ergospherical said:
Here's a first attempt, I wouldn't be surprised if there's a few (or a lot) of issues...
See below.
ergospherical said:
##\dfrac{1}{\sqrt{-g}}\partial_{\mu} \left(\sqrt{-g} \ j^{\mu}\right) = \nabla_{\mu} j^{\mu} = 0## implies that ##d\star \mathbf{J} = 0## therefore letting ##\Omega## be a "cylindrical-like" spacetime volume bounded by two spacelike hypersurfaces ##\Sigma_{1}## and ##\Sigma_{2}## and a timelike hypersurface ##\Pi##,$$0 = \int_{\Omega} d \star \mathbf{J} = \int_{\partial \Omega} \star \mathbf{J} = \int_{\Sigma_1} \star \mathbf{J} - \int_{\Sigma_2} \star \mathbf{J} + \int_{\Pi} \star \mathbf{J}$$Assuming ##\mathbf{J}## vanishes sufficiently fast at spatial infinity then we can put ##\mathbf{J} = 0## on ##\Pi## leading to ##0 = Q[\Sigma_1] - Q[\Sigma_2]##, therefore ##Q[\Sigma]## is time-independent.
That is correct and it is called “the fat-tube theorem”:
A conserved current \mathbf{J}
that vanishes outside some “fat” world tube will produce the same flux \int_{\Sigma} ~^*\!\mathbf{J}
through any closed hypersurface \Sigma intersected by the tube, i.e.,
the charge on a hypersurface will be the same for all hypersurfaces intersected by the tube.
ergospherical said:
For Lorentz invariance,\begin{align*}
\tilde{Q}[\Sigma] &= \int_{\Sigma} \dfrac{\sqrt{-g}}{3!} \tilde{j}^{\mu}(x) \epsilon_{\mu \nu \rho \sigma} d\tilde{x}^{\nu} \wedge d\tilde{x}^{\rho} \wedge d\tilde{x}^{\sigma} \\ \\
&= \int_{\Sigma} \dfrac{\sqrt{-g}}{3!} j^{\alpha}(x) (\epsilon_{\mu \nu \rho \sigma} {\Lambda^{\mu}}_{\alpha} {\Lambda^{\nu}}_{\beta} {\Lambda^{\rho}}_{\gamma} {\Lambda^{\sigma}}_{\delta}) dx^{\beta} \wedge dx^{\gamma} \wedge dx^{\delta} \\ \\
&= \int_{\Sigma} \dfrac{\sqrt{-g}}{3!} j^{\alpha}(x) (|\Lambda|\epsilon_{\alpha \beta \gamma \delta}) dx^{\beta} \wedge dx^{\gamma} \wedge dx^{\delta} \\ \\
&= Q[\Sigma]
\end{align*}since the determinant ##|\Lambda|## of a Lorentz transformation is unity.
If the current is
not conserved, the charge will not be Lorentz invariant. So, in which of the above steps have you used current conservation?
To see that Lorentz invariance of Q follows directly from \partial_{\mu}j^{\mu} = 0(or indirectly from the “tube theorem”), I will present you with 4 methods. To save space, I will drop the wedge signs, set \sqrt{-g} = 1 and write \begin{equation}Q[\Sigma_{1}] = \int_{\Sigma_{1}} \frac{1}{3!} \ j^{\mu}(x) \ \epsilon_{\mu\nu\rho\sigma} \ dx^{\nu} dx^{\rho} dx^{\sigma} , \end{equation} in the first method.
In the second method, I write eq(1) as \begin{equation}Q[\Sigma] = \int_{\Sigma} d\sigma_{\mu}(x) \ j^{\mu}(x) .\end{equation}
In the third method, I will choose \Sigma to be the hyperplane x^{0} = t = \mbox{const.} which is possible because of the tube theorem. So, in the third method I put \begin{equation}Q(t) = \int d^{3}\mathbf{x} \ j^{0}(x) .\end{equation}.
