0
$\begingroup$

Starting with the lagrangian density $$L=\frac{1}{2}((\partial_\lambda\phi)(\partial^\lambda \phi) + \mu^2\phi^2),$$ Chen and Li yield the Klein Gordon equation $$(\partial_\lambda \partial^\lambda + \mu^2)\phi=0$$ using Euler-Lagrange formalism.

Looking at the first term, we have $$\partial_{\lambda}\frac{\delta L}{\delta\left(\partial_{\lambda}\phi\right)} = \partial_\lambda \frac{1}{2}\frac{\delta((\partial_\lambda\phi)(\partial^\lambda \phi))}{\delta(\partial_\lambda \phi)}$$

And then I guess by assuming product rule we get $\partial_\lambda \partial^\lambda\phi$. I think I'm missing some understanding in definitions. Shouldn't the covariant and contravariant tensors be formally different objects, so that the partial derivative with respect to $\partial_\lambda \phi$ would consider $\partial^\lambda \phi$ as a constant?

I guess not, but I don't understand what the formal treatment of the partial derivative is, when we differentiate with respect to a tensor of one type and inspect a tensor of the other type.

For example, if we have $$f=a\cdot x_\mu$$for some scalar constant $a$, does $\delta f/\delta x^\mu = a$?

Specifically, the question also holds for the conjugate momentum $$\pi\left(x\right)=\frac{\delta L}{\delta\left(\partial_{0}\phi\right)}=?=\partial_0\phi$$

And how does this partial derivative generalize to a tensor of rank $(n,m)$?

$\endgroup$
3
$\begingroup$

Varying $\phi$ satisfies $$\delta\partial_\lambda\phi=\partial_\lambda\phi,\,\delta\partial^\lambda\phi=\partial^\lambda\delta\phi=g^{\lambda \mu}\delta\partial_\mu\phi$$ so $\frac{\partial\partial_\mu\phi}{\partial\partial_\lambda \phi}=\delta_\mu^\lambda, \,\frac{\partial\partial^\mu\phi}{\partial\partial_\lambda \phi}=g^{\mu\lambda}$. Hence $$\frac{\partial L}{\partial\partial_\lambda\phi}=\frac{1}{2}\left(\delta_\mu^\lambda\partial^\mu\phi+g^{\mu\lambda}\partial_\mu\phi\right)=\partial^\lambda\phi.$$The rest of your questions are addressed the same way.

$\endgroup$
3
$\begingroup$

They are different objects but not independent objects. In particular, they are related to each other via the metric as $$\partial_\mu\phi=\eta_{\mu\nu}\partial^\nu\phi$$$$\partial^\lambda\phi=\eta^{\lambda\kappa}\partial_\kappa\phi$$Thus, the way the variational differentiation of $\partial_\mu\phi\partial^\mu\phi$ with respect to $\partial_\lambda \phi$ works is a bit more multi-stepped than you seem to imagine. Also, if you are using $\lambda$ as a free index then use a different dummy index in the kinetic term, i.e., write $\partial_\mu\phi\partial^\mu\phi$ rather than $\partial_\lambda\phi\partial^\lambda\phi$ for the sake of clarity and also, because, while you can work with $\partial_\lambda\phi\partial^\lambda\phi$ if you are careful enough, it is really wrong. (Treat the dummy indices as if they are in a really spontaneous chemical reaction which just flares up the moment you write down the dummy indices and sums over them but a free index is like an inert element that just stably exists for eternity ;-) So the two cannot be the same as the dummy indices are actually burnt--it looks like they are there but they really are gone!)

Anyway, so here is how that calculation actually works out. $$\frac{\delta(\partial_\mu\phi\partial^\mu\phi)}{\delta(\partial_\lambda\phi)}=\frac{\delta(\partial_\mu\phi\eta^{\mu\nu}\partial_\nu\phi)}{\delta(\partial_\lambda\phi)}$$

$$=\eta^{\mu\nu}\frac{\delta(\partial_\mu\phi\partial_\nu\phi)}{\delta(\partial_\lambda\phi)}$$

$$=\eta^{\mu\nu}\frac{\delta(\partial_\mu\phi)}{\delta(\partial_\lambda\phi)}\partial_\nu\phi+\partial_\mu\phi\frac{\delta(\partial_\nu\phi)}{\delta(\partial_\lambda\phi)}\eta^{\mu\nu}$$ $$=\eta^{\mu\nu}\delta^\lambda_\mu\partial_\nu\phi+\partial_\mu\phi\delta^\lambda_\nu\eta^{\mu\nu}$$ $$=\eta^{\lambda\nu}\partial_\nu\phi+\partial_\mu\phi\eta^{\mu\lambda}$$ $$=\partial^\lambda\phi+\partial^\lambda\phi$$ $$=2\partial^\lambda\phi$$ Neat, right?!


To make the relevant point distinctly clear, I would reemphasize the part you were confused about. Since the covariant and the contravariant derivatives are different but not independent, $$\frac{\delta(\partial^\mu\phi)}{\delta(\partial_\lambda\phi)}=\eta^{\mu\nu}\frac{\delta(\partial_\nu\phi)}{\delta(\partial_\lambda\phi)}=\eta^{\mu\nu}\delta^\lambda_\nu=\eta^{\mu\lambda}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.