I want to extremize this well known action. $$S[\phi]=\int \mathcal{L}(\phi(t),\dot{\phi}(t)) dt $$ The result is also well known. It turns out to be E-L equation. The Action principle states that the functional variation or variation of action should be zero for a particular $\phi$ i.e., $$ \delta S=0 $$ So to vary $\phi$, Can I think of this $\phi$ parametrized by $\lambda$ like $\phi \longmapsto \phi_{\lambda}(t)$ and vary the action so that for different $\lambda$ different $\phi$ is assigned to check which $\phi$ extremizes the action? Also the expression of $\delta S$. Isn't $$\delta S=\frac{dS}{d \lambda}$$ in this case which is followed by $$\delta \mathcal{L}=\frac{d \mathcal{L}}{d \lambda}$$ and $$\delta \phi=\frac{d \phi}{d \lambda}~?$$

Another question is - What'd be the notation that represents the functional derivative of $S[\phi]$? Though I know it's something that resides in the integrand and after calculating $\delta \mathcal{L}$ we can get an expression for this like below. $$\frac{\partial \mathcal{L}}{\partial \phi}- \frac{d}{dt}(\frac{\partial \mathcal{L}}{\partial \dot{\phi}}) $$

  • $\begingroup$ @Qmechanic- I've corrected the post. Sorry for using wrong notation. $\lambda$ should have been given as an index. I actually meant $\phi$ is a function of $t$. And $\phi_{\lambda}(t)$ denotes a family of $\phi(t)$'s. $\endgroup$ Commented May 24, 2019 at 16:03

2 Answers 2


You are in fact correct as far as the definition goes. One can define a directional functional derivative as follows.

Given a functional $S$, a function $\phi_0$ (the "point") and a function $\alpha$ (the "direction"), we can consider the family of functions $\phi_\epsilon = \phi_0 + \epsilon \alpha$. Then $S[\phi_\epsilon]$ is just a regular function of $\epsilon$, and we define the functional derivative of $S$ at $\phi_0$ in the direction of $\alpha$ as

$$\frac{d}{d\epsilon}S[\phi_0 + \epsilon \alpha]\big|_{\epsilon=0}.$$

We say that the functional derivative of $S$ at $\phi_0$ is zero if the above vanishes for all $\alpha$. The problem is that to actually check that the derivative is zero you need to check all possible functions $\alpha$, which is clearly not very practical. That's why there is another, very closely related, definition: we say that $S$ is differentiable if we have that

$$S[\phi_0 + \alpha] = S[\phi_0] + F[\alpha] + \mathcal{O}(\alpha^2),$$

where $F$ is a linear functional and $\mathcal{O}(\alpha^2)$ goes to zero quadratically as $\alpha$ and its derivative go to zero uniformly (see Arnold's Mathematical Methods of Classical Mechanics). If we're lucky, and in physics we're often lucky, we can write the functional $F$ as

$$F[\alpha] = \int f(t) \alpha(t)\, dt,$$

(don't forget that $F$ and $f$ depend on $\phi_0$), and we call $f$ the functional derivative of $S$. $F[\alpha]$ is what I called the directional derivative above; the advantage of this definition is that everything reduces to the single function $f$.


In your notation $\phi$ is already parametrized by $t$, you can relabel it $\phi(t)\rightarrow\phi(\lambda)$ but these are just labels and thus not significant what we choose them to be. Your intuition is correct, you have to find the $\phi(t)$ that extremizes the action.

The derivative of the functional $\delta S$ means the total derivative

$\delta S = \displaystyle\int\delta\mathcal{L}(\phi(t),\dot{\phi}(t))dt=\int\left(\frac{\partial\mathcal{L}}{\partial\phi}\frac{d\phi}{dt}+\frac{\partial\mathcal{L}}{\partial\dot{\phi}}\frac{d\dot{\phi}}{dt}\right)dt$

Where I've used the chain rule.

  • $\begingroup$ Okay. I've corrected the post. $\lambda$ should have been given as an index. I didn't mean it in the place of $t$. $\endgroup$ Commented May 24, 2019 at 15:59
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    $\begingroup$ Indicating $\phi$ with a label $\phi\rightarrow\phi_{\lambda}$ again isn't very meaningful, you are just changing the "name" of the function. What exactly are the differences between the $\phi_{\lambda}$ ? The action principle states that there exist some trajectory in time $\phi(t)$ that a particle will follow and this trajectory is the one that minimizes the action. $\phi(t)$ already encapsulates all possible trajectories in time, you just don't know which one. By minimizing the action you get the constraints (Euler-Lagrange equations) that will help you find the correct trajectory. $\endgroup$
    – PhysicsMan
    Commented May 24, 2019 at 16:13
  • $\begingroup$ So I guess using $\lambda$ in here is redundant. Removing this redundancy, then $\delta S = \frac{dS}{dt}$ and $\delta \mathcal{L}$, $\delta \phi$ also take the same form. Isn't it? $\endgroup$ Commented May 24, 2019 at 16:23
  • $\begingroup$ Yes, following this recipe will lead you to the E-L equations. $\endgroup$
    – PhysicsMan
    Commented May 24, 2019 at 16:26
  • $\begingroup$ Okay. There are a few things I need to be clear about. Why are you calling $\delta S$ the functional derivative? It's a variation of the functional. The derivative should be with respect to $\phi$ !! And What would be the notation of functional derivative in this case? $\endgroup$ Commented May 24, 2019 at 16:37

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