How to Relate the Functional Derivative to Infinitesimal Change in Noether's Theorem

Question

When the Euler-Lagrange equation or the expression for Noether current are derived the term infinitesimal change is often used. For example, we write $\phi\rightarrow \phi + \delta\phi$ and say that $\delta\phi$ is an infinitesimal change in the field.

I have recently been introduced to the functional derivative, which made me think that I could now throw the notion of infinitesimal change away, and find the Euler-Lagrange equation and the expression for the Noether current "more rigorously" using the functional derivative. However, I have not been able to find a proof for for the Euler-Lagrange equation/Noether current without some mentioning of `infinitesimal change'. Is this possible?

The proof for Noether's current on Wikipedia (under Field-theoretic derivation) states something like:

\begin{align*} 0 &\overset{!}{=} \delta S = \int d^4 x \left(\frac{\partial \mathcal{L}}{\partial\phi}\delta\phi + \frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\delta\partial_{\mu}\phi + \frac{\partial\mathcal{L}}{\partial x^{\mu}}\delta x^{\mu}\right) \\ &= \int d^4x \left(\frac{\partial \mathcal{L}}{\partial\phi}\delta\phi + \partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)} \delta\phi\right) - \partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\right)\delta\phi + \partial_{\mu}\mathcal{L}\delta x^{\mu}\right) \\ &=\int d^4 x \underbrace{\left(\frac{\partial\mathcal{L}}{\partial\phi} - \partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\right)\right) \delta\phi}_{\text{Euler-Lagrange eq.}} + \partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\delta\phi + \mathcal{L}\delta x^{\mu}\right), \end{align*} where the second term is argued to vanish since it is a total derivative and by Stoke's Theorem. This second term is the Noether's current: \begin{align} j^{\mu} = \frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\delta\phi + \mathcal{L}\delta x^{\mu}. \tag*{(1)} \end{align} But I find myself confused as to how to interpret $\delta\phi$ and $\mathcal{L}\delta x^{\mu}$. What is $\delta x^{\mu}$? - Without saying it is an infinitesimal change in $x^{\mu}$. Usually, the Noether current is expressed like \begin{align} j^{\mu} = \frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\Delta\phi + J^{\mu}, \tag*{(2)} \end{align} where $\Delta\phi$ would be the infinitesimal change in $\phi\rightarrow \phi + \Delta\phi$ and $J^{\mu}$ an arbitrary surface term. How can I go from (1) to (2), while avoiding saying that $\delta f$ is something infinitesimal?

Answer 1

One way to do a calculation similar to OP's without a notion of infinitesimal is to consider a one-parameter deformation of the field $\phi^\epsilon$ (with $\phi = \phi^0$) and define $\delta$ to mean the derivative with respect to $\epsilon$, evaluated at $\epsilon = 0$, eg $\delta S = \frac{\text d}{\text d\epsilon} S[\phi^\epsilon]\Big|_{\epsilon= 0}$. Then, the calculation unfolds (without the $\delta x^\mu$) : \begin{align} \delta S &= \int \text d^4 x \delta \mathcal L \\ & = \int d^4 x \left(\frac{\partial \mathcal{L}}{\partial\phi}\delta\phi + \frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\delta\partial_{\mu}\phi\right) \\ & = \int d^4 x \left(\frac{\partial \mathcal{L}}{\partial\phi}\delta\phi + \frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\partial_{\mu}\delta\phi\right) \\ &= \int d^4x \left(\frac{\partial \mathcal{L}}{\partial\phi}\delta\phi + \partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)} \delta\phi\right) - \partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\right)\delta\phi \right) \\ &=\int d^4 x \underbrace{\left(\frac{\partial\mathcal{L}}{\partial\phi} - \partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\right)\right) \delta\phi}_{\text{Euler-Lagrange eq.}} + \partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\delta\phi \right) \end{align}

The second line follows from the first by interchanging the derivative with respect to $\epsilon$ and the integral, the fourth follows from the third by interchanging the derivatives with respect to $\epsilon$ and $x^\mu$.

If the deformation under consideration is a symmetry, then $\delta \mathcal L = \partial_\mu K^\mu$ so that the action is only changed by boundary terms. We find that for fields $\phi$ satisfying the Euler-Lagrange equation, the current $j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\delta\phi - K^\mu$ is conserved.

