NOTES ON GENERAL RELATIVITY (GR) AND GRAVITY ERNEST YEUNG
Abstract. These are notes on General Relativity (GR) and Gravity. As of March 23, 2015, I find that the Central Lectures given by Dr. Frederic P. Schuller for the WE Heraeus International Winter School to be, unequivocally, the best, most lucid, and well-constructed lecture series on General Relativity and Gravity. Instead of reinventing the wheel, I write these notes to build upon and supplement the video lectures and tutorials already created by them. This includes my corrections, comments, relations to other aspects of theoretical physics, and code implementing calculations in GR. It should be noted that for symbolic computation, I heavily use the SageManifolds v.0.7 package for Sage Math. My goal in this area is this: we see a concept or idea from GR and we go from the equation on the blackboard or textbook and into (Python/Sage Math) code that immediately computes a calculation. I keep these notes available online, openly accessible, and free for anyone, anytime (with your (financial) help and contribution at Tilt/Open, which is a subscription service). I want to keep these notes openly accessible because I want to encourage anyone to freely edit, copy, and make their own notes in the spirit of open-source software. I continuously update these notes and post them here ernestyalumni.wordpress.com The stated goal of the WE Heraeus International Winter School on Gravity and Light is to take the student from an introduction to the research frontier (cf. http://www.gravity-and-light.org/lectures). I want to get myself and other students or ambitious non-academic (maybe he or she is a working professional who had studied physics before in college, went to work in another field, maybe even, gasp, investment banking or mobile app developer, but still is curious and passionate about physics and want to contribute) equipped with all the tools available to do research, do calculations, to design experiments or collect data. Again, we’re not here to reinvent the wheel. I’m not trying to make a General Relativity appreciation class, but this is a serious attempt towards training to do research.
Part 1. WE Heraeus International Winter School on Gravity and Light Introduction (from EY) 1. Lecture 1: Topology Topology Tutorial Sheet 2. Lecture 2: Topological Manifolds Tutorial Topological manifolds 3. 4. Lecture 4: Differentiable Manifolds Tutorial 4 Differentiable Manifolds 5. Lecture 5: Tangent Spaces 6. 7. Lecture 7: Connections 8. Lecture 8: Parallel Transport & Curvature (International Winter School on Gravity and Light 2015) Tutorial 8 Parallel transport & Curvature 9. Lecture 9: Newtonian spacetime is curved! 10. Lecture 10: Metric Manifolds
2 2 2 4 4 5 7 7 10 12 16 16 20 21 23 26
Date: 23 mars 2015. 1991 Mathematics Subject Classification. General Relativity. Key words and phrases. General Relativity, Gravity, Differential Geometry, Manifolds, Integration, MIT OCW, education, mathematics, physics. I write notes, review papers, and code and make calculations for physics, math, and engineering to help with education and research. With your support, we can keep education and research material available online, openly accessible, and free for anyone, anytime. If you like what I’m trying to do for physics education research, please go to my Tilt/Open crowdfunding campaign, read the mission statement, share the page, and contribute financially if you can. ernestyalumni.tilt.com . 1
11. Symmetry 12. Integration 13. Lecture 13: Relativistic spacetime 14. Lecture 14: Matter 15. Lecture 15: Einstein gravity Tutorial 13 Schwarzschild Spacetime 16. 17. 18. 19. 20. 21. 22. Lecture 22: Black Holes
31 34 37 40 43 47 48 48 48 48 48 48 48
Part 2. Special Relativity References
51 51
Contents
Part 1. WE Heraeus International Winter School on Gravity and Light Introduction (from EY) The International Winter School on Gravity and Light held central lectures given by Dr. Frederic P. Schuller. These lectures on General Relativity and Gravity are unequivocally and undeniably, the best and most lucid and well-constructed lecture series on General Relativity and Gravity. The mathematical foundation from topology and differential geometry from which General Relativity arises from is solid, well-selected in rigor. The lectures themselves are well-thought out and clearly explained. Even more so, the International Winter School provided accompanying Tutorial Sessions for each of the lectures. I had given up hopes in seeing this component of the learning process ever be put online so that anyone and everyone in the world could learn through the Tutorial process as well. I was afraid that nobody would understand how the Tutorial or “Office Hours” session was important for students to digest and comprehend and work out-doing exercises-the material presented in the lectures. This International Winter School gets it and shows how online education has to be done, to do it in an excellent manner, moving forward. For anyone who is serious about learning General Relativity and Gravity, I would simply point to these video lectures and tutorials. What I want to do is to build upon the material presented in this International Winter School. Why it’s important to me, and to the students and practicing researchers out there, is that the material presented takes the student from an introduction to the research frontier. That is the stated goal of the International Winter School. I want to dig into and help contribute to the cutting edge in research and this entire program with lectures and tutorials appears to be the most direct and sensible route directly to being able to do research in General Relativity and Gravity. -EY 20150323 1. Lecture 1: Topology 1.1. Lecture 1: Topological Spaces. Definition 1. Let M be a set. A topology O is a subset O ⊆ P(M ), P(M ) power set of M : set of all subsets of M . satisfying (i) ∅ ∈ O, M ∈ O 2
(ii) U ∈ O, V ∈ O =⇒ U
T
V ∈O S (iii) Uα ∈ O, α ∈ A =⇒ α∈A Uα ∈ O O} utterly useless Definition 2. Ostandard ⊆ P(Rd ) EY : 20150524 I’ll fill in the proof that Ostandard is a topology. Proof. ∅ ∈ Ostandard since ∀ p ∈ ∅, ∃ r ∈ R+ : Br (p) ⊆ ∅ (i.e. satisfied “vacuously”)
Suppose U, V ∈ Ostandard . Let p ∈ U
T
V . Then ∃ r1 , r2 ∈ R+ s.t. Br1 (p) ⊆ U Br2 (p) ⊆ V
Let r = min {r1 , r2 }. T T Clearly Br (p) ⊆ U and Br (p) ⊆ V . Then Br (p) ⊆ U V . So U V ∈ Ostandard .
