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Suppose $F$ is independent of velocity,so Newton's law can be expressed as : $m \ddot{\mathbf{x}}(t)=\mathbf{F}(\mathbf{x}(t)) .$ Then an energy function of the form $$ E(\mathbf{x}, \dot{\mathbf{x}})=\frac{1}{2} m|\dot{\mathbf{x}}|^{2}+V(\mathbf{x}) $$ is conserved(i.e for any solution of Newton's equation $E$ is a constant independent of $t$) if and only if $V$ satisfies $$-\nabla V=\mathbf{F}. $$

The proof is by calculating: $$ \begin{aligned} \frac{d}{d t}\left(\frac{1}{2} m|\dot{\mathbf{x}}(t)|^{2}+V(\mathbf{x}(t))\right) &=m \sum_{j=i}^{n} \dot{x}_{j}(t) \ddot{x}_{j}(t)+\sum_{j=1}^{n} \frac{\partial V}{\partial x_{j}} \dot{x}_{j}(t) \\ &=\dot{\mathbf{x}}(t) \cdot[m \ddot{\mathbf{x}}(t)+\nabla V] \\ &=\dot{\mathbf{x}}(t) \cdot[\mathbf{F}(\mathbf{x})+\nabla V(\mathbf{x})]. \end{aligned} $$

The question here is if $ -\nabla V=\mathbf{F} $ easy to see energy is conserved

but how to prove the "only if" part?

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2 Answers 2

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You already have proved the "only if" part. What your derivation shows is that $$\frac{d}{d t}\left(\frac{1}{2} m|\dot{\mathbf{x}}(t)|^{2}+V(\mathbf{x}(t))\right)$$ equals zero only if $$\mathbf{F} = -\nabla V.$$ In other words, your work proves $$\mathbf{F} = -\nabla V \implies \frac{d}{d t}\left(\frac{1}{2} m|\dot{\mathbf{x}}(t)|^{2}+V(\mathbf{x}(t))\right) = 0.$$

Now, for the other direction, start with the assumption that $$\frac{d}{d t}\left(\frac{1}{2} m|\dot{\mathbf{x}}(t)|^{2}+V(\mathbf{x}(t))\right) = 0$$ and derive $$\mathbf{F} = -\nabla V.$$ The derivation will look very similar to what you've already done.

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  • $\begingroup$ The point here is $$\begin{aligned} \frac{d}{d t}\left(\frac{1}{2} m|\dot{\mathbf{x}}(t)|^{2}+V(\mathbf{x}(t))\right) &=m \sum_{j=i}^{n} \dot{x}_{j}(t) \ddot{x}_{j}(t)+\sum_{j=1}^{n} \frac{\partial V}{\partial x_{j}} \dot{x}_{j}(t) \\ &=\dot{\mathbf{x}}(t) \cdot[m \ddot{\mathbf{x}}(t)+\nabla V] \\ &=\dot{\mathbf{x}}(t) \cdot[\mathbf{F}(\mathbf{x})+\nabla V(\mathbf{x})]. \end{aligned}$$ the last expression is the scalar product,we need to show $\dot{x}(t)$ can reach any vector in $\mathbb{R}^n$ in order to prove $\mathbf{F} = -\nabla V.$ But I don't know how to show $\dot{x}$ is arbitrary $\endgroup$
    – yi li
    Aug 2, 2020 at 5:58
  • $\begingroup$ @yi_li You don't need to show that $\dot{x}(t)$ is arbitrary. There are no constraints on it, so it is already arbitrary. The only thing that matters is whether the last expression is identically zero. If you assume it is zero (energy is conserved), then $F = -\nabla V$ ($F$ is a conservative force) because dotting with a zero vector is the only way to guarantee a zero result. If you assume that $F = -\nabla V$ ($F$ is a conservative force), then the expression is zero (energy is conserved) for the same reason. That's the entire argument is establish "if and only if". $\endgroup$
    – Mark H
    Aug 2, 2020 at 7:25
  • $\begingroup$ $\dot{x}(t)$ is a curve generated by some initial condition in Newton's equation, we only know the initial condition is arbitrary, intuitively $\dot{x}(t) $ can be arbitrary with the proper chosen initial condition, maybe we should start from the free initial condition to prove $\dot{x}(t)$ is arbitrary when it is evolved with the Newton's law? $\endgroup$
    – yi li
    Aug 2, 2020 at 7:49
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    $\begingroup$ @yi_li The only way $\dot{x}(t)$ affects the derivation is if it is identically zero for all values of $t$. That would mean the particle doesn't move at all, in which case the conservation of energy is trivial. As long as the particle is moving at some point in time, you can conclude (or assume, for the other implication direction) that $F+\nabla V = 0$. $\endgroup$
    – Mark H
    Aug 2, 2020 at 7:57
  • $\begingroup$ sorry I made a bit mistake, I mean: maybe we should start from the initial condition(which can be chosen freely) to prove $\dot{x}(t)$ can arbitrary at fixed time $t$ when it is evolved with Newton's law. But It never hurt, since in the above proof if we set $t=0$ it's enough since $F(x)$ and $V(x)$ is independent of how the particle evolved, independent of $t$ , evaluate its relation with $t=0$ is enough) $\endgroup$
    – yi li
    Aug 2, 2020 at 8:05
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Let look at this example one dimensional

Kinetic Energy is

$$T=\frac{m}{2}\,\dot{x}^2$$ and Potential Energy is

$$ U=U(x)$$

with Euler Lagrange you get

$$m\,\ddot{x}+\frac{\partial}{\partial x}\,U(x)=0$$

thus according to Newton second law

$$F=-\frac{\partial}{\partial x}\,U(x)$$

and the total energy

$$E=T+U(x)=~\text{constant}$$

but if the potential energy is :

$$U=U(x,t)$$

you obtain the equation of motion

$$m\,\ddot{x}+\frac{\partial}{\partial x}\,U(x,t)=0$$

and your force is

$$F=-\frac{\partial}{\partial x}\,U(x,t)$$

$$E=T+U(x,t)\ne ~\text{constant}$$

thus only if the potential energy is $U=U(x)~,\text{or}~,F=F(x)$ the total energy is constant

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  • $\begingroup$ Very thanks for your help,You only show that $V$ potential is time-independent(we assume here already $F$ and $V$ only depend on $x$), Do you mean $F$ is conservative depend on Euler-Lagrangian, but the proof should not use Euler-Lagrangian.I have already proved the "only if" part.that is to show $\mathbf{F} = -\nabla V.$,we can evaluate the scalar production at $t = 0$ so $\dot{x}(0)$ can choose any value.In the same time any $x$ at $t=0$ is possible,since the initial configuration is arbitrary. $\endgroup$
    – yi li
    Aug 2, 2020 at 15:37

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