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Ch. 1 Solutions

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Basic

Medium

Advanced

Basic

B-1. Calculating Gravitational Acceleration at Earth's Surface

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Approach: Substitute numerical values into \(|\mathbf{g}| = GM_\oplus/R_\oplus^2\).

\[ |\mathbf{g}| = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{(6.37 \times 10^6)^2} \]

Numerator: \(6.67 \times 10^{-11} \times 5.97 \times 10^{24} = 3.98 \times 10^{14}\)

Denominator: \((6.37 \times 10^6)^2 = 4.06 \times 10^{13}\)

\[ |\mathbf{g}| = \frac{3.98 \times 10^{14}}{4.06 \times 10^{13}} \approx 9.80\ \text{m/s}^2 \]
\[ \boxed{|\mathbf{g}| \approx 9.8\ \text{m/s}^2} \]

This agrees with the gravitational acceleration \(g \approx 9.8\ \text{m/s}^2\) learned in high school physics. We have confirmed that the gravitational acceleration at the Earth's surface is correctly derived from Newton's law of universal gravitation.

Verification: Dimensional check: \([GM/R^2] = \text{m}^3\text{s}^{-2}\text{kg}^{-1} \cdot \text{kg} / \text{m}^2 = \text{m/s}^2\). Correct. ✓

B-2. Differentiation of the \(x\) Component of the Gravitational Potential

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Strategy: Apply the chain rule to \(\Phi = -GM/r\), \(r = \sqrt{x^2 + y^2 + z^2}\).

\[ \frac{\partial \Phi}{\partial x} = \frac{d\Phi}{dr}\cdot\frac{\partial r}{\partial x} \]

First, compute each factor:

\[ \frac{d\Phi}{dr} = \frac{d}{dr}\left(-\frac{GM}{r}\right) = \frac{GM}{r^2} \]
\[ \frac{\partial r}{\partial x} = \frac{\partial}{\partial x}\sqrt{x^2+y^2+z^2} = \frac{x}{r} \]

Therefore

\[ \frac{\partial \Phi}{\partial x} = \frac{GM}{r^2}\cdot\frac{x}{r} = \frac{GMx}{r^3} \]

Thus the \(x\)-component of the gravitational field is

\[ \boxed{g_x = -\frac{\partial \Phi}{\partial x} = -\frac{GMx}{r^3}} \]

Verification: At the point \((r,0,0)\) on the \(x\)-axis, \(g_x = -GM/r^2\), which is correct as an attractive force directed toward the origin.

B-3. Vector Representation of the Gravitational Field

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Computing the \(y\) and \(z\) components in the same manner as Problem B-2. Differentiation of the \(x\) Component of the Gravitational Potential:

\[ g_y = -\frac{\partial \Phi}{\partial y} = -\frac{GMy}{r^3}, \qquad g_z = -\frac{\partial \Phi}{\partial z} = -\frac{GMz}{r^3} \]

Combining these as a vector:

\[ \mathbf{g} = -\nabla\Phi = -\frac{GM}{r^3}(x,\,y,\,z) = -\frac{GM}{r^3}\,\mathbf{r} \]

Using the unit vector \(\hat{\mathbf{r}} = \mathbf{r}/r = (x/r,\,y/r,\,z/r)\), we obtain

\[ \boxed{\mathbf{g} = -\frac{GM}{r^2}\,\hat{\mathbf{r}}} \]

This is consistent with Eq. (1.3).

Verification: \(|\mathbf{g}| = GM/r^2\), and the direction is \(-\hat{\mathbf{r}}\) (attractive force directed toward the origin), which is physically correct.

B-4. Superposition of Two Point Masses

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Since the Poisson equation is linear, by the superposition principle

\[ \boxed{\Phi(\mathbf{r}) = -\frac{GM_1}{|\mathbf{r}|} - \frac{GM_2}{|\mathbf{r}-\mathbf{r}_0|}} \]

The gravitational field is given by \(\mathbf{g} = -\nabla\Phi\) as

\[ \boxed{\mathbf{g}(\mathbf{r}) = -\frac{GM_1}{|\mathbf{r}|^3}\,\mathbf{r} - \frac{GM_2}{|\mathbf{r}-\mathbf{r}_0|^3}\,(\mathbf{r}-\mathbf{r}_0)} \]

Verification: Setting \(M_2 = 0\) reduces to the case of a single point mass. ✓ Also, as \(\mathbf{r} \to \infty\), we have \(\Phi \to -G(M_1+M_2)/r\), which shows that at large distances the behavior is the same as if the total mass were concentrated at a single point. ✓

B-5. Calculating \(\nabla^2(r^n)\)

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Approach: Apply the radial part of the Laplacian to the spherically symmetric function \(f(r) = r^n\).

\[ \nabla^2(r^n) = \frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d(r^n)}{dr}\right) \]

Step 1: First derivative

\[ \frac{d(r^n)}{dr} = nr^{n-1} \]

