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I have confusion regarding the notation that is used for infinitesimal Lorentz transformations and the parameters that define the Lorentz transformation (used in various books such as Srednicki's and Weinberg's).

Both of the above sources first define an infinitesimal Lorentz transformation as $\Lambda^\mu_{~\nu}=\delta^\mu_\nu+\epsilon^\mu_{~\nu}$ which leads to the condition $\epsilon^{\mu\nu}=-\epsilon^{\nu\mu}$.

They then go on define a finite Lorentz transformation as $$\Lambda=\exp\left(\frac{i}{2}\epsilon_{\mu\nu}M^{\mu\nu}\right)$$ where $M^{\mu\nu}$ are the generators of the Lorentz group. This makes it seem like $\epsilon_{\mu\nu}$ are the parameters of the transformation.

Based on the first definition, what I understand is that a general Lorentz transformation is got by exponentiating $\epsilon_{\mu\nu}=\left(\frac{i}{2}\Omega_{\rho\sigma}M^{\rho\sigma}\right)_{\mu\nu}$ where $\epsilon_{\mu\nu}$ is the whole Lorentz transformation and $\Omega_{\rho\sigma}$ defines the parameters of the transformation.

It would be great if someone could clarify what is going on here.

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  • $\begingroup$ I’m confused as to whether epsilon is the infinitesimal transformation or just the parameters of the transformation as it seems like it’s being used both ways. $\endgroup$
    – adithya
    Jun 27, 2019 at 7:55
  • $\begingroup$ $\epsilon$ is indeed the parameter, the "infinitesimal angle" or however else people want to call it; from your notation it seems it is indeed used for both and you can just correct the latter notation with a new symbol. $\endgroup$
    – gented
    Jun 27, 2019 at 7:58
  • $\begingroup$ @adithya they're the same thing $\endgroup$
    – Avantgarde
    Jun 27, 2019 at 8:08
  • $\begingroup$ @Avantgarde I don’t understand. Can you please explain how they’re the same? Because as far I can see, in one case, epsilon would be a part of the Lie algebra while in the other it’s not. $\endgroup$
    – adithya
    Jun 27, 2019 at 8:54
  • $\begingroup$ @adithya The generators form a Lie algebra. They are $M_{\rho \sigma}$, not $\epsilon_{\mu \nu}$. $\endgroup$
    – Avantgarde
    Jun 27, 2019 at 10:42

2 Answers 2

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As I suspect many students of field theory will have the same question, I'll elaborate on @Rindler98 's answer: In the defining representation of the Lorentz group the matrix ${\omega^\mu}_\nu$ is different but closely related to $\omega_{\mu\nu}$.

To see this, first of all note that in the defining representation we can write $$ {\omega^\mu}_\nu = g^{\mu\alpha}\omega_{\alpha\nu} $$ where $g$ is the Minkowski / Lorentzian metric.

The trick Rindler98 mentions is to notice that \begin{align} g^{\mu\alpha}\omega_{\alpha\nu} &= g^{\mu\alpha} {\delta^\beta}_\nu \omega_{\alpha\beta} \\ &= \frac{1}{2} \omega_{\alpha\beta} \left(g^{\mu\alpha} {\delta^\beta}_\nu - g^{\mu\beta} {\delta^\alpha}_\nu \right) \\ &= -\frac{i}{2} \omega_{\alpha\beta} {(M^{\alpha\beta})^\mu}_\nu \end{align} where in the second equality the antisymmetry of $\omega_{\alpha\beta}$ has been used, and obviously $$ {(M^{\alpha\beta})^\mu}_\nu = i \left(g^{\mu\alpha} {\delta^\beta}_\nu - g^{\mu\beta} {\delta^\alpha}_\nu \right) $$ Hence $$ {\omega^\mu}_\nu = -\frac{i}{2} \omega_{\alpha\beta} {(M^{\alpha\beta})^\mu}_\nu $$

So the explicit form of the generators ${(M^{\alpha\beta})^\mu}_\nu$ depends on the representation (eg the one we found above), but the commutation relations do not.

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The thing is that by explicit computation, one can find a set of generators $M^{\mu\nu}$ (these are matrices, not matrix elements!) such that $\omega^\mu_{\;\nu}=\frac{1}{2}\omega_{\rho\sigma}(M^{\rho\sigma})^\mu_{\;\nu}$. I admit this is a little miraculous at first sight, but it’s really nothing more than a clever trick.

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