In classical mechanics we often define the action as the quantity

$$ \int_{0}^{T} \left[ T - V \right] dt$$

Which in many applications is some variant of

$$ \int_{0}^{T} \left[ \frac{1}{2}m \left( x' \right)^2 - V(x) \right] dt. $$

The usual justification for the principle of least action is the observation that if you take integrand above and put it into the euler lagrange equations you get back Newton's law.

I.E. if you believe

$$ \frac{\partial L}{\partial x} - \frac{d}{dt} \left( \frac{\partial L}{\partial x'} \right) + \frac{d^2}{dt^2} \left(\frac{\partial L}{\partial x''} \right) - ... = 0$$

With $L = \frac{1}{2}m \left( x' \right)^2 - V(x) $ you will find

$$ - \frac{dV}{dx} = mx'' $$

(i.e $F = ma$).

So this is old news that we consider quite well understood but then I realized the following, suppose we try to minimize this action instead:

$$ \int_{0}^{T} \left[ \frac{1}{2}mxx'' + V(x) \right] dt $$

I.E. $L = \frac{1}{2}mxx'' + V(x) $. If we plug this into the euler lagrange equations we ALSO end up deriving

$$ - \frac{dV}{dx} = mx'' $$


$$ \frac{\partial }{\partial x}[\frac{1}{2}mxx'' + V] + \frac{d^2}{dt^2}\frac{\partial}{\partial x''}\left[ \frac{1}{2}mxx'' + V \right] = 0 \rightarrow \frac{1}{2}mxx'' + \frac{dV}{dx} + \frac{1}{2}mxx'' = 0 \rightarrow mxx'' + \frac{dV}{dx} = 0 \rightarrow F = -\frac{dv}{dx}$$

I found this very curious, I recognize the physical significance of $(mx'')*x$ as the classical expression for work (Force x distance). But is there any deeper physical significance to this second lagrangian, or is this just a curious mathematical oddity/not a useful problem solving tool. Can this second lagrangian be used in place of the first in other contexts (ex: in the Feynman path integral).

So it seems that $\frac{1}{2}mxx'' - V$ is a conserved quantity. (I came to this conclusion after checking only one example involving a newtonian gravitational field between two bodies at 2 locations, so maybe this is wrong.)

  • 1
    $\begingroup$ you can obtain the first lagrangian from your second lagrangian with an integration by parts of the first term (up to a factor of minus) $\endgroup$ Commented Oct 28, 2021 at 1:51
  • $\begingroup$ @Filipe Miguel: I see what you mean. I guess its just another name for energy [shifted by some constant] but I am surprised I never learned about this thing, considering it too is a conserved quantity $\endgroup$ Commented Oct 28, 2021 at 1:54
  • $\begingroup$ @frogeyedpeas I am not sure about your claim that it is a conserved quantity. I mean't the lagrangian and not your last statement $\endgroup$ Commented Oct 28, 2021 at 1:55
  • $\begingroup$ Where did you get the third equation? It is not the standard Euler-Lagrange's equation en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation#Statement @frogeyedpeas $\endgroup$
    – SRS
    Commented Oct 28, 2021 at 8:31
  • $\begingroup$ @SRS the Euler Lagrange equation is actually originally in the form of the third equation when you first derive it but then they drop the higher order terms to keep matters simple. I will provide a derivation to the bottom of this question $\endgroup$ Commented Oct 28, 2021 at 11:05

4 Answers 4


It is wellknown that given a set of EOMs, the action $S$ is not necessarily unique, cf. e.g. this Phys.SE post. OP points out that the Euler-Lagrange (EL) equations are not affected if we add a boundary term, cf. e.g. this Phys.SE post. However, the caveat is that the boundary conditions (BCs) [which are necessary to impose in order to make the variational principle well-posed] may change!

