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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.

Qmechanic
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adithya
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  • 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. – adithya Jun 27 '19 at 07:55
  • $\epsilon$ 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. – gented Jun 27 '19 at 07:58
  • @adithya they're the same thing – Avantgarde Jun 27 '19 at 08:08
  • @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. – adithya Jun 27 '19 at 08:54
  • @adithya The generators form a Lie algebra. They are $M_{\rho \sigma}$, not $\epsilon_{\mu \nu}$. – Avantgarde Jun 27 '19 at 10:42
  • @Avantgarde How would they be the same thing? If they are the generators of the Lie algebra then you cannot expand $\Lambda = \delta + generators$, which is what we have in the first form of the equations. – gented Jun 27 '19 at 11:50
  • @gented I never said $\epsilon$ is a generator? – Avantgarde Jun 27 '19 at 13:02
  • @Avantgarde So if they aren't a generator how can they be the "same thing"? The same thing of what (maybe I didn't understand the original comment)? – gented Jun 27 '19 at 13:26
  • @gented I meant saying that $\epsilon$ is an infinitesimal transformation or a parameter is the same thing. – Avantgarde Jun 27 '19 at 13:34
  • @Avantgarde Well, is it, though? A parameter is a number, a transformation is an operator acting onto a manifold :). – gented Jun 27 '19 at 14:11
  • @gented Well, the road to pedantic city is a long, long one and I'm too tired! – Avantgarde Jun 27 '19 at 15:00
  • @Avantgarde Then you shouldn't give incorrect answers to the questions in the first place :) – gented Jun 27 '19 at 18:07
  • @gented It's standard to not be too fussy about self-evident things in physics. – Avantgarde Jun 27 '19 at 18:10
  • @Avantgarde The OP explicitly asked about the subtle difference between the two things and instead of explaining what the difference is you explicitly stated that there is no difference (which there is) after it was pointed out that there is indeed a subtlety; is this how you do physics? – gented Jun 27 '19 at 18:17
  • @gented Yes; like I said, it's self-evident. A related question was asked here:, for example: https://physics.stackexchange.com/q/99906/133418, where you can clearly understand if there are any differences or not, without even talking about manifolds. – Avantgarde Jun 27 '19 at 18:27
  • @Avantgarde I still don't see how it is self-evident: one thing is a matrix, another thing are the entries of the matrix - so not only is it not self-evident (as the link that you posted in fact correctly shows the difference of), it is also wrong to identify the two things (again, the answer in your link does make this difference). – gented Jun 27 '19 at 18:30

2 Answers2

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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.

Rindler98
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