Formalities of the variational integral

Question

Usually when the variational principle is introduced one starts by defining a Lagrangian density

$${\mathscr L}(x,\phi(x),\partial_{\mu}\phi(x))$$

and an action

$$S[\phi]=\int_R d(x) {\mathscr L}$$

Then one tries to find the form of $\phi(x)$ that makes the value of $S$ be an extremum (either maximum or minimum). To solve this problem one applies a "small" perturbation to $x$ and $\phi$ to see how the value of $S$ is modified. Thus by "taking the limit" to smaller and smaller perturbations results in an infinitesimal change $\delta S$ which is set to zero.

Now there are several ways to implement the "small" perturbation. One is to construct the homotopy between two functions that contains the solution. This means that we have a family of functions $\phi(x;\epsilon)$ parametrized continuously by $\epsilon$. However, one can argue that the homotopy cannot contain all possible variations because it only contains the "intermediate" functions of the deformation of the initial function to the final one.

Another way, which is contained in most physics textbooks, is to consider the transformations: $$x^\mu\to x'^\mu=x^\mu+\delta x^\mu$$ $$\phi(x)\to \phi'(x)=\phi(x)+\delta \phi(x)$$ Where both $\delta x^\mu$ and $\delta \phi(x)$ are arbitrary small functions of $x$. Now, the lagrangian density also depends on $\partial_\mu \phi(x)$ so it is straightforward to take another small function, say $\delta \partial_\mu\phi(x)$, and consider $$\partial_\mu \phi(x) \to \partial_\mu \phi'(x) = \partial_\mu \phi(x)+ \delta\partial_\mu \phi(x)$$ However, taking the derivative of the fourth equation gives $$\partial_\mu \phi(x)\to \partial_\mu \phi'(x)=\partial_\mu \phi(x)+\partial_\mu (\delta \phi(x))$$ Then it is tempting to identify $$\delta \partial_\mu\phi(x)=\partial_\mu (\delta \phi(x))$$ But here is the question: how can we guarantee that the identification is correct? i.e. that $\partial_\mu (\delta \phi(x))$ is small. For example I can think of a variation of the kind $$\delta \phi(x)=\epsilon \sin\left(\frac{1}{x-a}\right)$$ where $\epsilon$ is a small number and $a$ a constant. Then obviously even when $\delta \phi(x)$ is small, $$\partial_\mu (\delta \phi(x))=-\frac{\epsilon}{(x-a)^2}\cos\left(\frac{1}{x-a}\right)$$ so we cannot guarantee that this variation is small close to $a$.

Answer 1

As far as I understand, there is no "identification" of the manner you describe. There is just a single function, $\phi(x)$, on which the action functional depends. (The derivative terms are not independent.) Adding a small variation to this function $\phi(x) \rightarrow \phi(x) + \delta\phi(x)$, the derivative terms automatically transform as $\partial_{\mu} \phi(x) \rightarrow \partial_{\mu}\phi(x) + \partial_{\mu}\delta\phi(x)$ just by straightforward differentiation. I see no real need to consider the quantity $\delta\partial_{\mu}\phi$.

Regarding your question about smallness, remember you can always integrate $\partial_{\mu}\delta\phi$ by parts and drop boundary terms where the variation is zero anyway. This is the standard cure-all in variational calculus. So it doesn't matter if it's small or not: because it appears under an integral sign, any term $f(x) \partial_{\mu}\delta\phi(x)$ just turns into $-(\partial_{\mu}f(x))\delta\phi(x)$, which is small.

Answer 2

The real answer to your question requires a proper definition of the variation (which physicists rarely seem to give unfortunately). When we perform a variation, what we have in mind is a one-parameter transformation $F_\epsilon$ on the space of fields $\phi$ called a flow; $$ \phi\to F_\epsilon(\phi) $$ What makes this transformation a flow is that $$ F_0(\phi) = \phi, \qquad F_{\epsilon_1}(F_{\epsilon_2}(\phi)) = F_{\epsilon_1+\epsilon_2}(\phi) $$ Given a flow, we can essentially series expand the effect of the flow on the field in the small parameter $\epsilon$; $$ F_\epsilon(\phi) = F^{(0)}(\phi) + \epsilon F^{(1)}(\phi) + \mathcal O(\epsilon^2) $$ Determining the coefficient $F^{(1)}(\phi)$ is what physicists really mean when they take variational derivatives. Note that there is a simple formula for this coefficient; $$ F^{(1)}(\phi)(x) = \partial_\epsilon \Big(F_\epsilon(\phi)(x)\Big)\big|_{\epsilon=0} $$ To make contact with your notation, we would observe that if we write $$ \phi\to\phi+\delta\phi + \cdots $$ then we can identify $$ \delta\phi = \epsilon F^{(1)}(\phi) $$ To make things simpler, let's abuse notation a little and write $F^{(1)}\phi = \Delta\phi$, then we want to show that $$ \Delta(\partial_\mu\phi) = \partial_\mu(\Delta\phi) $$ This follows immediately if the flow of the derivative of the field is is defined as the derivative of the flowed field $$ F_\epsilon(\partial_\mu\phi) = \partial_\mu F_\epsilon(\phi) $$ which is an appropriate definition in this context because the idea is that we want to see what happens to the action to first order if everywhere we see a $\phi$, we replace it with its transformed counterpart $F_\epsilon(\phi)$.

In case you're interested, this stuff is intimately related to the following mathematical objects that I encourage you to look into:

Tangent space

Derivation

Lie Algebra

Let me know how it goes!

Cheers!

Answer 3

Let us for simplicity discuss this in the context of point mechanics with variables $q^i(t)$ [rather than field theory with variables $\phi^{\alpha}({\bf x},t)$].

We have the following short comments to the question (v1):

1. One is not necessarily looking for an extremal (=minimal or maximal) configurations. We are looking for stationary configurations.

2. When deriving the Euler-Lagrange equations, one doesn't make horizontal variation $\delta t=\ldots$. Only vertical variations $\delta q^i=\ldots$ (Horizontal variations, on the other hand, are used in the context of (off-shell) symmetry of the action, cf. Noether's Theorem.)

3. Let us consider OP's example. By redefining time $t\to t+a$, we may assume that $a=0$. Then OP's example reads $$\delta q(t)~=~\epsilon \sin\frac{1}{t}.$$ Now the Lagrangian is a function $L(q,\dot{q},t)$. The action $S[q]$ is therefore a priori only well-defined for differential$^1$ paths $q$. The variation $\delta S$ is by definition $S[q+\delta q] - S[q]$. We must therefore demand that $q+\delta q$ is also a differential path, and hence the variation $\delta q$ is a differential path. This means that the problematic instant $t=0$ cannot belong to the physical time domain $[t_i,t_f]$, and thus OP's example is not a problem.

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$^1$ Let us ignore for simplicity that it is possible to extend the definition of $S[q]$ to include e.g. continuous, piecewise differentiable paths $q$, etc.