Carroll, General Relativity

[2024/2/28]

Central intuition in GR

$\bullet$ acceleration due to gravity universal

Consider in classical mechanics: Newton's Second Law $$\vec{F}=m\vec{a}$$ where $m$ is inertial mass

Gravity is $$\vec{F}_g=q_g(-\vec{\nabla}\Phi)$$ and gravitational force for each other is same $$\frac{\vec{F}_{g,1}}{m_1}=\frac{q_{g,1}(-\vec{\nabla}\Phi)}{m_1}\ = \frac{\vec{F}_{g,2}}{m_2}=\frac{q_{g,2}(-\vec{\nabla}\Phi)}{m_2}$$ We find $\frac{q_g}{m}$ is constant for gravity case, which is $$m=q_g,$$ which called Weak Equivalence Principle, this means constant acceleration is same as constant gravity.

Then Strong Equivalence Principle is all laws of nature is same acceleration as gravity in any situation.

Then in freely-falling frame, there is no gravity. In each infinitesimal region of spacetime, the physics in local inertial frame is indistinguishable from that in special relativity.

Mathematical manifestation on idea

$\bullet$ Spacetime is a manifold ($\equiv$ $n$-dimensional manifold is a surface everywhere locally looks like $\mathbb{R}^n$.)

A manifold is given by "smoothly sewing together patches of $\mathbb{R}^n$"

Given some set $M$ a chart is an injective map from some subset $U\subset M$ $$\phi:U\rightarrow \mathbb{R}^n$$

A $C^\infty$ atlas of charts is an indexed collection of charts $\{(U_\alpha,\phi_\alpha)\}$ satisfiying

1. $\bigcup U_\alpha=M$

2. $U_\alpha\cap U_\beta\ne 0$ $$\phi_\alpha\cdot \phi_\beta^{-1}:\ \phi_\beta(U_\alpha\cap U_\beta)\rightarrow \phi_\alpha(U_\alpha \cap U_\beta)$$ onto and $C^\infty$

An $n$-dimensional manifold is a set $M$ with a maximal atlas.

$\bullet$ Tangent spaces vectors, tensors

Given $n$-dim $M$, the tangent space $T_p$ at a point $p\in M$ capturing notion of "tangent vectors"

Consider space $\mathcal{F}$ of all smooth functions on $M$ $$f:M\rightarrow \mathbb{R}$$

Each curve $\gamma(\lambda):\mathbb{R}\rightarrow M$ through $p$ defines a directional derivative $$f\mapsto \frac{d}{d\lambda}f(p)$$

The tangent space $T_p$ is the space of directional derivative operators along curves through $p$.

Can check derivative operators form a vector space closure given two curves $x^\mu(\lambda).x^\mu(\eta)$ $$\mathcal{D}\equiv (a\frac{d}{d\lambda}+b\frac{d}{d\eta}), \mathcal{D}(fg)=\mathcal{D}f\cdot g+f\cdot \mathcal{D}g$$

In practice, coordinate basis for $T_p$ chart with coordiantes $x^\mu$ $$\left\{ \partial_\mu=\frac{\partial}{\partial x^\mu}\right\}$$

Consider chart $\phi:M\rightarrow \mathbb{R}^n$ and curve $\gamma:\mathbb{R}\rightarrow M$ and function $f:M\rightarrow \mathbb{R}$ $$\frac{d}{d\lambda}f=\frac{d}{d\lambda}(f\circ \gamma(\lambda))=\frac{\partial(f\circ \phi^{-1})}{\partial x^\mu}\frac{d(\phi\circ \gamma)^\mu}{d\lambda}(f\circ \phi^{-1})\circ (\phi\circ \gamma)=\frac{dx^\mu}{d\lambda}\partial_\mu f$$

[Homework 1]

Problem 1 (Carroll, Exercise 2.2)

By clever choice of coordinate charts, can we make R2 look like a one-dimensional manifold? Can we make $R^1$ look like a two-dimensional manifold? If so, explicitly construct an appropriate atlas, and if not, explain why not. You might have to forget that you already know a definition of “open set” in the original $R^2$ or $R^1$, and define them as being appropriately inherited from the $R^1$ or $R^2$ to which they are being mapped.

Answer)

No. Chart's coordinate map can't be homeomorphism.

In definition of manifold, there are injective map from some subset $U\subset M$ $$\phi:U\rightarrow \mathbb{R}^n$$ For case of $M=\mathbb{R}^2$ with $n=1$, or $M=\mathbb{R}$ with $n=2$, map $\phi$ can't defined well. They can be some injective map with strange topology, but at least it is not in metric topology.

Problem 2 (Carroll, Exercise 2.3)

Show that the two-dimensional torus $T^2$ is a manifold, by explicitly constructing an appropriate atlas: (Not a maximal one, obviously.)

Answer)

$$T^2=\{(x,y,z)\in\mathbb{R}^3|(\sqrt{x^2+y^2}-1)^2+z^2=\frac{1}{4}\}$$ In cylindrical coordinates $x=r\cos\theta$, $y=r\sin\theta$, $z=z$, the equation of torus is $(r-1)^2+z^2=\frac{1}{4}$. We can make chart like $$(x,y,z)=((1+\frac{1}{2}\cos\varphi)\cos\theta,(1+\frac{1}{2}\cos\varphi)\sin\theta,\frac{1}{2}\sin\varphi)$$ when we set domain of chart like $\theta\times \varphi:(0,2\pi)\times (-\pi,\pi)$, region that we can't cover is

we need at least 3 chart to cover every region $(0,2\pi)\times (-\pi,\pi)$, $(-\pi,\pi)\times(-\frac{3}{2}\pi,\frac{1}{2}\pi)$, and $(-\frac{3}{2}\pi,\frac{1}{2}\pi)\times (-\frac{1}{2}\pi,\frac{3}{2}\pi)$, like

We can see the intersection of three color is $\varnothing$, which means charts cover all $T^2$, then became atlas.

Problem 3 (Carroll, Exercise 2.6)

Consider $R^3$ as a manifold with the flat Euclidean metric, and coordinates $(x, y, z)$. Introduce spherical polar coordinates $(r,\theta,\phi)$ related to $(x, y, z)$ by $$x=r\sin\theta\cos\phi\\ y=r\sin\theta\sin\phi\\ z=r\cos\theta,$$

(a) A particle moves along a parameterized curve given by $$x(\lambda)=\cos\lambda,\quad y(\lambda)=\sin\lambda,\quad z(\lambda)=\lambda.$$ Express the path of the curve in the $(r,\theta,\phi)$ system.

(b) Calculate the components of the tangent vector to the curve in both the Cartesian and spherical polar coordinate systems.

Answer)

(a)

$x=r\sin\theta\cos\phi=\cos\lambda$

$y=r\sin\theta\sin\phi=\sin\lambda$

Then, $\phi=\lambda$, $r\sin\theta=1$

$z=r\cos\theta=\lambda$

Then, $\tan\theta=1/\lambda$, $r=\sqrt{\lambda^2+1}$

Finally, $r(\lambda)=\sqrt{\lambda^2+1}, \phi(\lambda)=\lambda, \theta(\lambda)=\tan^{-1}\frac{1}{\lambda}$

(b)

[Cartesian]

location: $\cos\lambda\hat{x}+\sin\lambda\hat{y}+\lambda\hat{z}$

velocity: $-\sin\lambda\hat{x}+\cos\lambda\hat{y}+\hat{z}$

[Sperical]

location: $\sqrt{\lambda^2+1}\hat{r}$

basis ($t\equiv \lambda$) :

$\hat{r}=\sin\theta\cos\phi\hat{x}+\sin\theta\sin\phi\hat{y}+\cos\theta\hat{z}$

$\frac{d\hat{r}}{dt}=(\dot{\theta}\cos\theta\cos\phi-\dot{\phi}\sin\theta\sin\phi)\hat{x}+(\dot{\theta}\cos\theta\sin\phi+\dot{\phi}\sin\theta\cos\phi)\hat{y}\\ -\dot{\theta}\sin\theta\hat{z} = \dot{\phi}\sin\theta\hat{\phi}+\dot{\theta}\hat{\theta}$

$\hat{\theta}=\cos\theta\cos\phi\hat{x}+\cos\theta\sin\phi\hat{y}-\sin\theta\hat{z}$

$\frac{d\hat{\theta}}{dt}=(-\dot{\theta}\sin\theta\cos\phi-\dot{\phi}\cos\theta\sin\phi)\hat{x}+(-\dot{\theta}\sin\theta\sin\phi+\dot{\phi}\cos\theta\cos\phi)\hat{y}\\ -\dot{\theta}\cos\theta\hat{z}=-\dot{\theta}\hat{r}+\dot{\phi}\cos\theta\hat{\phi}$

$\hat{\phi}=-\sin\phi\hat{x}+\cos\phi\hat{y}$

$\frac{d\hat{\phi}}{dt}=-\dot{\phi}\cos\phi\hat{x}-\dot{\phi}\sin\phi\hat{y}=-\dot{\phi}\sin\theta\hat{r}-\dot{\phi}\cos\theta\hat{\theta}$

velocity: $\frac{\lambda}{\sqrt{\lambda^2+1}}\hat{r}+\sqrt{\lambda^2+1}(\frac{1}{\sqrt{\lambda^2+1}}\hat{\phi}-\frac{1}{\lambda^2+1}\hat{\theta})$

[2024/3/4]

The metric is a tensor which equips space of vectors ($T_p$) with inner product. $$g(V,W)=g_{\mu\nu}V^\mu W^\nu=V\cdot W \quad V,W\in T_p$$

The norm of $V$: $$\langle V,V\rangle=g_{\mu\nu}V^\mu V^\nu\quad \lVert v\rVert =\sqrt{|\langle V,V\rangle|}$$

Metric measures distance along (infinitesimal) displacement along $dx^\mu$ $$ds^2=g_{\mu\nu}dx^\mu dx^\nu$$

We usually specifying spacetime by giving metric in some coordinates.

In SR, spacetime is 4-dim Minkowski space $$ ds^2=-(dx^0)^2+(dx^1)^2+(dx^2)^2+(dx^3)^2$$

In GR, we consider more general spacetime

eg) 2-dim anti-de Sitter space $$ds^2=\frac{-d\phi^2+d\theta^2}{\cos^2\theta}\quad -\infty<\phi<\infty,\ -\frac{\pi}{2}<\theta<\frac{\pi}{2} $$

Recall notion of charts. We can switch between different charts "coordinate transformation" Given $\{x^\mu\}$ and $\{y^\alpha\}$, $$y^\alpha=\frac{\partial y^\alpha}{\partial x^\mu}x^\mu$$ (i think it's $dy^\alpha$ and $dx^\mu$.) $$g'_{\alpha\beta}(y)=g_{\mu\nu}(x)\frac{\partial x^\mu}{\partial y^\alpha}\frac{\partial x^\nu}{\partial y^\beta}$$ We can always perform some coordinate transformation in neighborhood around $p$ $$g_{\mu\nu}(p)=\mbox{diag}(-1,-1-1,1,1,0,\cdots,0)$$ We call the diagonal values the "signature" of the metric. Physical $n$-dim spacetimes have signature $(-1,1,\cdots,1)$

In Lorentzian signature, $$t\rightarrow i\tau\quad g_{\mu\nu}=(1,1,\cdots,1)$$

Let's consider trajectory of particle $x^\mu(\lambda)$. Its four-velocity is the vector $$U^\mu=\frac{dx^\mu}{d\tau}$$ when $d\tau=-\sqrt{-ds^2}$, $$U^\mu=\frac{dx^\mu(\lambda)}{d\lambda}\frac{d\lambda}{d\tau}$$ Which norm is $-1$.

Its four-momentum is the vector $$p^\mu=mU^\mu$$ which norm is $-m^2$. This is coordinate-independent expression.

Dual vectors (1form) and more general tensors

Vector is $$V=V^\mu\partial_\mu\in T_p$$ which coordinate index is in upper index, and dual vector (1-form) is $$\omega=\omega_\mu dx^\mu\in T_p^*$$ which coordinate index is in lower index. Mathematically, these form "cotangent space" at $p$, $T_p^*$, or space of all linear maps $T_p\rightarrow \mathbb{R}$ $$\omega:T_p\rightarrow \mathbb{R}\\ \omega(V)=\omega_\mu V^\mu$$ Double dual of vector is again vector.

