How Do Conservation Laws Apply to a 4 Current in General Relativity?

[fengqiu]

Consider a 4 current
J^\mu and a metric g then conservation laws will require \del_\mu J^\mu = 0
my lecturer gave me a brief problem and I think I'm missing some understanding of it

he writes

What I'm not understanding is, where he states, if we choose B to be the time slice between etc...
This doesn't make sense because to have Q(t1)-Q(t2) don't you need to integrate equation 4 over 2 times different times again? or at least evaluate at t1 and subtract by the value evaluated at t2?

Adam

 

[pervect]

Try imagining a group of charges flowing through space-time. Each charge will have a worldline, the group of worldlines is given the name "a congruence of worldlines", I believe you can regard it as a proper noun like "a flock of sheep".

Then $J^a$ is just a vector field that's tangent everywhere to the congruence. In the discreete approximation, the length of the tangent is always unity (or the value of the charge if the charge is not unity). In the distributed case, we replace the finite set of congruences with a vector field, the component of this vector that's pointing in the time direction is the charge density. The continuity condition $\nabla_a J^a=0$ says that no charges are created or destroyed, the worldlines are continuous. Along with the condition that $J^a$ goes to zero at infinity, this implies that the number of worldlines that pass through t=0 is the number of worldines that pass through t=a.

 

[samalkhaiat]

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-dimensional 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 $$Q(t_{2}) = Q(t_{1}) .$$