General Relativity * – Page 2

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 21, 2019January 19, 2023

201902210304 Exercise 6.3.1

Fill in the gap between equations (6.18) and (6.19).

Roughwork.

Equation (6.16):

$g^{\mu\alpha}g_{\alpha\nu}=\delta^\mu_{\enspace\nu}$ and $g'^{\mu\beta}g'_{\beta\nu}=\delta^\mu_{\enspace\nu}$

Equation (6.9):

$g'_{\mu\nu}=\displaystyle{\frac{\partial x^\alpha}{\partial x'^\mu}} \displaystyle{\frac{\partial x^\beta}{\partial x'^\nu}}g_{\alpha\beta}$

Equation (6.11):

$g'^{\mu\nu}= \displaystyle{\frac{\partial x'^\mu}{\partial x^\alpha}} \displaystyle{\frac{\partial x'^\nu}{\partial x^\beta}}g^{\alpha\beta}$

$\begin{aligned} g'^{\mu\nu} g_{\mu\nu} & = \bigg( \displaystyle{\frac{\partial x'^\mu}{\partial x^\alpha}} \displaystyle{\frac{\partial x'^\nu}{\partial x^\beta}} g^{\alpha\beta} \bigg) \bigg( \displaystyle{\frac{\partial x^\gamma}{\partial x'^\mu}} \displaystyle{\frac{\partial x^\delta}{\partial x'^\nu}} g_{\gamma\delta} \bigg) \\ & = \bigg( \displaystyle{\frac{\partial x^\gamma}{\partial x^\alpha}} \displaystyle{\frac{\partial x^\delta}{\partial x^\beta}} \bigg) g^{\alpha\beta}g_{\gamma\delta} \\ & = \delta^\gamma_{\enspace \alpha} \delta^\delta_{\enspace \beta} g^{\alpha\beta} g_{\gamma\delta} \\ & = g^{\alpha\beta} g_{\alpha\beta} \end{aligned}$

February 21, 2019January 19, 2023

201902210301 Exercise 6.2.1-6.2.3

Again consider polar coordinates on a flat plane. The transformation equations between polar coordinates $r$ , $\theta$ (the primed coordinate system) and Cartesian coordinates $x$ , $y$ (the unprimed coordinate system) are given by equation 6.15. Consider also the vector $\mathbf{v}$ whose components are $v^x=1$ and $v^y=0$ .

Lowering an index as given by Equation (6.5):

$A_\mu \equiv g_{\mu\nu}A^\nu$ .

In Cartesian coordinate system the metric tensor is

$\mathbf{g}=\begin{bmatrix} g_{xx} & g_{xy} \\ g_{yx} & g_{yy} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Thus

$v_x\equiv g_{x\nu}v^\nu = g_{xx}v^x+g_{xy}v^y = (1)(1) + (0)(0)=1$

$v_y\equiv g_{y\nu}v^\nu = g_{yx}v^x+g_{yy}v^y = (0)(1) + (1)(0)=0$

The definition of covector is given by Equation (6.2):

$B'_\nu = \displaystyle{\frac{\partial x^\mu}{\partial x'^\nu}}B_{\mu}$

So,

$v_r=\displaystyle{\frac{\partial x}{\partial r}}v_x + \displaystyle{\frac{\partial y}{\partial r}}v_y = (\cos\theta )(1) + (\sin\theta )(0) = \cos\theta$

$v_\theta =\displaystyle{\frac{\partial x}{\partial \theta}}v_x + \displaystyle{\frac{\partial y}{\partial \theta}}v_y = (-r\sin\theta )(1) + (r\cos\theta )(0) = -r\sin\theta$

The metric tensor for the polar coordinate basis is given by Equation (5.19):

$g_{\mu\nu}\equiv \mathbf{e}_{\mu}\cdot\mathbf{e}_\nu = \begin{bmatrix} \mathbf{e}_{r}\cdot\mathbf{e}_r & \mathbf{e}_{r}\cdot\mathbf{e}_\theta \\ \mathbf{e}_{\theta}\cdot\mathbf{e}_r & \mathbf{e}_{\theta}\cdot\mathbf{e}_\theta \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & r^2 \\ \end{bmatrix}$

Exercise 6.2.3.