More generally, a $(k,l)$ tensor is a multi-linear map from collection of $k$ dual vectors + $l$ vectors to $\mathbb{R}$. $$T:T^*_p\otimes \cdots \otimes $T^*_p\otimes T_p\otimes \cdots \otimes T_p\rightarrow \mathbb{R}$$ It's specified by its components $$T^{\mu_1,\cdots,\mu_k}_{\nu_1,\cdots,\nu_l}=T(dx^{\mu_1},\cdots,dx^{\mu_k},\partial_{\nu_1},\cdots,\partial_{\nu_l})$$ Under coordinate transformation, $$T'^{\alpha_1,\cdots,\alpha_k}_{\beta_1,\cdots,\beta_l}(y)=T^{\mu_1,\cdots,\mu_k}_{\nu_1,\cdots,\nu_l}(x)\frac{\partial y^{\alpha_1}}{\partial x^{\mu_1}}\cdots \frac{\partial y^{\alpha_k}}{\partial x^{\mu_k}}\frac{\partial x^{\nu_1}}{\partial y^{\beta_1}}\cdots \frac{\partial x^{\nu_l}}{\partial y^{\beta_l}}$$

metric tensor $g_{\mu\nu}$ is (0,2) tensor we can define a $(2,0)$ tensor which is the inverse on metric $$g_{\mu\nu}g^{\nu\rho}=\delta_\mu^\rho$$

[3/6]

We can raise or lower tensor's indices with $$g_{\mu\nu}\quad g^{\mu\nu}$$

Symmetrize indices $$A_{(\mu,\nu)}=\frac{1}{2}(A_{\mu\nu}+A_{\nu\mu})$$ $$A_{[\mu\nu]}=\frac{1}{2}(A_{\mu\nu}-A_{\nu\mu})$$

Jacobian is determinant of infinitesimal coordinate transformation $x^\mu\rightarrow y^\alpha$, $$J\equiv =\left| \frac{\partial y}{\partial x}\right|$$

$$W'^{\alpha_1\cdots \alpha_k}_{\beta_1\cdots\beta_l}(y)=W^{\mu_1\cdots\mu_k}_{\nu_1\cdots \nu_l}(x)\frac{\partial y^{\alpha_1}}{\partial x^{\mu_1}}\cdots \frac{\partial y^{\alpha_k}}{\partial x^{\mu_k}}\frac{\partial x^{\nu_1}}{\partial y^{\beta_1}}\cdots \frac{\partial x^{\nu_l}}{\partial y^{\beta_l}} \left| \frac{\partial y}{\partial x}\right|^P$$ when $P$ is weight (which needs for tensor density, not just tensor)

Levi-Civita symbol is $$\tilde{\epsilon}_{\mu_1\cdots\mu_n}=\begin{cases}1&(\mu_1\cdots\mu_n) \mbox{ even perm of }(0,1,\cdots,n-1)\\ -1 & \mbox{" odd " "}\\ 0&\mbox{otherwise}\end{cases}$$

Weight of Levi-Civita symbol is +1. We want to make this tensor.

Consider determinant of the metric $$g=|g_{\mu\nu}|$$ $$g'_{\alpha\beta}(y)=g_{\mu\nu}(x)\frac{\partial x^\mu}{\partial y^\alpha}\frac{\partial x^\nu}{\partial y^\beta}$$ $$g'(y)=g(x)\left| \frac{\partial x}{\partial y}\right|^2$$ So $g$ transforms as a scalar density with weight -2.

Levi-Civita tensor is $$\epsilon_{\mu_1\cdots\mu_n}=\sqrt{|g|}\tilde{\epsilon}_{\mu_1\cdots \mu_n}$$ transform as a tensor. In upper indices, $$\epsilon^{\mu_1\cdots\mu_n}=\frac{1}{\sqrt{|g|}}\tilde{\epsilon}^{\mu_1\cdots \mu_n}$$ when $\tilde{\epsilon}^{\mu_1\cdots\mu_n}=\mbox{sgn}(g)\tilde{\epsilon}_{\mu_1\cdots\mu_n}$

Volume, volume element

Integration in $\mathbb{R}^n$ is $\int d^nx\, f(x)$

Integration in spacetime $\int d^n\, \sqrt{|g|} f(x)$

when $$d^nx\sqrt{|g|}=\epsilon_{\mu_1\mu_2\cdots\mu_n}dx^{\mu_1}\otimes dx^{\mu_2}\otimes\cdots\otimes dx^{\mu_n}$$

Why? Idea: The volume element in $n$-dim is a completely antisymmetric $(0,n)$ tensor. It takes $n$ vectors to real number $$\epsilon(V_1,\cdots,V_n)$$ so volume element is $(0,n)$ tensor, and we want 'signed' volume, so is have to be completely antisymmetric.

We also get volume element is coordinate invariant. $$d^ny\sqrt{|g'(y)|}=d^nx\sqrt{|g(x)|}$$

Classical Field Theory (covariant formulation of electrodynamics)

Relativistic (respects covariance of spacetime) field theory: $$S=\int d^nx\sqrt{|g|}\mathcal{L}(\Phi^i,\nabla_\mu\Phi^i)$$

Extremizing action with fields gives Euler-Lagrange eq. solve for (classical) configuration $$\frac{\delta S}{\delta \Phi^i}=\frac{\partial \mathcal{L}}{\partial \Phi^i}-\nabla_\mu\left( \frac{\partial\mathcal{L}}{\partial (\nabla_\mu\Phi^i)}\right)=0$$ $$\delta \mathcal{L}=\frac{\partial \mathcal{L}}{\partial \Phi^i}\delta\Phi^i+\frac{\partial \mathcal{L}}{\partial (\nabla_\mu\Phi^i)}\nabla_\mu\delta \Phi^i$$ $$\Phi^i\rightarrow \Phi^i+\delta \Phi^i$$ $$\nabla_\mu\Phi^i\rightarrow \nabla_\mu\Phi^i+\nabla_\mu \delta\Phi^i$$ $$\delta S=\int d^nx\sqrt{|g|}\left(\frac{\partial\mathcal{L}}{\partial\Phi^i}\delta \Phi^i+\frac{\partial \mathcal{L}}{\partial(\nabla_\mu\Phi^i)}\nabla_\mu\delta\Phi^i\right)$$ By IBP of second term and $\nabla_\mu g_{\alpha\beta}=0$ (Compatible), $$\rightarrow \int d^nx\sqrt{|g|}\left(\frac{\partial\mathcal{L}}{\partial \Phi^i}-\nabla_\mu\left(\frac{\partial\mathcal{L}}{\partial (\nabla_\mu \Phi^i)}\right)\right)\delta\Phi^i$$

ex) EM as covariant field theory

Define EM field strength $$F_{\mu\nu}=\nabla_\mu A_\nu-\nabla_\nu A_\mu$$ $$A^\mu=(\Phi,A^i),\quad J^\mu=(\rho,J^i)$$ $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+A_\mu J^\mu$$ $$\mathcal{\partial \mathcal{L}}{\partial A_\nu}-\nabla_\mu \left( \frac{\partial\mathcal{L}}{\partial (\nabla_\mu A_\nu)}\right)=J^\nu -\nabla_\mu(-F_{\mu\nu})=0$$ $$J^\nu=\nabla_\mu F^{\nu\mu}$$

Energy-momentum tensor

Symmetric tensor $T_{\mu\nu}$: "flux of four-momentum $p^\mu$ across surface of constant $x^\nu$"

$T^{00}$ is energy density

$T^{i0}=T^{0i}$ is momentum density

$T^{ij}$ is momentum flux

EM tensor is always conserved.

$$\nabla_\mu T^{\mu\nu}=0$$

[Homework 2]

Problem 1 (Carroll, Exercise 1.8)

If $\partial_\nu T^{\mu\nu}=Q^\mu$, what physically does the spatial vector $Q^i$ represent? Use the dust energy momentum tensor to make your case.

Sol) Dust energy momentum tensor is $$T^{\mu\nu}=p^\mu N^\nu=mn U^\mu U^\nu=\rho U^\mu U^\nu,$$ when $U^\mu=\frac{dx^\mu}{d\tau}$, $p^\mu=mU^\mu$, and pressure $p_i=T^{ii}$ and number-flux four-vector $N^\mu=nU^\mu$. Then $$\partial _\mu T^{\mu\nu}=(\partial_\mu \rho) U^\mu U^\nu+\rho(U^\nu\partial_\mu U^\mu+U^\mu \partial_\mu U^\nu).$$ By normalization $U_\nu U^\nu=-1$ implies the useful identity $$U_\nu \partial_\mu U^\nu=\frac{1}{2}\partial_\mu(U_\nu U^\nu)=0.$$ And project like $$U_\nu \partial_\mu T^{\mu\nu}=-\partial_\mu(\rho U^\mu)$$

Also, we can project with projection tensor $$P^\sigma_\nu=\delta^\sigma_\nu+U^\sigma U_\nu.$$ which satisfy $$P^\sigma_\nu V^\nu_\parallel=0\\ P^\sigma_\nu W^\nu_\perp=W^\sigma_\perp.$$ when $V^\mu_\parallel$ is parallel to $U^\mu$ and $W^\mu_\perp$ is perpendicular to $U^\mu$. Result is $$P^\sigma_\nu \partial_\mu T^{\mu\nu}=\rho U^\mu \partial_\mu U^\sigma$$

$Q^\mu$ means $T^{\mu\nu}$ is not conserved current.

Problem 2 (Carroll, Exercise 1.10)

Using the tensor transformation law applied to $F_{\mu\nu}$, show how the electric and magnetic field 3-vectors $\mathbf{E}$ and $\mathbf{B}$ transform under

(a) a rotation about the $y$-axis,

(b) a boost along the $z$-axis.

Sol) $$F_{\mu\nu}=\begin{pmatrix}0&-E_1&-E_2&-E_3\\ E_1& 0&B_3&-B_2\\ E_2&-B_3&0&B_1\\E_3&B_2&-B_1&0\end{pmatrix}$$

(a)

rotation about the $y$-axis is $$\Lambda^\mu_\alpha=\begin{pmatrix}1 &0&0&0\\ 0&\cos\theta&0&-\sin\theta\\ 0&0&1&0\\ 0& \sin\theta&0&\cos\theta\end{pmatrix}$$

$$\Lambda_\alpha^\mu\Lambda_\beta^\nu F_{\mu\nu}=\\ \begin{pmatrix}1 &0&0&0\\ 0&\cos\theta&0&\sin\theta\\ 0&0&1&0\\ 0&- \sin\theta&0&\cos\theta\end{pmatrix} \begin{pmatrix}0&-E_1&-E_2&-E_3\\ E_1& 0&B_3&-B_2\\ E_2&-B_3&0&B_1\\E_3&B_2&-B_1&0\end{pmatrix} \begin{pmatrix}1 &0&0&0\\ 0&\cos\theta&0&-\sin\theta\\ 0&0&1&0\\ 0& \sin\theta&0&\cos\theta\end{pmatrix}=\\ \begin{pmatrix}0&-E_1\cos\theta-E_3\sin\theta& -E_2& -E_3\cos\theta+E_1\sin\theta\\ E_1\cos\theta+E_3\sin\theta&0&B_3\cos\theta-B_1\sin\theta&-B_2\\ E_2&-B_3\cos\theta+B_1\sin\theta&0&B_1\cos\theta+B_3\sin\theta\\ E_3\cos\theta-E_1\sin\theta& B_2& -B_1\cos\theta-B_3\sin\theta&0\end{pmatrix}$$

Finally, $$(E_1,E_2,E_3)\rightarrow (E_1\cos\theta+E_3\sin\theta,E_2,E_3\cos\theta-E_1\sin\theta)$$ which is rotation of 3-vector. Also, $$(B_1,B_2,B_3)\rightarrow (B_1\cos\theta+B_3\sin\theta,B_2,B_3\cos\theta-B_1\sin\theta)$$ which is rotation of 3-vector.

(b)

boost is $$\Lambda_\alpha^\mu=\begin{pmatrix}\cosh\phi&0&0&-\sinh\phi\\ 0&1&0&0\\ 0&0&1&0\\ -\sinh\phi&0&0&\cosh\phi \end{pmatrix}$$ when $\phi=\tanh^{-1}v$. $$\Lambda_\alpha^\mu\Lambda_\beta^\nu F_{\mu\nu}=\begin{pmatrix}0&-E_1\cosh\theta-B_2\sinh\theta&-E_2\cosh\theta+B_1\sinh\theta&-E_3\\ E_1\cosh\theta+B_2\sinh\theta&0&B_3&-B_2\cosh\theta-E_1\sinh\theta\\ E_2\cosh\theta B_1\sinh\theta& -B_3&0&B_1\cosh\theta -B_2\sinh\theta\\ E_3& B_2\cosh\theta +E_1\sinh\theta& -B_1\cosh\theta +E_2\sinh\theta &0\end{pmatrix}$$ which is $$(E_1,E_2,E_3)\rightarrow (E_1\cosh\theta+B_2\sinh\theta,E_2\cosh\theta-B_1\sinh\theta,E_3)$$ and $$(B_1,B_2,B_3)\rightarrow (B_1\cosh\theta-B_2\sinh\theta,B_2\cosh\theta+E_1\sinh\theta,B_3)$$ these means blend of electric field and magnetic field.

Problem 3 (Carroll, Exercise 1.13)

Consider adding to the Lagrangian for electromagnetism an additional term of the form $\mathcal{L}'=\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma}$.

(a) Express $\mathcal{L}'$ in terms of $\mathbf{E}$ and $\mathbf{B}$.

(b) Show that including $\mathcal{L}'$ does not affect Maxwell's equations. Can you think of a physical reason?

Sol)

(a) $$\frac{1}{4}\tilde{\epsilon}^{\mu\nu\rho\sigma}F_{\rho\sigma}F_{\mu\nu}=\frac{1}{c}\mathbf{E}\cdot\mathbf{B}$$ $$\frac{1}{4}F^{\mu\nu}F_{\mu\nu}=\frac{\epsilon_0}{2}\mathbf{E}^2-\frac{1}{2\mu_0}\mathbf{B}^2$$

(b) This action term can be written as a total dericative, $$S=\frac{1}{2}\int d^4x\, \partial_\mu(\epsilon^{\mu\nu\rho\sigma}A_\nu \partial_\rho A_\sigma)$$ So when boundary is very large and if we can ignore boundary effect, we can ignore this lagrangian term. Also this Lagrangian is only differ from boundary, instead of local metric, since there is no contraction with $g^{\mu\nu}$, but only $\epsilon^{\mu\nu\rho\sigma}$, so it doesn't get physics locally.

