# Calculating different "kinds" of variations

Tags:
1. Feb 6, 2019

### Markus Kahn

1. The problem statement, all variables and given/known data
Let $x$ and $x'$ be two points from the Minkowski space connected through a Poincare transformation such that $x'^\mu =\Lambda_{\nu}^\mu x^\nu+a^\mu$ and $u:\mathcal{M}\to \mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, $\mathcal{M}$ the Minkowski space. We define:
\begin{align*} \Delta u(x) &~:=~ u^{\prime}(x^{\prime})-u(x) \qquad\text{ total infinitesimal variation}\cr \delta u(x) &~:=~ u^{\prime}(x)-u(x) \qquad \text{ local/vertical infinitesimal variation},\cr \mathrm{d}u(x)&~:=~ u(x^{\prime})- u(x) \qquad\text{ differential/horizontal infinitesimal variation}, \end{align*}
where and $u'$ is the function after the transformation. We further know that we can write
$$\Lambda_\nu^\mu = \delta_\nu^\mu - \frac{i}{2} \delta \omega^{\rho\sigma}(S_{\rho\sigma})^\mu_\nu,$$
with $\omega^{\mu\nu}+\omega^{\nu\mu}=0$, $S_{ij}=S_{ij}^\dagger$ and $S_{0k}-S_{0k}^\dagger$ (Spin antisymmetric tensor).

Now, show that:
1. $\delta x^\mu = \delta \omega^{\mu\nu}g_{\nu\rho}x^\rho +\delta \omega^\mu = \frac{1}{2}\delta\omega^{\rho\sigma}L_{\rho\sigma}x^\mu - i\delta\omega^{\rho}P_\rho x^\mu,$ where $P_\mu = i\partial_\mu$ and $L_{\mu\nu}\equiv x_\mu P_\nu - x_\nu P_\mu.$
2. For any relativistic covariant vector wave field $V'_\mu(x')=\Lambda_\nu^\mu V_\nu(x)$, we have $\Delta V_\mu (x)= -\frac{1}{2}i\delta \omega^{\rho\sigma}(S_{\rho\sigma})_\mu^\nu V_\nu (x).$
2. Relevant equations
I'm not sure if I really mentioned everything needed to complete the two exercises (since I honestly don't understand how to approach the problem in the first place), so everything that I summerized here can be found in the lec notes of R. Soldati on relativistic quantum field theory (pages 57-63, where I try to explicitly prove the result between eq. 2.16 and 2.17 as well as eq. 2.22), see here: http://robertosoldati.com/archivio/news/107/Campi1.pdf

3. The attempt at a solution
What I've tried:
1. \begin{align*}\delta x^\mu &= x'^\mu-x^\mu = \Lambda_\nu^\mu x^\nu + a^\mu -x^\mu = \left(\delta_\nu^\mu - \frac{i}{2} \delta \omega^{\rho\sigma}(S_{\rho\sigma})^\mu_\nu\right)x^\nu -x^\mu + a^\mu\\ &= \frac{i}{2} \delta \omega^{\rho\sigma}(S_{\rho\sigma})^\mu_\nu x^\nu + a^\mu\ \neq \delta \omega^{\mu\nu}g_{\nu\rho}x^\rho +\delta \omega^\mu. \end{align*} Not sure how I'm supposed to get to the desired result... The last equality is a complete mystery to me. Here I can't even find a way to start calculating what I'm supposed to..
2. Similarly to above I just plugged in the definition: \begin{align*}\Delta V_\mu (x)&= V'_\mu (x') -V_\mu (x)=\Lambda_\mu^\nu V_\nu (x) - V_\mu(x)\\ &= - \frac{i}{2} \delta \omega^{\rho\sigma}(S_{\rho\sigma})^\nu_\mu V_\nu(x), \end{align*} which leads to the desired result. I would still appreciate if someone could give a quick glance over it since I'm rather unsure about the indices...

2. Feb 11, 2019 at 6:00 PM

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.