Modifying Euler-Lagrange equation to multivariable function

In summary, the conversation discusses deriving the multidimensional generalization of the Euler-Lagrange equation for a multivariable function in field theory. The equation is derived by taking the variation of the action with respect to the field and performing integration by parts, resulting in the equation: $$\frac{\delta S}{\delta \phi} = \frac{\partial \mathcal{L}}{\partial \phi} - \partial_{\mu} \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)}.$$
  • #1
offscene
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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.
Screen Shot 2023-06-06 at 5.15.42 PM.png


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.
 
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  • #2
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.
 
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Related to Modifying Euler-Lagrange equation to multivariable function

What is the Euler-Lagrange equation in the context of multivariable functions?

The Euler-Lagrange equation in the context of multivariable functions is a generalization of the classical Euler-Lagrange equation used in the calculus of variations. It is used to find the extrema of functionals, which depend on multiple variables and their partial derivatives. For a functional \( J \) defined as \( J = \int_{\Omega} L(x, y, u, u_x, u_y) \, dx \, dy \), where \( L \) is the Lagrangian and \( u \) is the function to be optimized, the Euler-Lagrange equation is given by \( \frac{\partial L}{\partial u} - \frac{\partial}{\partial x} \left( \frac{\partial L}{\partial u_x} \right) - \frac{\partial}{\partial y} \left( \frac{\partial L}{\partial u_y} \right) = 0 \).

How do you derive the Euler-Lagrange equation for a multivariable function?

To derive the Euler-Lagrange equation for a multivariable function, start by considering a functional \( J \) that depends on a function \( u \) and its partial derivatives. Introduce a small perturbation \( \eta \) to the function \( u \) and express the functional in terms of this perturbed function. Expand the functional to first order in \( \eta \) and integrate by parts to isolate \( \eta \) and its derivatives. By setting the first variation of the functional to zero, you obtain the Euler-Lagrange equation: \( \frac{\partial L}{\partial u} - \frac{\partial}{\partial x} \left( \frac{\partial L}{\partial u_x} \right) - \frac{\partial}{\partial y} \left( \frac{\partial L}{\partial u_y} \right) = 0 \).

What are the boundary conditions for the Euler-Lagrange equation in multiple dimensions?

The boundary conditions for the Euler-Lagrange equation in multiple dimensions depend on the specific problem being solved. Typically, they involve specifying the values of the function \( u \) and possibly its derivatives on the boundary of the domain \( \Omega \). For example, Dirichlet boundary conditions specify the value of \( u \) on the boundary, while Neumann boundary conditions specify the value of the normal derivative of \( u \) on the boundary. Mixed boundary conditions can also be used, where some parts of the boundary have Dirichlet conditions and others have Neumann conditions.

Can the Euler-Lagrange equation be applied to vector-valued functions?

Yes, the Euler

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