One can show that in the polar coordinate system, $v^r=\cos\theta$ and $v^\theta =-\sin\theta /r$ (see Problem P6.1). Show that $v'^\mu v'_{\mu}=1$ . Does this make sense?

Equation (6.5):

$A_{\mu} \equiv g_{\mu\nu}A^\nu$

Hence

$\begin{aligned} v_r & \equiv g_{r\nu}v^\nu \\ \cos\theta & = g_{rr}v^r + g_{r\theta} v^\theta \\ \cos\theta & = (1)(v^r) + (0)(v^\theta ) \\ v^r & = \cos\theta \end{aligned}$

$\begin{aligned} v_\theta & \equiv g_{\theta\nu}v^\nu \\ -r\sin\theta & = g_{\theta r}v^r + g_{\theta\theta} v^\theta \\ -r\sin\theta & = (0)(\cos\theta ) + (r^2)(v^\theta) \\ v^\theta & = \displaystyle{\frac{-\sin\theta }{r}} \\ \end{aligned}$

are checked.

$\begin{aligned} v'^\mu v'_{\mu} & = v^r v_r + v^\theta v_{\theta} \\ & = (\cos\theta )(\cos\theta ) + (\displaystyle{\frac{-\sin\theta}{r}})(-r\sin\theta ) \\ & = \cos^2\theta + \sin^2\theta \\ & = 1 \end{aligned}$ .

Remark. Invariant norm.

February 21, 2019January 19, 2023

201902210253 Exercise 6.1.1-6.1.3

Consider polar coordinates on a flat plane. The transformation equations between the polar coordinates $r$ , $\theta$ (the primed coordinate system) and cartesian coordinates $x$ , $y$ (the unprimed coordinate system) are

Equation (6.15a):

$x=r\cos\theta$ , $y=r\sin\theta$

Equation (6.15b):

$r=\sqrt{x^2+y^2}$ , $\theta =\tan^{-1}\bigg( \displaystyle{\frac{y}{x}} \bigg)$

Consider also the scalar function $\varPsi =bxy=br^2\cos\theta\sin\theta$ .

Calculate the four transformation partials $\displaystyle{\frac{\partial x^\mu}{\partial x'^\nu}}$ for the transformations given above:

$\displaystyle{\frac{\partial x}{\partial r}}=\cos\theta$ , $\displaystyle{\frac{\partial x}{\partial \theta}}=-r\sin\theta$ , $\displaystyle{\frac{\partial y}{\partial r}}=\sin\theta$ , $\displaystyle{\frac{\partial y}{\partial \theta}}=r\cos\theta$ .

It can be shown that the gradient of $\varPsi$ in cartesian coordinate system is

$\partial_x\varPsi =by$ and $\partial_y\varPsi =bx$

and that of $\varPsi$ in polar coordinate system is

$\partial_r\varPsi =2br\cos\theta\sin\theta$ and $\partial_\theta\varPsi =br^2[\cos^2\theta -\sin^2\theta]$ .

To make practice of the covector transformation rule:

Equation (6.3):

$\partial_\mu\varPsi \equiv \displaystyle{\frac{\partial \varPsi}{\partial x^\mu}}$

Equation (6.4):

$\partial'_\nu \equiv \displaystyle{\frac{\partial \varPsi}{\partial x'^\nu}} = \displaystyle{\frac{\partial x^\mu}{\partial x'^\nu}}\displaystyle{\frac{\partial \varPsi}{\partial x^\mu}} = \displaystyle{\frac{\partial x^\mu}{\partial x'^\nu}}\partial_\mu \varPsi$

