Methodus inveniendi by Leonhard Euler, translated by Wikisource Additamentum II

Appendix 2

Concerning the motion of particles in a non-resistant medium, determined by a method of maxima and minima

1. Since all natural phenomena obey a certain maximum or minimum law; there is no doubt that some property must be maximized or minimized in the trajectories of particles acted upon by external forces. However, it does not seem easy to determine which property is minimized from metaphysical principles known a priori. Yet if the trajectories can be determined by a direct method, the property being minimized or maximized by these trajectories can be determined, provided that sufficient care is taken. After considering the effects of external forces and the movements they generate, it seems most consistent with experience to assert that the integrated momentum (i.e., the sum of all momenta contained in the particle's movement) is the minimized quantity. This assertion is not sufficiently proven at present; however, if I can show it to be connected with some truth known a priori, it will carry such weight as to utterly vanquish every conceivable doubt. If indeed it's truth can be verified, this assertion will make it easier to investigate the deepest laws of Nature and their final causes, and also easier to identify a firmer rationale for this assertion.

2. Let the mass of a moving particle be $M$, and let its speed be $v$ while being moved over an infinitesimal distance ds. The particle will have a momentum $M v$ that, when multiplied by the distance ds, gives $M v ds$, the momentum of the particle integrated over the distance ds. Now I assert that the true trajectory of the moving particle is the trajectory to be described (from among all possible trajectories connecting the same endpoints) that minimizes $\int M v ds$ or (since M is constant) $\int v ds$. Since the speed $v$ resulting from the external forces can be calculated a posteriori from the trajectory itself, a method of maxima and minima should suffice to determine the trajectory a priori. The minimized integral can be expressed in terms of the momentum (as above), but also in terms of the kinetic energy. For, given an infinitesimal time $dt$ during which the element $ds$ is traversed, we have $ds = v dt$. Hence, $\int M v ds = \int M v^{2} dt$ is minimized, i.e., the true trajectory of a moving particle minimizes the integral over time of its instantaneous kinetic energies. Thus, this minimum principle should appeal both to those who favor momentum for mechanics calculations and to those who favor kinetic energy.

3. For our first example, consider a moving particle free of external forces, which has a constant speed, denoted $b$. By our principle, such a particle describes a trajectory that minimizes $\int b ds$ or $\int ds = s$. Hence, the true path of a free particle has the minimum length of all paths connecting the same endpoints; this path is a straight line, just as the first principles of Mechanics postulate. I do not present this example as evidence for the general principle, since the integral of any function of the constant speed $v=b$ would, upon minimization, produce a straight line. I begin with this simple case merely to illustrate the reasoning.

Figure 26

4. Let us proceed to the case of uniform gravity or, more generally, to the case in which a moving particle is acted upon by a downwards force of constant acceleration $g$. Let the trajectory of the particle under these conditions be AM (Figure 26), let Mm represent the infinitesimal distance $ds$, and let $x$ and $y$ represent the vertical coordinate along AP and the horizontal coordinate along the perpendicular axis PM, respectively. The external force produces an acceleration described by $dv^{2} = 2 g dy$, which integrates to $v^{2} = v_{0}^{2} + 2 g y$. Hence, we seek the trajectory that minimizes the integral $\int ds \sqrt{v_{0}^{2} + 2 g y}$. Defining $p \equiv \frac{dx}{dy}$, we have $ds = dy \sqrt{1 + p^{2}}$; in other words, we seek the trajectory that minimizes $\int dy \sqrt{v_{0}^{2} + 2 g y} \, \sqrt{1 + p^{2}}$. Comparing this expression with our general formula $\int Z dy$, we see that $Z = \sqrt{v_{0}^{2} + 2 g y} \, \sqrt{1 + p^{2}}$ and, given that $dZ \equiv M dx + N dy + P dp$, we obtain $M=0$ and $P = \frac{p\sqrt{v_{0}^{2} + 2 g y}}{\sqrt{1 + p^{2}}}$. Minimization requires that $M dy = dP$ and, since $M=0$ in this case, $dP = 0$ and, thus, $P = \sqrt{C}$, where $C$ is a constant. Hence, we have the differential equation $\sqrt{C} = \frac{p\sqrt{v_{0}^{2} + 2 g y}}{\sqrt{1 + p^{2}}} = \frac{dx\sqrt{v_{0}^{2} + 2 g y}}{ds}$. This may be rearranged to $C dx^{2} + C dy^{2} = dx^{2} \left( v_{0}^{2} + 2 g y \right)$ and separated $dx = \frac{dy \sqrt{C}}{\sqrt{v_{0}^{2} - C + 2 g y}}$. Integration yields the trajectory solution $x = \frac{1}{g} \sqrt{C \left( v_{0}^{2} - C + 2 g y \right)}$.

