- #1
MathematicalPhysicist
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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?!
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?!