User:Thepigdog/Play3

Assuming that displacements in one direction in one co-ordinate system, result only in displacements in the same direction in the primed system,

$\mu \neq \nu \implies n_{\mu ,\nu }=0$

Then equation 3 gives,

$1=\sum _{\nu }n_{\mu ,\nu }e_{\nu }=n_{\mu ,\mu }e_{\mu }$

Solve for $n_{\mu ,\mu }$ gives,

$n_{\mu ,\mu }={\frac {1}{e_{\mu }}}$

gives,

$x_{\mu }=n_{\mu ,0}(-vt'+{\frac {x'_{\mu }}{e_{\mu }}})$

which can be re-arranged as,

$x_{\mu }={\frac {n_{\mu ,0}}{e_{\mu }}}(x'_{\mu }-vt'e_{\mu })$ (equation 5a)

and from relativity,

$x'_{\mu }={\frac {n_{\mu ,0}}{e_{\mu }}}(x_{\mu }+vte_{\mu })$ (equation 5b)

Note: The divisor term $e_{\mu }$ will be cancelled out later on. There is no divide by zero problem in the final formula. A more careful but less easy to follow process would have avoided it.

No distortion at right angles to relative velocity

$\mathbf {x} \cdot \mathbf {e} =0\implies \mathbf {x} =\mathbf {x} '$

$\sum _{\nu }x_{\nu }e_{\nu }=0\implies x_{\mu }=x'_{\mu }$

From equation 1

$x_{\mu }=m_{\mu ,0}t'+\sum _{\nu }m_{\mu ,\nu }x'_{\nu }$

$x_{\mu }=(p_{\mu ,0}e_{\mu }+q_{\mu ,0}(1-e_{\mu }))t'+\sum _{\nu }(p_{\mu ,\nu }e_{\mu }+q_{\mu ,\nu }(1-e_{\mu }))x'_{\nu }$

$q_{\mu ,0}=0$ $q_{\mu ,\nu }=\delta _{\nu }^{\mu }$

$x_{\mu }=e_{\mu }(p_{\mu ,0}t'+\sum _{\nu }p_{\mu ,\nu }x'_{\nu })+(1-e_{\mu })x'_{\nu }$ (equation 1)