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,
 * $$\mu \ne \nu \implies n_{\mu,\nu} = 0$$

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,
 * $$x_\mu = n_{\mu,0} (-v t' + \frac{x'_\mu}{e_\mu})$$

which can be re-arranged as,
 * $$x_\mu = \frac{n_{\mu,0}}{e_\mu} (x'_\mu - v t' e_\mu)$$ (equation 5a)

and from relativity,
 * $$x'_\mu = \frac{n_{\mu,0}}{e_\mu} (x_\mu + v t e_\mu)$$ (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
 * $$x_\mu = m_{\mu,0} t' + \sum_\nu m_{\mu,\nu} x'_\nu$$
 * $$x_\mu = (p_{\mu,0} e_\mu + q_{\mu,0} (1 - e_\mu)) t' + \sum_\nu (p_{\mu,\nu} e_\mu + q_{\mu,\nu} (1 - e_\mu)) x'_\nu$$

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


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