Galilean Algebra in the low velocity limit of Poincare Algebra (Weinberg vol 1)

  • #1
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Main Question or Discussion Point

Hi,

Can someone please explain the following statement on page 62 of Weinberg's Vol 1 on QFT:

For a system of particles of typical mass m and typical velocity v, the (..) angular momentum operator is expected to be of order J ~ 1
(I understand the part for P ~ mv, so the "quote" is slightly distorted, intentionally).

Also how is

K of order 1/v
?

Thanks in advance!
 

Answers and Replies

  • #2
haushofer
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Mmm, I would say that looks a little odd. The Galilean algebra can be obtained by a contraction of the Poincare algebra. In this contraction you scale, in the Poincare algebra, the spatial P_i and K_i by a factor epsilon. Then you calculate again the Lie algebra and take the limit epsilon --> 0. This epsilon can be seen as 1/c, and the contraction can be regarded as sending c to infinity.

The J's constitute the SO(3) subgroup of Poincare which stays untouched, and the boosts will commute. I would say that K is of order 1/c, not 1/v.
 
  • #3
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Mmm, I would say that looks a little odd. The Galilean algebra can be obtained by a contraction of the Poincare algebra. In this contraction you scale, in the Poincare algebra, the spatial P_i and K_i by a factor epsilon. Then you calculate again the Lie algebra and take the limit epsilon --> 0. This epsilon can be seen as 1/c, and the contraction can be regarded as sending c to infinity.
Are you referring to the so called Inonu-Wigner contraction here? (I don't know what it is, save a reference in Weinberg on the same page.)

The J's constitute the SO(3) subgroup of Poincare which stays untouched, and the boosts will commute. I would say that K is of order 1/c, not 1/v.
What about J being of order 1? How does that come about?
 
  • #4
haushofer
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Are you referring to the so called Inonu-Wigner contraction here? (I don't know what it is, save a reference in Weinberg on the same page.)
Yes :) Would you ever find the tendency of wanting to know more about it in a pedagogical way: the notes of Robert Gilmore about Lie groups explain it quite well, in chapter 13.

What about J being of order 1? How does that come about?
Well, because in this particular contraction you leave the SO(3) untouched (because SO(3) is a subgroup of both the Poincare as the Galilei group, so you want to keep it), the generators J of SO(3) are not scaled by epsilon. I think that's what Weinberg means, but I'll check his statement.

But I assume you mean by J the [itex]J_i[/itex] which generate the rotations, not the [itex]J_{\mu\nu}[/itex], right?
 
  • #5
haushofer
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Ok, I have checked. My guess is that the 1/v of K is a typo, and should be read as 1/c. That would mean that K is of order [itex]\epsilon^{1}[/itex]. The J's are indeed the generators of SO(3), and because of the particular contraction you take they scale like [itex]\epsilon^{0}[/itex].

Otherwise it wouldn't make sense to me :)
 
  • #6
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Yes :) Would you ever find the tendency of wanting to know more about it in a pedagogical way: the notes of Robert Gilmore about Lie groups explain it quite well, in chapter 13.
Thanks. I'll check out the notes. Right now, I'm unable to reach his website.

But, Weinberg makes it look as if it is possible to get to the Galilean results 'by inspection' (and H = M + W).
 

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