Generalized force - Aetherwisp

The generalized forces $Q_{i}$ acting on a systems are coefficients that, when multiplied with virtual displacements $\delta q_{i}$ over the axes of the configuration space, give the virtual work of a system:

\delta W=\sum_{i=1}^{n} Q_{i}\delta q_{i}

$n$ is the number of degrees of freedom of the system. If the system is made up of $N$ point masses, each of which has a force $\mathbf{F}_{i}$ applied onto it, they are given by

Q_{j}=\sum_{i=1}^{N} \mathbf{F}_{i}\cdot \frac{ \partial \mathbf{r}_{i} }{ \partial q_{j} }

Definition#

We'll start by considering a system of $N$ point masses, each of which has a classical force $\mathbf{F}_{i}$ applied to it. The virtual work that these forces would do over virtual displacements $\delta \mathbf{r}_{i}$ is

\delta W=\sum_{i=1}^{N} \mathbf{F}_{i}\cdot \delta\mathbf{r}_{i}

If the position vectors $\mathbf{r}_{i}$ of the particles are functions of the generalized coordinates $q_{1},\ldots,q_{N}$ , we express $\delta \mathbf{r}_{i}$ as a Linear combination using the Chain rule:

\delta W=\sum_{i=1}^{N} \mathbf{F}_{i}\cdot\left( \sum_{j=1}^{n} \frac{ \partial \mathbf{r}_{i} }{ \partial q_{j} } \delta q_{j} \right)=\sum_{j=1}^{n} \underbrace{ \left( \sum_{i=1}^{N} \mathbf{F}_{i}\cdot \frac{ \partial \mathbf{r}_{i} }{ \partial q_{j} } \right) }_{ Q_{j} }\delta q_{j}=\sum_{j=1}^{n} Q_{j}\delta q_{j}

The quantity $Q_{j}$ is a coefficient that, when multiplied with a displacement, gives an amount of work. It must, then, be some kind of force: we call these generalized forces.

Conservative forces#

If the forces $\mathbf{F}_{i}$ are conservative, and thus permit a Potential $\tilde{V}(\mathbf{r}_{1},\ldots,\mathbf{r}_{N},t)$ such that $\mathbf{F}_{i}=-\nabla_{i}\tilde{V}$ , we can derive $Q_{j}$ just from the $\tilde{V}$ :

\begin{align} Q_{j}&=\sum_{i=1}^{N} \mathbf{F}_{i}\cdot \frac{ \partial \mathbf{r}_{i} }{ \partial q_{j} } \\ &=-\sum_{i=1}^{N} \nabla_{i}\tilde{V}\cdot \frac{ \partial \mathbf{r}_{i} }{ \partial q_{j} } \\ &=-\sum_{i=1}^{N} \left( \frac{ \partial \tilde{V} }{ \partial x_{i} } \frac{ \partial x_{i} }{ \partial q_{j} } +\frac{ \partial \tilde{V} }{ \partial y } \frac{ \partial y }{ \partial q_{j} } +\frac{ \partial \tilde{V} }{ \partial z } \frac{ \partial z }{ \partial q_{j} } \right) \\ &=-\frac{ \partial }{ \partial q_{j} } \tilde{V}(\mathbf{r}_{1}(q,t),\ldots,\mathbf{r}_{N}(q,t),t) \end{align}

where we used the fact that the sum was just the derivative of a composite function using the Chain rule. In fact, if we call $V(q,t)\equiv \tilde{V}(\mathbf{r}_{1}(q,t),\ldots,\mathbf{r}_{N}(q,t),t)$ the composition of $\tilde{V}$ with all the $\mathbf{r}_{i}$ , we can express the generalized force itself as the derivative of a potential

\boxed{Q_{j}=- \frac{ \partial V }{ \partial q_{j} } }

Examples#

> These are $N$ equations. If we project these over the configuration space by taking the [[Scalar product|scalar products]] with the tangent basis we get > $$\sum_{i=1}^{N} (m_{i}\mathbf{a}_{i}+\mathbf{F}_{i}-\Phi_{i})\cdot \frac{ \partial \mathbf{r}_{i} }{ \partial q_{j} } =0,\quad j=1,\ldots,n

This leads to $n$ equations to solve instead of $N$ . Specifically, $n$ ODEs for the $n$ unknowns $q_{1}(t),\ldots,q_{n}(t)$ . Finding these unknowns is sufficient to fully determine motion.

We start from the acceleration. The acceleration of a body is the time total derivative of its velocity:

> from which we can state > $$m \mathbf{a}_{i}\cdot \frac{ \partial \mathbf{r}_{i} }{ \partial q_{j} } =m \frac{d\mathbf{v}_{i}}{dt}\cdot \frac{ \partial \mathbf{r}_{i} }{ \partial q_{j} } =\ldots

The trick to move forward is to move the total time derivative from $\mathbf{v}_{i}$ to the whole scalar product. Of course, we can't just do that. What we can do, is move the derivative while subtracting the difference by using the product derivation rule:

> That'll end up with an actual equality. With this in mind, we get > $$\ldots=m\frac{d}{dt} \left( \mathbf{v}_{i}\cdot \frac{ \partial \mathbf{r}_{i} }{ \partial q_{j} } \right)-m\mathbf{v}_{i}\cdot\frac{d}{dt} \frac{ \partial \mathbf{r}_{i} }{ \partial q_{j} } \tag{1}

It's worth remembering that $\mathbf{r}_{i}\equiv \mathbf{r}_{i}(q(t),t)$ , so its time derivative is given to us by the chain rule:

> Now, let's step off of this equation for a moment and let's consider this other, simpler expression: > $$\sum_{k=1}^{N} \frac{ \partial \mathbf{r}_{i} }{ \partial q_{k} }(q,t)\dot{q}_{k}+\frac{ \partial \mathbf{r}_{i} }{ \partial t }(q,t)

This is some function of $q$ . I promise you'll see why this matter in just a second. Since it's a function of $q$ , we can take its partial derivative in $q_{j}$ . Then we can evaluate it over the motion, $q=q(t)$ and $\dot{q}=\dot{q}(t)$ :

> But this is just our total derivative from before. You might be wondering why this matters; the answer is that the "some function of $q$" that we "chose" is actually just $\mathbf{v}$. Really, just look at it, it's the total time derivative of $\mathbf{r}_{i}$. So what we have in the end is that $(2)$ is just a partial derivative of velocity *evaluated over the motion*: > $$\ldots=\frac{ \partial \mathbf{v} }{ \partial q_{j} } (q(t),\dot{q}(t),t)

If we go back to $(1)$ now we can write

> If we now take the derivative of $\mathbf{v}_{i}$ with respect to $\dot{q}_{j}$ we get > $$\frac{ \partial \mathbf{v}_{i} }{ \partial \dot{q}_{j} }(q,\dot{q},t) =\frac{ \partial }{ \partial \dot{q}_{j} } \left[ \sum_{k=1}^{N} \frac{ \partial \mathbf{r}_{i} }{ \partial q_{k} }(q,t)\dot{q}_{k}+\cancel{ \frac{ \partial \mathbf{r}_{i} }{ \partial t }(q,t) } \right]

The time derivative is independent of $\dot{q}_{j}$ so it vanishes. Meanwhile, the significant part of the other term is

> It's a [[Kronecker delta]], which leaves us with > $$\frac{ \partial \mathbf{v}_{i} }{ \partial \dot{q}_{j} } (q,\dot{q},t)=\frac{ \partial \mathbf{r}_{i} }{ \partial q_{j} }

As such, we can write

> This is *almost* the [[kinetic energy]].