Tutorials

The following chapters are the tutorials for the course. I haven’t included everything from the tutorials (e.g. the definitions etc are omitted), but I have tried to include the most helpful questions. Some of the answers to the tutorial questions have also been placed within the main content of the lecturers as remarks etc.

26.1 Topology

Topologies On A Simple Set

Solution: Check the conditions:

1. \(\emptyset\in\mathcal{O}_1\) and \(\mathcal{M} \in \mathcal{O}_1\).

2. \(\emptyset\cap\{1\} = \emptyset\in\mathcal{O}_1\), \(\emptyset\cap\{1,2,3,4\}=\emptyset\in\mathcal{O}_1\) and \(\{1\}\cap\{1,2,3,4\}=\{1\}\in\mathcal{O}_1\).

3. \(\emptyset\cup\{1\}=\{1\} \in\mathcal{O}_1\), \(\emptyset\cup\{1,2,3,4\}=\{1,2,3,4\}\in\mathcal{O}_1\), \(\{1\}\cup\{1,2,3,4\}=\{1,2,3,4\}\in\mathcal{O}_1\) and \(\emptyset\cup\{1\}\cup\{1,2,3,4\}=\{1,2,3,4\}\in\mathcal{O}_1\).

So the answer is yes.

Solution: The only new addition is \(\{2\}\) so just need to check its involvement. We see straight away that \(\{1\}\cup\{2\} = \{1,2\}\notin\mathcal{O}_2\) and so we conclude that it is not a topology.

Continuous Functions

Solution: The chaotic topology here is \[\mathcal{O}_{\text{chaotic}} = \big\{\emptyset,\{1,2,3,4\}\big\}.\] We see straight away that the preimage of \(\{1\}\in\mathcal{O}_{\text{target}}\) is \(\{1\}\) which is not in \(\mathcal{O}_{\text{chaotic}}\), and so the map is not continuous w.r.t. these topologies.

Solution: \[\text{id}^{-1}_{\mathcal{M}}(1) = 1, \quad \text{id}^{-1}_{\mathcal{M}}(2) = 2, \quad \text{id}^{-1}_{\mathcal{M}}(3) = 3, \quad \text{id}^{-1}_{\mathcal{M}}(4) = 4\] Now consider the preimages. \[\text{preim}_{\text{id}^{-1}_{\mathcal{M}}}(\emptyset) = \emptyset, \qquad \text{and} \qquad \text{preim}_{\text{id}^{-1}_{\mathcal{M}}}(\{1,2,3,4\}) = \{1,2,3,4\} = \mathcal{M},\] both of which are in our domain’s topology. Therefore the map is continuous w.r.t. to these topologies.

The Standard Topology On \(\mathbb{R}^d\)

26.2 Topological Manifolds

An Atlas From A Real World — the Möbius river

Solution: The Möbius strip can be represented by the following diagram

where the arrows indicate how we identify the two edges together (i.e. the bottom right corner goes to the top left corner). The blue line is the river. It is clear that we will need at least two charts in order to map the whole strip. We could choose them as follows

We then define maps from these shaded regions into \(\mathbb{R}^2\). For example we can just map them so that they are the shaded regions on the page, taking care to region the two parts of \(U_1\) properly. To save my writing more Tikz code, I’ll leave this to your imagination.

A Real World From An Atlas

Before The Invention Of The Wheel

Solution: The map is \(x: F^1 \to [-1,1]\) given by \(x:(m,n) \mapsto m\). Clearly this map is not injective as \(x\big((m,n)\big) = x\big((m,n')\big)\) where \(n' = -n\).

Solution: The problem was that \(n'=-n\) gave non-injectivity, so clearly we define \[\begin{split} U_{\uparrow} & := \{(m,n)\in\mathbb{R}^2 \, | \, m^4 + n^4 = 1, n>0\} \\ U_{\downarrow} & := \{(m,n)\in\mathbb{R}^2 \, | \, m^4 + n^4 = 1, n<0\} \end{split}\] and so have the maps \[x_{\uparrow} : U_{\uparrow} \to (-1,1), \qquad \text{and} \qquad x_{\downarrow} : U_{\downarrow} \to (-1,1).\] It is important that we take the inequalities for \(n\) (i.e. \(n\neq 0\)) so that our sets are open, as required by the question.

Solution: Same as above we simply define \[V_{\uparrow} := \{(m,n)\in\mathbb{R}^2 \, | \, m^4 + n^4 = 1, m>0\},\] and the map \[y_{\uparrow} : V_{\uparrow} \to (-1,1),\] given by \(y_{\uparrow}:(m,n)\mapsto n\).

Solution: Yes it is invertible as it is bijective. We construct the inverse as \[y_{\uparrow}^{-1} : (-1,1) \to V_{\uparrow},\] where the action is given by \[y_{\uparrow}^{-1} : n \mapsto \big( \sqrt[4]{1-n^4}, n\big).\]

Solution: Yes their domains overlap as \[U_{\uparrow}\cap V_{\uparrow} = \{(m,n)\in\mathbb{R}^2 \, | \, m^4 + n^4 = 1, n,m>0\}\] The transition map is \[\big(x_{\uparrow}\circ y_{\uparrow}^{-1}\big) : (0,1) \to (0,1)\] with \[\big(x_{\uparrow}\circ y_{\uparrow}^{-1}\big)(a) = x_{\uparrow}\big(\sqrt[4]{1-a^4}, a\big) = \sqrt[4]{1-a^4} \in (0,1).\]

Solution: The answer is obviously 4 maps, \(x_{\uparrow},x_{\downarrow},y_{\uparrow}\) and \(y_{\downarrow}\), and yes they form an atlas. From this we see that \(F^1\) is a topological manifold of dimension \(d=1\).

26.3 Multilinear Algebra

Vector Spaces

Solution: We need to check it meets the 8 axioms.

1. Commutative w.r.t. \(\oplus\): \[(a,b,c) \oplus (d,e,f) = (a+d,b+e,c+f) = (d+a,b+e,f+c) = (d,e,f) \oplus (a,b,c).\]

2. Associative w.r.t. \(\oplus\): \[\begin{split} \big[ (a,b,c)\oplus(d,e,f) \big] \oplus (h,i,j) & = (a+d,b+e,c+f) \oplus (h,i,j) \\ & = (a+d+h,b+e+i,c+f+j) \\ & = (a,b,c) \oplus (d+h,e+i,f+j) \\ & = (a,b,c)\oplus \big[ (d,e,f)\oplus(h,i,j) \big]. \end{split}\]

3. Neutral element w.r.t. \(\oplus\): this is clearly just \((0,0,0)\).

4. Inverse w.r.t. \(\oplus\): this is clearly just \((-a,-b,-c)\). Note for this to be true it is important that we take the whole real line, not just the positive reals (otherwise \((-a,-b,-c)\notin V\).

5. Associative w.r.t. \(\odot\): \[\begin{split} \lambda \odot \big[\mu\odot(a,b,c)\big] & = \lambda \odot (\mu\cdot a,\mu\cdot b, \mu\cdot c) \\ & = (\lambda \odot\mu\cdot a,\lambda \odot\mu\cdot b, \lambda \odot\mu\cdot c) \\ & = \big[\lambda \odot \mu] \odot (a,b,c). \end{split}\]

6. Distributive 1: \[\begin{split} (\lambda + \mu)\odot (a,b,c) & = \big( (\lambda + \mu)\cdot a, (\lambda + \mu)\cdot b,(\lambda + \mu)\cdot c\big) \\ & = (\lambda \cdot a, \lambda \cdot b, \lambda \cdot c) \oplus (\mu\cdot a,\mu\cdot b, \mu \cdot c) \\ & = \lambda \odot (a,b,c) \oplus \mu \odot (a,b,c). \end{split}\]

7. Distributive 2: \[\begin{split} \lambda \odot \big[(a,b,c)\oplus(d,e,f)\big] & = \lambda \odot (a+d,b+e,c+f) \\ & = \big( \lambda\cdot(a+d),\lambda\cdot(b+e), \lambda\cdot(c+f)\big) \\ & = (\lambda\cdot a, \lambda\cdot b, \lambda\cdot c) + (\lambda\cdot d, \lambda\cdot e, \lambda\cdot f) \\ & = \lambda \odot (a,b,c) \oplus \lambda \odot (d,e,f). \end{split}\]

8. Unitary w.r.t. \(\odot\): Clearly \(1\odot(a,b,c)=(a,b,c)\).

Solution: Consider \[\begin{split} d\big( \lambda\odot (a,b,c) \oplus (d,e,f)\big) & = d\big( ([\lambda \cdot a]+d, [\lambda \cdot b]+e,[\lambda \cdot c]+f)\big) \\ & = ([\lambda \cdot b]+e, 2[\lambda \cdot c]+2f, 0) \\ & = \lambda\odot(b, 2c,0) \oplus (e,2f) \\ & = \lambda\odot d\big((a,b,c)\big) \oplus d\big((d,e,f)\big), \end{split} so yes it's linear.\]

Solution: This follows immediately from Theorem 3.2.2, however if you want you can show it explicitly (we won’t here to save me writing.)

Solution: The linearity calculation is exactly the same method as above, so to save me typing I’m just going to say the answer: yes it is. Being a linear map from \(V\) to \(\mathbb{R}\) it is, by definition, an element of \(V^*\).

Solution: Recall that \(\delta:P\to P\) is defined as the derivative, i.e. \(\delta(p) = p'\). Clearly for this question we want to focus on the map \(\delta : P_2\to P_2\). Recall the definition \[P_2 := \{ p :\mathbb{R} \to \mathbb{R} \, | \, p(x) = a + bx + cx^2\},\] and so we have \[p'(x) = b + 2cx + 0x^2.\] We instantly see this resembles the map \(d: V \to V\) which acts as \(d\big((a,b,c)\big) := (b,2c,0)\). We just need to construct the map \(j:P_2\to \mathbb{R}^3\). With a bit of thought it is clear that the answer is simply \[j(p) := (a,b,c),\] where \(p = a + bx +cx^2\). The inverse map \(j^{-1}:\mathbb{R}^3\to P_2\) is then simply given by \[\big(j^{-1}(a,b,c)\big)(x) := a+bx+cx^2.\] Direct substitution then gives \[\big((j\circ \delta \circ j^{-1}\big)(a,b,c) := (b,2c,0) =: d\big((a,b,c)\big).\]

Similar exercises can be done to compare \(i:V\to\mathbb{R}\) to \(I:P\to\mathbb{R}\) given in the lecture (and also for the above exercise which I haven’t typed here.) See the video for more details.

