# Lorentz invariance of the electric charge. (Gravitation and Cosmology by Weinberg)

I was reading the Special Relativity chapter from Weinberg's Gravitation and Cosmology book and could use some help to prove that charge is a Lorentz scalar.

6 Currents and Densities
Suppose that we have a system of particles with position $$\,\mathbf x_{n}(t)\,$$ and charges $$\,e_{n}$$. The current and charge densities are usually defined by \begin{align} \mathbf J(\mathbf x,t) & \boldsymbol{\equiv} \sum\limits_{n} e_{n}\delta^3\left[\mathbf x\boldsymbol{-}\mathbf x_{n}(t)\right]\dfrac{\mathrm d \mathbf x_{n}(t)}{\mathrm d t} \tag{2.6.1}\label{2.6.1}\\ \boldsymbol{\varepsilon}(\mathbf x,t) & \boldsymbol{\equiv} \sum \limits_{n} e_{n}\delta^3\left[\mathbf x\boldsymbol{-}\mathbf x_{n}(t)\right] \tag{2.6.2}\label{2.6.2} \end{align} Here $$\,\delta^3\,$$ is the Dirac delta function, defined by the statement that for any smooth function $$\,f(x)$$, $$\begin{equation} \int\mathrm d^3 x f(\mathbf x)\delta^3\left(\mathbf x\boldsymbol{-}\mathbf y\right) \boldsymbol{=}f(\mathbf y) \nonumber \end{equation}$$ We can unite $$\,\mathbf J\,$$ and $$\,\boldsymbol{\varepsilon}\,$$ into a four-vector $$\,J^{\alpha}\,$$ by setting $$\begin{equation} J^{0}\boldsymbol{\equiv}\boldsymbol{\varepsilon} \tag{2.6.3}\label{2.6.3} \end{equation}$$ that is $$\begin{equation} J^{\alpha}(x) \boldsymbol{\equiv} \sum\limits_{n} e_{n}\delta^3\left[\mathbf x\boldsymbol{-}\mathbf x_{n}(t)\right]\dfrac{\mathrm d x^{\alpha}_{n}(t)}{\mathrm d t} \tag{2.6.4}\label{2.6.4} \end{equation}$$
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We also note that in four-dimensional language $$\begin{equation} \dfrac{\partial}{\partial x^{\alpha}}J^{\alpha}(x)\boldsymbol{=}0 \tag{2.6.6}\label{2.6.6} \end{equation}$$ The Lorentz invariance of this statement is evident.
Whenever any current $$\,J^{\alpha}(x)\,$$ satisfies the invariant conservation law \eqref{2.6.6}, we can form a total charge $$\begin{equation} Q\boldsymbol{\equiv} \int\mathrm d^3 x J^{0}(x) \tag{2.6.7}\label{2.6.7} \end{equation}$$ This quantity is time-independent, because \eqref{2.6.6} and Gauss's theorem give $$\begin{equation} \dfrac{\mathrm d Q}{\mathrm d t}\boldsymbol{=} \int\mathrm d^3 x \dfrac{\partial}{\partial x^{0}} J^{0}(x)\boldsymbol{=}\boldsymbol{-} \int\mathrm d^3 x \boldsymbol{\nabla\cdot}\mathbf J(x) \boldsymbol{=} 0 \nonumber \end{equation}$$ If $$\,J^{\alpha}(x)\,$$ is a four-vector, $$\,Q\,$$ is not only constant but a scalar. To see this, write $$\,Q\,$$ as $$\begin{equation} Q\boldsymbol{=}\int \mathrm d^4 x J^{\alpha}(x)\partial_{\alpha}\theta(n_{\beta}x^{\beta}) \tag{2.6.8}\label{2.6.8} \end{equation}$$ where $$\,\theta\,$$ is the step function $$\begin{equation} \theta(s)\boldsymbol{=} \begin{cases} 1\quad s>0\\ 0\quad s<0 \end{cases} \nonumber \end{equation}$$ and $$\,n_{\lambda}\,$$ is defined by $$\begin{equation} n_{1}\boldsymbol{\equiv}n_{2}\boldsymbol{\equiv}n_{3}\boldsymbol{\equiv}0,\quad n_{0}\boldsymbol{\equiv}\boldsymbol{+}1 \nonumber \end{equation}$$ The effect of a Lorentz transformation on $$\,Q\,$$ is then evidently simply to change $$\,n\,$$: $$\begin{equation} Q'\boldsymbol{=}\int \mathrm d^4 x J^{\alpha}(x)\partial_{\alpha}\theta(n'_{\beta}x^{\beta}) \nonumber \end{equation}$$ $$\begin{equation} n'_{\beta}\boldsymbol{\equiv}\Lambda^{\gamma}_{\hphantom{\gamma}\beta}n_{\gamma} \nonumber \end{equation}$$ and using \eqref{2.6.6}, the change in $$\,Q\,$$ is then $$\begin{equation} Q'\boldsymbol{-}Q\boldsymbol{=}\int \mathrm d^4 x \partial_{\alpha} \left[J^{\alpha}(x)\{\theta(n'_{\beta}x^{\beta})\boldsymbol{-}\theta(n_{\beta}x^{\beta})\}\right] \nonumber \end{equation}$$ The current $$\,J^{\alpha}(x)\,$$ can be presumed to vanish if $$\,\vert\mathbf x\vert\longrightarrow \boldsymbol{+}\infty\,$$ with $$\,t\,$$ fixed.Whereas the function $$θ(n′_β x^β)−θ(n_β x^β)$$ vanishes as |t|⟶+∞ with x fixed. Hence we can apply the four-dimensional Gauss theorem, and find $$\,Q'\boldsymbol{-}Q\boldsymbol{=}0$$; that is, $$\,Q\,$$ is a scalar.
(For the current density $$\,J^{0}\,$$ defined by \eqref{2.6.2} the charge \eqref{2.6.7} is $$\begin{equation} Q\boldsymbol{=}\sum\limits_{n}e_{n} \nonumber \end{equation}$$ which of course is a constant scalar; however, in dealing with the charge and current distributions of extended particles it is important to realize that \eqref{2.6.7} defines a time-independent scalar for any conserved four-vector $$\,J^{\alpha}$$.)