Problem 4 (Carroll, Exercise 2.7)

Prolate spheroidal coordinates can be used to simplify the Kepler problem in celestial mechanics. They are related to the usual cartesian coordinates $(x,y,z)$ of Euclidean three-space by $$x=\sinh \chi \sin \theta\cos\phi,\\ y=\sinh\chi \sin\theta\sin\phi,\\ z=\cosh\chi\cos\theta.$$ Restrict your attention to the plane $y=0$ and answer the following questions.

(a) What is the coordiante transformation matrix $\partial x^\mu/\partial x^{\nu'}$ relating $(x,z)$ to $(\chi,\theta)$?

(b) What does the line element $ds^2$ look like in porlate spheroidal coordinates?

Sol)

(a) Since $y=0$, $x^2+y^2=\sinh^2\chi\sin^2\theta$, $$x=\pm \sinh\chi\sin\theta$$ then $$\frac{\partial x}{\partial \chi}=\pm\cosh\chi\sin\theta$$ $$\frac{\partial z}{\partial\chi}=\sinh\chi\cos\theta$$ $$\frac{\partial x}{\partial \theta}=\pm \sinh\chi \cos\theta$$ $$\frac{\partial z}{\partial \theta}=-\cosh\chi \sin\theta$$ then $$\frac{\partial x^\mu}{\partial x^{\nu'}}=\begin{pmatrix}\pm \cosh\chi\sin\theta& \sinh\chi \cos\theta\\ \pm \sinh\chi \cos\theta & -\cosh\chi\sin\theta\end{pmatrix}$$

(b) $$ds^2=dx^2+dz^2=(\cosh^2\chi\sin^2\theta+\sinh^2\chi\cos^2\theta)(d\chi^2+d\theta^2)$$

[3/11]

So far, manifold, tangent space, metric $$V,W\in T_p$$ $$g(V,W)=g_{\mu\nu}V^\mu V^\nu$$ $g_{\mu\nu}$ specifies geometry, e.g. length of curve, volume of region.

Also want to define curvature at a point of spacetime. Curvature has to do with non-commutativity in transporting vectors.

Methods of transporting vectors along a path.

1. Covariant derivatives, parallel transport

Motivation: take derivatives in coordinate-independent way, derivative operator takes tensors to tensors.

Ordinary partial derivative $\partial_\mu V^\nu$: Does it transform as tensor?

$$\partial_\alpha V'^\beta=\partial_\nu V^\mu\frac{\partial y^\beta}{\partial x^\mu}\frac{\partial x^\nu}{\partial y^\alpha}$$ $$\partial_\alpha\left( V^\mu \frac{\partial y^\beta}{\partial x^\mu}\right)= \left( \frac{\partial x^\nu}{\partial y^\alpha}\partial_\nu V^\mu\right) +V^\mu\frac{\partial x^\nu}{\partial y^\alpha}\frac{\partial^2 y^\beta}{\partial x^\nu \partial x^\mu}$$ there is extra second term. Then we define $$\nabla_\mu V^\nu=\partial_\mu V^\nu+\Gamma^\nu_{\mu\lambda}V^\lambda$$ when $$\Gamma^\lambda_{\mu\nu}=\frac{1}{2}g^{\lambda\sigma}(\partial_\mu g_{\nu\sigma}+\partial_\nu g_{\mu\sigma}-\partial_\sigma g_{\mu\nu})$$ is the Christoffel symbol, $\nabla_\mu V^\nu$ indeed transforms as tensor.

Using the Christoffel symbol, we can take covariant derivatives of tensors as follows: $$\nabla_\mu \omega_\nu =\partial_\mu \omega_\nu -\Gamma^\lambda_{\mu\nu}\omega_\lambda$$ $$\nabla_\sigma T^{\mu_1\cdots\mu_k}_{\nu_1\cdots\nu_l}\\ =\partial_\sigma T^{\mu_1\cdots\mu_k}_{\nu_1\cdots\nu_l}+\Gamma^{\mu_1}_{\sigma\lambda}T^{\lambda\mu_2\cdots\mu_k}_{\nu_1\cdots\nu_l}+\Gamma^{\mu_2}_{\sigma\lambda}T^{\mu_1\lambda\cdots\mu_k}_{\nu_1\cdots\nu_l}+\cdots -\Gamma^\lambda_{\sigma\nu_1}T^{\mu_1\cdots\mu_k}_{\lambda\nu_2\cdots\nu_l}-\Gamma^\lambda_{\sigma\nu_2}T^{\mu_1\cdots\mu_k}_{\nu_1\lambda\cdots\nu_l}-\cdots.$$

Then covariant derivatives satisfies following nice properties

$\bullet$ Leibniz rule $$\nabla_\mu (V^{\rho\sigma}T_{\sigma\kappa})=(\nabla_\mu V^{\rho\sigma})T_{\sigma\kappa}+V^{\rho\sigma}(\nabla_\mu T_{\sigma\kappa})$$

$\bullet$ Commutes. e.g. $$\nabla_\mu (T^\lambda_{\lambda\rho})=(\nabla T)^\lambda_{\lambda\rho}$$

$\bullet$ on scalars $$\nabla_\mu \phi=\partial_\mu\phi$$

Covariant derivative specifies how to transport vectors on manifold

Coefficients $T^\sigma_{\mu\nu}$ are specifying a connection.

There are many connections on a manifold.

However Christoffel symbol gives unique connection that is

$\bullet$ torsion-free $$\Gamma^\lambda_{\mu\nu}=\Gamma^\lambda_{(\mu\nu)}$$

$\bullet$ metric-compatible, i.e. $$\nabla_\alpha g_{\mu\nu}=0$$ In GR, we use Christoffel connection.

Then e.g. we have covariant Stokes' Theorem $$\int_\Sigma d^nx\sqrt{|g|}\nabla_\mu V^\mu=\int_{\partial\Sigma}d^{n-1}x\sqrt{|\gamma|}n_\mu V^\mu$$ when $\gamma$ metric on surface.

We can define notion of parallel transport.

A tensor is parallel-transported along $x^\mu(\lambda)$ $$\left( \frac{D}{d\lambda}T\right)^{\mu_1\cdots\mu_k}_{\nu_1\nu_2\cdots\nu_l}=\frac{dx^\sigma}{d\lambda}\nabla_\sigma T^{\mu_1\cdots\mu_k}_{\nu_1\cdots\nu_l}=0$$

Note the metric is always parallel-transported $$\frac{D}{d\lambda}g_{\mu\nu}=\frac{dx^\sigma}{d\lambda}\nabla_\sigma g_{\mu\nu}=0$$ $\Rightarrow$ The inner product between two parallel-transported vectors is preserved.

2) Diffeomorphisms, Lie derivatives

There is another way of transporting vectors, does not require connection.

$\bullet$ Diffeomorphism $$\phi:M\rightarrow N$$ which is one-to-one and onto and $\phi$, $\phi^{-1}$ are both smooth.

$$\phi:M\rightarrow M',\quad $f:M'\rightarrow \mathbb{R},\quad \phi^*f:M\rightarrow \mathbb{R} $$ pull-back of $f$: $(\phi^* f)(p)\equiv f(\phi(p))$

push-forward of $V$: $(\phi_* V)(f)\equiv V(\phi^* f)$

Previously, discussed coordinate transformations involving two different charts on manifold $M$, $\{x^\mu\}$ and $\{y^\mu\}$

Alternative way of viewing coordinate transformations

$\bullet$ passive way: same manifold $M$, there is two coordinate $\{x^\mu\}$ and $\{y^\mu\}$.

$\bullet$ active way: $$\phi: M\rightarrow M',\quad x^\mu:M'\rightarrow \mathbb{R}^n,\quad \phi^*x^\mu=y^\mu$$ $$y^\mu(p)=(\phi^*x^\mu)(p)=x^\mu (\phi(p))$$

Tensor transformations we have discussed actually describe the pull-back of tensors $$\phi^* T)^{\alpha_1\cdots\alpha_k}_{\beta_1\cdots\beta_l}=\frac{\partial y^{\alpha_1}}{\partial x^{\mu_1}}\cdots \frac{\partial y{^\alpha_k}}{\partial x^{\mu_l}}\frac{\partial x^{\nu_1}}{\partial y^{\beta_1}}\cdots \frac{\partial x^{\nu_l}}{\partial y^{\beta_l}}T^{\mu_1\cdots\mu_k}_{\nu_1\cdots\nu_l}(\phi(p))$$

$\bullet$ Lie derivatives

A vector field $V^\mu$ on $M$ we can construct integral curves satisfying $$\frac{dx^\mu}{dt}=V^\mu$$ consider diffeomorphism $\phi_t$ representing "flow down integral curves" For a fixed $t>0$ $$\Delta_t T^{\mu_1\cdots\mu_k}_{\nu_1\cdots\nu_l}(p)=\phi^*_t[T^{\mu_1\cdots\mu_k}_{\nu_1\cdots\nu_l}(\phi_t(p))]-T^{\mu_1\cdots\mu_k}_{\nu_1\cdots\nu_l}(p)$$ Lie derivative $$\mathcal{L}_VT^{\mu_1\cdots\mu_k}_{\nu_1\cdots\nu_l}=\lim_{t\rightarrow 0}\frac{\Delta_t T^{\mu_1\cdots\mu_k}_{\nu_1\cdots\nu_l}}{t}$$

[Homework 3]

Problem 1 (Carroll, Exercise 3.4)

In Euclidean three-space, we can define paraboloidal coordinates $(u,v,\phi)$ via $$x=uv\cos\phi\quad y=uv\sin\phi\quad z=\frac{1}{2}(u^2-v^2)$$

(a) Find the coordinate transformation matrix between paraboloidal and Cartesian coordinates $\partial x^\alpha/\partial x^{\beta'}$ and the inverse transformation. Are there any singular points in the map?

(b) Find the basis vectors and basis one-forms in terms of Cartesian basis vectors and forms.

(d) Calculate the Christoffel symbols.

(e) Calculate the divergence $\nabla_\mu V^\mu$ and Laplacian $\nabla_\mu\nabla^\mu f$.

Sol)

(a) $$\frac{\partial x^\alpha}{\partial x^{\beta'}}=\begin{pmatrix}v\cos\phi& u\cos\phi& -uv\sin\phi\\ v\sin\phi& u\sin\phi& uv\cos\phi\\ u& -v& 0\end{pmatrix}$$

$$\det = (u)u^2v+(-v)(-uv^2)=uv(u^3+v^3)$$

$$u= \sqrt{\sqrt{z^2+x^2+y^2}+z},\ v=\pm\sqrt{\sqrt{z^2+x^2+y^2}-z} ,\ \phi=\tan^{-1}\frac{y}{x}$$

$$\frac{\partial x^{\beta'}}{\partial x^\alpha}=\begin{matrix}\frac{x}{2\sqrt{z^2+x^2+y^2}\sqrt{\sqrt{z^2+x^2+y^2}+z}}& \frac{y}{2\sqrt{z^2+x^2+y^2}\sqrt{\sqrt{z^2+x^2+y^2}+z}}& \frac{z+\sqrt{z^2+x^2+y^2}}{2\sqrt{z^2+x^2+y^2}\sqrt{\sqrt{z^2+x^2+y^2}+z}}\\ \frac{x}{2\sqrt{z^2+x^2+y^2}\sqrt{\sqrt{z^2+x^2+y^2}-z}}& \frac{y}{2\sqrt{z^2+x^2+y^2}\sqrt{\sqrt{z^2+x^2+y^2}-z}}& \frac{z-\sqrt{z^2+x^2+y^2}}{2\sqrt{z^2+x^2+y^2}\sqrt{\sqrt{z^2+x^2+y^2}-z}}\\ \frac{-y}{x^2+y^2}& \frac{x}{x^2+y^2}& 0\end{matrix}$$