Suppose, Uα ∈ Ostandard , ∀ α ∈ A. S Let p ∈ α∈A Uα . Then p ∈ Uα for at least 1 α ∈ A. S S ∃ rα ∈ R+ s.t. Brα (p) ⊆ Uα ⊆ α∈A Uα . So α∈A Uα ∈ Ostandard
1.2. 2. Continuous maps.
1.3. 3. Composition of continuous maps.
1.4. 4. Inheriting a topology. EY : 20150524 I’ll fill in the proof that given f continuous (cont.), then the restriction of f onto a subspace S is cont. If you want a reference, check out Klaus J¨ anich [2, pp. 13, Ch. 1 Fundamental Concepts, Sec. Continuous Maps] If cont. f : M → N , S ⊆ M , then f |S cont. Proof. Let open V ⊆ N , i.e. V ∈ ON i.e. V in the topology ON of N . −1
f |S (V ) = {m ∈ M | f |S (m) ∈ V } Now f −1 (V ) = {m ∈ M |f (m) ∈ V }. T −1 So f −1 (V ) S = f |S (V ) Now f cont. So f −1 (V ) ∈ ON . T and recall OS | := {U S|U ∈ OM }. T −1 −1 so f −1 (V ) S = f |S (V ) ∈ OS i.e. f |S (V ) open. =⇒ f |S cont.
3
Topology Tutorial Sheet filename : main.pdf The WE-Heraeus International Winter School on Gravity and Light: Topology
EY : 20150524 What I won’t do here is retype up the solutions presented in the Tutorial (cf. https://youtu.be/_XkhZQ-hNLs): the presenter did a very good job. If someone wants to type up the solutions and copy and paste it onto this LaTeX file, in the spirit of open-source collaboration, I would encourage this effort. Instead, what I want to encourage is the use of as much CAS (Computer Algebra System) and symbolic and numerical computation because, first, we’re in the 21st century, second, to set the stage for further applications in research. I use Python and Sage Math alot, mostly because they are open-source software (OSS) and fun to use. Also note that the structure of Sage Math modules matches closely to Category Theory. In checking whether a set is a topology, I found it strange that there wasn’t already a function in Sage Math to check each of the axioms. So I wrote my own; see my code snippet, which you can copy, paste, edit freely in the spirit of OSS here, titled topology.sage: gist github ernestyalumni topology.sage Download topology.sage Loading topology.sage, after changing into (with the usual Linux terminal commands, cd, ls) by sage : load ( ‘ ‘ topology . sage ’ ’)
Exercise 2: Topologies on a simple set. Question Does O1 := . . . constitute a topology . . . ?.
Solution: Yes, since we check by typing in the following commands in Sage Math: emptyset in O_1 Axiom2check ( O_1 ) # True Axiom3check ( O_1 ) # True
Question What about O2 . . . ?.
Solution: No since the 3rd. axiom fails, as can be checked by typing in the following commands in Sage Math: emptyset in O_2 Axiom2check ( O_2 ) # True Axiom3check ( O_2 ) # False
2. Lecture 2: Topological Manifolds Lecture 2: Manifolds. Topological spaces: ∃ so man that mathematicians cannot even classify them. For spacetime physics, we may focus on topological spaces (M, O) that can be charted, analogously to how the surface of the earth is charted in an atlas. 2.1. Topological manifolds. Definition 3. A topological space (M, O) is called a d-dimensional topological method if ∀ p ∈ M : ∃ U ∈ O, U 3 p : ∃ x : U ⊆ M → x(U ) ⊆ Rd (M, O), (Rd , Ostd ) (i) x invertible : x−1 : x(U ) → U 4
(ii) x continuous (iii) x−1 continuous 2.2. Terminology. 2.3. 3. Chart transition maps. Imagine 2 charts (U, x) and (V, y) with overlapping regions. 2.4. 4. Manifold philosophy. Often it is desirable (or indeed the way) to define properties (“continuity”) of real-world γ object (“R − → M ”) by judging suitable coordinates not on the “real-world” object itself, but on a chart-representation of that real world object. EY’s add-ons. This lecture gives me a good excuse to review Topology and Topological Manifolds from a mathematician’s point of view. I find John M. Lee’s Introduction to Topological Manifolds book good because it’s elementary and thorough and it’s fairly recent (2010) so it’s up to date [3]. See my notes and solutions for the book; it’s a file titled LeeJM_IntroTopManifolds_sol.pdf of which I’ll try to keep the pdf and LaTeX file available for download on my ernestyalumni Google Drive (so try to search for it on Google).
