Show that the Lorentz boosts, (γ−βγ−βγγ)form a one-parameter Lie group.

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Friegordigh7r7 · Accepted Answer

Step 1Set the components of the matrix equal to one another.On one hand, the product of the two boosts is equal to:[γ1γ2(1+β1β2)−γ1γ2(β1+β2)−γ1γ2(β1+β2)γ1γ2(1+β1β2)]on the other hand, for closure, you want this product to have the form of a boost for some unknown value of&amp;nbsp;β3&amp;nbsp;and&amp;nbsp;γ3(β3):[γ3−γ3β3−γ3β3γ3]Setting these two matrices equal to one another, you find constraints on&amp;nbsp;γ3&amp;nbsp;and&amp;nbsp;β3:γ3=γ1γ2(1+β1β2)γ3=γ1γ2(1+β1β2)−γ3β3=−γ1γ2(β1+β2)⇒β3=β1+β21+β1β2So you can solve for&amp;nbsp;γ3&amp;nbsp;and&amp;nbsp;β3&amp;nbsp;in terms of the original parameters&amp;nbsp;β1,&amp;nbsp;β2,&amp;nbsp;γ1,&amp;nbsp;γ2&amp;nbsp;and show that&amp;nbsp;γ3&amp;nbsp;is indeed the Lorenz factor corresponding to&amp;nbsp;β3, i.e. you can prove that, as required,γ3=11−β32so the product of two boosts is itself a boost corresponding to&amp;nbsp;bata3,&amp;nbsp;γ3

Show that the Lorentz boosts, \(\begin{pmatrix}\gamma & -\beta\gamma\\ -\beta\gamma &

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