A Derivatives of Lagrangian Terms: Why We Lower?

[TroyElliott]

In Lagrangians we often take derivatives ($\frac{\partial}{\partial (\partial_{\mu}\phi)}$) of terms like $(\partial_{\nu}\phi \partial^{\nu}\phi)$. We lower the $\partial^{\nu}$ term with the metric and do the usual product rule. My question is why do we do this? Isn't $\partial^{\nu}\phi$ an element of a different vector space? I would think that $\frac{\partial}{\partial (\partial_{\mu}\phi)}$ is really just shorthand for a tensor product of operators, with one being an identity operator in the $\partial^{\nu}$ space. Can anyone explain why this isn't the correct way of looking at doing such a derivative? Thanks!

 

[haushofer]

Well, your Lagrangian also contains implicitly a metric, which is a mapping between vector spaces and dual spaces. Write it out in full and use the chain rule :)

I don't get your "I would think"-remark, though.

 

[haushofer]

Mmm, perhaps I didn't get your question.

 

[Orodruin]

It is not a question of what vector spaces things belong to. It is a question about what things you can vary independently. Due to $\partial^\mu\phi$ being directly related to $\partial_\mu\phi$ through the metric, they do not vary independently (and neither does $\phi$, that is why you have a term $\partial/\partial\phi$.

Now, you could consider the Lagrangian as a function of $\partial^\mu\phi$ and $\partial_\mu\phi$ and obtain an extra derivative in your EL equations. The end result would be the same anyway so it is typically simpler to lower (or raise) all indices and work from there.

 

[TroyElliott]

Well, your Lagrangian also contains implicitly a metric, which is a mapping between vector spaces and dual spaces. Write it out in full and use the chain rule :)

I don't get your "I would think"-remark, though.

I was trying to say that if $\partial_{\nu}\phi$ is in some vector space and $\partial^{\nu}\phi$ is located in a different vector space, both connected through a linear map $g^{\mu \nu}$, I would think that we would have to explicitly choose the vector space that our operator $\partial_{\nu}$ is acting on i.e. either the original vector space or the dual space.

 

[vanhees71]

$\partial_{\nu} \phi$ and $\partial^{\nu} \phi$ are not in a vector space, they are components of vectors (or more precisely vector fields).

 

[Orodruin]

Also, the partial derivatives do not act on vector fields. You generally need a connection for that.

Regardless, I think the issue has been properly addressed by the first few posts.

I was trying to say that if $\partial_{\nu}\phi$ is in some vector space and $\partial^{\nu}\phi$ is located in a different vector space, both connected through a linear map $g^{\mu \nu}$, I would think that we would have to explicitly choose the vector space that our operator $\partial_{\nu}$ is acting on i.e. either the original vector space or the dual space.

See #4. It does not matter if you pick one or use both as long as you do your variation properly.