(b) $$\hat{u}=\frac{\partial }{\partial u}=\frac{\partial x}{\partial u}\hat{x}+\frac{\partial y}{\partial u}\hat{z}+\frac{\partial z}{\partial u}\hat{z}=v\cos\phi\hat{x}+u\cos\phi\hat{y}-uv\sin\phi\hat{z}\\ =\frac{x\sqrt{\sqrt{z^2+x^2+y^2}+z}}{\sqrt{x^2+y^2}}\hat{x}+\frac{x\sqrt{\sqrt{z^2+x^2+y^2}-z}}{\sqrt{x^2+y^2}}\hat{y}-y\hat{z}$$

$$\hat{v}=\frac{\partial }{\partial v}=\frac{\partial x}{\partial v}\hat{x}+\frac{\partial y}{\partial v}\hat{z}+\frac{\partial z}{\partial v}\hat{z}=v\sin\phi\hat{x}+u\sin\phi\hat{y}+uv\cos\phi\hat{z}$$

$$\hat{\phi}=\frac{\partial }{\partial \phi}=\frac{\partial x}{\partial \phi}\hat{x}+\frac{\partial y}{\partial \phi}\hat{z}+\frac{\partial z}{\partial \phi}\hat{z}=u\hat{x}-v\hat{y}$$

$$\check{u}=du=\frac{\partial u}{\partial x}\check{x}+\frac{\partial u}{\partial y}\check{y}+\frac{\partial u}{\partial z}\check{z}=\frac{\check{x}}{v\cos\phi}+\frac{\check{y}}{u\cos\phi}+\frac{\check{z}}{-uv\sin\phi}$$

$$\check{v}=dv=\frac{\partial v}{\partial x}\check{x}+\frac{\partial v}{\partial y}\check{y}+\frac{\partial v}{\partial z}\check{z}=\frac{\check{x}}{v\sin\phi}+\frac{\check{y}}{u\sin\phi}+\frac{\check{z}}{uv\cos\phi}$$

$$\check{\phi}=d\phi=\frac{\partial \phi}{\partial x}\check{x}+\frac{\partial \phi}{\partial y}\check{y}+\frac{\partial \phi}{\partial z}\check{z}=\frac{\check{x}}{u}-\frac{\check{y}}{v}$$

(c) $$g_{ij}dx^idx^j=dx^2+dy^2+dz^2=(v\cos\phi\check{u}+u\cos\phi\check{v}-uv\sin\phi\check{\phi})^2+(v\sin\phi\check{u}+u\sin\phi\check{v}+uv\cos\phi\check{\phi})^2+(u\check{u}-v\check{v})^2\\ =(u^2+v^2)\check{u}^2+(u^2+v^2)\check{v}^2+(u^2v^2)\check{\phi}^2$$

$$g_{ij}=\begin{pmatrix}u^2+v^2&0&0\\ 0& u^2+v^2&0\\ 0&0&u^2v^2\end{pmatrix}$$

$$g^{ij}=\frac{1}{u^6v^2+2u^4v^4+u^2v^6}\begin{pmatrix}u^4v^2+u^2v^4&0&0\\ 0&u^4v^2+u^2v^4&0\\ 0&0&(u^2+v^2)^2\end{pmatrix}$$

(d) In Levi-Civita connection, $$\Gamma^i_{kl}=\frac{1}{2}g^{im}\left(\frac{\partial g_{mk}}{\partial x^l}+\frac{\partial g_{ml}}{\partial x^k}-\frac{\partial g_{kl}}{\partial x^m}\right)$$

$$\frac{\partial g_{ij}}{\partial u}=\begin{pmatrix}2u&0&0\\ 0& 2u&0\\ 0&0&2uv^2\end{pmatrix}$$

$$\frac{\partial g_{ij}}{\partial v}=\begin{pmatrix}2v&0&0\\ 0& 2v&0\\ 0&0&2vu^2\end{pmatrix}$$

$$\frac{\partial g_{ij}}{\partial \phi}=0$$

$$\Gamma^1_{kl}=\frac{1}{u^2+v^2}\begin{pmatrix}u&v&0\\ v&-u&0\\ 0&0&-uv^2\end{pmatrix}$$

$$\Gamma^2_{kl}=\frac{1}{u^2+v^2}\begin{pmatrix}-v&u&0\\ u&v&0\\ 0&0&-u^2v\end{pmatrix}$$

$$\Gamma^3_{kl}=\frac{1}{uv}\begin{pmatrix}0&0&v\\ 0&0&u\\ v&u&0\end{pmatrix}$$

(e)

$$\nabla^2 f=\frac{1}{v^2+u^2}\left[ \frac{1}{v}\frac{\partial}{\partial v}\left( v\frac{\partial f}{\partial v}\right)+\frac{1}{u}\frac{\partial}{\partial u}\left( u\frac{\partial f}{\partial u}\right)\right] +\frac{1}{u^2v^2}\frac{\partial^2 f}{\partial \phi^2}$$

Problem 2 (Carroll, Exercise 3.5)

Consider a 2-sphere with coordinates $(\theta,\phi)$ and metric $$ds^2=d\theta^2+\sin^2 \theta d\phi^2.$$

(a) Show that lines of constant longitude ($\phi=$constant) are geodesics, and that the only line of constant latitude ($\theta$=constant) that is a geodesic is the equator ($\theta=\pi/2$).

(b) Take a vector with components $V^\mu=(1,0)$ and parallel-transport if once around a circle of constant latitude. What are the components of the resulting vector, as a function of $\theta$?

Sol)

(a) $$\ddot{x}^\mu +\Gamma^\mu_{\nu\rho}\dot{x}^\nu\dot{x}^\rho=0$$

$$\Gamma^1_{ij}=\begin{pmatrix}0&0\\ 0&-\cos\theta\sin\theta\end{pmatrix}$$

$$\Gamma^2_{ij}=\begin{pmatrix}0&\cot\theta\\ \cot\theta&0\end{pmatrix}$$

$$\ddot{\theta}-\cos\theta\sin\theta\dot{\phi}^2=0$$

$$\ddot{\phi}+2\cot\theta\dot{\phi}\dot{\theta}=0$$

$\phi$=const: $\dot{\phi}=0$, $\ddot{\phi}=0$, then $$\ddot{\theta}=0\\ 0=0$$ $$\theta=At+B$$

$\theta$=const: $\dot{\theta}=0$, $\ddot{\theta}=0$, then $$-\cos\theta\sin\theta\dot{\phi}^2=0\\ \ddot{\phi}=0$$ $$\phi=A't+B,\ A'\cos\theta\sin\theta=0$$

$A'=0$: $\phi$=const is same as above,

$A'\ne 0$: $\theta=0,\frac{\pi}{2}$, $\theta=0$ is not line, $\theta=\frac{\pi}{2}$.

(b) $$\dot{V^\mu}+\Gamma^\mu_{\nu\rho}V^\nu\dot{x^\rho}=0$$ when $\dot{\theta}=0$

$$\phi=t$$

$$\dot{V}^1-\cos\theta\sin\theta V^2\dot{\phi}=0$$

$$\dot{V}^2+\cot\theta V^1\dot{\phi}=0$$

$$\ddot{V}^1=\cos\theta\sin\theta \dot{V}^2=-\cos^2\theta V^1$$

$$\ddot{V}^2=-\cos^2\theta V^2$$

$$V^\mu=(\cos(t\cos\theta ),\sin(t\cos\theta ))$$

Problem 3 (Carroll, Exercise 3.6)

A good approximation to the metric outside the surface of the Earth is provided by $$ds^2=-(1+2\Phi)dt^2+(1-2\Phi)dr^2+r^2(d\theta^2+\sin^2\theta d\phi^2)$$ where $$\Phi=-\frac{GM}{r}$$ may be thought of as the familiar Newtonian gravitational potential. Here $G$ is Newton's constant and $M$ is the mass of the earth. For this problem $\Phi$ may be assumed to be small.

(a) Imagine a clock on the surface of the Earth at distance $R_1$ from the Earth's center. Calculate the time elapsed on each clock as a function of the coordinate time $t$. Which clock moves faster?

(b) Solve for a geodesic corresponding to a circular orbit around the equator of the Earth ($\theta=\pi/2$). What is $d\phi/dt$?

(c) How much proper time elapses while a satellite at radius $R_1$ (skimming along the surface of the earth, neglecting air resistance) completes one orbit? You can work to first order in $\Phi$ if you like. Plug in the actual numbers for the radius of the Earth and so on (don't forgoet to restore the speed of light) to get an answer in seconds. How does this number compare to the proper time elapsed on the clock stationary on the surface?

Sol)

(a)

Center: $ds^2=-dt^2+dr^2$

Surface: $ds^2=-(1+2\Phi)dt^2+(1-2\Phi)dr^2+R^2(d\theta^2+\sin^2\theta d\phi^2)$

$d\tau/dt$ is bigger in center, so center is faster.

(b)

$$\ddot{x}^\mu +\Gamma^\mu_{\nu\rho}\dot{x}^\nu\dot{x}^\rho=0$$

$$\Gamma^1_{\mu\nu}=\frac{GM}{2GMr-r^2}\begin{pmatrix}0&-1&0&0\\ -1&0&0&0\\ 0&0&0&0\\ 0&0&0&0\end{pmatrix}$$

$$\Gamma^2_{\mu\nu}=\frac{GM}{2GMr+r^2}\begin{pmatrix}1&0&0&0\\ 0&-1&0&0\\ 0&0&-\frac{r^3}{GM}&0\\ 0&0&0&-\frac{r^3\sin^2\theta}{GM}\end{pmatrix}$$

$$\Gamma^3_{\mu\nu}=\frac{1}{r}\begin{pmatrix}0&0&0&0\\ 0&0&1&0\\ 0&1&0&0\\ 0&0&0&-r\cos\theta\sin\theta\end{pmatrix}$$

$$\Gamma^4_{\mu\nu}=\frac{1}{r}\begin{pmatrix}0&0&0&0\\ 0&0&0&1\\ 0&0&0&r\cot\theta\\ 0&1&r\cot\theta&0\end{pmatrix}$$

When $r=R$, $\theta=\pi/2$,

$$\ddot{t}=0$$

$$0+\frac{GM}{2GMr+r^2}\dot{t}^2-\frac{r^2\sin^2\theta}{2GM+r}\dot{\phi}^2=0$$

$$0-0=0$$

$$\ddot{\phi}=0$$

Then $\dot{t}=A$, $\dot{\phi}=B$, $$GMA^2=r^3\sin^2\theta B^2$$

By chain rule,

$$\frac{d\phi}{dt}=\frac{\dot{\phi}}{\dot{t}}=\sqrt{\frac{GM}{r^3\sin^2\theta}}$$

(c)

$R=6.371km$

$c=3*10^8m/s$

Problem 4 (Carroll, Exercise 3.13)

Find explicit expressions for a complete set of Killing vector fields for the following spaces:

(a) Minkowski space, with metric $ds^2=-dt^2+dx^2+dy^2+dz^2$.

(b) A spacetime with coordinates $(u,v,x,y)$ and metric $$ds^2=-(du\, dv+dv\, du)+a^2(u)dx^2+b^2(u)dy^2$$ where $a$ and $b$ are unspecified functions of $u$. This represents a gravitational wave spacetime. (Hints, which you need not show: there are five Killing vectors in all, and all of them have a vanishing $u$ component $K^\mu$.)

Be careful, in all of there cases, about the distinction between upper and lower indices.

Sol)

(a) $$\nabla_\mu X_\nu+\nabla_\nu X_\mu=0$$

In Minkowski spacetime, $$\partial_\mu X_\nu+\partial_\nu X_\mu=0$$

4 translations $X^\mu_{(\nu)}=\delta^\mu_{(\nu)}$

3 rotations: $X^\mu=(0,-y,x,0),\ (0,-z,0,x),\ (0,0,-z,y)$

3 boosts: $X^\mu=(x,t,0,0),\ (y,0,t,0),\ (z,0,0,t)$

(b) $$\nabla_\mu X_\nu+\nabla_\nu X_\mu=0$$

$$\nabla_\mu X_\nu=\partial_\mu X_\nu-\Gamma^\rho_{\mu\nu}X_\rho$$

$$\Gamma^1_{\mu\nu}=\begin{pmatrix}0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\end{pmatrix}$$

$$\Gamma^2_{\mu\nu}=\begin{pmatrix}0&0&0&0\\ 0&0&0&0\\ 0&0&a[u]a'[u]&0\\ 0&0&0&b[u]b'[u]\end{pmatrix}$$

$$\Gamma^3_{\mu\nu}=\begin{pmatrix}0&0&\frac{a'[u]}{a[u]}&0\\ 0&0&0&0\\ \frac{a'[u]}{a[u]}&0&0&0\\ 0&0&0&0\end{pmatrix}$$

$$\Gamma^4_{\mu\nu}=\begin{pmatrix}0&0&0&\frac{b'[u]}{b[u]}\\ 0&0&0&0\\ 0&0&0&0\\ \frac{b'[u]}{b[u]}&0&0&0\end{pmatrix}$$

$\mu=1,\nu=1$

$$\partial_1X_1=0$$

$\mu=1,\nu=2$

$$\partial_1X_2+\partial_2X_1=0$$

$\mu=1,\nu=3$

$$\partial_1X_3+\partial_3X_1-2\frac{a'[u]}{a[u]}X_3=0$$

$\mu=1,\nu=4$

$$\partial_1X_4+\partial_4X_1-2\frac{b'[u]}{b[u]}X_4=0$$

$\mu=2,\nu=2$

$$\partial_2X_2=0$$

$\mu=2,\nu=3$

$$\partial_2X_3+\partial_3X_2=0$$

$\mu=2,\nu=4$

$$\partial_2X_4+\partial_4X_2=0$$

$\mu=3,\nu=3$

$$\partial_3X_3-a[u]a'[u]X_2=0$$

$\mu=3,\nu=4$

$$\partial_3X_4+\partial_4X_3=0$$

$\mu=4,\nu=4$

$$\partial_4X_4-b[u]b'[u]X_2=0$$

See Mathematica

[Homework 4]

Problem 1 (Carroll, Exercise 3.2)

You are familiar with the operations of gradient ($\nabla \varphi$), divergence ($\nabla\cdot V$) and curl ($\nabla\times V$) in ordinary vector analysis in three-dimensional Euclidean space. Using covariant derivatives, derive formulae for these operations in spherical polar coordinates ($r,\theta,\phi$) defined by $$x=r\sin\theta\cos\phi\\ y=r\sin\theta\sin\phi\\ z=r\cos\theta.$$ Compare your results to those in Jackson (1999) or an equivalent text. Are they identical? Should they be?

