Question : Show in the one-dimensional case, how for potential forces $F(x) = \dfrac{−dV (x)}{dx}$, energy conservation follows from Newton’s 2nd law

From Newton's second law we know $$F=ma=m\ddot{x}$$

How do we derive the conservation of energy equation from this?

So far I have:


$\implies \dfrac{−dV (x)}{dx}=m\ddot{x}$

Now I don't know what to do. I want to integrate, but they're both derivatives of different variables. Thanks in advance.



$m \ddot x = -\dfrac{dV(x)}{dx}, \tag{1}$

we have

$m \ddot x + \dfrac{dV(x)}{dx} = 0; \tag{2}$

multiplying by $\dot x$ yields

$m \ddot x \dot x + \dfrac{dV(x)}{dx} \dot x = 0, \tag{3}$

which, using the chain rule may be re-written as

$\dfrac{1}{2} \dfrac{d(m\dot x^2)}{dt} + \dfrac{dV(x)}{dt} = 0, \tag{4}$


$\dfrac{d}{dt}(\dfrac{1}{2}( m \dot x^2) + V(x)) = 0. \tag{5}$

Now integrating both sides with respect to $t$ reveals that

$\dfrac{1}{2}m \dot x^2 + V(x) = E, \tag{6}$

where $E$ is a constant. QED.

  • 1
    $\begingroup$ Hi, Robert, and thanks for your excellent answer. I've edited it, in accordance with the guidelines given in our help center, to remove the extra "signature" you appended to it. Your posts here are already automatically signed with your username, avatar and a link to your user page, which is where you may include any extra information about yourself if you want. Anyway, have a +1, and please do feel welcome to keep contributing to this site. $\endgroup$ – Ilmari Karonen Jun 11 '14 at 22:10

$$ -\frac{dV}{dx} = m\frac{d}{dt}\frac{dx}{dt} \\ -\frac{dV}{dx}\frac{dx}{dt} = m\frac{dx}{dt}\times \frac{d}{dt}\frac{dx}{dt} $$ Now LHS is $$ -\frac{d}{dt} V(x(t)) $$and RHS is $$ m\frac{d}{dt}\left(\frac 12 \left(\frac {dx}{dt} \right)^2 \right)$$

It means that, if $m$ does not depend on $t$ (already assumed when you wrote the law of dynamics): $$ \frac{d}{dt}\left(\frac 12 m\left(\frac {dx}{dt} \right)^2 + V(x(t)) \right)=0$$

  • $\begingroup$ Yup, got it. It was the multiplication by $\dot{x}$ and then "cancelling out" the variable in which we were differentiating with respect to. Thanks :) $\endgroup$ – Mr Croutini Jun 11 '14 at 16:48
  • $\begingroup$ actually the result remain true without the assumption on the mass. But in that case the equation of dynamics is wrong. $\endgroup$ – mookid Jun 11 '14 at 16:50

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