# Lorentz invariance of charge in classical electrodynamics - What volumes to look at?

There is an (plausible claim) that the charge is a lorentz invariant quantity (best example would be that the electron as a point charge has the same charge in every frame of reference). I'm trying to derive the property of charge density being the zero component of a 4-vector, and I'm having a hard time to formulate the claim for lorentz-invariance of charge for a finite volume ( a cube ).

I'd like to state the claim as "given a Volume in 3 dimensions", the charge contained in this volume will be equal in all inertial frames.

But here the problems begin: Given (for example) a cube in 3 dimensions, spanned by 3 vectors $\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$,$\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$, at a give time t= 0, what will the "same" volume be in another frame of reference?

If I boost to x-direction, then the vectors that I want to span the volume are now (in 4 dimensions) $\begin{bmatrix} -\beta \gamma \\ \gamma \\ 0 \\0 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 0 \\ 1 \\0 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 0 \\ 0 \\1 \end{bmatrix}$.

They happen at different points in time (and thus don't give the right volume anymore, because the volume is moving in time) (if it was at rest in system 1).

Of course,if I assume the volume I want to look at to be constant, I can calculate the vectors that would spann the volume in system 2 at equal times, but - Is this the volume that the statement about charge invariance applies to? If not, which volume will be the right one to look at in system 2, if one wants to calculate equal charges by integrating over the charge density?

• if charge density is a component of a 4-vector ($j_0$), why would you expect it to behave like a scalar? – JEB Sep 18 '18 at 22:51
• – Frobenius Sep 18 '18 at 23:13
• I don't think you really need to worry about what the "same" volume is. Just consider a charge distribution that's confined to some finite region of space. Then any observer in any state of motion will find that the total charge that exists on a surface of simultaneity is the same. – user4552 Sep 19 '18 at 0:09
• @JEB The integral of the charge density behaves like a scalar (at least according to some users here. As you can see at my question, I'm still trying to make sense of this statement. – Quantumwhisp Sep 19 '18 at 6:12
• @Frobenius the question you linked to is where my confusion stems from. In your answer, you just use that the volume-forms times charge density are invariant. The purpose of the question I posted here is how this invariance translates to the invariance of charge in a finite volume (and how to formulate this). – Quantumwhisp Sep 19 '18 at 6:15

The total charge $$\:\delta q=\rho\,\mathrm dx_1\mathrm dx_2\mathrm dx_3\:$$ within a small-volume element is an experimental invariant $$\begin{equation} \rho\,\mathrm dx_1\mathrm dx_2\mathrm dx_3=\rho'\,\mathrm dx'_1\mathrm dx'_2\mathrm dx'_3 \tag{01}\label{eq01} \end{equation}$$ To integrate on a finite volume $$\;V\;$$ has no sense since the simultaneous existence of charges at various points in this volume at a given instant $$\;t\;$$ is transformed to a set of not simultaneous events in the primed system. So, the charge invariance is valid only in differential form.

EDIT

responds to a comment by the OP owner :

The word "volume element" doesn't have that much meaning to me. By "Volume element", do you mean 2 little 3-dimensional cubes in the 4 dimensional space time, with edge-length $$dx_1, dx_2$$... and so on, tilted against each other, so each point in one of the cubes does have the same t-component, and each point in the other cube has the same t' component?

In the unprimed inertial frame $$\,S\,$$ let a finite 3-volume $$\,V$$. We find the charge $$\,Q_{_V}(t)\,$$ contained in $$\,V\,$$ at a given instant $$\,t\,$$ by integration of the lhs of equation \eqref{eq01} $$\begin{equation} Q_{_V}(t)=\iiint\limits_{V}\rho\left(\mathbf{x},t\right)\mathrm dx_1\mathrm dx_2\mathrm dx_3\,, \quad \mathbf{x}=\left(x_1,x_2,x_3\right) \tag{02}\label{eq02} \end{equation}$$ On a same way in the primed inertial frame $$\,S'\,$$ let a finite 3-volume $$\,V'$$. We find the charge $$\,Q'_{_{V'}}(t')\,$$ contained in $$\,V'\,$$ at a given instant $$\,t'\,$$ by integration of the rhs of equation \eqref{eq01} $$\begin{equation} Q'_{_{V'}}(t')=\iiint\limits_{V'}\rho'\left(\mathbf{x}',t'\right)\mathrm dx'_1\mathrm dx'_2\mathrm dx'_3\,, \quad \mathbf{x}'=\left(x'_1,x'_2,x'_3\right) \tag{03}\label{eq03} \end{equation}$$ Now let the inertial system $$\:\mathrm S'\:$$ be translated with respect to the inertial system $$\:\mathrm S\:$$ with constant velocity
$$\begin{equation} \boldsymbol{\upsilon}=\left(\upsilon_{1},\upsilon_{2},\upsilon_{3}\right)=\left(\upsilon \mathrm n_{1},\upsilon \mathrm n_{2},\upsilon \mathrm n_{3}\right)=\upsilon \mathbf n\,, \qquad \upsilon \in \left(-c,c\right) \tag{04}\label{eq04} \end{equation}$$ The Lorentz transformation in differential form is \begin{align} \mathrm d\mathbf{x}^{\boldsymbol{\prime}} & = \mathrm d\mathbf{x}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathrm d\mathbf{x})\mathbf{n}-\gamma\boldsymbol{\upsilon}\mathrm dt \tag{05a}\label{eq05a}\\ \mathrm d t^{\boldsymbol{\prime}} & = \gamma\left(\mathrm d t-\dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathrm d\mathbf{x}}{c^{2}}\right) \tag{05b}\label{eq05b} \end{align} Under above transformation there is no finite 3-volume $$\,V'\,$$ in $$\:\mathrm S'\:$$ at an instant $$\,t'\,$$ which is the image of the 3-volume $$\,V\,$$ in $$\:\mathrm S\:$$ at an instant $$\,t$$. So there is no way to relate the two integrals \eqref{eq02} and \eqref{eq03}.

