Can we motivate the principle of least action with the principle of inertia that causes a mass particle to resist changes in its momentum? After all, the principle of inertia is the starting point and the fundamental assumption of classical mechanics, like the Encyclopædia Britannica states.

Leonhard Euler gave the following consideration in his "Additamentum 2" (1744):

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.

We also have the following by Pierre Louis Maupertuis in his "Les Loix" (1746):

When a change occurs in Nature, the Quantity of Action necessary for that change is as small as possible. [...] The Quantity of Action is the product of the Mass of Bodies times their velocity and the distance they travel. When a Body is transported from one place to another, the Action is proportional to the Mass of the Body, to its velocity and to the distance over which it is transported.

When a mass particle changes its position by $\Delta s$ then, according to the Galileo's law of inertia, it tends to maintain its momentum $p$ with no change in direction or magnitude, ideally $\Delta p = 0$. That is, when the particle has no net external force acting upon it, a product $p (s_2-s_1)>0$ has the minimum value in the sense that the physically realized straight line is the shortest distance between the successive points $s_1$ and $s_2$. If we consider the influence of an external force, we may assume that the inertial resistance holds for every infinitesimal part of the physically realized overall path. The initial momentum $p_0$ is constant and all the change is due to the external force. Therefore, the inertia of motion causes a sum $\sum \overline{p} \cdot \Delta s$, where $\Delta s$ is an infinitesimal part of the overall path in 3 dimensions, during an infinitesimal time step $\Delta t$, and $\overline{p}$ is the average momentum on each $\Delta s$, to obtain a minimum value. This motivates the Euler's form of the least action principle, i.e.,

$$\delta \int p \cdot ds=0.$$

Moreover, in the presence of conservative force, the only possible source of momentum dependent kinetic energy $T$ is position dependent potential energy $V$, and vice versa. Now the inertia of motion resists the flow of energy between $T$ and $V$, ideally $\Delta T = 0$. The initial kinetic energy $T_0$ is constant and all the change is due to the external force. Therefore, the inertia of motion causes a sum $\sum \overline{T} \Delta t$, where $\Delta t$ is an infinitesimal time step during the overall energy flow process and $\overline{T}$ is the average kinetic energy on each $\Delta t$, to obtain a minimum value. This motivates the Maupertuis' form of the least action principle, i.e.,

$$\delta \int T dt=0.$$

The previous results are two sides of the same coin:

$$ \int p \cdot ds= \int p \cdot \dot{s} dt=\int 2Tdt.$$

On the other hand, the previous energy flow between $T$ and $V$ means that $\Delta T = -\Delta V$, and this suggests that we can replace $T$ by $-V$ in the Maupertuis' form. Since in the both cases we seek to find a minimum, we may consider the sum of the two terms and this motivates the Lagrange's form of the least action principle, i.e.,

$$\delta \int (T-V) dt=0.$$

  • $\begingroup$ The title question (v10) seems like a non-starter: E.g. the stationary action principle evidently knows about Newton's 2nd law. $\endgroup$
    – Qmechanic
    Commented Jan 21, 2023 at 9:29
  • 1
    $\begingroup$ @Qmechanic Do you see this approach as pedagogically/philosophically meaningful? $\endgroup$
    – Hulkster
    Commented Jan 25, 2023 at 11:16
  • $\begingroup$ Hi @Hulkster. The title proposal seems incomplete at best. $\endgroup$
    – Qmechanic
    Commented Jan 25, 2023 at 11:21
  • 1
    $\begingroup$ the principle of minimal action is valid also for EM waves and massless fields in general. I do not understand what inertia has to do with it ? One can define an action for any sort of physical system where a dynamical evolution takes place and a Lagrangian can be defined. $\endgroup$
    – drgetwrekt
    Commented Dec 10, 2023 at 9:48
  • $\begingroup$ @drgetwrekt Inertia of moving bodies is something tangible and thus it provides a good stepping stone. After all, we can't pull the action principle out of the magic hat. $\endgroup$
    – Hulkster
    Commented Dec 10, 2023 at 10:43

2 Answers 2


We have the following two concepts:

The relation between force, inertial mass, and acceleration: $$ F=ma \tag{1.1} $$

