Physicspupil – Page 40

March 3, 2019January 19, 2023

201903030504 Problem 2.12

From $\displaystyle{\frac{\mathrm{d}V}{1-V^2/c^2}}=f(\tau)\,\mathrm{d}\tau$ ,

upon integration it becomes

$V(\tau)=c\tanh \phi(\tau)$

where $c\,\phi (\tau)=\displaystyle{\int_0^\tau f(\tau)\,\mathrm{d}\tau}$ .

$\begin{aligned} \displaystyle{\frac{\mathrm{d}V}{1-\frac{V^2}{c^2}}} & = f(\tau)\,\mathrm{d}\tau \\ \displaystyle{\int\frac{1}{1-x^2}}(c\,\mathrm{d}x) & = \displaystyle{\int_0^\tau}f(\tau)\,\mathrm{d}\tau \\ c\,\mathrm{argtanh} x & =\displaystyle{ \int_0^\tau f(\tau)\,\mathrm{d}\tau} \\ \mathrm{argtanh} x & = \displaystyle{\frac{1}{c}\int_0^\tau f(\tau)\,\mathrm{d}\tau} \\ \tanh \bigg(\displaystyle{\frac{1}{c}\int_0^\tau f(\tau)\,\mathrm{d}\tau} \bigg) & = x = \displaystyle{\frac{V}{c}} \\ V(\tau) & = c\tanh \phi(\tau) \\ \end{aligned}$

Similarly one can obtain

$\dot{t}\equiv \displaystyle{\frac{\mathrm{d}t}{\mathrm{d}\tau}}=\displaystyle{\frac{1}{\sqrt{1-V^2/c^2}}}=\cosh \phi (\tau)$ ,

$\dot{x}\equiv \displaystyle{\frac{\mathrm{d}x}{\mathrm{d}\tau}}=\displaystyle{\frac{V}{\sqrt{1-V^2/c^2}}}=c\sinh \phi(\tau)$ .

March 3, 2019January 19, 2023

201903030456 Exercise 4.6.1

Exercise 4.6.1. Use a similar approach (involving renaming indices and grouping like terms) to prove equation 4.20.

Eq. (4.20):

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}\tau}}(A^2)=2\eta_{\mu\nu}A^\mu \displaystyle{\frac{\mathrm{d}A^\nu}{\mathrm{d}\tau}}$

Eq. (4.10):

$A^2=\mathbf{A}\cdot\mathbf{A}=\eta_{\mu\nu}A^\mu A^\nu$

Then,

$\begin{aligned} \displaystyle{\frac{\mathrm{d}}{\mathrm{d}\tau}}(A^2) & = \displaystyle{\frac{\mathrm{d}}{\mathrm{d}\tau}}(\eta_{\mu\nu}A^\mu A^\nu )\\ & = \eta_{\mu\nu} \bigg[ A^{\mu}\displaystyle{\frac{\mathrm{d}}{\mathrm{d}\tau}(A^\mu )} + \displaystyle{\frac{\mathrm{d}}{\mathrm{d}\tau}}(A^\mu )\cdot A^\nu \bigg] \\ & = 2\eta_{\mu\nu}A^\mu \displaystyle{\frac{\mathrm{d}A^\nu}{\mathrm{d}\tau}} \end{aligned}$

March 3, 2019January 19, 2023

201903030450 Exercise 9.1.1

The Schwarzschild metric is given by Eq. (9.3):

$\mathrm{d}s^2=-\bigg( 1-\displaystyle{\frac{r_s}{r}} \bigg) \mathrm{d}t^2+\bigg( 1-\displaystyle{\frac{r_s}{r}} \bigg)^{-1}\,\mathrm{d}r^2+r^2\,\mathrm{d}\theta +r^2\sin^2\theta\,\mathrm{d}\phi^2$

whereas the metric for spherical coordinates in flat spacetime is given by Eq. (9.2):

$\mathrm{d}s^2=-\mathrm{d}t^2+\mathrm{d}r^2+r^2\,\mathrm{d}\theta^2+r^2\sin^2\theta\,\mathrm{d}\phi^2$

Along a purely radial worldline $\mathrm{d}t=0$ , $\mathrm{d}\theta =0$ and $\mathrm{d}\phi =0$ , of the Schwarzschild metric will become

$\mathrm{d}s^2=\bigg( 1-\displaystyle{\frac{r_s}{r}} \bigg)^{-1}\,\mathrm{d}r^2$ .

Now that the Schwarzschild radius $r_s$ is $2GM$ , there is Eq. (9.15):

$\mathrm{d}s=\displaystyle{\frac{\mathrm{d}r}{\sqrt{1-2GM/r}}}$ .

