(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Derive the commutation relations for the generators of the Galilean group directly from group multiplication law (without using our results for the Lorentz group). Include the most general set of central charges that cannot be eliminated by redefinition of the group generators.

The general Galilean transformation is given by

[tex]{\mathbf{\bar x}} = {\mathbf{Gx}} + {\mathbf{a}}[/tex]

where [tex]{\mathbf{a}} \in \mathbb{R}^4 [/tex] and

[tex]

{\mathbf{G}} = \left( {\begin{array}{*{20}c}

1 & 0 & 0 & 0 \\

{v_x } & {r_{xx} } & {r_{xy} } & {r_{xz} } \\

{v_y } & {r_{yx} } & {r_{yy} } & {r_{yz} } \\

{v_z } & {r_{zx} } & {r_{zy} } & {r_{zz} } \\

\end{array} } \right)

[/tex] where [tex]\vec v \in \mathbb{R}^3 [/tex] is our velocity of our coordinate system and

[tex]\left( {\begin{array}{*{20}c}

1 & 0 & 0 & 0 \\

0 & {r_{xx} } & {r_{xy} } & {r_{xz} } \\

0 & {r_{yx} } & {r_{yy} } & {r_{yz} } \\

0 & {r_{zx} } & {r_{zy} } & {r_{zz} } \\

\end{array} } \right) \in SO(3,\mathbb{R})

[/tex]

Taking the Unitary representation of this coordinate transformation [tex]{\mathbf{U}}\left( {{\mathbf{G}},{\mathbf{a}}} \right)[/tex] we have a multiplication rule of

[tex]

{\mathbf{U}}\left( {{\mathbf{\bar G}},{\mathbf{\bar a}}} \right){\mathbf{U}}({\mathbf{G}},{\mathbf{a}}) = \exp \left( {i\phi } \right){\mathbf{U}}\left( {{\mathbf{\bar GG}},{\mathbf{\bar Ga}} + {\mathbf{\bar a}}} \right)[/tex] (equation 1) , in our projective space.

[tex]

{\mathbf{U}}\left( {{\mathbf{G}},{\mathbf{a}}} \right) = 1 + i\frac{1}

{2}\omega _{ij} {\mathbf{J}}^{ij} + i\delta v_i {\mathbf{K}}^i + i\varepsilon _i {\mathbf{P}}^i - i\varepsilon _0 {\mathbf{H}} + \cdots

[/tex] (equation 2)

infinitesimally

[tex]

\bar G_\mu ^\alpha G_\beta ^\mu = (\delta _\mu ^\alpha + \bar \omega _\mu ^\alpha )(\delta _\beta ^\mu + \omega _\beta ^\mu )

[/tex]

and

[tex]

\bar G_\beta ^\alpha a^\beta + \bar a^\alpha = \left( {\delta _\beta ^\alpha + \omega _\beta ^\alpha } \right)\varepsilon ^\beta + \bar \varepsilon ^\alpha

[/tex]

using this we get

[tex]

\begin{gathered}

{\mathbf{U}}({\mathbf{\bar GG}},{\mathbf{\bar Ga}} + {\mathbf{\bar a}}) = 1 + i\frac{1}

{2}\left( {\omega _{ij} + \bar \omega _{ij} + \bar \omega _{ir} \delta ^{rs} \omega _{sj} } \right){\mathbf{J}}^{ij} + i\left( {\delta v_i + \delta \bar v_i + \bar \omega _{ir} \delta ^{rs} \delta v_s } \right){\mathbf{K}}^i + \hfill \\

i\left( {\varepsilon _i + \bar \varepsilon _i + \delta \bar v_i \varepsilon _0 + \bar \omega _{ir} \delta ^{rs} \varepsilon _s } \right){\mathbf{P}}^i - i\left( {\varepsilon _0 + \bar \varepsilon _0 } \right){\mathbf{H}} + \cdots \hfill \\

\end{gathered}

[/tex] (equation 3)

[tex]

\phi = f^{ij} \bar \varepsilon _i \varepsilon _j + f^{ij,k} \bar \omega _{ij} \varepsilon _k + f^{ij,rs} \bar \omega _{ij} \omega _{rs} + \cdots

[/tex] (equation 4)

Plugging equations 2, 3, and 4 into equation 1 and working out the commutation relations and applying all the possible Jacobi identities I get:

