Consider an oscillator at \(r=r_{0}\) emitting a pulse of light (null geodesic) at \(t=t_{0}\). If this is received by an observer at \(r=r_{1}\) at \(t=t_{1}\), show that $$ t_{1}=t_{0}+\int_{r_{0}}^{r_{1}} \frac{d r}{c(I-2 m / r)} $$ By considering a signal emitted at \(t_{0}+\Delta t_{0}\), received at \(t_{1}+\Delta t_{1}\) (assuming the radial positions \(r_{0}\) and \(r_{1}\) to be constant), shou that \(t_{0}=t_{1}\) and the gravitational redsbift found by comparing proper times at cmission and reception is given by $$ 1+z=\frac{\Delta t_{1}}{\Delta \tau_{0}}=\sqrt{\frac{1-2 m / r_{1}}{1-2 m / r_{0}}} $$ Show that for two clocks at different heights \(h\) on the Earth's surface, this reduces to $$ z \approx \frac{2 G M}{c^{2}} \frac{h}{R} $$ where \(M\) and \(R\) are the mass and radius of the Earth.

Step by step solution

01

Establish Basic Equations

Given that light travels in null geodesics in spacetime, under the Schwarzschild metric, the relation between \(dr\) and \(dt\) is given by \(c dt = dr/(1 - 2m/r)\). If we integrate both sides from \(r_0\) to \(r_1\) and from \(t_0\) to \(t_1\) respectively, we can establish that \(t_1 = t_0 + \int_{r_0}^{r_1} dr/(c(1 - 2m/r))\).

02

Calculate Redshift

To show the second equation, consider a pulse of light emitted from \(r_0\) at \(t_0 + dt_0\) and reaches \(r_1\) at \(t_1 + dt_1\). The increase in time at the source and destination can be written as \(dt_0\) and \(dt_1 = dt_0(1 - 2m/r_1)/(1 - 2m/r_0)\) respectively. Comparing these two times gives the gravitational redshift formula as \(1 + z = dt_1/dt_0 = sqrt{(1 - 2m/r_1)/(1 - 2m/r_0)}\).

03

Apply To Earth

We know that on Earth, \(2m/r << 1\), so we can use the linear approximation \((1 - x)^{-1/2} = 1 + x/2\) for small \(x\), which gives the redshift as \(z \approx (2m/r) h2/R\). As \(2GM/c^2 = 2m\), and \(r\) can be replaced with Earth's radius \(R\), we finally get \(z \approx 2GM/c^2 h/R\).

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