How Do Lorentz Transformations Relate to SL(2,ℂ) Boosts?

latentcorpse
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Define B( \theta, \vec{n} ) \in SL( 2 , \mathbb{C} ) by

B( \theta , \vec{n}) = \cosh { \frac{1}{2} \theta} + \vec{\sigma} \cdot \vec{n} \sinh{ \frac{1}{2} \theta} where \vec{n}^2 =1

Show that this corresponds to a Lorentz boost with velocity \vec{v}=\tanh{ \theta} \vec{n}. Show that

( 1 + \frac{1}{2} \vec{sigma} \cdot \delta \vec{v}) B(\theta, \vec{n}) = B( \theta' , \vec{n}' ) R

where, to 1st order in \delta \vec{v},

\theta' = \theta + \delta \vec{v} \cdot \vec{n}, \vec{n}'=\vec{n} \coth{\theta} (\delta \vec{v} - \vec{n} \vec{n} \cdot \delta \vec{v})

and R is an infinitesimal rotation given by

R= 1 + \tanh{\frac{1}{2} \theta} \frac{1}{2}i ( \delta \vec{v} \times \vec{n} ) \cdot \vec{\sigma} = 1 + \frac{\gamma}{\gamma + 1} \frac{1}{2} i ( \delta \vec{v} \times \vec{v} ) \cdot \vec{\sigma}
and \gamma = ( 1 - \vec{v}^2) ^{-\frac{1}{2}}

Show that \vec{v}' = \vec{v} + \delta \vec{v} - \vec{v} \vec{v} \cdot \delta \vec{v}

Note that \vec{\sigma} \cdot \vec{a} \vec{\sigma} \cdot \vec{b} = \vec{a} \cdot \vec{b} 1 + i \vec{\sigma} \cdot ( \vec{a} \times \vec{b} )I don't understand how to go about the first bit here - what do I need to do in order to show this is a Lorentz boost?

Thanks!
 
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latentcorpse said:
I don't understand how to go about the first bit here - what do I need to do in order to show this is a Lorentz boost?

Thanks!

B( \theta, \vec{n} ) is an element of a representation of SL( 2 , \mathbb{C} ). What vector space does the representation act on? Is there a map from Minkowski space to this vector space that you know of?
 


fzero said:
B( \theta, \vec{n} ) is an element of a representation of SL( 2 , \mathbb{C} ). What vector space does the representation act on? Is there a map from Minkowski space to this vector space that you know of?

well SL(2, \mathbb{C}) is just matrices so surely they can act on any vector in \mathbb{R}^2?
 
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