5. This solution is obviously a parabola, but we should consider more carefully whether it agrees with experience. The initial tangent to the trajectory is clearly horizontal, i.e., $dy=0$ at the point where $v_{0}^{2} - C + g y = 0$. Since the origin (0,0) of the coordinate system may be chosen at will, we set it at that tangent point, (i.e., at the highest point of the trajectory), which is equivalent to setting $C = v_{0}^{2}$. In this coordinate frame, the trajectory solution is $x = v_{0} \sqrt{\frac{2y}{g}}$, which is a parabola. Moreover, since the initial speed is evidently $v_{0}$, the height CA from which the falling body acquires the same speed from the same force $g$ is $\frac{v_{0}^{2}}{2g}$, just as predicted by the direct methods of mechanics.

Figure 27

6. Now let the particle be acted upon by a downwards vertical force that depends on the height CP (Figure 27). We set $y$ equal to the vertical coordinate along CP, and let $F_{y}(y)$ be a downwards vertical force that is a function of $y$. Similarly, let the horizontal coordinate along PM be denoted as $x$ and let Mm represent the infinitesimal distance $ds$. Defining $p \equiv \frac{dx}{dy}$, we have $dv^{2} = 2 F_{y} \, dy$ and, thus, $v^{2} = v_{0}^{2} + \int 2 F_{y} \, dy$. Hence, we seek the trajectory that minimizes the integral $\int dy \sqrt{v_{0}^{2} + \int 2 F_{y} \, dy} \, \sqrt{1 + p^{2}}$. By arguments analogous to those used in paragraph 4, we obtain the trajectory equation $\sqrt{C} = \frac{p\sqrt{v_{0}^{2} + \int 2 F_{y} \, dy}}{\sqrt{1 + p^{2}}}$ or, equivalently, $p = \frac{\sqrt{C}}{\sqrt{v_{0}^{2} - C + \int 2 F_{y} \, dy}}$, which may be simplified to $x = \int \frac{dy\sqrt{C}}{\sqrt{v_{0}^{2} - C + \int 2 F_{y} \, dy}}$. The tangent will be horizontal whenever $\int 2 F_{y} \, dy = C - v_{0}^{2}$. These results agree with the trajectories predicted by direct methods of mechanics.

7. Now let the body be acted upon by two external forces, denoted as $F_{x}$ in the horizontal direction MP and $F_{y}$ in the vertical direction MQ (Figure 27). Let $F_{x}(x)$ be a function of the horizontal distance PM = $x$ and $F_{y}(y)$ be a function of the vertical height MQ = CP = $y$. Defining $p \equiv \frac{dx}{dy}$ as before, we have $dv^{2} = -2 F_{x} dx - 2 F_{y} dy$, which integrates to $v^{2} = v_{0}^{2} - \int 2 F_{x} dx - \int 2 F_{y} dy$. Hence, we seek the trajectory that minimizes the integral $\int dy \sqrt{1 + p^{2}} \sqrt{v_{0}^{2} - \int 2 F_{x} dx - \int 2 F_{y} dy}$. Differentiating the integrand $\sqrt{1 + p^{2}} \sqrt{v_{0}^{2} - \int 2 F_{x} dx - \int 2 F_{y} dy}$ yields

$\frac{-F_{x} dx \sqrt{1 + p^{2}}}{\sqrt{v_{0}^{2} - \int 2 F_{x} dx - \int 2 F_{y} dy}} - \frac{F_{y} dy \sqrt{1 + p^{2}}}{\sqrt{v_{0}^{2} - \int 2 F_{x} dx - \int 2 F_{y} dy}} + \frac{p dp \sqrt{v_{0}^{2} - \int 2 F_{x} dx - \int 2 F_{y} dy}}{\sqrt{1 + p^{2}}}$

