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.