How is 4-current a 4-vector?

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?

• 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 6 '16 at 20:52
• – Qmechanic May 6 '16 at 21:07
• @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 6 '16 at 21:14
• @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 6 '16 at 21:57
• 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 6 '16 at 22:05

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

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 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).

Now we 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 $J^{\mu}$:

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

protected by Qmechanic♦Jul 28 '17 at 12:12

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