Effects of a Lorentz boost on the normalization of a probability density (Dirac equation) My question regards the probability densities of the Dirac equation.
As is well known, the Dirac equation implies a continuity equation
$$
\partial_\mu j^\mu = 0
$$
for $j^\mu = c\overline\psi \gamma^\mu\psi$. This is interpreted as a probability current, with the zeroth component $j^0/c = \psi^\dagger \psi = \rho$ being a locally conserved probability density.
When applying a Lorentz boost in the x direction (i.e. changing our description to the POV of a reference frame moving in the x direction), the following things change in the following ways:
An arbitrary volume in space becomes
$$
V \rightarrow V' = \frac{V}{\gamma}
$$
because of the Lorentz contraction in the direction of motion. Same thing for an infinitesimal volume element:
$$
dV \rightarrow dV' = \frac{dV}{\gamma}
$$
The probability density $\rho$ transforms as the zeroth component of a four-vector:
$$
\rho \rightarrow \rho'=\gamma \left(\rho - \frac{\beta J_x}{c}\right)
$$
and the x component of the probability current, $J_x$, becomes
$$
J_x \rightarrow J_x' = \gamma \left(J_x - \beta c \rho \right)
$$
The way I interpret the above transformation laws for $\rho$ and $J_x$ is that, when changing to a moving reference frame, part of what was seen as probability density is now seen as probability current, and vice versa. That makes intuitive sense to me, and it doesn't violate the covariance of the continuity equation $\partial_\mu j^\mu = 0$.
My confusion starts when I consider normalization. We must have (in the original frame, for example)
$$
\int_V \rho\ dV = 1
$$
where V is the volume of whatever domain I have set up for my system.
I would expect that, in the boosted frame, we should have
$$
\int_{V'}\rho'dV' = 1
$$
and that this could be seen just from the previous transformation laws. When I try substituting for $\rho'$, $V'$ and $dV'$, however, I get:
$$
\int_{V'}\rho'dV' = \int_{V/\gamma} \gamma \left(\rho - \frac{\beta J_x}{c}\right) \frac{dV}{\gamma} = \int_{V/\gamma} \rho dV - \int_{V/\gamma} \frac{\beta J_x}{c} dV
$$
I was hoping to get something like $\int_{V'} \rho'\ dV' = \int_V \rho\ dV$, so that normalization in one frame implies normalization in the other. But I'm not sure that's possible. Is it?
In fact, we don't even get $\rho dV = \rho' dV'$, which is totally something I would expect, but I guess it's not necessary. Would love some clarification. Thanks in advance.
 A: For the $\psi$ in the Dirac equation, interpreting the quantity $|\psi(x)|^2$ as a probability density doesn't work for other reasons, but the quantity $|j^0(x)|^2\propto |\psi(x)|^2$ is still meaningful as a charge density. The total charge should be Lorentz invariant, so the question still stands.
To answer the question, we need to address two things:

*

*First, length contraction is not the right concept to apply here. A Lorentz boost mixes spatial coordinate(s) with the time coordinate, even when transforming the volume element $dV$.


*Second, to show that the total charge is Lorentz invariant, the continuity equation $\partial_\mu j^\mu=0$ must be used. If $j^\mu$ didn't satisfy the continuity equation, then the total charge wouldn't be Lorentz invariant.
To see how this works in detail, consider a boost in the $t$-$x$ plane, so that the other two coordinates $y$ and $z$ are not affected. The boost can be written
\begin{gather}
 \hat t = ct+sx
\\
 \hat x = st+cx
\tag{1}
\end{gather}
with $c^2-s^2=1$, and
\begin{gather}
 \hat j^0 = cj^0  +sj^1
\\
 \hat j^1 = sj^0 + cj^1.
\tag{2}
\end{gather}
Therefore,
$$
 \int_{\hat\Sigma} d\hat x\,\hat j^0
 = \int (s\,dt + c\,dx)\,(cj^0+s j^1).
\tag{3}
$$
where the integral is over a hyperplane $\hat\Sigma$ defined by $\hat t=$ constant. Compare this to the quantity
$$
 \int_\Sigma dx\,j^0,
\tag{4}
$$
where the integral is over a hyperplane $\Sigma$ defined by $t=$ constant. The quantities (3) and (4) are not equal to each other in general, but they are if $j^\mu$ satisfies the continuity equation. To prove this, first consider the integral $\int_R d^4x\ \partial_\mu j^\mu$ over any finite region $R$ of spacetime. Don't assume $\partial_\mu j^\mu=0$ yet. The general version of Stokes' theorem says
$$
 \int_R d^4x\ \partial_\mu j^\mu =
 \int_{\partial R} dS_\mu j^\mu,
\tag{5}
$$
where the second integral is over the boundary $\partial R$ of $R$ and where $dS_\mu$ is the "volume" element tangent to the boundary. Working through the proof of (5) is left as an exercise. It's worth the investment, because the general version of Stokes' theorem is used for lots of different things in physics.
To continue, suppose that $j^\mu$ satisfies the continuity equation
$$
 \partial_\mu j^\mu=0.
$$
Then (5) implies
$$
 \int_{\partial R} dS_\mu j^\mu = 0
\tag{6}
$$
for the boundary $\partial R$ of any finite region $R$ of spacetime. Equation (6) is the key to showing that (3) and (4) are equal to each other. To finish the proof, we need to acknowledge a tacit assumption, namely that $j^\mu$ falls to zero sufficiently rapidly at spacelike infinity so that the integrals (3) and (4) are well-defined. For simplicity, let $S$ be some finite region of spacetime, and suppose that $j^\mu$ is zero everywhere in the causal complement $S$. The causal complement consists of all points that are spacelike-separated from all points in $S$, so this implements the requirement that $j^\mu$ falls to zero sufficiently rapidly at spatial infinity. Now let $R$ be the part of $S$ that is between the two hyperplanes $\Sigma$ and $\hat\Sigma$, so these two hyperplanes make up part of the boundary of $R$. (These two hyperplanes intersect each other, and that's okay. That just means that $R$ consists of two "wedges" that touch only along their sharp edges.) We assumed that $j^\mu=0$ on the other parts of the boundary of $R$, so equation (6) says
$$
 \int_{\Sigma} dS_\mu j^\mu + \int_{\hat\Sigma} dS_\mu j^\mu = 0,
\tag{7}
$$
The first term is the quantity (4), and the second term is the negative of the quantity (3). The negative sign comes from the fact that the "volume" element in equation (6) is oriented, and its orientation changes smoothly as we move around the boundary $\partial R$, so we get a relative sign between the "top" and "bottom" hypersurfaces. Altogether, this gives
$$
 \int_\Sigma dx\,j^0 -
 \int_{\hat\Sigma} d\hat x\,\hat j^0 = 0,
\tag{8}
$$
which is what we wanted to prove.
