My question is regarding a proof given in Greiner's "Relatavistic Quantum Mechanics", 3rd Edition textbook. On pg 148, he proves that the current density $j^{\nu}(x)$ is invariant to a Lorentz transformation so that $$ j'^{\mu}(x')=a_{\nu}^{\mu} j^{\nu}(x) \tag{3.65} $$ where $a_{\nu}^{\mu}$ is the Lorentz transformation matrix. He then writes the continuity equation in Lorentz invariant form: $$ \frac{\partial j^{\mu}(x)}{\partial x^{\mu}}=0 \tag{3.66} $$ He also defines $j^0(x)$ as the probability density: $$ j^{0}(x)=c \varrho(x)=c \psi^{\dagger} \psi(x) \tag{3.66a} $$ The next step is the one I don't understand. He then claims that an invariant probability is guaranteed, because it holds for the Lorentz system of observer A: $$ \frac{\partial}{\partial t} \int j^{0}(x) \mathrm{d}^{3} x=0 \quad \Rightarrow \quad \int j^{0}(x) \mathrm{d}^{3} x=1 \tag{3.66b} $$ and for observer B: $$ \frac{\partial}{\partial t^{\prime}} \int j^{\prime 0}\left(x^{\prime}\right) \mathrm{d}^{3} x^{\prime}=0 \Rightarrow \int j^{\prime 0}\left(x^{\prime}\right) \mathrm{d}^{3} x^{\prime}=1 \tag{3.66c} $$ Please, can someone explain how Eq. (3.66c) follows from the previous equations?
1 Answer
By the properties of Lorentz transformations, one has that $$ \eta_{\mu \nu} a_{\gamma}^{\mu} a_{\sigma}^{\nu}=\eta_{\gamma\sigma} $$ where $\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)$. It follows that $$ \partial x_{\mu}^{\prime}=\eta_{\mu \gamma} \partial x^{\prime \gamma}=\eta_{\mu \rho} a^{\rho}_{\nu} \partial x^{\nu}=\eta_{\mu \gamma} \eta^{\nu \sigma} a_{\nu}^{\gamma} \partial x_{\sigma}=(a^{-1})_{\mu}^{\gamma} \partial x_{\gamma} $$ where the last step holds since $ \eta_{\mu \alpha} \eta^{\gamma\beta} a_{\gamma}^{\mu}=(a^{-1})_{\alpha}^{\beta}$ which can be immediately obtained from $\eta_{\mu \nu} a_{\gamma}^{\mu} a_{\sigma}^{\nu}=\eta_{\gamma\sigma}$. Now your new current density $j^{\prime \mu}(x^{\prime})=a_{\nu}^{\mu} j^{\nu}(x)$ in equation (3.65) has derivatives given by $$ \frac{\partial j^{\prime \mu}(x^{\prime})}{\partial x^{\prime \nu}} =a_{\gamma}^{\mu} \frac{\partial x^{\sigma}}{\partial x^{\prime \nu}} \frac{\partial j^{\gamma}(x)}{\partial x^{\sigma}} =a_{\gamma}^{\mu}(a^{-1})_{\nu}^{\sigma} \frac{\partial j^{\gamma}(x)}{\partial x^{\sigma}} $$ and therefore (unsurprisingly) satisfies the same continuity equation as $j^{\mu}(x)$ in equation (3.66) i.e. $$ \frac{\partial j'^{\gamma}(x')}{\partial x'^{\gamma}}=0 $$ Now equation (3.66c) follows from this conservation law in the same way that (3.66b) follows from (3.66), which is a routine exercise.
