(CORRECTED) Noether current for incremental Lorentz transformation.
Posted by peeterjoot on September 8, 2009
Had logic errors in previous post on the same. Corrected here (replacing the pdf version, but retaining the previous mistaken notes).
Let’s assume that we can use the exponential generator of rotations
to alter a Lagrangian density.
In particular, that we can use the first order approximation of this Taylor series, applying the incremental rotation operator to transform the Lagrangian.
Suppose that we parametrize the rotation bivector using two perpendicular unit vectors , and . Here perpendicular is in the sense so that . For the bivector expressed this way our incremental rotation operator takes the form
The operator is reduced to a pair of torque-like scaled directional derivatives, and we’ve already examined the Noether currents for the translations induced by the directional derivatives. It’s not unreasonable to take exactly the same approach to consider rotation symmetries as we did for translation. We found for incremental translations
So for incremental rotations the change to the Lagrangian is
Since the choice to make and both unit vectors and perpendicular has been made, there is really no loss in generality to align these with a pair of the basis vectors, say and .
The incremental rotation operator is reduced to
Similarly the change to the Lagrangian is
Subtracting the two, essentially forming , we have
We previously wrote
for the Noether current of spacetime translation, and with that our conservation equation becomes
As is, this doesn’t really appear to say much, since we previously also found . We appear to need a way to pull the x coordinates into the derivatives to come up with a more interesting statement. A test expansion of to see what is left over compared to shows that there is in fact no difference, and we actually have the identity
This suggests that we can pull the coordinates into the derivatives of (31) as in
However, expanding this derivative shows that this is fact not the case. Instead we have
So instead of a Noether current, following the procedure used to calculate the spacetime translation current, we have only a mediocre compromise
Jackson () ends up with a similar index upper expression
and then uses a requirement for vanishing 4-divergence of this quantity
to symmetrize this tensor by subtracting off all the antisymmetric portions. The differences compared to Jackson with upper verses lower indexes are minor for we can follow the same arguments and arrive at the same sort of result as we had in (31)
The only difference is that our not-really-a-conservation equation becomes
An example of the symmetry.
While not a proof that application of the incremental rotation operator is a symmetry, an example at least provides some comfort that this is a reasonable thing to attempt. Again, let’s consider the Coulomb Lagrangian
For this we have
If the variational derivative of the incremental rotation contribution is zero, then we have a symmetry.
As found in (32), we have , so we have
for this specific Lagrangian as expected.
Note that the test expansion I used to state (32) was done using only the bivector . An expansion with shows that this is also the case in shows that this is true more generally. Specifically, this expansion gives
(since the metric tensor is symmetric).
Loosely speaking, the geometric reason for this is that takes its maximum (or minimum) when is colinear with and is zero when is perpendicular to . The vector is a combined projection and 90 degree rotation in the plane of the bivector, and the divergence is left with no colinear components to operate on.
While this commutation of the with the divergence operator didn’t help with finding the Noether current, it does at least show that we have a symmetry. Demonstrating the invariance for the general Lagrangian (at least the single field variable case) likely follows the same procedure as in this specific example above.
 JD Jackson. Classical Electrodynamics Wiley. 2nd edition, 1975.