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I am looking at Jackson sec 11.9, where he states that the $\rho,\bf{J}$ form the 4-current $$J^\alpha=(c\rho,\bf{J})$$ Jackson says this is from the invariant of the 4-divergence $\partial^\alpha J_\alpha$ is invariant (which is 0 for the 4-current).

So I want to understand this in terms of a point charge, where $$\rho= q\delta^3({\bf r}-{\bf r}(t))$$ $${\bf J} = q{\bf v}(t)\delta^3({\bf r}-{\bf r}(t))$$ Is there a way to understand why this transforms as a 4-vector?

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    $\begingroup$ This is called the quotient theorem: If $A^{\mu\nu\cdots}$ is a tensor and the quantity $A^{\mu\nu\cdots}B_{\mu\alpha\cdots}$ is a tensor, then $B$ must transform as a tensor. As the continuity equation dictates that $\partial_\mu J^\mu = 0$ is a tensor, $J^\mu$ must transform as a tensor (if you already know, that $\partial_\mu$ is a tensor). $\endgroup$ May 6 '16 at 20:52
  • $\begingroup$ Hint: physics.stackexchange.com/q/57191/2451 $\endgroup$
    – Qmechanic
    May 6 '16 at 21:07
  • $\begingroup$ @Qmechanic I looked at that for a while before I posted and I could not quite find the connection. When my professor just jumped over the fact that this is a four vector, he said it could be shown with coordinate transforms and addition of velocities, which is why I am getting confused here. $\endgroup$ May 6 '16 at 21:14
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    $\begingroup$ @SebastianRiese's comment is probably the best way to go in this case. More generally, whether an object is a tensor or not can be understood through its transformation properties under symmetry transformations (that's basically the very definition). So, if you can show that $J^\mu$ transforms like a tensor should under Lorentz transformation, you have shown that $J^\mu$ is a tensor. $\endgroup$
    – Prahar
    May 6 '16 at 21:57
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    $\begingroup$ Note that the $\rho$ in your first equation, is not the invariant $\rho_0$. It's the charge density in the specific frame you're dealing with. $\rho=\gamma \rho_0$ $\endgroup$ May 6 '16 at 22:05
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As per, http://en.wikipedia.org/wiki/Four-velocity, we can define four-current density as: $J = \rho_0 U$, where $U$ is the four-velocity. Since it's a scalar times a four-vector, it's another four-vector.

$$J = \gamma(v)(\rho_0 c,\rho_0 \vec{v})$$ $$J = (\gamma(v)\rho_0 c,\gamma(v)\rho_0 \vec{v})$$

Now it remains to show that this fits the definition you gave:

$$J=(c\rho,\mathbf{J})$$

ie: We need to show that $$\rho = \gamma(v)\rho_0$$ $$\mathbf{J} = \gamma(v)\rho_0 \vec{v}=\rho \vec{v}$$

Suppose we have an infinitesimal volume of charge moving with velocity $\vec{v}$. Suppose its dimensions in the rest frame are $\Delta x'$,$\Delta y'$, $\Delta z'$. Its volume in the rest frame $V' = \Delta x'\Delta y'\Delta z'$. Total charge within this volume is $\rho_0V'$. We know by length contraction that $\Delta x = \dfrac{\Delta x'}{\gamma(v)}$, $\Delta y = \Delta y'$, $\Delta z = \Delta z'$.

So in the original frame the volume of this charge is: $V = \Delta x \Delta y \Delta z = \dfrac{V'}{\gamma(v)}$.

Total charge is the same in both frames (why? we define charge as being measured in the rest frame, making it invariant).

So charge density in original frame, $\rho = \dfrac{\rho_0 V'}{\left(\dfrac{V'}{\gamma(v)}\right)} = \gamma (v)\rho_0 $

so that takes care of the first relation. The second relation $$\mathbf{J} = \rho \vec{v}$$ just follows from the definition of current density. Going back to our infinitesimal volume of charge, suppose the charge crosses some boundary perpendicular to the x-axis over some time $\Delta t$. $$I = \dfrac{Q}{\Delta t} = \dfrac{\rho \Delta x \Delta y \Delta z}{\Delta t}$$ Cross sectional area $$A = \Delta y \Delta z$$ So magnitude of current density = $\dfrac{I}{A} = \dfrac{\rho \Delta x}{\Delta t}$. Taking infintesimals we get $\rho \left|\left|\dfrac{dx}{dt}\right|\right|$. Multiply this by a unit vector in the direction of motion and we get $$\mathbf{J} = \rho \vec{v}$$

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    $\begingroup$ Although you first define $\,J = \rho_0 U\,$ (so a priori a Lorentz 4-vector) and then prove that is identical to $\,J=(c\rho,\mathbf{J})\,$, your answer is correct while mine and Arturo don Juan's are false. $\endgroup$
    – Frobenius
    May 11 '21 at 17:28
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enter image description here

I use other symbols in order to prevent confusion in the following.

