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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\dot{x}^{i}}{\partial\dot{q}^{j}}=\frac{\partial 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?

J.G.
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Steven Thomas Hatton
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    Possible duplicates: http://physics.stackexchange.com/q/885/2451 , http://physics.stackexchange.com/q/419/2451 , http://physics.stackexchange.com/q/233073/2451 , http://physics.stackexchange.com/q/285250/2451 and links therein. – Qmechanic Oct 28 '16 at 20:42
  • 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." – Steven Thomas Hatton Oct 29 '16 at 15:49
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    One way to justify the title eq. (4) is to assume that all the constraints are holonomic. – Qmechanic Oct 29 '16 at 16:14
  • 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. – Steven Thomas Hatton Nov 01 '16 at 17:14
  • 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 $\dot{q}{j}$. $$x{1}=r\cos[\theta]=q_{1}\cos[q_{2}]$$ $$x_{2}=r\sin[\theta]=q_{1}\sin[q_{2}]$$ $$q_{1}=r$$ $$q_{2}=\theta=\frac{1}{2}\alpha t^{2}$$ $$\dot{q}{1}=0$$ $$\dot{q}{2}=\omega=\alpha t$$ $$t=\frac{\omega}{\alpha}=\frac{\dot{q}{2}}{\alpha}$$ $$q{2}=\theta=\frac{1}{2\alpha}\dot{q}{2}^{2}$$ $$x{1}=q_{1}\cos[\frac{1}{2\alpha}\dot{q}{2}^{2}]$$ $$x{2}=q_{1}\sin[\frac{1}{2\alpha}\dot{q}_{2}^{2}]$$ – Steven Thomas Hatton Nov 02 '16 at 18:32
  • I posted the above comment as a counter example which might occur to the many who struggle with this alternative way of modeling the physical world. The sin in the example is that I added a "constraint" on velocity. – Steven Thomas Hatton Nov 02 '16 at 18:37
  • The 4th equation in the example is the culprit. As a definition of a generalized coordinate, it makes no sense. As a solution to an EOM, it makes sense. The title eq. (4) is understood without the use of EOMs.(The acronym EOM (Equations Of Motion) here refer to a coupled set of differential equations that governs time evolution, i.e. Newton's 2nd law.) – Qmechanic Nov 02 '16 at 18:50
  • I'm not sure the best way to ask this. If I posted the question as a new topic, it would require considerable redundancy with this topic. So here it is: does $\frac{\partial\dot{q}^{j}}{\partial x^{i}}=0$ follow from the assumption that our "equation of motion" is a functional rather than an ordinary function? By functional I intend: $Y(x)=\varepsilon \eta (x)+y(x)$. – Steven Thomas Hatton Nov 22 '16 at 16:23
  • In your last comment, the meaning of the content after the word "our" is not clear. It is also a red herring. – Qmechanic Nov 22 '16 at 17:53
  • What I meant by "equation of motion" is the curve parameterized by time describing the path (career, trajectory, world line, etc.,) of the configuration point. My example above could describe the behavior of a real physical system. It seems the same symbols and words have different meanings when used in different contexts. I don't see how the introduction of holonomic constraints induces a mapping between velocity as an invertable function of position and velocity independent of position. – Steven Thomas Hatton Dec 01 '16 at 17:29
  • A parametrized curve (with time as parameter) is conventionally not referred to as an EOM. – Qmechanic Dec 01 '16 at 19:17
  • What we call those equations is not the issue. What is at issue is what is mean by such terms as "the configuration point" and "the velocity of the configuration point". I assume, with reasonable justification, that the actual system under investigation is accurately described by a unique time-parameterized curve. In some cases (at least), it is possible to express the velocity as a function of position. Now, if we construct a mathematical space where velocity and position are decoupled and fiddle around with hypothetical paths we might arrive at that same parameterized curve. – Steven Thomas Hatton Dec 02 '16 at 18:03
  • I stand by my original confusion. Hamill makes this assertion before introducing the notion of constraints. "But $\frac{\partial x_i}{\partial \dot{q}_j}=0$ because $x$ does not depend on the generalized velocity $\dot{q}_j$."

    Hamill, Patrick. A Student's Guide to Lagrangians and Hamiltonians (p. 8). Cambridge University Press. Kindle Edition.

