User:Thepigdog/Play

Distance and the moving frames V2

The two reference systems are moving with a relative constant velocity v in a direction e (given by a unit vector). So at, $\mathbf {x} =(0,0,0)$, $\mathbf {x} '=vt'\mathbf {e}$. This equation may be written,

$x'_{\nu }=vt'e_{\nu }$

Note: There is an assumption here that $\mathbf {e}$ and $\mathbf {e} '$ are the same. This assumes that the co-ordinate axis are parallel.

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

(Equation 1) which applies for any $x'_{\mu }$. The general formula may be applied to describe the origin of the co-ordinate systems. Substituting,

$x_{\nu }$ for $0$

$x'_{\nu }$ for $vt'e_{\nu }$

in (equation 1) gives,

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

which simplifies to,

$0=p_{\mu ,0}+v((1-e_{\mu })+\sum _{\nu }p_{\mu ,\nu }e_{\nu })$

and then,

$-{\frac {p_{\mu ,0}}{v}}=(1-e_{\mu })-\sum _{\nu }m_{\mu ,\nu }e_{\nu }$

This suggests a more natural constant for $(\mu ,0)$. Define, $n_{\mu ,0}$ as,

$n_{\mu ,0}=-{\frac {p_{\mu ,0}}{v}}$ (definition 1)

then,

$n_{\mu ,0}=(1-e_{\mu })+\sum _{\nu }p_{\mu ,\nu }e_{\nu }$

Then define $n_{\mu ,\nu }$ as,

$p_{\mu ,\nu }=n_{\mu ,0}n_{\mu ,\nu }$ (definition 2)

then,

$n_{\mu ,0}=(1-e_{\mu })+\sum _{\nu }n_{\mu ,0}n_{\mu ,\nu }e_{\nu }$

$1-{\frac {1-e_{\mu }}{n_{\mu ,0}}}=\sum _{\nu }n_{\mu ,\nu }e_{\nu }$ (equation 3)

Substituting the (definition 1) and (definition2) into (equation 1) gives,

$x_{\mu }=e_{\mu }n_{\mu ,0}(-vt'+\sum _{\nu }n_{\mu ,\nu }x'_{\nu })+(1-e_{\mu })x'_{\mu }$

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,

$1-{\frac {1-e_{\mu }}{n_{\mu ,0}}}=n_{\mu ,\mu }e_{\mu }$

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

$n_{\mu ,\mu }={\frac {1-{\frac {1-e_{\mu }}{n_{\mu ,0}}}}{e_{\mu }}}=1+(1-{\frac {1}{n_{\mu ,0}}})({\frac {1}{e_{\mu }}}-1)$

gives,

$x_{\mu }=e_{\mu }n_{\mu ,0}(-vt'+{\frac {1-{\frac {1-e_{\mu }}{n_{\mu ,0}}}}{e_{\mu }}}x'_{\mu })+(1-e_{\mu })x'_{\mu }$

$x_{\mu }=e_{\mu }n_{\mu ,0}(-vt'+(1+(1-{\frac {1}{n_{\mu ,0}}})({\frac {1}{e_{\mu }}}-1))x'_{\mu })+(1-e_{\mu })x'_{\mu }$

$x_{\mu }=e_{\mu }n_{\mu ,0}(-vt'+x'_{\mu }+(1-{\frac {1}{n_{\mu ,0}}})({\frac {1}{e_{\mu }}}-1)x'_{\mu })+(1-e_{\mu })x'_{\mu }$

$x_{\mu }=e_{\mu }n_{\mu ,0}(-vt'+x'_{\mu })+e_{\mu }n_{\mu ,0}(1-{\frac {1}{n_{\mu ,0}}})({\frac {1}{e_{\mu }}}-1)x'_{\mu }+(1-e_{\mu })x'_{\mu }$

$x_{\mu }=e_{\mu }n_{\mu ,0}(-vt'+x'_{\mu })+(n_{\mu ,0}-1)(1-e_{\mu })x'_{\mu }+(1-e_{\mu })x'_{\mu }$

$x_{\mu }=e_{\mu }n_{\mu ,0}(-vt'+x'_{\mu })+(n_{\mu ,0}-1+1)(1-e_{\mu })x'_{\mu }$

$x_{\mu }=e_{\mu }n_{\mu ,0}(-vt'+x'_{\mu })+n_{\mu ,0}(1-e_{\mu })x'_{\mu }$

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

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

???? same result - seems broken ????