Sol)

$ds^2=dx^2+dy^2+dz^2=(\sin\theta\cos\phi dr+r\cos\theta\cos\phi d\theta-r\sin\theta\sin\phi d\phi)^2+(\sin\theta\sin\phi dr+r\cos\theta\sin\phi d\theta+r\sin\theta\cos\phi d\phi)^2+(\cos\theta dr-r\sin\theta d\theta)^2$

$=dr^2+r^2d\theta^2+r^2\sin\theta^2d\phi^2$

$$g=\begin{pmatrix}1&0&0\\ 0&r^2&0\\ 0&0&r^2\sin^2\theta\end{pmatrix}$$

$$g^{-1}=\begin{pmatrix}1&0&0\\ 0&\frac{1}{r^2}&0\\ 0&0&\frac{1}{r^2\sin^2\theta}\end{pmatrix}$$

$\sqrt{|g|}=r\sqrt{1+\sin^2\theta}$

$\Gamma^\sigma_{\mu\nu}=\frac{1}{2}g^{\sigma\rho}(\partial_\mu g_{\nu\rho}+\partial_\nu g_{\rho\mu}-\partial_\rho g_{\mu\nu})$

$$\Gamma^r=\begin{pmatrix}0&0&0\\ 0&-r&0\\ 0&0&-r\sin^2\theta\end{pmatrix}$$ (third terms alive)

$$\Gamma^\theta=\begin{pmatrix}0&\frac{1}{r}&0\\ \frac{1}{r}&0&0\\0&0&-\sin\theta\cos\theta\end{pmatrix}$$ (first&second term alive)

$$\Gamma^\phi\begin{pmatrix}0&0&\frac{1}{r}\\ 0&0&\cot\theta\\ \frac{1}{r}& \cot\theta&0\end{pmatrix}$$

$$\nabla \varphi=\partial \phi=\frac{\partial \varphi}{\partial r}\hat{r}+\frac{1}{r}\frac{\partial \varphi}{\partial \theta}\hat{\theta}+\frac{1}{r\sin\theta}\frac{\partial \varphi}{\partial \phi}\hat{\phi}$$

$$\nabla\cdot V=\nabla_\mu V^\mu=\partial_\mu V^\mu+\Gamma^\mu_{\mu\lambda}V^\lambda\\ =\left( \frac{\partial }{\partial r}+\frac{2}{r} \right) V^r+\left( \frac{1}{r}\frac{\partial }{\partial \theta}+\cot\theta \right) V^\theta+\left( \frac{1}{r\sin\theta}\frac{\partial }{\partial \phi}\right) V^\phi$$

$$\nabla\times V=\frac{1}{r\sin\theta}\left[ \frac{\partial}{\partial\theta}(V^\phi\sin\theta)-\frac{\partial V^\theta}{\partial\phi}\right]\hat{r}+\frac{1}{r}\left[ \frac{1}{\sin\theta}\frac{\partial V^r}{\partial \phi}-\frac{\partial}{\partial r}(rV^\phi)\right]\hat{\theta}+\frac{1}{r}\left[ \frac{\partial}{\partial r}(rV^\theta)-\frac{\partial V^r}{\partial \theta}\right]\hat{\phi}$$

Same as ordinary spherical coordinate differentiation.

Problem 2 (Carroll, Exercise 3.8)

The metric for the three-sphere in coordinates $x^\mu=(\psi,\theta,\phi)$ can be written $$ds^2=d\psi^2+\sin^2\psi(d\theta^2+\sin^2\theta d\phi^2).$$

(a) Calculate the Christoffel connection coefficients. Use whatever method you like, but it is good practice to get the connection coefficients by varying the integral (3.49).

(b) Calculate the Riemann tensor, Ricci tensor, and Ricci scalar.

(c) Show that (3.191) is obeyed by this metric, confirming that the three-sphere is a maximally symmetric space (as you would expect).

Sol)

(a)

$$g=\begin{pmatrix}1&0&0\\ 0&\sin^\psi&0\\ 0&0&\sin^2\psi \sin^2\theta\end{pmatrix}$$

$$\Gamma^\psi=\begin{pmatrix}0&0&0\\ 0&-\cos\psi\sin\psi&0\\ 0&0& -\cos\psi\sin\psi\sin^2\theta\end{pmatrix}$$

$$\Gamma^\theta=\begin{pmatrix}0&\cot\psi&0\\ \cot\psi &0&0\\ 0&0&-\cos\theta\sin\theta\end{pmatrix}$$

$$\Gamma^\phi=\begin{pmatrix}0&0&\cot \psi\\ 0&0&\cot\theta\\ \cot\psi&\cot\theta&0\end{pmatrix}$$

(b) $$R^\rho_{\sigma\mu\nu}=\partial_\mu\Gamma^\rho_{\nu\sigma}-\partial_\nu\Gamma^\rho_{\mu\sigma}+\Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma}-\Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma}.$$

$$R^\psi_{\sigma\psi\nu}=\begin{pmatrix}0&0&0\\ 0&\sin^2\psi&0\\ 0&0&\sin^2\psi\sin^2\theta\end{pmatrix}$$

$$R^\psi_{\sigma\theta\nu}=\begin{pmatrix}0&0&0\\ -\sin^2\psi&0&0\\ 0&0&0\end{pmatrix}$$

$$R^\psi_{\sigma\phi\nu}=\begin{pmatrix}0&0&0\\ 0&0&0\\ -\sin^2\psi\sin\theta^2&0&0\end{pmatrix}$$

$$R^\theta_{\sigma\psi\nu}=\begin{pmatrix}0&-1&0\\ 0&0&0\\ 0&0&0\end{pmatrix}$$

$$R^\theta_{\sigma\theta\nu}=\begin{pmatrix}1&0&0\\ 0&0&0\\ 0&0&\sin^2\psi\sin^2\theta\end{pmatrix}$$

$$R^\theta_{\sigma\phi\nu}=\begin{pmatrix}0&0&0\\ 0&0&0\\ 0&-\sin^2\psi\sin^2\theta&0\end{pmatrix}$$

$$R^\phi_{\sigma\psi\nu}=\begin{pmatrix}0&0&-1\\ 0&0&0\\ 0&0&0\end{pmatrix}$$

$$R^\phi_{\sigma\theta\nu}=\begin{pmatrix}0&0&0\\ 0&0&-\sin^2\psi\\ 0&0&0\end{pmatrix}$$

$$R^\phi_{\sigma\phi\nu}=\begin{pmatrix}1&0&0\\ 0&\sin^2\psi&0\\ 0&0&0\end{pmatrix}$$

$$R_{\sigma\nu}=R^\lambda_{\mu\lambda\nu}=\begin{pmatrix}2&0&0\\ 0&2\sin^2\psi&0\\ 0&0&2\sin^2\psi\sin^2\theta\end{pmatrix}$$

$$R=g^{\mu\nu}R_{\mu\nu}=6$$

$$\frac{R}{n(n-1)}=1$$

$$R_{11\mu\nu}=\begin{pmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{pmatrix}$$

$$g_{1\mu}g_{1\nu}-g_{1\nu}g_{1\mu}=\begin{pmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{pmatrix}$$

$$R_{12\mu\nu}=\begin{pmatrix}0&\sin^2\psi&0\\ -\sin^2\psi&0&0\\ 0&0&0\end{pmatrix}$$

$$g_{1\mu}g_{2\nu}-g_{1\nu}g_{2\mu}=\begin{pmatrix}0&\sin^2\psi&0\\ -\sin^2\psi&0&0\\ 0&0&0\end{pmatrix}$$

$$R_{13\mu\nu}=\begin{pmatrix}0&0&\sin^2r\sin^2\theta\\ 0&0&0\\ -\sin^2\psi\sin^2\theta&0&0\end{pmatrix}$$

$$g_{1\mu}g_{3\nu}-g_{1\nu}g_{3\mu}=\begin{pmatrix}0&0&\sin^2r\sin^2\theta\\ 0&0&0\\ -\sin^2\psi\sin^2\theta&0&0\end{pmatrix}$$

$$R_{21\mu\nu}=\begin{pmatrix}0&-\sin^2\psi&0\\ \sin^2\psi&0&0\\ 0&0&0\end{pmatrix}$$

$$g_{2\mu}g_{1\nu}-g_{2\nu}g_{1\mu}=\begin{pmatrix}0&-\sin^2\psi&0\\ \sin^2\psi&0&0\\ 0&0&0\end{pmatrix}$$

$$R_{22\mu\nu}=\begin{pmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{pmatrix}$$

$$g_{2\mu}g_{2\nu}-g_{2\nu}g_{2\mu}=\begin{pmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{pmatrix}$$

$$R_{23\mu\nu}=\begin{pmatrix}0&0&0\\ 0&0&\sin^4\psi\sin^2\theta\\ 0&-\sin^4\psi\sin^2\theta&0\end{pmatrix}$$

$$g_{2\mu}g_{3\nu}-g_{2\nu}g_{3\mu}=\begin{pmatrix}0&0&0\\ 0&0&\sin^4\psi\sin^2\theta\\ 0&-\sin^4\psi\sin^2\theta&0\end{pmatrix}$$

$$R_{13\mu\nu}=\begin{pmatrix}0&0&-\sin^2r\sin^2\theta\\ 0&0&0\\ \sin^2\psi\sin^2\theta&0&0\end{pmatrix}$$

$$g_{3\mu}g_{1\nu}-g_{3\nu}g_{1\mu}=\begin{pmatrix}0&0&-\sin^2r\sin^2\theta\\ 0&0&0\\ \sin^2\psi\sin^2\theta&0&0\end{pmatrix}$$

$$R_{23\mu\nu}=\begin{pmatrix}0&0&0\\ 0&0&-\sin^4\psi\sin^2\theta\\ 0&\sin^4\psi\sin^2\theta&0\end{pmatrix}$$

$$g_{3\mu}g_{2\nu}-g_{3\nu}g_{2\mu}=\begin{pmatrix}0&0&0\\ 0&0&-\sin^4\psi\sin^2\theta\\ 0&\sin^4\psi\sin^2\theta&0\end{pmatrix}$$

$$R_{33\mu\nu}=\begin{pmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{pmatrix}$$

$$g_{3\mu}g_{3\nu}-g_{1\nu}g_{3\mu}=\begin{pmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{pmatrix}$$

This space is maximally symmetric space.

Problem 3 (Carroll, Exercise 3.9)

Show that the Weyl tensor $C^\mu_{\nu\rho\sigma}$ is left invariant by a conformal transformation.

Sol)

Guidline

$$C_{\rho\sigma\mu\nu}=R_{\rho\sigma\mu\nu}-\frac{2}{(n-2)}(g_{\rho[\mu}R_{\nu]\sigma}-g_{\sigma[\mu}R_{\nu]\rho})+\frac{2}{(n-1)(n-2)}g_{\rho[\mu}g_{\nu]\sigma}R$$

$C_{\rho\sigma\mu\nu}=C_{[\rho\sigma][\mu\nu]}$, $C_{\rho\sigma\mu\nu}=C_{\mu\nu\rho\sigma}$, $C_{\rho[\sigma\mu\nu]}=0$.

$$\tilde{g}_{\mu\nu}=\Omega^2(x)g_{\mu\nu}$$

$$\tilde{\Gamma}^\rho_{\mu\nu}=\Gamma^\rho_{\mu\nu}+C^\rho_{\mu\nu}$$

$$\nabla_\rho g_{\mu\nu}=0,\quad \tilde{\nabla}_\rho \tilde{g}_{\mu\nu}=0$$

$$\tilde{\nabla}_\rho\tilde{g}_{\mu\nu}=\nabla_\rho (\Omega^2 g_{\mu\nu})-C^\sigma_{\rho\mu}(\Omega^2g_{\sigma\nu})-C^\sigma_{\rho\nu}(\Omega^2g_{\mu\sigma})\\ =g_{\mu\nu}\nabla_\rho\Omega^2-C^\sigma_{\rho\mu}(\Omega^2g_{\sigma\nu})-C^\sigma_{\rho\nu}(\Omega^2g_{\mu\sigma})$$

$$\tilde{\nabla}_\rho\tilde{g}_{\mu\nu}-\tilde{\nabla}_\mu\tilde{g}_{\nu\rho}-\tilde{\nabla}_\nu\tilde{g}_{\rho\mu}=0\\ =g_{\mu\nu}2\Omega\nabla_\rho\Omega-C^\sigma_{\rho\mu}(\Omega^2g_{\sigma\nu})-C^\sigma_{\rho\nu}(\Omega^2g_{\mu\sigma})\\-g_{\rho\mu}2\Omega\nabla_\nu\Omega+C^\sigma_{\nu\rho}(\Omega^2g_{\sigma\mu})+C^\sigma_{\nu\mu}(\Omega^2g_{\rho\sigma})\\-g_{\nu\rho}2\Omega\nabla_\mu\Omega+C^\sigma_{\mu\nu}(\Omega^2g_{\sigma\rho})+C^\sigma_{\mu\rho}(\Omega^2g_{\nu\sigma})$$

