Typical presentations of the Euler-Lagrange equations in field theory assume that the Lagrangian density is a function of the fields and their derivatives, so that the action $S$ is the integral over space and time of: $$S[\phi] = \int L(\phi(x),\partial_\mu \phi(x),x)dx\tag{1}$$ I am curious about the case where the Lagrangian density may itself depend on a functional of the field, such as: $$S[\phi] = \int L(\phi(x),\partial_\mu \phi(x),I[\phi](x),x)dx\tag{2}$$ $$I[\phi](x)=\int f(\phi(y),y,x)dy\tag{3}$$

For example, a non-relativistic Schrodinger field with a Coulomb self-interaction: $$S[\psi] = \int i\psi^*(x,t)\partial_t \psi(x,t) - \frac{1}{2m}\partial_i \psi^*(x,t) \partial_i \psi(x,t) - \frac{q}{2}V[\psi](x,t)\psi^*(x,t)\psi(x,t) dxdt\tag{4}$$ $$V[\psi](x,t) = q\int \frac{\psi^*(y,t)\psi(y,t)}{|x-y|}dy\tag{5}$$

How can the Euler-Lagrange equation be derived in a case like this?

My initial thought is to proceed by calculating $\delta S$ as follows (ignoring boundary terms): $$\delta S=\int (\frac{\partial L}{\partial\phi}-\partial_\mu \frac{\partial L}{\partial\partial_\mu \phi})(x)\delta \phi(x)+\frac{\partial L}{\partial I}(x)\delta I(x)dx \tag{6}$$ $$\delta I=\int \frac{\partial f}{\partial\phi}(y)\delta \phi(y)dy \tag{7}$$ Then we can bring everything into a double integral and make it all proportional to the same variation: $$\delta S=\int\int ((\frac{\partial L}{\partial\phi}-\partial_\mu \frac{\partial L}{\partial\partial_\mu \phi})(x)\delta(x-y)+\frac{\partial L}{\partial I}(x)\frac{\partial f}{\partial\phi}(y))\delta \phi(y)dydx \tag{8}$$ And since the variation is arbitrary, $\delta S=0$ implies: $$(\frac{\partial L}{\partial\phi}-\partial_\mu \frac{\partial L}{\partial\partial_\mu \phi})(x)\delta(x-y)+\frac{\partial L}{\partial I}(x)\frac{\partial f}{\partial\phi}(y)=0\tag{9}$$ This doesn't really make sense, since the $\delta (x-y)$ isn't a proper function, but we can integrate it back out: $$\frac{\partial L}{\partial\phi}-\partial_\mu \frac{\partial L}{\partial\partial_\mu \phi}+\frac{\partial L}{\partial I}\int\frac{\partial f}{\partial\phi}(\phi(y),y,x)dy=0\tag{10}$$ (Where, to be clear, every instance of $L$ above should be evaluated at the arguments $(\phi(x),\partial_\mu \phi(x),I[\phi](x),x)$.) But I have no idea if this is correct.


1 Answer 1


OP has the right idea. We start with the property $$\frac{\delta\phi(x)}{\delta\phi(y)}~=~\delta^n(x-y).\tag{A}$$ If $$ S[\phi]~=~\int \! d^nx~L(x), \qquad L(x)~=~L(\phi(x),\partial\phi(x), I[\phi;x],x), \tag{2}$$ with $$ I[\phi;x]~=~\int \! d^ny~f(\phi(y),y;x),\tag{3} $$ then the functional derivative of $I$ is $$ \begin{align} \frac{\delta I[\phi;x]}{\delta\phi(z)} ~\stackrel{(3)}{=}~&\int \! d^ny~\frac{\partial f(\phi(y),y;x)}{\partial\phi(y)} \frac{\delta\phi(y)}{\delta\phi(z)} \cr ~\stackrel{(A)}{=}~&\ldots~=~\frac{\partial f(\phi(z),z;x)}{\partial\phi(z)},\end{align} \tag{B}$$ and therefore the functional derivative of $S$ is $$\begin{align} \frac{\delta S[\phi]}{\delta\phi(z)}~\stackrel{(2)}{=}~&\int \! d^nx~\left(\frac{\partial L(x)}{\partial\phi(x)}\frac{\delta\phi(x)}{\delta\phi(z)} +\frac{\partial L(x)}{\partial(\partial_{\mu}\phi(x))}\frac{\delta\partial_{\mu}\phi(x)}{\delta\phi(z)} +\frac{\partial L(x)}{\partial I[\phi;x]}\frac{\delta I[\phi;x]}{\delta\phi(z)}\right)\cr ~\stackrel{(A)}{=}~&\ldots~\stackrel{(B)}{=}~\frac{\partial L(z)}{\partial\phi(z)}-\frac{d}{dz^{\mu}}\frac{\partial L(z)}{\partial(\partial_{\mu}\phi(z))}+\int \! d^nx~\frac{\partial L(x)}{\partial I[\phi;x]}\frac{\partial f(\phi(z),z;x)}{\partial\phi(z)}, \end{align}\tag{C}$$ which should be compared with OP's last eq. (10). (Note that in order for functional derivatives to exist, one has to imposes adequate boundary conditions.)


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