Tutorial Topological manifolds filename: Sheet_1.2.pdf Exercise 4: Before the invention of the wheel.
Another one-dimensional topological manifold. Another one? Consider set F 1 := {(m, n) ∈ R2 |m4 + n4 = 1}, equipped with subset topology Ostd |F 1 Question x : F 1 → R is what?. Solution . EY : 20150525 The tutorial video https://youtu.be/ghfEQ3u_B6g is really good and this solution is how I’d
write it, but it’s really the same (I needed the practice).
x : F1 → R (m, n) 7→ m If m = 0, n4 = 1 so n = ±1 so it’s not injective. Let the closed n-dim. upper half-space Hn ⊆ R1 . Then Hn = {(x1 . . . xn ) ∈ Rn |xn ≥ 0} intHn = {(x1 . . . xn ) ∈ Rn |xn > 0} −Hn = {(x1 . . . xn ) ∈ Rn |xn ≤ 0} −intHn = {(x1 . . . xn ) ∈ Rn |xn < 0} Question This map x may be made injective by restricting its domain to either of 2 maximal open subsets of F 1 . Which ones?. Solution .
Let U+ = F 1 ∩ intH2 U− = F 1 ∩ −intH2 5
Look at x4 = 1 − n4 =⇒x = ±(1 − n4 )1/4 Then for x−1 + : (−1, 1) ⊆ R → U+ m 7→ (m, (1 − m4 )1/4 ) x−1 − : (−1, 1) ⊆ R → U− m 7→ (m, −(1 − m4 )1/4 ) x+ ,x− injective (since left inverse exists). Question Construct injective y. Solution .
Let V+ = F 1 ∩ intH1 V− = F 1 ∩ −intH1 Then y+ : V+ → (−1, 1) ⊆ R (m, n) 7→ n y− : V− → (−1, 1) ⊆ R (m, n) 7→ n Question Construct inverse y −1 .
Solution .
For −1 : (−1, 1) ⊆ R → V+ y+
n 7→ ((1 − n4 )1/4 , n) −1 y− : (−1, 1) ⊆ R → V−
n 7→ (−(1 − n4 )1/4 , n) y+ ,y− injective (since left inverse exists).
/ U+ , U− Note (−1, 0) ∈ (1, 0) ∈ / U+ , U− and
(0, 1) ∈ / V+ , V− (0, −1) ∈ / V+ , V− Question construct transition map x ◦ y −1 . Solution . 6
−1 x+ y+ : (0, 1) ⊆ R → (0, 1) ⊆ R
n 7→ (1 − n4 )1/4 −1 x− y+ : (−1, 0) ⊆ R → (0, 1) ⊆ R −1 y+
x−
n −−→ ((1 − n4 )1/4 , n) −−→ (1 − n4 )1/4 −1 x+ y− : (0, 1) ⊆ R → (−1, 0) ⊆ R
n 7→ −(1 − n4 )1/4 −1 x− y− : (−1, 0) ⊆ R → (−1, 0) ⊆ R
n 7→ −(1 − n4 )1/4 Question . . . Does the collection of these domains and maps form an atlas of F 1 ?.
Yes, with atlas
A={
(U+ , x+ ) (V+ , y+ ) , } (U− , x− ) (V− , y− )
Clearly U+ ∪ U− ∪ V+ ∪ V− = (F 1 ∩ intH2 ) ∪ (F 1 ∩ −intH2 ) ∪ (F 1 ∩ intH1 ) ∪ (F 1 ∩ −intH1 ) = = F 1 ∩ R2 \{(0, 0)} = F 1 and (the point is that) x± , y± are homeomorphisms of open sets of F 1 onto open sets of 1 dim. R1 (namely (−1, 1) ⊆ R1 ), and so A is an atlas of F 1 .
3. 4. Lecture 4: Differentiable Manifolds cf. https://youtu.be/HSyTEwS4g80?list=PLFeEvEPtX_0S6vxxiiNPrJbLu9aK1UVC_
so far: top. mfd.
(M, O) dimM = d
we wish to define a notion of differentiable curves R → M function M → R maps M → N 4.1. 1. Strategy. choose a chart (U, x) γ : R → M portion of curve in chart domain γ:R x◦γ
U x x(U ) ⊆ Rd
idea. try to “lift” the undergraduate notion of differentiability of a curve on Rd to a notion of differentiability of a curve on M Problem Can this be well-defined under change of chart? 7
y(U ∩ V ) ⊆ Rd y◦γ
y U ∩ V 6= ∅ y ◦ x−1
γ:R x◦γ
x x(U ∩ V ) ⊆ Rd
x ◦ γ undergraduate differentiable (“as a map R → Rd ”) Rd →Rd
y◦γ | {z }
maybe only continuous, but not undergraduate differentiable
z }| { = (y ◦ x−1 ) ◦ | {z } continuous
R→Rd
z }| { (x ◦ γ ) | {z }
= y ◦ (x−1 ◦ x) ◦ γ
undergrad differentiable
At first sight, strategy does not work out. 4.2. 2. Compatible charts. In section 1, we used any imaginable charts on the top. mfd. (M, O). To emphasize this, we may say that we took U and V from the maximal atlas A of (M, O). Definition 4. Two charts (U, x) and (V, y) of a top. mfd. are called `-compatible if either (a) U ∩ V = ∅ or (b) U ∩ V 6= ∅ chart transition maps have undergraduate ` property. EY : 20151109 e.g. since Rd → Rd , can use undergradate ` property such as continuity or differentiability. y ◦ x−1 : x(U ∩ V ) ⊆ Rd → y(U ∩ V ) ⊆ Rd x ◦ y −1 : y(U ∩ V ) ⊆ Rd → x(U ∩ V ) ⊆ Rd Philosophy: Definition 5. An atlas A‘ is a `-compatible atlas if any two charts in A` are `-compatible. Definition 6. A `-manifold is a triple ( M, O , A` ) | {z }
A` ⊆ Amaximal
top. mfd.
