Andrew Magalich edited When_such_observer_measures_the__.tex  almost 8 years ago

Commit id: 7f9d07e29d83a1c93fd4a11110ea1fb9b399a14e

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When such observer with 4-velocity $u^\mu$  measures the energy of the radial lightray, light ray,  it is equivalent to the product $E_i = -\frac{dx_i^\mu}{d\tau} p_\mu -u_i^\mu \frac{dx_\mu}{d\lambda}  = - g_{\mu\nu} \frac{dx_i^\mu}{d\tau} p^\mu$ u_i^\mu \frac{dx^\nu}{d\lambda}$  $$  E_i = \left(1-\frac{2M}{R_i}\right)^{\frac12} \frac{dt}{d_\lambda}  $$  The energy of the photon is defined by a product of its 4-velocity and the time-like Killing vector:  $$  E_\gamma = -K_\mu \frac{dx^\mu}{d\lambda} = \left(1-\frac{2M}{r}\right) \frac{dt}{d\lambda}  $$  Then  $$\frac{dt}{d\lambda} = \left(1-\frac{2M}{r}\right)^{-1} E_\gamma$$  and   $$  E_i = \left(1-\frac{2M}{R_i}\right)^{-\frac12} E_\gamma  $$  Finally,  $$  \frac{\nu_1}{\nu_2} = \sqrt{\frac{1-\frac{2M}{R_2}}{1-\frac{2M}{R_1}}}  $$