Can't one always define potential energy $U=-m\ddot xx$?

Question

Can't one always define potential energy $$U=-m\ddot xx$$ so that $$F=m\ddot x=-\frac{\partial U}{\partial x}?$$ An example is gravitational potential energy $U=mgh$ where $h=x$ and $g=-\ddot x$. I know this has to be wrong somehow because not all force fields are conservative, but I can't see it.

Answer 1

Answers from the comments, as suggested here. Answers should be posted as answers, and comments should be used for their intended purpose.

@Refik Mansuroglu

You are assuming that $\partial\ddot x/\partial x=0$, which you cannot always do.

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@Ján Lalinský

You can define such function $U(x,\ddot x)$, but it won't be recognized as potential energy, because the important defining feature of a potential energy function is that it depends only on coordinates, not velocities or higher derivatives.

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@Maximal Ideal

Something else to note is that if you want a conservative force law, you would want an equation of the form $$m\ddot x=\text{something else as a function of position}$$ In your case, you get $m\ddot x=m\ddot x$ which gives no information about the force law that the mass is subjected to. In fact, the equation of motion, if you can call it that, reduces to $0=0$.

Answer 2

In the expression $U(x)$, $x$ is meant to be a coordinate describing space. You can evaluate the function $U$ at any spatial point $x$. This is important since it means that the force $F$ is defined everywhere.

Meanwhile, in an expression like $x(t)$, $\dot{x}(t)$, or $\ddot{x}(t)$, $x$ is meant to be understood as a trajectory of a particle. In particular, a time derivative like $\dot{x}$ makes sense if you think of the velocity of a moving particle following a path $x(t)$. But, if you want to think of $x$ as simply a coordinate -- a label for a fixed point in space -- then the time derivative of that label doesn't really mean much. So a function like $f(\ddot{x}(t))$ best understood as a function of a particular trajectory. You need more information than the labels of individual points in space, you also need to know how that particle moves in time, to evaluate $f$.

To say this differently, recall that Newton's laws are really a second order differential equation. We are meant to solve it to determine $x(t)$. In an expression like $m\ddot{x}=-\frac{\partial U}{\partial x}$, the right hand side is a function of $x$, so we have a well-defined differential equation. If we used $U=m \ddot{x} x$, then we no longer have a differential equation to solve, but simply a definition of $U$. The equation becomes a tautology. We need to set $m\ddot{x}$ equal to something independent in order to have a differential equation to solve.

Answer 3

1. In a nutshell, the Newton's 2nd law$^{1}$ $$m{\bf a}~\approx~{\bf F}\tag{1}$$ ("EOM") equates kinematics ("mass times acceleration") with dynamics ("forces").

2. The (generalized) potential energy $U$ [if it exists] satisfies by definition that the force $$\begin{align}-{\bf F}~=~\frac{\delta U}{\delta{\bf r}}~\equiv~&\frac{\partial U}{\partial{\bf r}}-\frac{d}{dt}\frac{\partial U}{\partial{\bf v}}+\left(\frac{d}{dt}\right)^2\frac{\partial U}{\partial{\bf a}}\cr &-\left(\frac{d}{dt}\right)^3\frac{\partial U}{\partial{\bf j}}+\ldots\end{align}\tag{2}$$ is minus the (functional) derivative of $U$.

3. The (generalized) potential energy $U({\bf r},{\bf v},{\bf a},{\bf j},\ldots,t)$ may in principle depend on velocity, acceleration, jerk, etc. [Think e.g. on the Abraham-Lorentz force.]

4. The potential energy $U$ should always be associated [via eq. (2)] with the dynamics ("forces"); not the kinematics ("mass times acceleration"). This is where OP's suggestion$^2$ fails: At best it produces a useless trivial identity rather than a useful non-trivial EOM $$m{\bf a}~\approx~-\frac{\delta U}{\delta{\bf r}}.\tag{3}$$ Kinematics and dynamics should not be conflated.

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

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$^1$ The $\approx$ symbol means equality modulo eqs. of motion (EOM).

$^2$ It might be helpful to point out that if "mass times acceleration" $m{\bf a}$ is interpreted as a force, its corresponding (generalized) potential energy (2) is not surprisingly the kinetic energy $\frac{m}{2}{\bf v}^2$ (modulo total time derivative terms). Which can be rewritten to e.g. $\frac{m}{2}{\bf v}^2-\frac{d}{dt}(\frac{m}{2}{\bf r}\cdot{\bf v})=-\frac{m}{2}{\bf r}\cdot{\bf a}$ (modulo total time derivative terms). The latter is a half times OP's proposal.

Answer 4

Can't one always define $U=-m\ddot xx$?

One can make such a definition, but one should not do so. A function $U$ defined as $-m\ddot x x$ would not be the correct potential energy for any case beside a one-dimensional constant force.

The potential energy is usually defined for a conservative force, up to a constant, via $$ \vec F = -\nabla U\;, $$

If you choose to make some other ad hoc definition like $$ U = -m\ddot x x $$ then no one will understand what you are talking about, since we already have a well-defined definition of potential energy (up to a constant).

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Consider, for example, a Hooke's law spring force. The usual meanings of force and potential energy are well known: $$ F = -kx $$ $$ U = \frac{1}{2}kx^2\;. $$

If you try to use your flawed definition, you wind up with $$ U_{your\;proposal} = kx^2\;, $$ which is clearly wrong by a factor of 2.

Answer 5

Instead of the dot notation for time derivatives I will use the trio

$s$ position
$v$ velocity
$a$ acceleration

And of course $t$ for time

By definiton:

$$ v = \frac{ds}{dt} \ \Leftrightarrow \ ds = v \ dt \tag{1} $$

$$ a = \frac{dv}{dt} \ \Leftrightarrow \ dv = a \ dt \tag{2} $$

Next: evaluate the integral of acceleration with respect to position coordinate, $\int_{s_0}^s \ a \ ds$

$$ \int_{s_0}^s a \ ds \tag{3} $$ $$ \int_{t_0}^t a \ v \ dt \tag{4} $$ $$ \int_{t_0}^t v \ a \ dt \tag{5} $$ $$ \int_{v_0}^v v \ dv \tag{6} $$

First (1) was used to change the differential from $ds$ to $dt$, with corresponding change of limits. Next (2) was used - with change of limits - to arrive at (6).

(6) evaluates to $\tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2$, so we have the following mathematical relation:

$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \tag{7} $$

This is a general theorem; the definitions (1) and (2) are sufficient.

Multiplying both sides with $m$ for mass turns (7) into a dynamics statement:

$$ \int_{s_0}^s ma \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \tag{8} $$

And of course: if $F=ma$ holds good we can substitute $ma$ with $F$:

$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \tag{9} $$

(9) informs us: If a force field has the property that its corresponding potential is a function of the position coordinate only then:

$$ ma = \frac{dU}{dx} \tag{10} $$