Thus,

$\begin{aligned} &\quad\enspace \displaystyle{\frac{\partial x}{\partial r}}\partial_x\varPsi + \displaystyle{\frac{\partial y}{\partial r}}\partial_y\varPsi \\ & = (\cos\theta )(by) + (\sin\theta )(bx) \\ & = (\cos\theta )(br\sin\theta ) + (\sin\theta )(br\cos\theta) \\ & = 2br\cos\theta\sin\theta \\ & = \partial_r \varPsi \end{aligned}$

and

$\begin{aligned} &\quad\enspace \displaystyle{\frac{\partial x}{\partial \theta}}\partial_x\varPsi + \displaystyle{\frac{\partial y}{\partial \theta}}\partial_y\varPsi\\ & = (-r\sin\theta )(by) + (r\cos\theta )(bx) \\ & = (-r\sin\theta )(br\sin\theta ) + (r\cos\theta )(br\cos\theta ) \\ & = br^2[\cos^2\theta -\sin^2\theta ] \\ & = \partial_\theta \varPsi \end{aligned}$

February 21, 2019January 19, 2023

201902210247 Exercise 5.5.1

Check the cases where $\alpha =t$ and $\beta =x$ , and where $\alpha =\beta =x$ .

Roughwork.

This exercise is to check the transformation law for the metric tensor of flat spacetime, given by Equation (5.16):

$\eta'_{\alpha\beta}=\displaystyle{\frac{\partial x^\mu}{\partial x'^\alpha}}\displaystyle{\frac{\partial x^\nu}{\partial x'^\beta}} \eta_{\mu\nu}=\eta_{\mu\nu}(\varLambda^{-1})^\mu_{\enspace \alpha}(\varLambda^{-1})^\nu_{\enspace \beta}=\eta_{\alpha\beta}$

(where the second equality is by Equation (5.13), and the third equality by Equation 4.19)

The textbook has already had a checking for $\eta'_{tt}=\eta_{tt}$ through Equations (5.29) and (5.30). I will do the remaining.

Proof. ( $\eta'_{tx}=\eta_{tx}$ )

$\begin{aligned} \eta'_{tx} & = (\varLambda^{-1})^\mu_{\enspace t}(\varLambda^{-1})^\nu_{\enspace x}\eta_{\mu\nu} \\ & = (\varLambda^{-1})^t_{\enspace t}(\varLambda^{-1})^\nu_{\enspace x}\eta_{t\nu} + (\varLambda^{-1})^x_{\enspace t}(\varLambda^{-1})^\nu_{\enspace x}\eta_{x\nu} + (\varLambda^{-1})^y_{\enspace t}(\varLambda^{-1})^\nu_{\enspace x}\eta_{y\nu} \\ & \qquad\quad + (\varLambda^{-1})^z_{\enspace t}(\varLambda^{-1})^\nu_{\enspace x}\eta_{z\nu}\\ \end{aligned}$

But the metric tensor given by Equation (4.6):

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

tells that

$\eta_{t\nu}$ is nonzero only when $\nu =t$ ;

$\eta_{x\nu}$ is nonzero only when $\nu =x$ ;

$\eta_{y\nu}$ is nonzero only when $\nu =y$ ;

$\eta_{z\nu}$ is nonzero only when $\nu =z$ .

So, the summation continues as follows:

$\begin{aligned} \eta'_{tx} & = (\varLambda^{-1})^t_{\enspace t}(\varLambda^{-1})^t_{\enspace x}\eta_{tt} + (\varLambda^{-1})^x_{\enspace t}(\varLambda^{-1})^x_{\enspace x}\eta_{xx} + (\varLambda^{-1})^y_{\enspace t}(\varLambda^{-1})^y_{\enspace x}\eta_{yy} \\ &\qquad\quad + (\varLambda^{-1})^z_{\enspace t}(\varLambda^{-1})^z_{\enspace x}\eta_{zz}\\ & = (\gamma )(\gamma\beta)(-1) + (\gamma\beta)(\gamma)(1) + (0)(0)(1) + (0)(0)(1) \\ & = -\gamma^2\beta + \gamma^2\beta \\ & = 0 \\ & \equiv \eta_{tx} \end{aligned}$

February 21, 2019January 19, 2023

201902210242 Exercise 5.4.1

Similarly, check that $\displaystyle{\frac{\partial x'^\mu}{\partial x^\nu}}=\varLambda^\mu_{\enspace\nu}$ when $\mu =x$ and $\nu =t$ and when $\mu =\nu =y$ .