Let a point charge $\:q\:$ moving with position vector $\:\boldsymbol{\xi}\left(t\right)\:$ as in above Figure. Then the volume charge density and the charge current density are expressed via Dirac $\:\delta$-function as follows

\begin{align} \rho\left(\mathbf{x},t\right) & =q\cdot\delta^{3}\bigl(\mathbf{x}-\boldsymbol{\xi}\left(t\right)\bigr) \tag{01a}\\ \mathbf{j}\left(\mathbf{x},t\right) & =q\cdot\delta^{3}\bigl(\mathbf{x}-\boldsymbol{\xi}\left(t\right)\bigr) \cdot\dfrac{d\boldsymbol{\xi}\left(t\right)}{dt}=q\cdot\delta^{3}\bigl(\mathbf{x}-\boldsymbol{\xi}\left(t\right)\bigr)\cdot\mathbf{v}\left(t\right) \tag{01b} \end{align} where \begin{equation} \mathbf{v}\left(t\right)= \bigl(\upsilon_{1}\left(t\right),\upsilon_{2}\left(t\right),\upsilon_{3}\left(t\right)\bigr)= \biggl(\dfrac{d\xi_{1}\left(t\right)}{dt},\dfrac{d\xi_{2}\left(t\right)}{dt},\dfrac{d\xi_{3}\left(t\right)}{dt}\biggr)= \dfrac{d\boldsymbol{\xi}\left(t\right)}{dt} \tag{02} \end{equation} the velocity of the charge. Under the assumption that the electric charge $\:q\:$ is invariant (observers in different inertial systems agree on the same value) we must show that the 4-quantity \begin{equation} \dfrac{\mathbb{J}}{q} \equiv \left[\delta^{3}\bigl(\mathbf{x}-\boldsymbol{\xi}\left(t\right)\bigr), \:\delta^{3}\bigl(\mathbf{x}-\boldsymbol{\xi}\left(t\right)\bigr)\cdot\dfrac{d\boldsymbol{\xi}\left(t\right)}{dt} \right] \tag{03} \end{equation} is a 4-current. So we must prove that it satisfies the continuity equation \begin{equation} \dfrac{\partial \left[\delta^{3}\bigl(\mathbf{x}-\boldsymbol{\xi}\left(t\right)\bigr) \right]}{\partial t}+ \boldsymbol{\nabla}_{\mathbf{x}}\boldsymbol{\cdot} \left[\delta^{3}\bigl(\mathbf{x}-\boldsymbol{\xi}\left(t\right)\bigr)\cdot\dfrac{d\boldsymbol{\xi}\left(t\right)}{dt} \right]=0 \tag{04} \end{equation} or \begin{equation} \dfrac{\partial \left[\delta^{3}\bigl(\mathbf{x}-\boldsymbol{\xi}\left(t\right)\bigr) \right]}{\partial t}+ \boldsymbol{\nabla}_{\mathbf{x}}\boldsymbol{\cdot} \left[\delta^{3}\bigl(\mathbf{x}-\boldsymbol{\xi}\left(t\right)\bigr)\cdot\mathbf{v}\left(t\right)\right]=0 \tag{04a} \end{equation} If proved, this 4-current would be a 4-vector also.

Now \begin{equation} \delta^{3}\bigl(\mathbf{x}-\boldsymbol{\xi}\left(t\right)\bigr) =\delta\bigl(x_{1}-\xi_{1}\left(t\right)\bigr)\cdot\delta\bigl(x_{2}-\xi_{2}\left(t\right)\bigr)\cdot\delta\bigl(x_{3}-\xi_{3}\left(t\right)\bigr) \tag{05} \end{equation} Using the following property of Dirac $\:\delta$-function \begin{equation} z\delta\left( z \right)=0 \Rightarrow \dfrac{\partial \delta\left(z\right)}{\partial z} = - \dfrac{ \delta\left(z\right)}{ z} \tag{06} \end{equation} we have \begin{equation} \dfrac{\partial \left[\delta\bigl(x_{k}-\xi_{k}\left(t\right)\bigr) \right]}{\partial t}=\:+\:\dfrac {\dfrac{d \xi_{k}}{dt}}{x_{k}-\xi_{k}\left(t\right)}\cdot\delta\bigl(x_{k}-\xi_{k}\left(t\right)\bigr)=\:+\:\dfrac {v_{k}\left(t\right)}{x_{k}-\xi_{k}\left(t\right)}\cdot\delta\bigl(x_{k}-\xi_{k}\left(t\right)\bigr) \tag{07} \end{equation}

