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  1. Are the principle of least action and the principle of minimum potential energy equivalent? How does one show that?

  2. Also, are Newton's laws of motion equivalent to the principle of least action? How can one show that? Is this totally empirical?

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    $\begingroup$ 1) No, the latter is an equilibrium statement. Least action will tell you how a ball moves when thrown through the air. Minimum potential energy says that when it stops, it'll be on the ground. 2) Yes, in the context of mechanics. You show it by deriving the Euler-Lagrange equations and seeing they are the same equations as what you get from Newton's laws. $\endgroup$ – knzhou Jan 18 '16 at 15:40
  • $\begingroup$ Subquestion 2 is essentially a duplicate of physics.stackexchange.com/q/78138/2451and links therein. (It is trivial to show that Newtons 2nd law follows from the Lagrange equations for the Lagrangian $L=\frac{1}{2}\sum_im_iv_i^2-U$. The other way is non-trivial.) $\endgroup$ – Qmechanic Jan 18 '16 at 18:42
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Extremum of potential vs. extremum of action

Yes, the action principle is in a special case equivalent to the principle of extremum of potential energy (the maximum of a potential also presents an equilibrium, even though it is an unstable one!).

Consider the action principle of a point particle in a potential $V(\vec{x})$, then the Hamilton's action principle reads ($\dot{\vec{x}}$ is the velocity $\vec{v}$) $$ S= \int_{t_1}^{t_2} \! \frac{1}{2} m \dot{\vec{x}}^2 - V(\vec{x}) \,\mathrm{d}t\,. $$ We now want to find an equilibrium, that is, a solution for which $\vec{x}(t)=\vec{x}_0={\rm const.,\;} \dot{\vec{x}}(t)=0$. We thus impose a small variation $\delta \vec{x},\, \delta \dot{\vec{x}}$ around this trajectory to check under which conditions the variations of the action $\delta S=0$. We do this by comparing the action of the original trajectory $\vec{x}(t)=\vec{x}_0 ,\;\dot{\vec{x}}(t)=0$ with a trajectory shifted to $\vec{x}(t)=\vec{x}_0 + \delta \vec{x}(t),\; \dot{\vec{x}}(t)=0 + \delta \dot{ \vec{x}}(t)$ $$ \delta S= \int_{t_1}^{t_2} \! \frac{1}{2} m \delta \dot{\vec{x}}^2 - V(\vec{x}_0 + \delta \vec{x}) \,\mathrm{d}t - \int_{t_1}^{t_2} \! 0 - V(\vec{x}_0) \,\mathrm{d}t\,. $$ But since the variation $\delta \vec{x}$ is infinitely small, we only consider terms to linear order in $\delta \vec{x}$ and neglect higher order terms such as $\delta \dot{\vec{x}}^2,\, \delta {\vec{x}}^2$. Hence, we get $$ \delta S = -\int_{t_1}^{t_2} \sum_i\frac{ \partial V}{\partial x^i}|_{\vec{x}=\vec{x}_0} \delta x^i \,\mathrm{d}t, $$ where we have made a Taylor expansion of the potential $V$ around $\vec{x}=\vec{x}_0$ and neglected higher-order terms in the expansion. You can now see that for a fully general and unconstrained $\delta \vec{x}$ the integral will vanish only if $$ \frac{ \partial V}{\partial x^i}|_{\vec{x}=\vec{x}_0} =0\,,\, \forall i $$ which is precisely the condition for the extremum of the potential $V$ at the point $\vec{x}=\vec{x}_0$. Note, however, that there is no way to derive the action principle from the principle of extremal potential energy, as the latter is only a very special case of the former.


Equivalence of Newton's law and principle of action

Yes, the Hamilton's action principle is equivalent to the Newton's law of motion, at least for motion in a potential field or e.g. under the influence of the electromagnetic force. Let's treat the potential-field case. There the action principle is once again $$ S= \int_{t_1}^{t_2} \! \frac{1}{2} m \dot{\vec{x}}^2 - V(\vec{x}) \,\mathrm{d}t\,. $$ A textbook derivation which can be found e.g. in the canonical Classical mechanics by Goldstein or on wikipedia gives you the equations of motion which are necessary for $\delta S=0$ ($\ddot{x}^i$ is the acceleration $a^i$) $$m \ddot{x}^i = - \frac{ \partial V}{\partial x^i}$$ which is equivalent to the Newton's law of motion because for a potential force field we have $F_i =- \partial V /\partial x^i$.

There are special cases which the action principle is not able to describe and the Newton's law can, such as when the force field is not conservative. However, it is a more or less unspoken assumption throughout theoretical physics that all fundamental dynamics are in fact, in one way or another, described by an action principle. The special cases which cannot be described by an action such as the mentioned non-conservative force field are generally considered as non-fundamental.

However, ACuriousMind rightly commented that I forgot about one important example of a non-conservative force which is quite crucial for many models - friction. The reason why friction is difficult to describe by a fundamental action principle is the fact that friction is in fact a consequence of an immense amount of degrees of freedom of the molecules of the material. (You can find more about that by studying statistical mechanics.)

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