Use symmetric property of $g$ and $C$, $$C^\sigma_{\mu\nu}=\Omega^{-1}g^{\sigma\rho}(g_{\rho\mu}\nabla_\nu\Omega+g_{\nu\rho}\nabla_\mu\Omega-g_{\mu\nu}\nabla_\rho\Omega)\\= \Omega^{-1}(\delta^\sigma_\mu\nabla_\nu\Omega+\delta^\sigma_\nu\nabla_\mu\Omega-g_{\mu\nu}g^{\sigma\rho}\nabla_\rho\Omega)$$

$$R^\rho_{\sigma\mu\nu}=\partial_\mu\Gamma^\rho_{\nu\sigma}-\partial_\nu\Gamma^\rho_{\mu\sigma}+\Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma}-\Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma}$$

$$\tilde{R}^\rho_{\sigma\mu\nu}=R^\rho_{\sigma\mu\nu}+\partial_\mu C^\rho_{\nu\sigma}-\partial_\nu C^\rho_{\mu\sigma}+C^\rho_{\mu\lambda}C^\lambda_{\nu\sigma}-C^\rho_{\nu\lambda}C^\lambda_{\mu\sigma}\\+ C^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma}+\Gamma^\rho_{\mu\lambda}C^\lambda_{\nu\sigma}-C^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma}-\Gamma^\rho_{\nu\lambda}C^\lambda_{\mu\sigma}$$

..Too complicated.

$$R^\rho_{\sigma\mu\nu}V^\sigma=\nabla_\mu\nabla_\nu V^\rho-\nabla_\nu\nabla_\mu V^\rho$$

$$\tilde{R}^\rho_{\sigma\mu\nu}V^\sigma=R^\rho_{\sigma\mu\nu}V^\sigma+\nabla_\mu(C^\rho_{\nu\lambda}V^\lambda)-\nabla_\nu(C^\rho_{\mu\lambda}V^\lambda)\\+C^\rho_{\mu\lambda}\nabla_\nu V^\lambda-C^\lambda_{\mu\nu}\nabla_\lambda V^\rho-C^\rho_{\nu\lambda}\nabla_\mu V^\lambda+C^\lambda_{\nu\mu}\nabla_\lambda V^\rho$$

$$=R^\rho_{\sigma\mu\nu}+V^\lambda\nabla_\mu C^\rho_{\nu\lambda}-V^\lambda\nabla_\nu C^\rho_{\mu\lambda}$$

$$\nabla_\mu C^\rho_{\nu\lambda}=\Omega^{-1}(\delta^\rho_\nu\nabla_\mu\nabla_\lambda\Omega+\delta^\rho_\lambda\nabla_\mu\nabla_\nu\Omega-g_{\nu\lambda}g^{\rho\kappa}\nabla_\mu\nabla_\kappa\Omega)\\-\Omega^{-2}\nabla_\mu\Omega(\delta^\rho_\nu\nabla_\lambda\Omega+\delta^\rho_\lambda\nabla_\nu\Omega-g_{\nu\lambda}g^{\rho\kappa}\nabla_\kappa\Omega)$$

$$\tilde{R}^\rho_{\sigma\mu\nu}=R^\rho_{\sigma\mu\nu}-2(\delta^\rho_{[\mu}\delta^\alpha_{\nu]}\delta^\beta_\sigma-g_{\sigma[\mu}\delta^\alpha_{\nu]}g^{\rho\beta})\Omega^{-1}(\nabla_\alpha\nabla_\beta\Omega)\\ +2(2\delta^\rho_{[\mu}\delta^\alpha_{\nu]}\delta^\beta_\sigma-2g_{\sigma[\mu}\delta^\alpha_{\nu]}g^{\rho\beta}+g_{\sigma[\mu}\delta^\rho_{\nu]}g^{\alpha\beta})\Omega^{-2}(\nabla_\alpha\Omega)(\nabla_\beta\Omega)$$

$$\tilde{R}_{\sigma\nu}=R_{\sigma\nu}-[(n-2)\delta^\alpha_\sigma\delta^\beta_\nu+g_{\sigma\nu}g^{\alpha\beta}]\Omega^{-1}(\nabla_\alpha\nabla_\beta\Omega)+[2(n-2)\delta^\alpha_\sigma\delta^\beta_\nu-(n-3)g_{\sigma\nu}g^{\alpha\beta}]\Omega^{-2}(\nabla_\alpha\Omega)(\nabla_\beta\Omega)$$

$$\tilde{R}=\Omega^{-2}R-2(n-1)g^{\alpha\beta}\Omega^{-3}(\nabla_\alpha\nabla_\beta\Omega)-(n-1)(n-4)g^{\alpha\beta}\Omega^{-4}(\nabla_\alpha\Omega)(\nabla_\beta\Omega)$$

Now apply this to Weyl tensor.

$$C_{\rho\sigma\mu\nu}=R_{\rho\sigma\mu\nu}-\frac{2}{(n-2)}(g_{\rho[\mu}R_{\nu]\sigma}-g_{\sigma[\mu}R_{\nu]\rho})+\frac{2}{(n-1)(n-2)}g_{\rho[\mu}g_{\nu]\sigma}R$$

$$\tilde{C}_{\rho\sigma\mu\nu}=\Omega^2 C_{\rho\sigma\mu\nu}-2\Omega^2 g_{\rho\kappa}(\delta^\kappa_{[\mu}\delta^\alpha_{\nu]}\delta^\beta_\sigma-g_{\sigma[\mu}\delta^\alpha_{\nu]}g^{\kappa\beta})\Omega^{-1}(\nabla_\alpha\nabla_\beta\Omega)\\ +2\Omega^2 g_{\rho\kappa}(2\delta^\kappa_{[\mu}\delta^\alpha_{\nu]}\delta^\beta_\sigma-2g_{\sigma[\mu}\delta^\alpha_{\nu]}g^{\kappa\beta}+g_{\sigma[\mu}\delta^\kappa_{\nu]}g^{\alpha\beta})\Omega^{-2}(\nabla_\alpha\Omega)(\nabla_\beta\Omega)\\ -\frac{2}{(n-2)}(\Omega^2 g_{\rho[\mu}\left(-[(n-2)\delta^\alpha_{\nu]}\delta^\beta_\sigma+g_{\nu]\sigma}g^{\alpha\beta}]\Omega^{-1}(\nabla_\alpha\nabla_\beta\Omega)+[2(n-2)\delta^\alpha_{\nu]}\delta^\beta_\sigma-(n-3)g_{\nu]\sigma}g^{\alpha\beta}]\Omega^{-2}(\nabla_\alpha\Omega)(\nabla_\beta\Omega)\right)+\Omega^2 g_{\sigma[\mu}\left(-[(n-2)\delta^\alpha_{\nu]}\delta^\beta_\rho+g_{\nu]\rho}g^{\alpha\beta}]\Omega^{-1}(\nabla_\alpha\nabla_\beta\Omega)+[2(n-2)\delta^\alpha_{\nu]}\delta^\beta_\rho-(n-3)g_{\nu]\rho}g^{\alpha\beta}]\Omega^{-2}(\nabla_\alpha\Omega)(\nabla_\beta\Omega)\right))\\ +\Omega^4\frac{2}{(n-1)(n-2)}g_{\rho[\mu}g_{\nu]\sigma}\left(-2(n-1)g^{\alpha\beta}\Omega^{-3}(\nabla_\alpha\nabla_\beta\Omega)-(n-1)(n-4)g^{\alpha\beta}\Omega^{-4}(\nabla_\alpha\Omega)(\nabla_\beta\Omega)\right)$$

Comare each term of $(\nabla_\alpha\nabla_\beta\Omega)$, $(\nabla_\alpha\Omega)(\nabla_\beta\Omega)$, $\frac{1}{n-2}(\nabla_\alpha\nabla_\beta\Omega)$, $\frac{1}{n-2}(\nabla_\alpha\Omega)(\nabla_\beta\Omega)$, these vanishes.

$$\tilde{C}^\rho_{\sigma\mu\nu}=C^\rho_{\sigma\mu\nu}$$

Problem 4 (Carroll, Exercise 3.11)

Since the Poincare half-plane with metric (3.192) is maximally symmetric, we might expect that it is rotationally symmetric around any point, although this certainly isn’t evident in the $(x,y)$ coordinates. If that is so, it should be possible to put the metric in a form where the rotational symmetry is manifest, such as $$ds^2=f^2(r)(dr^2+r^2d\theta^2).$$ To show that this work, calculate the curvature scalar for this metric and solve for the function $f(r)$ subject to the condition $R = −2/a^2$ everywhere. What is the range of the coordinate $r$?

Sol)

$$\Gamma^1=\begin{pmatrix}\frac{k'(r)}{k(r)}&0\\ 0&-\frac{r(k(r)+rk'(r))}{k(r)}\end{pmatrix}$$

$$\Gamma^2=\begin{pmatrix}0& \frac{1}{r}+\frac{k'(r)}{k(r)}\\ \frac{1}{r}+\frac{k'(r)}{k(r)}&0\end{pmatrix}$$

$$R^1_{1\mu\nu}=\begin{pmatrix}0&0\\ 0&0\end{pmatrix}$$

$$R^2_{1\mu\nu}=\frac{r}{k(r)^2}(rk'(r)^2-k(r)(k'(r)+rk''(r)))\begin{pmatrix}0&1\\ -1 &0\end{pmatrix}$$

$$R^1_{2\mu\nu}=\frac{r}{k(r)^2}(rk'(r)^2-k(r)(k'(r)+rk''(r)))\begin{pmatrix}0&-1\\ 1 &0\end{pmatrix}$$

$$R^2_{2\mu\nu}=\begin{pmatrix}0&0\\ 0&0\end{pmatrix}$$

$$R_{\mu\nu}=\frac{rf'(r)^2-f(r)(f'(r)+rf''(r))}{f(r)^2}\begin{pmatrix}\frac{1}{r}&0\\ 0&r\end{pmatrix}$$

$$R=\frac{2rf'(r)^2-2f(r)(f'(r)+rf''(r))}{rf(r)^4}$$

$$f(r)=\pm\frac{aC_1}{r}\sqrt{\tanh(C_1(C_2-\log (r)))-1}$$

$$r>0$$

[Homework 5]

Problem 1 (Carroll, Exercise 4.1)

The Lagrange density for electromagnetism in curved space is $$\mathcal{L}=\sqrt{-g}\left(-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+A_\mu J^\mu\right)$$ where $J^\mu$ is the conserved current.

(a) Derive the energy-momentum tensor by functional differentiation with respect to the metric.

(b) Consider adding a new term to the Lagrangian, $$\mathcal{L}'=\beta R^{\mu\nu}g^{\rho\sigma}F_{\mu\rho}F_{\nu\sigma}.$$ How are Maxwell’s equations altered in the presence of this term? Einstein’s equation? Is the current still conserved?

Sol)

(a) Hilbert Stress energy tensor (This is different from (Belinfante) canonical stress-energy tensor from Noether's theorem, but same)$$T_{\mu\nu}=-2\frac{1}{\sqrt{-g}}\frac{\delta S_M}{\delta g^{\mu\nu}}.$$ $$(\delta S)=\int d^nx\left[ \sqrt{-g}\left(-\frac{1}{4}\delta(g^{\rho\alpha}g^{\sigma\beta})F_{\alpha\beta}F_{\rho\sigma}+\delta g^{\rho\sigma}A_\rho J_\sigma\right)+\delta\sqrt{-g}\left(-\frac{1}{4}g^{\rho\alpha}g^{\sigma\beta}F_{\alpha\beta}F_{\rho\sigma}+ g^{\rho\sigma}A_\rho J_\sigma\right)\right]$$

Use (from carroll) $$\delta g_{\mu\nu}=-g_{\mu\rho}g_{\nu\sigma}\delta g^{\rho\sigma}$$ $$\delta g=g(g^{\mu\nu}\delta g_{\mu\nu})=-g(g_{\mu\nu}\delta g^{\mu\nu})$$ $$\delta\sqrt{-g}=-\frac{1}{2}\sqrt{-g}g_{\mu\nu}\delta g^{\mu\nu}$$

Use for any rank 2 tensor satisfy $$\frac{\partial}{\partial G^{\alpha\beta}}G^{\sigma\gamma}=\delta^\sigma_\alpha\delta^\gamma_\beta$$ Then $$\delta(g^{\rho\alpha}g^{\sigma\beta})=\delta g^{\rho\alpha}g^{\sigma\beta}+g^{\rho\alpha}\delta g^{\sigma\beta}=\delta g^{\mu\nu}(g^{\sigma\beta}\delta^\rho_\mu\delta^\alpha_\nu+g^{\rho\alpha}\delta^\sigma_\mu\delta^\beta_\nu)$$

Then (careful. i confused dummey index like $\rho,\sigma$) $$T_{\mu\nu}=2\frac{1}{\sqrt{-g}}\frac{\delta S}{\delta g^{\mu\nu}}\\=F_\mu^\sigma F_{\nu\sigma}-2A_\mu J_\nu-\frac{1}{4} g_{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}+g_{\mu\nu}A^\rho J_\rho $$