To the contrary, we could relate the integral on a finite 4-volume $$\:\mathcal V\:$$ in the uprimed system $$\:\mathrm S\:$$ $$\begin{equation} F_{_{\mathcal V}}=\iiiint\limits_{\mathcal V}f\left(\mathbf{X}\right)\mathrm dx_1\mathrm dx_2\mathrm dx_3\mathrm dx_4\,, \quad \mathbf{X}=\left(x_1,x_2,x_3,x_4\right)=\left(x_1,x_2,x_3,c\,t\right) \tag{06}\label{eq06} \end{equation}$$ to that in the primed system $$\:\mathrm S'\:$$ $$\begin{equation} F'_{_{\mathcal V'}}=\iiiint\limits_{\mathcal V'}f'\left(\mathbf{X'}\right)\mathrm dx'_1\mathrm dx'_2\mathrm dx'_3\mathrm dx'_4\,, \quad \mathbf{X}'=\left(x'_1,x'_2,x'_3,x'_4\right)=\left(x'_1,x'_2,x'_3,c\,t'\right) \tag{07}\label{eq07} \end{equation}$$ since $$\:\mathcal V'\:$$ is a well-defined 4-volume, image of $$\:\mathcal V\:$$ under the Lorentz transformation \eqref{eq05a}-\eqref{eq05b}.

Moreover in the 4-dimensional case the infinitesimal 4-volume is a Lorentz invariant scalar(1) $$\begin{equation} \mathrm dx'_1\mathrm dx'_2\mathrm dx'_3\mathrm dx'_4=\mathrm dx_1\mathrm dx_2\mathrm dx_3\mathrm dx_4=\text{invariant scalar} \tag{08}\label{eq08} \end{equation}$$

Finally note that the experimental charge invariance in differential form \eqref{eq01} is used by Jackson, Landau and others to prove that the 4-dimensional quantity $$\:\mathbf{J}=(c\rho,\mathbf{j})\:$$ is a Lorentz 4-vector.

(1) For a proof see the footnote under my $$\color{blue}{\textbf{ANSWER A}}\:$$ here : How do we prove that the 4-current jμ transforms like xμ under Lorentz transformation?.

• The word "volume element" doesn't have that mutch meaning to me. By "Volume element", do you mean 2 little 3-dimensional cubes in the 4 dimensional space time, with edge-length $dx_1$, $dx_2$... and so on, tilted against each other, so each point in one of the cubes does have the same t-component, and each point in the other cube has the same t' component? – Quantumwhisp Sep 20 '18 at 19:16

If $\partial_t \rho = \nabla \cdot \vec j$ and $(\rho, \vec j)$ transforms as a Lorentz vector then $\int d^3 x \rho$ is a Lorentz scalar or invariant.

• This doesn't adress my question. – Quantumwhisp Sep 19 '18 at 6:18
• Charge invariance is not proved from charge conservation (continuity equation). – Frobenius Sep 19 '18 at 11:23
• @Frobenius I will give you a reference. – my2cts Sep 19 '18 at 17:52
• Although it doesn't address the question, I'd as well like to know the prove of the statement in your answer. I tried to calculate it, but it didn't work out yet – Quantumwhisp Sep 21 '18 at 22:34

I will give an answer to my own question by showing the inverse case (as claimed by user my2cts): Given that there is a 4-vector $$j$$ that satisfies $$\partial_{\mu} j^{\mu}$$, one can show that the integral of the component of $$j$$, which is perpendicular to a spacelike 3-dimensional hypersurface, performed over this surface, will always be the same. The Proof only holds if all compononents of $$j$$ become 0 when we spatially go to infinity.