Indices

Solution: We simply uses the duality of the basis elements, namely \(e_i(\epsilon^j) = \delta^j_i = \epsilon^j(e_i)\). So we have \[A^{ab} = A(\epsilon^a,\epsilon^b), \qquad\text{and}\qquad B_{ab} = B(e_a,e_b).\]

Solution: From direct calculation we have \[A^{[ab]} := \frac{1}{2}\big(A^{ab}-A^{ba}\big) = - \frac{1}{2} \big( A^{ba} - A^{ab}\big) =: - A^{[ba]}.\] We also have \[\begin{split} A^{[ab]}B_{ab} & := \frac{1}{2}\big(A^{ab}-A^{ba}\big)B_{ab} \\ & = \frac{1}{2} \big( A^{ab}B_{ab} - A^{ba}B_{ab}\big) \\ & = \frac{1}{2} \big( A^{ab}B_{ab} - A^{ab}B_{ba}\big) \\ & = A^{ab}\frac{1}{2}\big(B_{ab}-B_{ba}\big) \\ & =: A^{ab}B_{[ab]}, \end{split}\] where we have used the Einstein summation convention to relabel the dummy indices of the second term to get to the third line.

Solution: Follows exactly like the previous solution.

Solution: We have \[\begin{split} A^{[ab]}B_{(ab)} & := \frac{1}{2}\big( A^{ab}B_{(ab)} - A^{ba}B_{(ab)} \big) \\ & = \frac{1}{2}\big( A^{(ab)}B_{ab} - A^{(ba)}B_{ab} \big) \\ & = \frac{1}{2}\big( A^{(ab)}B_{ab} - A^{(ab)}B_{ab}\big) \\ & = 0. \end{split}\]

Linear Maps As Tensors

Solution: We know a \((1,1)\)-tensor is a linear map \(T_{\phi}: V^*\times V \xrightarrow{\sim} \mathbb{R}\). We also know that \(\psi\in V^*\) is a map \(\psi: V\xrightarrow{\sim} \mathbb{R}\). So it’s clear that we just define \[T_{\phi}(\sigma, v) := \big(\phi(\sigma)\big)(v),\] where \(\sigma\in V^*\) and \(v\in V\).

Solution: This is just the previous exercise in reverse. That is we define \[\big(\phi_T(\sigma)\big)(v) := T(\sigma,v),\] which we write as \[\phi_T(\sigma) := T(\sigma, \cdot).\]

Solution:

1. We have \[T_{\phi_T}(\sigma,v) := \big(\phi_T(\sigma)\big)(v) =: T(\sigma,v).\]

2. We have \[\big(\phi_{T_{\phi}}(\sigma)\big)(v) := T_{\phi}(\sigma,v) =: \big(\phi(\sigma)\big(v),\] and so \(\phi_{T_{\phi}} = \phi\).

Solution: The above exercises have just shown us that there is a unique link between \(T\) and \(\phi\) and so we can think of them as being isomorphic as maps and as such can be viewed as the same object.

26.4 Differential Manifolds

Restricting The Atlas

Solution: We have \(x^{-1}:\mathbb{R} \to \mathbb{R}\); \(a\mapsto x^{-1}(a) = a\), and so \(y\circ x^{-1}:\mathbb{R}\to\mathbb{R}\); \(a\mapsto a^3\). This is \(C^{\infty}(\mathbb{R})\).

Solution: We have \(x\circ y^{-1}: \mathbb{R}\to\mathbb{R}\); \(a\mapsto \sqrt[3]{a}\), but this function is not even \(C^1(\mathbb{R}\to\mathbb{R})\) as the first derivative is \(\frac{1}{3}a^{-2/3}\), which blows up as \(a\to 0\). So no it is not a differentiable manifold.

Solution: The problem above came from the chart transition maps. Both chart maps themselves are \(C^{\infty}(\mathbb{R}\to\mathbb{R})\) and their domain is the whole manifold (namley \(\mathbb{R}\)) and so if we just remove one of the two charts we get a smooth manifold.

Soft Squares on \(\mathbb{R}\times\mathbb{R}\)

Solution: For \((U_n,x_n)\) to be charts, we need to show that \(U_n\in \mathcal{O}_{ssq}\) and that the \(x_n\)s are homeomorphisms (i.e. inevitable and continuous).

The soft square topology is rather self explanatory, it is the set of squares around the origin without the boundary. This is exactly the definition of the \(U_n\)s where the sides of consecutive \(U_n\) increase by \(2n\). So we have \(U_n\in\mathcal{O}_{ssq}\).

Next we need to show that the chart maps are homeomorphisms. It is clear that \(x_n\) is continuous w.r.t. \(\mathcal{O}_{ssq}\) and \(\mathcal{O}_{\text{standard}}\). So we just need to check that \(x_n^{-1}\) exists and is continuous. We have \[x_n^{-1}(a,b) = \big(n(a+b), n(a-b)\big),\] which is again clearly continuous.

Therefore we know that the \((U_n,x_n)\) are indeed charts.

Solution: The chart transition maps are given by \[\begin{split} x_m \circ x_n^{-1} : x_n(U_n\cap U_m) & \to x_m(U_n\cap U_m) \\ (a,b) & \mapsto \bigg(\frac{na}{m}, \frac{nb}{m}\bigg) = \frac{n}{m}(a,b). \end{split}\] This tells us that \(x_m\circ x_n^{-1} = \frac{n}{m}\beta1_{\mathbb{R}^2}\), which together with \(m\neq 0\) tells us \(k=\infty\).

Solution: There are many, but we could take the map \(\widetilde{x}:U_5 \to \widetilde{U_5}\subset\mathbb{R}^2\); \((a,b) \mapsto (a,b)\), i.e. it is the identity map restricted to \(U_5\). It is clear that this will be \(C^{\infty}\) compatible with any overlapping charts and so it lies in the atlas.

Undergraduate Multi-Dimensional Analysis

Differentiability On A Manifold

26.5 Tangent Spaces

Virtuoso Use Of The Symbol \(\big(\frac{\partial}{\partial x^i}\big)_p\)

Solution: We have \[\begin{split} \delta^a_b & = \bigg(\frac{\partial}{\partial x^b}\bigg)_p(x^a) \\ & := \partial_b\big(x^a\circ x^{-1}\big)\big(x(p)\big) \\ & = \partial_b\big(x^a\circ (y^{-1}\circ y) \circ x^{-1}\big)\big(x(p)\big) \\ & = \partial_b\big((x^a\circ y^{-1}) \circ (y \circ x^{-1})\big)\big(x(p)\big) \\ & = \partial_b\big(y^m\circ x^{-1}\big)\big(x(p)\big) \cdot \partial_m\big(x^a\circ y^{-1}\big) \big(y(p)\big) \\ & =: \bigg(\frac{\partial y^m}{\partial x^b}\bigg)_p \bigg(\frac{\partial x^a}{\partial y^m}\bigg)_p, \end{split}\] where we have inserted the identity, used the associativity of the composition of maps, and the multidimensional chain rule along with \(\big(y\circ x^{-1}\big)\big(x(p)\big) = y(p)\).

Solution: This follows in a similar manner to the previous question and so I won’t type it here.

Solution: The above expression is clearly of the form of \[T_{(x)i}(p) = \bigg(\frac{\partial y^m}{\partial x^i}\bigg)_p T_{(y)m}(p),\] which is the transformation law for the components of a \((0,1)\)-tensor. So the answer is "yes" and the valence is \((0,1)\) and the rank is \(1\).

Transformation Of Vector Components

Solution: We have \(\big(x \circ y^{-1}\big)\big((u,v)\big) = (u, v-u^3)\), and so direct calculation gives \[\bigg(\frac{\partial x^1}{\partial y^1}\bigg)_p := \partial_1 \big(x^1 \circ y^{-1}\big)\big(y(p)\big) = 1,\] where we have used \(\partial_1\big(x^\circ y^{-1}\big)\big((u,v)\big) = 1\), irrespective of the value \(u\) takes.

Next we have, \[\bigg(\frac{\partial x^2}{\partial y^1}\bigg)_p := \partial_1\big( x^2 \circ y^{-1}\big)\big(y(p)\big) = \partial_1\big( x^2 \circ y^{-1}\big)\big((a,b+a^3)\big) = -3a^2,\] where we have used \(\partial_2\big(x^1\circ y^{-1}\big)\big((u,v)\big) = -2u^2\) along with \(p=(a,b)\).

Similar calculations give \[\bigg(\frac{\partial x^1}{\partial y^2}\bigg)_p = 0, \qquad\text{and}\qquad \bigg(\frac{\partial x^2}{\partial y^2}\bigg)_p = 1\]

Solution: We have \((x\circ\gamma)(\lambda) = (\lambda, -\lambda)\) and \((y\circ\gamma)(\lambda) = (\lambda, -\lambda+\lambda^3)\), so we have \[\begin{split} \dot{\gamma}_x^1(\lambda_0) & := \big(x\circ \gamma)^{1\prime} (\lambda_0) = 1, \\ \dot{\gamma}_x^2(\lambda_0) & := \big(x\circ \gamma)^{2\prime} (\lambda_0) = -1, \\ \dot{\gamma}_y^1(\lambda_0) & := \big(y\circ \gamma)^{1\prime} (\lambda_0) = 1, \\ \dot{\gamma}_y^2(\lambda_0) & := \big(x\circ \gamma)^{2\prime} (\lambda_0) = -1+3\lambda_0^2. \end{split}\]

Solution: The answer is clearly to just use the transformation property \[\dot{\gamma}_x^i(\lambda_0) = \bigg(\frac{\partial x^i}{\partial y^m}\bigg)_p \dot{\gamma}_y^m(\lambda_0).\] That is, we have \[\begin{split} \dot{\gamma}_x^1(\lambda_0) & = \bigg(\frac{\partial x^1}{\partial y^1}\bigg)_p \dot{\gamma}_y^1(\lambda_0) + \bigg(\frac{\partial x^1}{\partial y^2}\bigg)_p \dot{\gamma}_y^2(\lambda_0) \\ & = 1\cdot 1 + 0\cdot(-1+3\lambda_0)^2 \\ & = 1, \end{split}\] and \[\begin{split} \dot{\gamma}_x^2(\lambda_0) & = \bigg(\frac{\partial x^2}{\partial y^1}\bigg)_p \dot{\gamma}_y^1(\lambda_0) + \bigg(\frac{\partial x^2}{\partial y^2}\bigg)_p \dot{\gamma}_y^2(\lambda_0) \\ & = -3\lambda_0^2 \cdot 1 + 1\cdot (-1+3\lambda_0^2) \\ & = -1. \end{split}\]

The Gradient

Solution: Clearly we just want \((f\circ \gamma)(\lambda)=c\) for all \(\lambda\in\mathbb{R}\). We can formulate this as \(f\circ \gamma\) being equivalent to the constant function, which simply obeys the above property.