     – Steven Thomas Hatton Dec 23 '16 at 16:52
  • Hamill write on p. 7 that $x^i = x^i (q^1 , q^2 , q^3 , t).$ Be aware that Hamill uses $x$ in two meanings: As a variable and as a function. Hamill's sentence on p. 8 should be read as follows: "But $\frac{\partial x^i}{\partial \dot{q}^j}=0$ because the function $x$ does not depend on the generalized velocity $\dot{q}^j$." – Qmechanic Dec 24 '16 at 12:33
  • While I agree that it is somewhat non-standard, I note the following from Introduction to the Theory of Relativity, by Peter Gabriel Bergmann, page 88 (a and b are constants):

    "The equations of motion before the collision are, thus

    $\underset{1}{x}=at=-\underset{2}{x}$

    $\underset{1}{y}=bt=-\underset{2}{y}$

    $\underset{1}{z}=\underset{2}{z}$ "

     – Steven Thomas Hatton Dec 29 '16 at 16:33
  • That quote is indeed non-standard. – Qmechanic Dec 29 '16 at 17:51

2 Answers2

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I'm pretty sure this is the same kind of problem as that which led to covariant differentiation. In generalized coordinates our coordinate transformations are position dependent. Assuming our constraints are time-independent, we can think of the (differential) coordinate transformations as persisting in time. At each point along the configuration path, we use the pre-existing coordinate differential (derivative matrix) at that point. The tangent vectors to the coordinate curves form a coordinate basis of the tangent plane at that point. The differential matrix of the coordinate transformation maps infinitesimal displacements linearly to the "embedding" rectangular Cartesian system.

At the point of evaluation, the velocity vector lies in the tangent plane. From the perspective of the Cartesian system, the configuration point is instantaneously moving in a straight line lying in the tangent plane. It is true that the tangent plane at the configuration point is changing as the point moves, but this amounts to a second order derivative since it is the directional derivative of the coordinate differential with respect to velocity. Since velocity is a first order derivative, the effect of the changing tangent plane vanishes when taking the limit as $\Delta t$ goes to zero.

And now I will stop waving my hands vigorously.

Well, not just yet.

There has to be more to this, since we have the situation where

$$\frac{d}{dt}\left[\frac{\partial x^{i}}{\partial q^{j}}\right]=\frac{\partial^{2}x^{i}}{\partial q^{k}\partial q^{j}}\dot{q}^{k}=\frac{\partial}{\partial q^{j}}\left[\frac{\partial x^{i}}{\partial q^{k}}\dot{q}^{k}\right]=\frac{\partial\dot{x}^{i}}{\partial q^{j}}.$$

This means we are varying the $q^{j}$ while holding the $\dot{q}^{k}$ fixed, and finding the amount by which the $\dot{x}^{i}$ vary as a result. That is, we are “hypothetically” varying the path and asking what the consequences of doing so are. For example, we might ask how $\dot{x}$ changes if angular velocity is held fixed while the radius is changed.

For a real mythical point mass, all of the dynamic and kinematic values are determined at each point in the path. All we can ask is how these quantities change along the path. In analytical dynamics we are investigating vector fields formed over a congruence of possible paths. This break from the mythical reality of a lone point mass happens when we decouple generalized position and generalized velocity.

I believe this is an extremely important distinction to understand if we are to properly interpret the statements of quantum mechanics.

Steven Thomas Hatton
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The condition in the title follows from assuming that the constraints are holonomic. That is, the conditions of constraint can be expressed as equations connecting the coordinates of the particles (and possibly the time) having the form $$f(\mathbf{r_1}, \mathbf{r_2}, \mathbf{r_3}, ... ,t)=0$$ If the constrains contain time as an explicit variable they are known as "rheonomous". If they are not explicitly dependent on time then they are "scleronomous." Constraint equations make the original coordinates dependent. If the $k$ constraints are holonomic and we have $N$ particles, we may use the constraint equations to introduce independent coordinates: $$\mathbf{r_1} = \mathbf{r_1}(q_1, q_2, ..., q_{3N-k}, t)$$ $$\vdots$$ $$\mathbf{r}_N = \mathbf{r}_N(q_1, q_2, ..., q_{3N-k}, t)$$ Putting aside the question of the practicality of inverting these expressions we may invert; $$q_1 = q_1(\mathbf{r_1}, \mathbf{r_2}, \mathbf{r_3}, ... ,t)$$ $$\vdots$$ $$q_k = q_k(\mathbf{r_1}, \mathbf{r_2}, \mathbf{r_3}, ... ,t)$$ In other words, the condition $\frac{\partial r^i}{\partial \dot{q}^j} = 0$ and $\frac{\partial q^i}{\partial \dot{r}^j} = 0$ are just assumptions/follow by construction. (Reference: Goldstein)

It is not clear why (5) holding and (12) holding is a contradiction. What is the issue?

Jbag1212
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