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}
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 $\,J^{\alpha}$.)


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?.}


1 Answer 1


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 $$Q=\int d^4x J^0\delta(t)=\int d^3x J^0\int dt\delta (t)=\int d^3 J^0.$$

  • $\begingroup$ 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. '' $\endgroup$ Aug 3, 2021 at 3:52
  • $\begingroup$ Interesting question. I will think about this later. $\endgroup$
    – Kksen
    Aug 3, 2021 at 5:25
  • 1
    $\begingroup$ This requires a rather elaborate explanation. For a reference,you can read Chapter 5 of Core Principles of Special and General Relativity by James Luscombe, which contains explanation of integration on Minkowski space. $\endgroup$
    – Kksen
    Aug 3, 2021 at 5:41
  • $\begingroup$ 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? $\endgroup$
    – James
    Sep 10, 2023 at 12:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.