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## Noether currents for incremental Lorentz transformation.

Posted by peeterjoot on September 8, 2009

[Click here for a (CORRECTED) 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)

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.

FIXME: bother showing this explicitly?

The end result is that we should be able to bring the $x$ coordinates into the derivatives of (31) provided both are brought in. That gives us a more interesting conservation statement, something that has the looks of field angular momentum

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

The conservation identity could be summarized using

\begin{aligned}{M^{\alpha}}_{\mu\nu} &\equiv x_\nu {T^\alpha}_\mu - x_\mu {T^\alpha}_\nu \\ 0 &= \partial_\alpha {M^{\alpha}}_{\mu\nu} \end{aligned} \quad\quad\quad(34)

FIXME: Jackson ([4]) states a similar index upper expression

\begin{aligned}M^{\alpha\mu\nu} &\equiv x_\nu T^{\alpha\mu} - x_\mu T^{\alpha\nu} \\ 0 &= \partial_\alpha M^{\alpha\mu\nu} \end{aligned} \quad\quad\quad(36)

should try to show that these are identical or understand the difference.

# References

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