Moore’s A General Relativity Workbook – Page 2

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):

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

February 21, 2019January 19, 2023

201902210102 Exercise 3.1.1

$\mathbf{A}=\begin{bmatrix} A'^t\\ A'^x\\ A'^y\\A'^z \end{bmatrix}=\begin{bmatrix} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\\end{bmatrix}\begin{bmatrix} A^t \\ A^x \\ A^y \\ A^z \end{bmatrix}$

$\mathbf{B}=\begin{bmatrix} B'^t\\ B'^x\\ B'^y\\B'^z \end{bmatrix}=\begin{bmatrix} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\\end{bmatrix}\begin{bmatrix} B^t \\ B^x \\ B^y \\ B^z \end{bmatrix}$

are two four-vectors.

$\begin{aligned} A'^tB'^t & =(\gamma A^t-\gamma\beta A^x)(\gamma B^t-\gamma\beta B^x)\\ & = \gamma^2A^tB^t-\gamma^2\beta A^tB^x-\gamma^2\beta A^xB^t+\gamma^2\beta^2A^xB^x \end{aligned}$

$\begin{aligned} A'^xB'^x & = (-\gamma\beta A^t+\gamma A^x)(-\gamma\beta B^t+\gamma B^x)\\ & = \gamma^2\beta^2A^tB^t-\gamma^2\beta A^tB^x-\gamma^2\beta A^xB^t+\gamma^2A^xB^x\\ \end{aligned}$

$\begin{aligned} A'^yB'^y & = A^yB^y \end{aligned}$

$\begin{aligned} A'^zB'^z & = A^zB^z \end{aligned}$

$\begin{aligned} &\quad\enspace -A'^tB'^t+A'^xB'^x+A'^yB'^y+A'^zB'^z \\ & = (-\gamma^2+\gamma^2\beta^2)A^tB^t+(-\gamma^2\beta^2+\gamma^2)A^xB^x+A^yB^y+A^zB^z\\ & = -\gamma^2(1-\beta^2)A^tB^t+\gamma^2(1-\beta^2)A^xB^x+A^yB^y+A^zB^z\\ & = -\bigg( \frac{1}{\sqrt{1-\beta^2}} \bigg)^2(1-\beta^2)A^tB^t+\bigg( \frac{1}{\sqrt{1-\beta^2}} \bigg)^2(1-\beta^2)A^xB^x+A^yB^y+A^zB^z\\ & = -A^tB^t+A^xB^x+A^yB^y+A^zB^z \end{aligned}$

Scalar product of any two four-vectors is frame-independent.