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My question has to do with the justification for the assertion that $\partial x^{i}/\partial\dot{q}^{j}=0$ when working with generalized coordinates. The following is an example of where this appears, and why it is confusing to me. Summation on like, upper and lower index pairs is assumed.

We shall rewrite D'Alembert's following equation.

\begin{equation} \sum_{i=1}^{3N}\left(F^{i}-m^{i}\ddot{x}^{i}\right)\delta x^{i}=0 \tag{1}\label{1}\end{equation} The next two equations follow. The second requires our coordinate transformations to be time-independent.

\begin{equation} \delta x^{i}=\frac{\partial x^{i}}{\partial q^{j}}\delta q^{j} \tag{2}\label{2}\end{equation}

\begin{equation} \dot{x}^{i}=\frac{\partial x^{i}}{\partial q^{j}}\dot{q}^{j} \tag{3}\label{3}\end{equation} This next equivalence is a commonly promulgated assertion which I have never been comfortable with.

\begin{equation} \frac{\partial x^{i}}{\partial\dot{q}^{j}}=0 \tag{4}\label{4}\end{equation} It says the $x^{i}$ are independent of the $\dot{q}^{j}$. That implies that

\begin{equation} \frac{\partial\dot{q}^{j}}{\partial x^{i}}=0 \tag{5}\label{5}\end{equation} also holds.

We now make use of \eqref{4}

\begin{equation} \frac{\partial\dot{x}^{i}}{\partial\dot{q}^{j}}=\frac{\partial}{\partial\dot{q}^{j}}\left(\frac{\partial x^{i}}{\partial q^{k}}\dot{q}^{k}\right)=\frac{\partial^{2}x^{i}}{\partial\dot{q}^{j}\partial q^{k}}\dot{q}^{k}+\frac{\partial x^{i}}{\partial q^{k}}\frac{\partial\dot{q}^{k}}{\partial\dot{q}^{j}}=0+\frac{\partial x^{i}}{\partial q^{k}}\delta_{j}^{k}=\frac{\partial x^{i}}{\partial q^{j}} \tag{6}\label{6}\end{equation} to arrive at this useful relationship.

\begin{equation} \frac{\partial^{j}\dot{x}^{i}}{\partial\dot{q}^{j}}=\frac{\partial^{j}x^{i}}{\partial q^{j}} \tag{7}\label{7}\end{equation}

We replace $\delta x^{i}$ in D'Alembert's equation \eqref{1} with the hight-hand side of \eqref{2}

\begin{equation} \sum_{i=1}^{3N}\left(F^{i}-m^{i}\ddot{x}^{i}\right)\frac{\partial x^{i}}{\partial q^{j}}\delta q^{j}=0 \tag{8}\label{8}\end{equation}

Now we shall find an alternative form for the following terms which appear in \eqref{8}.

\begin{equation} m^{i}\ddot{x}^{i}\frac{\partial x^{i}}{\partial q^{j}}\delta q^{j} \tag{9}\label{9}\end{equation}

Use the product rule for differentiation.

\begin{equation} \ddot{x}^{i}\frac{\partial x^{i}}{\partial q^{j}}=\frac{d}{dt}\left[\dot{x}^{i}\frac{\partial x^{i}}{\partial q^{j}}\right]-\dot{x}^{i}\frac{d}{dt}\left[\frac{\partial x^{i}}{\partial q^{j}}\right] \tag{10}\label{10}\end{equation}

Using \eqref{7} we find the following form for the first term on the right-hand side of \eqref{10}.

\begin{equation} \frac{d}{dt}\left[\dot{x}^{i}\frac{\partial x^{i}}{\partial q^{j}}\right]=\frac{d}{dt}\left[\dot{x}^{i}\frac{\partial\dot{x}^{i}}{\partial\dot{q}^{j}}\right]=\frac{d}{dt}\left[\frac{1}{2}\frac{\partial\left(\dot{x}^{i}\right)^{2}}{\partial\dot{q}^{j}}\right] \tag{11}\label{11}\end{equation}

Use the chain rule and reverse the order of mixed partial differentiation.

