User:Thepigdog/General Relativity - 02 - Lorentz transforms

The assumptions are,

• The two axis of the reference frames are parallel.
• The transformation between frames depends only the velocity.
  • The transformation from frame to the primed frame uses a velcoity v.
  • The transformation from primed frame to the frame uses a velcoity -v.
• The speed of light as measured in the two frames is constant.
  • The difference in position divided by the difference in time of a light particle recorded at two positions is always the constant c.
• In the transformation, each component in the reference frame is a linear combination of the components in the primed frame.

The last condition is expressed as,

$x'_{\mu }=\sum _{\nu }m_{\mu ,\nu }x_{\nu }$

where $m_{\mu ,\nu }$ are the components of the transformation matrix.

One approach would be two put this transformation into the Minkowski metric,

$\sum _{\mu }g_{\mu ,\mu }x_{\mu }^{2}=\sum _{\mu }g_{\mu ,\mu }{x'}_{\mu }^{2}$

$\sum _{\mu }g_{\mu ,\mu }x_{\mu }^{2}=\sum _{\mu }g_{\mu ,\mu }(\sum _{\nu }m_{\mu ,\nu }x_{\nu })^{2}$

$=\sum _{\mu }g_{\mu ,\mu }(\sum _{\nu }m_{\mu ,\nu }x_{\nu })(\sum _{\rho }m_{\mu ,\rho }x_{\rho })$

$=\sum _{\mu }\sum _{\nu }\sum _{\rho }g_{\mu ,\mu }m_{\mu ,\nu }x_{\nu }m_{\mu ,\rho }x_{\rho }$

$=\sum _{\nu }\sum _{\rho }x_{\nu }x_{\rho }\sum _{\mu }g_{\mu ,\mu }m_{\mu ,\nu }m_{\mu ,\rho }$

$=\sum _{\nu ,\rho }x_{\nu }x_{\rho }\sum _{\mu }g_{\mu ,\mu }m_{\mu ,\nu }m_{\mu ,\rho }$

$=(\sum _{\nu ,\nu }x_{\nu }^{2}\sum _{\mu }g_{\mu ,\mu }m_{\mu ,\nu }^{2})+(\sum _{\nu ,\rho }^{\nu \neq \rho }x_{\nu }x_{\rho }\sum _{\mu }g_{\mu ,\mu }m_{\mu ,\nu }m_{\mu ,\rho })$

Gives two equations,

$\sum _{\nu }g_{\nu ,\nu }x_{\nu }^{2}=\sum _{\nu ,\nu }x_{\nu }^{2}\sum _{\mu }g_{\mu ,\mu }m_{\mu ,\nu }^{2}$

$0=\sum _{\nu ,\rho }^{\nu \neq \rho }x_{\nu }x_{\rho }\sum _{\mu }g_{\mu ,\mu }m_{\mu ,\nu }m_{\mu ,\rho })$

From the first equation,

$g_{\nu ,\nu }=\sum _{\mu }g_{\mu ,\mu }m_{\mu ,\nu }^{2}$

From the second equation,

$0=\sum _{\nu ,\rho }^{\nu >\rho }x_{\nu }x_{\rho }\sum _{\mu }g_{\mu ,\mu }m_{\mu ,\nu }m_{\mu ,\rho }$

$\nu >\rho \implies 0=\sum _{\mu }g_{\mu ,\mu }m_{\mu ,\nu }m_{\mu ,\rho }$