Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/20

 is subject are such that certain functions of the coordinates cannot exceed certain limits, either constant or variable with the time. If certain values of $$\delta \ddot{x}, \delta \ddot{y}, \delta \ddot{z}$$ (with unvaried values of $$x, y, z$$, and $$\dot{x}, \dot{y}, \dot{z}$$) are simultaneously possible at a given instant, equal or proportional values with the same signs must be possible for $$\delta x, \delta y, \delta z$$ immediately after the instant considered, and must satisfy formula (1), and therefore (6), in connection with the values of $$\ddot{x}, \ddot{y}, \ddot{z}, X, Y, Z$$ immediately after that instant. The values of $$\ddot{x}, \ddot{y}, \ddot{z}$$, thus determined, are of course the very quantities which we wish to obtain, since the acceleration of a point at a given instant does not denote anything different from its acceleration immediately after that instant.

For an example of a somewhat different class of cases, we may suppose that in a system, otherwise free, $$\dot{x}$$ cannot have a negative value. Such a condition does not seem to affect the possible values of $$\delta x$$, as naturally interpreted in a dynamical problem. Yet, if we should regard the value of $$\delta x$$ in (7) as arbitrary, we should obtain which might be erroneous. But if we regard $$\delta x$$ as expressing a velocity of which the system, if at rest, would be capable (which is not a natural signification of the expression), we should have $$\delta x \geqq 0$$, which, with (7), gives This is not incorrect, but it leaves the acceleration undetermined. If we should regard $$\delta x$$ as denoting such a variation of the velocity as is possible for the system when it has its given velocity (this also is not a natural signification of the expression), formula (7) would give the correct value of $$\ddot{x}$$ except when $$\dot{x} = 0$$. In this case (which cannot be regarded as exceptional in a problem of this kind), we should have $$\delta x \geqq 0$$, which will leave $$\ddot{x}$$ undetermined, as before.

The application of formula (6), in problems of this kind, presents no difficulty. From the condition   which is the only limitation on the value of $$\delta \ddot{x}$$. With this condition, we deduce from (6) that either