# Action principle and Functional derivative in CM

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. $$\frac{\partial \mathcal{L}}{\partial \phi}- \frac{d}{dt}(\frac{\partial \mathcal{L}}{\partial \dot{\phi}})$$

• @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. – SaidurRahman May 24 at 16:03

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.

• Okay. I've corrected the post. $\lambda$ should have been given as an index. I didn't mean it in the place of $t$. – SaidurRahman May 24 at 15:59
• 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. – PhysicsMan May 24 at 16:13
• 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? – SaidurRahman May 24 at 16:23
• Yes, following this recipe will lead you to the E-L equations. – PhysicsMan May 24 at 16:26
• 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? – SaidurRahman May 24 at 16:37