` C0 C1 Ck Dk .. . C∞
undergraduate ` C 0 (Rd → Rd ) = C 1 (Rd → Rd ) =
continuous maps w.r.t. O differentiable (once) and is continuous k-times continuously differentiable k-times differentiable
C ∞ (Rd → Rd )
⊇ Cω C∞
∃ multi-dim. Taylor exp. satisfy Cauchy-Riemann equations, pair-wise
EY : 20151109 Schuller says: C k is easy to work with because you can judge k-times cont. differentiability from existence of all partial derivatives and their continuity. There are examples of maps that partial derivatives exist but are not Dk , k-times differentiable. 8
Theorem 1 (Whitney). Any C k≥1 -atlas, AC k≥1 of a topological manifold contains a C ∞ -atlas. Thus we may w.l.o.g. always consider C ∞ -manifolds, “smooth manifolds”, unless we wish to define Taylor expandibility/complex differentiability . . . EY : 20151109 Hassler Whitney
1
Definition 7. A smooth manifold ( M, O , | {z } top. mfd.
A ) |{z}
C ∞ −atlas
γ M
R x◦γ
x Rd
EY: 20151109 Schuller was explaining that the trajectory is real in M ; the coordinate maps to obtain coordinates is x ◦ γ φ
→N 4.3. 4. Diffeomorphisms. M − If M, N are naked sets, the structure preserving maps are the bijections (invertible maps). e.g. {1, 2, 3} → {a, b} Definition 8. M ∼ =set N (set-theoretically) isomorphic if ∃ bijection φ : M → N Examples. N ∼ =set Z N∼ =set Q (EY: 20151109 Schuller says from diagonal counting) ∼ R N =set Now (M, OM ) ∼ =top (N, ON ) (topl.) isomorphic = “homeomorphic” ∃ bijection φ : M → N φ, φ−1 are continuous. (V, +, ·) ∼ =vec (W, +w , ·w ) (EY: 20151109 vector space isomorphism) if ∃ bijection φ : V → W linearly finally Definition 9. Two C ∞ -manifolds (M, OM , AM ) and (N, ON , AN ) are said to be diffeomorphic if ∃ bijection φ : M → N s.t. φ:M →N φ−1 : N → M are both C ∞ -maps Rd
ye ◦ φ ◦ x e−1
Re ye
x e C∞
M ⊇U
φ
V ⊆N y
x Rd
y◦φ◦x−1 undergraduate C ∞
Re
1http://mathoverflow.net/questions/8789/can-every-manifold-be-given-an-analytic-structure 9
Theorem 2. # = number of C ∞ -manifolds one can make out of a given C 0 -manifolds (if any) - up to diffeomorphisms. dimM 1 2 3 4 5 6 .. .
# 1 Morse-Radon theorems 1 Morse-Radon theorems 1 Morse-Radon theorems uncountably infinitely many finite surgery theory finite surgery theory finite
surgery theory
EY : 20151109 cf. http://math.stackexchange.com/questions/833766/closed-4-manifolds-with-uncountably-many-diff The wild world of 4-manifolds Tutorial 4 Differentiable Manifolds EY : 20151109 The gravity-and-light.org website, where you can download the tutorial sheets and the full length videos for the tutorials and lectures, are no longer there. = ( Hopefully, the YouTube video will remain: https://youtu.be/FXPdKxOq1KA?list=PLFeEvEPtX_0RQ1ys-7VIsKlBWz7RX-FaL Exercise 1: True or false?. These basic questions are designed to spark discussion and as a self-test.
Tick the correct statements, but not the incorrect ones! (a) The function f : R → R, . . . • • • . . . , defined by f (x) = |x3 |, lies in C 3 (R → R). EY : 20151109 Solution 1a3. For f : R → R, f (x) = |x3 | =
f 0 (x) =
3x2
−3x2 6x f 00 (x) = −6x
x 3
if x ≥ 0
−x3
if x < 0
if x ≥ 0 if x < 0 if x ≥ 0 if x < 0
Thus, f (x) = |x3 | ∈ C 1 (R) but f (x) ∈ / C 2 (R) ⊆ C 3 (R) • • (b) (c) Short Exercise 4: Undergraduate multi-dimensional analysis . A good notation and basic results for partial differentiation. For a map f : Rd → R we denote by the map ∂i f : Rd → R the partial derivative with respect to the i-th entry. 10
Question :. Given a function
f : R3 → R; (α, β, δ) 7→ f (α, β, δ) := α3 β 2 + β 2 δ + δ calculate the values of the following derivatives: Solution :.
