Noether's currents under dilatations (scaling transformations)

[askalot]

Hello,
1. Homework Statement
Suppose we have the following Lagrangian density, in $3 + 1$ dimensions:
$$L = \frac{1}{2}\partial_{\mu}\phi \partial^{\mu}\phi - g \phi^4$$
Under the dilatation (scaling transformation): $x \rightarrow \lambda x^{\mu}, \phi (x) \rightarrow \lambda^{-1} \phi(\lambda^{-1} x)$, we should calculate the resulting conserved currents.

Homework Equations

Noether's currents' definition formula:
$$D^{\mu} = \frac{\partial L}{\partial (\partial_{\mu} \phi)}\delta\phi - F^{\mu}$$
$$\partial_{\mu} F^{\mu} = \delta L$$

The Attempt at a Solution

The answer would be a simple application of the general Noether's currents' definition formula, if I knew how to calculate: $\delta\phi$ and $F^{\mu}$.

Thank you in advance,
askalot.

 

[DrDu]

I guess $\delta \phi= (\partial \phi/\partial \lambda) d\lambda$.

 

[samalkhaiat]

Hello,
1. Homework Statement
Suppose we have the following Lagrangian density, in $3 + 1$ dimensions:
$$L = \frac{1}{2}\partial_{\mu}\phi \partial^{\mu}\phi - g \phi^4$$
Under the dilatation (scaling transformation): $x \rightarrow \lambda x^{\mu}, \phi (x) \rightarrow \lambda^{-1} \phi(\lambda^{-1} x)$, we should calculate the resulting conserved currents.

Homework Equations

Noether's currents' definition formula:
$$D^{\mu} = \frac{\partial L}{\partial (\partial_{\mu} \phi)}\delta\phi - F^{\mu}$$
$$\partial_{\mu} F^{\mu} = \delta L$$

The Attempt at a Solution

The answer would be a simple application of the general Noether's currents' definition formula, if I knew how to calculate: $\delta\phi$ and $F^{\mu}$.

Thank you in advance,
askalot.

If the coordinates transform like
\bar{x}^{\mu} = e^{-\epsilon} x^{\mu} \approx (1-\epsilon) x^{\mu} ,
i.e.,
\delta x^{\mu} = -\epsilon \ x^{\mu} , the fields transform according to
\bar{\varphi}(\bar{x}) = e^{\epsilon \Delta} \ \varphi(x) , where \Delta is the scaling dimension of the field (\Delta = 1 for scalar field). If you expand both sides to first order in \epsilon, you find
\delta \varphi (x) \equiv \bar{\varphi}(x) - \varphi (x) = \epsilon \left( \Delta + x^{\mu} \partial_{\mu} \right) \varphi (x) .
Okay, now you do the substitution into the Noether current of dilatation D^{\mu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \varphi)} \delta \varphi + \delta x^{\mu} \mathcal{L} .