Question

Equation \((2-30)\) is an approximation correct only if the gravitational time- dilation effect is small. In this exercise, it is also assumed to be small. but we still allow for a nonuniform gravitational field. We stan with (2-29), based on the Doppler effect in the acceleraing frame. Consider two elevations, the lower at \(r_{1}\) and the upper at \(r_{1}+d r\). Equation \((2-29)\) becomes $$ \frac{f\left(r_{1}+d r\right)}{f\left(r_{i}\right)}=\left(1-\frac{g\left(r_{1}\right) d r}{c^{2}}\right) $$ Similarly, if we consider elevations \(r_{1}+d r\) and \(r_{1}+\) \(2 d r\). we have $$ \frac{f\left(r_{1}+2 d r\right)}{f\left(r_{1}+d r\right)}=\left(1-\frac{g\left(r_{1}+d r\right) d r}{c^{2}}\right) $$ We continue the process, incrementing \(r\) by \(d r\), until we reach \(r_{2}\) $$ \frac{f\left(r_{2}\right)}{f\left(r_{2}-d r\right)}=\left(1-\frac{g\left(r_{2}-d r\right) d r}{c^{2}}\right) $$ Now imagine multiplying the left sides of all the equations and setting the product equal to the product of all the right sides. (a) Argue that the left side of the product is simply \(f\left(r_{2}\right) / f\left(r_{1}\right)\). (b) Assuming that the term \(g d r / c^{2}\) in each individual equation is very small. so that products of such terms can be ignored, argue that the right side of the product is $$ 1-\frac{1}{c^{2}} \int g(r) d r $$ (c) Deduce \(g(r)\) from Newton's universal law of gravitation, then argue that equation \((2-31)\) follows from the result, just as \((2 \cdot 30)\) does from \((2-29)\).

Short answer

The key to solving these problems relies on treating \(gdr/c^2\) as very small so we can use the first order approximation to simplify the expression. Particularly critical was our ability to deduce \(g(r)\) from Newton's law of gravitation and apply it within the integral expression. This exercise demonstrates the fascinating interplay of gravitational fields, the Doppler effect, frequency changes, and acceleration within mathematical and physics problems.

Step-by-step solution

Simplify the Left-side

Starting with the given equations, multiply the left sides together to get the product \(f(r_{2})/f(r_{1})\). This is derived by canceling out the common terms in numerator and denominator in the sequence of fractions as \(f(r_{1}+dr)/f(r_{1})\), \ldots, \(f(r_{2})/f(r_{2}-dr)\). The final product will contain only the first term of the first equation in the numerator and the last term of the last equation in the denominator, which simplifies to \(f(r_{2})/f(r_{1})\).

Simplify the Right-side

For the right sides, since we assume \(gdr/c^2\) is very small, this allows us to use the first order approximation: \(1-x \approx e^{-x}\) for small \(x\). Thus, each \(1-gdr/c^2\) can be approximated as \(e^{-gdr/c^2}\). Multiply these exponential terms together for each increment of \(dr\) from \(r_{1}\) to \(r_{2}\) together gives \(exp(-\int_{r_{1}}^{r_{2}} g(r)dr/c^2)\). By using integral rules, we get \(1-1/c^2 \int g(r) dr\).

Connect to Newton's Gravitational Law

From Newton's universal law of gravitation, we know \(g(r) = GM/r^2\) where \(G\) is the gravitational constant, \(M\) is the object mass which exerts the gravitation, and \(r\) is the distance from the centre of the object. Substitute \(g(r)\) into the integral to obtain Equation (2-31). This validation connects the Doppler effect in an accelerating frame, revealing how it aligns with Newton's gravitation law.

Key concepts

Newton's universal law of gravitation

Newton's universal law of gravitation is a foundational principle in physics that explains how every pair of objects in the universe attracts each other with a force. This force, known as gravitational force, is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This can be mathematically described by the formula: \( F = \frac{G \cdot m_1 \cdot m_2}{r^2} \) Where:

• \( F \) is the gravitational force between two objects,
• \( m_1 \) and \( m_2 \) are the masses of the objects,
• \( r \) is the distance between the centers of the two masses,
• \( G \) is the gravitational constant \( (6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2) \).

This law allows us to calculate the gravitational pull that affects all objects, from apples falling to the ground to the orbits of planets. It's crucial for understanding phenomena from the Earth’s gravity to galactic interactions. In the problem at hand, Newton's law helps us determine the gravitational field \( g(r) \), which is expressed as \( g(r) = \frac{GM}{r^2} \), where \( M \) is the mass exerting the gravitational force.

Doppler effect

The Doppler effect refers to the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source. This is most commonly experienced with sound waves, such as when a passing ambulance sounds different as it approaches than when it recedes.
In the context of the exercise, the Doppler effect is extended to light waves in a gravitational field. When a wave moves in a gravitational field, its frequency changes, much like how sound changes due to movement. This is crucial in an accelerating frame, where changes in frequency are due to differences in gravitational potential.
Gravitational time dilation adds another layer, where time flows differently at different heights in a gravitational field—all stemming from the Doppler effect. By observing these frequency shifts, physicists can understand gravitational fields and motion effects.

Nonuniform gravitational field

A nonuniform gravitational field means that the gravitational force changes over different distances or directions. Unlike a uniform field, where the gravitational pull is the same everywhere, a nonuniform field varies, often noticeably with changes in position.
In space, as you move further from or closer to a large mass, the gravitational pull does not remain constant. It becomes weaker as the distance increases. Earth experiences this slightly, which is why objects weigh a tiny bit less at higher altitudes.
In the problem, as we consider transitions between heights \( r_1 \) and \( r_2 \), this concept helps understand how gravitational effects like force and potential change. Nonuniformity is crucial to account for in calculations, particularly for long-range forces and astronomical measurements.

Approximation in physics

Approximation is an invaluable tool in physics, simplifying complex equations when exact solutions are difficult or impossible to find. Particularly, in problems involving small quantities, approximations make it easier to manage and solve equations.
For instance, when a term is very small, such as \( \frac{g \cdot dr}{c^2} \), physics principles allow us to approximate complex functions using simpler ones. In this exercise, we used the first-order approximation \(1 - x \approx e^{-x}\) for small \( x \). This reduces the complexity substantially, making integrals and other calculations more tractable.
Approximations are everywhere in physics, from calculating orbits to estimating time dilations. They allow scientists to address real-world problems with manageable solutions, pushing forward our understanding of the universe.