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The problem is from Classical mechanics by Goldstein, 2nd Edition.

When two billiard balls collide the instantaneous forces between them are very large but act only in an infinitesimal time ${\Delta}t$, in such a manner that the quantity $$\int_{{\Delta}t}Fdt$$ remains finite, the integral is known as the impulse of the force. Show that $$\bigg(\frac{\partial{L}}{\partial{\dot{q_j}}}\bigg)_f-\bigg(\frac{\partial{L}}{\partial{\dot{q_j}}}\bigg)_i=S_j.$$

So I started with Euler Lagrange equation $$\frac{d}{dt}\bigg(\frac{\partial{L}}{\partial{\dot{q}}}\bigg)-\bigg(\frac{\partial{L}}{\partial{q}}\bigg)=Q_j$$ where $Q$ is the generalised force not derivable from the potential, $$d\bigg(\frac{\partial{L}}{\partial{\dot{q_j}}}\bigg)-\bigg(\frac{\partial{L}}{\partial{q_j}}\bigg)dt=Q_jdt$$

Integrating boths sides with difference of limits is infinitesimal ${\Delta}t$,

$$\int_{\Delta{t}}{d}\bigg(\frac{\partial{L}}{\partial{\dot{q_j}}}\bigg)-{\int_{\Delta{t}}}\bigg(\frac{\partial{L}}{\partial{q_j}}\bigg)dt=\int_{\Delta{t}}Q_jdt$$

the first integral on the LHS is $$\bigg(\frac{\partial{L}}{\partial{\dot{q_j}}}\bigg)_f-\bigg(\frac{\partial{L}}{\partial{\dot{q_j}}}\bigg)_i$$

The integral on RHS is impulse of the force, the second integral on the LHS is zero because the difference of limit is infinitesmial. Is my reasoning correct? I am able to show the above relation but I dont know whether my reasoning is correct.

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  • $\begingroup$ What you have done is absolutely right. But can you give reasons for the vanishing of LHS 2nd integral? $\endgroup$ – Icchyamoy Aug 31 '17 at 18:37
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This reasoning is correct, but it is also necessary to prove that the LHS 2nd integral is zero.

If the primitive $f$ of the partial derivative $ \frac{\partial L}{\partial q} $, i.e., $ \frac{df}{dt} = \frac{\partial L}{\partial q} $ did not satisfy $f_{i}=f_{f}$, then the LHS 2nd integral would not be zero.

Since the impulsive forces are supposed to produce only a non continuous change on the generalized velocities in the infinitesimal time $\Delta t$, but not on the generalized coordinates, then $\frac{\partial L}{\partial q}$ remains continuous if the non-impulsive forces are only space dependent.

This is the case in this particular example of two billiard balls in a x-y plane. The $x_{i}$ and $y_{i}$ coordinates are not constrained and they are the generalized coordinates $q_{i}$ also. So the relation $$ \frac{\partial L}{\partial q_{i}} =\frac{\partial}{\partial q_{i}} (T - V) = - \frac{\partial V}{\partial q_{i}} = \tilde{F}_{i} \,\, ,$$

is valid, where $\tilde{F}_{i}$ is the space dependent non-impulsive force.

Note that the billiard balls are not irrelevant to the problem. If the balls were subjected to a velocity dependent force, or the kinetic energy depended also on the space coordinates $q_{i}$, then this result would not be valid as the LHS 2nd integral could be non vanishing. For instance, this result is not valid for two colliding pendulums, because the kinetic energy $T = m/2 (\dot{r}^{2}+r^{2}\dot{\theta}^{2})$ depends explicitly on the generalized coordinate $r$. Physically, this means that just after the collision a non continuous change occurs on the tension on the massless rod which supports the mass.

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