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  1. Conservation of a charge $$ Q = \int dV \, j^0 $$ follows from current conservation $$ \partial_\mu j^\mu = 0 $$ and $$ \dot{Q} = - \oint dS \, j = 0 $$ where the Gauß divergence theorem has been applied (and where sufficiently localized charge and current distributions are assumed).

  2. Lorentz invariance of charge should follow from an appropriate definition $$ Q[V] = \int_V dV \, j^0 \to \int_V dV_\mu \, j^\mu $$ and using the invariance of $$ dV_\mu \, j^\mu. $$

The integral can also be defined using the trick

$$ Q[V] = \int_M d^4x \, j^\mu \, \partial_\mu \theta(v_\nu x^\nu) $$

where the resulting $\delta$-function projects to the 3-manifold $V$.

Usually (e.g. Weinberg „Gravitation“ 2.6) the calculation of the integral is done using

$$ \partial_\mu j^\mu = 0 $$

and therefore writing

$$ j^\mu \, \partial_\mu \theta = \partial_\mu (j^\mu \, \theta) $$

The last trick means that the proof of the Lorentz invariance of a charge presupposes the conservation of the same.

  1. Questions:
  • Is there a way to proof Lorentz invariance of a charge without using charge conservation?
  • If not, what does it mean in case of charges that are not conserved? (in quantum field theories a conserved classical charge may be no longer conserved due to quantum anomalies; such an anomaly does not break Lorentz symmetry)

Remark: The fact that for each $V$ the charge $Q[V]$ is a Lorentz scalar is trivial b/c for fixed $V$ the term $dV_\mu \, j^\mu$ is a scalar. The question is not about this trivial fact but about the proof of

$$Q[V^\prime] = Q[V]$$

that means, which charge is measured by two different observers O, O’ represented by $V, V^\prime$.

TomS
  • 883

1 Answers1

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Is there a way to proof Lorentz invariance of a charge without using charge conservation?

No. It's impossible to prove this statement, because it's not true. If the current is not conserved, the associated total charge is not a Lorentz scalar.

If not, what does it mean in case of charges that are not conserved?

Here's a very simple example that may give some intuition. If the current $J^\mu$ were not conserved, then charge could just appear and disappear out of nowhere. Suppose that at time $t = 0$ in some reference frame, a charge $q$ and another charge $-q$ instantly appear, at separate locations. In this reference frame, the total charge is always equal to zero. Therefore, if charge is to be a Lorentz scalar, it must also always be zero in any other reference frame.

But the notion of simultaneity differs between reference frames. Thus, in another frame you could have one charge appear before the other, so that the total charge is momentarily nonzero; thus it can't be a scalar.

The key step in your derivation is showing that the integral that defines charge is independent of the hypersurface $V$. That's the step that requires current conservation; without it, it fails because of examples like the one above. A similar conclusion holds for other conserved currents/tensors. For example, if the stress-energy tensor $T^{\mu\nu}$ isn't conserved, then the total four-momentum $P^\mu = \int dV \, T^{\mu 0}$ is not a four-vector.

knzhou
  • 101,976
  • Thanks, sounds reasonable. What does that mean for a charge Q that is conserved classically but not conserved due to a quantum anomaly. The l.h.s. of $\partial_\mu j^\mu = \ldots$ is still a scalar (or a pseudo-scalar), so is the r.h.s., i.e. the anomalous term. So what is the transformation of the integral taking the anomaly into account? – TomS Nov 14 '21 at 21:41