion with initial conditions $q^k(t_0)$ and $(\dot{q}^1(t_0), \ldots, \dot{q}^n(t_0))$ of length $1$ and parallel to $(\delta q^1, \ldots, \delta q^k)$. Finally $\Delta t = \sqrt{\sum_k (\delta q^k)^2}$. Commented Feb 22, 2022 at 20:53
• Yes it is so. Virtual displacements only consider the admissible configurations restricted by contraints. Real displacements also consider the solutions of the equation of motion. Commented Feb 22, 2022 at 21:01

If you want to think of a virtual displacement as a curve $$s\mapsto q(s)$$, since time $$t$$ is frozen, you cannot pick time $$t$$ as a curve parameter, you have to pick something else, say $$s$$. Hence $$\delta q=\frac{dq}{ds}\delta s$$. See also e.g. this, this & this Phys.SE posts and links therein.

• I find a bit misleading to introduce another parameter $s$. As far as i know, a virtual displacement $\delta {\bf r}_i$ stays in the real space $\delta {\bf r}_i = \sum_k \frac{\partial {\bf r}_i}{\partial q^k} \delta q^k$. Here the $\delta q^k$ are numbers small or big it does not matter. Commented Feb 21, 2022 at 20:13
• Yes, I agree. $s$ is strictly speaking unnecessary information from a mathematical perspective. Commented Feb 21, 2022 at 20:26
• So if I get it right, we look at a specific instance of time say $t=5$ and we imaging an infinitesimal displacement compatible with the constraints. Because $q$'s in that case are "imaginary" i.e. doesn't need to satisfy the equations of motion we can drop the time dependence? Commented Feb 21, 2022 at 20:51
• Why time is frozen is explained in the linked posts. Commented Feb 21, 2022 at 21:04