Solution: We have \[(df)_p (v_{\gamma,p}) := v_{\gamma,p}(f) := \big(f\circ \gamma)^{\prime}(\lambda_0) =0,\] as the derivative of a constant vanishes.

Is There A Well-Defined Sum Of Curves?

Solution: We need to construct the chart transition maps \[\begin{split} (x\circ y^{-1})(u,v) & = x\big((u, v \cdot e^{-u})\big) = (u,v\cdot e^{-u}) \\ (y\circ x^{-1})(u,v) & = y\big((u,v)\big) = (u, v\cdot e^u). \end{split}\] These are both infinitely times continuously differentiable and so the answer is "yes".

Solution: The answer is "no" because our manifold is just the set \(\mathbb{R}^2\) (with a topology) and so carries no vector space structure so we cannot talk about the addition on \(\mathbb{R}^2\).

Solution: First consider the \(x\) chart. We have \[(x\circ \gamma) (\lambda) = (\lambda, 1), \qquad\text{and}\qquad (x\circ \delta) (\lambda) = (1,\lambda).\]

In the \(y\) chart we have \[(y\circ \gamma) (\lambda) = (\lambda, e^{\lambda}), \qquad\text{and}\qquad (y\circ \delta) (\lambda) = (1,\lambda\cdot e)\] The illustrations are as follows

The intersection points are \((1,1)\) in the \(x\) chart and \((1,e)\) in the \(y\) chart. These both correspond to \(\lambda=1\), which it must from the definition of the curves.

Solution: Using the previous question, direct calculation gives: for the \(x\) chart \[\begin{split} \sigma_x(\lambda) & = x^{-1}\big( (\lambda+1,1) + (1,\lambda+1) - (1,1)\big) \\ & = x^{-1}(\lambda+1,\lambda+1) \\ & = (\lambda+1,\lambda+1). \end{split}\] The calculation i the \(y\) chart gives \[\sigma_y(\lambda) = (\lambda+1, 1+ \lambda \cdot e^{-\lambda}).\]

Solution: The intersection point is \((1,1)\) in \(x\) and so we require \(\lambda=0\). Using this, we have \[\dot{\sigma}_{(x)x}^1 = 1 \qquad\text{and}\qquad \dot{\sigma}_{(x)x}^2 = 1,\] as both are just the derivative w.r.t. \(\lambda\) of \(\lambda+1\).

In the \(y\) chart we have (note we use the \(x\) chart in to find these components in order to compare it to the above ones) \[\dot{\sigma}_{(x)y}^1 = 1 \qquad\text{and}\qquad \dot{\sigma}_{(x)y}^2 = 1 \cdot e^0 - 0\cdot e^0 = 1.\]

So we have \[v_{\sigma_x,(1,1)} = \dot{\sigma}_{(x)x}^i \bigg(\frac{\partial}{\partial x^i}\bigg)_p = \dot{\sigma}_{(x)y}^i\bigg(\frac{\partial}{\partial x^i}\bigg)_p = v_{\sigma_y,(1,1)}.\]

26.6 Fields

Vector Fields For Practitioners

Solution: We need to check that it is a section. That is \(\pi\circ \frac{\partial}{\partial x^i} = \beta1_{\mathcal{M}}\). This is clearly true as \(\pi:T_p\mathcal{M} \to \mathcal{M}\) is defied as \[\pi : \bigg(\frac{\partial}{\partial x^i}\bigg)_p \mapsto p.\] We now need to check if this section is smooth. If we denote the chart map on \(TU\) as \(\xi_x\), that is we need to check that \[\xi_x \circ \frac{\partial}{\partial x^i} \circ x^{-1} : x(U) \to \xi(TU)\] is smooth. Recalling the definition of \(\xi_x\), and using the fact that the only non-vanishing component of \(\frac{\partial}{\partial x^i}\) is the \(i^{\text{th}}\) entry, we have \[\bigg(\xi_x \circ \frac{\partial}{\partial x^i} \circ x^{-1}\bigg)(\alpha^1,...,\alpha^d) = (\alpha^1,...,\alpha^d, 0,...,1,...,0),\] where the \(1\) appears at the \((d+i)^{\text{th}}\) entry. This is clearly smooth (w.r.t. the standard topologies on \(\mathbb{R}^d\) and \(\mathbb{R}^{2d}\)), and so \(\frac{\partial}{\partial x^i}\) is a vector field on \(U\).

The Cotangent Bundle \(T^*\mathcal{M} \xrightarrow{\pi} \mathcal{M}\)

Solution: We check the three conditions for a topology in the order given in the definition.

1. We have \[\text{preim}_{\pi}(\emptyset) = \emptyset, \qquad\text{and}\qquad \text{preim}_{\pi}(\mathcal{M}) = T^*\mathcal{M},\] and so \(\emptyset,T^*\mathcal{M}\in\mathcal{O}_{T^*\mathcal{M}}\).

2. This just follows from properties of the preimage, namely \[\text{preim}_f(U\cap V) = \text{preim}_f(U)\cap \text{preim}_f(V).\]

3. This just follows from another property of the preimage, namely \[\bigcup_i \text{preim}_f(U_i) = \text{preim}_f\Big(\bigcup_i U_i\Big).\]

26.7 Connections

Practical Rules For How \(\nabla\) Acts

Solution: We have \[\begin{split} \big(\nabla_{[m} A\big)_{n]} & := \frac{1}{2}\Big[\big(\nabla_mA\big)_n - \big(\nabla_nA\big)_m\Big] \\ & = \frac{1}{2} \big( A_{n,m} - {\Gamma^r}_{nm}A_r - A_{m,n} + {\Gamma^r}_{mn}A_r\big) \\ & = A_{[n,m]} + {\Gamma^r}_{[mn]} \\ & = \frac{1}{2}F_{mn}, \end{split}\] where we have used the fact that \(\nabla\) is torsion free and so \({\Gamma^a}_{[bc]}=0\).

Next, we have \[\big(\nabla_{[m}\omega\big)_{nr]} = \frac{1}{3!}\Big[ \big(\nabla_{m}\omega\big)_{nr} - \big(\nabla_{m}\omega\big)_{rn} + \big(\nabla_{r}\omega\big)_{mn} - \big(\nabla_{r}\omega\big)_{nm} + \big(\nabla_{n}\omega\big)_{rm} - \big(\nabla_{n}\omega\big)_{mr}\Big].\] If we expand this out we will see that the \(\Gamma\)s all cancel, e.g. \[\begin{split} \big(\nabla_{m}\omega\big)_{nr} - \big(\nabla_{n}\omega\big)_{mr} & = \omega_{nr,m} - {\Gamma^s}_{nm}\omega_{sr} - {\Gamma^s}_{rm}\omega_{ns} - \omega_{mr,n} + {\Gamma^s}_{mn}\omega_{sr} + {\Gamma^s}_{rn}\omega_{ms} \\ & = \omega_{nr,m} - \omega_{mr,n} + 2{\Gamma^s}_{[mn]}\omega_{sr} + 2{\Gamma^s}_{r[n}\omega_{m]s} \\ & = \omega_{nr,m} - \omega_{mr,n} + 2{\Gamma^s}_{r[n}\omega_{m]s}, \end{split}\] and the other \(\Gamma\) will cancel with another term’s expansion. We are therefore left with \[\big(\nabla_{[m}\omega\big)_{nr]} = \omega_{[nr,m]}.\]

Connection Coefficients

Solution: Recall that the transformation law is \[\Gamma^a_{(y)bc} = \frac{\partial y^a}{\partial x^k} \frac{\partial^2 x^k}{\partial y^b\partial y^c} + \frac{\partial y^a}{\partial x^k} \frac{\partial x^n}{\partial y^b}\frac{\partial x^m}{\partial y^c} \Gamma^{k}_{(x)mn}.\] The assumption is \(\Gamma^q_{(x)sp}=0\), and so we just need to find the first term. We have \[\bigg(\frac{\partial x^k}{\partial y^c}\bigg)_{y^{-1}(r,\varphi)} := \partial_c\big(x^k\circ y^{-1}\big)(r,\varphi) = \begin{pmatrix} \cos\varphi & -r\sin\varphi \\ \sin\varphi & r\cos\varphi \end{pmatrix}^k_{\,\,\, c}.\] Now, using the first question in the tangent spaces tutorial above, we have \[\begin{split} \bigg(\frac{\partial y^a}{\partial x^k}\bigg)_p & = \bigg(\frac{\partial x^k}{\partial y^a}\bigg)^{-1} \\ & = \frac{1}{r} \begin{pmatrix} r\cos\varphi & r\sin\varphi \\ -\sin\varphi & \cos\varphi \end{pmatrix}^k_{\,\,\, a} \\ & = \begin{pmatrix} \cos\varphi & \sin\varphi \\ -\frac{1}{r}\sin\varphi & \frac{1}{r}\cos\varphi \end{pmatrix}^k_{\,\,\, a}, \end{split}\] where we have used the fact that the determinant of the first matrix is \(r\). We can also calculate \[\begin{split} \bigg(\frac{\partial^2 x^1}{\partial y^1 \partial y^1}\bigg)_{y^{-1}(r,\varphi)} & = 0 \\ \bigg(\frac{\partial^2 x^1}{\partial y^2 \partial y^1}\bigg)_{y^{-1}(r,\varphi)} & = -\sin\varphi \\ \bigg(\frac{\partial^2 x^1}{\partial y^1 \partial y^2}\bigg)_{y^{-1}(r,\varphi)} & = -\sin\varphi \\ \bigg(\frac{\partial^2 x^1}{\partial y^2 \partial y^2}\bigg)_{y^{-1}(r,\varphi)} & = -r\cos\varphi \\ \bigg(\frac{\partial^2 x^2}{\partial y^1 \partial y^1}\bigg)_{y^{-1}(r,\varphi)} & = 0 \\ \bigg(\frac{\partial^2 x^2}{\partial y^2 \partial y^1}\bigg)_{y^{-1}(r,\varphi)} & = \cos\varphi \\ \bigg(\frac{\partial^2 x^2}{\partial y^2 \partial y^2}\bigg)_{y^{-1}(r,\varphi)} & = -r\sin\varphi. \end{split}\]

