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Of Bernoullied Euler and Eulered Lagrange and Other Gianted Giants - Part II

Posted on January 2, 2021

Previously, we looked at the principle of virtual work.

\[\sum\limits_{i=i}^k\boldsymbol{f}_i^T\delta \boldsymbol{r}_i=0\]

Now recall the linear independence style of reasoning which is prevalent in linear algebra. It is tempting to wonder if the same style of reasoning can be applied to the above equation. In particular, recall that for \(k\) vectors \((\boldsymbol{b_1},\cdots,\boldsymbol{b_k})\) and for \(k\) scalars \(a_1,\cdots,a_k\), linear independence asserts that the following equation

\[a_1\boldsymbol{b_1} + a_2\boldsymbol{b_2} + \cdots + a_k\boldsymbol{b_k} = \boldsymbol{0}\]

implies that all the scalars must be zero. It is easy to see that if the \(k\) vectors \(\delta \boldsymbol{r}_1,\cdots,\delta \boldsymbol{r}_k\) were linearly independent, then for the first component of the \(k\) vectors \((\boldsymbol{f}_1)_1,\cdots,(\boldsymbol{f}_k)_1\), we have the following equation

\[\sum\limits_{i=i}^k(\boldsymbol{f}_i)_1^T\delta \boldsymbol{r}_i=0\]

The linear independence assumption implies that all \(k\) scalars \((\boldsymbol{f}_1)_1,\cdots,(\boldsymbol{f}_k)_1\) are zero, and this can be extended to the other components of the vectors \(\boldsymbol{f}_i\), implying that \(\boldsymbol{f}_i=0\) which is absurd. This is because clearly the virtual displacements are not independent as the position coordinates \(\boldsymbol{r}_i\) are not independent which the discussion of generalized coordinates above had already established, so we cannot use the above reasoning to conclude that each coefficient individually equals zero. However, we can rewrite the above equations in terms of the independent generalized coordinates to apply the same reasoning which we do in the next section.

Finally, we can also relax the first assumption. So if the system is not in equilibrium, then we apply fictitious pseudo force, \(-\boldsymbol{\dot{p}}_i\), on each particle \(i\) where \(\boldsymbol{p}_i\) is the momentum of each particle. So for such a system, the principle of virtual work translates to

\[\sum\limits_{i=i}^k\boldsymbol{f}_i^T\delta \boldsymbol{r}_i-\sum\limits_{i=i}^k\dot{\boldsymbol{p}}_i^T\delta \boldsymbol{r}_i=0\]

Tricks!

Classical mathematics is characterised by the use of clever tricks to arrive at remarkable insights but mainly to solve enticing problems. Contrarily, the conceptual school of abstract mathematics adopted largely by professional mathematicians of today largely avoids the classical kind of mathematics. This stark division can be slightly understood in terms of what Vladimir Arnold said, "There are two principal ways to formulate mathematical assertions (problems, conjectures, theorems, \(\cdots\)): Russian and French. The Russian way is to choose the most simple and specific case (so that nobody could simplify the formulation preserving the main point). The French way is to generalize the statement as far as nobody could generalize it further." And when the foundations of mathematics seemed to come under threat, even the conceptual school ended up getting divided between Hilbert's Formalists and Brouwer's Intuitionists. However, for most engineering and technological purposes, the tricks-based classical mathematics not only suffices but sometimes may be the only option left.

Recall that virtual displacements \(\delta \boldsymbol{r}_i\) can be expressed in terms of virtual generalized displacements \(\delta q_i\) as follows

\[\delta \boldsymbol{r}_i = \sum_{j=1}^n \frac {\partial \boldsymbol {r}_i} {\partial q_j} \delta q_j\]

The principle of virtual work for systems not in equilibrium states that

\[\sum\limits_{i=i}^k\boldsymbol{f}_i^T\delta \boldsymbol{r}_i-\sum\limits_{i=i}^k\dot{\boldsymbol{p}}_i^T\delta \boldsymbol{r}_i=0\]

The first summand can be rewritten as

\[\sum\limits_{i=1}^{k} \boldsymbol{f}_{i}^{T} \delta \boldsymbol{r}_{i}=\sum\limits_{i=1}^{k} \sum\limits_{j=1}^{n} \boldsymbol{f}_{i}^{T} \frac{\partial \boldsymbol{r}_{i}}{\partial q_{j}} \delta q_{j}\]

\[=\sum\limits_{j=1}^{n} \left(\sum\limits_{i=1}^{k} \boldsymbol{f}_{i}^{T} \frac{\partial \boldsymbol{r}_{i}}{\partial q_{j}}\right) \delta q_{j}\]

