Problem with Lagrange's Equations

[Erland]

I have problem with Lagrange's equations and their derivation, the way it is presented in Goldstein's "Classical Mechanics". I have never seen this problem mentioned anywhere, so I wonder if I am the only one who see this problem.

To see the problem, consider a simple case where the motion of a system can be described by just one generalized coordinate $q$, plus time explicitly. Let $x$ be a cartesian coordinate of one particle of the system. Then $x=x(q,t)$.
Then, $$\dot x=\frac{\partial x}{\partial q}\dot q+ \frac{\partial x}{\partial t}.$$
(In Goldstein we have several generalized coordinates, but the reasoning is the same.)

Now, reasoning as Goldstein, we could without problem differentiate $\dot x$ wrt $\dot q$, obtaining $\partial \dot x / \partial \dot q =\partial x/\partial q$, and also plug in $\dot q$ in the expression for the total kinetic energy $T$ and the Lagrangian $L=T-V$, and then without problem differentiate $L$ wrt $\dot q$, obtaning Lagrange's Equations (one for each generalized coordinate):
$$\frac d {dt}\frac{\partial L}{\partial\dot q}-\frac{\partial L}{\partial q}=0.$$
(Differentiating $L$ wrt $q$ seems to be no problem, either.)

But how can we differentiate $\dot x$ (and then $L$) wrt $\dot q$? In the formula we obatined for $\dot x$, we have only one indepentent variable: $t$, and therefore it shouldn't make sense to differentiate wrt any other variable.

To be able to differentiate wrt $\dot q$, we must actually view $\dot x$ (and similarly, $L$) as a composite function, $\dot x= f(q,\dot q,t)$, where $f(u_1,u_2,u_3)=g_1(u_1,u_3)u_2+g_2(u_1,u_3)$ and $g$ is the function which gives the coordinate transformation: $x=x(q,t)=g(q,t)$, and $f$ and $g$ can be differentiated wrt any of their arguments. ($g_1$ and $g_2$ are the partial derivatives of $g$.)
The expression $\partial \dot x /\partial \dot q$ will then make sense if we recognize it as the same as $f_2(q,\dot q,t)$, and then it will be equal to $\partial x/\partial q$, just as it should be.

However, this will be consistent only if the choice of the function $f$ is unique. If there was another function $h(u_1,u_2,u_3)\ne f(u_1,u_2,u_3)$, which also satisfies $\dot x = h(q,\dot q,t)$, but for which $h_2(q,\dot q,t)\ne f_2(q,\dot q,t)$, then $\partial \dot x/\partial \dot q$ cannot be uniquely defined (and neither can $\partial L/\partial \dot q$).

This problem is not discussed in Goldstein, and I have not seen it mentioned anywhere (although I did not search very much). Am I really the only person who see this problem?

Fortunately, the problem can be solved. It turns out that the function $f$ above must be unique. The reason is that the equation $\dot x= f(q,\dot q,t)$ must hold for all possible paths $q(t)$, and, for every triplet of values $(u_1,u_2.u_3)$, it is always possible to find a path $q(t)$ such that $q(t)=u_1$ and $q'(t)=u_2$, for $t=x_3$. This implies that $f$ is uniquely determined by the requirement $\dot x= f(q,\dot q,t)$, it must be the function given above. This argument can be generalized to any number of generalized coordinates, and it also follows that $\partial L/\partial \dot q$ will be meaningful and unique.

But it is not trivial to find and prove this. I was very confused when I read Goldstein the first time, and I wondered how on Earth one could differentiate wrt $\dot q$. It took a while until I found the "proof" above. I think it is the author's job to do this, not the reader's.

 

[Bill_K]

Erland, The reason you're confused is that you've made a crucial change in the notation. The aim is to describe the transformation between Cartesian coordinates and generalized coordinates. To do this you need to express x and v in terms of q and q^·. The transformation of the position is given by x(q, t). The transformation of the velocity is v(q^·, q, t):

v = (∂x/∂q) q^· + ∂x/∂t

The mistake you've made is calling the left hand side x^·, which it is not.

 

[Erland]

Erland, The reason you're confused is that you've made a crucial change in the notation. The aim is to describe the transformation between Cartesian coordinates and generalized coordinates. To do this you need to express x and v in terms of q and q^·. The transformation of the position is given by x(q, t). The transformation of the velocity is v(q^·, q, t):

v = (∂x/∂q) q^· + ∂x/∂t

The mistake you've made is calling the left hand side x^·, which it is not.

But $v=\dot x$, by definition (at least according to Goldstein, 3rd ed, p. 18, eq. 1.46), so I don't understand what difference it makes to use v instead.

 

[Bill_K]

The difference is the functional dependence. In the Lagrangian formulation, position and velocity are regarded as independent variables. x(q, t) and x^· are functions of two variables q and t, and therefore do not depend on q^·. But v(q^·, q, t) is a function of three variables and does depend on q^·.

Similarly, when you get to the Lagrangian L(q, q^·, t) one could argue it's all just a function of one variable q(t). But the functional form is crucial. One must treat q^· as an independent variable and L as a function of all three variables.

 

[Erland]

OK, but then the question arises: what does the functional dependence $v=v(q, \dot q, t)$ look like? The only way to determine this is to use the condition $v(q(t),\dot q(t),t)=\dot x(t)$. But then, what if there is a different function $w=(q, \dot q,t)$ which also satisfies $w(q(t),\dot q(t),t)=\dot x(t)$, but for which
$\partial v/ \partial \dot q\ne \partial w/\partial\dot q$. Whch of the two latter quantities is then the correct one to use? There seems to be no way to decide this.
It is therefore important to prove that $v=v(q, \dot q, t)$ is uniquely defined by the condition $v(q(t),\dot q(t),t)=\dot x(t)$. I do that in my first post, but I think Goldstein (and other textbook writers) should do this job.