The presence of the $\delta x^\mu$ in some derivations and not others is, I believe, the result of the different choices between passive and active transformations. In principle, one could also consider, simultaneous to the deformation of $\phi$ a one parameter family of change of variables $x\mapsto x^\epsilon$ and the define the variation $\delta$ as before. I am unsure how to make this work properly though.

Answer 2

I will be quoting a lot from the functional derivative page on wikipedia, since it provides a great reference.

Recall that a function is an object which takes as input a function and has as output a number. In physics functionals are usually integrals.

Example: $$F[\rho]=\int\mathrm d x\, \rho(x)^2$$ I will now write $F[\rho]$ as $F[\rho(x)]$. This is not good notation, because $x$ is a dummy variable and, as far $F$ is concerned, it does not exist. But it will make some things more clear.

One handwavy way of calculating the functional derivative would be \begin{align} \frac{\delta F[\rho(x)]}{\delta\rho(y)}=\lim_{\epsilon\rightarrow0}\frac{F[\rho(x)+\epsilon\delta(x-y)]-F[\rho(x)]}{\epsilon} \end{align}

Let's apply this to our $\rho^2$ functional. \begin{align} \frac{\delta F[\rho(x)]}{\delta\rho(y)}&=\lim_{\epsilon\rightarrow0}\frac 1\epsilon\left(\int\mathrm d x\,\left( \rho(x)+\epsilon\delta(x-y)\right)^2-\int\mathrm d x\,\rho(x)^2\right)\\ &=\lim_{\epsilon\rightarrow0}\frac 1\epsilon\left(\int\mathrm d x\,\left(\rho(x)^2+2\epsilon\rho(x)\delta(x-y)+\epsilon^2\delta^2(x-y)\right)-\int\mathrm d x\,\rho(x)^2\right)\\ &=\lim_{\epsilon\rightarrow0}\int\mathrm d x\,\left(2\rho(x)\delta(x-y)+\epsilon\delta^2(x-y)\right)\\ &=\int\mathrm d x\,2\rho(x)\delta(x-y)\\ &=2\rho(y) \end{align} So we see the functional derivative of an integral often consumes the integral and then takes the derivative of the integrand. The derivation above was pretty sketchy, for example it cancelled a $\int \delta^2$ agains the limit $\epsilon\rightarrow 0$ to get zero. Is there a better way to do this? There is!

On the same wikipedia page, the Gateaux derivative is defined as \begin{align} \delta F[\rho,\phi]&=\lim_{\epsilon\rightarrow 0}\frac{F[\rho+\epsilon\phi]-F[\rho]}{\epsilon}\\ &=\left.\frac{d}{d\epsilon}F[\rho+\epsilon\phi]\right|_{\epsilon=0} \end{align} What does the $\phi$ do here? It is a sort of test function, and later it will take on the role $\delta \rho$, which you are used to. If your functional is defined as an integral, you can define

\begin{align} \delta F[\rho,\phi]&=\int \mathrm d x\,\frac{\delta F}{\delta\rho}(x)\phi(x) \end{align} where $\frac{\delta F}{\delta\rho}(x)$ is now defined as the functional derivative of $F$. If you plug in $\phi=\delta\rho$, this will become the more familiar

\begin{align} \delta F[\rho,\delta\rho]&=\int \mathrm d x\,\frac{\delta F}{\delta\rho}(x)\delta\rho(x) \end{align}

Let's now see how the weird juggling with infinitessimals relates to this new definition, by deriving the $\rho^2$ functional derivative one more time.

\begin{align} F[\rho+\epsilon\delta \rho]&=\int\mathrm d x\, (\rho+\epsilon\delta\rho)^2\\ &=\int\mathrm d x\,\rho^2+\epsilon\int\mathrm d x\, 2\rho\,\delta\rho+\epsilon^2\int \mathrm d x\, \delta\rho^2\\ &=F[\rho]+\epsilon\,\delta F[\rho,\delta\rho]+\mathcal O(\epsilon^2) \end{align} Using one or two more lines, you can find $\frac{\delta F}{\delta\rho}=2\rho(x)$. Note that $\delta\rho$ always has the same order as $\epsilon$, so you can take a mental shortcut and just don't write down the epsilons and use the $\delta\rho$ as a bookkeeping device. Which is what you often see in introductory physics textbooks.

Answer 3

1. It is not entirely clear what OP is trying to achieve. If OP is trying to avoid using infinitesimal variations and infinitesimals, one can in principle equivalently use vector fields and Lie derivatives.

2. More mathematically, calculus of variations of an action functional can be formulated on a variational bi-complex using jet bundles.