Roughwork.

Equation (5.12):

$\begin{aligned} t' & = \gamma (t-\beta x) \\ x' & = \gamma (-\beta t+x) \\ y' & = y \\ z' & = z \end{aligned}$

and

$\begin{aligned} t & = \gamma (t'+\beta x') \\ x & = \gamma (\beta t'+x') \\ y & = y' \\ z & = z' \end{aligned}$

Equation (5.13a):

$\displaystyle{\frac{\partial x'^\mu}{\partial x^\nu}}=\varLambda^\mu_{\enspace\nu} =\begin{bmatrix} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$

Proof.

$\displaystyle{\frac{\partial t'}{\partial t}}=\gamma$ , $\displaystyle{\frac{\partial t'}{\partial x}}=-\gamma\beta$ , $\displaystyle{\frac{\partial t'}{\partial y}}=0$ , $\displaystyle{\frac{\partial t'}{\partial z}}=0$

$\displaystyle{\frac{\partial x'}{\partial t}}=-\gamma\beta$ , $\displaystyle{\frac{\partial x'}{\partial x}}=\gamma$ , $\displaystyle{\frac{\partial x'}{\partial y}}=0$ , $\displaystyle{\frac{\partial x'}{\partial z}}=0$

$\displaystyle{\frac{\partial y'}{\partial t}}=0$ , $\displaystyle{\frac{\partial y'}{\partial x}}=0$ , $\displaystyle{\frac{\partial y'}{\partial y}}=1$ , $\displaystyle{\frac{\partial y'}{\partial z}}=0$

$\displaystyle{\frac{\partial z'}{\partial t}}=0$ , $\displaystyle{\frac{\partial z'}{\partial x}}=0$ , $\displaystyle{\frac{\partial z'}{\partial y}}=0$ , $\displaystyle{\frac{\partial z'}{\partial z}}=1$

February 21, 2019March 29, 2024

201902210108 Exercise 5.3.2

Calculate all eight partial derivatives $\displaystyle{\frac{\partial x'^\mu}{\partial x^\nu}}$ and $\displaystyle{\frac{\partial x^\mu}{\partial x'^\nu}}$ .

Roughwork.

Equation (5.23):

$p(x,y)=x$ , $q(x,y)=y-cx^2$ ;

Equation (5.24):

$x(p,q)=p$ , $y(p,q)=cp^2+q$ .

Then, the eight partial derivatives are

$\displaystyle{\frac{\partial p}{\partial x}}=1$ , $\displaystyle{\frac{\partial p}{\partial y}}=0$ , $\displaystyle{\frac{\partial q}{\partial x}}=-2cx$ , $\displaystyle{\frac{\partial q}{\partial y}}=1$ ,

$\displaystyle{\frac{\partial x}{\partial p}}=1$ , $\displaystyle{\frac{\partial x}{\partial q}}=0$ , $\displaystyle{\frac{\partial y}{\partial p}}=2cp$ , $\displaystyle{\frac{\partial y}{\partial q}}=1$ .

Equation (5.10):

$\mathrm{d}s^2=g_{pq}\,\mathrm{d}p\,\mathrm{d}q$ .

