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We know that we can write the energy of a particle as a function of position $\mathbf{x}$ like this: $$E(\mathbf{x}) = \frac{\mathbf{p}^2(\mathbf{x})}{2m} + V(\mathbf{x})$$ Where $\mathbf{p}$ is the momentum and $V(\mathbf{x})$ is the potential energy. ($m$ stands form mass and $E$ for energy). And we know that $E(\mathbf{x})$ must be a constant function, and so, $\frac{\mathbb{d}E}{\mathbb{d}\mathbf{x}} = 0$. For simplicity, let's assume that we are in a one dimensional movement and so we can treat the vector $\mathbf{x}$ as a scalar $x$. Differentiating both sides with respect to $x$: $$0 = \frac{1}{2m}2p(x)p'(x) - \mathbf{F}(x)$$ Here i used the fact that for some vector field $\mathbf{F}$ we have that $\mathbf{F} = - \nabla V$, where $V$ is the potential energy. And so we get to: $$\mathbf{F}(x) = \frac{1}{2m} 2m\mathbf{v}(x)m\mathbf{a}(x)$$ Which turns out to be $$\mathbf{F}(x) = m\mathbf{v}(x)\mathbf{a}(x)$$ Which is obviously not equal to the classic formula. Where is my mistake? Is the treatment correct at all? Thanks in advance.

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  • $\begingroup$ $p'(x)\neq ma(x)$ $\endgroup$ – Ali Jul 28 '17 at 19:44
  • $\begingroup$ Why is $p$ a function of $x$? $\endgroup$ – WillO Jul 28 '17 at 19:46
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    $\begingroup$ You're differentiating with respect to $x$ to get $F$ but with respect to $t$ to get $v^\prime=a$... $\endgroup$ – lemon Jul 28 '17 at 19:46
  • $\begingroup$ Care to explain the downvote? $\endgroup$ – Vitor C Goergen Jul 28 '17 at 23:46
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Simple: $\vec{p}^{\prime}(\vec{x})$ is not $m\vec{a}$. In particular, $\vec{p}^{\prime}$ is the space derivative of $\vec{p}$, but $m\vec{a}=m\partial_t\vec{v}=\partial_t(m\vec{v})=\partial_t\vec{p}$.

You can correct your argument by noting that $\partial_t\vec{x}=\vec{v}$ and applying chain rule.

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It should be noted that the Hamiltonian (what you called $E(\textbf{x})$) is not constant in space, but is only constant along the path your particle takes. The real requirement, then, is that $E$ is constant in time, ie $\dot{E}=0$. If you are to follow through taking time derivatives, you would get

$$\dot{E}=\frac{1}{m}\textbf{p}\cdot\dot{\textbf{p}}+\textbf{v}\cdot\boldsymbol{\nabla}U=0.$$

Thus, if we have that $\dot{\textbf{p}}=-\boldsymbol{\nabla}U\equiv F$, then the equations of motion $\dot{E}=0$ are satisfied.

I hope this helps!

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    $\begingroup$ I think this answer is remarkable because it doesn't answer my question, however it clarifies the concepts which for me is a fundamental idea. $\endgroup$ – Vitor C Goergen Jul 28 '17 at 21:33
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Hint:

$$ \frac{d}{dx}\left(\frac {p^2}{2m}\right) \\ =\frac{p}{m}\frac{d}{dx}p \\ =\frac{p}{m}\frac{dp}{dt} \frac{dt}{dx} $$

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