Now switching from position coordinates \(\boldsymbol{r}_j\) to generalized coordinates \(q_j\) allows us to switch other physical quantities which were expressed in one set of coordinates to another. This is referred to as coordinate change. The change of coordinates has been rigorously worked out in mathematical analysis and differential geometry. The external forces \(\boldsymbol{f}_{j}\) are expressed in terms of the position coordinates \(\boldsymbol{r}_j\) and can be expressed in terms of generalized coordinates as follows

\[\psi_{j}=\sum\limits_{i=1}^{k}\boldsymbol{f}_{i}^{T} \frac{\partial \boldsymbol{r}_{i}}{\partial q_{j}}\]

These new \(\psi_{j}\) terms are called the generalized forces. So finally,

\[\sum\limits_{i=1}^{k} \boldsymbol{f}_{i}^{T} \delta \boldsymbol{r}_{i}=\sum\limits_{j=1}^{n} \psi_{j} \delta q_{j}\]

Noting that momentum is defined in classical mechanics as \(\boldsymbol{p}_i=m_i\dot{\boldsymbol{r}}_{i}\), the second summand can be rewritten as follows

\[\sum\limits_{i=1}^{k} \dot{p}_{i}^{T} \delta \boldsymbol{r}_{i}=\sum\limits_{i=1}^{k} m_{i} \ddot{\boldsymbol{r}}_{i}^{T} \delta \boldsymbol{r}_{i}=\sum\limits_{i=1}^{k} \sum\limits_{j=1}^{n} m_{i} \ddot{\boldsymbol{r}}_{i}^{T} \frac{\partial \boldsymbol{r}_{i}}{\partial q_{j}} \delta q_{j}\]

\[=\sum\limits_{j=1}^{n} \left(\sum\limits_{i=1}^{k} m_{i} \ddot{\boldsymbol{r}}_{i}^{T} \frac{\partial \boldsymbol{r}_{i}}{\partial q_{j}}\right) \delta q_{j}\]

On the other hand using the product rule of differentiation, we note that

\[\frac{d}{d t}\left[m_{i} \dot{\boldsymbol{r}}_{i}^{T} \frac{\partial \boldsymbol{r}_{i}}{\partial q_{j}}\right]=m_{i} \ddot{\boldsymbol{r}}_{i}^{T} \frac{\partial \boldsymbol{r}_{i}}{\partial q_{j}}+m_{i} \dot{\boldsymbol{r}}_{i}^{T} \frac{d}{d t}\left[\frac{\partial \boldsymbol{r}_{i}}{\partial q_{j}}\right]\]

Substituting this appropriately gives

\[\sum\limits_{i=1}^{k} m_{i} \ddot{\boldsymbol{r}}_{i}^{T} \frac{\partial \boldsymbol{r}_{i}}{\partial q_{j}}=\sum\limits_{i=1}^{k}\left\{\frac{d}{d t}\left[m_{i} \dot{\boldsymbol{r}}_{i}^{T} \frac{\partial \boldsymbol{r}_{i}}{\partial q_{j}}\right]-m_{i} \dot{\boldsymbol{r}}_{i}^{T} \frac{d}{d t}\left[\frac{\partial \boldsymbol{r}_{i}}{\partial q_{j}}\right]\right\}\]

We shall derive a couple of equations to help simplify the above expression. Recall that

\[\boldsymbol{r}_i=\boldsymbol{r}_i(q_1,\cdots,q_n,t)=0,\quad i=1,\cdots,k\]

\[d\boldsymbol{r}_i = \frac {\partial \boldsymbol{r}_i}{\partial t} d t + \sum\limits_{h=1}^n \frac {\partial \boldsymbol{r}_i} {\partial q_h} d q_h\]

Thus,

\[\boldsymbol{v}_i \equiv \frac{\mathrm{d}\boldsymbol{r}_i}{\mathrm{d}t} = \dot{\boldsymbol{r}}_i = \sum\limits_{h=1}^n \frac {\partial \boldsymbol{r}_i} {\partial q_h} \dot{q_h} + \frac{\partial \boldsymbol{r}_i}{\partial t}\]

From the above, it is obvious that

\[\boldsymbol{v}_i \equiv \boldsymbol{v}_i(q_h, \dot{q_h}, t)\]

We can apply the partial derivative operator \(\partial/\partial\dot{q_j}\) to \(\boldsymbol{v}_i\) as follows

\[\frac{\partial\boldsymbol{v}_i}{\partial\dot{q_j}} = \sum\limits_h \frac{\partial^2\boldsymbol{r}_i}{\partial\dot{q_j}\partial q_h}\dot{q_h}+\sum\limits_h\frac{\partial \boldsymbol{r}_i}{\partial q_h}\frac{\partial\dot{q}_h}{\partial \dot{q_j}} + \frac{\partial^2\boldsymbol{r}_i}{\partial\dot{q_j}\partial t}\]

Let us change the order of the partial derivatives in the first and third terms.

