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?
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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