  1. OP's 1st example: $$L_1 ~=~ \frac{1}{2}m\dot{q}^2-V(q).\tag{1a}$$ The infinitesimal variation reads $$ \delta S_1 ~=~\int_{t_i}^{t_f} \! dt~ {\rm EOM}~\delta q + \left[m\dot{q}\delta q \right]^{t=t_f}_{t=t_i}. \tag{1b}$$ If we focus$^1$ on the initial condition (IC), we have to impose either

    • $\color{red}{\rm weak}$ Dirichlet IC: $ q(t_i)=q_i $,


    • Neumann IC: $ \dot{q}(t_i)=0, $

    in order to make the boundary term disappear [which is necessary in order to derive the EL equation from the variational principle]. See also e.g. my Math.SE answer here.

  2. OP's 2nd example: $$L_2 ~=~ -\frac{1}{2}mq\ddot{q}-V(q)~=~L_1 - \frac{d^2}{dt^2}(\frac{m}{4}q^2).\tag{2a}$$ The infinitesimal variation reads $$ \delta S_2 ~=~\int_{t_i}^{t_f} \! dt~ {\rm EOM}~\delta q + \frac{m}{2}\left[\dot{q}\delta q - q\delta \dot{q}\right]^{t=t_f}_{t=t_i}. \tag{2b}$$ We have to impose either

    • $\color{red}{\rm strong}$ Dirichlet IC: $ q(t_i)=0 $,


    • Neumann IC: $ \dot{q}(t_i)=0 $.

    There are no other possibilities!

TL;DR: The lesson is that depending on the physical system and the physically relevant BCs, we might have to choose a specific action for the variational principle.

See also e.g. this related Phys.SE post.


$^1$The final condition (FC) is similar.


First, the Lagrangian is not a conserved quantity, so $\frac{1}{2} m x \ddot{x} + V$ is not conserved. For example, for a particle in a constant gravitational potential with $V = m g x$, taking the solution $x(t) = - \frac{1}{2} g t^2 $ would yield for this quantity $\frac{1}{4} m g^2 t ^2 - \frac{1}{2} m g^2 t^2 = -\frac{1}{4} m g^2 t^2$, which is obviously not conserved.

Second, if you go through the derivation of the Euler-Lagrange equation, you will find that the non-uniqueness of the Lagrangian follows from the assumption that the variation vanishes at the end points of the path. This allows you to ignore terms in the Lagrangian that are total derivatives; since the difference between the standard kinetic term and the one you wrote down is a total derivative: \begin{equation} \frac{1}{2} m \dot{x}^2 - \left(-\frac{1}{2} m x \ddot{x}\right) = \frac{{\rm d}}{{\rm d}t} \left(\frac{1}{2} m \dot x x\right) \end{equation} you can treat the Lagrangians as equivalent. Assuming the boundaries of the path are fixed is a common assumption, which is frequently sufficient for what we are interested in.

However, there is information in the variation of the boundary of the action, and in some cases it is important to properly fix the boundary terms. For example, you can derive the canonical momentum from requiring that the variation of the path at the boundary vanishes. In these cases, the Lagrangian is fixed, with no ambiguity about the boundary term, by having a well-defined variational principle. In GR, the so-called Gibbons-Hawking-York boundary term needs to be added to the action, and is important to describe quantum effects. [1]

Finally, in your question you implicitly changed the sign of the action. So long as you are speaking classically and only looking at the system you wrote down, this doesn't create a problem. But if you added your Lagrangian (with the wrong sign) to a standard Lagrangian $\frac{1}{2} M \dot{X}^2 - V(X)$, and added a term coupling $x$ and $X$, then you would find your system had a ghost instability. So, it's a good idea to be careful about the sign of the Lagrangian, and therefore a better way to write your term would have been $-\frac{1}{2} m x \ddot{x} - V$ (even though in your case it wouldn't matter).