So \begin{equation} \dfrac{\partial \left[\delta^{3}\bigl(\mathbf{x}-\boldsymbol{\xi}\left(t\right)\bigr) \right]}{\partial t}=\:+\left(\sum_{k=1}^{k=3}\dfrac {v_{k}\left(t\right)}{x_{k}-\xi_{k}\left(t\right)}\right)\cdot\delta^{3}\bigl(\mathbf{x}-\boldsymbol{\xi}\left(t\right)\bigr) \tag{08} \end{equation} On the same footing we can prove that \begin{equation} \dfrac{\partial \left[\delta\bigl(x_{k}-\xi_{k}\left(t\right)\bigr)\cdot v_{k}\left(t\right)\right]}{\partial x_{k}}=\:-\:\dfrac {v_{k}\left(t\right)}{x_{k}-\xi_{k}\left(t\right)}\cdot\delta\bigl(x_{k}-\xi_{k}\left(t\right)\bigr) \tag{09} \end{equation} that is \begin{equation} \boldsymbol{\nabla}_{\mathbf{x}}\boldsymbol{\cdot} \left[\delta^{3}\bigl(\mathbf{x}-\boldsymbol{\xi}\left(t\right)\bigr)\cdot\mathbf{v}\left(t\right)\right]=\:-\left(\sum_{k=1}^{k=3}\dfrac {v_{k}\left(t\right)}{x_{k}-\xi_{k}\left(t\right)}\right)\cdot\delta^{3}\bigl(\mathbf{x}-\boldsymbol{\xi}\left(t\right)\bigr) \tag{10} \end{equation} proving the continuity equation (04).


EDIT : A strange invariant

Realizing that the 4-quantity $\left(\mathbb{J} /\right)q$ of equation (03) is a contravariant 4-vector, say $\mathbb{V}$
\begin{equation} \mathbb{V} \equiv \delta^{3}\bigl(\mathbf{x}-\boldsymbol{\xi}\left(t\right)\bigr)\cdot \left[c, \:\dfrac{d\boldsymbol{\xi}\left(t\right)}{dt} \right]=\delta^{3}\bigl(\mathbf{x}-\boldsymbol{\xi}\left(t\right)\bigr)\cdot\Biggl[\:\:c\:\:,\:\:\mathbf{v}\:\:\Biggr] \tag{11} \end{equation} and having in mind- (and comparing it with-) the contravariant 4-vector for velocity \begin{equation} \mathbb{U} \equiv \gamma_{v}\cdot \left[c, \:\dfrac{d\boldsymbol{\xi}\left(t\right)}{dt} \right]=\gamma_{v}\cdot\Biggl[\:\:c\:\:,\:\:\mathbf{v}\:\:\Biggr] \tag{12} \end{equation} I was wondering which would be the relation between the Dirac $\:\delta$-function $\delta^{3}\bigl(\mathbf{x}-\boldsymbol{\xi}\left(t\right)\bigr)$, a function of $\:\left(\mathbf{x},\:t\:\right)$, and $\:\gamma_{v}\:$, a function of $\:t\:$ \begin{equation} \gamma_{v}= \left[1-\left(\dfrac{v}{c}\right)^{2}\right]^{-1/2}=\left[1-\left\Vert\dfrac{d\boldsymbol{\xi}\left(t\right)}{c dt}\right\Vert ^{2}\right]^{-1/2} \tag{13} \end{equation} We know that the inner product of two 4-vectors (in Minkowski space) is Lorentz-invariant, so \begin{equation} \mathbb{U}\boldsymbol{\circ} \mathbb{V }= c^{2}\left[1-\left\Vert\dfrac{d\boldsymbol{\xi}\left(t\right)}{c dt}\right\Vert ^{2}\right]^{1/2}\cdot \delta^{3}\bigl(\mathbf{x}-\boldsymbol{\xi}\left(t\right)\bigr) = \text{invariant} \tag{14} \end{equation} If we see this invariant in the rest frame of the particle, then \begin{equation} \bbox[#FFFF88,12px]{\left[1-\left\Vert\dfrac{d\boldsymbol{\xi}\left(t\right)}{c dt}\right\Vert ^{2}\right]^{1/2}\cdot \delta^{3}\bigl(\mathbf{x}-\boldsymbol{\xi}\left(t\right)\bigr) = \text{invariant}=\delta^{3}\bigl(\mathbf{x}_{rf}\bigr)} \tag{15} \end{equation} where $\:\mathbf{x}_{rf}\:$ the position vector of a reference point with respect to the rest frame of the particle.