We then simply have to pick the relevant expressions to find the \(\Gamma^a_{(y)bc}\)s. For example \[\Gamma^1_{(y)11} = \frac{\partial y^1}{\partial x^1} \frac{\partial^2 x^1}{\partial y^1\partial y^1} + \frac{\partial y^1}{\partial x^2} \frac{\partial^2 x^2}{\partial y^1\partial y^1} = 0,\] and \[\begin{split} \Gamma^1_{(y)22} & = \frac{\partial y^1}{\partial x^1} \frac{\partial^2 x^1}{\partial y^2\partial y^2} + \frac{\partial y^1}{\partial x^2} \frac{\partial^2 x^2}{\partial y^2\partial y^2} \\ & = \cos\varphi (-r\cos\varphi) + \sin\varphi(-r\sin\varphi) \\ & = -r(\cos^2\varphi + \sin^2\varphi) \\ & = -r. \end{split}\]

26.8 Parallel Transport and Curvature

Where Connection Coefficients Appear

Solution: Recall the definition \[\text{Riem}(\omega,Z,X,Y) := \omega : \Big( \nabla_X\nabla_YZ - \nabla_Y\nabla_XZ - \nabla_{[X,Y]}Z\Big).\] We have already seen² that Riem is \(C^{\infty}\)-linear in all its entries and so we can just consider basis elements. That is, we can set (using the notation \(\partial_j := \frac{\partial}{\partial x^j}\)) \[\omega = dx^i, \qquad Z = \partial_j, \qquad X = \partial_k, \qquad\text{and}\qquad Y = \partial_m.\] So we have \[\begin{split} \text{Riem}(dx^i,\partial_j,\partial_k,\partial_m) & := dx^i : \Big(\nabla_k\nabla_m\partial_j - \nabla_m\nabla_k\partial_j - \nabla_{[\partial_k,\partial_m]}\partial_j\Big) \\ & = dx^i : \Big[ \nabla_k \big(\Gamma^r_{(x)jm}\partial_r\big) - \nabla_m\big( \Gamma^r_{(x)jk}\partial_r\big) \Big] \\ & = dx^i : \Big[ \Gamma^r_{(x)jm,k}\partial_r + \Gamma^r_{(x)jm}\Gamma^s_{(x)rk}\partial_s - \Gamma^r_{(x)jk,m}\partial_r - \Gamma^r_{(x)jk}\Gamma^s_{(x)rm}\partial_s \Big] \\ {\text{Riem}^i}_{jkm} & = \Gamma^i_{(x)jm,k} - \Gamma^i_{(x)jk,m} + \Gamma^r_{(x)jm}\Gamma^i_{(x)rk} - \Gamma^r_{(x)jk}\Gamma^i_{(x)rm}. \end{split}\] We can see the antisymmetry in the last two entries immediatley from the above. That is \({\text{Riem}^i}_{jkm} = -{\text{Riem}^i}_{jmk}\).

Solution: No, as if it is one-dimensional we only have one \(\Gamma\), namely \({\Gamma^1}_{11}\), and if we put this into the above definition, all the only component \({\text{Riem}^1}_{111}\) vanishes and so there is no curvature.

Geometrically this makes sense if we think about embedding the one-dimensional manifold into a higher dimensional space. We can always ‘pull’ the one-dimensional manifold straight and thus show that it has no intrinsic curvature.

The Round Sphere

How Not To Define Parallel Transport

Solution: Consider the transformation to another chart with the same domain \((U,y)\), \[X^i_{(y)} = \bigg(\frac{\partial y^i}{\partial x^j}\bigg)_p X^j_{(x)} = \bigg(\frac{\partial y^i}{\partial x^j}\bigg)_p Y^j_{(x)} = \bigg(\frac{\partial y^i}{\partial x^j}\bigg)_p \bigg(\frac{\partial x^j}{\partial y^k}\bigg)_q Y^k_{(y)},\] but because \(p\neq q\) we cannot, in general, use \[\bigg(\frac{\partial y^i}{\partial x^j}\bigg)_p \bigg(\frac{\partial x^j}{\partial y^k}\bigg)_q = \delta^i_k,\] and so \[X^i_{(y)} \neq Y^i_{(y)}.\]

26.9 Metric Manifolds

Recognising & Dealing With Different Signatures

Solution: W.l.o.g. let’s assume the vector space is 3-dimensional and introduce a basis \(\{e_1,e_2,e_3\}\). The relevant equations giving the correct surfaces are

1. \(g(X,X) = -(X^1)^2 + (X^2)^2 + (X^3)^2 = 0\), so we have \((-,+,+)\), or equivalently \((+,-,-)\).

2. \(g(X,X)=0\) only for the zero vector, and so we need \((+,+,+)\) or \((-,-,-)\).

3. From the above case, we just need to have one of the entries to be 0, i.e. \((0,+,+)\)/\((0,-,-)\), as then all vectors with only an \(e^1\) component are null.

4. Extending the previous case, we have \((0,0,+)\)/\((0,0-)\).

Levi-Civita Connection

Solution: We just do the first one, as the other two are obtained by simply relabeling the indices. We have \[\big(\nabla_ag\big)_{bc} = g_{bc,a} - {\Gamma^m}_{ba}g_{mc} - {\Gamma^m}_{ca}g_{bm}.\]

Solution: Consider (i)+(ii)-(iii), \[g_{bc,a} - {\Gamma^m}_{ba}g_{mc} - {\Gamma^m}_{ca}g_{bm} + g_{ca,b} - {\Gamma^m}_{cb}g_{ma} - {\Gamma^m}_{ab}g_{cm} - g_{ab,c} + {\Gamma^m}_{ac}g_{mb} + {\Gamma^m}_{bc}g_{am}.\] Now if we consider a metric compatible connection and vanishing torsion, we have that the above vanishes (as each of (i), (ii) and (iii) vanish themselves) and that the \(\Gamma\)s are symmetric in the lower two indices. Also using the fact that the metric is symmetric, we have \[0 = g_{bc,a} + g_{ac,b} - g_{ab,c} - 2{\Gamma^m}_{ba}g_{mc},\] which after rearranging and using \(g_{mc}g^{-1})^{cn} = \delta^n_m\) we have \[{\Gamma^n}_{ba} = \frac{1}{2}\big(g^{-1}\big)^{cn}\big(g_{bc,a} + g_{ac,b} - g_{ab,c}\big),\] which, relabelling \(n\to a \to c \to m\) gives \[{\Gamma^a}_{bc} = \frac{1}{2}\big(g^{-1}\big)^{ma}\big(g_{bm,c} + g_{cm,b} - g_{cb,m}\big)\] which is the result (when you us the symmetries). So we see the connection coefficient functions are uniquely determined by the metric components given the above conditions.

Massaging The Length Functional

Solution: Using \[\mathcal{L}[\gamma] = \sqrt{g(v_{\gamma},v_{\gamma})}, \qquad\text{and}\qquad \dot{\widetilde{\gamma}}^a(\lambda) = (x^a\circ \gamma)'(\lambda),\] and introducing the notation \[\widetilde{\lambda} := \sigma(\lambda), \qquad \implies \qquad \widetilde{\gamma}(\lambda) = \gamma(\widetilde{\lambda}),\] we have \[\begin{split} L[\widetilde{\gamma}] & := \int_0^1 d\lambda \sqrt{g(v_{\widetilde{\gamma}},v_{\widetilde{\gamma}})} \\ & = \int_0^1 d\lambda \sqrt{g_{ab}\big(\widetilde{\gamma}(\lambda)\big)\cdot\dot{\widetilde{\gamma}}^a(\lambda) \cdot \dot{\widetilde{\gamma}}^b}(\lambda) \\ & = \int_0^1 d\lambda \sqrt{g_{ab}\big(\gamma(\widetilde{\lambda})\big) \cdot (x^a\circ \gamma \circ \sigma)'(\lambda) \cdot (x^b\circ \gamma \circ \sigma)'(\lambda)} \\ & = \int_0^1 d\lambda \sqrt{g_{ab}\big(\gamma(\widetilde{\lambda})\big) \cdot (x^a\circ \gamma)'(\widetilde{\lambda})\cdot \dot{\sigma}(\lambda) \cdot (x^b\circ \gamma)'(\widetilde{\lambda)}\cdot\dot{\sigma}(\lambda)} \\ & = \int_0^1 d\lambda \dot{\sigma}(\lambda)\sqrt{g_{ab}\big(\gamma(\widetilde{\lambda})\big) \cdot \dot{\gamma}^a(\widetilde{\lambda}) \cdot \dot{\gamma}^b(\widetilde{\lambda})} \\ & = \int_0^1 d\widetilde{\lambda} \sqrt{g_{ab}\big(\gamma(\widetilde{\lambda})\big) \cdot \dot{\gamma}^a(\widetilde{\lambda}) \cdot \dot{\gamma}^b(\widetilde{\lambda})} \\ & = L[\gamma], \end{split}\] where we have used the chain rule, and the result \(d\widetilde{\lambda} = \dot{\sigma}d\lambda\).

Solution: Let’s denote the canonical variables at \(t\) and \(q\), then the Euler-Lagrange equations for \(\mathcal{L}\) read \[\frac{d}{dt}\bigg(\frac{\partial \mathcal{L}}{\partial \dot{q}^a}\bigg) - \frac{\partial \mathcal{L}}{\partial q^a} = 0.\] Substituting in \(\mathcal{L} := \sqrt{\mathcal{T}}\), and using the assumption that we only consider solutions were \(\mathcal{T}=1\), and so it is a constant w.r.t. \(t\), we have \[\begin{split} 0 & = \frac{d}{dt}\bigg(\frac{\partial \sqrt{\mathcal{T}}}{\partial \dot{q}^a}\bigg) - \frac{\partial \sqrt{\mathcal{T}}}{\partial q^a} \\ & = \frac{d}{dt}\bigg(\frac{1}{2\sqrt{\mathcal{T}}}\frac{\partial \mathcal{T}}{\partial \dot{q}^a}\bigg) - \frac{1}{2\sqrt{\mathcal{T}}} \frac{\partial \mathcal{T}}{\partial q^a} \\ & = \frac{1}{2\sqrt{\mathcal{T}}} \bigg[ \frac{d}{dt}\bigg(\frac{\partial \mathcal{T}}{\partial \dot{q}^a}\bigg) - \dfrac{\partial \mathcal{T}}{\partial q^a} \bigg], \end{split}\] which multiplying out the \(1/2\sqrt{\mathcal{T}}\) gives the answer.

