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Gravitational redshift discrepancy?

Physics Asked by Pancio on August 31, 2020

I want to compute the redshift of a signal emitted by a static observer in $r=R_1$, $phi=phi_1$and recieved by another static observ at $r=R_2$, $phi=phi_2$ with $R_2>R_1$, in Schwarzschild metric.
So i determined it in two different manners obtaing different results.
First i considered the metric for a static observer

$$ds^2=-(1-frac{2m}{r})dt^2=-dtau^2$$
$$dt=frac{dtau_1}{(1-frac{2m}{R_1})^{1/2}}=frac{dtau_2}{(1-frac{2m}{R_2})^{1/2}}$$
So results

$$frac{lambda_2}{lambda}=frac{(1-frac{2m}{R_2})^{1/2}}{(1-frac{2m}{R_1})^{1/2}}$$

Instead using the simmetry under timereversal of the metric we have

$$frac{dt}{dtau}(1-frac{2m}{r})=constant$$
$$dt=frac{dtau_1}{(1-frac{2m}{R_1})}=frac{dtau_2}{(1-frac{2m}{R_2})}$$

Giving

$$frac{lambda_2}{lambda}=frac{(1-frac{2m}{R_2})}{(1-frac{2m}{R_1})}$$
What i’m doing wrong?

One Answer

Stationary observer
The procedure for the gravitational redshift is correct.

Time independence of the metric
The time independence of Schwarzschild allows for the energy as a conserved quantity. However it is defined for a geodesic. It is not applicable to a stationary observer.

Answered by Michele Grosso on August 31, 2020

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