[1] http://quark.itp.tuwien.ac.at/~grumil/pdf/lecture7_2018.pdf

  • $\begingroup$ wait the lagrangian is $\frac{1}{2}mxx'' + V$, $\frac{1}{2}mxx'' - V$ isn't the lagrangian, and I still think it (the second expression not the lagrangian) is conserved $\endgroup$ Commented Oct 28, 2021 at 2:01
  • $\begingroup$ ok I see what you mean so this new lagrangian is just the classical shifted by a constant that the boundary terms would yield. $\endgroup$ Commented Oct 28, 2021 at 2:03
  • $\begingroup$ @frogeyedpeas I fixed the sign and added an example where this is not conserved. $\endgroup$
    – Andrew
    Commented Oct 28, 2021 at 2:09
  • 1
    $\begingroup$ @frogeyedpeas The logic is not that the boundary terms shift the action by a constant. The logic is that when you vary a total derivative (T.D.) term like $\mathcal{L}_{\rm T.D.} = \frac{d f}{dt}$, where $f$ is some function of $x$, you get $\delta L_{\rm T.D.} = \frac{\partial f}{\partial x} \delta x \Big|^{x=x(t_f)}_{x=x(t_i)}$, where $t_i$ is the initial time and $t_f$ is the final time. If we assume $\delta x(t_i) = \delta x(t_f) =0$, then $\delta L_{\rm T.D.} = 0$. $\endgroup$
    – Andrew
    Commented Oct 28, 2021 at 2:20
  • $\begingroup$ @frogeyedpeas (a) With these simple examples sometimes you get accidental cancellations; originally I was going to use the harmonic oscillator as a counter example, but it doesn't work since $\frac{1}{2} m x \ddot{x} + V = \frac{1}{2} x \left(m \ddot{x} + \frac{2 V}{x}\right)$, and it just happens by chance for a harmonic oscillator that $\frac{2 V}{x} = V'$, so this combination vanishes using the equations of motion. It's just an artifact of an example that's too easy. (b) However, neither of the two quantities you wrote are conserved in my example. $\endgroup$
    – Andrew
    Commented Oct 28, 2021 at 2:28

In the case of Hamilton's stationary action there is no significance in the non-uniqueness you refer to.

The point where the rubber meets the road is the constraint that as an object is moving along, subject to acceleration due to a potential gradient, the rate of change of kinetic energy must match the rate of change of potential energy.

Inserting the Lagrangian ($E_k-E_p$) in the Euler-Lagrange equation achieves the goal of satisfying that constraint.

You can build more convoluted expressions that also satisfy the constraint that rate-of-change-of-kinetic- energy-must-match-rate-of-change-of-potential-energy, but those convoluted expressions do not add anything. They're just ($E_k-E_p$) with unnecessary luggage added.

Hamilton's stationary action does one thing and one thing only: it expresses the constraint that the rate of change of kinetic energy must match the rate of change of potential energy. For a demonstration of that: I refer to an answer I submitted here on physis.SE:
Hamilton's stationary action

(The nature of the demonstration is graphical; the core points are visualized in diagrams.)


About the expression $\tfrac{1}{2}m x x''$ that you offer.

In the following I will be using expressions with squaring, so for readability I will use dot notation:
$ x' = \dot{x} $
$ x'' = \ddot{x} $

The question now is: how does the expression you offered relate to the expression for kinetic energy: $\tfrac{1}{2} m \dot{x}^2$ ?

Well, for one thing, the two are dimensionally the same.

About these three quantities:

I first give the case where $\ddot{x}$ is a constant, and then the case where $\ddot{x}$ is an arbitrary function of time:

With $\ddot{x}$ a constant:

$$ \dot{x} = \ddot{x} t \tag{1} $$

$$ x = \tfrac{1}{2} \ddot{x} t^2 \tag{2}$$

Multiply both sides of (2) with $\ddot{x}$, proceed, use (1):

$$ \ddot{x} x = \ddot{x} \tfrac{1}{2} \ddot{x} t^2 = \tfrac{1}{2} \ddot{x}^2 t^2 = \tfrac{1}{2} (\ddot{x} t)^2 = \tfrac{1}{2} \dot{x}^2 \tag{3} $$

