Problem 25 - A Course in Modern Mathematical Physics: Groups, Hilbert Space and Differential Geometry

Question

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

Answer

The change in time for light to travel from one position \(r_0\) to another \(r_1\) in a gravitational field is given by the integral \(t_1 = t_0 + \int_{r_0}^{r_1} dr/(c(1 - 2m/r))\). The gravitational redshift is found to be \(1 + z = dt_1/dt_0 = sqrt{(1 - 2m/r_1)/(1 - 2m/r_0)}\). Applying this formula to a situation on Earth, the redshift between two clocks at different heights \(h\) on the Earth is approximately \(z \approx 2GM/c^2 h/R\).

Step 1: 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 \(t_1 = t_0 + \int_{r_0}^{r_1} dr/(c(1 - 2m/r))\).

Step 2: 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 \(1 + z = dt_1/dt_0 = sqrt{(1 - 2m/r_1)/(1 - 2m/r_0)}\).

Step 3: 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 \(z \approx 2GM/c^2 h/R\).

Key Concepts

Schwarzschild Metric

The Schwarzschild metric is a solution to Einstein's field equations in general relativity. It describes the gravitational field outside a spherical, non-rotating mass, like a planet or a star.
This metric is crucial in understanding how gravity affects spacetime.
Some key points include:

• The Schwarzschild metric accounts for the way gravity warps the fabric of spacetime, affecting both time and space.
• It provides a mathematical description of how objects move in this curved space.
• The metric is essential in calculating orbits, light paths, and gravitational effects near massive bodies.

In mathematical terms, the metric can be expressed as:\[ ds^2 = - \left(1 - \frac{2m}{r}\right)c^2 dt^2 + \left(1 - \frac{2m}{r}\right)^{-1} dr^2 + r^2 (d\theta^2 + \sin^2 \theta d\phi^2)\]Here, \(m = GM/c^2\), and it captures how distances and time intervals change due to gravity.
This metric is used in the exercise to determine the travel of light beams, which are affected by gravity.

Null Geodesics

In general relativity, a geodesic is the path that an object follows under the influence of gravity alone.For light, this path is called a null geodesic.
A few simple points on null geodesics include:

• They are the paths taken by light in curved spacetime.
• Unlike timelike geodesics (for massive objects), null geodesics have zero spacetime interval.
• These paths are crucial in understanding how light signals travel in a gravitational field.

In the context of the Schwarzschild metric, null geodesics follow the equation:\[c^2 dt^2 = \frac{dr^2}{(1 - 2m/r)}\]This equation shows how light bends around a massive object due to the curvature of spacetime.
It illustrates that light's path isn't straight when viewed in a gravitational field but rather follows a curved trajectory.

Gravitational Time Dilation

Gravitational time dilation occurs because time runs slower in stronger gravitational fields.
It's a phenomenon predicted by general relativity and has several key aspects:

• Time passes differently at different distances from a massive object due to its gravitational pull.
• The closer an object is to the source of gravity, the slower time moves for it relative to an object farther away.
• This effect is not just theoretical; it's been experimentally confirmed using precise clocks.

For two clocks at different heights, the time dilation is expressed as:\[1 + z = \frac{\Delta t_1}{\Delta \tau_0} = \sqrt{\frac{1 - 2m/r_1}{1 - 2m/r_0}}\]This formula shows how time 'stretches' differently at different radial positions because of gravity.
In simpler terms, time runs slower on Earth's surface than it does at a higher altitude.
For small differences in height on Earth, this simplifies to approximately:\[z \approx \frac{2GM}{c^2} \frac{h}{R}\]This is important in understanding satellite communications and GPS technology, where precision timing is crucial.