I am trying to derive Newton's second law from the principle of least action, that is, setting the functional derivative $\frac{\delta S}{\delta x(t)}$ equal to 0. $$S = \int dt' \left[ \frac{m}{2} \left( \frac{dx}{dt'} \right)^2 - V(x(t')) \right] \tag{1} $$ So, \begin{align} \frac{\delta S}{\delta x(t)} &= \int dt' \left[ \frac{m}{2} \frac{\delta}{\delta x(t)} \left( \frac{dx}{dt'} \right)^2 -\frac{\delta V(x(t'))}{\delta x(t)} \right] \tag{2} \\ &= \int dt' \left[ m \frac{dx}{dt'}\frac{d}{dt'}\delta(t-t') - \frac{\delta V(x(t'))}{\delta x(t')}\frac{\delta x(t')}{\delta x(t)} \right] \tag{3} \\ &= - \int dt' \left[ m \frac{d^2x}{dt'^2}\delta(t-t') + \frac{\delta V(x(t'))}{\delta x(t')} \delta (t-t') \right] \tag{4} \\ &= -\left[ m \frac{d^2x}{dt^2} + \color{Red}{\frac{\delta V(x(t))}{\delta x(t)}} \right]. \tag{5} \\ \end{align}

Now that I have calculated $(5)$, and then set the variation of the action equal to zero, I know that $\frac{\delta V(x(t))}{\delta x(t)}$ must be the same as $\frac{\partial V(x(t))}{\partial(x(t))}$ in order to reproduce Newton's second law. How does the functional derivative turn into the partial derivative in this case?

Note: to get the second term in $(3)$, I used chain rule, but for functional derivatives.


2 Answers 2


The definition of the (integral of the) functional derivative (at least a definition that's good enough for physics level rigor) is the difference of the functional evaluated on a path $x(t)$ plus an arbitrary variation $\epsilon(t)$ and the functional evaluated on the path, to leading order in $\epsilon$. In other words \begin{equation} S[x(t)+\epsilon(t)]-S[x]=\int dt \frac{\delta S}{\delta x} \epsilon(t) + O(\epsilon^2) \end{equation} The fact that this definition puts the functional derivative inside of an integral is a reflection of the fact that the functional derivative is a distribution, like a Dirac delta function, it is only well defined inside of an integral.

Now define \begin{equation} S_V[x(t)]=\int dt V(x(t)) \end{equation} Then \begin{equation} S_V[x(t)+\epsilon(t)]=\int dt V(x+\epsilon)=\int dt\left( V(x) + \frac{\partial V}{\partial x}\epsilon+O(\epsilon^2)\right) \end{equation} Comparing with the definition of the functional derivative, we see we can identify \begin{equation} \frac{\delta S_V}{\delta x} = \frac{\partial V}{\partial x} \end{equation} which is the statement you need.


Here's how I think about it. An action is a functional: It eats a function and returns a number. The functional derivative asks: "For very small changes in the function fed to the functional, how how the functionals value change?"

First let's think of a trajectory, $x(t)$. This is what we will feed to the functional. Now let's consider a smooth family of such trajectories, $x_\lambda (t)$. That is, for each $\lambda$ we have a different trajectory, with small changes in $\lambda$ leading to small changes in $x_\lambda (t)$. Assume, in fact that there is a function $\delta x (t)$ such that

$$\delta x (t) = \lim_{\lambda \to 0} \frac{x_\lambda (t) - x_0 (t)}{\lambda}.$$

If each $\lambda$ gives a trajectory, each trajectory gives a real number when fed to a functional, then composition gives a function

$$S[x_\lambda]: \mathbb{R} \to \mathbb{R}$$.

This is just a real function, so we can take its derivative without any navel gazing.

If $S$ is nice enough, then there is a function which we will tantalizingly refer to as $\frac{\delta S}{\delta x}$ such that for any family $x_\lambda$, we have

$$\left.\frac{d S[x_\lambda]}{d \lambda}\right|_{\lambda = 0} = \int \frac{\delta S}{\delta x} \delta x \,dt.$$

So let's deal with a really simple "action" that is all potential:

$$S[x] = \int_{t_i}^{t_f} (V \circ x)(t) \, dt$$

I give you a function $x(t)$, you compose it with V, integrate it, and out pops a real number. If I give you a family of $x_\lambda$, then we have a function

$$S[x_\lambda] = \int_{t_i}^{t_f} (V \circ x_\lambda)(t) \, dt$$

Each $\lambda$ gives a different function, and therefore a different number. It's just a vanilla $\mathbb{R} \to \mathbb{R}$ function. Taking its derivative gives

$$\left.\frac{d S[x_\lambda]}{d \lambda}\right|_{\lambda = 0} = \frac{d }{d \lambda}\int_{t_i}^{t_f} (V \circ x_\lambda)(t) \, dt\\ = \int_{t_i}^{t_f} \frac{d }{d \lambda} (V \circ x_\lambda)(t) \, dt\\ = \int_{t_i}^{t_f} (V^\prime \circ x) \left( \left.\frac{d x_\lambda}{d \lambda}\right|_{\lambda = 0} \right) \, dt\\ = \int_{t_i}^{t_f} (V^\prime \circ x) \delta x \, dt$$

So that by looking at our definition we see that

$$\frac{\delta S}{\delta x} = V^\prime \circ x.$$

Note that the penultimate line follows just from chain rule.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.