It is known that classical electrodynamics is time reversal invariant if one assumes that the transformation laws under such operation are $$\mathbf E(t,\mathbf x)\mapsto\mathbf E(-t,\mathbf x)$$ $$\mathbf B(t,\mathbf x)\mapsto -\mathbf B(-t,\mathbf x)$$ $$\rho(t,\mathbf x)\mapsto \rho(-t,\mathbf x)$$ $$\mathbf J(t,\mathbf x)\mapsto -\mathbf J(-t,\mathbf x)$$
How are these transformations related to the time reversal $T$ of the full Lorentz group $O(1,3)$? Here I would like to assume matrix notation for simplicity, so that $$T = \begin{bmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}$$ which is the Jacobian of the transformation $(t,\mathbf x)\mapsto(-t,\mathbf x)$. If you think of the 4-current as a 1-form $J$ over space-time, and you assume this transformation, that is, time reversal, to be passive, i.e. just a change of coordinates, then $$(\rho,-\mathbf J)\mapsto(-\rho,-\mathbf J),$$ from which one actually deduces $\rho(t,\mathbf x)\mapsto-\rho(-t,\mathbf x)$ and $\mathbf J(t,\mathbf x)\mapsto\mathbf J(-t,\mathbf x)$. One has a similar situation when transforming the electromagnetic tensor $F$ with $T$, which then gives $\mathbf E\mapsto -\mathbf E$ and $\mathbf B\mapsto\mathbf B$, but on the other hand the constitutive tensor $\star F$ gives the expected transformation laws for the fields, namely the ones given above. Is this just a mere coincidence?
Answer
The problem here arises because the 4-current in the OP is assumed to be a 1-form, and after many years of accumulated rust on the subject I completely forgot that this is, strictly speaking, not the right geometrical object that can describe current density. Indeed, being a density, it must be a 3-form, and therefore the correct geometrical object is $$J = \rho\ \text dx\wedge\text dy\wedge\text dz + \text dt\wedge(\mathbf J\cdot\star\text d\mathbf x)$$ where $\star\text d\mathbf x$ is the Hodge dual in $\mathbb R^3$ of the formal vector $(\text dx,\text dy,\text dz)$. This object has now the correct transformation law under time reversal, since $\text dt\mapsto -\text dt$ and $\text d\mathbf x\mapsto\text d\mathbf x$, and therefore $$J\mapsto\rho\ \text dx\wedge\text dy\wedge\text dz - \text dt\wedge(\mathbf J\cdot\star\text d\mathbf x).$$
No comments:
Post a Comment