(b)

$$\frac{\widehat{\mathcal{L}}}{\partial A_\nu}-\nabla_\mu\left( \frac{\partial \widehat{\mathcal{L}}}{\partial(\nabla_\mu A_\nu)}\right)=0$$ When $B\equiv\beta/\sqrt{-g}$ $$\widehat{\mathcal{L}}=\left(-\frac{1}{4}g^{\alpha\beta}+B R^{\alpha\beta}\right) g^{\rho\sigma}F_{\alpha\rho}F_{\beta\sigma}+g^{\rho\sigma}A_\rho J_\sigma$$ Then $$\frac{\partial}{\partial(\nabla_\mu A_\nu)}(F_{\alpha\beta}F_{\rho\sigma})=\frac{\partial}{\partial (\nabla_\mu A_\nu)}[(\nabla_\alpha A_\beta-\nabla_\beta A_\alpha)(\nabla_\rho A_\sigma-\nabla_\sigma A_\rho)]\\ =(\delta^\mu_\alpha\delta^\nu_\beta-\delta^\mu_\beta\delta^\nu_\alpha)(\nabla_\rho A_\sigma-\nabla_\sigma A_\rho)+(\nabla_\alpha A_\beta-\nabla_\beta A_\alpha)(\delta^\mu_\rho \delta^\nu_\sigma-\delta^\mu_\sigma\delta^\nu_\rho)\\ =\frac{1}{2}(-F^{\mu\nu}-F^{\nu\mu})+2B(R^{\mu\lambda}F_\lambda^\nu-R^{\nu\lambda}F_\lambda^\mu)=-F^{\mu\nu}+4BR^{\lambda[\nu}F^{\mu]}_\lambda$$

Then $$\frac{\partial \widehat{\mathcal{L}}}{\partial (\nabla_\mu A_\nu)}=-\frac{1}{2}(g^{\mu\beta}g^{\nu\sigma}F_{\beta\sigma}-g^{\nu\beta}g^{\mu\sigma}F_{\beta\sigma})+2B(R^{\mu\beta}g^{\nu\sigma}F_{\beta\sigma}-R^{\nu\beta}g^{\mu\sigma}F_{\beta\sigma})$$

Finally, $$\frac{\partial\widehat{\mathcal{L}}}{\partial A_\nu}=\delta^\nu_\rho g^{\rho\sigma} J_\sigma=J^\nu$$

Maxwell's equation is $$\nabla_\mu(-F^{\mu\nu}+4BR^{\lambda[\nu}F^{\mu]}_\lambda)=J^\nu$$

Since $F^{\mu\nu}+4BR^{\lambda[\nu}F^{\mu]}_\lambda$ is antisymmetric, in Minkowski spacetime, $\nabla_\nu J^\nu=0$, but in general spcaetime, conservation is not guaranteed.

To get Einstein equation, we have to derivative $R_{\mu\nu}$.

$$R^\rho_{\mu\lambda\nu}=\partial_\lambda\Gamma^\rho_{\nu\mu}+\Gamma^\rho_{\lambda\sigma}\Gamma^\sigma_{\nu\mu}-(\lambda\leftrightarrow \nu)$$

$$\nabla_\lambda(\delta \Gamma^\rho_{\nu\mu})=\partial_\lambda(\delta \Gamma^\rho_{\nu\mu})+\Gamma^\rho_{\lambda\sigma}\delta \Gamma^\sigma_{\nu\mu}-\Gamma^\sigma_{\lambda\nu}\delta\Gamma^\rho_{\sigma\mu}-\Gamma^\sigma_{\lambda\mu}\delta\Gamma^\rho_{\nu\sigma}$$

Then we can know $$\delta R^\rho_{\mu\lambda\nu}=\nabla_\lambda(\delta \Gamma^\rho_{\nu\mu})-\nabla_\nu(\delta \Gamma^\rho_{\lambda\mu})$$

$$\delta R_{\mu\nu}=\nabla_\lambda(\delta\Gamma^\lambda_{\nu\mu
})-\nabla_\nu(\delta \Gamma^\lambda_{\lambda\mu})$$

$$\delta \Gamma^\sigma_{\mu\nu}=-\frac{1}{2}[g_{\lambda\mu}\nabla_\nu(\delta g^{\lambda\sigma})+g_{\lambda\nu}\nabla_\mu(\delta g^{\lambda\sigma})-g_{\mu\alpha}g_{\nu\beta}\nabla^\sigma(\delta g^{\alpha\beta})]$$

Set $$g^{\rho\sigma}R^{\eta\theta}=g^{\rho\sigma}g^{\eta\alpha}g^{\theta\beta}R_{\alpha\beta}$$

Then variance of $R_{\alpha\beta}$:

$$g^{\rho\sigma}g^{\eta\alpha}g^{\theta\beta}\delta R_{\alpha\beta}=g^{\rho\sigma}g^{\eta\alpha}g^{\theta\beta}[\nabla_\lambda(\delta \Gamma^\lambda_{\beta\alpha})-\nabla_\beta(\delta \Gamma^\lambda_{\lambda\alpha})]$$

$$=\nabla_\lambda(g^{\rho\sigma}g^{\eta\alpha}g^{\theta\beta}\delta \Gamma^\lambda_{\beta\alpha})-\nabla_\beta(g^{\rho\sigma}g^{\eta\alpha}g^{\theta\beta}\delta \Gamma^\lambda_{\lambda\alpha})$$

$$=-\frac{1}{2}\nabla_\lambda\left[g^{\rho\sigma}g^{\eta\alpha}g^{\theta\beta}(g_{\gamma\beta}\nabla_\alpha(\delta g^{\gamma\lambda})+g_{\gamma\alpha}\nabla_\beta(\delta g^{\gamma\lambda})-g_{\beta\xi}g_{\alpha\pi}\nabla^\lambda(\delta g^{\xi\pi}))\right]\\ +\frac{1}{2}\nabla_\beta\left[ g^{\rho\sigma}g^{\eta\alpha}g^{\theta\beta}(g_{\gamma\lambda}\nabla_\alpha(\delta g^{\gamma\lambda})+g_{\gamma\alpha}\nabla_\lambda(\delta g^{\gamma\lambda})-g_{\lambda\xi}g_{\alpha\pi}\nabla^\lambda(\delta g^{\xi\pi})) \right]$$

$$=-\frac{1}{2}\nabla_\lambda\left[g^{\rho\sigma}g^{\eta\alpha}\nabla_\alpha(\delta g^{\theta\lambda})+g^{\rho\sigma}g^{\theta\beta}\nabla_\beta(\delta g^{\eta\lambda})-g^{\rho\sigma}\nabla^\lambda(\delta g^{\theta\eta})\right]\\ +\frac{1}{2}\nabla_\beta\left[g^{\rho\sigma}g^{\eta\alpha}g^{\theta\beta}g_{\gamma\lambda}\nabla_\alpha(\delta g^{\gamma\lambda})+ g^{\rho\sigma}g^{\theta\beta}\nabla_\lambda(\delta g^{\eta\lambda})-g^{\rho\sigma}g^{\theta\beta}g_{\lambda\xi}\nabla^\lambda (\delta g^{\xi\eta}) \right]$$

$$2\times g^{\rho\sigma}g^{\eta\alpha}g^{\theta\beta}\delta R_{\alpha\beta}F_{\rho\eta}F_{\sigma\theta}\\ =-F^{\sigma\alpha}F_{\sigma\theta}\nabla_\lambda\nabla_\alpha(\delta g^{\theta\lambda})-F_{\rho\eta} F^{\rho\beta}\nabla_\lambda\nabla_\beta(\delta g^{\eta\lambda})+F^\sigma_\eta F_{\sigma\theta}\nabla_\lambda\nabla^\lambda(\delta g^{\theta\eta}) \\ +F^{\sigma\alpha}F^\beta_\sigma g_{\gamma\lambda}\nabla_\beta\nabla_\alpha(\delta g^{\gamma\lambda})+F^\sigma_\eta F^\beta_\sigma\nabla_\beta\nabla_\lambda(\delta g^{\eta\lambda})-F^\sigma_\eta F^\beta_\sigma\nabla_\beta\nabla_\xi(\delta g^{\xi\eta})$$

Last two terms are same, ($\xi\rightarrow\lambda$) $$=-F^{\sigma\alpha}F_{\sigma\theta}\nabla_\lambda\nabla_\alpha(\delta g^{\theta\lambda})-F_{\rho\eta} F^{\rho\beta}\nabla_\lambda\nabla_\beta(\delta g^{\eta\lambda})\\ +F^\sigma_\eta F_{\sigma\theta}\nabla_\lambda\nabla^\lambda(\delta g^{\theta\eta}) +g_{\gamma\lambda}F^{\sigma\alpha}F^\beta_\sigma \nabla_\beta\nabla_\alpha(\delta g^{\gamma\lambda})$$

Use IBP at first term, $$F^{\sigma\alpha}F_{\sigma\theta}\nabla_\lambda\nabla_\alpha(\delta g^{\theta\lambda})=\nabla_\lambda( F^{\sigma\alpha}F_{\sigma\theta}\nabla_\alpha(\delta g^{\theta\lambda}))-\nabla_\lambda(F^{\sigma\alpha}F_{\sigma\theta})\nabla_\alpha(\delta g^{\theta\lambda})$$

Use IBP again, $$\nabla_\lambda(F^{\sigma\alpha}F_{\sigma\theta})\nabla_\alpha(\delta g^{\theta\lambda})=\nabla_\alpha\nabla_\lambda(F^{\sigma\alpha}F_{\sigma\theta}\delta g^{\theta\lambda})-\nabla_\alpha\nabla_\lambda(F^{\sigma\alpha}F_{\sigma\theta})\delta g^{\theta\lambda}$$

By Stock's theorem $$\int_\Sigma \nabla_\mu V^\mu \sqrt{|g|}d^nx=\int_{\partial \Sigma}n_\mu V^\mu \sqrt{|\gamma|}d^{n-1}x$$ we can ignore total derivative term when set 0 variation.

Then every term changes $$2\times g^{\rho\sigma}g^{\eta\alpha}g^{\theta\beta}\delta R_{\alpha\beta}F_{\rho\eta}F_{\sigma\theta}\\ =-\delta g^{\theta\lambda}\nabla_\alpha\nabla_\lambda(F^{\sigma\alpha}F_{\sigma\theta})-\delta g^{\eta\lambda}\nabla_\beta\nabla_\lambda(F_{\rho\eta} F^{\rho\beta})\\ +\delta g^{\theta\eta}\nabla^\lambda\nabla_\lambda(F^\sigma_\eta F_{\sigma\theta}) +g_{\gamma\lambda}\delta g^{\gamma\lambda} \nabla_\alpha\nabla_\beta(F^{\sigma\alpha}F^\beta_\sigma)$$

First two term is same, $$=\delta g^{\mu\nu}\left[ -2\nabla_\alpha\nabla_\nu(F^{\sigma\alpha}F_{\sigma\mu})+\nabla^\lambda\nabla_\lambda(F_\nu^\sigma F_{\sigma\mu})+g_{\mu\nu}\nabla_\alpha\nabla_\beta(F^{\sigma\alpha}F_\sigma^\beta) \right]$$

Then variation of the Lagrangian is $$\delta \widehat{\mathcal{L}}=B\delta(R^{\eta\theta}g^{\rho\sigma})F_{\rho\eta}F_{\sigma\theta}=B\delta(g^{\rho\sigma}g^{\eta\alpha}g^{\theta\beta}R_{\alpha\beta})F_{\rho\eta}F_{\sigma\theta}\\= B\delta g^{\mu\nu}\left(R^{\eta\theta}F_{\mu\eta}F_{\nu\theta}+R_{\nu\beta} F_{\rho\mu} F^{\rho\beta}+R_{\alpha\nu}F^{\sigma\alpha}F_{\sigma\mu}\\ -\nabla_\alpha\nabla_\nu(F^{\sigma\alpha}F_{\sigma\mu})+\frac{1}{2}\nabla^\lambda\nabla_\lambda(F_\nu^\sigma F_{\sigma\mu})+\frac{1}{2}g_{\mu\nu}\nabla_\alpha\nabla_\beta(F^{\sigma\alpha}F_\sigma^\beta) \right)$$

Change dummy index of third term ($\sigma\rightarrow \rho$), second and third term are same. $$=B\delta g^{\mu\nu}\left(R^{\eta\theta}F_{\mu\eta}F_{\nu\theta}+2R_{\nu\beta}F_{\rho\mu} F^{\rho\beta}-\nabla_\alpha\nabla_\nu(F^{\sigma\alpha}F_{\sigma\mu}) \\ +\frac{1}{2}\nabla^\lambda\nabla_\lambda(F_\nu^\sigma F_{\sigma\mu})+\frac{1}{2}g_{\mu\nu}\nabla_\alpha\nabla_\beta(F^{\sigma\alpha}F_\sigma^\beta) \right)$$

