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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

$$ds^2=-(1-\frac{2m}{r})dt^2=-d\tau^2$$ $$dt=\frac{d\tau_1}{(1-\frac{2m}{R_1})^{1/2}}=\frac{d\tau_2}{(1-\frac{2m}{R_2})^{1/2}}$$ 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

$$\frac{dt}{d\tau}(1-\frac{2m}{r})=constant$$ $$dt=\frac{d\tau_1}{(1-\frac{2m}{R_1})}=\frac{d\tau_2}{(1-\frac{2m}{R_2})}$$

Giving

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

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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.

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  • $\begingroup$ But should not a stationary observer still have the time indipendece? The metric for it (first equation) seems recover the invariance under time-reversal and so the conservation of energy. $\endgroup$ – Pancio Jan 8 at 16:03
  • $\begingroup$ The energy as a conserved quantity is given by the inner product of the time Killing vector and the momentum applied to a geodesic, i.e. to a particle in free fall. A stationary observer is not following a geodesic, so it is not applicable. $\endgroup$ – Michele Grosso Jan 8 at 16:51
  • $\begingroup$ So maybe i misunderstood the text: A static observer at r = R1, θ = π/2, φ = φ1 in Schwarzschild spacetime sends a light signal with wavelength λ towards another static observer at r = R2 > R1, θ = π/2, φ = φ2. Compute the wavelength measured by the second observer (using energy conservation along geodesic motion). It says eplicitly to use energy conservation $\endgroup$ – Pancio Jan 8 at 17:41
  • $\begingroup$ The energy conservation allows to express the energy (frequency) of a free falling photon vs. the energy at infinity (far away from the massive object) in Schwarzschild coordinates. However that is not the energy (frequency) measured by a stationary observer. Reason is the time dilation between Schwarzschild coordinates and a stationary observer. $\endgroup$ – Michele Grosso Jan 9 at 17:23

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