A Practical Way To Quickly Determine Christoffel Symbols

Solution: We have \(\mathcal{L}[\gamma] :=\sqrt{g(v_{\gamma},v_{\gamma})}\), but the previous question showed us we can instead consider \(\mathcal{T}:= g(v_{\gamma},v_{\gamma}) = g_{ab}\dot{\gamma}^a\dot{\gamma}^b\) if we restrict ourselves to \(\mathcal{T}=1\) on the parameterisation. For our metric components, we have \[\mathcal{T} = R^2 \cdot \dot{\vartheta}(\lambda) \cdot \dot{\vartheta}(\lambda) + R^2\sin^2\vartheta \cdot \dot{\phi}(\lambda) \cdot \dot{\phi}(\lambda).\] Plugging this into our Euler-Lagrange equations, we have \[\begin{split} \frac{d}{d\lambda}\bigg(\frac{\partial \mathcal{T}}{\partial \dot{\vartheta}}\bigg) - \frac{\partial \mathcal{T}}{\partial \vartheta} & = 2R^2 \big(\ddot{\vartheta}(\lambda) - \sin\vartheta\cos\vartheta \cdot \dot{\phi}^2(\lambda)\big) \\ \frac{d}{d\lambda}\bigg(\frac{\partial \mathcal{T}}{\partial \dot{\phi}}\bigg) - \frac{\partial \mathcal{T}}{\partial \phi} & = 2R^2\sin\vartheta \big( \sin\vartheta\cdot \ddot{\phi}(\lambda) + 2\cos\vartheta \cdot \dot{\vartheta}\cdot \dot{\phi}\big) \end{split}\] which simplify to \[\begin{split} \ddot{\vartheta}(\lambda) - \sin\vartheta\cos\vartheta \cdot \dot{\phi}^2(\lambda) & = 0 \\ \ddot{\phi} + 2\cot\vartheta \cdot \dot{\vartheta}\cdot \dot{\phi} & = 0. \end{split}\] These are the geodesic equations for our round sphere of radius \(R\).

Solution: Comparing the above result to the geodesic equation for the metric induced connection coefficients, \[\ddot{\gamma}^a + {\Gamma^a}_{bc}\dot{\gamma}^b\dot{\gamma}^c,\] we see straight away that \[{\Gamma^1}_{22} = -\sin\vartheta\cos\vartheta, \qquad\text{and}\qquad {\Gamma^2}_{12} = {\Gamma^2}_{21} = \cot\vartheta,\] with all other \(\Gamma\)s vanishing. Note in the second expression there is no \(2\) as we distribute it across the \({\Gamma^2}_{12}\) and \({\Gamma^2}_{21}\).

Properties Of The Riemann-Christoffel Tensor

Solution: Using \((g^{-1})^{ab}g_{bs} = \delta^a_s\), we have \[\begin{split} 0 & = \frac{\partial}{\partial x^c} \delta^a_s \\ & = \frac{\partial}{\partial x^c}\Big( \big(g^{-1}\big)^{ab}g_{bs}\Big) \\ & = \bigg(\frac{\partial}{\partial x^c}\big(g^{-1}\big)^{ab}\bigg)g_{bs} + \big(g^{-1}\big)^{ab}\bigg(\frac{\partial}{\partial x^c}g_{bs}\bigg) \\ & = \frac{\partial}{\partial x^c}\big(g^{-1}\big)^{ar} + \big(g^{-1}\big)^{sr}\big(g^{-1}\big)^{ab}\bigg(\frac{\partial}{\partial x^c}g_{bs}\bigg) \end{split}\] which, after relabelling and rearranging gives the result.

Solution: This is just a long calculation involving the product rule and using the fact that in normal coordinates all the \(\Gamma\)s vanish. The full calculation is given in the video. The result is \[R_{abcd} = \frac{1}{2}\big( g_{ad,bc} - g_{bd,ac} + g_{ac,bd} - g_{bc,ad} \big),\] where \[g_{ab,cd} := \frac{\partial^2 g_{ab}}{\partial x^c\partial x^d}.\]

Solution: Switching the indices \(a\leftrightarrow b\) in the result of the previous exercise we have \[R_{bacd} = \frac{1}{2}\big( g_{bd,ac} - g_{ad,bc} + g_{bc,ad} - g_{ac,bd}\big) = -\frac{1}{2}\big( g_{ad,bc} - g_{bd,ac} + g_{ac,bd} - g_{bc,ad} \big) = - R_{abcd}.\]

Solution: Again just switch the indices and get the result.

Solution: We have \[R_{a[bcd]} := \frac{1}{3!}\big( R_{abcd} - R_{abdc} + R_{acdb} - R_{acbd} + R_{adbc} - R_{adcb}\big).\] Then using the previous two questions, we also have \[R_{abcd} = R_{cdab} = -R_{dcab} = -R_{abdc}.\] Following this with the other indice arrangments, we get \[R_{a[bcd]} = \frac{1}{3}\big( R_{abcd} + R_{acdb} + R_{adbc}\big).\] If you then plug in the expansion in terms of \(g_{ab}\) and its derivatives, you can show everything cancels and you get the result.

26.10 Symmetry

Pull-Back & Push-Forward

Solution: By definition, we have \[(\phi_*X)\langle f \rangle = X \langle f\circ \phi \rangle,\] for \(X\in T\mathcal{M}\), giving \[\begin{split} \phi^*(df) : X & := df : \phi_*(X) \\ & = \phi_* X \langle f \rangle \\ & = X \langle f\circ \phi \rangle \\ & = d(f\circ \phi) : X \\ & =: d(\phi^*f) :X, \end{split}\] which holds for arbitrary \(X\) and therefore proves the result.

Solution: If we considered a general vector and gradient, we would have \[df : \phi_*X = \phi_*X\langle f \rangle = X\langle f \circ \phi \rangle = X^i \bigg(\frac{\partial(f\circ\phi)}{\partial x^i}\bigg).\] Now we have \[\begin{split} \bigg(\frac{\partial (f\circ\phi)}{\partial x^i}\bigg)_p & := \partial_i\big(f\circ\phi\circ x^{-1}\big)\big|_{x(p)} \\ & = \partial_i\big(f\circ y^{-1}\circ y\circ \phi\circ x^{-1}\big)\big|_{x(p)} \\ & = \partial_j\big(f\circ y^{-1}\big)\big|_{(y\circ\phi\circ x^{-1}\circ x)(p)} \cdot \partial_i\big(y^i\circ \phi\circ x^{-1}\big)\big|_{x(p)} \\ & =: \bigg(\frac{\partial f}{\partial y^j}\bigg)_q \cdot \bigg(\frac{\partial (y^j\circ\phi)}{\partial x^i}\bigg)_p, \end{split}\] where \(q:=\phi(p)\in\mathcal{N}\). Now put in \(f=y^a\) and \(X=\frac{\partial}{\partial x^b}\), giving \[\begin{split} \phi_{*\,\, b}^{\,\, a}(p) & = \bigg(\frac{\partial y^a}{\partial y^j}\bigg)_q \cdot \bigg(\frac{\partial (y^j\circ \phi)}{\partial x^b}\bigg)_p \\ & = \delta^a_j \cdot \bigg(\frac{\partial (y^j\circ \phi)}{\partial x^b}\bigg)_p \\ & = \bigg(\frac{\partial (y\circ \phi)^a}{\partial x^b}\bigg)_p. \end{split}\]

Solution: With the definition of the induced metric in mind, we have \[\begin{split} (\phi^*g)(X,Y) & = g\big(\phi_*X,\phi_*Y\big) \\ & = g_{ab}\big(\phi_*X\big)^a\big(\phi_*Y\big)^b \\ & = g_{ab} \big(dy^a:\phi_*X\big) \big(dy^b:\phi_*Y\big). \end{split}\] Then if we use \(g_{ab}=g\big(\frac{\partial}{\partial x^a},\frac{\partial}{\partial x^b}\big)\) and the results of the previous exercise we get \[\begin{split} (\phi^*g)_{ab}(p) & = g_{mn}(q) \cdot \bigg[dy^m:\phi_*\bigg(\frac{\partial}{\partial x^a}\bigg)\bigg](p) \cdot \bigg[dy^n:\phi_*\bigg(\frac{\partial}{\partial x^b}\bigg)\bigg](p) \\ & = \bigg(\frac{\partial(y\circ\phi)^m}{\partial x^a}\bigg)_p \bigg(\frac{\partial(y\circ\phi)^n}{\partial x^b}\bigg)_p g_{mn}\big(\phi(p)\big). \end{split}\]

Lie Derivative — The Pedestrian Way

Solution: Nothing as in order to talk about shape you need either a covariant derivative or a metric, neither of which we have. I guess you could say it is some closed and compact 2-dimensional shape, but you could not specify which one, i.e. if its a round sphere or an ellipsoid or a potato.

Solution: Using the result of the previous questions, we need to find \[\bigg(\frac{\partial (y\circ \iota)^m}{\partial x^a}\bigg)_p := \partial_a\big(y^m\circ \iota \circ x^{-1}\big)\big|_{x(p)}.\] Using the definition given, we have \[\begin{split} \bigg(\frac{\partial (y\circ \iota)^1}{\partial x^1}\bigg)_p & = a\cos\varphi\cos\vartheta, \qquad \bigg(\frac{\partial (y\circ \iota)^1}{\partial x^2}\bigg)_p = -a\sin\varphi\sin\vartheta \\ \bigg(\frac{\partial (y\circ \iota)^2}{\partial x^1}\bigg)_p & = b\sin\varphi\cos\vartheta, \qquad \bigg(\frac{\partial (y\circ \iota)^2}{\partial x^2}\bigg)_p = b\cos\varphi\sin\vartheta \\ \bigg(\frac{\partial (y\circ \iota)^3}{\partial x^1}\bigg)_p & = -c\sin\vartheta, \qquad \qquad \bigg(\frac{\partial (y\circ \iota)^3}{\partial x^2}\bigg)_p = 0. \end{split}\] You then just plug in the relevant terms, giving \[\begin{split} g^{\text{ellipsoid}}_{11} & = a^2 \cos^2\varphi\cos^2\vartheta + b^2\sin^2\varphi\cos^2\vartheta + c^2\sin^2\vartheta \\ g^{\text{ellipsoid}}_{22} & = a^2\sin^2\varphi\sin^2\vartheta + b^2\cos^2\varphi\sin^2\vartheta, \end{split}\] and \(g^{\text{ellipsoid}}_{12}=0=g^{\text{ellipsoid}}_{21}\).