Step 2: Multiply by \(r^2\)

\[ r^2 \cdot nr^{n-1} = nr^{n+1} \]

Step 3: Differentiate with respect to \(r\) once more

\[ \frac{d}{dr}(nr^{n+1}) = n(n+1)r^n \]

Step 4: Multiply by \(1/r^2\)

\[ \nabla^2(r^n) = \frac{n(n+1)r^n}{r^2} = n(n+1)\,r^{n-2} \]
\[ \boxed{\nabla^2(r^n) = n(n+1)\,r^{n-2}} \]

Summary for various values of \(n\):

\(n\) \(\nabla^2(r^n)\)
\(0\) \(0\)
\(1\) \(2r^{-1}\)
\(2\) \(6\)
\(-1\) \(0\)
\(-2\) \(6r^{-4}\)

Verification: For \(n=2\), we have \(f = r^2 = x^2+y^2+z^2\), so \(\nabla^2 f = 2+2+2 = 6\). This agrees. ✓

B-6. Laplace Equation Outside a Point Mass

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Since \(\Phi = -GM\,r^{-1}\), setting \(n = -1\) in the result from Problem B-5. Calculating \(\nabla^2(r^n)\) gives

\[ \nabla^2(r^{-1}) = (-1)(-1+1)\,r^{-1-2} = (-1)(0)\,r^{-3} = 0 \]

Therefore

\[ \boxed{\nabla^2\Phi = -GM\cdot\nabla^2(r^{-1}) = 0 \quad (r \neq 0)} \]

Consistency with the Poisson equation:

The Poisson equation is \(\nabla^2\Phi = 4\pi G\rho\). The mass density of a point mass is

\[ \rho(\mathbf{r}) = M\,\delta^3(\mathbf{r}) \]

For \(r \neq 0\), \(\delta^3(\mathbf{r}) = 0\) so \(\rho = 0\), and \(\nabla^2\Phi = 0\) is consistent with the Poisson equation. At \(r = 0\), there is a singularity due to the delta function, and the equality \(\nabla^2(1/r) = -4\pi\delta^3(\mathbf{r})\) holds in the distributional sense.

B-7. Potential Constant Inside a Uniform Density Sphere

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Substitute \(\Phi(r) = Ar^2 + B\) into the Poisson equation \(\nabla^2\Phi = 4\pi G\rho_0\).

From the result of Problem B-5. Calculating \(\nabla^2(r^n)\), \(\nabla^2(r^2) = 6\) and \(\nabla^2(B) = 0\), so

\[ \nabla^2\Phi = A\cdot 6 + 0 = 6A \]

From the Poisson equation:

\[ 6A = 4\pi G\rho_0 \]
\[ \boxed{A = \frac{2\pi G\rho_0}{3}} \]

Verification: The dimensions of \(A\) are \([G][\rho_0] = (\mathrm{m^3\,kg^{-1}\,s^{-2}})(\mathrm{kg\,m^{-3}}) = \mathrm{s^{-2}}\). The dimensions of \(\Phi = Ar^2\) are \(\mathrm{s^{-2}\cdot m^2} = \mathrm{m^2\,s^{-2}}\) (dimensions of potential), which is correct. ✓

B-8. Divergence of the Gravitational Field and Poisson's Equation

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\[ \nabla\cdot\mathbf{g} = \nabla\cdot(-\nabla\Phi) = -\nabla^2\Phi \]

Substituting Poisson's equation \(\nabla^2\Phi = 4\pi G\rho\):

\[ \boxed{\nabla\cdot\mathbf{g} = -4\pi G\rho} \]

This is the differential form of Gauss's law for the gravitational field.

Verification: Where mass is present (\(\rho > 0\)), we have \(\nabla\cdot\mathbf{g} < 0\), meaning the gravitational field "converges" toward the mass. This is physically correct. ✓

B-9. Contradiction Between Instantaneous Propagation and Special Relativity

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Solution Strategy: Calculate the light travel time between the Sun and Earth, and compare with Newton's instantaneous propagation.

Calculation

The mean distance between the Sun and Earth is \(d = 1\;\text{AU} \approx 1.50 \times 10^{11}\;\text{m}\). The speed of light is \(c = 3.00 \times 10^8\;\text{m/s}\).

The time for light to travel from the Sun to Earth is:

\[ \Delta t = \frac{d}{c} = \frac{1.50 \times 10^{11}}{3.00 \times 10^8} = 500\;\text{s} \approx 8.3\;\text{min} \]
\[ \boxed{\Delta t \approx 500\;\text{s} \approx 8\;\text{min}\;20\;\text{s}} \]

Contradiction Between Newton's Model and Special Relativity

In Newton's gravitational model, gravity obeys the Poisson equation \(\nabla^2 \Phi = 4\pi G\rho\), and changes in the source \(\rho\) are reflected instantaneously in the potential \(\Phi\) (since there is no time-derivative term). Therefore, if the Sun were to suddenly disappear, the gravitational force felt by Earth would become zero at that very instant, and Earth would immediately begin moving in a straight line (uniform motion in the tangential direction).

