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In Weinberg's Gravitation and Cosmology it's stated that

$\delta^4(x - x(\tau))$ is a scalar (because Det $\Lambda$ = 1)

(just after eqn 2.6.5, for reference.)

I cannot make sense of this statement. What is the relationship between the determinant of a Lorentz transformation and the invariance of the 4-delta function?

Metropolis
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    Have you tried working out, or looking up, how the delta transforms under coordinate changes? – Richard Myers May 09 '21 at 03:18
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    Perhaps this answer of mine Physical meaning of the Jacobian in relation to Dirac delta function will help you a little. The (proper homogeneous) Lorentz transformation $,\Lambda,$ is the Jacobi matrix of the space-time 4-dimensional coordinate transformation and $,\det(\Lambda),$ is the Jacobian determinant. – Frobenius May 09 '21 at 05:53
  • @RichardMyers I'll do so now. I had assumed that it was a trivial connection that I should be able to just see from the way it's stated. – Metropolis May 09 '21 at 15:11
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    Scalar functions satisfy $f(\Lambda x) = f(x)$. Dirac delta function satisfies $\delta^4(\Lambda x) = \frac{1}{| \det \Lambda | },, \delta^4(x) = \delta^4(x)$ since $| \det \Lambda| = 1$. Thus, $\delta^4(x)$ is a scalar. – Prahar May 15 '21 at 21:31

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So we want to prove that the expression $\,\delta^4\left[\boldsymbol{x-x}\left(\tau\right)\right]\,$ is a Lorentz invariant scalar, that is \begin{align} & \delta^4\left[\boldsymbol{x'}\boldsymbol{-}\boldsymbol{x'}_n\left(\tau\right)\right] \boldsymbol{=} \: \delta^4\left[\boldsymbol{x-x}_n\left(\tau\right)\right] \tag{01a}\label{01a}\\ & \boldsymbol{x'} \boldsymbol{=} \Lambda \boldsymbol{x}\,,\quad \boldsymbol{x'}_n\left(\tau\right) \boldsymbol{=} \Lambda \boldsymbol{x}_n\left(\tau\right) \tag{01b}\label{01b}\\ & \Lambda \boldsymbol{=} \texttt{ proper homogeneous Lorentz transformation } \left(\det \Lambda \boldsymbol{=+} 1\right) \tag{01c}\label{01c} \end{align} Note that in above equations \begin{align} \boldsymbol{x} & \boldsymbol{=}\left(x_1,x_2,x_3,x_4\right)\boldsymbol{=}\left(x_1,x_2,x_3,c\,t\right) \tag{02a}\label{02a}\\ \boldsymbol{x'} & \boldsymbol{=}\left(x'_1,x'_2,x'_3,x'_4\right)\boldsymbol{=}\left(x'_1,x'_2,x'_3,c\,t'\right) \tag{02b}\label{02b} \end{align} are the space-time position 4-vectors of a point while \begin{align} \boldsymbol{x}_n\left(\tau\right) & \boldsymbol{=}\left[x_{n1}\left(\tau\right),x_{n2}\left(\tau\right),x_{n3}\left(\tau\right),x_{n4}\left(\tau\right)\right]\boldsymbol{=}\left[x_{n1}\left(\tau\right),x_{n2}\left(\tau\right),x_{n3}\left(\tau\right),c\tau\right] \tag{03a}\label{03a}\\ \boldsymbol{x'}_n\left(\tau\right) & \boldsymbol{=}\left[x'_{n1}\left(\tau\right),x'_{n2}\left(\tau\right),x'_{n3}\left(\tau\right),x'_{n4}\left(\tau\right)\right]\boldsymbol{=}\left[x'_{n1}\left(\tau\right),x'_{n2}\left(\tau\right),x'_{n3}\left(\tau\right),c\tau\right] \tag{03b}\label{03b} \end{align} is the parametric representation of the 4-dimensional path of the $n-$particle with parameter the proper time $\tau\,$ of this same particle.

As a general remark, equality between Dirac $\delta-$functions like \eqref{01a} is not literally valid as it is since here we have improper functions. For example the well-known equation \begin{equation} \delta\left(\alpha x\boldsymbol{+}\beta\right) \boldsymbol{\doteq} \dfrac{1}{\boldsymbol{\vert}\alpha\boldsymbol{\vert}} \delta\left(x\boldsymbol{+}\frac{\,\beta\,}{\alpha}\right)\,\qquad x,\alpha,\beta \in \mathbb{R} \tag{04}\label{04} \end{equation} is valid in the following sense \begin{equation} \int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\delta\left(\alpha x\boldsymbol{+}\beta\right)\mathrm{h}\left(x\right)\mathrm dx\boldsymbol{=}\dfrac{1}{\boldsymbol{\vert}\alpha\boldsymbol{\vert}} \int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\delta\left(x\boldsymbol{+}\frac{\,\beta\,}{\alpha}\right)\mathrm{h}\left(x\right)\mathrm dx \tag{05}\label{05} \end{equation} where $\,\mathrm{h}\left(x\right)\,$ any proper real scalar function.

