Let us define the functional
Q[\Sigma]=\int_{\Sigma}d\sigma_{a}(x) V^{a}(x).
where \Sigma denotes an arbitrary space-like hypersurface in space-time (M^{4},\eta^{ab}), and
d\sigma_{a}(x) = \frac{1}{3!}\epsilon_{abcd} dx^{b}dx^{c}dx^{d},
is a 4-vector differential at x. The functional derivative at some point x is defined by
\frac{\delta Q[\Sigma]}{\delta \sigma (x)} = \lim_{\omega (x)\rightarrow0}\frac{Q[\bar{\Sigma}]-Q[\Sigma]}{\omega (x)},
where \omega (x) is the volume enclosed between \bar{\Sigma} and \Sigma. Therefore, according to Gauss' theorem, we have
\frac{\delta Q[\Sigma]}{\delta\sigma (x)}= \partial_{a}V^{a}.
Now, if V^{a}(x) is a conserved vector field, then \delta Q / \delta\sigma = 0 and therefore Q[\Sigma] is independent of \sigma (x). This means that we are free to pick a particular hypersurface to evaluate Q. So, we choose the hyperplane \Sigma : x^{0}= t =\mbox{const.} to evaluate Q;
Q(t) = \int_{t = \mbox{const.}} d^{3}x\, V^{0}(t,\mathbf{x}).
Clearly, this integral is time-independent iff the conserved vector field satisfies the boundary condition
|\mathbf{x}|^{2}V^{i}(x) \rightarrow 0 \ \ \mbox{as} \ |\mathbf{x}| \rightarrow \infty . \ \ (1)
Indeed; dQ/dt = \int d^{3}x\ \partial_{0}V^{0} = -\int d^{3}x \ \partial_{i}V^{i}(x) = 0.
So let us summarize what we have done by the following: If V^{a}(x) is a conserved vector field (i.e., \partial_{a}V^{a}=0) satisfying the boundary condition in eq(1), then Q is independent of the hypersurface on which it is evaluated, i.e., time independent;
<br />
Q = \int_{\Sigma}d\sigma_{a}(y) \ V^{a}(y) = \int_{t = \mbox{const.}}d\sigma_{a}(y) \ V^{a}(y), \ \ (2)<br />
This result will be used below to prove that the charge Q is a Lorentz invariant quantity.
The Lorentz transform of Q is obtained by conjugating it with U(\Lambda) \in SO(1,3);
<br />
\bar{Q} = U^{-1}QU = \int_{t = \mbox{const.}} d\sigma_{a}(x) U^{-1}(\Lambda)V^{a}(x)U(\Lambda).<br />
Since (vector representation of SO(1,3))
<br />
U^{-1}(\Lambda)V^{a}(x)U(\Lambda) = \Lambda^{a}{}_{c}V^{c}(\Lambda^{-1}x).<br />
Thus
<br />
\bar{Q} = \int_{t = \mbox{const.}}d\sigma_{a}(x) \ \Lambda^{a}{}_{c}V^{c}(\Lambda^{-1}x). \ \ (3)<br />
Now, we change integration variables according to
x = \Lambda y \ \ (4)
To find the Jacobian, we note that
<br />
d\sigma_{a}(x) = \frac{1}{3!} \epsilon_{abcd} \Lambda^{b}{}_{p} \Lambda^{c}{}_{q} \Lambda^{d}{}_{r}dy^{p}dy^{q}dy^{r}.<br />
Now, we use the identity
<br />
\epsilon_{abcd}\Lambda^{b}{}_{p}\Lambda^{c}{}_{q} \Lambda^{d}{}_{r} = (\Lambda^{-1})^{s}{}_{a}\epsilon_{spqr}\det \Lambda<br />
Since \det \Lambda = 1, we find
<br />
d\sigma_{a}(x) = \frac{1}{3!}(\Lambda^{-1})^{s}{}_{a} \epsilon_{spqr} \ dy^{p}dy^{q}dy^{r} = (\Lambda^{-1})^{s}{}_{a}d\sigma_{s}(y). \ \ (5)<br />
Inserting eq(4) and eq(5) in eq(3) (with the new integration domain \Sigma) we find
<br />
\bar{Q} = \int_{\Sigma}d\sigma_{s}(y) (\Lambda^{-1})^{s}{}_{a} \Lambda^{a}{}_{c}V^{c}(y) = \int_{\Sigma}d\sigma_{a}(y)V^{a}(y).<br />
Thus, using eq(2), we arrive at
\bar{Q} = \int_{t = \mbox{const.}} d\sigma_{a}(y)V^{a}(y) = Q
This proves that Q is invariant under SO(1,3). qed
Sam