On the other hand, according to special relativity, no physical information or influence can propagate faster than the speed of light \(c\). The information that the Sun has disappeared should take at least \(\Delta t \approx 500\;\text{s}\) (approximately 8 minutes and 20 seconds) to reach Earth.

This contradiction manifests concretely as follows:

  • Newton's model: Sun disappears at \(t = 0\) → Earth's orbit changes at \(t = 0\) (zero delay)
  • A theory consistent with special relativity: Sun disappears at \(t = 0\) → Earth continues its normal orbit until \(t \approx 500\;\text{s}\) → orbit changes only after \(t \approx 500\;\text{s}\)

Newton's gravitational model involves instantaneous action at a distance, which fundamentally contradicts the causality of special relativity (information propagation speed \(\leq c\)). To resolve this contradiction, a theory in which gravitational changes propagate at a finite speed (\(c\)) is required — namely, general relativity.

Verification

Dimensional check: \([d/c] = \text{m}/(\text{m/s}) = \text{s}\). Correct. ✓

Reasonableness of the result: The fact that sunlight takes approximately 8 minutes to reach Earth is a widely known basic fact in astronomy, consistent with our calculated result. ✓

B-10. Estimating Relativistic Effects at the Solar Surface

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Solution strategy: Substitute numerical values into \(GM_\odot/(R_\odot c^2)\) and evaluate the magnitude of relativistic effects.

Calculation

Given values:

  • \(G = 6.67 \times 10^{-11}\;\text{N}\cdot\text{m}^2/\text{kg}^2\)
  • \(M_\odot = 1.99 \times 10^{30}\;\text{kg}\)
  • \(R_\odot = 6.96 \times 10^{8}\;\text{m}\)
  • \(c = 3.00 \times 10^{8}\;\text{m/s}\)

Numerator:

\[ GM_\odot = 6.67 \times 10^{-11} \times 1.99 \times 10^{30} = 1.327 \times 10^{20}\;\text{m}^3/\text{s}^2 \]

Denominator:

\[ R_\odot c^2 = 6.96 \times 10^{8} \times (3.00 \times 10^{8})^2 = 6.96 \times 10^{8} \times 9.00 \times 10^{16} = 6.264 \times 10^{25}\;\text{m}^3/\text{s}^2 \]

Ratio:

\[ \frac{GM_\odot}{R_\odot c^2} = \frac{1.327 \times 10^{20}}{6.264 \times 10^{25}} = 2.12 \times 10^{-6} \]
\[ \boxed{\frac{GM_\odot}{R_\odot c^2} \approx 2.1 \times 10^{-6}} \]

Estimating the Magnitude of Relativistic Effects

This dimensionless quantity \(GM_\odot/(R_\odot c^2) \sim 10^{-6}\) serves as an indicator of the relative magnitude of general relativistic effects near the solar surface.

  1. Gravitational redshift: The frequency of light emitted from the solar surface is redshifted by \(\Delta\nu/\nu \sim GM_\odot/(R_\odot c^2) \sim 10^{-6}\) when observed at infinity. This is a magnitude detectable with high-resolution spectroscopic observations.

  2. Deflection of light: Light passing near the Sun is deflected by general relativity on the order of \(\delta\theta \sim GM_\odot/(R_\odot c^2) \sim 10^{-6}\;\text{rad} \approx 0.4''\) (the actual value is \(4GM_\odot/(R_\odot c^2) \approx 1.75''\)). This was confirmed during the 1919 solar eclipse observation.

  3. Perihelion precession of Mercury: Corrections of order \(GM_\odot/(ac^2) \sim 10^{-8}\) (where \(a\) is Mercury's orbital semi-major axis) accumulate with each orbit, resulting in a discrepancy of approximately 43 arcseconds per century.

The value of \(10^{-6}\) means that Newtonian mechanics is an extremely good approximation at everyday scales, but general relativistic corrections are detectable—and have indeed been detected—in precision astronomical observations and high-accuracy experiments.

Verification

Dimensional check: $[GM/(Rc^2)] = (\text{m}^3\text{s}^{-2}\text{kg}^{-1} \cdot \text{kg})/(\text{m} \cdot \text{m}^2\text{s}^{-2}) = $ dimensionless. Correct. ✓

Consistent with the value \(GM_\odot/(R_\odot c^2) \sim 10^{-6}\) for the Sun given in the prologue table. ✓

B-11. Estimating the Relativistic Parameter of a Neutron Star

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Given values:

  • \(M = 1.4\,M_\odot\)
  • \(R = 10\;\mathrm{km}\)
  • \(GM_\odot/c^2 \approx 1.48\;\mathrm{km}\)
\[ \frac{GM}{Rc^2} = \frac{G\cdot 1.4\,M_\odot}{R\,c^2} = \frac{1.4\,GM_\odot/c^2}{R} = \frac{1.4\times 1.48\;\mathrm{km}}{10\;\mathrm{km}} \]
\[ = \frac{2.07\;\mathrm{km}}{10\;\mathrm{km}} \approx 0.21 \]
\[ \boxed{\frac{GM}{Rc^2} \sim 10^{-1} \approx 0.2} \]

This value is close to 1, meaning that near the surface of a neutron star, Newtonian gravity receives significant corrections, and general relativistic effects cannot be neglected.

