Talk:Shapiro time delay/to do

The beginning formula is wrong.

xp = 760 and xe = 500 produces a delay: dt = 206us;

and for exactly the same transmission but reversed only, produces result: xe = 760 and xp = 500: dt = 215us

This is a nonsense!

Author of this fallacy should correct this immediately, or delete completely this erroneous formula - if he is unable to do math.

additional note:
 * $$\Delta t \approx \frac{4GM}{c^3} \left\{\ln\left[\frac{x_p + (x_p^2 + d^2)^{1/2}}{-x_e + (x_e^2 + d^2)^{1/2}} \right] - \frac{1}{2}\left[\frac{x_p}{(x_p^2 + d^2)^{1/2}} + \frac{2x_e + x_p}{(x_e^2 + d^2)^{1/2}}\right]\right\} + O\left(\frac{G^2M^2}{c^6}\right)$$

hence an approximated form - up to second order is:
 * $$\Delta t\approx\frac{4GM}{c^3}\left\{\ln\left[\frac{x_p + x_p(1 + d^2/2x_p^2)}{-x_e + x_e(1 + d^2/2x_e^2)} \right] - \frac{1}{2}\left[3 + \frac{x_p}{x_e}\right]\right\}=\frac{4GM}{c^3}\left\{\ln\left[\frac{4x_p\cdot x_e}{d^2} \right] - 1/2\left[3 + \frac{x_p}{x_e}\right]\right\}$$

so, this is inconsistent with the next equation. You lost this: :$$-0.5(3 + \frac{x_p}{x_e})$$, what is a quite big number.