The relation between mechanical potential energy and kinetic energy:

$$ \Delta E_k + \Delta E_p = 0 \tag{1.2} $$

You grant the concept of conservation of mechanical energy, but it appears you think of that as independent from $F=ma$. That is in fact not the case:

$ ds = v \ dt \qquad \qquad (2.1) $

$ a \ dt = dv \qquad \qquad (2.2) $

$ F = ma \qquad \qquad (2.3) $

$ \int_{s_0}^s F \ ds = \int_{s_0}^s ma \ ds \qquad \qquad (2.4) $

$ \int_{s_0}^s a \ ds = \int_{t_0}^t a \ v \ dt = \int_{t_0}^t v \ a \ dt = \int_{v_0}^v v \ dv = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \qquad (2.5) $

$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \qquad \qquad (2.6) $

$ \Delta E_p = -\int_{s_0}^s F \ ds \qquad \qquad (2.7) $

$ \Delta E_k = -\Delta E_p \quad \Leftrightarrow \quad \Delta E_k + \Delta E_p = 0 \qquad \qquad (2.8) $

From (2.3) to (2.4):
To both sides of (2.3) the same operation is applied: integration with respect to the position coordinate.

In (2.5) the expression $ \int_{s_0}^s a \ ds $ is developed. The acceleration is unspecified, but we can develop nonetheless because position and acceleration are not independent of each other. In developing the differential is changed twice, according to (2.1) and (2.2), with corresponding change of limits.

(2.7) states the definition of potential energy.


We have in general that in physics we can shift from one representation to another by application of a mathematical transformation. In the case of $F=ma$ and $\Delta E_k + \Delta E_p = 0$: those two representations are related by integration.

Conversely: to recover $F=ma$ from $\Delta E_k = -\Delta E_p$ you take the derivative with respect to the position coordinate on both sides.

My point here is: from a physics point of view: to grant $F=ma$ or to grant $\Delta E_k = -\Delta E_p$ is one and the same thing.

We have: in cases where $\Delta E_k + \Delta E_p = 0$ holds good: Hamilton's stationary action will hold good also (no additional assumption required).

Discussion of that is in an answer by me (oktober 2021) to a question about motivation for Lagrangian formalism

  • $\begingroup$ I don't agree with this answer. $F=ma$ allows for arbitrary formulas of force. Not all of them can be derived from a scalar potential energy. $\endgroup$
    – Ryder Rude
    Commented Jan 21, 2023 at 19:16
  • $\begingroup$ @RyderRude In the context of static load I use the following operational definition of force: force is that which is measured with a load cell For example, when a rocket engine is secured to a test stand the thrust of the engine can be measured. In dynamic context it can be seen whether the observed acceleration of the rocket upon ascent is consistent with the known thrust capability. I acknowledge that procedures must be developed to correlate observed acceleration with direct measurement. $\endgroup$
    – Cleonis
    Commented Jan 21, 2023 at 20:31
  • $\begingroup$ @RyderRude Can you point me to a resource with more information about the 'allows for arbitrary formulas of force'? I'm interested. Our theories are assemblies of interconnected inferences. There is room for ambiguity - I'm interested in that. $\endgroup$
    – Cleonis
    Commented Jan 21, 2023 at 20:40
  • $\begingroup$ @Cleonis I don't have a resource on this. But aren't theories defined axiomatically? It would be circular to define a force in terms of what a load cell measures, when that mechanism itself relies on how interactions work in the universe. You have to construct axiomatic models. $\endgroup$
    – Ryder Rude
    Commented Jan 22, 2023 at 2:12
  • 1
    $\begingroup$ @RyderRude Obvously this whole discussion should move to chat. Stackexchange is giving me the prompt for that, but clicking it fails to move the discussion to chat $\endgroup$
    – Cleonis
    Commented Jan 22, 2023 at 15:52

The basic flaw in your argument is that the sum is not zero. It tends to resist but it always doesn't, if you have some infinitesimal movement $\delta x$, you will have some associated change in momentum.

But your final equation is correct, then how would you go about deriving it? You see in the path of least action, you know the Lagrangian is zero for any path at the beginning and end points; so you can impose boundary conditions.

  • $\begingroup$ Please be careful with the grammar and the readability of your text, while posting your answer. Also make habit to use MathJax for the mathematical notations. $\endgroup$
    – SCh
    Commented Jan 21, 2023 at 8:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.