Since action is $$S=\int d^nx\sqrt{-g}\widehat{\mathcal{L}}$$ variation of action is $$\delta S'=\int d^nx\delta\sqrt{-g}\widehat{\mathcal{L}}'+\int d^nx\sqrt{-g}\delta\widehat{\mathcal{L}}'$$ $$=\int d^nx\sqrt{-g}B\delta g^{\mu\nu}\left( -\frac{1}{2}g_{\mu\nu}g^{\rho\sigma}R^{\eta\theta}F_{\rho\eta}F_{\sigma\theta}\\ + R^{\eta\theta}F_{\mu\eta}F_{\nu\theta}+2R_{\nu\beta}F_{\rho\mu} F^{\rho\beta} -\nabla_\alpha\nabla_\nu(F^{\sigma\alpha}F_{\sigma\mu})\\ +\frac{1}{2}\nabla^\lambda\nabla_\lambda(F_\nu^\sigma F_{\sigma\mu})+\frac{1}{2}g_{\mu\nu}\nabla_\alpha\nabla_\beta(F^{\sigma\alpha}F_\sigma^\beta)\right)$$

Finally, Einstein equation is $$\frac{1}{16\pi G}\left( R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}\right)+\frac{1}{\sqrt{-g}}\frac{\delta S}{\delta g^{\mu\nu}}=0$$

Additional term arise in $T_{\mu\nu}$. $$R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=8\pi G\left( T^{(EM)}_{\mu\nu}\\+B\left(g_{\mu\nu}g^{\rho\sigma}R^{\eta\theta}F_{\rho\eta}F_{\sigma\theta}- 2R^{\eta\theta}F_{\mu\eta}F_{\nu\theta}-4R_{\nu\beta}F_{\rho\mu} F^{\rho\beta}\\ +2\nabla_\alpha\nabla_\nu(F^{\sigma\alpha}F_{\sigma\mu})-\nabla^\lambda\nabla_\lambda(F_\nu^\sigma F_{\sigma\mu})-g_{\mu\nu}\nabla_\alpha\nabla_\beta(F^{\sigma\alpha}F_\sigma^\beta)\right)\right)$$

Problem 2 (Carroll, Exercise 4.2)

We showed how to derive Einstein’s equation by varying the Hilbert action with respect to the metric. They can also be derived by treating the metric and connection as independent degrees of freedom and varying separately with respect to them; this is known as the Palatini formalism. That is, we consider the action $$S=\int d^4x\sqrt{-g}g^{\mu\nu}R_{\mu\nu}(\Gamma)$$ where the Ricci tensor is thought of as constructed purely from the connection, not using the metric. Variation with respect to the metric gives the usual Einstein’s equations, but for a Ricci tensor constructed from a connection that has no a priori relationship to the metric. Imagining from the start that the connection is symmetric (torsion free), show that variation of this action with respect to the connection coefficients leads to the requirement that the connection be metric compatible, that is, the Christoffel connection. Remember that Stokes’s theorem, relating the integral of the covariant divergence of a vector to an integral of the vector over the boundary, does not work for a general covariant derivative. The best strategy is to write the connection coefficients as a sum of the Christoffel symbols $\Gamma^\lambda_{\mu\nu}=\tilde{\Gamma}^\lambda_{\mu\nu}+C^\lambda_{\mu\nu}$ and then show that $C^\lambda_{\mu\nu}$ must vanish.

Sol)

$$\delta S=\int d^n \sqrt{-g}g^{\mu\nu}\delta R_{\mu\nu}(\Gamma)\\ =\int d^nx\sqrt{-g}g^{\mu\nu}\left(\nabla_\lambda(\delta \Gamma^\lambda_{\nu\mu}\right)-\nabla_\nu(\delta \Gamma^\lambda_{\lambda\mu}))$$ Stockes's theorem gives $$=-\int d^nx\left( \nabla_\lambda (\sqrt{-g}g^{\mu\nu})\delta\Gamma^\lambda_{\nu\mu}-\nabla_\nu (\sqrt{-g}g^{\mu\nu})\delta \Gamma^\lambda_{\lambda_\mu} \right)+\mbox{(boundary)}\\ =-\int d^nx(\nabla_\lambda(\sqrt{-g}g^{\mu\nu})-\delta^\nu_\lambda\nabla_\sigma(\sqrt{-g}g^{\mu\sigma}))\delta \Gamma^\lambda_{\nu\mu}$$ Use torsion free ($\Gamma^\lambda_{\nu\mu}=\Gamma^\lambda_{\mu\nu}$), antisymmetric part vanish. $$\nabla_\lambda(\sqrt{-g}g^{\mu\nu})-\frac{1}{2}\delta^\nu_\lambda\nabla_\sigma(\sqrt{-g}g^{\mu\sigma})-\frac{1}{2}\delta^\mu_\lambda\nabla_\sigma(\sqrt{-g}g^{\nu\sigma})=0$$

We can see separately $\lambda=\mu,\lambda=\nu,\lambda\ne \mu,\nu$. $$\nabla_\lambda (\sqrt{-g}g^{\mu\nu})=0$$

To get $\nabla_\lambda\sqrt{-g}=0$, differentiate Levi-Civita symbol. $$\nabla_\lambda \epsilon_{0123}=\nabla_\lambda(\sqrt{-g}\tilde{\epsilon}_{0123})=\partial_\lambda\epsilon_{0123}-\Gamma^\sigma_{0\lambda}\epsilon_{\sigma123}-\Gamma^\sigma_{1\lambda}\epsilon_{0\sigma23}-\Gamma^\sigma_{2\lambda}\epsilon_{01\sigma3}-\Gamma^\sigma_{3\lambda}\epsilon_{012\sigma}\\ =\partial_\lambda(\sqrt{-g}\tilde{\epsilon}_{0123})-\Gamma^\sigma_{\sigma\lambda}(\sqrt{-g}\tilde{\epsilon}_{0123})$$

$\tilde{\epsilon}_{0123}$ is number, so $$\nabla_\lambda\sqrt{-g}=\partial_\lambda\sqrt{-g}-\Gamma^\sigma_{\sigma\lambda}\sqrt{-g}$$

$$\partial_\lambda\sqrt{-g}=-\frac{1}{2}\sqrt{-g}g_{\mu\nu}\partial_\lambda g^{\mu\nu}=-\frac{1}{2}\left( g_{\mu\nu}\partial_\lambda(\sqrt{-g}g^{\mu\nu})-g_{\mu\nu}g^{\mu\nu}\partial_\lambda\sqrt{-g}\right)\\ =-\frac{1}{2}g_{\mu\nu}\partial_\lambda (\sqrt{-g}g^{\mu\nu})+2\partial_\lambda\sqrt{-g}$$ Then $$\partial_\lambda\sqrt{-g}=\frac{1}{2}g_{\mu\nu}\partial_\lambda (\sqrt{-g}g^{\mu\nu})$$ Finally, $$\nabla_\lambda\sqrt{-g}=\partial_\lambda\sqrt{-g}-\Gamma^\sigma_{\sigma\lambda}\sqrt{-g}\\ =\frac{1}{2}g_{\mu\nu}\partial_\lambda(\sqrt{-g}g^{\mu\nu})-\Gamma^\sigma_{\sigma\lambda}\sqrt{-g}=-\frac{1}{2}g_{\mu\nu}(\Gamma^\mu_{\lambda\sigma}(\sqrt{-g}g^{\sigma\nu})+\Gamma^\nu_{\lambda\sigma}(\sqrt{-g}g^{\mu\sigma})-\Gamma^\sigma_{\sigma\lambda}(\sqrt{-g}g^{\mu\nu}))-\Gamma^\sigma_{\sigma\lambda}\sqrt{-g}=-\frac{1}{2}\sqrt{-g}(\Gamma^\mu_{\lambda\sigma}\delta^\sigma_\mu+\Gamma^\nu_{\lambda\sigma}\delta^\sigma_\nu-4\Gamma^\sigma_{\sigma\lambda})-\Gamma^\sigma_{\sigma\lambda}\sqrt{-g}=0$$

Finally, $$\nabla_\lambda(\sqrt{-g}g^{\mu\nu})=0$$ Imply $$\nabla_\lambda g^{\mu\nu}=0$$ And this conclude $$\nabla_\lambda g^{\mu\nu}=\nabla_\lambda(g^{\mu\alpha}g^{\nu\beta}g_{\alpha\beta})=g^{\mu\alpha}g^{\nu\beta}\nabla_\lambda g_{\alpha\beta}=0$$ So $$\nabla_\lambda g_{\alpha\beta}=0$$

Problem 3 (Carroll, Exercise 5.4)

Consider Einstein’s equations in vacuum, but with a cosmological constant, $G_{\mu\nu}+\Lambda g_{\mu\nu}=0$.

(a) Solve for the most general spherically symmetric metric, in coordinates $(t, r)$ that reduce to the ordinary Schwarzschild coordinates when $\Gamma=0$

(b) Write down the equation of motion for radial geodesics in terms of an effective potential, as in (5.66). Sketch the effective potential for massive particles.

Sol)

(a)

Spherically symmetric metric is $$ds^2=-e^{2\alpha(r)}dt^2+e^{2\beta(r)}dr2+r^2d\Omega^2$$

$$e^{2(\beta-\alpha)}R_{tt}+R_{rr}=\frac{2}{r}(\partial_r\alpha+\partial_r\beta)\\ =\Lambda\left(e^{2(\beta-\alpha)}g_{tt}+g_{rr}\right)=\Lambda\left(-e^{2(\beta-\alpha)}e^{2\alpha}+e^{2\beta}\right)=0$$ Then $$\alpha=-\beta$$ Also, from $R_{\theta\theta}=\Lambda g_{\theta\theta}$$, $$\partial_r(re^{2\alpha})=1-\Lambda r^2$$ $$e^{2\alpha}=1-\frac{R}{r}-\frac{1}{3}\Lambda r^2$$

$$ds^2=-\left(1-\frac{2GM}{r}-\frac{1}{3}\Lambda r^2\right) dt^2+\left( 1-\frac{2GM}{r}-\frac{1}{3}\Lambda r^2\right)^{-1}dr^2+r^2 d\Omega^2$$

(b) $$V(r)=\frac{1}{2}\epsilon -\epsilon \frac{GM}{r}+\frac{L^2}{2r^2}-\frac{GML^2}{r^3}=\frac{1}{2}\left( 1-\frac{2GM}{r}\right)\left( \frac{L^2}{r^2}+\epsilon\right)$$

Replace $(1-\frac{2GM}{r})$ with $(1-\frac{2GM}{r}-\frac{1}{3}\Lambda r^2)$. $$V(r)=\frac{1}{2}\left( 1-\frac{2GM}{r}-\frac{1}{3}\Lambda r^2\right) \left( \frac{L^2}{r^2}+\epsilon\right) \\ =\frac{1}{2}\epsilon -\epsilon \frac{GM}{r}-\frac{1}{6}\epsilon \Lambda r^2+\frac{L^2}{2r^2}-\frac{GML^2}{r^3}-\frac{1}{3}\Lambda L^2$$

Problem 4 (Carroll, Exercise 5.5)

Consider a comoving observer sitting at constant spatial coordinates $(r_*,\theta_*,\phi_*)$, around a Schwarzschild black hole of mass $M$. The observer drops a beacon into the black hole (straight down, along a radial trajectory). The beacon emits radiation at a constant wavelength $\lambda_{em}$ (in the beacon rest frame).

(a) Calculate the coordinate speed $dr/dt$ of the beacon, as a function of $r$.

(b) Calculate the proper speed of the beacon. That is, imagine there is a comoving observer at fixed $r$, with a locally inertial coordinate system set up as the beacon passes by, and calculate the speed as measured by the comoving observer. What is it at $r = 2GM$?

(c) Calculate the wavelength $\lambda_{obs}$, measured by the observer at $r_∗$, as a function of the radius rem at which the radiation was emitted.

(d) Calculate the time tobs at which a beam emitted by the beacon at radius rem will be observed at $r_∗$.

(e) Show that at late times, the redshift grows exponentially: $\lambda_{obs}/\lambda_{em}\propto e^{t_{obs}/T}$. Give an expression for the time constant $T$ in terms of the black hole mass $M$.

Sol)

(a) $$d\tau^2=\left(1-\frac{2GM}{r}\right)dt^2-\left( 1-\frac{2GM}{r}\right)^2dr^2$$ Since metric is independent to $t$, we can set time direction Killing vector with energy $$E=\left(1-\frac{2GM}{r}\right)\frac{dt}{d\tau}$$ Then $$\frac{dr}{dt}=\pm \left(1-\frac{2GM}{r}\right)\sqrt{1-\left(1-\frac{2GM}{r}\right)E^{-2}}$$

(b) $$0=\frac{dr}{dt}=\left(1-\frac{2GM}{r_*}\right)\sqrt{1-\left(1-\frac{2GM}{r_*}\right)E^{-2}}$$ or $$E^2=\left(1-\frac{2GM}{r_*}\right)$$

$$\frac{dr}{d\tau}=\pm\sqrt{E^2-\left(1-\frac{2GM}{r}\right)}=\pm \sqrt{\frac{2GM(r_*-r)}{rr_*}}$$

At $r=2GM$, $$\frac{dr}{d\tau}=\sqrt{1-\frac{2GM}{r_*}}$$

By (5.103), $$\frac{\lambda_{ob}}{\lambda_e'}=\sqrt{\frac{1-R_s/r_e}{1-R_s/r_*}}$$

(d)

(e)

Reference

(Suh, sunok) [PH489] Topics on Quantum Gravity