Omitting the intermediate steps:

$$ \ddot{x} x = \tfrac{1}{2} \dot{x}^2 \tag{4} $$

The case where $\ddot{x}$ is an arbitrary function of time:
Integration from initial point $x_i$ to end point $x_f$

$$ \int_{x_i}^{x_f} \ddot{x} \ dx \tag{5} $$

Change the differential, using $dx = \dot{x} dt$, with corresponding change of limits:

$$ \int_{x_i}^{x_f} \ddot{x} \ dx = \int_{t_i}^{t_f} \ddot{x} \ \dot{x} \ dt \tag{6} $$

Final change of differential, using $\ddot{x} dt = d\dot{x}$, with corresponding change of limits:

$$ \int_{t_i}^{t_f} \dot{x} \ \ddot{x} \ dt = \int_{\dot{x}_i}^{\dot{x}_f} \dot{x} \ d\dot{x} = \tfrac{1}{2}{\dot{x}_f}^2 - \tfrac{1}{2}{\dot{x}_i}^2 \tag{7} $$

So in all we have:

$$ \int_{x_i}^{x_f} \ddot{x} \ dx = \tfrac{1}{2}{\dot{x}_f}^2 - \tfrac{1}{2}{\dot{x}_i}^2 \tag{8} $$

And when $x_i$ is zero the expression simplifies to:

$$ \int_{o}^{x} \ddot{x} \ dx = \tfrac{1}{2}\dot{x}^2 \tag{9} $$

The striking thing is: (4) and (9) end up as the same expression:

$$ \ddot{x} x = \tfrac{1}{2} \dot{x}^2 \tag{4} $$

$$ \int_{o}^{x} \ddot{x} \ dx = \tfrac{1}{2}\dot{x}^2 \tag{9} $$

Here is why I'm pointing that out: (9) is a clue to why there is room for an error to not have consequences.

Returning to the expression $\tfrac{1}{2}m x x''$ that you offered.

We have that (8) is what the work-energy theorem hinges on.

To derive the work-energy theorem: start with $F=m\ddot{x}$, and on both sides evaluate the integral with respect to position coordinate.

The left hand side of $F=m\ddot{x}$:
As you mentioned: $\int F dx$ is the expression for work done. Potential energy is defined as the negative of work done.

The right hand side of $F=m\ddot{x}$:

$$ \int_{x_i}^{x_f} m \ \ddot{x} \ dx = \tfrac{1}{2}m{\dot{x}_f}^2 - \tfrac{1}{2}m{\dot{x}_i}^2 \tag{10} $$

The above derivations explain why the expression $\tfrac{1}{2}m x x''$ has the same effect as the correct expression $\tfrac{1}{2}m\dot{x}^2$.

Using $m x x''$ directly is an error, it skips an integration step; $m x x''$ must be evaluated as an integral with respect to position coordinate.

However, we have the relations:
Because of those relations: putting a factor $\tfrac{1}{2}$ in front of $m x x''$ was enough to make it have the same effect as the correct $\tfrac{1}{2}m\dot{x}^2$

Phrased differently: using $\tfrac{1}{2}m x x''$ as part of the Lagrangian is an error, but along the way the error vanishes; you still end up with the correct equation of motion.

  • $\begingroup$ @SidharthGhoshal About the origin of kinetic energy; the biggest clue is that factor $\tfrac{1}{2}$. No two ways about it: kinetic energy comes out of an integration step. Recapitulating: we start with $F=ma$ and then both sides are integrated with respect to position: $$ \int_{s_i}^{s_f} F \ ds = \int_{s_i}^{s_f} m \ a \ ds $$ $$ \int_{s_i}^{s_f} F \ ds = \tfrac{1}{2}m{v_f}^2 - \tfrac{1}{2}m{v_i}^2$$ What that shows is: there is a direct connection between $F=ma$ and the property that the sum of kinetic energy and potential energy is a conserved quantity. $\endgroup$
    – Cleonis
    Commented Jan 6 at 17:49

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