Lorentz Invariance of Q in Weinberg: Justifying Transformation

[ergospherical]

If $\partial_{\alpha} J^{\alpha}(x) = 0$ then $Q \equiv \displaystyle{\int} d^3 x J^t(x)$ is time-invariant. To show that if $J^{\alpha}(x)$ is a four-vector then $Q$ is also Lorentz-invariant, he re-writes it as \begin{align*}
Q = \int d^4 x J^{\alpha}(x) \partial_{\alpha} H(n_{\beta} x^{\beta})
\end{align*}where $n_{\alpha} = (1,0,0,0)$ and $H(x)$ is the unit step function. I think this makes sense because $n_{\beta} x^{\beta} = t$ and \begin{align*}
Q = \int d^4 x J^{\alpha}(x) \partial_{\alpha} H(t) &= \int d^4 x J^t(x) \partial_t H(t) \\
&= \int d^4 x J^t(x) \delta(t) \\
&= \int d^3 x J^t(x)
\end{align*}The next part is to find $Q'$ under a Lorentz transformation, which he claims is done by changing $n_{\beta} \rightarrow n'_{\beta} = {{\Lambda}^{\gamma}}_{\beta} n_{\gamma}$. That is, $Q' - Q = \displaystyle{\int} d^4 x \left[ J^{\alpha}(x)\{H(n_{\beta}' x^{\beta}) - H(n_{\beta} x^{\beta}) \}\right]$. What is the justification for this step - why, for example, is $n_{\beta}$ the only quantity that needs to be transformed in the integrand, and not $x$ or $J(x)$?

 

[ergospherical]

Maybe\begin{align*}
Q' = \int d^4 x' J^{\alpha}(x') \partial'_{\alpha}H(n'_{\beta} x'^{\beta}) \overset{x' \rightarrow x}{=} \int d^4 x J^{\alpha}(x) \partial_{\alpha}H(n'_{\beta} x^{\beta})
\end{align*}because $x'$ is a dummy variable of integration, yes?

 

[PeterDonis]

Maybe\begin{align*}
Q' = \int d^4 x' J^{\alpha}(x') \partial'_{\alpha}H(n'_{\beta} x'^{\beta}) \overset{x' \rightarrow x}{=} \int d^4 x J^{\alpha}(x) \partial_{\alpha}H(n'_{\beta} x^{\beta})
\end{align*}because $x'$ is a dummy variable of integration, yes?

$x$ is not really a dummy variable, no, because of the step function and the contraction with the normal $n_\beta$ in the argument to the step function.

I think the term "Lorentz invariance" may be a bit of a misnomer here. "Lorentz invariance" is usually used to refer to quantities at a single spacetime point, but $Q$ is an integral over a spacelike hypersurface. The "Lorentz transformation" here is really changing which hypersurface is being integrated over (heuristically, by changing the "angle" of the hypersurface in the spacetime). So $Q'$ and $Q$ are integrals over two different hypersurfaces.

The purpose the step function (or more precisely its derivative) serves is to pick out the spacelike hypersurface (or more precisely its "angle"); Lorentz transforming the normal vector $n_\beta$ changes the "angle" of the hypersurface that gets picked out by the delta function derivative of the step function. So the "Lorentz transformation" here is only supposed to operate on $n_\beta$; it would make no sense to have it also operate on $x^\beta$ or $J^\alpha$, because those operations would change things other than which hypersurface we are integrating over. See further comments below.

why, for example, is $n_{\beta}$ the only quantity that needs to be transformed in the integrand, and not $x$ or $J(x)$?

I think it's the reason I just gave above. However, I'm not sure that's the only step that was left out in obtaining the equation for $Q' - Q$. I think the fact that $\partial_\alpha J^\alpha = 0$ has to come into play somewhere, because without some additional constraint on $j^\alpha$, beyond it just being a 4-vector, I don't see why $Q$ would be Lorentz invariant--why would an integral over a different spacelike hypersurface at a different "angle" have to give the same answer for any 4-vector? But if $J^\alpha$ has to be divergence free, that additional condition might be enough.