• (∂2 f )(x, y, z) = • (∂1 f )(, ◦, ∗) = • (∂1 ∂2 f )(a, b, c) = • (∂32 f )(299, 1222, 0) = EY: 20151110 For f (α, β, δ) := α3 β 2 + β 2 δ + δ, or f (x, y, z) = x3 y 2 + y 2 z + z, (∂2 f ) = 2(x3 y + yz) (∂1 f ) = 3x2 y 2 (∂1 ∂2 f ) = 6x2 y (∂32 f ) = 0 and so • (∂2 f )(x, y, z) = 2(x3 y + yz) • (∂1 f )(, ◦, ∗) = 32 ◦2 • (∂1 ∂2 f )(a, b, c) = 6a2 b • (∂32 f )(299, 1222, 0) = 0 Exercise 5: Differentiability on a manifold.
How to deal with functions and curves in a chart Let (M, O, A) be a smooth d-dimensional manifold. Consider a chart (U, x) of the atlas A together with a smooth curve γ : R → U and a smooth function f : U → R on the domain U of the chart. Draw a commutative diagram containing the chart domain, chart map, function, curveand the respective representatives of the function and the curve in the chart. Question :.
Solution :.
x◦γ x◦γ
γ U
R
x R
d
γ
τ ∈R
x−1 (f ◦ x−1 )
f
p∈U f
x x−1
x(p) = (x ◦ γ)(τ ) ∈ Rd (f ◦ x−1 )
f (p) ∈ R
R Question :. Consider, for d = 2,
(x ◦ γ)(λ) := (cos (λ), sin (λ)) and (f ◦ x−1 )((x, y)) := x2 + y 2 Using the chain rule, calculate (f ◦ γ)0 (λ) 11
explicitly. Solution :.
x◦γ
γ U
R f ◦γ
EY : 20151109 Indeed, the domains and codomains of this f γ mapping makes sense, from R → R for x◦γ
γ
τ ∈R
p∈U f ◦γ
x
x− f R
x(p) = (x ◦ γ)(τ ) ∈ Rd
x−1
f
(f ◦ x−1 )
f (p) ∈ R
(f ◦ γ)0 (λ) = (Df ) · γ(λ) ˙ =
∂f j γ˙ (λ) = 2x(− sin λ) + 2y cos λ = 2(− cos λ sin λ + sin λ cos λ) = 0 ∂xj 5. Lecture 5: Tangent Spaces
lead question: “what is the velocity of a curve γ point p? 5.1. Velocities. Definition 10. (M, O, A) smooth mfd. curve γ : R → M at least C 1 . Suppose γ(λ0 ) = p The velocity of γ p is the linear map ∼
vγ,p : C ∞ (M ) − →R C ∞ (M ) := {f : M → R|f smooth function } equipped with (f ⊕ g)(p) := f (p) + g(p) (λ ⊗ g)(p) := λ · g(p) ∼ denotes linear map on top of → −.
f 7→ vγ,p (f ) := (f ◦ γ)0 (λ0 ) γ R
f M
R
f ◦γ intuition Schuller says: children run around the world. Temperature function as temperature contour lines. You feel the temperature. You observe the rate of change of temperature as you run around. f is temperature. 12
x
past: “ |{z} v i (∂i f ) = ( v i ∂i )f |{z} vector
5.2. Tangent vector space. Definition 11. For each point p ∈ M def the set “tangent space 6=0 M p “ Tp M := {vγ,p |γ smooth curves } picture: rather M than (embedded) p Tp M EY : 20151109 see https://youtu.be/pepU_7NJSGM?t=12m38s for the picture Observation: Tp M can be made into a vector space. ⊕ :Tp M × Tp M → (vγ,p ⊕ vδ,p )(
f ) := vγ,p (f ) +R vδ,p (f ) |{z}
∈C ∞ (M )
:R × Tp M → Hom(C ∞ (R), R) (α vγ,p )(f ) := α ·R vγ,p (f ) Remains to be shown that (i) ∃ σ curve : vγ,p ⊕ vδ,p = vσ,p (ii) ∃ τ curve : α vγ,p = vτ,p
Claim: τ : R → M
where µα :R → R
7→ τ (λ) := γ(αλ + λ0 ) = (γ ◦ µα )(λ)
, does the trick.
r 7→ α · r + λ0
τ (0) = γ(λ0 ) = p vτ,p := (f ◦ τ )0 (0) = (f ◦ γ ◦ µα )0 (0) = (f ◦ γ)0 (λ0 ) · α = = α · vγ,p Now for the sum: ?