Setting $M=\frac{-F_{x} \sqrt{1 + p^{2}}}{\sqrt{v_{0}^{2} - \int 2 F_{x} dx - \int 2 F_{y} dy}}$, and $P=\frac{p \sqrt{v_{0}^{2} - \int 2 F_{x} dx - \int 2 F_{y} dy}}{\sqrt{1 + p^{2}}}$, the minimal trajectory must satisfy $M dy = dP$, i.e.,

$\frac{-F_{x} dy \sqrt{1 + p^{2}}}{\sqrt{v_{0}^{2} - \int 2 F_{x} dx - \int 2 F_{y} dy}} = \frac{dp \sqrt{v_{0}^{2} - \int 2 F_{x} dx - \int 2 F_{y} dy}}{\left( 1 + p^{2} \right) \sqrt{1 + p^{2}}} - \frac{p F_{y} dy - p F_{x} dx}{\sqrt{1 + p^{2}}\sqrt{v_{0}^{2} - \int 2 F_{x} dx - \int 2 F_{y} dy}}$

which may be written as

$\frac{dp \sqrt{v_{0}^{2} - \int 2 F_{x} dx - \int 2 F_{y} dy}}{\left( 1 + p^{2} \right) \sqrt{1 + p^{2}}} = \frac{F_{y} dx - F_{x} dy}{\sqrt{1 + p^{2}}\sqrt{v_{0}^{2} - \int 2 F_{x} dx - \int 2 F_{y} dy}}$

or, more simply,

$\frac{dp}{1 + p^{2}} = \frac{F_{y} dx - F_{x} dy}{v_{0}^{2} - \int 2 F_{x} dx - \int 2 F_{y} dy}$

This trajectory agrees with experience, as shown by replacing $v_{0}^{2} - \int 2 F_{x} dx - \int 2 F_{y} dy$ with $v^{2}$,

$\frac{v^{2} dp}{\left( 1 + p^{2} \right)^{3/2}} = \frac{F_{y} dx - F_{x} dy}{\sqrt{1 + p^{2}}}$

The instantaneous radius of curvature of the trajectory is defined $R \equiv - \frac{\left( 1 + p^{2} \right)^{3/2} dy}{dp}$, yielding the equation $\frac{v^{2}}{R} = \frac{F_{x} dy - F_{y} dx}{ds}$. Here, $\frac{v^{2}}{R}$ is the centripetal acceleration required to produce the trajectory's curvature, whereas $\frac{F_{x} dy - F_{y} dx}{ds}$ is the force (perpendicular to the trajectory) that provides that centripetal acceleration. Their equality is evidence that our method agrees with experience.

8. The resulting equation $\frac{dp}{1 + p^{2}} = \frac{F_{y} dx - F_{x} dy}{v_{0}^{2} - \int 2 F_{x} dx - \int 2 F_{y} dy}$ is generally integrable. Multiplication by $\frac{p \left( v_{0}^{2} - \int 2 F_{x} dx - \int 2 F_{y} dy \right)}{1 + p^{2}}$ yields the equation

$\frac{p dp \left( v_{0}^{2} - \int 2 F_{x} dx - \int 2 F_{y} dy \right)}{\left( 1 + p^{2} \right)^{2}} + \frac{F_{x} dx - p^{2} F_{y} dy}{1 + p^{2}} = 0$

which may be integrated as

$\frac{\int 2 F_{x} dx - p^{2} \int 2 F_{y} dy - v_{0}^{2}}{1 + p^{2}} = C$

or, more simply,

$\int 2 F_{x} dx - p^{2} \int 2 F_{y} dy = v_{0}^{2} + C + C p^{2}$

Hence, $p = \sqrt{\frac{B + \int 2 F_{x} dx}{C + \int 2 F_{y} dy}}$, if we define $B \equiv -v_{0}^{2} - C$. Then, since $p = \frac{dx}{dy}$, we may separate the variables $x$ and $y$

$\int \frac{dx}{\sqrt{B + \int 2 F_{x} dx }} = \int \frac{dy}{\sqrt{C + \int 2 F_{y} dy }}$

in the equation for the desired trajectory. Converting the constants $B$ and $C$ into their negatives, we obtain

$\int \frac{dx}{\sqrt{B - \int 2 F_{x} dx }} = \int \frac{dy}{\sqrt{v_{0}^{2} - \int 2 F_{y} dy }}$

