Lagrangian of Klein-Gordon

1. The problem statement, all variables and given/known data

I'm trying to derive the Klein-Gordon equation from its lagrangian density

[tex] \mathcal{L} = - \frac{1}{2} \partial^{\mu} \varphi \partial_{\mu} \varphi - \frac{1}{2} m^2 \varphi^2 + \Omega_0 [/tex]

(Srednicki p.24)


2. Relevant equations

[tex] S = \int d^4x \mathcal{L} [/tex]

[tex] \delta S = 0 [/tex]


3. The attempt at a solution

So here is what I got so far,

[tex] \int d^4x \left[ -\frac{1}{2} \partial^{\mu} (\varphi + \delta\varphi) \partial_{\mu} (\varphi + \delta\varphi)- \frac{1}{2} m^2 (\varphi+\delta\varphi)^2 + \Omega_0 \right] = 0[/tex]


[tex] \int d^4x \left[ -\frac{1}{2} \partial^{\mu} \delta\varphi \partial_{\mu} \varphi -\frac{1}{2} \partial^{\mu} \varphi \partial_{\mu} \delta\varphi - m^2 \varphi \delta\varphi \right] + \int d^4x \left[\partial^{\mu}\varphi \partial_{\mu} \varphi + \partial^{\mu}\delta\varphi \partial_{\mu} \delta\varphi + \Omega_0 \right] = 0[/tex]

The answer in the book is the first integral on the left. But that would mean that the second integral has to vanish. If this is correct why does the second integral is zero?

 

I get it now, didn't realize we have to subtract that off, now it works.

The second part requires integration by parts of the first two terms,

[tex] \int d^4x \left[ -\frac{1}{2} \partial^{\mu} \delta\varphi \partial_{\mu} \varphi -\frac{1}{2} \partial^{\mu} \varphi \partial_{\mu} \delta\varphi - m^2 \varphi \delta\varphi \right] = 0 [/tex]

to obtain

[tex] \int d^4x \left[+\partial^{\mu}\partial_{\mu} \varphi - m^2 \varphi \right] \delta\varphi = 0 [/tex]

If I do this to the first term

[tex] u = \partial^{\mu} \delta\varphi [/tex] and [tex] v' = \partial_{\mu} \varphi [/tex]

[tex] du = 2 \partial^{\mu} \partial_{\mu} \delta\varphi [/tex] and [tex] v = \varphi [/tex]

then

[tex] (\partial^{\mu} \delta\varphi)\varphi - \int 2 \partial_{\mu}\partial^{\mu} \delta\varphi [/tex]

and the same to the second term and add it up, then I guess the constants with [itex] \delta\varphi [/itex] should vanish, but I will get a constant of two in the integrand, which can't seem to get rid off.

 

That's interesting. So if the boundary integral vanishes, this integral is antisymmetric with respect to u, and v.

[tex] \int_R d^4x \; u \partial_{\mu} v = -\int_R d^4x \; v \partial_{\mu} u [/tex]

So when applying this to the original problem

[tex] \int d^4x \left[ -\frac{1}{2} \partial^{\mu} \delta\varphi \partial_{\mu} \varphi -\frac{1}{2} \partial^{\mu} \varphi \partial_{\mu} \delta\varphi - m^2 \varphi \delta\varphi \right] = 0 [/tex]


this kind of works by flipping u and v in the second term of the integral,

but when flipping u and v in the first term of the integral, [itex] \delta\varphi [/itex] is still in front of the derivate,

[tex] u = \partial^{\mu} \delta\varphi [/tex]

[tex] v = \varphi [/tex]