Consistency check: For comparison, in the case of the Sun, \(GM_\odot/(R_\odot c^2) \approx 1.48\;\mathrm{km}/7\times10^5\;\mathrm{km} \sim 10^{-6}\), which is consistent with Newtonian gravity being a sufficiently good approximation. ✓

Medium

M-1. Derivation of Poisson's Equation from Gauss's Law

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Starting point: Gauss's law for the gravitational field (integral form)

\[ \oint_S \mathbf{g}\cdot d\mathbf{A} = -4\pi G\,M_{\mathrm{enc}} \tag{i} \]

Step 1: Apply the divergence theorem to the left-hand side.

\[ \oint_S \mathbf{g}\cdot d\mathbf{A} = \int_V \nabla\cdot\mathbf{g}\;dV \tag{ii} \]

Step 2: Express \(M_{\mathrm{enc}}\) on the right-hand side as an integral over density.

\[ M_{\mathrm{enc}} = \int_V \rho\;dV \tag{iii} \]

Step 3: Substitute (ii) and (iii) into (i).

\[ \int_V \nabla\cdot\mathbf{g}\;dV = -4\pi G\int_V \rho\;dV \]
\[ \int_V \left[\nabla\cdot\mathbf{g} + 4\pi G\rho\right]dV = 0 \tag{iv} \]

Step 4: Equation (iv) holds for arbitrary volumes \(V\). If the integrand is continuous, this implies that the integrand itself is zero:

\[ \nabla\cdot\mathbf{g} + 4\pi G\rho = 0 \]
\[ \nabla\cdot\mathbf{g} = -4\pi G\rho \tag{v} \]

Step 5: Substitute \(\mathbf{g} = -\nabla\Phi\).

\[ \nabla\cdot(-\nabla\Phi) = -4\pi G\rho \]
\[ -\nabla^2\Phi = -4\pi G\rho \]
\[ \boxed{\nabla^2\Phi = 4\pi G\rho} \]

This is Poisson's equation.

Verification: When \(\rho = 0\), it reduces to Laplace's equation \(\nabla^2\Phi = 0\). For a point mass \(\rho = M\delta^3(\mathbf{r})\), it has been confirmed in Problem B-6. Laplace Equation Outside a Point Mass that the solution \(\Phi = -GM/r\) is obtained. ✓

M-2. Complete Potential Solution for a Uniform Density Sphere

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Consider a sphere of radius \(R\), uniform density \(\rho_0\), and total mass \(M = \frac{4}{3}\pi R^3\rho_0\).

(a) Exterior (\(r > R\))

In the exterior region \(\rho = 0\), so the spherically symmetric Poisson equation becomes the Laplace equation:

\[ \frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d\Phi}{dr}\right) = 0 \]

Integrating gives \(r^2\,d\Phi/dr = C_1\) (constant), so

\[ \frac{d\Phi}{dr} = \frac{C_1}{r^2} \]

Integrating once more:

\[ \Phi_{\mathrm{out}}(r) = -\frac{C_1}{r} + C_2 \]

Boundary condition: \(\Phi\to 0\) as \(r\to\infty\) gives \(C_2 = 0\).

By the shell theorem (or Gauss's law), the exterior potential is the same as that of the total mass \(M\) concentrated at the center, so \(C_1 = GM\).

\[ \boxed{\Phi_{\mathrm{out}}(r) = -\frac{GM}{r}} \]

(b) Interior (\(r < R\))

In the interior \(\rho = \rho_0\), so the Poisson equation is

\[ \frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d\Phi}{dr}\right) = 4\pi G\rho_0 \]

From the result of Problem B-7. Potential Constant Inside a Uniform Density Sphere, the particular solution is \(\Phi_p = Ar^2\) with \(A = \frac{2\pi G\rho_0}{3}\).

The spherically symmetric solution to the homogeneous equation \(\nabla^2\Phi = 0\) is \(\Phi_h = \alpha/r + \beta\). Regularity (finiteness) at \(r = 0\) requires \(\alpha = 0\).

