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## (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).

# Guts

[Click here for a PDF of this sequence of posts with nicer formatting]

Let’s assume that we can use the exponential generator of rotations

\begin{aligned}e^{(i \cdot x) \cdot \nabla} = 1 + (i \cdot x) \cdot \nabla + \cdots \end{aligned} \quad\quad\quad(25)

to alter a Lagrangian density.

In particular, that we can use the first order approximation of this Taylor series, applying the incremental rotation operator $(i \cdot x) \cdot \nabla = i \cdot (x \wedge \nabla)$ to transform the Lagrangian.

\begin{aligned}\mathcal{L} \rightarrow \mathcal{L} + (i \cdot x) \cdot \nabla \mathcal{L} \end{aligned} \quad\quad\quad(26)

Suppose that we parametrize the rotation bivector $i$ using two perpendicular unit vectors $u$, and $v$. Here perpendicular is in the sense $u v = -v u$ so that $i = u \wedge v = u v$. For the bivector expressed this way our incremental rotation operator takes the form

\begin{aligned}(i \cdot x) \cdot \nabla &=((u \wedge v) \cdot x) \cdot \nabla \\ &=(u (v \cdot x) - v (u \cdot x)) \cdot \nabla \\ &=(v \cdot x) u \cdot \nabla - (u \cdot x)) v \cdot \nabla \\ \end{aligned}

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

\begin{aligned}a \cdot \nabla \mathcal{L}&=\partial_\alpha \left(\frac{\partial {\mathcal{L}}}{\partial {(\partial_\alpha A^\beta)}} (a \cdot \nabla) {A^\beta}\right) \end{aligned} \quad\quad\quad(27)

So for incremental rotations the change to the Lagrangian is

\begin{aligned}(i \cdot x) \cdot \nabla \mathcal{L}&=(v \cdot x)\partial_\alpha \left(\frac{\partial {\mathcal{L}}}{\partial {(\partial_\alpha A^\beta)}} (u \cdot \nabla) {A^\beta}\right) -(u \cdot x)\partial_\alpha \left(\frac{\partial {\mathcal{L}}}{\partial {(\partial_\alpha A^\beta)}} (v \cdot \nabla) {A^\beta}\right) \end{aligned} \quad\quad\quad(28)

Since the choice to make $u$ and $v$ 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 $u = \gamma_\mu$ and $v = \gamma_\nu$.

The incremental rotation operator is reduced to

\begin{aligned}(i \cdot x) \cdot \nabla &=(\gamma_\nu \cdot x) \gamma_\mu \cdot \nabla - (\gamma_\mu \cdot x)) \gamma_\nu \cdot \nabla \\ &=x_\nu \partial_\mu - x_\mu \partial_\nu \\ \end{aligned}

Similarly the change to the Lagrangian is

\begin{aligned}(i \cdot x) \cdot \nabla \mathcal{L}&=x_\nu\partial_\alpha \left(\frac{\partial {\mathcal{L}}}{\partial {(\partial_\alpha A^\beta)}} \partial_\mu {A^\beta}\right) -x_\mu\partial_\alpha \left(\frac{\partial {\mathcal{L}}}{\partial {(\partial_\alpha A^\beta)}} \partial_\nu {A^\beta}\right) \end{aligned} \quad\quad\quad(29)

Subtracting the two, essentially forming $(i \cdot x) \cdot \nabla \mathcal{L} - (i \cdot x) \cdot \nabla \mathcal{L} = 0$, we have

\begin{aligned}0 =x_\nu\partial_\alpha \left(\frac{\partial {\mathcal{L}}}{\partial {(\partial_\alpha A^\beta)}} \partial_\mu {A^\beta}- {\delta^\alpha}_\mu \mathcal{L}\right) -x_\mu\partial_\alpha \left(\frac{\partial {\mathcal{L}}}{\partial {(\partial_\alpha A^\beta)}} \partial_\nu {A^\beta}- {\delta^\alpha}_\nu \mathcal{L}\right) \end{aligned} \quad\quad\quad(30)

We previously wrote

\begin{aligned}{T^\alpha}_\nu &= \frac{\partial {\mathcal{L}}}{\partial {(\partial_\alpha A^\beta)}} \partial_\nu A^\beta - {\delta^\alpha}_\nu \mathcal{L} \\ \end{aligned}

for the Noether current of spacetime translation, and with that our conservation equation becomes

\begin{aligned}0 = x_\nu \partial_\alpha {T^\alpha}_\mu - x_\mu \partial_\alpha {T^\alpha}_\nu \end{aligned} \quad\quad\quad(31)

As is, this doesn’t really appear to say much, since we previously also found $\partial_\alpha {T^\alpha}_\nu = 0$. 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 $\nabla \cdot (i \cdot x) \mathcal{L}$ to see what is left over compared to $(i \cdot x) \cdot \nabla \mathcal{L}$ shows that there is in fact no difference, and we actually have the identity

\begin{aligned}i \cdot (x \wedge \nabla) \mathcal{L} = (i \cdot x) \cdot \nabla \mathcal{L} = \nabla \cdot (i \cdot x) \mathcal{L} \end{aligned} \quad\quad\quad(32)

