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Given a scalar field $\phi_{(t,x,y,z)}$, an action S is defined as:

$$S = \int L_{(\phi, \partial_\mu \phi)}d^4x$$

Does this notation mean explicitly the quadruple integration over all $\{t, x, y, z\}$ all the way to infinity, as in:

$$S = (\int_{t=-\infty}^\infty (\int_{x=-\infty}^\infty (\int_{y=-\infty}^\infty (\int_{z=-\infty}^\infty L_{(\phi, \partial_\mu \phi)} dt)dx)dy)dz)$$

and that the Euler-Lagrange equation is now after applying Minkowsky $g^{\mu\nu}$(edited):

$$\frac{\partial}{\partial t} (\frac{\partial L}{\partial \phi_{,t}}) - \frac{\partial}{\partial x} (\frac{\partial L}{\partial \phi_{,x}}) - \frac{\partial}{\partial y} (\frac{\partial L}{\partial \phi_{,y}}) - \frac{\partial}{\partial z} (\frac{\partial L}{\partial \phi_{,z}}) = \frac{\partial L}{\partial \phi}$$

Would this be the right definition of S for a scalar field?

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    $\begingroup$ The definition of $S$ is correct, but the E-L equation is not of that form: see the wiki page $\endgroup$ Mar 21, 2023 at 17:00
  • $\begingroup$ Can you fix the latex? $\endgroup$ Mar 21, 2023 at 21:59
  • $\begingroup$ @ArcticChar i have edited the E-L into the question, does it look more correct? $\endgroup$
    – James
    Mar 21, 2023 at 22:26

2 Answers 2

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Given a first-order action functional of the form $$ S[\phi]~=~\int d^nx ~{\cal L}(\phi(x),\partial\phi(x),x), \tag{1}$$ the Euler-Lagrange (EL) equations is given by the functional/variational derivative $$ 0~=~\frac{\delta S[\phi]}{\delta \phi(x)} ~=~\frac{\partial {\cal L}(\phi(x),\partial\phi(x),x)}{\partial\phi(x)}-\sum_{\mu}\frac{d}{dx^{\mu}} \frac{\partial {\cal L}(\phi(x),\partial\phi(x),x)}{\partial\phi_{,\mu}(x)}.\tag{2}$$

Notice in particular that the Minkowski metric does not enter eq. (2) explicitly.

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$$\frac{\partial}{\partial t} (\frac{\partial L}{\partial \phi_{,t}}) - \frac{\partial}{\partial x} (\frac{\partial L}{\partial \phi_{,x}}) - \frac{\partial}{\partial y} (\frac{\partial L}{\partial \phi_{,y}}) - \frac{\partial}{\partial z} (\frac{\partial L}{\partial \phi_{,z}}) - \frac{\partial L}{\partial \phi} = 0$$ is the general form of the 4-dimensional Euler-Lagrange condition.

Taking the Klein-Gordon L as an example,

$$L = \frac{1}{2}[(\frac{d\phi}{dt})^2-(\frac{d\phi}{dx})^2-(\frac{d\phi}{dy})^2-(\frac{d\phi}{dz})^2 - k^2\phi^2]$$

Consider now a slightly deformed field $$L_{\phi + \epsilon} = \frac{1}{2}[(\frac{d(\phi+\epsilon)}{dt})^2-(\frac{d(\phi+\epsilon)}{dx})^2-(\frac{d(\phi+\epsilon)}{dy})^2-(\frac{d(\phi+\epsilon)}{dz})^2 - k^2(\phi+\epsilon)^2]$$

After much algebraic manipulation, we end up with

$$L_{\phi+\epsilon}-L_{\phi} = \epsilon(\frac{d^2\phi}{dt^2}-\frac{d^2\phi}{dx^2}-\frac{d^2\phi}{dy^2}-\frac{d^2\phi}{dz^2} - k^2\phi) + higher.order. terms$$

Since extremization of L means

$$\int_t\int_x\int_y\int_z(L_{\phi-\epsilon}-L_{\epsilon}) dtdxdydz = 0$$

it follows that

$$\int_t\int_x\int_y\int_z\epsilon(\frac{d^2\phi}{dt^2}-\frac{d^2\phi}{dx^2}-\frac{d^2\phi}{dy^2}-\frac{d^2\phi}{dz^2} - k^2\phi) dtdxdydz = 0$$

Since $\epsilon$ itself can be any arbitrary field, this result is only possible when the entire multiplying factor for $\epsilon$ is identically zero, ie.

$$\frac{d^2\phi}{dt^2}-\frac{d^2\phi}{dx^2}-\frac{d^2\phi}{dy^2}-\frac{d^2\phi}{dz^2} - k^2\phi = 0$$

which can be checked to conform to the general form of the Euler-Lagrange equation

$$\frac{\partial}{\partial t} (\frac{\partial L}{\partial \phi_{,t}}) - \frac{\partial}{\partial x} (\frac{\partial L}{\partial \phi_{,x}}) - \frac{\partial}{\partial y} (\frac{\partial L}{\partial \phi_{,y}}) - \frac{\partial}{\partial z} (\frac{\partial L}{\partial \phi_{,z}}) - \frac{\partial L}{\partial \phi} = 0$$

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