Lorentz transformation of continuity equation

Question

In a particular reference frame with coordinates $x^mu$, we can define a current density 4-vector $J^mu=(crho,vec{J})$ where $rho$ is the charge density and $vec{J}$ is the current density.
The continuity equation in this frame is then $$frac{partial J^mu}{partial x^mu}=0.$$
How can it be shown that this equation holds in a Lorentz boosted frame with coordinates $x'^mu=Lambda^mu_nu x ^nu$,
i.e. $$frac{partial J'^mu}{partial x'^mu}=0$$ is true?

Frank · Accepted Answer

From $J'^mu=Lambda^mu_nu J^nu$ take derivative on both sides
$$frac{partial J'^mu}{partial x'^mu}=Lambda^mu_nu frac{partial J^nu}{partial x'^mu}$$
Note that $frac{partial}{partial x'^mu}=Lambda^sigma_mufrac{partial}{partial x^sigma}$ Then we have
$$frac{partial J'^mu}{partial x'^mu}=Lambda^mu_nu frac{partial J^nu}{partial x'^mu}=Lambda^mu_nuLambda^sigma_mufrac{partial J^nu}{partial x^sigma}=delta^sigma_nufrac{partial J^nu}{partial x^sigma}=frac{partial J^nu}{partial x^nu}=0$$

Frobenius · Answer

Generally
begin{align}
&texttt{The 4-gradient contravariant vector operator}
nonumber
& qquad qquad qquad partial^{mu}boldsymbol{equiv}dfrac{partial }{partial x_{mu}}boldsymbol{=}left(dfrac{partial }{partial x^{0}},,boldsymbol{-nabla}right) 
tag{01a}label{01a}
&texttt{The 4-gradient covariant vector operator}
nonumber
&qquad qquad qquad partial_{mu}boldsymbol{equiv}dfrac{partial }{partial x^{mu}}boldsymbol{=}left(dfrac{partial }{partial x^{0}},,boldsymbol{+nabla}right) 
tag{01b}label{01b}
end{align}
The 4-divergence of a 4-vector $A^{mu}boldsymbol{=}left(A^{0},mathbf{A}right),A_{mu}boldsymbol{=}left(A^{0},boldsymbol{-}mathbf{A}right) $ is the invariant
begin{equation}
partial^{mu}A_{mu}boldsymbol{=}partial_{mu}A^{mu}boldsymbol{=}dfrac{partial A^{0} }{partial x^{0}}boldsymbol{+nablacdot}mathbf{A}
tag{02}label{02} 
end{equation}
$=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=$
Note : May be in the future you ask yourself if the converse of eqref{02} is true. No, it's not : if $A^{mu}boldsymbol{=}left(A^{0},mathbf{A}right)$ is a 4-dimensional quantity then
begin{equation}
partial_{mu}A^{mu}boldsymbol{=}texttt{invariant}quad boldsymbol{=!ne!Rightarrow}quad A^{mu}boldsymbol{=}texttt{contravariant Lorentz 4-vector}
tag{03}label{03} 
end{equation}
$-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!-!!$
Counter example
Consider that $Y^{mu}left(mathbf{x},tright)boldsymbol{=}left[!!left[ Y^{0}left(mathbf{x},tright),mathbf{Y}left(mathbf{x},tright)right]!!right]$ is a (contravariant) 4-vector function of the space-time coordinates $left(mathbf{x},tright)$.  According to equation eqref{02} its 4-divergence is invariant
begin{equation}
partial_{mu}Y^{mu}boldsymbol{=}dfrac{partial Y^{0} }{partial x^{0}}boldsymbol{+nablacdot}mathbf{Y}boldsymbol{=}texttt{invariant}
tag{04}label{04} 
end{equation}
Now consider a 4-dimensional vector $rm a^{mu}boldsymbol{=}left(a^{0},mathbf{a}right)boldsymbol{=}left(a^{0},a^{1},a^{2},a^{3}right)$ with components 4 arbitrary real numbers constants, that is not depending on the  space-time coordinates $left(mathbf{x},tright)$ and form the 4-dimensional vector
begin{equation}
A^{mu}boldsymbol{=}Y^{mu}boldsymbol{+}rm a^{mu}
tag{05}label{05} 
end{equation}
Then
begin{equation}
partial_{mu}A^{mu}boldsymbol{=}partial_{mu}Y^{mu}boldsymbol{+}overbrace{partial_{mu}rm a^{mu}}^{0}boldsymbol{=}partial_{mu}Y^{mu}boldsymbol{=}texttt{invariant}
tag{06}label{06} 
end{equation}
But $A^{mu}$ as defined by equation eqref{05} in not a 4-vector due mainly to the arbitrariness of $rm a^{mu}$.

Lorentz transformation of continuity equation

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