The total radial distance between two events differing only by $r$ -coordinates, i.e., $r_A$ and $r_B$ is calculated by definite integration, given by Eq. (9.16):

$\Delta s=\displaystyle{\int \mathrm{d}s=\int_{r_A}^{r_B}\frac{\mathrm{d}r}{\sqrt{1-2GM/r}}}$

Trying binomial approximation:

$(1+x)^n =1+nx+\displaystyle{\frac{n(n-1)}{2!}}x^2 +\displaystyle{\frac{n(n-1)(n-2)}{3!}}x^3+O(x^4)$

then,

$\begin{aligned} &\quad\enspace (1-2GM/r)^{-\frac{1}{2}} \\ & =1+\bigg(-\frac{1}{2}\bigg)(-2GM/r) +\displaystyle{\frac{(-\frac{1}{2})(-\frac{1}{2}-1)}{2!}}(-2GM/r)^2+O(r^2) \\ \end{aligned}$

Considering only first-order approximation:

$1+\displaystyle{\frac{GM}{r}}$ .

Upon integration, there is Eq. (9.17):

$\begin{aligned} \Delta s & \approx \displaystyle{\int_{r_A}^{r_B}}\bigg( 1+\displaystyle{\frac{GM}{r}}\bigg) \,\mathrm{d}r \\ & = \bigg[ r+GM\ln \bigg( \displaystyle{\frac{r_B}{r_A}} \bigg) \bigg]\bigg|_{r_A}^{r_B} \\ & = (r_B-r_A) + GM\ln \bigg( \frac{r_B}{r_A} \bigg) \end{aligned}$

March 3, 2019January 19, 2023

201903030446 Exercise 14.1.1 & 14.1.2

Eq. (14.7):

$\mathrm{d}s=\displaystyle{\frac{\mathrm{d}r}{\sqrt{1-2GM/r}}}\Longrightarrow \Delta s=\displaystyle{\int \frac{\mathrm{d}r}{\sqrt{1-2GM/r}}}$

Let Eq.(14.8):

$u\equiv \sqrt{1-\displaystyle{\frac{2GM}{r}}}\Longrightarrow r=\displaystyle{\frac{2GM}{1-u^2}}$

From the event horizon $r=r_s=2GM$ to $r=R$ , it corresponds to $u=0$ and $u(R)=\sqrt{1-\displaystyle{\frac{2GM}{R}}}$ .

From $r=\displaystyle{\frac{2GM}{1-u^2}}$ , by quotient rule:

$\displaystyle{\frac{\mathrm{d}r}{\mathrm{d}u}}=\displaystyle{\frac{(1-u^2)(0)-(2GM)(-2u)}{(1-u^2)^2}}$ ,

and

$\mathrm{d}r=\displaystyle{\frac{4GMu}{(1-u^2)^2}}\,\mathrm{d}u$ .

Substituting these into Eq. (14.7), I get Eq. (14.9):

$\Delta s=\displaystyle{\int_0^{u(R)}}\displaystyle{\frac{4GM}{(1-u^2)^2}}\,\mathrm{d}u$

March 3, 2019January 19, 2023

201903030442 Exercise 17.1.1

$\begin{aligned} \mathrm{d}\mathbf{A} & =\mathrm{d} (A^\mu \mathbf{e}_\mu )\\ & = \bigg( \displaystyle{\frac{\partial A^\mu}{\partial x^\sigma}}\, \mathrm{d}x^\sigma \bigg) \,\mathbf{e}_\mu + A^\mu \displaystyle{\frac{\partial \mathbf{e}_\mu}{\partial x^\alpha}} \,\mathrm{d}x^\alpha \\ & = \bigg( \displaystyle{\frac{\partial A^\mu}{\partial x^\sigma}} \, \mathrm{d}x^\sigma \bigg) \,\mathbf{e}_\mu + A^\mu \Big( \Gamma^\nu_{\alpha\mu} \mathbf{e}_\nu \Big) \,\mathrm{d}x^\alpha \\ \dots &\, \textrm{changing dummy variables: }\mu\Leftrightarrow \nu\textrm{ and }\alpha\Rightarrow \sigma\,\dots \\ & = \bigg[ \displaystyle{\frac{\partial A^\mu}{\partial x^\sigma}} + \Gamma^\mu_{\sigma\nu}A^\nu \bigg] \,\mathbf{e}_\mu \,\mathrm{d}x^\sigma \\ & \stackrel{\mathrm{def}}{=} (\nabla_\sigma A^\mu )\,\mathbf{e}_\mu \,\mathrm{d}x^\sigma \end{aligned}$