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$

$\:\color{red}{\textbf{THIS ANSWER OF MINE IS WRONG !!!}}$ because

Proofs that the 4-dimensional electric charge current density $\,\mathbf J\,$ is transformed as a Lorentz 4-vector based upon the conservation law (continuity equation) 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).

\begin{equation} \partial_{\mu}A^{\mu}\boldsymbol{=}\texttt{invariant}\quad \boldsymbol{=\!\ne\!\Rightarrow}\quad A^{\mu}\boldsymbol{=}\texttt{contravariant Lorentz 4-vector} \tag{16}\label{16} \end{equation}

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    $\begingroup$ @Arturo don Juan : \begin{equation} \require{cancel} \partial_{\mu}A^{\mu}\boldsymbol{=}0\quad \cancel{=\!=\!=\!\Longrightarrow}\quad A^{\mu}\boldsymbol{\equiv}0 \end{equation} $\endgroup$
    – Frobenius
    May 12 '21 at 6:14
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    $\begingroup$ @Arturo don Juan --- The proof that the 4-dimensional current charge density $\:\mathbf J=(c\varrho,\mathbf j)\:$ is a Lorentz 4-vector is based on two facts : (1) the invariance of the electric charge and (2) the invariance of the infinitesimal 4-dimensional $''$volume$''$ $\mathrm dx_0\mathrm dx_1\mathrm dx_2\mathrm dx_3$. $\endgroup$
    – Frobenius
    May 12 '21 at 6:29
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    $\begingroup$ Wow I am so stupid, for some reason I confused divergence and gradient, so was thinking if $\partial_{\mu}f(x)=0$ then $f=0$. That was so so dumb I wish I could take it back. $\endgroup$ May 12 '21 at 15:15
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    $\begingroup$ I read the answers to that question you linked, and now I see what you're talking about. I'll edit my answer to reflect that. $\endgroup$ May 12 '21 at 15:57
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    $\begingroup$ @Arturo don Juan --- I will not delete my answer. I want to leave as it is since we learn from our mistakes. If possible please downvote it. $\endgroup$
    – Frobenius
    May 12 '21 at 16:21
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Edit: I now understand that this answer is incorrect. See @Frobenius's answer and comments.


Here is a quick (standard) proof of the quotient rule in the context of the four-current being Lorentz covariant, as mentioned by @SebastianRiese. We assume that in any inertial frame (where the metric is Minkowski, as expressed in the coordinates we are using) the physical continuity equation holds:

$$\frac{\partial\rho}{\partial t}+\nabla\cdot \vec{J}=0\tag{1}$$

Knowing that the partial derivatives $\partial_{\mu}$ do transform as Lorentz vectors, we write the continuity equation in a suggestive form.

$$\partial_{\mu}J^{\mu}=0 \tag{2}$$

where as usual I have defined:

$$\partial_{\mu}\equiv \frac{\partial}{\partial x^{\mu}}=\left(\frac{\partial}{\partial (ct)},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\tag{3}$$

$$J^{\mu}=\begin{pmatrix}c\rho\\ J^x\\J^y\\J^z\end{pmatrix}\tag{4}$$

What is important to remember here is that although we know how the derivative four-vector changes under general coordinate transformations (among them Lorentz transformations), we do not yet know how this four-component $J^{\mu}$ changes under coordinate transformations - i.e. we do not really know if it is actually a four-vector. We do know that under general under rotations $\vec{J}$ will physically transform as a Cartesian vector and $\rho$ a Cartesian scalar, but it is not immediately obvious that under boosts $\vec{J}$ and $\rho$ will mix as they would if they were unified into the four-vector $J^{\mu}$ from (4).

The trick is to realize that on the RHS of eq. (2), we have $0$ which is a Lorentz scalar. Lorentz transform to another inertial coordinate system, where the continuity equation is now:

$$\begin{align} \partial'_{\mu}J'^{\mu}&=\left(\Lambda_{\mu}^{\,\,\nu} \partial_{\nu}\right)J'^{\mu}\\ &=\partial_{\nu}\left(\Lambda_{\mu}^{\,\,\nu}J'^{\mu}\right)\\ &=\partial_{\nu}J^{\nu}\\ &=0 \end{align}$$

where in passing from the first line to the second I have used that Lorentz transformations are constant in coordinates. This implies the transformation rule for $J^{\mu}$:

$$J'^{\mu}=\Lambda_{\nu}^{\,\,\mu}J^{\nu}=\Lambda^{\mu}_{\,\,\nu}J^{\nu}\tag{5}$$

So the 4-current is indeed a Lorentz vector.

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  • $\begingroup$ Our answers are false. See the reasoning in the end of my answer. $\endgroup$
    – Frobenius
    May 11 '21 at 17:22
  • $\begingroup$ @Frobenius I responded to your answer. $\endgroup$ May 11 '21 at 22:48

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