Although the construction of the trajectory is relatively easy in principle, it may be difficult to express that trajectory in algebraic equations. For example, let $F_{x}$ and $F_{y}$ be proportional to the same power $n$ of $x$ and $y$, respectively

$\int \frac{dx}{\sqrt{b^{n} - x^{n}}} = \int \frac{dy}{\sqrt{a^{n} - y^{n}}}$

If $n=1$, this equation describes a parabola whereas, if $n=2$, it describes an ellipse centered on C. (Even in this case, each integration requires trigonometric quadratures.) Similarly, in other cases, neither integration succeeded in producing trajectories described by algebraic functions. Nevertheless, this method for finding these trajectories is still desirable.

Figure 27

9. Now let the particle at M be acted upon by a central force, i.e., one that always lies along the direction MC and is a function $F_{r}(r)$ only of the distance MC = $r$ (Figure 27). As before, let CP = $y$, PM = $x$, and $dx = p dy$; thus, CM = $\sqrt{x^{2} + y^{2}} = r$. Let $F_{r}$ be resolved into perpendicular components along the lines MQ and MP; the force along MQ is $\frac{F_{r} y}{r}$, whereas the force along MP equals $\frac{F_{r} x}{r}$. These forces give rise to an acceleration $dv^{2} = - \frac{2 F_{r} x dx}{r} - \frac{2 F_{r} y dy}{r} = - 2 F_{r} dr$ (because $x dx + y dy = r dr$), from which we have $v^{2} = v_{0}^{2} - \int 2 F_{r} \, dr$. Thus, the expression to be minimized is the integral $\int dy \sqrt{1 + p^{2}} \sqrt{v_{0}^{2} - \int 2 F_{r} \, dr}$. Following our protocol, we must differentiate the integrand $\sqrt{1 + p^{2}} \sqrt{v_{0}^{2} - \int 2 F_{r} \, dr}$, yielding

$-\frac{F_{r} dr \sqrt{1 + p^{2}}}{\sqrt{v_{0}^{2} - \int 2 F_{r} \, dr}} + \frac{p dp \sqrt{v_{0}^{2} - \int 2 F_{r} \, dr}}{\sqrt{1 + p^{2}}}$

Since $dr = \frac{x dx + y dy}{r}$, we may assign $M = \frac{- F_{r} x \sqrt{1 + p^{2}}}{r \sqrt{v_{0}^{2} - \int 2 F_{r} \, dr}}$ and $P = \frac{p \sqrt{v_{0}^{2} - \int 2 F_{r} \, dr}}{\sqrt{1 + p^{2}}}$; using the general law for optimal curves $M dy = dP$, we obtain

$\frac{- F_{r} x dy \sqrt{1 + p^{2}}}{r \sqrt{v_{0}^{2} - \int 2 F_{r} \, dr}} = \frac{dp \sqrt{v_{0}^{2} - \int 2 F_{r} \, dr}}{\left( 1 + p^{2} \right) \sqrt{1 + p^{2}}} - \frac{p F_{r} dr}{\sqrt{1 + p^{2}} \sqrt{v_{0}^{2} - \int 2 F_{r} \, dr}}$

which may be simplified to

$\frac{F_{r} \left( y dx - x dy\right)}{r \left( v_{0}^{2} - \int 2 F_{r} \, dr \right)} = \frac{dp}{1 + p^{2}}$

10. Although this equation contains four separate variables ($x, y, p$ and $r$), it can be integrated (with sufficient skill). For we know that $x dx + y dy = r dr = p x dy + y dy$; thus, $dy = \frac{r dr}{y + px}$ and $dx = \frac{p r dr}{y + px}$. Substituting these values into the equation yields

$\frac{\left( p y - x\right) F_{r} dr}{\left( y + p x \right) \left( v_{0}^{2} - \int 2 F_{r} \, dr \right)} = \frac{dp}{1 + p^{2}}$

which may be written as

$\frac{F_{r} dr}{v_{0}^{2} - \int 2 F_{r} \, dr} = \frac{dp \left( p x + y \right)}{\left( 1 + p^{2} \right) \left( p y - x\right)}$