Therefore the general solution is

\[ \Phi_{\mathrm{in}}(r) = \frac{2\pi G\rho_0}{3}\,r^2 + \beta \]

Boundary condition 1: Continuity of the potential at \(r = R\)

\[ \frac{2\pi G\rho_0}{3}R^2 + \beta = -\frac{GM}{R} \]

Using \(M = \frac{4}{3}\pi R^3\rho_0\) gives \(\frac{GM}{R} = \frac{4\pi G\rho_0 R^2}{3}\), so

\[ \frac{2\pi G\rho_0}{3}R^2 + \beta = -\frac{4\pi G\rho_0}{3}R^2 \]
\[ \beta = -\frac{4\pi G\rho_0}{3}R^2 - \frac{2\pi G\rho_0}{3}R^2 = -2\pi G\rho_0 R^2 \]

Boundary condition 2: Continuity of \(d\Phi/dr\) at \(r = R\)

\[ \left.\frac{d\Phi_{\mathrm{in}}}{dr}\right|_{r=R} = \frac{4\pi G\rho_0}{3}R, \qquad \left.\frac{d\Phi_{\mathrm{out}}}{dr}\right|_{r=R} = \frac{GM}{R^2} = \frac{4\pi G\rho_0}{3}R \]

Both sides agree. ✓ (Since \(\beta\) does not affect the derivative, boundary condition 1 alone determines \(\beta\), and condition 2 is automatically satisfied.)

Substituting \(\rho_0 = \frac{3M}{4\pi R^3}\) to rewrite \(\beta\) in terms of \(M\) and \(R\):

\[ \beta = -2\pi G\cdot\frac{3M}{4\pi R^3}\cdot R^2 = -\frac{3GM}{2R} \]

Similarly \(A = \frac{2\pi G}{3}\cdot\frac{3M}{4\pi R^3} = \frac{GM}{2R^3}\)

\[ \boxed{\Phi_{\mathrm{in}}(r) = \frac{GM}{2R^3}\,r^2 - \frac{3GM}{2R} = -\frac{GM}{2R}\left(3 - \frac{r^2}{R^2}\right)} \]

(c) Comparison of values at \(r = 0\) and \(r = R\)

\[ \Phi(0) = -\frac{3GM}{2R} \]
\[ \Phi(R) = -\frac{GM}{2R}(3-1) = -\frac{GM}{R} \]
\[ \boxed{\Phi(0) = -\frac{3GM}{2R}, \qquad \Phi(R) = -\frac{GM}{R}} \]

The potential at the center is \(3/2\) times deeper (in absolute value) than its value at the surface:

\[ \frac{\Phi(0)}{\Phi(R)} = \frac{3}{2} \]

Verification: At \(r = R\), \(\Phi_{\mathrm{in}}(R) = \frac{GM}{2R^3}R^2 - \frac{3GM}{2R} = \frac{GM}{2R} - \frac{3GM}{2R} = -\frac{GM}{R} = \Phi_{\mathrm{out}}(R)\). ✓

M-3. Scale Estimation of Mercury's Perihelion Precession

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Calculation of the dimensionless quantity:

\[ \frac{GM_\odot}{ac^2} = \frac{1.48\;\mathrm{km}}{5.79\times10^{7}\;\mathrm{km}} = 2.56\times10^{-8} \]
\[ \boxed{\frac{GM_\odot}{ac^2} \approx 2.6\times10^{-8}} \]

Dimensional analysis argument:

The general relativistic correction is expected to appear as a relative deviation from Newtonian gravity at the order of the dimensionless quantity \(GM_\odot/(ac^2)\). This sets the scale for the perihelion precession angle (in radians) per orbit.

Let us estimate the number of orbits in 100 years. Mercury's orbital period is approximately 88 days, so

\[ N = \frac{100\;\mathrm{yr}}{88\;\mathrm{day}} = \frac{100\times365.25}{88} \approx 415\;\text{orbits} \]

The order of the cumulative deviation over 100 years is

\[ \Delta\varphi_{100} \sim N\cdot\frac{GM_\odot}{ac^2} \sim 415\times2.6\times10^{-8} \approx 1.1\times10^{-5}\;\mathrm{rad} \]

On the other hand, converting the observed value of 43 arcseconds to radians gives

\[ 43'' = 43\times\frac{\pi}{180\times3600}\;\mathrm{rad} \approx 2.1\times10^{-4}\;\mathrm{rad} \]

Comparing the two:

\[ \frac{\text{observed value}}{\text{estimate}} = \frac{2.1\times10^{-4}}{1.1\times10^{-5}} \approx 19 \approx 6\pi \]

In other words, the dimensional analysis estimate \(\sim GM_\odot/(ac^2)\) agrees with the observed value to within a numerical factor of \(6\pi \approx 19\). Since dimensional analysis cannot determine \(O(1)\) numerical factors (such as \(2\pi\) or \(6\pi\)), this can be said to correctly correspond in order of magnitude.

