Is the Lagrange Equation Valid for All Holonomic Systems?

[Petar015]

Homework Statement

Show that for an arbitrary ideal holonomic system (n degrees of freedom)

<br /> \frac{1}{2} \frac{\partial \ddot T}{\partial\ddot q_j} - \frac{3}{2} \frac{\partial T}{\partial q_j} = Q_j <br />

where T is kinetic energy and q_j generalized coordinates.[/B]

Homework Equations

Lagrange's equation
<br /> \frac{d}{dt} \frac{\partial T}{\partial\dot q_j} - \frac{\partial T}{\partial q_j} = Q_j <br />[/B]

The Attempt at a Solution

We know that T(q_1,...q_n,\dot q_1,...,\dot q_n,t)
The idea is to express \dot T and \ddot T and then plug it into initial equation in order to obtain equivalence with Lagrange's equation.

So we write

\frac {dT}{dt}=\dot T=\frac{\partial T}{\partial \dot q_j} \ddot q_j + \frac{\partial T}{\partial q_j} \dot q_j + \frac{\partial T}{\partial t}

So I figure that I should express \ddot T
in the same manner, but I'm stuck at doing the chain rule for the first 2 terms.
[/B]

 

[Dr.D]

As you say, T = T(q1,q2,...,q1d,q2d,...) so the partial of T with respect to qidd is necessarily zero.

I really don't believe the result that you are trying to prove. I can say with confidence that in almost 60 years of doing dynamics, I have never seen this expression anywhere, and it looks entirely bogus to me. I look forward to whatever light others may bring to this issue. Maybe there is something new under the sun after all!