vγ,p ⊕ vδ,p = vσ,p make a choice of chart (|{z} U , x) In cloud: ill definition alarm bells. 3p
and define: Claim: σ:R→M σ(λ) := x−1 ((x ◦ γ)(λ0 + λ) +(x ◦ δ)(λ1 + λ) − (x ◦ γ)(λ0 )) | {z } R→Rd
does the trick. Proof. Since: σx (0) = x−1 ((x ◦ γ)(λ0 ) + (x ◦ δ)(λ1 ) − (x ◦ γ)(λ0 )) = δ(λ1 ) = p 13
Now: vσx ,p (f ) := (f ◦ σx )0 (0) = = ((f ◦ x−1 ) ◦ (x ◦ σx ))0 (γ) = | {z } | {z } Rd →R
R→Rd
(x ◦ σx )0 (0) | {z }
(x◦γ)0 (λ0 )+(x◦δ)0 (λ1 )
· ∂i (f ◦ x−1 ) (x(σ(0))) = |{z} p
= (x ◦ γ)0 (λ0 )(∂i (f ◦ x−1 ))(x(p)) + (x ◦ δ)(λ1 )(∂i (f ◦ x−1 ))(x(p)) = (f ◦ γ)0 (λ0 ) + (f ◦ δ)0 (λ1 ) = = vγ,p (f ) + vδ,p (f )
∀ f ∈ C ∞ (M )
vγ,p ⊕ vδ,p = vσ,p f
σ
M
R x◦σ
x
R f ◦ x−1
Rd picture: (cf. https://youtu.be/pepU_7NJSGM?t=39m5s) γ:R→M δ:R→M (γ⊕)(λ) := γ(λ) + δ(λ) EY : 20151109 Schuller says adding trajectories is chart dependent, bad. Adding velocities is good. 5.3. Components of a vector wrt a chart. Definition 12. Let (U, x) ∈ Asmooth . γ:R→U Let . γ(0) = p Calculate vγ,p (f ) := (f ◦ γ)0 (0) = ((f ◦ x−1 ) ◦ (x ◦ γ))0 (0) | {z } | {z } Rd →R
i0
R→Rd
−1
= (x ◦ γ) (0) · (∂i (f ◦ x ))(x(p)) | {z } | {z } i (0) γ˙ x =:( ∂fi ) ∂x p think cloud f : M → R =
γ˙ xi (0)
·
∂ ∂xi
f
∀ f ∈ C ∞ (M )
p
∴ as a map.
vγ,p
= |{z}
∂ ∂xi | {z }
γxi (0) | {z }
use of chart “components of the velocity v γ,p ”
basis elements of the Tp M wrt which the components need to be understood. “chart induced basis of Tp M ”
Picture: https://youtu.be/pepU_7NJSGM?t=1h16s 14
5.4. 4. Chart-induced basis. Definition 13. (U, x) ∈ Asmooth ∂ the ∂x , . . . , ∂x∂ d p ∈ Tp U ⊆ Tp M 1 p constitute a basis of Tp U
Proof. remains: linearly independent λi =⇒ λi
∂ ∂xi
p
∂ ∂xi
!
=0 p
xj ◦ x−1 : Rd → R
j −1 (xj ) = λi ∂i (x | ◦{zx })(x(p)) =
= λi δij = λj
(α1 , . . . , αd ) 7→ αj
j = 1, . . . , d
in cloud: xj : U → R differentiable Corollary 1. dimTp M = d = dimM Terminology: X ∈ Tp M → ∃ γ : R → M : X = vγ,p and ∂ ∃ X11 , . . . , X d : X = X i ∂x i p | {z } ∈R
5.5. 5. chart.
Change of vector components under a change of chart. 8 vector does not change under change of
Let (U, x) and (V, y) be overlapping charts and p ∈ U ∩ V . Let X ∈ Tp M
i X(y) ·
∂ ∂y i
p
= X |{z} = Xxi |{z}
(V,y)
∂ ∂xi
(U,x)
p
to study change of components formula: ∂ f = ∂i (f ◦ x−1 )(x(p)) = ∂xi p = ∂i ((f ◦ y −1 ) ◦(y ◦ x−1 )(x(p)) | {z } | {z } Rd →R i −1
Rd →Rd
= (∂i (y ◦ x ))(x(p)) · (∂j (f ◦ y −1 ))(y(p)) = p ∂y ∂f = · f ∂xi p ∂y j p
=⇒
i X(x)
∂y j ∂xi
p
j =⇒ X(y)
∂ ∂ j = X(y) ∂y j p ∂y j p j ∂y = Xi ∂xi p (x) 15
5.6. 6. Cotangent spaces. Tp M = V ∼
trivial (Tp M )∗ := {ϕ : Tp M − → R} Example: f ∈ C ∞ (M ) ∼
(df )p :Tp M − →R X 7→ (df )p (X) i.e. (df )p ∈ Tp M ∗ (df )p called the gradient of f p ∈ M . Calculate components of gradient w.r.t. chart-induced basis (U, x) ((df )p )j := (df )p =
∂f ∂xj
∂ ∂xj
! p
= ∂j (f ◦ x−1 )(x(p))
p
Theorem 3. Consider chart (U, x) =⇒ xi : U → R Claim: (dx1 )p , (dx2 )p , . . . , (dxd )p basis of Tp∗ M =⇒ In fact: dual basis:
a
(dx )p
∂ ∂xb
!
=
p
∂xa ∂xb
= · · · = δba p
5.7. 7. Change of components of a covector under a change of chart: Tp∗ M with ω(y) (dy j )p = ω = ω(x)i (dxi )p | {z } 3ω
=⇒ ω(y)i =
∂xj ω(x)j ∂y i
6. 7. Lecture 7: Connections
∇X f = Xf = (df )(X) but (not quite) X : C ∞ (M ) → C ∞ (M ) df : Γ(T M ) → C ∞ (M ) ∇X : C ∞ (M ) → C ∞ (M ) ∇X : C ∞ (M ) .. .
