1. Sep 20, 2008

### flix

ok, quick and dirty and stupid question about calculation rules with 4 gradients:

consider the Klein Gordon Lagrangian $$L_{KG} = \frac{1}{2} \partial_{\mu}\Phi\partial^{\mu} \Phi - \frac{1}{2} m^2 \Phi^2$$.

Why is

$$\partial_{\mu} \left( \frac{\partial L_{KG} }{\partial(\partial_{\mu} \Phi)} \right) = \partial_{\mu}\partial^{\mu} \Phi$$

Where does the factor 2 come from that cancels out the 1/2 ?

2. Sep 20, 2008

### malawi_glenn

have you taken the lagrangian and 4gradient from same source?

I have always written KG lagrangian (density) as: $$L_{KG} = (\partial_{\mu}\Phi) ^{\dagger}\partial^{\mu} \Phi - m^2 |\Phi |^2$$

Then the 4gradient is the one you have written.

3. Sep 20, 2008

### flix

same source.

the factors 1/2 are there throughout, and it certainly makes sense for the mass term where a factor 2 comes from differentiating.

But where does the factor 2 come from when differentiating by $$\partial_{\mu} \Phi$$ ?? Probably I miss out a very simple thing...

4. Sep 20, 2008

### cristo

Staff Emeritus
I can't see where it comes from either, but then I often miss basic things.

Is there some reason you feel the 2 should be there?

5. Sep 20, 2008

### flix

well yes, since applying the Euler Lagrange equation on the KG Lagrangian should produce the KG equation:

EL: $$\frac{\partial L}{\partial \Phi} - \partial_{\mu} \left( \frac{\partial L}{\partial(\partial_{\mu} \Phi} \right) = 0$$

KG equation: $$(\square + m^2) \Phi(x, t) = 0$$

6. Sep 20, 2008

### haushofer

Maybe I'm overlooking something, but as far as I can see the factor 2 comes from the product rule. It gives you 2 delta functions.

7. Sep 20, 2008

### cristo

Staff Emeritus
Ok, I see. Well, as I said above, I always miss obvious things: note that $\partial^{\mu}\varphi$ and $\partial_{\mu}\varphi$ are not independent, thus your derivative will include two terms. We can rewrite the Lagrangian as $$\mathcal{L}=\frac{1}{2}g^{\mu\nu}\partial_{\mu}\varphi\partial_{\nu}\varphi-\frac{1}{2}m^2\varphi^2$$. Differentiating wrt $\partial_{\mu}\varphi$ then yields $$\frac{1}{2}\left[\partial_{\nu}\varphi g^{\mu\nu}+\delta_{\mu\nu}\partial_{\mu}\varphi g^{\mu\nu}\left]=\frac{1}{2}\left[2\partial^{\mu}\varphi\left]$$, which yields the result.

Does that make sense?

Edit: Looks like I was beaten to it!

8. Sep 21, 2008

### flix

Thank you so much!

I never really liked the covariant picture, although it looks very elegant. It always leads to me missing out basic things.
I really have to dig into it now...