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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.

Sito
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2 Answers2

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  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 $\partial_{\mu}$.

Qmechanic
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$\delta x^\mu$ 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).

Photon
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