∇X :
T M p ⊗ T ∗ M q i.e. p tensor field q
C ∞ (M ) .. . T M p ⊗ T ∗ M q i.e. p tensor field q 16
7.1. Directional derivatives of tensor fields. manifold with connection is quadruple (M, O, A, ∇) topology O atlas A Consider chart (U, x) ∈ A Definition 14. ∀ pair (X, (p, q) − tensor field) ≡ (X, (p, q) − T F ), connection ∇ on smooth manifold (M, O, A) ∇ : (X, (p, q) − T F ) → (p, q) − T F s.t. (i) ∇X f = Xf (ii) ∇X (T + S) = ∇X T + ∇X S (iii) ∇X (T (ω, Y )) = (∇X T )(ω, T ) + T (∇X ω, Y ) + T (ω, ∇X Y ) “Leibnitz” rule. As T ⊗ S(ω(1) . . . ω(p+r) , Y(1) . . . Y(q+s) ) = T (ω(1) . . . ω(p) , Y(1) . . . Y(q) ) · S(ω(p+1) . . . ω(p+r) , Y(q+1) . . . Y(q+s) ) so ∇X (T ⊗ S) = (∇X T ) ⊗ S + T ⊗ ∇X S (iv) ∇f X+Z T = f ∇X T + ∇Z T C ∞ -linear 7.2. New structure on (M, O, A) required to fix ∇. There are (dimM )3 many Γijk Γijk : U → R p 7→ dxi (∇ ∂
∂ ) (p) ∂x ∂xj
Now ∇
∂ ∂xm
(dxi ) =?
∂ ∂ )= (δ i ) = 0 ∇ ∂m (dx j ∂x ∂x ∂xm j {z } | i
δ ij
=
|
{z (iii)
= ∇
}
∂ ∂xm
dxi
∂ ∂ + dxi (∇ ∂m j ) = 0 j ∂x ∂x | {z ∂x } ∂ Γqjm ∂x q
=⇒ ∇ ∇
∂ ∂xm
∂ ∂xm
dxi
∂ ∂xj
= −Γijm
dxi = −Γijm dxj
Hence (∇X Y )i = X(Y i ) + Γij
m |{z}
last entry goes in direction of X
(∇X ω)i = X(ωi ) +
−Γjim ωj X m 17
Y jXm
Note that for the immediately above expression for (∇X Y )i , in the second term on the right hand side, Γijm has the last entry at the bottom, m going in the direction of X, so that it matches up with X m . This is a good mnemonic to memorize the index positions of Γ. summary so far: (∇X Y )i = X(Y i ) + Γijm Y j X m (∇X ω)i = X(ωi ) + −Γjim ωj X m similarly, by further application of Leibnitz T a (1, 2)-TF (tensor field) (∇X T )ijk = X(T ijk ) + Γism T sjk X m − Γsjm T isk X m − Γskm T ijs X m What is a Euclidean space: (M = Rn , Ost , A) smooth manifold. Assume (Rn , idRn ) ∈ A and
(Γi(x) )jk = dxi (∇E )
∂ ∂xj
∂ ∂xk
!
=0
7.3. Change of Γ’s under change of chart. (U, x), (V, y) ∈ A and U ∩ V 6= ∅ Γijk (y) := dy i ∇
∂ ∂y k
∂ ∂y j
=
∂xs ∂ ∂y i q dx ∇ ∂xp ∂ p j s ∂y k ∂x ∂y ∂x ∂xq
Note ∇f X is C ∞ -linear for f X covector dy i is C ∞ -linear in its argument =⇒ Γijk (y) = =
∂y i q dx ∂xq
∇
∂ ∂xp
∂xs ∂y j
∂ ∂xs + j s ∂x ∂y
∇
∂ ∂xp
∂ ∂xs
=
∂y i ∂xp ∂ ∂xs q ∂y i ∂xp ∂xs q δ + Γ (x) s ∂xq ∂y k ∂xp ∂y j ∂xq ∂y k ∂y j sp Γijk (y) =
(7.1)
∂xp ∂y k
∂y i ∂ 2 xq ∂y i ∂xs ∂xp q + Γ (x) q j k ∂x ∂y ∂y ∂xq ∂y j ∂y k sp
Eq. (7.1) is the change of connection coefficient function under the change of chart (U ∩ V, x) → (U ∩ V, y) 7.4. Normal Coordinates. Tutorial 7 Connections. Exercise 1. : True or false? (a)
• ∇f X Y = f ∇X Y by definition so ∇f X = f ∇X i.e. ∇X is C ∞ (M )-linear in X • f ∈ C ∞ (M ) is a (0, 0)-tensor field. ∇X f = Xf ≡ X(f ) by definition. • If the manifold is flat, I’m assuming that means that the manifold is globally a Euclidean space, and by definition, Γ = 0. ∂ ∂ ∂ ∂Y i ∂ (Y i ) i + Γijk Y k X k i = X j j +0 j ∂x ∂x ∂x ∂x ∂xi and similarly for any (p, q)-tensor field, i.e. ∇X Y = X j
i ...i
∇X T = X j • ∇X f = X j
∂Tj11...jqp ∂xj
∂f = X · grad(f ) ∂xj 18
• ∀ (U, x) ∈ A, locally (after working out the first few cases, and doing induction, one can look up the expression for the local form; I found it in Nakahara’s Geometry, Topology and Physics, Eq. 7.26, and it needs to be modified for the convention of order of bottom indices for Γ: λ1 ...λp λ1 κλ2 ...λp λp λ1 ...λp−1 κ p p p ∇ν tλµ11 ...λ − Γκµ1 ν tλκµ1 ...λ − · · · − Γκµq ν tµλ11 ...λ ...µq = ∂ν tµ1 ...µq + Γ κν tµ1 ...µq + · · · + Γκν tµ1 ...µq ...µq−1 κ 2 ...µq
Clearly, ∇X is uniquely fixed ∀ p ∈ M by choosing each of the (dimM )3 many connection coefficient functions Γ.