[tex]

\begin{gathered}

=[{\mathbf{K}}^i ,{\mathbf{H}}] = i(C^i {\mathbf{1}} + {\mathbf{P}}^i ) \hfill \\

[{\mathbf{K}}^i ,{\mathbf{K}}^j ] = [{\mathbf{J}}^{ij} ,{\mathbf{H}}] = [{\mathbf{P}}^i ,{\mathbf{H}}] = [{\mathbf{P}}^i ,{\mathbf{P}}^j ] = 0 \hfill \\

[{\mathbf{J}}^{ij} ,{\mathbf{K}}^k ] = i\left( {\delta ^{ik} (L^j {\mathbf{1}} + {\mathbf{K}}^j ) - \delta ^{jk} (L^i {\mathbf{1}} + {\mathbf{K}}^i )} \right) \hfill \\

[{\mathbf{J}}^{ij} ,{\mathbf{P}}^k ] = i\left( {\delta ^{ik} (C^j {\mathbf{1}} + {\mathbf{P}}^j ) - \delta ^{jk} (C^j {\mathbf{1}} + {\mathbf{P}}^j )} \right) \hfill \\

[{\mathbf{J}}^{ij} ,{\mathbf{J}}^{rs} ] = i\left( {\delta ^{ir} (D^{js} {\mathbf{1}} + {\mathbf{J}}^{js} ) + \delta ^{is} (D^{rj} {\mathbf{1}} + {\mathbf{J}}^{rj} ) - \delta ^{js} (D^{ri} {\mathbf{1}} + {\mathbf{J}}^{ri} ) - \delta ^{jr} (D^{is} {\mathbf{1}} + {\mathbf{J}}^{is} )} \right) \hfill \\

[{\mathbf{K}}^i ,{\mathbf{P}}^j ] = 0 \hfill \\

\end{gathered}

[/tex]

where the stuff next to the [tex]{\mathbf{1}}[/tex]'s are my central charges. Now this is fine and dandy and matches what's on wikipedia for Galilean transformations http://en.wikipedia.org/wiki/Galilean_transformation, but have a problem with commutation relation [tex][{\mathbf{K}}^i ,{\mathbf{P}}^j ] = 0[/tex] it's lacking the central extension [tex]{\mathbf{M}}[/tex], the mass of the system. Now, from what I got, I can redefine my variables to eliminate my central charges. Namely:

[tex]

\begin{gathered}

{\mathbf{\tilde P}}^i \equiv C^i {\mathbf{1}} + {\mathbf{P}}^i \hfill \\

{\mathbf{\tilde K}} \equiv L^i {\mathbf{1}} + {\mathbf{K}}^i \hfill \\

{\mathbf{\tilde J}}^{ij} \equiv D^{ij} {\mathbf{1}} + {\mathbf{J}}^{ij} \hfill \\

\end{gathered}

[/tex]

giving

[tex]

\begin{gathered}

= [{\mathbf{\tilde K}}_i ,{\mathbf{\tilde K}}_j ] = [{\mathbf{\tilde J}}_i ,{\mathbf{H}}] = [{\mathbf{\tilde P}}_i ,{\mathbf{H}}] = [{\mathbf{\tilde P}}_i ,{\mathbf{\tilde P}}_j ] = 0 \hfill \\

[{\mathbf{\tilde J}}_i ,{\mathbf{\tilde J}}_j ] = i\varepsilon _{ijk} {\mathbf{\tilde J}}_k \hfill \\

[{\mathbf{\tilde J}}_i ,{\mathbf{\tilde K}}_j ] = i\varepsilon _{ijk} {\mathbf{\tilde K}}_k \hfill \\

[{\mathbf{\tilde J}}_i ,{\mathbf{\tilde P}}_j ] = i\varepsilon _{ijk} {\mathbf{\tilde P}}_k \hfill \\

[{\mathbf{\tilde K}}_i ,{\mathbf{\tilde P}}_j ] = 0 \hfill \\

\end{gathered}

[/tex]

where

[tex]

{\mathbf{\vec J}} = \{ {\mathbf{J}}^{23} ,{\mathbf{J}}^{31} ,{\mathbf{J}}^{12} \}

[/tex]

My question is where does [tex]{\mathbf{M}}[/tex] arise in the commutation relation [tex][{\mathbf{\tilde K}}_i ,{\mathbf{\tilde P}}_j ][/tex]

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Lie algebras of the Galilean group: Problem three chapter 2 of Wienberg vol. I

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