This suggests that we can pull the $x$ coordinates into the derivatives of (31) as in

\begin{aligned}0 = \partial_\alpha \left( {T^\alpha}_\mu x_\nu - {T^\alpha}_\nu x_\mu \right) \end{aligned} \quad\quad\quad(33)

However, expanding this derivative shows that this is fact not the case. Instead we have

\begin{aligned}\partial_\alpha \left( {T^\alpha}_\mu x_\nu - {T^\alpha}_\nu x_\mu \right) &={T^\alpha}_\mu \partial_\alpha x_\nu - {T^\alpha}_\nu \partial_\alpha x_\mu \\ &={T^\alpha}_\mu \eta_{\alpha\nu}- {T^\alpha}_\nu \eta_{\alpha\mu} \\ &=T_{\nu\mu} - T_{\mu\nu} \end{aligned}

So instead of a Noether current, following the procedure used to calculate the spacetime translation current, we have only a mediocre compromise

\begin{aligned}{M^{\alpha}}_{\mu\nu} &\equiv {T^\alpha}_\mu x_\nu - {T^\alpha}_\nu x_\mu \\ \partial_\alpha {M^{\alpha}}_{\mu\nu} &= T_{\nu\mu} - T_{\mu\nu} \end{aligned} \quad\quad\quad(34)

Jackson ([4]) ends up with a similar index upper expression

\begin{aligned}M^{\alpha\beta\gamma} &\equiv T^{\alpha\beta} x^\gamma - T^{\alpha\gamma} x^\beta \\ \end{aligned} \quad\quad\quad(36)

and then uses a requirement for vanishing 4-divergence of this quantity

\begin{aligned}0 &= \partial_\alpha M^{\alpha\beta\gamma} \end{aligned} \quad\quad\quad(38)

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 $0 - 0 = 0$ result as we had in (31)

\begin{aligned}0 = x^\nu \partial_\alpha T^{\alpha\mu} - x^\mu \partial_\alpha T^{\alpha\nu} \end{aligned} \quad\quad\quad(39)

The only difference is that our not-really-a-conservation equation becomes

\begin{aligned}\partial_\alpha M^{\alpha\mu\nu} = T^{\nu\mu} - T^{\mu\nu} \end{aligned} \quad\quad\quad(40)

## 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

\begin{aligned}\mathcal{L} = \frac{1}{{2}} (\boldsymbol{\nabla} \phi)^2 - \frac{1}{{\epsilon_0}}\rho \phi \end{aligned}

For this we have

\begin{aligned}\mathcal{L}' &= \mathcal{L} + (i \cdot \mathbf{x}) \cdot \boldsymbol{\nabla} \mathcal{L} \\ &= \mathcal{L} - (i \cdot \mathbf{x}) \cdot \frac{1}{{\epsilon_0}} \left( \rho \boldsymbol{\nabla} \phi + \phi \boldsymbol{\nabla} \rho \right) \end{aligned}

If the variational derivative of the incremental rotation contribution is zero, then we have a symmetry.

\begin{aligned}\frac{\delta }{\delta \phi} (i \cdot \mathbf{x}) \cdot \boldsymbol{\nabla} \mathcal{L} \\ &=(i \cdot \mathbf{x}) \cdot \frac{1}{{\epsilon_0}} \boldsymbol{\nabla} \rho - \sum_m \partial_m \left( (i \cdot \mathbf{x}) \cdot \frac{1}{{\epsilon_0}} \rho \mathbf{e}_m \right) \\ &=(i \cdot \mathbf{x}) \cdot \frac{1}{{\epsilon_0}} \boldsymbol{\nabla} \rho - \boldsymbol{\nabla} \cdot \left( (i \cdot \mathbf{x}) \frac{1}{{\epsilon_0}} \rho \right) \\ \end{aligned}

As found in (32), we have $(i \cdot \mathbf{x}) \cdot \boldsymbol{\nabla} = \boldsymbol{\nabla} \cdot (i \cdot \mathbf{x})$, so we have

\begin{aligned}\frac{\delta }{\delta \phi} (i \cdot \mathbf{x}) \cdot \boldsymbol{\nabla} \mathcal{L} = 0 \end{aligned} \quad\quad\quad(41)

for this specific Lagrangian as expected.

Note that the test expansion I used to state (32) was done using only the bivector $i = \gamma_\mu \wedge \gamma_\nu$. An expansion with $i = u^\alpha u^\beta \gamma_\alpha \wedge \gamma_\beta$ shows that this is also the case in shows that this is true more generally. Specifically, this expansion gives

\begin{aligned}\nabla \cdot (i \cdot x) \mathcal{L} &= (i \cdot x) \cdot \nabla \mathcal{L} + (\eta_{\alpha\beta} - \eta_{\beta\alpha}) u^\alpha v^\beta \mathcal{L} \\ &= (i \cdot x) \cdot \nabla \mathcal{L} \end{aligned}

(since the metric tensor is symmetric).

Loosely speaking, the geometric reason for this is that $\nabla \cdot f(x)$ takes its maximum (or minimum) when $f(x)$ is colinear with $x$ and is zero when $f(x)$ is perpendicular to $x$. The vector $i \cdot x$ 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 $i \cdot \mathbf{x}$ 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.

# References

[4] JD Jackson. Classical Electrodynamics Wiley. 2nd edition, 1975.
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