Each of these two terms can be integrated using logarithms. Specifically, $\int \frac{F_{r} dr}{v_{0}^{2} - \int 2 F_{r} \, dr } = - \frac{1}{2} \log \left( v_{0}^{2} - \int 2 F_{r} \, dr \right)$; expanding in partial fractions, $\int \frac{dp \left( p x + y \right)}{\left( 1 + p^{2} \right) \left( p y - x\right)}$ resolves into $\int \frac{y dp}{p y - x} - \int \frac{p dp}{1 + p^{2}} = \log \frac{p y - x}{\sqrt{1 + p^{2}}}$. Thus, we have $\frac{C}{\sqrt{v_{0}^{2} - \int 2 F_{r} \, dr}} = \frac{p y - x}{\sqrt{1 + p^{2}}}$. This equation states that the particle speed $v = \sqrt{v_{0}^{2} - \int 2 F_{r} \, dr}$ is inversely proportional to the component of the radius that is perpendicular to the trajectory, a handy property of these motions.

Figure 28

11. This same problem may be solved more conveniently by choosing different variables (Figure 28). The method described above does not require that the coordinates be orthogonal to each other, merely that they suffice to define the trajectory uniquely. For this reason, we cannot take as our two variables the radius CM and its component perpendicular to the trajectory, since some trajectories cannot be described using these variables. However, we can take as our coordinates the radius CM and the arc BP of a unit circle centered on C; these two coordinates suffice to determine every point exactly, just as for orthogonal coordinates. This ability to change variables extends the applicability of our Method more broadly than might be imagined otherwise.

12. Let the radius of the particle be CM = $r$, and let the central force $F_{r}(r)$ be some function of the radius $r$. We take as our second variable $\theta$, the arc-length BP of a unit circle centered on C; the infinitesimal arc $d\theta$ corresponds to Pp in Figure 28. Defining $p \equiv \frac{d\theta}{dr}$, the applied force results in an acceleration $dv^{2} = - 2 F_{r} \, dr$, from which we obtain $v^{2} = v_{0}^{2} - \int 2 F_{r} \, dr$. The trajectory sweeps out an infinitesimal circular arc Mn of radius CM = $r$ about the center C, whereas mn represents the infinitesimal change in radius $dr$. From the geometrical similarity CP:Pp = CM:Mn follows the equation Mn = $p r dr$, and the infinitesimal distance Mm = $ds = dr \sqrt{1 + p^{2} r^{2}}$. Thus, we seek the trajectory that minimizes the integral $\int dr \sqrt{v_{0}^{2} - \int 2 F_{r} \, dr} \sqrt{1 + p^{2} r^{2}}$. From this, we may obtain the differential $\frac{d}{dr} \left[ \frac{p r^{2} \sqrt{v_{0}^{2} - \int 2 F_{r} \, dr}}{\sqrt{1 + p^{2} r^{2}}} \right]$, which equals zero, following our general protocol. This yields the equation $\sqrt{C} = \frac{p r^{2} \sqrt{v_{0}^{2} - \int 2 F_{r} \, dr}}{\sqrt{1 + p^{2} r^{2}}}$, or rather $C + C p^{2} r^{2} = \left( v_{0}^{2} - \int 2 F_{r} \, dr \right) p^{2} r^{4}$, from which follows

$p = \frac{\sqrt{C}}{\sqrt{\left( v_{0}^{2} - \int 2 F_{r} \, dr \right) r^{4} - C r^{2}}} = \frac{\sqrt{C}}{r \sqrt{\left( v_{0}^{2} - \int 2 F_{r} \, dr \right) r^{2} - C }}$

which may be written as

$d\theta = \frac{dr \sqrt{C}}{r \sqrt{\left( v_{0}^{2} - \int 2 F_{r} \, dr \right) r^{2} - C }}$

This same equation may also be derived using direct methods of mechanics.

13. In these examples, our principle agrees perfectly with experience; and the suspicion that it will fail in more complicated cases can likely be overcome. However, our principle should be studied carefully to determine the limits of its validity. To explain what I mean, all motion of particles can be divided into two classes, one in which position alone determines the speed and another in which it does not. In the first case, the particle always has the same speed when it reaches the same position; for example, the speed of a particle acted upon by one or more centers of force will be a function only of the distances to those center(s). In the second case, the speed is not determined solely by the position of the particle; for example, the centers of force themselves may be moving or the particle may be in a resistant medium. Our principle pertains only to the first class of motion; in particular, even when a particle is attracted to multiple fixed centers of force, its momentum integrated along its trajectory should be minimal.