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I have trouble understanding the part where it says "the effect of a Lorentz transformation on Q is simply to change $$n$$." I know the dot product between a covariant and a contravariant vector is a Lorentz scalar. So the dot product of current density and partial derivative function is a Lorentz scalar. But why change $$n$$ to $$n'$$? The term inside the step function is also in the form of a dot product between a covariant and contravariant vector. so shouldn't it be a Lorentz scalar? If the volume element is also invariant can't we say right away that charge is a Lorentz scalar?

{And in the equation of $$Q'-Q$$, how did the current density vector go inside the partial derivative function?.}

Let us denote $$\{x^\alpha\}$$ as the original coordinates and $$\{x'^\alpha\}$$ as the transformed coordinates. There are 4 quantities that undergo Lorentz transformation in total. First is the volume element $$d^4x$$. It transforms as $$d^4y=Jd^4x,$$ where $$J=\sqrt{g/g'}$$ is the Jacobian of the transformation and $$g$$, as well as $$g'$$, denote the determinant of the metric tensor in both coordinate systems respectively. In Special Relativity (SR), both determinants are $$-1$$, since the metric tensor is always the Minkowski metric in all Lorentz frames. So the 4-volume $$d^4x$$ is a Lorentz scalar.
The second quantity and the third quantities are the 4-current vector and the partial derivative respectively. You are correct in that the product of these two is a Lorentz scalar, so we need not to change them. The final quantity is the variable in the Heaviside step function $$\theta(x)$$. Expanding the expression $$\theta(n_{\beta}x^{\beta})$$ gives $$\theta(t)$$ and under a Lorentz transformation, the Heaviside step function becomes $$\theta(at'+bx')$$ for some scalars $$a$$ and $$b$$. Since the same variable $$x$$ is used in the expression in $$Q'$$, then we can express $$a t'+b x'$$ as $$n'_\beta x^\beta.$$ in which the variable $$x$$ now denotes the transformed coordinates.
The 4-current vector can be inserted inside the partial derivative in virtue of expression $$(2.2.6)$$. Using chain rule, we have $$\partial_\alpha J^\alpha\{\theta(n'_\beta x^\beta)-\theta(n_\beta x^\beta)\}+J^\alpha\partial_\alpha\{\theta(n'_\beta x^\beta)-\theta(n_\beta x^\beta)\},$$ where by $$eq.(2.2.6),$$ the left hand side vanishes and we recover the original expression for $$Q$$.
The term in the Heaviside step function is not a dot product between a contravariant vector and a covariant vector as $$n_\beta$$ is certainly not a vector. It is just a convenient way to write up the expression. Finally, it might be instructive to see how the $$eq.(2.6.8)$$ can be converted to the usual expression for electric charge $$Q=\int d^3xJ^0(x).$$ Expanding the sum in step function $$\theta$$ gives $$Q=\int d^4xJ^\alpha\partial_\alpha\theta(t).$$ Using $$\frac{d\theta}{dx}=\delta(x)$$, we have $$Q=\int d^4x J^0\delta(t)=\int d^3x J^0\int dt\delta (t)=\int d^3 J^0.$$
• Understood. Now an explanation to this part will be great. ''The current Jα(x) can be presumed to vanish if |x|⟶+∞ with t fixed. Whereas the function $θ(n'_β x^β)−θ(n_β x^β)$ vanishes as |t|⟶+∞ with x fixed. Hence we can apply the four-dimensional Gauss theorem, and find Q′−Q=0; that is, Q is a scalar. '' Aug 3, 2021 at 3:52
• Weinberg does assume that $n_\beta$ is a four vector when he assumes that $n'_\beta=\Lambda^\alpha_\beta n_\alpha$. Can't we use that to show that $n'_\beta x^\beta$ is Lorentz invariant? Sep 10 at 12:22