Solution: Let \(f\in C^{\infty}(\mathcal{M})\) be an arbitrary smooth function. We need to consider the action of the Lie bracket expressions on \(f\), e.g. \([X_1,X_2]\langle f\rangle\). We use the clever trick that in this expansion only the terms where a derivative acts on a term in the \(X\)s will remain. That is, any terms that are second order derivative of \(f\) will vanish because it will appear in both \(X_1\langle X_2\langle f\rangle \rangle\) and \(X_2\langle X_1\langle f\rangle \rangle\) which the order switched, but partial derivative commute and so these terms cancel.

We then have (dropping the \(p\)s for notational reasons) \[\begin{split} [X_1,X_2]\langle f \rangle & := X_1\big\langle X_2\langle f\rangle \big\rangle - X_2\big\langle X_1\langle f\rangle \big\rangle \\ & = \bigg[\big(-\mathrm{cosec}^2\vartheta \sin^2\varphi + \cot^2\vartheta\cos^2\varphi\big) \bigg(\frac{\partial f}{\partial\varphi}\bigg) + \cot\vartheta\cos\varphi\sin\varphi \bigg(\frac{\partial f}{\partial \vartheta}\bigg)\bigg] \\ & \qquad - \bigg[\big(\mathrm{cosec}^2\vartheta\cos^2\varphi - \cot^2\vartheta\sin^2\varphi\big) \bigg(\frac{\partial f}{\partial \varphi}\bigg) + \cot\vartheta\sin\varphi\cos\varphi \bigg(\frac{\partial f}{\partial \vartheta}\bigg)\bigg] \\ & = \big(\cot^2\vartheta-\mathrm{cosec}^2\vartheta\big)\bigg(\frac{\partial f}{\partial \varphi}\bigg) \\ & = -\bigg(\frac{\partial }{\partial \varphi}\bigg)\langle f \rangle \\ & = - X_3\langle f \rangle \\ \implies [X_2,X_1] & = X_3, \end{split}\] where on the last line we have used the antisymmetry of the Lie bracket.

Next we have \[\begin{split} [X_1,X_3]\langle f \rangle & = 0 - \bigg[-\cos\varphi \bigg(\frac{\partial f}{\partial \vartheta}\bigg) + \cot\vartheta\sin\varphi\bigg(\frac{\partial f}{\partial \varphi}\bigg)\bigg] \\ & = X_2\langle f \rangle \\ \implies [X_1,X_3] & = X_2, \end{split}\] and \[\begin{split} [X_3,X_2] & = \bigg[ -\sin\varphi\bigg(\frac{\partial f}{\partial \vartheta}\bigg) -\cot\vartheta\cos\varphi\bigg(\frac{\partial f}{\partial \varphi}\bigg) \bigg] - 0 \\ & = X_1\langle f\rangle \\ \implies [X_3,X_2] = X_1. \end{split}\] So we see that \(\{X_1,X_2,X_3\}\) is closed under the Lie bracket, and so forms a Lie subalgebra. The structure constants are \[{C^3}_{21} = {C^2}_{13} = {C^1}_{32} = 1,\] and all other non-related (i.e. not \({C^3}_{12} = - {C^3}_{21}\), etc.) structure constants vanish.

Note this result tells us that \(\{X_1,X_2,X_3\}\) is a 3-dimensional rotation algebra, as defined in the lecture. We therefore expect it to be a symmtry of \(S^2\), which we show explictly below for \(X_3\).

Solution: Using \[v_{\gamma_p,\gamma_p(\lambda)} = (X_3)_{\gamma_p(\lambda)} \qquad \iff \qquad \dot{\gamma}^i_{p(x)}\bigg(\frac{\partial}{\partial x^i}\bigg)_{\gamma_p(\lambda)} = \bigg(\frac{\partial}{\partial \varphi}\bigg)_{\gamma_p(\lambda)},\] we have \[\dot{\gamma}^1_{p(x)}(\lambda) := \big(x^1\circ \gamma_p\big)'(\lambda) = 0, \qquad\text{and}\qquad \dot{\gamma}^2_{p(x)}(\lambda) := \big(x^2\circ \gamma_p\big)'(\lambda) = 1,\] from which is follows that \[\vartheta(\gamma_p) = a, \qquad\text{and}\qquad \varphi(\gamma_p) = \lambda + b,\] for constants \(a\) and \(b\). We see from the question that \(a=\vartheta_0\) and \(b=\varphi_0\). So we have \[\gamma_{p(x)}(\lambda) = \big(\vartheta_0, \lambda + \varphi_0\big),\] which satisfies \[\gamma_p(0) = x^{-1}\big(\gamma_{p(x)}(0)\big) = x^{-1}(\vartheta_0,\varphi_0) = p.\]

Solution: The flow is \[h^{X_3}_{\lambda} : p \mapsto \gamma_p(\lambda),\] so we have \[\Big(x^m\circ h^{X_3}_{\lambda}\circ x^{-1}\Big) : (\vartheta,\varphi) \mapsto \gamma^m_p(\lambda),\] and so \[\begin{split} \bigg(\frac{\partial (x\circ h^{X_3}_{\lambda})^1}{\partial x^1}\bigg)_p & := \partial_1\Big( x^1 \circ h^{X_3}_{\lambda}\circ x^{-1}\Big)\Big|_{x(p)} = \vartheta_0 \\ \bigg(\frac{\partial (x\circ h^{X_3}_{\lambda})^1}{\partial x^2}\bigg)_p & := \partial_2\Big( x^1 \circ h^{X_3}_{\lambda}\circ x^{-1}\Big)\Big|_{x(p)} = 0 \\ \bigg(\frac{\partial (x\circ h^{X_3}_{\lambda})^2}{\partial x^1}\bigg)_p & := \partial_1\Big( x^2 \circ h^{X_3}_{\lambda}\circ x^{-1}\Big)\Big|_{x(p)} = 0 \\ \bigg(\frac{\partial (x\circ h^{X_3}_{\lambda})^2}{\partial x^2}\bigg)_p & := \partial_2\Big( x^2 \circ h^{X_3}_{\lambda}\circ x^{-1}\Big)\Big|_{x(p)} = \lambda + \varphi_0. \end{split}\] This gives us \[\begin{split} \Big[\Big(h^{X_3}_{\lambda}\Big)^*g^{\text{ellipsoid}}\Big]_{11}(p) & = \vartheta_0^2 g^{\text{ellipsoid}}_{11}\big(\gamma_p(\lambda)\big) \\ \Big[\Big(h^{X_3}_{\lambda}\Big)^*g^{\text{ellipsoid}}\Big]_{22}(p) & = (\lambda+\varphi_0)^2 g^{\text{ellipsoid}}_{22}\big(\gamma_p(\lambda)\big) \end{split}\] and again the other two components vanish. If we then take the coordinate transformation \[\vartheta \to \frac{1}{\vartheta_0}\vartheta, \qquad\text{and}\qquad \varphi \to \frac{1}{\lambda+\varphi_0}\varphi,\] which we can do as \(\vartheta_0,\varphi_0 >0\) (as the ranges of \(\vartheta\) and \(\varphi\) are positive), we then get \[\Big[\Big(h^{X_3}_{\lambda}\Big)^*g^{\text{ellipsoid}}\Big]_{ab}(p) = g^{\text{ellipsoid}}_{ab}\big(\gamma_p(\lambda)\big),\] or more nicely \[\Big(h^{X_3}_{\lambda}\Big)^*g^{\text{ellipsoid}} = g^{\text{ellipsoid}}.\] This tells us that \(X_3\) is a symmetry of the metric, and so \(\mathcal{L}_{X_3}g^{\text{ellipsoid}} = 0\).

26.11 Integration

Integrals & Volumes

Solution: The first thing we have to not is that ‘volume’ here does not mean what we intuitively think, i.e. the Euclidean 3-volume, but it means what we would normally call the ‘surface area’. This distinction comes from which metric we are using, the metric on \(S^2\) itself or the Euclidean metric with a sphere of radius \(R\) embedded into it. Once this distinction is made the calculation is trivial, consider the chart with \((x^1,x^2) = (\vartheta,\varphi)\), then \[g_{(x)ab}\big(x^{-1}(\vartheta,\varphi)\big) = \begin{pmatrix} R^2 & 0 \\ 0 & R^2\sin^2\vartheta \end{pmatrix}, \qquad \implies \qquad g := \det\big(g_{(x)ab}\big)\big(x^{-1}(\vartheta,\varphi)\big) = R^4\sin^2\vartheta\] and so \[\begin{split} \int_{S^2}1 & := \int_{x(S^2)} d^2x \, \sqrt{g} 1 \\ & = \int^{\pi}_0 d\varphi \int^{2\pi}_0 d\vartheta \, \big|R^2\sin^2\vartheta\big| \\ & = 4\pi R^2, \end{split}\] which is what we expect.

Technically we need to include another chart, as \(x(S^2)\) will miss two antipodal points and a geodesic connecting them, however this will contribute nothing to the volume as, we would only consider this line (the partition of unity removing the overlap region), which has no ‘thickness’ and so no volume. For example if the line missing was the line of longitude connecting the North and South poles, we would have \[\int^{\varphi_0}_{\varphi_0}d\varphi \int^{\pi}_0 d\vartheta R^2\sin\vartheta = 0,\] where \(\varphi_0\) is the value of \(\varphi\) along the line of longitude.

26.12 Schwarzschild Spacetime

Geodesics In A Schwarzschild Spacetime

Solution: Using the metric given in the question (see the video if you don’t know it) we have \[\mathcal{L} = \bigg(1-\frac{2GM}{r}\bigg) \dot{t}^2 - \bigg(1-\frac{2GM}{r}\bigg)^{-1} \dot{r}^2 - r^2\dot{\theta}^2 - r^2\sin^2\theta \dot{\varphi}^2.\]

Solution: We see straight away that \[\frac{\partial \mathcal{L}}{\partial t} = 0.\] We also have \[\frac{d}{d\lambda}\bigg(\frac{\partial \mathcal{L}}{\partial \dot{t}}\bigg) = 2\bigg(1-\frac{2GM}{r}\bigg) \ddot{t} + \frac{4GM}{r^2}\dot{r}\dot{t},\] which gives the Euler-Lagrange equation \[\ddot{t} + \frac{2GM}{r^2\Big(1-\frac{2GM}{r}\Big)}\dot{r}\dot{t} = 0\]

Solution: We have \[(\mathcal{L}_{K_t} g)_{ab} = K_t\langle g_{ab}\rangle + g_{mb}\frac{\partial}{\partial x^a}(K_t)^m + g_{ab}\frac{\partial}{\partial x^b}(K_t)^m = 0,\] as all three terms vanish. This tells us that \(K_t\) is a symmetry of the metric. Indeed the Schwarzschild spacetime is stationary (and even static), the definitions for which are given in lecture 16. We will see this symmetry is the conservation of energy.

