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Imagine a four-velocity $U^\mu(x)$ defined all over spacetime. Define the number-flux four-vector as $$N^\mu = nU^\mu$$ where $n$ is the number density of a collection of particles in their rest frame. So in this rest frame, the number-flux four-vector reduces to $(n, 0, 0, 0)$. My question is how this vector transforms under a general Lorentz boost with velocity $\vec{v}$. Inspecting how other four-vectors, such as the four-velocity and four-momentum transform, one would guess that the number-flux four vector transforms as $(\gamma n, \gamma n v^1, \gamma n v^2, \gamma n v^3),$ where $v^1 = \frac{dx^1}{dt}$ etc. Is this correct? If so, what is the intuition behind how the number density of the particles, $n$, transforms to $\gamma n$?

A derivation of the correct Lorentz boosted number-flux four-vector would be very much appreciated, as this is a topic I still find quite confusing.

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  • $\begingroup$ Number density changes that way because the volume change the same way. $\endgroup$
    – vats dimri
    Commented Feb 3, 2023 at 7:52

1 Answer 1

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I think that the question has its exact analog in Classical Electrodynamics. Consider that in an inertial system $\,S\,$ we have a system of particles with \begin{align} n\left(\mathbf x,t\right) & \boldsymbol{=}\texttt{particle number volume density at } \mathbf x,t \tag{01a}\label{01a}\\ &\text{moving with 3-velocity} \nonumber\\ \mathbf u\left(\mathbf x,t\right) & \boldsymbol{=}\texttt{velocity of particles at } \mathbf x,t \tag{01b}\label{01b} \end{align} Suppose now that we $''$charge$''$ each particle with the unit quantum charge $\,\texttt{e}\,$ of the electron. Then we have the following quantities of Classical Electrodynamics (omitting for convenience the space-time dependence) \begin{align} \varrho & \boldsymbol{=}\texttt{e}\, n\boldsymbol{=}\texttt{electric charge volume density} \tag{02a}\label{02a}\\ \boldsymbol{j} & \boldsymbol{=}\varrho\,\mathbf u\boldsymbol{=}\texttt{e}\, n\,\mathbf u\boldsymbol{=}\texttt{electric charge current density} \tag{02b}\label{02b} \end{align} If particles are neither created nor destroyed then in the electrodynamics analog the electric charge obeys the easily proved conservation law \begin{equation} \dfrac{\partial \varrho}{\partial t}\boldsymbol{+}\boldsymbol{\nabla\cdot}\boldsymbol{j} \boldsymbol{=}0 \tag{03}\label{03} \end{equation} In electrodynamics we define the 4-dimensional electric charge current density $\,\mathbf J\,$ \begin{equation} \mathbf J \stackrel{\texttt{def}}{\boldsymbol{=\!=}}\left(\varrho\,c,\boldsymbol{j}\right)\boldsymbol{=}\varrho\left(c,\mathbf u\right) \tag{04}\label{04} \end{equation} so that the conservation law is expressed in terms of the 4-divergence \begin{equation} \boldsymbol{\square\!\!\!\!\square\,\cdot} \mathbf J \boldsymbol{=}\partial_{\mu}\mathrm J^{\mu}\boldsymbol{=}\partial^{\mu}\mathrm J_{\mu}\boldsymbol{=}0 \tag{05}\label{05} \end{equation} By analogy we define the 4-dimensional particle flux $\,\mathbf N\,$ \begin{equation} \mathbf N \boldsymbol{=}\dfrac{\mathbf J }{\texttt{e}}\boldsymbol{=}\dfrac{\left(\varrho\,c,\boldsymbol{j}\right)}{\texttt{e}}\boldsymbol{=}n\left(c,\mathbf u\right)\boldsymbol{=}\left(n\,c,\boldsymbol{f}\right) \tag{06}\label{06} \end{equation} where \begin{equation} \boldsymbol{f}\boldsymbol{=}n\,\mathbf u\boldsymbol{=}\texttt{the 3-dimensional particle flux} \tag{07}\label{07} \end{equation} The particle flux obeys the particle number conservation law corresponding to equation \eqref{03} \begin{equation} \dfrac{\partial n}{\partial t}\boldsymbol{+}\boldsymbol{\nabla\cdot}\boldsymbol{f} \boldsymbol{=}0 \tag{08}\label{08} \end{equation} or its 4-dimensional version corresponding to equation \eqref{05} \begin{equation} \boldsymbol{\square\!\!\!\!\square\,\cdot} \mathbf N \boldsymbol{=}\partial_{\mu}\mathrm N^{\mu}\boldsymbol{=}\partial^{\mu}\mathrm N_{\mu}\boldsymbol{=}0 \tag{09}\label{09} \end{equation}

Now, in electrodynamics it has been proved that the 4-dimensional electric charge current density $\,\mathbf J\,$ of equation \eqref{04} is a Lorentz 4-vector expressed also as the Lorentz 4-velocity $\mathbf U\boldsymbol{=}\gamma_{\rm u}\left(c,\mathbf u\right)$ times the invariant scalar rest charge density $\varrho_{0}\boldsymbol{=}\varrho/\gamma_{\rm u}$ \begin{equation} \mathbf J \boldsymbol{=}\varrho_{0}\mathbf U \tag{10}\label{10} \end{equation} By analogy the 4-dimensional particle flux $\,\mathbf N\,$ is a Lorentz 4-vector expressed also as the Lorentz 4-velocity $\mathbf U\boldsymbol{=}\gamma_{\rm u}\left(c,\mathbf u\right)$ times the invariant scalar rest particle number $n_{0}\boldsymbol{=}n/\gamma_{\rm u}$ \begin{equation} \mathbf N \boldsymbol{=}n_{0}\mathbf U \tag{11}\label{11} \end{equation}

So the proof that $\,\mathbf N\,$ is a Lorentz 4-vector is essentially the proof that the 4-dimensional electric charge current density $\,\mathbf J\,$ of equation \eqref{04} is a Lorentz 4-vector.

Proofs that $\,\mathbf J\,$ is transformed as a Lorentz 4-vector based upon the conservation law \eqref{03} are false. That electric charge is constant in an inertial system doesn't provide any information about how it is transformed between inertial systems. It's a confusion between what is a constant (it concerns what happens in a system) and what is an invariant (it concerns what happens between two systems).

For an elegant proof by L.D.Landau and E.M.Lifshitz see my $\color{blue}{\textbf{ANSWER A}}$ here How do we prove that the 4-current jμ transforms like xμ under Lorentz transformation?. The proof is based on one hand upon the fact that the charge on a particle is, from its very definition, an invariant quantity and on the other hand upon the fact that the 4-dimensional infinitesimal $''$volume$''$ $\mathrm dV=\mathrm dx^0\mathrm dx^1\mathrm dx^2\mathrm dx^3$ is a Lorentz invariant scalar. A proof of the latter is given in footnote (1) of the aforementioned answer.

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