fengqiu said:
my lecturer gave me a brief problem and I think I'm missing some understanding of it
I don’t know why you are confused. It look to me that your lecturer did a reasonable job explaining the issue. So, I will try to fill in some fine details which may help you understand things better.
Okay, we start by writing the covariant divergence of the vector field J^{\mu}(x) in the form
\sqrt{|g|} \nabla_{\mu} J^{\mu} = \partial_{\mu} \left(\sqrt{|g|} \ J^{\mu}\right) . \ \ \ \ \ \ \ \ \ (1)
Integrating (1) over some
bounded, 4-dimensional, region B of space-time, we get
\int_{B} d^{4}x \ \sqrt{|g|} \ \nabla_{\mu}J^{\mu} = \int_{B} d^{4}x \ \partial_{\mu} \left(\sqrt{|g|} \ J^{\mu}\right) . \ \ \ \ \ \ \ \ (2)
Now, if the boundary, \partial B, of the region B is
continuous,
piecewise differentiable and
orientable, we can apply the divergence theorem (Stokes/Gauss) to the RHS of (2) and obtain
\int_{B} d^{4}x \ \sqrt{|g|} \nabla_{\mu}J^{\mu}(x) = \int_{\partial B} dS_{\mu} \ \sqrt{|g^{(3)}|} \ J^{\mu} (x) = \int_{\partial B} d^{3}x \ \sqrt{|g^{(3)}|} \ n_{\mu}J^{\mu}(x) , \ \ \ \ (3)
where the 4-vector differential at x, dS_{\mu}(x) = d^{3}x \ n_{\mu}, denotes a 3-dimentional hyper-surface element on \partial B, n_{\mu} denotes the unit
outward pointing
normal vector on \partial B, and g^{(3)} is the determinant of the
induced metric on \partial B.
Due to the covariant conservation, \nabla_{\mu}J^{\mu} = 0, equation (3) becomes
\int_{\partial B} d^{3}x \ \sqrt{|g^{(3)}|} \ n_{\mu}J^{\mu}(x) = 0 . \ \ \ \ \ \ \ \ \ \ \ (4)
Assuming that the spacetime is foliated into (topologically same)
spacelike hypersurfaces \Sigma, defined by x^{0} = \mbox{const}, we can take \partial B to be the union of two such surfaces (\Sigma_{1} , \Sigma_{2}) with a
timelike surface \mathcal{T} which connects them at spatial infinity: \partial B = \Sigma_{1} \cup \Sigma_{2} \cup \mathcal{T} . This allows you to write (4) as \left( \int_{\Sigma_{1}} + \int_{\Sigma_{2}} + \int_{\mathcal{T}}\right) \left( d^{3}x \ \sqrt{|g^{(3)}|} \ n_{\mu}J^{\mu}(x^{0}, \vec{x}) \right) = 0 . \ \ \ (5) Before we evaluate this equation, let us “picture” what we said as follows: the region B is like a “fat” 4-dimensional world tube, \Sigma_{1} and \Sigma_{2} are 3-dimensional
cross-sections of the tube at x^{0} = t_{1} and x^{0} = t_{2} respectively, with
opposite orientations : n_{\mu}(\Sigma_{1}) = (-1, \vec{0}), n_{\mu}(\Sigma_{2}) = (+1, \vec{0}), and finally \mathcal{T} = [t_{1} , t_{2}] \times S^{2}_{\infty} is the “cylindrical” wall of the tube, n_{\mu}(\mathcal{T}) = (0 , \vec{n}), that joins \Sigma_{1} and \Sigma_{2} at spatial infinity. Now, I hope, you can easily evaluate (5) as follows
\int_{\Sigma_{1}} d^{3} \vec{x} \sqrt{|g(\Sigma)|} J^{0}(t_{1}, \vec{x}) = \int_{\Sigma_{2}} d^{3} \vec{x} \sqrt{|g(\Sigma)|} J^{0}(t_{2}, \vec{x}) + \int_{t_{1}}^{t_{2}} dx^{0} \int_{S_{\infty}^{2}} d\Omega |\vec{x}|^{2} \sqrt{|g(\mathcal{T})|} \ n_{i}J^{i}(x). If the spatial components of the current satisfy |\vec{x}|^{2} J^{i}(x) \to 0 as |\vec{x}| \to \infty, the third integral will give zero contribution, and you end up with
\int_{\Sigma_{1}} d^{3}\vec{x} \sqrt{|g(\Sigma)|} \ J^{0}(t_{1}, \vec{x}) = \int_{\Sigma_{2}} d^{3}\vec{x} \sqrt{|g(\Sigma)|} \ J^{0}(t_{2}, \vec{x}) , which is a time-independent (or charge conservation) statement Q(t_{2}) = Q(t_{1}) .
