I'm currently reading these$^1$ lec. notes as an introduction to relativistic QFT. In chapter two (pp.57-61) he introduces the concept of field variations along with some formulas for the different kind of variations (I think at least that is what he is doing...). I'm having a really hard time with the math that he writes down there.

He discusses the example of a Poincare-transformation that sends $x$ to $x'$. He states that we can then write $$x'^\mu \approx x^\mu + \delta x^\mu = x^\mu+\delta \omega^{\mu\nu}g_{\nu\lambda}x^\lambda+\delta \omega^\mu,\tag{2.12}$$ where $\delta\omega^{\mu\nu}=-\delta\omega^{\nu\mu}, |\delta\omega^{\mu\nu}|\ll 1$, $|\delta\omega^\mu|\ll 1$ and $g_{\mu\nu}$ the metric tensor. I can understand the first approxmiation, why $\delta\omega^{\mu\nu}$ has to be antisymmetric and much smaller than one, but I fail to see why $\delta x^\mu = \delta \omega^{\mu\nu}g_{\nu\lambda}x^\lambda+\delta \omega^\mu$ should make sense as an approximation. He then goes on to say that $$\begin{align*}\Delta u(x) &\equiv u'(x+\delta x)-u(x) \equiv \delta u(x+\delta x) + du(x)\\&=\delta u(x)+\delta x^{\mu} \partial_{\mu} \delta u(x)+\cdots+\mathrm{d} u(x) \\&= \delta u( x) +\delta x^\mu\partial_\mu u(x) + \mathcal{O}(\delta u \delta x),\end{align*}\tag{2.13}$$ where $u:\mathcal{M}\to\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$ is a function on the Minkowski space and $u'$ the function after the Poincare-transformation. I honestly don't understand any part of the above calculations (mostly because he doesn't even mention what $\delta$ and $d$ are supposed to be) and can even less imagine what $\delta u(x)$ is supposed to be.

Can somebody maybe explain to me what exactly is going on here and why assumptions like $[\delta,\partial_\mu]=0$ make sense? If you know of a better introduction to the topic feel free to suggest some books, paper, etc.

Math background: I'm not particularly familiar with the calculus of variations. I had some exposure to it in a classical machanics course, where we defined the variation of a functional ($\delta S[f]=\frac{d}{d \epsilon}\left.S[f+\epsilon \delta f]\right|_{\epsilon=0}$), but other than that I don't really now anything on the subject matter.

$^1$ Introduction to relativistic quantum field theory, R. Soldati, 2019.

  • $\begingroup$ You can recover their definitions in equations (2.2), (2.3) and (2.4) at the end of page 58 and using Qmechanic answer $\endgroup$ Feb 5, 2019 at 17:48
  • $\begingroup$ Regarding the request of other introductions to QFT, check the big list here. Be careful: if you have to take an exam with the professor and the professors gives you notes, it is much easier to pass the exam using his notes, especially for QFT, where everybody has a different approach and notations and it can be confusing. Using other resources for clarifications can be good though. Note that in the notes of your professor at the end of each chapter there's a little bibliography. $\endgroup$ Feb 5, 2019 at 18:48
  • $\begingroup$ @Runlikehell Thanks for the recommendations! I'm not taking the course of that specific professor, just got a recommendation to use to notes form a tutor. $\endgroup$
    – Sito
    Feb 5, 2019 at 19:20

2 Answers 2

  1. The "three kinds of infinitesimal variations" are defined as follows. $$\begin{align} \Delta u(x) &~:=~ u^{\prime}(x^{\prime})-u(x) \qquad\text{ total infinitesimal variation}, \tag{2.2}\cr \delta u(x) &~:=~ u^{\prime}(x)-u(x) \qquad \text{ local/vertical infinitesimal variation},\tag{2.3}\cr \mathrm{d}u(x)&~:=~ u(x^{\prime})- u(x) \qquad\text{ differential/horizontal infinitesimal variation}.\tag{2.4} \end{align}$$ Here the words horizontal and vertical spaces refer to spacetime and $u$-target space, respectively. While the terminology/names & notation vary from author to author, these three infinitesimal variations are often introduced in physics field theory textbooks.

  2. The statement $[\delta,\partial_\mu]=0$ means that vertical infinitesimal variations $\delta$ commute with spacetime-derivatives $\partial_{\mu}\equiv\frac{\partial}{\partial x^{\mu}}$. Be aware that total infinitesimal variations $\Delta$ do not necessarily commute with spacetime-derivatives $\partial_{\mu}$.

  • $\begingroup$ Can I ask you the origin of this terminology? It's different from the terminology used in the notes, where the author calls them total, local and differential variations $\endgroup$ Feb 5, 2019 at 17:46
  • 1
    $\begingroup$ I read it somewhere a long time ago. $\endgroup$
    – Qmechanic
    Feb 5, 2019 at 18:11
  • $\begingroup$ I'll accept the answer since after some messing around with the definitions that you provided I was able to recover most of the results! Thanks. $\endgroup$
    – Sito
    Feb 6, 2019 at 13:21

$\delta x^\mu$ has one upper index, so you just write down all possible terms which give some object with an upper index. Also $\delta x^\mu$ should only depend on $x^\mu$ and constants, because what else should it depend on? So you could come up with the idea to classify the contributions to $\delta x^\mu$ with respect to the order in $x$.

Zeroth order in $x$ is just a constant. Remember, it should carry an upper index, so it is a constant vector. $\delta x^\mu$ is infinitesimal, so each contribution should be infinitesimal. So you end up with an infinitesimal constant vector $$(\delta x^\mu)^{(0)}=\delta \omega^\mu$$ to zeroth order in $x$.

First order in $x$ is something of the form "constant times x". Since $x$ is a vector, the most general form would be contracting it with a constant tensor. The tensor should be infinitesimal (since $x$ itself is not infinitesimal). To get back an object with one upper index, the index structure should be as follows

$$(\delta x^\mu)^{(1)}=\delta\omega^\mu_\lambda x^\lambda$$

We can lift one of the indices using the metric tensor: $\delta\omega^\mu_\lambda=\delta\omega^{\mu\nu}g_{\nu\lambda}$ and will end up with

$$(\delta x^\mu)^{(1)}=\delta\omega^{\mu\nu}g_{\nu\lambda} x^\lambda$$

In total, up to first order in $x$ we get exactly (2.12).


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