Lorentz Transformation Matrices

Express the vector \[\left( \begin{matrix} \begin{matrix} x' \\ y' \end{matrix} \\ \begin{matrix} z' \\ t' \end{matrix} \end{matrix} \right) =\left( \begin{matrix} \begin{matrix} \gamma (x-vt) \\ y \end{matrix} \\ \begin{matrix} z \\ \gamma (t-vx/{ c }^{ 2 }) \end{matrix} \end{matrix} \right)\]

as a 4×44\times4 matrix for scalars γ,v,c\gamma, v, c. Show that the determinant of this matrix is 1. Find the inverse matrix. Verify that the inverse matrix is precisely the Inverse Lorentz Transformation (with no boost in the y and z axes).

Solution

The Lorentz Transformation: x′=γ(x−vt)y′=yz′=zt′=γ(t−vx/c2) \begin{aligned} x' &= \gamma (x-vt) \\ y' &= y \\ z' &= z \\ t' &= \gamma (t - vx / {c}^{2}) \\ \end{aligned}

or

(x′y′z′t′)=(γ00−γv01000010−γv/c200γ)(xyzt) \begin{pmatrix} x' \\ y' \\ z' \\ t' \\ \end{pmatrix} = \begin{pmatrix} \gamma & 0 & 0 & -\gamma v \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\gamma {v}/{c}^{2} & 0 & 0 & \gamma \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ t \\ \end{pmatrix}

Since the determinant is 1, we can just find the adjugate matrix. The adjugate matrix can be found by transposing the cofactor matrix. Hence,

(xyzt)=(γ00γv01000010γv/c200γ)(x′y′z′t′) \begin{pmatrix} x \\ y \\ z \\ t \\ \end{pmatrix} = \begin{pmatrix} \gamma & 0 & 0 & \gamma v \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \gamma {v}/{c}^{2} & 0 & 0 & \gamma \\ \end{pmatrix} \begin{pmatrix} x' \\ y' \\ z' \\ t' \\ \end{pmatrix}

We retrieve the Inverse Lorentz Transformation: x=γ(x′+vt′)y=y′z=z′t=γ(t′+vx′/c2) \begin{aligned} x &= \gamma (x'+vt') \\ y &= y' \\ z &= z' \\ t &= \gamma (t' + vx' / {c}^{2}) \\ \end{aligned}

Check out my other notes at Proof, Disproof, and Derivation

#Algebra #LinearAlgebra #SpecialRelativity #LorentzTransformation

Note by Steven Zheng
6 years, 10 months ago

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