March 3, 2019January 19, 2023

201903030439 Exercise 14.1.3

Eq. (14.9):

$\Delta s=4GM \displaystyle{\int_{0}^{u(R)}\frac{\mathrm{d}u}{(1-u^2)^2}}$

Since the indefinite integral evaluated by Eq. (14.10):

$\displaystyle{\int \frac{\mathrm{d}u}{(1-u^2)^2}=\frac{u}{2(1-u^2)}+\frac{1}{4}\ln \bigg| \frac{1+u}{1-u} \bigg|}$

and the identity given by Eq. (14.11):

$\tanh^{-1}u=\displaystyle{\frac{1}{2}\ln \bigg| \frac{1+u}{1-u} \bigg|}$

are combined to give

$\displaystyle{\int \frac{\mathrm{d}u}{(1-u^2)^2}=\frac{u}{2(1-u^2)}+\frac{1}{2}\tanh^{-1}u}$ .

From $u(R)=\sqrt{1-\displaystyle{\frac{2GM}{R}}}$ ,

$\begin{aligned} \displaystyle{\frac{u(R)}{2(1-u(R)^2)}} & = \displaystyle{\frac{1}{2\Big( 1-\Big( 1-\frac{2GM}{R} \Big) \Big)}} \sqrt{1-\displaystyle{\frac{2GM}{R}}}\\ & = \displaystyle{\frac{R}{4GM}} \sqrt{1-\displaystyle{\frac{2GM}{R}}} \end{aligned}$ .

$\Delta s=4GM \Bigg( \displaystyle{\frac{R}{4GM}} \sqrt{1-\displaystyle{\frac{2GM}{R}}} +\displaystyle{\frac{1}{2}}\tanh^{-1}\bigg( \sqrt{1-\displaystyle{\frac{2GM}{R}}}\bigg) \Bigg)$

and it checks with Eq. (14.3):

$\Delta s=R\sqrt{1-2GM/R}+2GM\tanh^{-1}\sqrt{1-2GM/R}$ .

March 3, 2019January 19, 2023

201903030435 Exercise 14.1.4

Check that the physical distance from $r=3GM$ to $r=2GM$ is indeed $3.05GM$ . (If your calculator cannot handle inverse hyperbolic functions, use equation 14.11 to eliminate $\tanh^{-1}u$ .)

Eq. (14.3):

$\Delta s=R\sqrt{1-2GM/R}+2GM\tanh^{-1}\sqrt{1-2GM/R}$

and the identity given by Eq. (14.11):

$\tanh^{-1}u=\displaystyle{\frac{1}{2}\ln \bigg| \frac{1+u}{1-u} \bigg|}$

are combined to give the following

$\begin{aligned} \Delta s(r) & =r\sqrt{1-2GM/r}+GM\ln \bigg| \displaystyle{\frac{1+u}{1-u}} \bigg| \\ & =r\sqrt{1-2GM/r}+GM\ln \Bigg(\Bigg| \displaystyle{\frac{1+\sqrt{1-2GM/R}}{1-\sqrt{1-2GM/R}}}\Bigg| \Bigg) \\ \end{aligned}$

which measures the physical distance $\Delta s$ from the event horizon $r=2GM$ .

Now that $r=3GM$ , the physical distance will be

$\begin{aligned} \Delta s(3GM) & =(3GM)\sqrt{\displaystyle{1-\frac{2GM}{(3GM)}}}+GM\ln \Bigg| \frac{1+\sqrt{1-\frac{2GM}{(3GM)}}}{1-\sqrt{1-\frac{2GM}{(3GM)}}} \Bigg| \\ & = \sqrt{3} GM + 1.3169 GM \\ & \approx 3.05GM\qquad (\textrm{3 s.f.}) \end{aligned}$

March 3, 2019January 19, 2023

201903030417 Exercise 8.5.2

Eq. (8.41):

$0=\displaystyle{\frac{\mathrm{d}^2\theta}{\mathrm{d}s^2}}-\sin\theta\cos\theta\bigg( \displaystyle{\frac{\mathrm{d}\phi}{\mathrm{d}s}} \bigg)^2$

Eq. (8.42):

$0=\displaystyle{\frac{\mathrm{d}}{\mathrm{d}s}}\bigg(\sin^2\theta \displaystyle{\frac{\mathrm{d}\phi}{\mathrm{d}s}}\bigg)$

Integrating on Eq. (8.42) gives Eq. (8.43):

$\sin^2\theta \displaystyle{\frac{\mathrm{d}\phi}{\mathrm{d}s}}=\textrm{Const.}\equiv \displaystyle{\frac{c}{R}}\Rightarrow \displaystyle{\frac{\mathrm{d}\phi}{\mathrm{d}s}=\frac{R}{\sin^2\theta}}$ .

