# Lagrangian of Klein-Gordon

## Homework Statement

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

$$\mathcal{L} = - \frac{1}{2} \partial^{\mu} \varphi \partial_{\mu} \varphi - \frac{1}{2} m^2 \varphi^2 + \Omega_0$$

(Srednicki p.24)

## Homework Equations

$$S = \int d^4x \mathcal{L}$$

$$\delta S = 0$$

## The Attempt at a Solution

So here is what I got so far,

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

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

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?

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Ben Niehoff
Gold Member
The variation operator $\delta$ is supposed to act like a derivative operator:

$$\delta(ab) = (\delta a)b + a \delta b$$

It looks like what happened is that you forgot two things. First, you need to subtract off the original function:

$$\delta f(\phi) = f(\phi + \delta \phi) - f(\phi)$$

and second, you forgot that $\delta \phi[/tex], because it is an infinitesimal, is nilpotent: $$\delta \phi \delta \phi = 0$$ 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, $$\int d^4x \left[ -\frac{1}{2} \partial^{\mu} \delta\varphi \partial_{\mu} \varphi -\frac{1}{2} \partial^{\mu} \varphi \partial_{\mu} \delta\varphi - m^2 \varphi \delta\varphi \right] = 0$$ to obtain $$\int d^4x \left[+\partial^{\mu}\partial_{\mu} \varphi - m^2 \varphi \right] \delta\varphi = 0$$ If I do this to the first term $$u = \partial^{\mu} \delta\varphi$$ and $$v' = \partial_{\mu} \varphi$$ $$du = 2 \partial^{\mu} \partial_{\mu} \delta\varphi$$ and $$v = \varphi$$ then $$(\partial^{\mu} \delta\varphi)\varphi - \int 2 \partial_{\mu}\partial^{\mu} \delta\varphi$$ and the same to the second term and add it up, then I guess the constants with [itex] \delta\varphi$ should vanish, but I will get a constant of two in the integrand, which can't seem to get rid off.

Ben Niehoff
Gold Member
When integrating by parts, you need to move the derivatives off of $\delta \phi$, not the other way around.

I'm not sure where you got the factor of 2 from in the first place. Integration by parts doesn't work exactly the same way in multi-dimensional integrals. What you need is a generalized divergence theorem, which you can derive from the product rule:

$$\partial_{\mu}(uv) = v \partial_{\mu} u + u \partial_{\mu} v$$

So

$$\int_R d^4x \; u \partial_{\mu} v = \int_R d^4x \; \partial_{\mu}(uv) - \int_R d^4x \; v \partial_{\mu} u = \int_{\partial R} d^3x \; uv - \int_R d^4x \; v \partial_{\mu} u$$

where $\partial R$ means the boundary of the region of integration R. Generally the fields are assumed to vanish on the boundary, so the boundary integral also vanishes.

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

$$\int_R d^4x \; u \partial_{\mu} v = -\int_R d^4x \; v \partial_{\mu} u$$

So when applying this to the original problem

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

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, $\delta\varphi$ is still in front of the derivate,

$$u = \partial^{\mu} \delta\varphi$$

$$v = \varphi$$

Ben Niehoff
Try looking at it again. I think you've confused yourself. You want to move the derivative off of $\delta \phi$, not the other way around.