Figure 27

14. Here is the essence of our hypothesis. We seek the trajectory between two specified endpoints that minimizes the integral $\int ds \, v$ where, by assumption, the speed $v$ of the particle at the endpoints does not depend on the path travelled. Even if there are many fixed centers of force, the speed $v$ can be expressed as a well-defined function of the variables CP = $y$ and PM = $x$ (Figure 27). Thus, the squared speed $v^{2}$ is also a function of $x$ and $y$, and we may write $dv^{2} = 2 F_{x} dx + 2 F_{y} dy$, from which we may see that our principle agrees with the true trajectory of the particle. Since $dv^{2} = 2 F_{x} dx + 2 F_{y} dy$, the forces acting on the particle at a point M can be grouped into two forces, a force $F_{x}$ that acts along the $x$ direction and another force $F_{y}$ that acts in the $y$ direction. These forces can themselves be resolved into a tangential force along the trajectory $\frac{F_{x} dx + F_{y} dy}{ds}$, and a normal force perpendicular to the trajectory $\frac{-F_{y} dx + F_{x} dy}{ds}$. The normal force provides the local centripetal acceleration $\frac{v^{2}}{R} = \frac{-F_{y} dx + F_{x} dy}{ds} = \frac{F_{x} - p F_{y}}{\sqrt{1 + p^{2}}}$, where $R$ is the local radius of curvature of the trajectory. Thus, our principle of deriving mechanics from a principle of maxima and minima conforms with experience.

15. We seek the trajectory that minimizes the integral $\int dy \, v \sqrt{1 + p^{2}}$; therefore, by our established protocol, we must differentiate the quantity $v \sqrt{1 + p^{2}}$. Since $dv^{2} = 2 F_{x} dx + 2 F_{y} dy$, the differentiation yields

$\frac{F_{x} dx \sqrt{1 + p^{2}}}{v} + \frac{F_{y} dy \sqrt{1 + p^{2}}}{v} + \frac{p dp v}{\sqrt{1 + p^{2}}}$

From our general law $Mdy = dP$, we obtain the desired trajectory

$\frac{F_{x} dy \sqrt{1 + p^{2}}}{v} = d\left[\frac{p v}{\sqrt{1 + p^{2}}} \right] = \frac{dp v}{\left( 1 + p^{2} \right)^{3/2}} + \frac{p F_{x} dx + p F_{y} dy}{v \sqrt{1 + p^{2}}}$

which may be written as

$\frac{- dp v}{\left( 1 + p^{2} \right)^{3/2}} = \frac{F_{y} p dy - F_{x} dy}{v \sqrt{1 + p^{2}}}$

The instantaneous radius of curvature of the trajectory at M equals $R = \frac{- \left( 1 + p^{2} \right) dy \sqrt{1 + p^{2}}}{dp}$, from which follows $\frac{v^{2}}{R} = \frac{F_{y} p - F_{x}}{\sqrt{1 + p^{2}}}$, just as predicted from the direct methods of mechanics. Thus, provided that the external forces can be resolved along the coordinate directions into two forces $F_{x}$ and $F_{y}$ that are functions of these variables $x$ and $y$, then every trajectory will represent the minimum of the integrated momentum over the path.

16. Therefore, this principle is broadly applicable, except to the case of motion in a resistant medium. The reason for this exception is easy to see, because the speed of the particle at the endpoints will depend on the path taken. Hence, neglecting any resistance to the particle's motion, the momentum integrated along the path should be a minimum. Moreover, this minimum law is true not only for the motion of single particles, but also for systems of particles bound together. No matter what their reciprocal interactions are, the path integral of their momenta is always minimal. Compared to traditional mechanics methods, the motion may be more difficult to calculate using our new method; however, it seems easier to grasp from first principles. Because of their inertia, bodies are reluctant to move, and obey applied forces as though unwillingly; hence, external forces generate the smallest possible motion consistent with the endpoints. A rigorous proof for this principle is lacking, I realize. Nevertheless, it agrees with experiment and I do not doubt that it will be verified by stronger proofs that use the principles of a complete Metaphysics. But such proofs I leave to the professors of Metaphysics.