 

[ergospherical]

The Lorentz transformation $Q \rightarrow Q'$ should keep the domain fixed, but a change of variables e.g. $x^{\alpha} \rightarrow x'^{\alpha} = {\Lambda^{\alpha}}_{\beta} x^{\beta}$ in the integration (but not a Lorentz transformation) should take the domain $D \rightarrow D'$. To put the answer back in terms of $D$ requires, say, the tube theorem (assuming $J(x)$ vanishes sufficiently quickly at large distance), right?

 

[PeterDonis]

The Lorentz transformation $Q \rightarrow Q'$ should keep the domain fixed

That's one possible interpretation (keep the 3-surface the same but change coordinates on the spacetime), but I'm not sure it's the one Weinberg intended. We would expect any integral of the same quantity over the same 3-surface to give the same result regardless of the choice of coordinates, just based on general principles of tensor analysis, so I would not see the point of making a big effort to prove it in a textbook. But if the integrals are over two different 3-surfaces, then it seems reasonable to investigate in a textbook whether they both give the same answer.

a change of variables e.g. $x^{\alpha} \rightarrow x'^{\alpha} = {\Lambda^{\alpha}}_{\beta} x^{\beta}$ in the integration (but not a Lorentz transformation) should take the domain $D \rightarrow D'$.

No. Remember that the integral over the 3-surface is being obtained from an integral over all of spacetime by using the derivative of a step function. So whatever determines the domain of the integral over the 3-surface has to appear in the argument of the step function. And the thing in that argument that obviously would do that is the normal $n_\beta$, since by definition that's the normal to the 3-surface, so to change the 3-surface, that is what you would change.

(Note that I say "the 3-surface", but actually $n_\beta$ is a vector field on the spacetime, so what it's actually defining is an entire foliation of the spacetime by 3-surfaces, the ones everywhere orthogonal to $n_\beta$. The derivative of the step function just picks out the one 3-surface in the foliation whose "time coordinate" in the obvious inertial coordinate chart implied by the foliation is $0$.)

 

[TSny]

The total charge in the unprimed frame is $Q = \int d^3x J^0(x)$, where the integral is carried out at one fixed time $t$ in the unprimed frame. Weinberg shows that $Q$ is independent of the choice of the fixed time $t$, so we might as well choose $t = 0$. This picks out the particular hypersurface in spacetime that defines the domain of integration for $Q$.

Likewise, the total charge according to the primed frame is $Q' = \int d^3x' J'^0(x')$, where we may choose $t' = 0$ as the hypersurface for this integral. This hypersurface is a different set of spacetime points than the hypersurface $t = 0$.

Defining $n_{\alpha} \equiv (1, 0, 0, 0)$, we have

$Q = \int d^4x J^{\alpha}(x)\partial_{\alpha} H(n_{\beta}x^{\beta})$

$Q' = \int d^4x' J'^{\alpha}(x')\partial'_{\alpha} H(n_{\beta}x'^{\beta})$

It's important to note that in these expressions for $Q$ and $Q'$, $n_\beta$ is the same set of numbers (1, 0, 0, 0) in both integrals. Thus, there is no prime on $n_\beta$ in the expression for $Q'$ above.

Next, we may write the integral for $Q'$ in terms of the unprimed coordinates $x^\mu$ and the unprimed $J^\mu(x)$ via $x'^\mu = \Lambda^\mu\, _\nu x^\nu$, $J'^\mu(x') = \Lambda^\mu\, _\nu J^\nu (x)$, and $\partial'_\alpha = \Lambda_\alpha \, ^\beta \partial_\beta$. It is well-known that $d^4x' = d^4x$ and it is easy to see (or show) that $J'^\alpha(x') \partial '_\alpha = J^\beta(x)\partial_\beta$. So,

$Q' = \int d^4x J^\alpha(x) \partial_\alpha H(n_\beta \Lambda^\beta \, _\gamma x^\gamma)$

$n_\beta$ here still represents the numbers (1, 0, 0, 0). Next, Weinberg defines the quantities

$n'_\gamma \equiv \Lambda^\beta \, _\gamma n_\beta$

This is just a definition. Weinberg is not treating $n_\beta$ as a covariant 4-vector. If $n_\beta$ were a 4-vector then the transformation law would be $n'_\gamma = \Lambda _\gamma \, ^\beta n_\beta$ , which is not the same as Weinberg's $n'_\gamma = \Lambda^\beta \, _\gamma n_\beta$.

