Energy of Scalar Field: Evaluating Rubakov's Expression

[quantum_smile]

Homework Statement

My question is just about a small mathematical detail, but I'll give some context anyways.
(From Rubakov Sec. 2.2)
An expression for energy is given by
$$E= \int{}d^3x\frac{\delta{}L}{\delta{}\dot{\phi}(\vec{x})}\dot{\phi}(\vec{x}) - L,$$
where L is the Lagrangian,
$$L=\int{}d^3{}x(\frac{1}{2}\dot{\phi}^2-\frac{1}{2}\partial_i\phi\partial_i\phi-\frac{m^2}{2}\phi^2).$$
To derive the expression for energy, Rubakov says that
$$\frac{\delta{}L}{\delta{}\dot{\phi}(\vec{x})}=\dot{\phi}(\vec{x}).$$
What I want to know is, simply, how does he get this expression for
$$\frac{\delta{}L}{\delta{}\dot{\phi}(\vec{x})}$$?

Homework Equations

The Attempt at a Solution

If I evaluate the expression, I just get
$$\delta{}L=\int{}d^3x(\dot{\phi}).$$

Where'd the integral go in Rubakov's expression?

 

[Matterwave]

I may be wrong, but I believe what is happening is a confusion between the Lagrangian and the Lagrangian density. Look at the expression for the energy, it has an integral in it, so probably the $L$ which appears in there should actually be the Lagrangian density $\mathcal{L}$ defined by $L=\int d^3x \mathcal{L}$

 

[quantum_smile]

Ah, that would make a lot of sense (and fix the weird unit problem). Maybe there's a tiny typo in the text.

 

[Matterwave]

A lot of field theory texts refer to the Lagrangian density as simply the "Lagrangian", so the language might be confusing. Usually the notation is used so that the Lagrangian density is in a calligraphic font though.

 

[paranormal]

The integral is still there in Rubakov's expression, but it has been absorbed into the functional derivative \frac{\delta{}L}{\delta{}\dot{\phi}(\vec{x})}. This is a common practice in field theory to simplify expressions and make them more compact. In this case, the integral over space has been taken into account in the functional derivative, which is defined as the derivative of the Lagrangian with respect to the field at a specific point in space. So the expression is still valid, but it has been rewritten in a more compact form.