The definition of pathlength along the curve is given by Eq. (8.16):

$g_{\mu\nu}\displaystyle{\frac{\mathrm{d}x^\mu}{\mathrm{d}s}} \displaystyle{\frac{\mathrm{d}x^\nu}{\mathrm{d}s}} = + \bigg( \displaystyle{\frac{\mathrm{d}s}{\mathrm{d}s}} \bigg) = +1$ .

Previously to Exercise 8.5.2., the metric tensor is given by Eq. (8.40):

$g_{\mu\nu}=\begin{bmatrix} R^2 & 0 \\ 0 & R^2\sin^2\theta \end{bmatrix}$ .

Eq. (8.45) begins:

$\begin{aligned} 1 & = R^2\bigg( \displaystyle{\frac{\mathrm{d}\theta}{\mathrm{d}s }} \bigg)^2 +R^2\sin^2\theta \bigg( \displaystyle{\frac{\mathrm{d}\phi}{\mathrm{d}s}} \bigg)^2 \\ 1 & = R^2\bigg( \displaystyle{\frac{\mathrm{d}\theta}{\mathrm{d}s }} \bigg)^2 +R^2\sin^2\theta \bigg( \displaystyle{\frac{c}{R\sin^2\theta}} \bigg)^2 \\ 1 & = R^2\bigg( \displaystyle{\frac{\mathrm{d}\theta}{\mathrm{d}s}} \bigg)^2 + \displaystyle{\frac{c^2}{\sin^2\theta}} \\ \bigg( \displaystyle{\frac{\mathrm{d}\theta}{\mathrm{d}s}} \bigg)^2 & = \displaystyle{\frac{1}{R^2}\bigg( 1-\frac{c^2}{\sin^2\theta} \bigg)} \\ \displaystyle{\frac{\mathrm{d}\theta}{\mathrm{d}s}} & = \pm \displaystyle{\frac{1}{R}\sqrt{1-\frac{c^2}{\sin^2\theta}}} \\ & = \pm \displaystyle{\frac{1}{R\sin\theta}}\sqrt{\sin^2\theta -c^2} \end{aligned}$

and this is Eq. (8.46).

March 3, 2019January 19, 2023

201903030411 Exercise 4.3.1

Write out the implied sum in $\mathrm{d}p^\mu /\mathrm{d}\tau =qF^{\mu\nu}\eta_{\nu\alpha}u^\alpha$ for $\mu =x$ and show that it is equivalent to the $x$ component of equation (4.22) at low velocities.

Eq. (4.22):

$\mathbf{F}_{\mathrm{em}}=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$

Eq. (4.15):

$\displaystyle{\frac{\mathrm{d}p^\mu}{\mathrm{d}\tau}}=qF^{\mu\nu}\eta_{\nu\alpha}u^\alpha$

the electromagnetic field tensor $F^{\mu\nu}$ of which is given by Eq. (4.14):

$\begin{bmatrix} F^{tt} & F^{tx} & F^{ty} & F^{tz} \\ F^{xt} & F^{xx} & F^{xy} & F^{xz} \\ F^{yt} & F^{yx} & F^{yy} & F^{yz} \\ F^{zt} & F^{zx} & F^{zy} & F^{zz} \end{bmatrix} = \begin{bmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & B_z & -B_y \\ -E_y & -B_z & 0 & B_x \\ -E_z & B_y & -B_x & 0 \end{bmatrix}$

and the Minkowski metric tensor $\eta_{\nu\alpha}$ of which is given by Eq. (4.6):

$\begin{bmatrix} \eta_{tt} & \eta_{tx} & \eta_{ty} & \eta_{tz} \\ \eta_{xt} & \eta_{xx} & \eta_{xy} & \eta_{xz} \\ \eta_{yt} & \eta_{yx} & \eta_{yy} & \eta_{yz} \\ \eta_{zt} & \eta_{zx} & \eta_{zy} & \eta_{zz} \end{bmatrix} = \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$

In the case $\mu =x$ , beginning with Eq. (4.15):

