Modifying Euler-Lagrange equation to multivariable function

[offscene]

Homework Statement

Not exactly homework but I was reading through the book "QFT for the gifted amateur by Lancaster and Blundell" and I was confused about how the line just above equation 1.33 is derived (Image attached below).

Relevant Equations

Euler Lagrange equation, the principle of least action.

I'm confused on how to derive the multidimensional generalization for a multivariable function. Everything makes sense here except the line,

$$
\frac{\delta S}{\delta \psi} = \frac{\partial L}{\partial \psi} - \frac{d}{dx} \frac{\partial L}{\partial(\frac{\partial \psi}{\partial x})} - \frac{d}{dt} \frac{\partial L}{\partial(\frac{\partial \psi}{\partial t})}
$$ and I'm confused about which rule I can use to derive this form of the Euler-Lagrange equation.

 

[vanhees71]

In field theory your Lagrangian is given by the spaticial integral of the Lagrange density ##\mathcal{L}(\phi,\partial_{\mu} \phi)## and thus the action as an integral over spacetime
$$S[\phi]=\int \mathrm{d}^4 x \mathcal{L}(\phi,\partial_{\mu} \phi).$$
In the above example oviously the author considers a field theory in (1+1)-dimensional spacetime, but that doesn't change much. So for clarity I assume here the full case of a (1+3)-dimensional spacetime.

Now note that the gradient of the field, ##\partial_{\mu} \phi=\frac{\partial \phi}{\partial x^{\mu}}## has four components. Now you take the variation of the action wrt. the field,
$$\delta S[\phi]=S[\phi+\delta \phi]-S[\phi]=\int \mathrm{d}^4 x \left [\delta \phi \frac{\partial \mathcal{L}}{\partial \phi} + \partial_{\mu} \delta \phi \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)} \right ].$$
Note that here the Einstein summation convention has been used, i.e., in the second term in the bracket you sum over ##\mu## from ##0## to ##3##. That's because the Lagrange density depends on all four components of the field gradient, ##\partial_{\mu} \phi##, and it's just the chain rule of multivariable calculus used here. It's pretty much the same as in point-particle mechanics, only that in this case the independent variables ##q(t)## are functions only of ##t##.

To get the Euler-Lagrange equations you just have to do an integration by parts in this 2nd term, using the assumption that ##\delta \phi## vanishes at the boundaries of the integration domain. Then you get
$$\delta S=\int \mathrm{d}^4 x \delta \phi \left [\frac{\partial \mathcal{L}}{\partial \phi} - \partial_{\mu} \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)} \right ],$$
i.e.,
$$\frac{\delta S}{\delta \phi} = \frac{\partial \mathcal{L}}{\partial \phi} - \partial_{\mu} \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)}.$$
If you write this out and restricting yourself to (1+1)D spacetime you get the result given in the book.