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

${\ddot {x}}={\frac {X}{m}},$

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

${\ddot {x}}\geqq {\frac {X}{m}}.$

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

${\dot {x}}\geqq 0,$

we obtain, first, if${\dot {x}}=0,$ ${\ddot {x}}\geqq 0,$

then, if${\dot {x}}=0$and${\ddot {x}}=0,$ $\delta {\ddot {x}}\geqq 0,$

which is the only limitation on the value of $\delta {\ddot {x}}$. With this condition, we deduce from (6) that either

${\dot {x}}=0,$${\ddot {x}}=0$and${\ddot {x}}\geqq {\frac {X}{m}};$

or${\ddot {x}}={\frac {X}{m}}\cdot$