In fact, the exact general relativistic calculation (derived in Ch. 8) gives the perihelion precession per orbit as

\[ \delta\varphi = \frac{6\pi GM_\odot}{a(1-e^2)c^2} \]

With \(e \approx 0.206\), we have \(1/(1-e^2) \approx 1.04\), and

\[ \Delta\varphi_{100} = 415\times\frac{6\pi\times2.56\times10^{-8}}{0.958} = 415\times5.03\times10^{-7} = 2.09\times10^{-4}\;\mathrm{rad} \approx 43'' \]

This agrees precisely with the observed value.

\[ \boxed{\text{The deviation from the Newtonian model is due to a relativistic correction of order } \frac{GM_\odot}{ac^2} \sim 10^{-8},} \]
\[ \boxed{\text{which accumulates over 100 years to } \sim 10^{-5}\text{–}10^{-4}\;\mathrm{rad}, \text{ consistent with the observed } 43''.} \]

M-4. Comparison of the Wave Equation and Poisson's Equation

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Reduction to the electrostatic field:

When the source \(\rho_e\) does not vary in time, \(\varphi\) is also independent of time, so \(\partial^2\varphi/\partial t^2 = 0\). The wave equation becomes

\[ \left(\nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\varphi = -\frac{\rho_e}{\varepsilon_0} \]

\(\Downarrow \quad \partial/\partial t = 0\)

\[ \boxed{\nabla^2\varphi = -\frac{\rho_e}{\varepsilon_0}} \]

This is the Poisson equation for the electrostatic field.

Summary of structural similarities and differences:

Poisson equation for Newtonian gravity Poisson equation for the electrostatic field
Equation \(\nabla^2\Phi = 4\pi G\rho\) \(\nabla^2\varphi = -\rho_e/\varepsilon_0\)
Source Mass density \(\rho\) (always positive) Charge density \(\rho_e\) (positive or negative)
Sign of force Always attractive Both attractive and repulsive
Coupling constant \(4\pi G\) (positive) \(-1/\varepsilon_0\) (negative)
Point source solution \(\Phi = -GM/r\) \(\varphi = q/(4\pi\varepsilon_0 r)\)

Similarities:

  • Both have the same mathematical structure \(\nabla^2(\cdot) = (\text{source})\) (elliptic partial differential equation)
  • Both yield a \(1/r\)-type potential for a point source (inverse-square law)
  • The superposition principle holds (linear equations)
  • Neither contains time derivatives, so changes in the source propagate instantaneously

Differences:

  • The signs are opposite: gravity is always attractive, while the electrostatic force is repulsive for like charges and attractive for unlike charges
  • The Poisson equation for the electrostatic field is obtained as the static limit of a more fundamental wave equation. In contrast, no corresponding "parent" wave equation exists for Newton's Poisson equation within Newtonian theory (general relativity is required)
  • In electromagnetism, the equation extends to a wave equation in the dynamic case, and electromagnetic waves propagate at the speed of light. Newtonian gravity has no such natural extension

Advanced

A-1. An Attempt at a Scalar Theory of Gravity

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(a) Derivation of the Dispersion Relation

Setting \(c_g = c\) and substituting the plane wave solution \(\Phi = \Phi_0\,e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}\) into the source-free (\(\rho = 0\)) equation

\[ \nabla^2\Phi - \frac{1}{c^2}\frac{\partial^2\Phi}{\partial t^2} = 0 \]

Computing each derivative:

\[ \nabla^2\Phi = -|\mathbf{k}|^2\,\Phi, \qquad \frac{\partial^2\Phi}{\partial t^2} = -\omega^2\,\Phi \]

Substituting gives

\[ -|\mathbf{k}|^2\,\Phi + \frac{\omega^2}{c^2}\,\Phi = 0 \]

Since \(\Phi \neq 0\):

\[ \boxed{\omega^2 = c^2|\mathbf{k}|^2 \quad \Longleftrightarrow \quad \omega = c|\mathbf{k}|} \]

This is the dispersion relation for a massless field (identical to that of the photon), with both phase velocity and group velocity equal to \(c\). Changes in gravity propagate at the speed of light.

Verification: This has the same form as the electromagnetic wave dispersion relation \(\omega = c|\mathbf{k}|\), which is expected since the wave equations have the same structure. ✓

(b) Limitations of the Scalar Theory of Gravity

Although this modification resolves the problem of instantaneous propagation, a theory describing gravity with only a scalar potential \(\Phi\) has the following serious problems.

1. The Source Problem:

In special relativity, energy and momentum are unified as the 4-momentum \(p^\mu = (E/c,\,\mathbf{p})\). Furthermore, for continuous matter, the source is described not only by energy density but by the energy-momentum tensor \(T^{\mu\nu}\) (a rank-2 symmetric tensor with 10 independent components), which includes momentum density, pressure, and stress.

In the Newtonian equation \((\ast)\), the source is only the mass density \(\rho\) (a scalar quantity). In special relativity, \(\rho c^2\) corresponds to the energy density and is merely the \(T^{00}\) component. The remaining components of \(T^{\mu\nu}\) (momentum flux, pressure, shear stress) would not contribute as gravitational sources, which is inconsistent under Lorentz transformations.

2. The Field Degrees of Freedom Problem:

In electromagnetism, the field is described by a vector potential \(A^\mu\) (4 components), which is consistent with the source current density \(J^\mu\) (4 components). In general relativity, the field is described by the metric tensor \(g_{\mu\nu}\) (a symmetric rank-2 tensor with 10 independent components), consistent with the source \(T^{\mu\nu}\) (10 components). A scalar \(\Phi\) (1 component) cannot adequately receive the information contained in \(T^{\mu\nu}\).

