Question

(a) Determine the Lorentz transformation matrix giving position and time in framefrom${\mathit{S}}^{\mathbf{\text{'}}}$those in framein the classical limitlocalid="1657533931071" $\mathit{v}\mathbf{<}\mathbf{<}\mathit{c}$. (b) Show that it yields equations (2-1).

Short answer

The representation of the relation of two reference frames moving relative to each other in the form of a matrix is known as the Lorentz transformation matrix. This matrix approximates to Galilean transformation for the classical limit.

Step-by-step solution

Define Lorentz transformation matrix.

The relation between two coordinate systems corresponding to two reference frames that are moving relative to each other can be represented in the form of a matrix known as the Lorentz transformation matrix. It is expressed as,

$\left[\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\\ z^{\text{'}}\\ c{t}^{\text{'}}\end{array}\right]=\left[\begin{array}{cccc}{\gamma }_v& 0& 0& -{\gamma }_v\frac{v}{c}\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -{\gamma }_v\frac{v}{c}& 0& 0& {\gamma }_v\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\\ ct\end{array}\right]$

 Step 2: Express the transformation matrix for the classical limit.

The transformation matrix relation for classical limit (i.e.$\frac{v}{c}<<1\Rightarrow {\gamma }_v\cong 1$,) can be expressed as,

localid="1659092294568" $\left[\begin{array}{c}{x}^{\text{'}}\\ y^{\text{'}}\\ z^{\text{'}}\\ c{t}^{\text{'}}\end{array}\right]=\left[\begin{array}{cccc}1& 0& 0& -\frac{v}{c}\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -\frac{v}{c}& 0& 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\\ ct\end{array}\right]$

The resulting equations that we get,

$\begin{array}{c}{x}^{\text{'}}=x-vt\\ y^{\text{'}}=y\\ z^{\text{'}}=z\\ t^{\text{'}}=t\end{array}$

The last equation for$t^{\text{'}}$, $\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$c$}\right.$is approximated to zero because of the classical limit. The Lorentz transformation matrix thus, results in the Galilean transformation matrix for the classical limit.