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

## 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
?

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haushofer
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.

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?

haushofer
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 $J_i$ which generate the rotations, not the $J_{\mu\nu}$, right?

haushofer
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 $\epsilon^{1}$. The J's are indeed the generators of SO(3), and because of the particular contraction you take they scale like $\epsilon^{0}$.