Use 2 sets of base vectors ($$\{ e_0, e_1, e_2, e_3 \}$$ and $$\{ \tilde{e_0}, \tilde{e_1}, \tilde{e_2}, \tilde{e_3} \}$$, that are connected by an active lorentz transformation $$\Lambda^{-1}$$ with $$\Lambda^{-1}(e_{\mu}) = \tilde{e_{\mu}}$$.

Both sets of base vectors give rise to 2 spatial 3 dimensional hypersurfaces, $$S_1$$ and $$S_2$$, which will be spanned by $$\{e_1, e_2, e_3 \}$$ and $$\{, \tilde{e_1}, \tilde{e_2}, \tilde{e_3} \}$$. Imagine hypersurface $$S_1$$ to be anchored at the point located at (using the first base) the coordinates (0, 0, 0, 0), while the 2nd one, $$S_2$$, is supposed to be located at (also using the first base) (T, 0, 0, 0). Now look at a 4-dimensional Volume, which on its "top" and its "bottom" side is bordered by $$S_1$$ and $$S_2$$, while its Sides" lie outside the region where $$j\neq0$$. The picture shows how it's meant to be, the black linings show the area where $$j$$ deviates from 0.

Now (indexed quantities like $$j^{\mu}$$ refer to components of the base-indepedent objects in the first base. $$\omega$$ is supposed to be the 4-dimensional volume form.

\begin{align} \int_V 0 \omega = \int_{V} 0 dx^0 \wedge dx^1 \wedge dx^2 \wedge dx^3 = 0 = \int_{V} \partial_{\mu} j^{\mu} dx^0 \wedge dx^1 \wedge dx^2 \wedge dx^3 \end{align}

Because of the (general) stokes-theorem for integration of differential forms, this is the same as

\begin{align} 0=\int_{\partial V} j^0 dx^1 \wedge dx^2 \wedge dx^3 + j^1 dx^0 \wedge dx^2 \wedge dx^3 + j^2 dx^0 \wedge dx^1 \wedge dx^3 + j^3 dx^0 \wedge dx^1 \wedge dx^2 \end{align}

We note at this point that $$f = \omega(j, \cdot, \cdot, \cdot)= j^0 dx^1 \wedge dx^2 \wedge dx^3 + j^1 dx^0 \wedge dx^2 \wedge dx^3 + j^2 dx^0 \wedge dx^1 \wedge dx^3 + j^3 dx^0 \wedge dx^1 \wedge dx^2$$ gives a 3-form, whiches exterior derivative $$d f = \partial_{\mu} j^{\mu} dx^0 \wedge dx^1 \wedge dx^2 \wedge dx^3$$ is zero.

Because the regions at the sides don't contribute to the integral anyways, the integration must only be performed over the hyper surfaces $$S_1$$ and $$S_2$$. So we know:

\begin{align} \int_{S_1} f = \int_{S_2} f \end{align}

Let us now calculate those integrals:

\begin{align} \int_{S_1} f = \int dx dy dz f(e_1, e_2, e_3) = \int dx dy dz j^0 = c Q_1 \end{align}

It is important to note here that the components of f, with respect to the base ($$\tilde{dx}^1 \wedge \tilde[{dx}^2 \wedge \tilde{dx}^3$$) transform like the components of a vector (which is intuitive, since the $$j^{\mu}$$ ARE the components of a vector!). So we have \begin{align} f = \omega(j, \dot, \dot, \dot)= \tilde{j}^0\tilde{ dx}^1 \wedge \tilde{dx}^2 \wedge \tilde{dx}^3 + \tilde{j}^1 \tilde{dx}^0 \wedge \tilde{dx}^2 \wedge \tilde{dx}^3 + \tilde{j}^2 \tilde{dx}^0 \wedge \tilde{dx}^1 \wedge tilde{dx}^3 +\tilde{ j}^3 \tilde{dx}^0 \wedge \tilde{dx}^1 \wedge \tilde{dx}^2 \end{align}

Then the second integral computes to: \begin{align} \int_{S_2} f = \int dx dy dz f(\tilde{e_1},\tilde{ e_2}, \tilde{e_3}) = \int dx dy dz \tilde{j}^0 = c Q_2 \end{align}

The total charge contained in the two hyper-surfaces is the same! Since charge conservation holds, this is not only true for this special choice of hyper-surfaces, but as well for $$S_2$$ to be anchored at any point. To give an answer to my own question, the invariance of charge holds for any spacelike hypersurface that covers the region with $$j \neq 0$$. This hypersurface can (in the right frame of reference) be regarded as a spatial volume.