$\begin{aligned} \displaystyle{\frac{\mathrm{d}p^x}{\mathrm{d}\tau}} & =qF^{x\nu}\eta_{\nu\alpha}u^\alpha \\ & = q \big( F^{xt}\eta_{t\alpha}u^\alpha + F^{xx}\eta_{x\alpha}u^\alpha + F^{xy}\eta_{y\alpha}u^\alpha + F^{xz}\eta_{z\alpha}u^\alpha \big) \\ & = q \big( -E_x \eta_{tt} u^t + (0)(\eta_{xx})(u^x) + B_{z}\eta_{yy}u^y -B_y\eta_{zz}u^z \big) \\ \dots\, & \textrm{when }v\ll c\textrm{, }u^t\approx 1\textrm{, }u^x\approx v_x\textrm{, }u^y\approx v_y\textrm{, and }u^z\approx v_z\,\dots \\ & = q\big( (-E_x)(-1)(1) + (B_z)(1)(v_y) - (B_y)(1)(v_z)\big) \\ & = q(E_x + B_{z}v_{y} - B_{y}v_{z}) \\ & = q(\mathbf{E}+\mathbf{v}\times\mathbf{B})_x \\ & = \mathbf{F}_{\mathrm{em},\,x}\end{aligned}$

February 25, 2019January 19, 2023

201902250313 Solution to 1987-AL-PHY-I-30

The electrostatic force $\mathbf{F}_{XW}$ acting on the charge $-Q$ at $X$ due to another charge $+Q$ at $W$ is given by Coulomb’s Law:

$\mathbf{F}_{XW}=\displaystyle{\frac{(-Q)(+Q)}{4\pi\epsilon_0r^2}}\,\hat{\mathbf{i}}=-\displaystyle{\frac{Q^2}{4\pi\epsilon_0r^2}}\,\hat{\mathbf{i}}$ ,

whereas the electrostatic force $\mathbf{F}_{XY}$ due to another charge $-Q$ at $Y$ is given similarly by:

$\mathbf{F}_{XY}=\displaystyle{\frac{(-Q)(-Q)}{4\pi\epsilon_0r^2}}\,\hat{\mathbf{j}}=\displaystyle{\frac{Q^2}{4\pi\epsilon_0r^2}}\,\hat{\mathbf{j}}$ .

When a fourth charge $q$ is placed on $Z$ , which is at a distance of $\sqrt{2}r$ from $X$ where $r$ is the side length of the square, the magnitude of electrostatic force $F_{XZ}$ is given by:

$F_{XZ}=\displaystyle{\frac{(-Q)(q)}{4\pi\epsilon_{0}(\sqrt{2}r)^2}}=\displaystyle{\frac{-Qq}{8\pi\epsilon_0r^2}}$ .

The electrostatic force $\mathbf{F}_{XZ}$ is then given by:

$\begin{aligned} \mathbf{F}_{XZ} & = F_{XZ}(\cos 45^\circ )\,\hat{\mathbf{i}} +F_{XZ}(\sin 45^\circ )\,\hat{\mathbf{j}} \\ & = \bigg( \frac{-Qq}{8\pi\epsilon_0r^2} \bigg) \frac{\sqrt{2}}{2} \,\hat{\mathbf{i}} + \bigg( \frac{-Qq}{8\pi\epsilon_0r^2} \bigg) \frac{\sqrt{2}}{2} \,\hat{\mathbf{j}} \\ & = \frac{-\sqrt{2}Qq}{16\pi\epsilon_0r^2} \,\hat{\mathbf{i}} + \frac{-\sqrt{2}Qq}{16\pi\epsilon_0r^2} \,\hat{\mathbf{j}} \end{aligned}$

When the net electrostatic force acting on position $X$ is in the left direction only, by summing over the above-calculated three forces, i.e.,

$\begin{aligned} \sum \mathbf{F} & =\mathbf{F}_{XW}+\mathbf{F}_{XY}+\mathbf{F}_{XZ} \\ & = \bigg( -\displaystyle{\frac{Q^2}{4\pi\epsilon_0r^2}}+\frac{-\sqrt{2}Qq}{16\pi\epsilon_0r^2}\bigg) \,\hat{\mathbf{i}} + \bigg(\displaystyle{\frac{Q^2}{4\pi\epsilon_0r^2}} + \frac{-\sqrt{2}Qq}{16\pi\epsilon_0r^2}\bigg) \,\hat{\mathbf{j}} \end{aligned}$

the vertical component of the electrostatic force must be zero, i.e.,

$\begin{aligned} \displaystyle{\frac{Q^2}{4\pi\epsilon_0r^2}} + \frac{-\sqrt{2}Qq}{16\pi\epsilon_0r^2} & =0 \\ q & = +2\sqrt{2}\,Q \end{aligned}$

And the answer is E.