The basic equation of Lagrange is given by, $$\frac{\mathrm d}{\mathrm dt} \frac{\partial L}{\partial \dot{q_j}} - \frac{\partial L}{\partial q_j} = Q_j \tag{1}$$

where $T$ is the kinetic energy, $V$ is the potential energy, $L = T-V$. But at any given instant $$T + V = \text{Constant}\tag{2}$$ (Conservation of energy)

I understand that the criteria for Lagrange application is that the coordinates that define the system are to be,

  • Independent

  • Complete

  • Holonomic

and for a mechanical system $V \neq f(\dot{q_j})$.

Thus, for mechanical systems, the above relation reduces to

$$\frac{\mathrm d}{\mathrm dt} \frac{\partial T}{\partial \dot{q_j}} - \dfrac{\partial T}{\partial q_j} + \frac{\partial V}{q_j} = Q_j . \tag{3}$$

Thus, as the coordinates are to be independent,

The effect of a force in any particular direction (component) will not give a displacement in the other directions, thus the energies in that particular direction can be expressed by isolated equations that are independent of each other (This is what I understood by the coordinates being independent).

Then, by applying

$$V = Constant - T = C - T\tag{4}$$

the above equation would just become

$$\frac{\mathrm d}{\mathrm dt} \frac{\partial T}{\partial \dot{q_j}} - \frac{\partial T}{\partial q_j} + \frac{\partial (C - T)}{\partial q_j} = Q_j \tag{5}$$

and eventually boil down to

$$\frac{\mathrm d}{\mathrm dt} \frac{\partial T}{\partial \dot{q_j}} - 2\frac{\partial T}{\partial q_j} = Q_j \tag{6}$$

as the derivative of a constant is always zero.

I am new to dynamics and so like to know where I am wrong conceptually.

  • $\begingroup$ If $Q\neq 0$ the energy function $T + V$ is not constant, so $V \neq C - T$ . $\endgroup$ – Futurologist Oct 15 '18 at 16:56
  • $\begingroup$ So if the system is free, then is the above expression valid? $\endgroup$ – Raptor Oct 15 '18 at 17:04
  • $\begingroup$ No it is not. The equality $ V = C - T$ does not hold for any $(q, \dot{q})$, i.e. $$\{ \, (q,\dot{q}) \, : \,\, T(q,\dot{q}) + V(q) = C\} \neq T\mathbb{R}^n$$ It is an equation and not an identity. $\endgroup$ – Futurologist Oct 15 '18 at 17:51
  • $\begingroup$ But, if there are no non-conservative forces acting on the body. Doesn't that mean that the energy is conserved in an ideal system. $\endgroup$ – Raptor Oct 15 '18 at 17:57
  • $\begingroup$ Energy is conserved means that for any solution $q(t)$ of the Lagrange equations, $$T(q(t),\dot{q}(t)) + U(q(t)) = C$$ for all $t$. Not for all pairs $(q,v)$ $\endgroup$ – Futurologist Oct 15 '18 at 18:14

Think of the difference between the expression $\frac{1}{2}\,\dot{q}^2 - \cos{q} = 1$, which is a special case of the pendulum's total energy, versus ${q}^2\sin^2(\dot{q}) + {q}^2 \cos^2(\dot{q}) + (1- {q})(1+ {q}) \equiv 1$. In the first case it is not true that $$\cos{q} = \frac{1}{2} \, \dot{q}^2 - 1$$ for all possible entries $(q,\dot{q})$ while $${q}^2\sin^2(\dot{q})+ {q}^2 \cos^2(\dot{q}) \equiv 1- (1- {q})(1+ {q}) $$ always, for all entries $(q,\dot{q})$.

If you check the derivatives for the first equation, $\frac{\partial }{\partial q} \,\cos{q} = - \sin{q}$ but $\frac{\partial }{\partial q} \, \left(\ \frac{1}{2} \, \dot{q}^2 - 1 \right) = 0$ so they are not equal. For the second identity, the equality will holds.


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