3. Specific Physical Consequences:

  • A scalar theory cannot correctly predict the deflection of light (it yields only half the prediction of general relativity)
  • Since pressure does not act as a gravitational source, the description of neutron star structure and cosmological expansion becomes inaccurate
  • Gravitational wave polarization modes cannot be correctly described (scalar waves have only a scalar mode, whereas general relativity predicts tensor modes)
\[ \boxed{\text{A scalar field } \Phi \text{ alone cannot incorporate all components of } T^{\mu\nu} \text{ as sources, making it impossible to construct a Lorentz-invariant theory of gravity}} \]

(c) The \(c_g \to \infty\) Limit

Taking \(c_g \to \infty\) in equation \((\ast)\):

\[ \frac{1}{c_g^2}\frac{\partial^2\Phi}{\partial t^2} \to 0 \]

Therefore

\[ \nabla^2\Phi - \underbrace{\frac{1}{c_g^2}\frac{\partial^2\Phi}{\partial t^2}}_{\to\,0} = 4\pi G\rho \quad \Longrightarrow \quad \nabla^2\Phi = 4\pi G\rho \]
\[ \boxed{c_g \to \infty \text{ reduces to the Poisson equation}} \]

This is consistent with the assertion that "Newtonian gravity is the limit in which the propagation speed of gravity is infinite, i.e., the \(c \to \infty\) approximation." In situations where special relativistic effects are negligible (\(v \ll c\), \(GM/(Rc^2) \ll 1\)), \(c\) can be effectively treated as infinite, so the Poisson equation serves as a good approximation.

Verification: Confirmed by dimensional analysis. The dimensions of each term in equation \((\ast)\) are \([\nabla^2\Phi] = \mathrm{m^2\,s^{-2}/m^2} = \mathrm{s^{-2}}\) and \([\partial^2\Phi/(c_g^2\partial t^2)] = \mathrm{m^2\,s^{-2}/(m^2\,s^{-2}\cdot s^2)} = \mathrm{s^{-2}}\), which are consistent. ✓

A-2. Shell Theorem and Tidal Force

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Derivation That the Potential Is Constant Inside a Spherical Shell

Consider a spherical shell of uniform density (inner radius \(R_1\), outer radius \(R_2\)). In the inner cavity (\(r < R_1\)), since \(\rho = 0\),

\[ \nabla^2\Phi = 0 \]

Assuming spherical symmetry, the Laplace equation in the radial direction only is

\[ \frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d\Phi}{dr}\right) = 0 \]

The general solution is

\[ \Phi(r) = -\frac{\alpha}{r} + \beta \]

Boundary condition: For the potential to be regular (finite) at \(r = 0\), we require \(\alpha = 0\).

\[ \boxed{\Phi(r) = \beta = \text{const} \quad (r < R_1)} \]

The value of the constant \(\beta\) is determined by the matching condition at the inner surface of the shell \(r = R_1\), but within the cavity the potential is constant.

(Specifically, in the exterior of the shell (\(r > R_2\)) \(\Phi = -GM_{\mathrm{shell}}/r\), one solves the Poisson equation inside the shell (\(R_1 < r < R_2\)), and uses the matching conditions at \(r = R_1\) and \(r = R_2\) to determine \(\beta\). The result is not simply \(\beta = -GM_{\mathrm{shell}}/R_1\), but rather a value that depends on the mass distribution of the shell.)

(a) Gravitational Force at an Arbitrary Position Inside the Spherical Shell

Since the potential is constant inside the cavity,

\[ \mathbf{g} = -\nabla\Phi = \mathbf{0} \]

Therefore, even if a mass \(m\) is placed at a position \(\mathbf{r}_0\) slightly displaced from the center of the shell, the gravitational force on the object is zero.

\[ \boxed{\mathbf{F} = m\mathbf{g} = \mathbf{0} \quad \text{(at any position within the cavity)}} \]

This is a consequence of Newton's shell theorem. Physically, from a displaced position, the nearer side of the shell is closer but subtends a smaller solid angle, while the farther side is more distant but subtends a larger solid angle. The \(1/r^2\) law and the solid angle effect cancel exactly, resulting in zero net force.

(b) Gravitational Field and Tidal Force Inside a Cavity When the Shell Is Deformed into an Ellipsoid

When the shell is slightly deformed from perfect spherical symmetry into an ellipsoid, the premise of the shell theorem (spherical symmetry) breaks down.

