Day: November 22, 2022

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202211221451 Problem 4.10

A sphere of radius R carries a polarization

\mathbf{P}(\mathbf{r})=k\mathbf{r},

where k is a constant and \mathbf{r} the vector from the centre.
(a) Calculate the bound charges \sigma_b and \rho_b.
(b) Find the field inside and outside the sphere.

Extracted from David J. Griffiths. (1999). Introduction to Electrodynamics.

Roughwork.

The surface charge is defined by Eq. (4.11):

\sigma_b\equiv \mathbf{P}\cdot\hat{\mathbf{n}}

whereas the volume charge by Eq. (4.12):

\rho_b\equiv -\nabla \cdot \mathbf{P}.

Text on pp. 167-168

On the surface r=|\mathbf{r}|=R:

\begin{aligned} \sigma_b & = P(R\,\hat{\mathbf{r}})\cdot\hat{\mathbf{r}} \\ & = k(R\,\hat{\mathbf{r}})\cdot\hat{\mathbf{r}} \\ & = kR \end{aligned}

Under the surface \mathbf{r}=\mathbf{r}(x,y,z):

\begin{aligned} \rho_b & = -\nabla\cdot\mathbf{P} \\ & = -\bigg(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\bigg)\cdot k(x,y,z) \\ & = -k\bigg(\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z}\bigg) \\ & = -3k \end{aligned}

Notice \displaystyle{(-3k)\bigg(\frac{4\pi R^3}{3}\bigg) +(kR)(4\pi R^2)=0}, so the sphere has zero net charge, i.e., Q_\textrm{net}=0, and thus produces no electric field outside, i.e., \mathbf{E}(\mathbf{r}:r>R)=\mathbf{0}.

Eq. (4.13):

V(\mathbf{r})=\displaystyle{\frac{1}{4\pi\epsilon_0}\iint_S\frac{\sigma_b}{r}\,\mathrm{d}a'+\frac{1}{4\pi\epsilon_0}\iiint_V\frac{\rho_b}{r}\,\mathrm{d}\tau'}

and the rest is left as an exercise to the reader.

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202211221415 Exercise 29.4.1

Use this fact, that the difference between a vacuum energy density \rho_v and \rho_m or \rho_r is that \rho_v does not change as the universe expands (though it may change for other physical reasons), the basic Friedman equation Eq. (29.13):

\displaystyle{\dot{a}^2-\frac{8\pi G}{3}(\rho_m+\rho_r+\rho_v)a^2=K}

with K=0, and the assumption of vacuum dominance (\rho_m=\rho_r\approx 0) to show that Eq. (29.9):

a(t)=a(t_s)\exp \Big(\sqrt{\frac{8}{3}\pi G\rho_v}(t-t_s)\Big).

This form of the equation is valuable because it no longer refers to the present state of the universe.

Extracted from Thomas A. Moore. (2013). A General Relativity Workbook.

Roughwork.

This solution is not mine. It was found on the Internet some years ago, to whose author(s) I lost references.

\begin{aligned} 0 & = \dot{a}^2-\frac{8\pi G}{3}\rho_va^2\\ \bigg(\frac{\mathrm{d}a}{\mathrm{d}t}\bigg)^2 & = \frac{8\pi G}{3}\rho_va^2 \\ \frac{1}{a}\frac{\mathrm{d}a}{\mathrm{d}t} & = \sqrt{\frac{8\pi G}{3}\rho_v} \\ \ln \big( a(t)\big) & = \bigg(\sqrt{\frac{8\pi G}{3}\rho_v}\bigg)\cdot t + C \\ a(t)& =a(t_s)\exp \bigg(\sqrt{\frac{8}{3}\pi G\rho_v}(t-t_s)\bigg) \\ \end{aligned}