Questions from Schwinger's particles, sources and fields

[MathematicalPhysicist]

My first question from his first volume.

On page 254, he writes down the action expression:

[tex](3-10.1)W=\int (dx)[K\phi+K^{\mu}\phi_{\mu}+\mathcal{L}][/tex]
Where the lagrangian is:
[tex] \mathcal{L}=-\phi^{\mu}\partial_{\mu} \phi +1/2 \phi^{\mu}\phi_{\mu} -1/2 m^2 \phi^2 [/tex]The consideration of infinitesimal, variable phase transformations of the sources:

[tex](3-10.2)\delta K(x) = ieq \delta \varphi (x) K(x) \ \ \delta K^{\mu}(x) = ieq \delta \varphi (x) K^{\mu}(x)[/tex]

and of the compensating field transformations:
[tex] \delta \phi (x) = ieq \delta \varphi (x) \phi(x) \ \ \delta \phi ^{\mu} (x) = ieq \delta \varphi (x) \phi^{\mu} [/tex]

gives directly (I don't see it how exactly, can someone show me?):
[tex](3-10.4) \delta W = \int (dx) [\phi((x) ieq K(x) + ieq \phi^{\mu} K_{\mu} ] \delta \varphi (x)[/tex]

I don't see how did he get (3-10.4), can someone elighten me.

Shouldn't he also take varaition of the lagrangian?!

 

[eljose79]

Hello,

Thank you for your question. I am happy to help clarify this for you.

In order to understand how the author obtained equation (3-10.4), we need to understand the concept of infinitesimal transformations. In physics, infinitesimal transformations are small changes made to a system that are so small that they can be considered to be infinitely small. These transformations are used to study the behavior of a system under small changes.

In this case, the author is considering infinitesimal, variable phase transformations of the sources, denoted by \delta K(x) and \delta K^{\mu}(x), and compensating field transformations, denoted by \delta \phi (x) and \delta \phi^{\mu} (x). These transformations are related to each other through the parameter \delta \varphi (x), which represents the small change made to the system.

Now, in order to derive equation (3-10.4), the author starts with the action expression (3-10.1). This action represents the total energy of the system, which is given by the integral of the Lagrangian over all space (dx). The Lagrangian, denoted by \mathcal{L}, is a function of the fields \phi and \phi^{\mu}, and their derivatives.

Next, the author substitutes the infinitesimal transformations into the action expression. This results in equation (3-10.3), which is the action expression with the infinitesimal transformations included.

To get to equation (3-10.4), the author then uses the fact that the infinitesimal transformations are infinitesimally small, and therefore, can be neglected in the expression (3-10.3). This allows the author to simplify the expression and obtain equation (3-10.4).

So, to answer your question, the author did not need to take the variation of the Lagrangian in order to obtain equation (3-10.4). The variation of the Lagrangian is used to obtain the equations of motion, but in this case, the author is simply deriving the action expression with the infinitesimal transformations included.

I hope this helps to clarify things for you. If you have any further questions, please don't hesitate to ask.