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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:

$F=ma$

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

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

Mr Croutini
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2 Answers2

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Writing

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

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

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

or

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

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.

Ilmari Karonen
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Robert Lewis
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    Hi, Robert, and thanks for your excellent answer. I've edited it, in accordance with the guidelines given in our [help/behavior], 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. – Ilmari Karonen Jun 11 '14 at 22:10
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$$ -\frac{dV}{dx} = m\frac{d}{dt}\frac{dx}{dt} \\ -\frac{dV}{dx}\frac{dx}{dt} = m\frac{dx}{dt}\times \frac{d}{dt}\frac{dx}{dt} $$ 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): $$ \frac{d}{dt}\left(\frac 12 m\left(\frac {dx}{dt} \right)^2 + V(x(t)) \right)=0$$

mookid
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  • 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 :) – Mr Croutini Jun 11 '14 at 16:48
  • actually the result remain true without the assumption on the mass. But in that case the equation of dynamics is wrong. – mookid Jun 11 '14 at 16:50