Solution: Using \((K_t)_a := g_{ab}(K_t)^b\), we have \[g_{ab}(K_t)^b(x^a)'(\lambda) = g_{00}t'(\lambda) = \bigg(1-\frac{2GM}{r}\bigg) t'(\lambda) = \text{const.}\] Letting \(\sqrt{E}\) be the constant, we have \[t'(\lambda) = \frac{r\sqrt{E}}{r-2GM}\]

Solution: Angular momentum.

Solution: From two questions above, we have \[\text{const.} = g_{ab}X_3^b(x^a)'(\lambda) = g_{33}\varphi'(\lambda) = -r^2\sin^2\theta\varphi'(\lambda),\] so using \(\theta=\frac{\pi}{2}\) and labelling the constant \(J\), we have \[\varphi'(\lambda) = \frac{J}{r^2}.\]

Solution: First note that we have replaced the dots with primes, and also note that \(\theta'=0\) as \(\theta\) is a constant. We therefore have \[1 = \bigg(1-\frac{2GM}{r}\bigg)\frac{E}{\Big(1-\frac{2GM}{r}\Big)^2} - \frac{1}{1-\frac{2GM}{r}}(r')^2 - r^2\frac{J^2}{r^4},\] which can be rearranged to \[E = (r')^2 + 1 - \frac{2GM}{r} + \frac{J^2}{r^2} - \frac{2GMJ^2}{r^3}.\]

Solution: If we then consider a particle of mass \(m=1\), we see the above formula as representing

• \(E\) is total energy,

• \((r')^2\) is the kinetic energy,

• \(1\) is the mass,

• \(-\frac{2GM}{r}\) is the Newtonian gravitational potential,

• \(\frac{J^2}{r^2}\) is some angular momentum contribution, and

• \(-\frac{2GMJ^2}{r^3}\) is some GR correction term.

Gravitational Redshift

Solution: Using \(1=\mathcal{L} =g_{ab}\dot{\gamma}^a\dot{\gamma}^b\), we get \[t'(\lambda) = \bigg(1-\frac{2GM}{r}\bigg)^{-1/2}.\]

Solution: We have \[\begin{split} \Delta t_1 & = \bigg(1-\frac{2GM}{r_1}\bigg)^{-1/2} \Delta \lambda_1 \\ \Delta t_2 & = \bigg(1-\frac{2GM}{r_2}\bigg)^{-1/2} \Delta \lambda_2. \end{split}\] Then, we use the fact that \(K_t\) was a Killing vector field, and so the path taken by the two emitted photons is the same, apart from a constant time shift.

As the above diagram shows, the Killing condition basically tells us that \(\Delta t_1 = \Delta t_2\), and so we get \[\Delta \lambda_2 = \sqrt{\frac{1-\frac{2GM}{r_2}}{1-\frac{2GM}{r_1}}} \Delta \lambda_1\]

Solution: We can think of \(\Delta \lambda\) being the time period (time between photons emitted) and so the frequency (which is the reciprocal of the time period) is \[\frac{\omega_1}{\omega_2} = \frac{\Delta \lambda_2}{\Delta \lambda_1} = \sqrt{\frac{1-\frac{2GM}{r_2}}{1-\frac{2GM}{r_1}}}.\] As \(r_1<r_2\), the above tells us that \(\omega_2<\omega_1\). In terms of wavelength, this is \(\mu_1<\mu_2\) (we have used \(\mu\) for wavelength as \(\lambda\) is already used above), which tells us that the light has been red-shifted. This is seen also in the definition of redshift, \[1+z = \frac{\omega_1}{\omega_2},\] where \(z>0\) is redshift and \(z<0\) is blueshift.

For \(r_2\to \infty\) we have \[\frac{\omega_1}{\omega_2} \to \bigg(1-\frac{2GM}{r_1}\bigg)^{-1/2}.\] For \(r_1\to r_s\), we have \[\frac{\omega_1}{\omega_2} \to \infty,\] which tells us \(\omega_1\to \infty\). In terms of wavelengths, this says \(\mu_2\to\infty\), and so the light is infinitely redshifted. This is a so-called infinite redshift surface.

This result is actually misleading as it really only holds because of the choice of reference frame. We took the coordinate time to be that of the black hole’s rest frame. If we use an in-falling observer’s clock to define the coordinate time, this infinite redshift behaviour disappears. I have discussed this in more detail in the notes I have put on my blog site.

26.13 Relativistic Spacetime, Matter & Gravitation

Lorentz Force Law

Solution: The observer \((\delta,e)\) has a frame such that \[g(e_a,e_b) = \eta_{ab}, \qquad\text{and}\qquad e_0(\lambda) = v_{\delta,\delta(\lambda)}.\] We thus have \[m\big(\nabla_{v_{\gamma}}v_{\gamma}\big)^a = qF^{ac}\eta_{cb}{v_{\gamma}}^b,\] and so \[\begin{split} m\big(\nabla_{v_{\gamma}}v_{\gamma}\big)^0 & = qF^{00}{v_{\gamma}}^0 + \sum_{\beta=1}^3 qF^{0\beta}{v_{\gamma}}^{\beta} \\ m\big(\nabla_{v_{\gamma}}v_{\gamma}\big)^{\alpha} & = -qF^{\alpha0}{v_{\gamma}}^0 - \sum_{\beta=1}^3 qF^{\alpha\beta}{v_{\gamma}}^{\beta}. \end{split}\]

Solution: We set \[\mathcal{F}^{\alpha} = m \big(\nabla_{v_{\delta}}v_{\delta}\big)^{\alpha} = q{F^{\alpha}}_b{v_{\delta}}^b,\] where \(\mathcal{F}\) is the Lorentz force, up to some factors. We can split the right-hand side into two terms, one for \(b=0\) and the other for \(b=\beta\). For the former we note that \[\eta^{\alpha\beta}E_{\beta} = \eta^{\alpha\beta}F_{\beta0} = {F^{\alpha}}_0,\] so, using \(\epsilon^0:(v_{\delta}) = {v_{\delta}}^0\), the first term is simply \[q\eta^{\alpha\beta}E_{\beta}\big(\epsilon^0:v_{\delta}\big).\] The second term need a little more work, but we start using the hint: using \(g^{ab}=\eta^{ab}\) in the observers frame, we have \[(v_{\delta}\times B)^{\alpha} = \eta^{\alpha\mu}\varepsilon_{\mu\rho\sigma} {v_{\delta}}^{\rho}B^{\sigma} := \frac{1}{2}\eta^{\alpha\mu}\varepsilon_{\mu\rho\sigma}\varepsilon^{\sigma\nu\tau} {v_{\delta}}^{\rho}F_{\nu\tau}.\] We see that the \(\eta^{\alpha\mu}\) tells us that \(\mu=\alpha\) in the Levi-Civita symbol. By definition, the only non-vanishing terms on the right, then, have \(\rho\neq\sigma\neq\alpha\). Let’s consider the case for \(\alpha=1\) (the other two follow analogously) \[\begin{split} \frac{1}{2}\eta^{1\mu}\varepsilon_{\mu\rho\sigma}\varepsilon^{\sigma\nu\tau}F_{\nu\tau}{v_{\delta}}^{\rho} & = \frac{1}{2}\varepsilon_{123}\big(\varepsilon^{312}F_{12} + \varepsilon^{321}F_{21}\big){v_{\delta}}^2 + \frac{1}{2}\varepsilon_{132}\big(\varepsilon^{213}F_{13} + \varepsilon^{231}F_{31}\big){v_{\delta}}^3 \\ & = \frac{1}{2} \varepsilon_{123} \big( \varepsilon^{123}F_{12} + (-\varepsilon^{123})(-F_{12})\big) {v_{\delta}}^2 + \frac{1}{2} (-\varepsilon_{123}) \big( (-\varepsilon^{123})F_{13} + \varepsilon^{123}(-F_{13})\big){v_{\delta}}^3 \\ & = F_{12}{v_{\delta}}^2 + F_{13}{v_{\delta}}^3, \end{split}\] where we have indicated where the antisymmetries of \(\varepsilon\) and \(F\) have been used, and we have also used \(\varepsilon_{123}=1=\varepsilon^{123}\). This generalises to \[\frac{1}{2}\eta^{\alpha\mu}\varepsilon_{\mu\rho\sigma}\varepsilon^{\sigma\nu\tau}F_{\nu\tau}{v_{\delta}}^{\rho} = \eta^{\alpha\sigma}F_{\sigma\beta}{v_{\delta}}^{\beta} = {F^{\alpha}}_{\beta}{v_{\delta}}^{\beta},\] where we note the fact that \(F_{\alpha\alpha}=0\) due to antisymmetry.

So the second term in our Lorentz force equation is simply \[q{F^{\alpha}}_{\beta}{v_{\delta}}^{\beta} = q\big( v_{\delta} \times B \big)^{\alpha},\] giving us \[\mathcal{F}^{\alpha} = q\eta^{\alpha\beta} E_{\beta} \big( \epsilon^0:v_{\delta}\big) + q\big(v_{\delta}\times B\big)^{\alpha}.\] Finally, we note that \[\frac{1}{(\epsilon^0:v_{\delta})} \big(v_{\delta}\times B\big)^{\alpha} = \big(\mathbf{v}\times B\big)^{\alpha},\] as everywhere in the calculation above we’ll get terms like \[\frac{v_{\delta}^\rho}{(\epsilon^0:v_{\delta})} = \frac{(\epsilon^\rho:v_{\delta})}{(\epsilon^0:v_{\delta})} =: \mathbf{v}^{\rho}.\] This gives us the form we want on the right-hand side, i.e. \[\frac{1}{(\epsilon^0:v_{\delta})}\mathcal{F}^{\alpha} = q\eta^{\alpha\beta}E_{\beta} + q\big(\mathbf{v}\times B\big)^{\alpha}.\]

Which Curvature Can Feature In Einstein’s Equations?

