For the Noether theorem for pseudoeuclidean 4-spacetime a-current $J_{a}^{\mu}$ is equal to

$$J_{a}^{\mu} = \frac{\partial L}{\partial (\partial_{\mu}\Psi_{k})}Y_{k, a} - \left( \frac{\partial L}{\partial (\partial_{\mu}\Psi_{k})}\partial_{\nu}\Psi_{k} - \delta_{\nu}^{\mu}L \right) X_{a}^{\nu},$$

$$\quad Y_{k, a} = \left(\frac{\partial F_{k}(\Psi , \omega)}{\partial \omega^{a}}\right)_{\omega = 0}, \quad X_{a}^{\nu} = \left(\frac{\partial f^{\nu}(x , \omega)}{\partial \omega^{a}}\right)_{\omega = 0},$$

$F_{k}$ - function of field transformation, $f^{\nu}$ - function of coordinates transformation, $\omega^{a}$ - parameter of transformation.

The charge is equal to $$\partial_{\mu}J_{a}^{\mu} = 0 \Rightarrow Q_{a} = \int J_{a}^{0} (t, \mathbf r)d^{3}\mathbf r .$$

How to "reduce" the number of dimentions in this theorem for euclidean 3-space (i.e. get the formulation of theorem for euclidean space)?

-
Comment to the question (v1): Noether's theorem can be formulated in any number of spacetime dimensions independently. Why do you want to perform a dimensional reduction? – Qmechanic Mar 18 '13 at 19:51
I want to see what the integral corresponds of symmetry of the Galilean transformations. – user8817 Mar 18 '13 at 20:45
And I don't want to formulate theorem for 3-dimensional euclidean space. – user8817 Mar 18 '13 at 20:57
The Noether charges for Lorentz transformations are e.g. treated in this Phys.SE post. The Noether charges for Galilean transformations are e.g. mentioned on Wikipedia. – Qmechanic Mar 18 '13 at 21:27