So, with this definition of $n'_\gamma$, we finally arrive at Weinberg's result

$Q' = \int d^4x J^\alpha(x) \partial_\alpha H(n'_\gamma x^\gamma)$

This integral still represents integration over the hypersurface $t' = 0$, but it is expressed in terms of the unprimed coordinates and values of the unprimed current $J^\alpha(x)$.

Finally we have,

$Q'-Q = \int d^4x J^\alpha(x) \partial_\alpha\left[H(n'_\beta x^\beta) - H(n_\beta x^\beta) \right]$

$\partial_\alpha H(n'_\beta x^\beta)$ is nonzero only on the hypersurface $t' = 0$ while $\partial_\alpha H(n_\beta x^\beta)$ is nonzero only on the hypersurface $t = 0$.

Weinberg proceeds from here, with the help of $\partial_\alpha J^\alpha = 0$, to show that $Q' = Q$.

 

[ergospherical]

Great, thanks! The approach makes sense after having understood the definition of $n'$.

 

[TSny]

The notation might be less confusing if we introduce two four-vectors $n$ and $N$ associated with the two hypersurfaces. $n$ is defined by having $n_\alpha = (1,0,0,0)$ in the unprimed frame and $N$ is defined to have $N'_\alpha = (1, 0, 0, 0)$ in the primed frame. Then

$Q = \int d^4x J^\alpha(x) \partial_\alpha H(n_\beta x^\beta) \,\,\,\,$ and $\,\,\,\, Q' = \int d^4x' J'^\alpha(x') \partial'_\alpha H(N'_\beta x'^\beta)$

Writing $Q'$ in terms of the unprimed coordinates and current yields

$Q' = \int d^4x J^\alpha(x) \partial_\alpha H(N_\beta x^\beta)$

Here, $N_\beta = (\Lambda^{-1})_\beta \, ^\gamma N'_\gamma = \Lambda ^\gamma \, _\beta N'_\gamma$

The components $N_\beta$ have the same values as Weinberg's $n'_\beta$.

Then $Q'-Q = \int d^4x J^\alpha(x) \partial_\alpha \left[ H(N_\beta x^\beta) - H(n_\beta x^\beta)\right]$

 

[vanhees71]

What Weinberg writes down is just the volume integral over $\partial_{\mu} J^{\mu}$ over a "world tube" enclosing the world line of the particle with two space-like hypersurfaces with the time-like normal vectors $N_{\beta}$ and $n_{\beta}$ as bottom and top boundary surface. Then you use the 4D Gauss's integral theorem to see that $Q'-Q=0$, where $Q$ ($Q'$) is the charge as calculated in reference frames with $n^{\beta}$ ($N^{\beta}$). This shows that the total charge is a scalar when calculated in this way, i.e.,
$$Q'(t')=\int_{\R^3} \mathrm{d}^3 x' \rho'(t',\vec{x'}) = \int_{\R^3} \mathrm{d}^3 x \rho(t,\vec{x})=Q(t).$$
Using the world tube with boundaries such that $N_{\beta}=n_{\beta}$ this proves that the total charge is conserved. Both results together show that electric charge is conserved and a Lorentz scalar.

It's important to keep in mind that the naive definition of the charge as spatial integrals at fixed time in an arbitrary inertial frame leads only to a Lorentz invariant charge if the local conservation law $\partial_{\mu} J^{\mu}=0$ holds, which is known as "von Laue's theorem".

For a more detailed discussion (applied to a conserved particle-number current rather than the em. current, but the math is of course identical) see pp. 18-19 in

https://itp.uni-frankfurt.de/~hees/publ/kolkata.pdf .

If you naively calculate such integral quantities from non-conserved densities, you easily run into trouble with interpretation. A famous example is the infamous "$4/3$ problem" in the definition of electromagnetic energy and momentum at presence of charge and current densities, where of course $\partial_{\mu} T_{\text{em}}^{\mu \nu}=-F^{\nu \rho} j_{\rho} \neq 0$.

 

[dextercioby]

I am pretty sure this particular problem at hand has been previously addressed by @samalkhaiat here (well, I do not recall if there was the exact argument of Weinberg explained/expanded upon, but the overall „Q is constant” issue).