So, we must view equation \eqref{01a} in the following sense \begin{equation} \!\!\!\!\!\!\!\iiiint\limits_{\texttt{space-time}}\!\!\!\!\delta^4\left[\boldsymbol{x'}\boldsymbol{-}\boldsymbol{x'}_n\left(\tau\right)\right]\mathrm{M}\left(\boldsymbol{x}\right) \mathrm d^4 \boldsymbol{x} \boldsymbol{=}\!\!\!\!\iiiint\limits_{\texttt{space-time}}\!\!\! \delta^4\left[\boldsymbol{x}\boldsymbol{-}\boldsymbol{x}_n\left(\tau\right)\right]\mathrm{M}\left(\boldsymbol{x}\right) \mathrm d^4 \boldsymbol{x}\boldsymbol{=}\mathrm{M}\left[\boldsymbol{x}_n\left(\tau\right)\right] \tag{06}\label{06} \end{equation} where $\,\mathrm{M}\left(\boldsymbol{x}\right)\,$ any proper real scalar function of the 4-vector $\,\boldsymbol{x}$.

In the first integral in the lhs of above equation we apply the following 4-dimensional variable change from $\,\boldsymbol{x}\,$ to $\,\boldsymbol{u}\,$(1) \begin{equation} \boldsymbol{u} \boldsymbol{=}\boldsymbol{x'}\boldsymbol{-}\boldsymbol{x'}_n\left(\tau\right)\boldsymbol{=}\Lambda \boldsymbol{x}\boldsymbol{-}\Lambda \boldsymbol{x}_n\left(\tau\right) \tag{07}\label{07} \end{equation} Note that this variable change is invertible since \begin{equation} \boldsymbol{x} \boldsymbol{=}\Lambda^{\boldsymbol{-}1}\boldsymbol{u}\boldsymbol{+} \boldsymbol{x}_n\left(\tau\right) \tag{08}\label{08} \end{equation}

Now, the relation between the infinitesimal 4-volumes is(1) \begin{equation} \!\!\!\!\!\!\!\mathrm d^4 \boldsymbol{x}\!\boldsymbol{=}\!\mathrm d x_1\mathrm d x_2\mathrm d x_3\mathrm d x_4\!\boldsymbol{=}\!\left\vert\dfrac{\partial\left(u_1,u_2,u_3,u_4\right)}{\partial\left(x_1,x_2,x_3,x_4\right)}\right\vert^{\boldsymbol{-}1}\!\!\!\mathrm d u_1\mathrm d u_2\mathrm d u_3\mathrm d u_4\boldsymbol{=}\left(\det\Lambda\right)^{\boldsymbol{-}1}\mathrm d^4\boldsymbol{u}\boldsymbol{=}\mathrm d^4\boldsymbol{u} \tag{09}\label{09} \end{equation} Inserting the expressions \eqref{07} and \eqref{09} in the lhs of \eqref{06} we have \begin{align} \iiiint\limits_{\texttt{space-time}}\delta^4\left[\boldsymbol{x'}\boldsymbol{-}\boldsymbol{x'}_n\left(\tau\right)\right]\mathrm{M}\left(\boldsymbol{x}\right) \mathrm d^4 \boldsymbol{x} & \boldsymbol{=}\iiiint\limits_{\texttt{space-time}} \delta^4\left(\boldsymbol{u}\right)\mathrm{M}\left[\Lambda^{\boldsymbol{-}1}\boldsymbol{u}\boldsymbol{+} \boldsymbol{x}_n\left(\tau\right)\right] \mathrm d^4 \boldsymbol{u} \nonumber\\ & \boldsymbol{=}\mathrm{M}\left[\Lambda^{\boldsymbol{-}1}\boldsymbol{0}\boldsymbol{+} \boldsymbol{x}_n\left(\tau\right)\right]\boldsymbol{=}\mathrm{M}\left[ \boldsymbol{x}_n\left(\tau\right)\right] \nonumber\\ & \boldsymbol{=} \iiiint\limits_{\texttt{space-time}} \delta^4\left[\boldsymbol{x}\boldsymbol{-}\boldsymbol{x}_n\left(\tau\right)\right]\mathrm{M}\left(\boldsymbol{x}\right) \mathrm d^4 \boldsymbol{x} \tag{10}\label{10} \end{align}

In summary : the proof is based on the invariant 4-dimensional infinitesimal volume under a proper homogeneous Lorentz transformation, equation \eqref{09}.

$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$

NOTE :

By equations (2.6.1)-(2.6.5) in Weinberg's $\boldsymbol{\S\:2.6 }\textbf{ Currents and Densities}$ it is proved that in the discontinuous case of finite charged moving particles the 4-dimensional current charge density is a Lorentz 4-vector.

For the continuous case we refer to "The Classical Theory of Fields", L.D.Landau and E.M.Lifshitz, Fourth Revised English Edition (see my $\color{blue}{\textbf{ANSWER A}}\:$ here How do we prove that the 4-current jμ transforms like xμ under Lorentz transformation?).

$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$

(1) REFERENCE : Physical meaning of the Jacobian in relation to Dirac delta function.

Frobenius
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