\[\frac{\partial\boldsymbol{v}_i}{\partial\dot{q_j}} = \sum\limits_h \frac{\partial}{\partial q_h}\left(\frac{\partial\boldsymbol{r}_i}{\partial\dot{q_j}}\right)\dot{q_h}+\sum\limits_h\frac{\partial \boldsymbol{r}_i}{\partial q_h}\frac{\partial\dot{q}_h}{\partial \dot{q_j}} + \frac{\partial}{\partial t}\left(\frac{\partial\boldsymbol{r}_i}{\partial\dot{q_j}}\right)\]

But \(\boldsymbol{r}_i\) does not depend on \(\dot{q_j}\). So the first and third terms vanish and we have

\[\frac{\partial\boldsymbol{v}_i}{\partial\dot{q_j}} = \sum\limits_h\frac{\partial \boldsymbol{r}_i}{\partial q_h}\frac{\partial\dot{q}_h}{\partial \dot{q_j}}\]

The generalized position coordinates \(q_h\) are independent and so are the generalized velocities \(\dot{q}_h\), therefore \(\partial\dot{q}_h/\partial \dot{q_j}=0\) if \(h\neq j\). Thus, the only non-zero term occurs when \(h=j\).

\[\frac{\partial\boldsymbol{v}_i}{\partial\dot{q_j}} = \frac{\partial\boldsymbol{r}_i}{\partial q_j}\]

This is sometimes referred to as the cancellation of dots. Similarly, we can apply the partial derivative operator \(\partial/\partial q_j\) to \(\boldsymbol{v}_i\) as follows

\[\frac{\partial\boldsymbol{v}_i}{\partial q_j} = \sum\limits_h \frac{\partial^2\boldsymbol{r}_i}{\partial q_j\partial q_h}\dot{q_h} +\sum\limits_h \frac{\partial\boldsymbol{r}_i}{\partial q_h}\frac{\partial \dot{q_h}}{\partial q_j} + \frac{\partial^2\boldsymbol{r}_i}{\partial q_j\partial t}\]

The generalized position coordinates \(q_h\) and the generalized velocities \(\dot{q}_h\) are independent¹ of each other so the middle term vanishes. As before, we change the order of the partial derivatives in the first and third term.

\[\frac{\partial\boldsymbol{v}_i}{\partial q_j} = \sum\limits_h \frac{\partial}{\partial q_h}\left(\frac{\partial\boldsymbol{r}_i}{\partial q_j}\right)\dot{q_h}+ \frac{\partial}{\partial t}\left(\frac{\partial\boldsymbol{r}_i}{\partial q_j}\right)\]

Let us compute another term without motivating any intuition. We compute the total derivative of \(\partial\boldsymbol{r}_i/\partial q_j\) just like we did for \(\boldsymbol{r}_i\) previously because both are functions of \(q_j\) and \(t\).

\[\frac{d}{d t}\left(\frac{\partial\boldsymbol{r}_i}{\partial q_j}\right) = \sum\limits_h \frac{\partial}{\partial q_h}\left(\frac{\partial\boldsymbol{r}_i}{\partial q_j}\right)\dot{q_h}+ \frac{\partial}{\partial t}\left(\frac{\partial\boldsymbol{r}_i}{\partial q_j}\right)\]

Comparing the two equations above shows that

\[\frac{d}{d t}\left(\frac{\partial\boldsymbol{r}_i}{\partial q_j}\right)=\frac{\partial\boldsymbol{v}_i}{\partial q_j}\]

Rest shall be continued in a later article. Let us again end by quoting Koestler.

\(``\)Never since the Thirty Years' War has the Church oppressed freedom of thought and expression to an extent comparable to the terror based on the "scientific" ideologies of Nazi Germany or Soviet Russia. The contemporary divorce between faith and reason is not the result of a contest for power or for intellectual monopoly, but of a progressive estrangement without hostility or drama, and therefore all the more deadly.\("\)

\(``\)As for Newton, who was a better scientist and hence a more muddled metaphysician than Galileo or Descartes, he assigned to God a two-fold function as Creator of the universal clockwork, and as its Supervisor for maintenance and repair.\("\)

\(``\)Freed from mystical ballast, science could sail ahead at breathtaking speed to its conquest of new lands beyond every dream. Within two centuries it transformed the mental outlook of homo sapiens and transformed the face of his planet. But the price paid was proportionate: it carried the species to the brink of physical self-destruction, and into an equally unprecedented spiritual impasse.\("\)

\(``\)It is appropriate that this survey of man's ideas about the universe should end with Newton for, in spite of more than two centuries that have passed since his death, our vision of the world is by and large still Newtonian.\("\)

\(``\)What he achieved was rather like an explosion in reverse. When a projectile blows up, its shiny, smooth, symmetrical body is shattered into jagged, irregular fragments. Newton found fragments and made them fly together into a simple, seamless, compact body, so simple that it appears as self-evident, so compact that any grammar-schoolboy can handle it.\("\)

It is now impossible to resist the temptation of citing the exemplary book by Subrahmanyan Chandrasekhar.²

References