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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?

Qmechanic
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yankeefan11
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    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). – Sebastian Riese May 06 '16 at 20:52
  • Hint: http://physics.stackexchange.com/q/57191/2451 – Qmechanic May 06 '16 at 21:07
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    @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. – yankeefan11 May 06 '16 at 21:14
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    @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. – Prahar May 06 '16 at 21:57
  • Look at the derivation for four-current density here: https://en.wikipedia.org/wiki/Four-velocity it's a scalar $\rho_0$ times a four vector $U$, hence it's another four-vector. – Ameet Sharma May 06 '16 at 22:01
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    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. $\rho=\gamma \rho_0$ – Ameet Sharma May 06 '16 at 22:05
  • @SebastianRiese: I think you should make that an answer. – ACuriousMind May 07 '16 at 11:27
  • @AmeetSharma: That, too, seems to be an answer, not a comment (if you add the bit where you show why the density transforms like that). – ACuriousMind May 07 '16 at 11:28
  • The 3 (till now) answers are all incorrect (including mine) : that the 4-current is a 4-vector could not in any case be proved from the continuity equation. All of us, with an answer or comment in favor to the contrary, have been mislead by the mistakes included in the question. – Frobenius Jun 23 '18 at 09:28
  • Jackson proves that the 4-current is a 4-vector based on the invariance of the charge not on the conservation of charge (continuity equation) as the OP post in the question. OP identifies incorrectly the charge invariance (same value observed by all inertial system) with charge conservation (the total charge is conserved in one inertial system). Jackson's proof is identical to that of Landau's. – Frobenius Jun 23 '18 at 12:47

3 Answers3

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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 $$\mathbf{J} = \rho \vec{v}$$

Banach space fan
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Ameet Sharma
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    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. – Frobenius May 11 '21 at 17:28
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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}

Frobenius
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  • Although I don't quite understand the wording of the last paragraph, I understand why you say that just because the continuity equation is invariant, that doesn't mean $J^{\mu}$ is a contravariant Lorentz 4-vector. It seems like this implication is impulsively and unjustifiably reached. However, the implication really does follow. Here's the reasoning... – Arturo don Juan May 11 '21 at 22:45
  • Consider two inertial coordinate systems, $x$ and $x'(x)$, where the transformation matrix between them is $\Lambda$ (constant!). We don't yet know how $J^{\mu}$ will change under this coordinate transformation, so simply call it $J'^{\mu}$. Subtract the continuity equation expressed in one coordinate system from another, and get $\partial_{\mu}\left(J'^{\mu}-\Lambda^{\mu}{\nu} J^{\nu}\right)=0$. In any coordinate system, a physical $J^\mu$ must vanish at spatial infinity, so the only solution to this is $J'^{\mu}=\Lambda^{\mu}{\nu}J^{\mu}$, which says that $J^{\mu}$ is indeed a 4-vector. – Arturo don Juan May 11 '21 at 22:45
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    @Arturo don Juan : \begin{equation} \require{cancel} \partial_{\mu}A^{\mu}\boldsymbol{=}0\quad \cancel{=!=!=!\Longrightarrow}\quad A^{\mu}\boldsymbol{\equiv}0 \end{equation} – Frobenius May 12 '21 at 06:14
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    @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$''$ $\mathrm dx_0\mathrm dx_1\mathrm dx_2\mathrm dx_3$. – Frobenius May 12 '21 at 06:29
  • @Arturo don Juan --- For a proof in the discontinuous case of finite charged moving particles see in Weinberg's Gravitation and Cosmology equations from (2.6.1)-(2.6.5), $\boldsymbol{\S:2.6 }\textbf{ Currents and Densities}$. But there the proof ends with the statement (without proof) that $,\delta^4\left[\boldsymbol{x-x}\left(\tau\right)\right],$ is a Lorentz invariant scalar.... – Frobenius May 12 '21 at 06:57
  • @Arturo don Juan --- ...I have proved the latter based on the invariance of the infinitesimal 4-dimensional $''$volume$''$ $\mathrm dx_0\mathrm dx_1\mathrm dx_2\mathrm dx_3$ here Why does it follow that the Dirac delta function is a scalar “because determinant of the Lorentz transformation is 1”. – Frobenius May 12 '21 at 06:57
  • @Arturo don Juan --- For a proof in the continuous case see "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?). – Frobenius May 12 '21 at 07:02
  • @Arturo don Juan --- For a proof also for the continuous case see Jackson's Classical Electrodynamics. It's identical to that of Landau's. – Frobenius May 12 '21 at 07:11
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    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. – Arturo don Juan May 12 '21 at 15:15
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    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. – Arturo don Juan May 12 '21 at 15:57
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    @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. – 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.

Arturo don Juan
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