Solution: First use the antisymmetry to give \[{R^w}_{zab;c} + {R^w}_{zbc;a} + {R^w}_{zca;b} = {R^w}_{zab;c} + {R^w}_{zbc;a} - {R^w}_{zac;b}.\] Now, we can use the fact that \(\nabla\) is metric-compatible and the result \[g^{av}g_{vw} = \delta^a_w\] to give \[\begin{split} g^{av}g_{vw}\big({R^w}_{zab;c} + {R^w}_{zbc;a} + {R^w}_{zca;b}\big) & = {R^a}_{zab;c} + {R^a}_{zbc;a} - {R^a}_{zac;b} \\ & = R_{zb;c} - R_{zc;b} + {R^a}_{zbc;a}. \end{split}\] Then contract with \(g^{bz}\) along with the results \(R^{\,\,\,a}_{z\,\,\,bc} = -{R^a}_{zbc}\)³ and \(R:= g^{ab}R_{ab}\) to give \[\begin{split} g^{bz}\big(R_{zb;c} - R_{zc;b} + {R^a}_{zbc;a}\big) & = R_{;c} - {R^b}_{c;b} - {R^{ba}}_{bc;a} \\ & = R_{;c} - {R^b}_{c;b} - {R^a}_{c;a} \\ & = R_{;c} - 2{R^b}_{c;b}. \end{split}\] Finally using the fact that the Ricci tensor is symmetric,⁴ and contracting with \(g^{ac}\), we have (after relabelling) \[\big(\nabla_aG)^{ab} = {R^{ab}}_{;a} - \bigg(\frac{1}{2}g^{ab}R\bigg)_{;a} = 0.\]

26.14 Cosmology

Killing’s Equation

Solution: If you have done the exercise in the notes to show \[g\big(\nabla_XK,Y\big) + g\big(X,\nabla_YK\big)=0,\] this question follows trivially by setting \(X=\partial_a\) and \(Y=\partial_b\): \[0 = g_{cb}\big(\nabla_aK)^c + g_{ac}\big(\nabla_bK)^c =: (\nabla_aK)_b + (\nabla_bK)_a.\]

If you didn’t do that exercise, go back and do it, but also you can see the video for a method considering the components.

This result is known as Killing’s equation.

Age Of The Universe...

Solution: We have \[\big(\nabla_aT\big)^{ab}u_b = \big(\nabla_aT\big)^{a0},\] as \(u_b=(1,0,0,0)_b\). Then using \(T^{a0}=0\) and \(T^{00} = \rho(t)\) we have \[\begin{split} \big(\nabla_aT\big)^{ab}u_b & = {T^{a0}}_{,a} + {\Gamma^a}_{na}T^{n0} + {\Gamma^0}_{na}T^{an} \\ & = \dot{\rho}(t) + {\Gamma^a}_{0a}\rho(t) + {\Gamma^0}_{na}T^{an}. \end{split}\] We now use the results from the lecture, \[{\Gamma^{\alpha}}_{0\alpha} = \frac{\dot{a}}{a}\delta^{\alpha}_{\alpha} = 3\frac{\dot{a}}{a}, \qquad {\Gamma^0}_{\alpha\beta} = a\dot{a}\gamma_{\alpha\beta},\] and the other \(\Gamma\)s vanishing. This gives \[\big(\nabla_aT\big)^{ab}u_b = \dot{\rho}(t) + 3\frac{\dot{a}}{a}\rho(t) + a\dot{a}\gamma_{\alpha\beta}T^{\alpha\beta}.\] Then using \(T^{\alpha\beta}=p(t)\) and \(g^{\alpha\beta}=\frac{1}{a^2}\gamma^{\alpha\beta}\) we get \[0 = \dot{\rho}(t) + 3\frac{\dot{a}}{a}\rho(t) + \frac{\dot{a}}{a}\gamma_{\alpha\beta}\gamma^{\alpha\beta}p(t),\] which gives the result.

Solution: We have \[\begin{split} \dot{\rho}(t) & = - 3\frac{\dot{a}(t)}{a(t)} (1+\omega)\rho(t) \\ \frac{\dot{\rho}(t)}{\rho(t)} & = -3\frac{\dot{a}(t)}{a(t)}(1+\omega) \\ \frac{d}{dt}\ln\big(\rho(t)\big) & = -3(1+\omega)\frac{d}{dt}\ln\big(a(t)\big) \\ \rho(t) & = Ba^{-3(1+\omega)}, \end{split}\] for some constant \(B=e^{A}\), where \(A\) is the constant of integration. If we plug this into the expression given in the question we have \[\bigg(\frac{\dot{a}}{a}\bigg)^2 = \frac{8\pi GB}{3}a^{-3(1+\omega)},\] which is an autonomous differential equation for \(a\).

Solution: By direct calculation, we have \[\alpha^2 t^{-2} = \frac{8\pi GBC}{3} t^{-3\alpha(1+\omega)},\] from which we see \[\alpha = \frac{2}{3(1+\omega)}.\]

Solution: We have \[a(t) = C\cdot t^{\frac{2}{3(1+\omega)}}, \qquad \implies \qquad H(t) = \frac{2}{3(1+\omega)}t^{-1}.\] So the age is given by \[t_0 = \frac{2}{3(1+\omega)}\frac{1}{H_0}.\] For dust \(\omega=0\) and so we have \[t^{\text{dust}}_0 \approx 8.6\times 10^9yrs.\] For radiation, \(\omega=\frac{1}{3}\), giving \[t^{\text{radiation}}_0 \approx 6.5 \times 10^9 yrs.\]

Solution: We have \[\ddot{a}(t) = \bigg(\frac{2}{3(1+\omega)}\bigg)\bigg(\frac{2}{3(1+\omega)}-1\bigg) \cdot a\cdot t^{-2}.\] An expanding universe means \(a(t)>0\) (or, equivalently, \(C>0\)), and so the turning point for accelerated expansion is the condition \[\frac{2}{3(1+\omega)}-1 = 0 \qquad \implies \qquad \omega = -\frac{1}{3}.\] We therefore get accelerated expansion for \(\omega<-\frac{1}{3}\) and decelerated expansion for \(\omega >-\frac{1}{3}\).

26.15 Diagrams

Penrose Diagram Of A Radiation-Filled Universe

Solution: A spatially flat FRW universe has \(\kappa=0\) and a metric with components \[g_{ab}(t,r,\vartheta,\varphi) = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & a^2(t) & 0 & 0 \\ 0 & 0 & a^2(t)r^2 & 0 \\ 0 & 0 & 0 & a^2(t)r^2\sin^2\vartheta \end{pmatrix}_{ab},\] in the chart given. If we have a radiation-filled universe, we have \(\omega=-\frac{1}{3}\), and from the last tutorial we have \[a(t) = C \cdot t^{1/2}.\] Radially null geodesics have \(\dot{\vartheta}=0=\dot{\phi}\), so our Lagrangian reads \[0 = -\dot{t}^2 + a^2\cdot \dot{r}^2,\] which, subbing in our \(a(t)\) expression and using the chain rule backwards, gives the differential equation \[\frac{dt}{dr} = \pm C\sqrt{t}.\]

We have seen that in the FRW universe, there is a beginning time (the Big Bang), and so our \(t\) coordinate must be lower bounded. We can choose to parameterise it such that \(t=0\) is the Big Bang value, giving us the coordinate ranges \[t \in (0,\infty), \qquad r \in (0, \infty), \qquad \vartheta \in (0,\pi), \qquad\text{and}\qquad \phi \in (0,2\pi),\] where the \(0\) point is removed from \(t\)’s range as it is not actually a point in our spacetime (it’s a singularity).

Solution: Solving the differential equation from the previous question gives us \[t_{\pm}(r) = \frac{1}{4}\big(A \pm Cr\big)^2,\] for some integration constant \(A\). So we have a series of squared curves, shifted for different values of \(A\).

The white circles indicated that these points are not part of our diagram as \(t\) and \(r\) cannot take the value \(0\). As these parts are not included, the lines to either side actually represent two separate geodesics, just as we saw with the Schwarzschild drawings in the lectures.

Solution: We can rearrange the expression for \(t_{\pm}\) in terms of \(r\) to give \[r = \pm\frac{2}{C}\sqrt{t_+} - \frac{A}{C}, \qquad r = \mp\frac{2}{C}\sqrt{t_-} + \frac{A}{C}.\] If we then define \[\Bar{t}_{\pm} := \frac{2}{C}\sqrt{t_{\pm}},\] we get \[r = \pm \Bar{t}_+ - B, \qquad r = \mp \Bar{t}_- + B,\] where \(B=\frac{A}{C}\). All of these plots are just lines of constant slope \(\pm1\). With a bit of thought, its clear that we can just consider either \(\bar{t}_+\) or \(\bar{t}_-\) and obtain the other results using different values of \(B\). So we shall just write \[r_{\pm} = \pm \bar{t} - B.\] The ranges are \[\bar{t} \in (0,\infty), \qquad r \in (0, \infty), \qquad \vartheta \in (0,\pi), \qquad\text{and}\qquad \phi \in (0,2\pi),\]

Solution: We define \[u := \bar{t} + r, \qquad\text{and}\qquad v := \bar{t}-r,\] and their ranges are \[u \in (0,\infty), \qquad\text{and}\qquad v\in(-\infty,\infty).\] We then have the conditions \(u+v=2\bar{t}>0\) and \(u-v=2\bar{r}>0\), so we need to exclude the regions \(u<0\), \(v+u<0\) and \(u-v<0\) from our diagram. We therefore get

where the shaded area is the only area we consider.

The blue lines are the geodesics in this chart. Note that by going to this chart we can compactify without ruining the 90-degree nature of the cone structure. This is exactly why this step is included, and is what Remark 23.1.1 is talking about.

Solution: We define \[p := \arctan(u), \qquad\text{and}\qquad q := \arctan(v),\] which have ranges \[p \in (0, \pi/2), \qquad\text{and}\qquad q \in (-\pi/2,\pi/2).\] Our other range conditions still hold, namely \(p+q>0\) and \(p-q>0\).

Our plot looks the same as in the previous question, apart from now the right-hand edge is bounded at value \(p=\pi/2\).

Solution: We define \[T := p+q, \qquad\text{and}\qquad R := p-q,\] whose ranges are \[T\in (0,\pi), \qquad\text{and}\qquad R\in(0,\pi).\] We have the further constraint \(T+R =2U <\pi\), so our Penrose-Carter diagram looks like

We have used the snake-like line to indicate the Big Bang singularity, and a dashed line on the left-hand side to remind us that there is nothing wrong here (i.e. lines that go off to the left just come back form the dashed line — think about rotating the diagram by reinstating \(\phi\)).

26.16 Perturbation Theory

To come later.