# Virtual Work: How is the applied force related to the coordinates chosen?

I have a question after reading a section from Goldstein's Classical Mechanics. The question deals with equation 1.43 in the text (given below):

$$\tag{1.43} \sum\limits_{i} {\bf F}_i^{(a)}\cdot \delta{\bf r}_i ~=~ 0.$$

Just below the equation in the text, Goldstein says

[...] in general ${\bf F}_i^{(a)} \neq 0$, since the $\delta {\bf r}_i$ are not completely independent but are connected by the constraints. In order to equate the coefficients to zero, we must transform the principle into a form involving the virtual displacements of the $q_i$, which are independent.

I do not understand what the fact that the $\delta {\bf r}_i$ are not completely independent has to do with the applied force ${\bf F}_i^{(a)}$. Moreover, I'd like to see how transforming into generalized coordinates $q_i$ can send the applied forces to zero.

A general remark.

Goldstein is not saying that the applied forces vanish when one "transforms to generalized coordinates," he is simply saying that the equation \begin{align} \sum_i \mathbf F_i^{(a)} \cdot \delta\mathbf r_i = 0 \end{align} does not necessarily imply that the applied forces are zero. The virtual infinitesimal displacements must respect the constraints, so the equation above doesn't hold for all $\delta \mathbf r_i\in\mathbb R^3$, it holds only for infinitesimal displacements satisfying the constraints. If there were no such constraints, then that equation would imply that the applied forces vanish.

Simple example - particle on the sphere.

Think of a simplified example -- a single particle constrained to move on the surface of a sphere. In that case, the allowed virtual infinitesimal displacements are all tangent to the sphere, so the equation above which reduces to \begin{align} \mathbf F^{(a)}\cdot\delta \mathbf r=0, \end{align} and it can be read as follows:

The dot product of the net applied force with any vector tangent to the sphere is zero.

This only implies that the applied force must be normal to the surface of the sphere, it does not mean that it vanishes.

Another way to think about this, which is what Goldstein is talking about when he refers to "independence" of coordinates, is as follows. When the particle is moving on the sphere, it's cartesian coordinates satisfy \begin{align} x^2 + y^2 + z^2 = R^2, \tag{sphere} \end{align} where $R$ is the radius of the sphere. It follows that \begin{align} x\,\delta x + y\,\delta y + z\,\delta z = 0 \tag{$\star$} \end{align} or more succinctly \begin{align} \mathbf r \cdot \delta \mathbf r = 0. \end{align} This means that the coordinates $x,y,z$ cannot be varied independently of one another when the particle moves on the sphere; indeed they are related by the equation $(\star)$. However, we can "solve" the constraint $(\mathrm{sphere})$ above by writing the cartesian coordinates in terms of two angles; \begin{align} x(\theta, \phi) &= R\sin\theta\cos\phi \\ y(\theta, \phi) &= R\sin\theta\sin\phi \\ z(\theta, \phi) &= R\cos\theta. \end{align} When we do this, the two coordinates $\theta$ and $\phi$ can be varied independently because they are "well-adapted" to the sphere, the surface of constraint.

• Thanks. However, I am still confused over Goldstein's insistence on $\it independence$ of the coordinates chosen. (see Goldstein's quote above) Sep 8, 2014 at 3:38
• @Joebevo I have edited the answer. Let me know if the added discussion helps. Sep 8, 2014 at 3:58
• I have two more questions: (1) How does $\mathbf r \cdot \delta \mathbf r = 0$ imply that $x$, $y$, $z$ cannot be varied independently of one another? (2) How does the fact that $\theta$ and $\phi$ can be varied independently allow us to set ${\bf F_i^{(a)}} = 0$? Regarding the first question, does it have anything to do with it Linear Algebra? Sep 8, 2014 at 16:13
• @Joebevo (1) That equation tells us that $\delta \mathbf r$ must be tangent to the surface of the sphere, so if you displace a point a certain amount in the $x$- and $y$-directions, then you must make sure that its displacement in the $z$-direction is such that the $\delta\mathbf r$ is tangent. See also the equation $(\star)$ which is a more explicit version. (2) The fact that $\theta$ and $\phi$ can be varied independently only tells us that the applied force must be normal to the sphere, not that it vanishes. Sep 8, 2014 at 17:03
• +1 but I think you need to emphasise for the benefit of the OP that if the δris were independent of one another, you could choose to set all of them to zero apart from δr1; this would then imply Fa1 MUST = 0 since the equation is true for all F δr. You'd also be able to do the same for δr2 etc which then implies that Fai=0. Sep 9, 2014 at 23:32