\begin{equation} \frac{d}{dt}\left[\frac{\partial x^{i}}{\partial q^{j}}\right]=\frac{\partial^{2}x^{i}}{\partial q^{k}\partial q^{j}}\dot{q}^{k}=\frac{\partial}{\partial q^{j}}\left[\frac{\partial x^{i}}{\partial q^{k}}\dot{q}^{k}\right]=\frac{\partial\dot{x}^{i}}{\partial q^{j}} \tag{12}\label{12}\end{equation}

This provides an equivalent form for the second term on the right-hand side of \eqref{10}.

\begin{equation} \dot{x}^{i}\frac{d}{dt}\left[\frac{\partial x^{i}}{\partial q^{j}}\right]=\frac{\partial}{\partial q^{j}}\left[\frac{1}{2}\left(\dot{x}^{i}\right)^{2}\right] \tag{13}\label{13}\end{equation}

We can now express \eqref{9} in this very suggestive form.

\begin{equation} m^{i}\ddot{x}^{i}\delta x^{i}=\left(\frac{d}{dt}\frac{\partial}{\partial\dot{q}^{j}}\left[\frac{1}{2}m^{i}\left(\dot{x}^{i}\right)^{2}\right]-\frac{\partial}{\partial q^{j}}\left[\frac{1}{2}m^{i}\left(\dot{x}^{i}\right)^{2}\right]\right)\delta q^{j} \tag{14}\label{14}\end{equation}

If we compare \eqref{5} with \eqref{12} we see that they are symmetrical opposites of one another. The coordinate transformations to and from generalized coordinates are typically given as $q^{i}=q^{i}[x^{1},x^{2},\ldots,x^{n}]$ and $x^{i}=x^{i}[q^{1},q^{2},\ldots,q^{m}]$. The $x^{i}$ are rectangular Cartesian coordinates, and the $q^{i}$may be any suitable coordinate system. There is nothing that says the $q^{i}$ are not also rectangular Cartesian coordinates. Even the identity transformation $q^{i}=x^{i}$ would satisfy this definition. How is it possible that \eqref{5} and \eqref{12} could both hold if the coordinate systems are both rectangular Cartesian? Is one coordinate system given special status which the others lack?

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  • $\begingroup$ I think the short answer to my question is "Yes. We really did tear apart your concept of reality by asserting (5) as a definition." The long answer is "Go fish V.I. Arnold's Ordinary Differential Equations out of your storage unit." $\endgroup$ – Steven Thomas Hatton Oct 29 '16 at 15:49
  • $\begingroup$ One way to justify the title eq. (4) is to assume that all the constraints are holonomic. $\endgroup$ – Qmechanic Oct 29 '16 at 16:14
  • $\begingroup$ Now that I think about it, I was mistaken in believing that $\frac{\partial x^{i}}{\partial\dot{q}^{j}}=0$ implies $\frac{\partial\dot{q}^{j}}{\partial x^{i}}=0$, because $0\neq1/0$. My deeper failure to fully understand how this works has to do with the concept of dependent and independent variables. I shall soldier on through the fog in pursuit of the Light. $\endgroup$ – Steven Thomas Hatton Nov 1 '16 at 17:14
  • $\begingroup$ A particle confined to a circle, its speed increasing constantly. The geometric constraints are $x_{3}=0$ and $\sqrt{x_{1}^{2}+x_{2}^{2}}-r=0$. Express $x_{i}$ in terms $\dot{q}_{j}$. $$x_{1}=r\cos[\theta]=q_{1}\cos[q_{2}]$$ $$x_{2}=r\sin[\theta]=q_{1}\sin[q_{2}]$$ $$q_{1}=r$$ $$q_{2}=\theta=\frac{1}{2}\alpha t^{2}$$ $$\dot{q}_{1}=0$$ $$\dot{q}_{2}=\omega=\alpha t$$ $$t=\frac{\omega}{\alpha}=\frac{\dot{q}_{2}}{\alpha}$$ $$q_{2}=\theta=\frac{1}{2\alpha}\dot{q}_{2}^{2}$$ $$x_{1}=q_{1}\cos[\frac{1}{2\alpha}\dot{q}_{2}^{2}]$$ $$x_{2}=q_{1}\sin[\frac{1}{2\alpha}\dot{q}_{2}^{2}]$$ $\endgroup$ – Steven Thomas Hatton Nov 2 '16 at 18:32

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