Equation (5.11):

$g'_{\mu\nu}=\displaystyle{\frac{\partial x^\alpha}{\partial x'^\mu}}\displaystyle{\frac{\partial x^\beta}{\partial x'^\nu}}\, g_{\alpha\beta}$

The metric tensor for the cartesian coordinates is given by equation (5.25):

$g_{\alpha\beta}=\begin{bmatrix} 1&0\\0&1 \end{bmatrix}$

Using Equations (5.11) and (5.25):

$\begin{aligned} g_{pp} & =\displaystyle{\frac{\partial x^\alpha}{\partial p}}\displaystyle{\frac{\partial x^\beta}{\partial p}}g_{\alpha\beta} \\ & = \displaystyle{\frac{\partial x}{\partial p}}\displaystyle{\frac{\partial x}{\partial p}}g_{xx}+\displaystyle{\frac{\partial x}{\partial p}}\displaystyle{\frac{\partial y}{\partial p}}g_{xy}+\displaystyle{\frac{\partial y}{\partial p}}\displaystyle{\frac{\partial x}{\partial p}}g_{yx}+\displaystyle{\frac{\partial y}{\partial p}}\displaystyle{\frac{\partial y}{\partial p}}g_{yy}\\ & = (1)(1)(1) + (1)(2cp)(0) + (2cp)(1)(0) + (2cp)(2cp)(1) \\ &= 1+4c^2p^2 \end{aligned}$

$\begin{aligned} g_{pq} & =\displaystyle{\frac{\partial x^\alpha}{\partial p}}\displaystyle{\frac{\partial x^\beta}{\partial q}}g_{\alpha\beta} \\ & = \displaystyle{\frac{\partial x}{\partial p}}\displaystyle{\frac{\partial x}{\partial q}}g_{xx}+\displaystyle{\frac{\partial x}{\partial p}}\displaystyle{\frac{\partial y}{\partial q}}g_{xy}+\displaystyle{\frac{\partial y}{\partial p}}\displaystyle{\frac{\partial x}{\partial q}}g_{yx}+\displaystyle{\frac{\partial y}{\partial p}}\displaystyle{\frac{\partial y}{\partial q}}g_{yy}\\ & = (1)(0)(1) + (1)(1)(0) + (2cp)(0)(0) + (2cp)(1)(1) \\ &= 2cp \end{aligned}$

$\begin{aligned} g_{qp} & =\displaystyle{\frac{\partial x^\alpha}{\partial q}}\displaystyle{\frac{\partial x^\beta}{\partial p}}g_{\alpha\beta} \\ & = \displaystyle{\frac{\partial x}{\partial q}}\displaystyle{\frac{\partial x}{\partial p}}g_{xx}+\displaystyle{\frac{\partial x}{\partial q}}\displaystyle{\frac{\partial y}{\partial p}}g_{xy}+\displaystyle{\frac{\partial y}{\partial q}}\displaystyle{\frac{\partial x}{\partial p}}g_{yx}+\displaystyle{\frac{\partial y}{\partial q}}\displaystyle{\frac{\partial y}{\partial p}}g_{yy}\\ & = (0)(1)(1) + (0)(2cp)(0) + (2cp)(0)(0) + (1)(2cp)(1) \\ &= 2cp \end{aligned}$

$\begin{aligned} g_{qq} & =\displaystyle{\frac{\partial x^\alpha}{\partial q}}\displaystyle{\frac{\partial x^\beta}{\partial q}}g_{\alpha\beta} \\ & = \displaystyle{\frac{\partial x}{\partial q}}\displaystyle{\frac{\partial x}{\partial q}}g_{xx}+\displaystyle{\frac{\partial x}{\partial q}}\displaystyle{\frac{\partial y}{\partial q}}g_{xy}+\displaystyle{\frac{\partial y}{\partial q}}\displaystyle{\frac{\partial x}{\partial q}}g_{yx}+\displaystyle{\frac{\partial y}{\partial q}}\displaystyle{\frac{\partial y}{\partial q}}g_{yy}\\ & = (0)(0)(1) + (0)(1)(0) + (1)(0)(0) + (1)(1)(1) \\ &= 1 \end{aligned}$

The metric tensor for $p$ – $q$ coordinate system is given by Equation (5.26):

$g'_{\mu\nu}=\begin{bmatrix} 1+4c^2p^2 & 2cp \\ 2cp & 1 \\ \end{bmatrix}$ .