Assuming 1 particle ($i$=1), and the mutually orthogonal Cartesian co-ordinates ($x$,$y$,$z$), the applied force is given by :

$${\bf F^{a}=\bf F_x^{a}+\bf F_y^{a}+\bf F_z^{a}}\tag1$$

Now, we have as given: $${\bf F^{(a)}}\cdot \delta{\bf r} = 0\tag2$$

Since the coordinates are mutually orthogonal, we can vary them independently, which means that, for example, we can vary the coordinates purely along the $x$-axis: $${\bf \delta r}={\bf \delta x}\tag3$$

Combining (1), (2) and (3), we get:$${\left[\bf F_x^{a}+\bf F_y^{a}+\bf F_z^{a} \right ]}\cdot \delta{\bf x} = 0\tag4$$

which gives us ${\bf F_x^{a}}=0$. Using similar reasoning, we can show that the $y$ and $z$ components of the applied force $\bf F^a$ are also both equal to zero.

• Comment to the answer (v1): In the absence of constraints, you are right. But the point Goldstein is trying to make is in the presence of constraints. E.g. say the particle is confined to the surface $x=0$. Then $\delta x=0$, and one cannot conclude anything about the $x$-component $F_x^{(a)}$ from the principle of virtual work. Sep 12, 2014 at 11:24

I) Perhaps an example is in order: Consider a curtain ring on a curtain rod.

The curtain ring$^1$ is constrained to move along the $x$-axis. The system has one degree of freedom. The generalized coordinate is $q\equiv x$. Note in particular that the generalized coordinate $q$ is unconstrained, while the position ${\bf r}$ is confined to the $x$-axis. The virtual displacements $\delta {\bf r}=\vec{\imath}\delta q$ are therefore also along the $x$-axis.

The applied$^2$ force

$$\tag{1} {\bf F}^{(a)}~=~-mg\vec{\jmath} ~\neq~ 0$$

on the ring is gravity in the $y$-direction, which is perpendicular to the $x$-direction. The principle of virtual work is therefore satisfied

$$\tag{2} {\bf F}^{(a)}\cdot \delta {\bf r} ~=~0.$$

II) Conversely, since we are not free to vary $\delta {\bf r}$ arbitrarily in eq. (2), we cannot deduce that ${\bf F}^{(a)}$ is zero. In components the principle of virtual work (2) becomes

$$\tag{3} F_x^{(a)} \delta q ~=~0$$

Since the generalized coordinate $q$ is unconstrained, we deduce deduce from (3) that

$$\tag{4} F_x^{(a)}~=~0.$$

Note in particular that the principle of virtual work (1) says nothing about the other force components $F_y^{(a)}$ and $F_z^{(a)}$. In other words, we can only deduce from (2) that

$$\tag{5} {\bf F}^{(a)} \perp \delta {\bf r},$$

i.e. that ${\bf F}^{(a)}$ is perpendicular to the $x$-direction.

III) When Goldstein below eq. (1.43) says

In order to equate the coefficients to zero, we must transform the principle into a form involving the virtual displacements of the $q_i$, which are independent,

he means that (after the transformation) we can equate the new coefficients to zero. He doesn't mean that we can equate the old coefficients to zero.

--

$^1$ Let us treat the ring as a point particle and ignore friction for simplicity.

$^2$ For the definition of applied force, see e.g. my Phys.SE answer here and links therein. The constraint force on the ring is the normal force from the rod, which ensures that the ring does not fall down.

• Thanks. What about the second part :"In order to equate the coefficients to zero, we must transform the principle into a form involving the virtual displacements of the $q_i$, which are independent." According to me, the vector directions of the constrained displacements won't change. They will still be given by the directions of the $\bf \delta r$'s.Then why is he saying that? Sep 12, 2014 at 3:28
• I updated the answer. Sep 12, 2014 at 11:32
• Thanks. It seems a lot clearer now. BTW, do you think Goldstein is guilty of some imprecise language here? It almost looks like he's saying that if you use the $q_i$'s you can set $\bf F_i^{(a)}$ to zero, which is obviously not the case. Sep 12, 2014 at 13:34
• Possibly. But in his defence, almost any single sentence written can be misunderstood in some way or another. And Goldstein's sentence below eq. (1.43) can also be interpreted in a way so it is correct. Sep 12, 2014 at 14:56