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I want to calculate the gravitational redshift for schwarzchild spacetime. If we take the emitter(source of light) and observer stationary, how can be redshift found?

Frequency seen by observer : $w=-u^ik_i$ u-four velocity, k-four wave vector

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If the emitter and the observer both are stationary, residing at different levels in the same spherically symmetric gravitational field you can calculate the redshift (or blueshift) from the expression:

$\lambda_o=\lambda_e\frac{\sqrt{1-\frac{2GM}{r_oc^2}}}{\sqrt{1-\frac{2GM}{r_ec^2}}}$

here $\lambda_o$ is perceived wavelength by the observer, $\lambda_e$ is the perceived wavelength at the emitter $r_o$ is distance from the center of the gravitational field do the observer and $r_e$ is distance from the center of the gravitational field to the emitter.


In general relativity the energy of an object at rest in a spherically symmetric gravitational field can be written as: $E=mc^2\sqrt{1-\frac{2GM}{rc^2}}$. You can look at the graviational redshift/blueshift as a consequence of this. I guess you can derive the blueshift/redshift formally somehow.

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  • $\begingroup$ thank you for your answer. I also want to see how can I get this expression (derivation). $\endgroup$ – Ali Oz May 28 '19 at 13:22
  • $\begingroup$ You get it from the metric. Note that stationary means dr = 0 in both cases. The time lapse dt then yields the redshift respectively the blueshift. $\endgroup$ – timm May 29 '19 at 8:23

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