Tensor Question (Klein–Gordon equation) [closed]

Question

I have a question following the derivation of the Klein-Gordon equation from a lagrangian. From Eq. (13d), where does $\delta^\mu_\nu$ come from? I guess it's a conversion factor of some sort.

Answer 1

Formally when you derive by the derivative of the field you have to choose a DIFFERENT index from the ones already used in the Lagrangian.

So for instance for the following massless Lagrangian you have

$$ \mathcal{L}=g^{\mu\nu}\partial_{\nu}\phi \partial_{\mu}\phi$$

And the derivative should be:

$$\frac{\partial \mathcal{L}}{\partial\left(\partial_{\alpha}\phi\right)}=... $$

Now when you consider $$ \frac{\partial \mathcal{L}}{\partial \partial_{\alpha}\phi}= g^{\mu\nu}\frac{\partial_{\mu} \phi}{\partial_{\alpha} \phi}\partial_{\nu} \phi \\ + g^{\mu\nu}\frac{\partial_{\nu} \phi}{\partial_{\alpha} \phi}\partial_{\mu} \phi $$

The $\frac{\partial_{\mu} \phi}{\partial_{\alpha} \phi}$ term amounts to the term $\delta^{\alpha}_{\mu}$.

So you get a term with one index which is $\alpha$, which you than have to derive with respect to $\alpha$ like so: $$ \partial_{\alpha}\left( \frac{\partial \mathcal{L}}{\partial \left(\partial_{\alpha}\phi\right)}\right)=\frac{\partial \mathcal{L}}{\partial \phi}$$

Answer 2

If you consider $\partial_\mu\phi$ as a coordinate $y_\mu$ then the second term on the third line reads as $$y_\mu\left(\frac{\partial}{\partial y_\mu}y_\nu\right).$$ By definition of (partial) derivatives this is just $$y_\mu\delta^\mu_\nu$$ with $\delta^\mu_\nu$ the Kronecker delta.