And finally in the fourth method, I will start from the expression \begin{equation}Q = \int d^{3}\mathbf{x} \ j^{0}(0 , \mathbf{x}) .\end{equation}
In all methods, we will be using the following Lorentz transformation law for the vector current: \begin{equation}\bar{j}^{\mu}(x) = U^{\dagger}(\Lambda) j^{\mu}(x) U(\Lambda) = \Lambda^{\mu}{}_{\nu} j^{\nu}(\Lambda^{-1}x) , \end{equation} where U(\Lambda) = e^{- i \epsilon M}, \ \ \ \ \ M = \omega_{\mu\nu}M^{\mu\nu} \in \mathfrak{so}(1,3), and \epsilon \ll 1. In the second method we will use the following (Schwinger) identity \begin{equation}\int_{\Sigma} \ d\sigma_{\mu}(x) \ \partial^{\nu}T^{A}(x) = \int_{\Sigma} \ d\sigma^{\nu}(x) \ \partial_{\mu}T^{A}(x) ,\end{equation} which can be proved for the well-behaved object T^{A}(x)
Method 1: This is your method but properly done. The Lorentz transform of Q[\Sigma_{1}] is obtained by conjugating it with the unitary operator U(\Lambda). So, doing this to eq(1) and using eq(5), we obtain \bar{Q}[\Sigma_{1}] = U^{\dagger} Q[\Sigma_{1}] U = \int_{\Sigma_{1}} \frac{1}{3!} \ \Lambda^{\mu}{}_{\tau} j^{\tau}(\Lambda^{-1}x) \epsilon_{\mu\nu\rho\sigma} \ dx^{\nu}dx^{\rho}dx^{\sigma} . That is it. We have done the Lorentz transformation. Now, in the integral, we let x \to \Lambda x, which is a simple change of integration variables,
not Lorentz transformation. After calculating the Jacobian and using \Lambda^{\mu}{}_{\tau}\Lambda^{\nu}{}_{\alpha}\Lambda^{\rho}{}_{\beta}\Lambda^{\sigma}{}_{\gamma} \ \epsilon_{\mu\nu\rho\sigma} = \epsilon_{\tau\alpha\beta\gamma}, we find (going to the new integration domain \Sigma_{2}) \bar{Q}[\Sigma_{1}] = \int_{\Sigma_{2}} \frac{1}{3!} \ j^{\tau}(x) \epsilon_{\tau\alpha\beta\gamma} \ dx^{\alpha}dx^{\beta}dx^{\gamma} \equiv Q[\Sigma_{2}]. But, according to the tube theorem, Q[\Sigma_{2}] = Q[\Sigma_{1}]. Thus \bar{Q}[\Sigma_{1}] = U^{\dagger}(\Lambda) Q[\Sigma_{1}] U(\Lambda) = Q[\Sigma_{1}] \ \ \ \ \ \ \ \mbox{qed}.
Method 2: In eq(5), put U = 1 - \epsilon \frac{i}{2}\omega_{\alpha\beta}M^{\alpha\beta},\Lambda^{\mu}{}_{\nu} = \delta^{\mu}_{\nu} + \epsilon \omega^{\mu}{}_{\nu}, then expand to first order in \epsilon and factor out the parameters \omega_{\alpha\beta} from both sides of eq(5). If you don’t make a mistake, you obtain the following commutator equation (the infinitesimal version of 5): \big[ iM^{\alpha\beta} ,j^{\mu}(x) \big] = x^{[\alpha}\partial^{\beta ]}j^{\mu} + \eta^{\mu [ \alpha}j^{\beta ]} , where A^{\rho [ \mu}B^{\nu ]} = A^{\rho \mu}B^{\nu} - A^{\rho \nu}B^{\mu}. Using \partial^{\alpha}x^{\beta} = \eta^{\alpha\beta}, we rewrite the above as \big[ iM^{\alpha\beta} ,j^{\mu}(x) \big] = \partial^{\beta} (x^{\alpha}j^{\mu}) - \partial^{\alpha}(x^{\beta}j^{\mu}) + \eta^{\mu [ \alpha}j^{\beta ]} . We now operate with \int_{\Sigma} d\sigma_{\mu}(x) on both sides, then we use eq(2) on the LHS, and apply Schwinger identity ,eq(6), to the first 2 terms on the RHS. This gives us \big[ iM^{\alpha\beta} ,Q\big] = \int d\sigma^{\beta} \partial_{\mu}(x^{\alpha}j^{\mu}) - \int d\sigma^{\alpha} \partial_{\mu}(x^{\beta}j^{\mu}) + \int d\sigma^{[ \alpha} j^{\beta ]} . Using current conservation in the first two terms on the RHS, we get \big[ iM^{\alpha\beta} ,Q[\Sigma]\big] = \int_{\Sigma} d\sigma^{[\beta}j^{\alpha ]} + \int_{\Sigma} d\sigma^{[\alpha}j^{\beta ]} = 0. This means that Q is Lorentz invariant scalar (for it commutes with all the generators of the Lorentz group).
Now my time is up, so I leave remaining two methods for another day.
Good day
