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flix
Sep20-08, 07:49 AM
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 ?

malawi_glenn
Sep20-08, 10:27 AM
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.

flix
Sep20-08, 03:42 PM
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...

cristo
Sep20-08, 03:50 PM
But where does the factor 2 come from when differentiating by \partial_{\mu} \Phi ?? Probably I miss out a very simple thing...
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?

flix
Sep20-08, 04:03 PM
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

haushofer
Sep20-08, 04:24 PM
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.

cristo
Sep20-08, 04:31 PM
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!

flix
Sep21-08, 08:24 AM
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...