Potential inside the cavity:

When the deformation is small, the potential can be decomposed into a spherically symmetric part and a perturbation:

\[ \Phi(\mathbf{r}) = \Phi_0 + \delta\Phi(\mathbf{r}) \]

where \(\Phi_0\) is a constant (the spherically symmetric part) and \(\delta\Phi(\mathbf{r})\) is the correction due to the deformation, which is spatially non-uniform. For an ellipsoidal deformation, \(\delta\Phi\) typically contains quadratic terms proportional to \(r^2\) (in the form of spherical harmonics \(Y_2^m\)). Specifically, the solution inside the cavity that satisfies \(\nabla^2(\delta\Phi) = 0\) and is regular at \(r = 0\) takes the form

\[ \delta\Phi(\mathbf{r}) = \sum_{\ell,m} C_{\ell m}\,r^\ell\,Y_\ell^m(\theta,\varphi) \]

When the ellipsoidal deformation (\(\ell = 2\) mode) is dominant,

\[ \delta\Phi \propto r^2\,Y_2^m(\theta,\varphi) \]

Gravitational field inside the cavity:

\[ \mathbf{g} = -\nabla\Phi = -\nabla(\delta\Phi) \neq \mathbf{0} \]

Since the potential is no longer constant, a non-zero gravitational field arises inside the cavity. For \(\delta\Phi \propto r^2 Y_2^m\), the field \(\mathbf{g} = -\nabla(\delta\Phi)\) depends linearly on position \(\mathbf{r}\). That is, the direction and magnitude of the gravitational field differ at different points within the cavity.

Relation to tidal force:

Tidal force arises from the spatial non-uniformity of the gravitational field—that is, the fact that the gravitational field varies from place to place. When two nearby point masses experience different gravitational accelerations, the difference is observed as a tidal force.

In Newtonian gravity, tidal forces are characterized by the second derivatives of the gravitational potential. The tidal tensor is defined as

\[ T_{ij} = -\frac{\partial^2\Phi}{\partial x^i\partial x^j} = \frac{\partial g_i}{\partial x^j} \]

This represents the spatial rate of change of the gravitational field (the gradient of the gravitational field).

  • Perfectly spherically symmetric shell: Inside the cavity \(\Phi = \text{const}\), so \(\partial^2\Phi/\partial x^i\partial x^j = 0\). The tidal tensor is zero, and no tidal force exists.
  • Ellipsoidally deformed shell: When terms like \(\delta\Phi \propto r^2 Y_2^m\) appear, \(\partial^2(\delta\Phi)/\partial x^i\partial x^j \neq 0\) (a constant tensor). A uniform tidal force arises within the cavity.

Specifically, along the major axis of the ellipsoid there is a stretching force (positive eigenvalues of \(T_{ij}\)), while along the minor axes there is a compressive force (negative eigenvalues of \(T_{ij}\)). This is essentially the same mechanism as the ocean tides on Earth caused by the Moon's gravity.

Constraint on the trace of the tidal tensor:

Since \(\rho = 0\) inside the cavity, the Poisson equation \(\nabla^2\Phi = 0\) holds. This means

\[ \mathrm{Tr}(T_{ij}) = -\nabla^2\Phi = 0 \]

That is, the tidal tensor is traceless, and the tidal force causes only volume-preserving deformations (stretching and compression balance each other).

Correspondence with general relativity:

In general relativity, tidal forces are described as curvature of spacetime. Specifically, the Riemann curvature tensor \(R^\alpha{}_{\beta\gamma\delta}\) gives the relative acceleration between two nearby geodesics (= tidal force) through the geodesic deviation equation:

\[ \frac{D^2\xi^\alpha}{d\tau^2} = -R^\alpha{}_{\beta\gamma\delta}\,u^\beta\,\xi^\gamma\,u^\delta \]

where \(\xi^\alpha\) is the deviation vector between geodesics and \(u^\beta\) is the 4-velocity.

In the Newtonian limit (weak gravitational field, low velocity), this equation reduces to

\[ \frac{d^2\xi^i}{dt^2} = -\frac{\partial^2\Phi}{\partial x^i\partial x^j}\,\xi^j = T_{ij}\,\xi^j \]

Therefore, the Newtonian tidal tensor \(\partial^2\Phi/\partial x^i\partial x^j\) corresponds to specific components of the Riemann tensor \(R^0{}_{i0j}\) (in the Newtonian limit).

\[ \boxed{\text{Newton: } \frac{\partial^2\Phi}{\partial x^i\partial x^j} \quad \longleftrightarrow \quad \text{General relativity: } R^0{}_{i0j} \quad (\text{tidal force} = \text{spacetime curvature})} \]

Furthermore, the traceless condition in vacuum \(\nabla^2\Phi = 0\) (\(\Leftrightarrow T_{ii} = 0\)) corresponds to the vacuum Einstein equation \(R_{\mu\nu} = 0\) (vanishing Ricci tensor) in general relativity. Even when the Riemann tensor itself is non-zero (tidal forces exist), the condition that its trace part (the Ricci tensor) vanishes characterizes the properties of the gravitational field in vacuum.

Consistency check: Inside the cavity \(\rho = 0\), so \(\nabla^2\Phi = 0\), i.e., \(T_{ii} = 0\). The tidal force is volume-preserving (stretching and compression balance). This corresponds to the vanishing Ricci tensor \(R_{\mu\nu} = 0\) in vacuum in general relativity, which is consistent. ✓