(b)
• ∇ : X(M ) → X(M ) ∇ : (p, q)-tensor field 7→ (p, q)-tensor field • By definition, ∇ satisfies the Leibniz rule. • • •
Exercise 2. : Practical rules for how ∇ acts Torsion-free covariant derivative boils down to a connection coefficient
function Γ that is symmetric in the bottom indices. • ∇X f = X(f ) = X i
∂f ∂xi
• (∇X Y )a = X i
∂Y a + Γajk Y j X k ∂xi
(∇X ω)a = X i
∂ωa − Γiak ωi X k ∂xj
•
• (∇m T )abc =
∂ i a a (T a ) + Γaim Tbc − Γibm Tic − Γjcm Tbj ∂xm bc
• (∇[m A) n] = (∇m A)n − (∇n A)m
∂An = − Γinm Ai − ∂xm
∂Am − Γimn Ai ∂xn
=
∂Am ∂Am − m ∂x ∂xn
• (∇m ω)nr =
∂ωnr − Γinm ωir − Γirm ωni ∂xm
Exercise 3. : Connection coefficients Question .
The connection coefficient functions Γ in chart (U ∩ V, y) is given, in terms of chart (U ∩ V, x) as follows: Recall Eq. (7.1) Γijk (y) =
∂y i ∂ 2 xq ∂y i ∂xs ∂xp q + Γ (x) ∂xq ∂y j ∂y k ∂xq ∂y j ∂y k sp 19
8. Lecture 8: Parallel Transport & Curvature (International Winter School on Gravity and Light 2015) 8.1. Parallelity of vector fields. Definition 15.
(1) parallely transported along smooth curve γ : R → M
if ∇vγ X = 0
(8.1) (2) A slightly weaker condition is “parallel”
(∇vγ,γ(λ) X)γ(λ) = µ(λ)Xγ(λ) 8.2. Autoparallely transported curves. Definition 16. curve γ : R → M is called autoparallely transported if !
∇v γ vγ = 0
(8.2) 8.3. Autoparallel equation.
∇v γ vγ = 0 in summary: m a b γ¨(x) (λ) + (Γm ˙ (x) (λ)γ˙ (x) (λ) = 0 (x) )ab (γ(λ))γ
(8.3) 8.4. Torsion.
Definition 17. torsion of a connection ∇ is the (1, 2)-tensor field T (ω, X, Y ) := ω(∇X Y − ∇Y X − [X, Y ])
(8.4) (Inside a cloud) [X, Y ] vector field defined by
[X, Y ]f := X(Y f ) − Y (Xf ) Proof. check T is C ∞ -linear in each entry
T (ω, f X, Y ) = ω(∇f X Y − ∇Y (f X) − [f X, Y ]) Definition 18. A (M, O, A, ∇) is called torsion-free if T = 0 In a chart T iab
∂ ∂ i := T dx , a , b = dxi (. . . ) ∂x ∂x = Γiab − Γiba = 2Γi[ab]
From now on, in these lectures, we only use torsion-free connections. 20
8.5. 4. Curvature. Definition 19. Riemann curvature of a connection ∇ is the (1, 3)-tensor field Riem(ω, Z, X, Y ) := ω(∇X ∇Y Z − ∇Y ∇X Z − ∇[X,Y ] Z)
(8.5)
Proof. do it: C ∞ -linear in each slot.
Tutorials Riemijab = . . . Tutorial 8 Parallel transport & Curvature Exercise 1. Exercise 2. : Where connection coefficients appear
It was suggested in the tutorial sheets and hinted in the lecture that the following should be committed to memory. Question : Recall the autoparallel equation for a curve γ.
(a) ∇v γ vγ = 0 (b) ∇vγ vγ = ∇γ˙
∂ ∂xµ
ν
vγ = γ˙ ∇∂ν vγ = γ˙
ν
ρ ∂vγµ ∂ ˙ ∂ ρ µ ν ∂γ ρ µ + Γµν vγ = γ˙ + Γµν γ˙ =0 ∂xν ∂xρ ∂xν ∂xρ
=⇒ γ¨ ρ + Γρµν γ˙ µ γ˙ ν as, for example, for F (x(t)), dF (x(t)) ∂F d = x˙ = F dt ∂x dt so that γ˙ ν
∂vγµ d µ d2 µ = v = γ γ ∂xν dλ dλ2
Question : Determine the coefficients of the Riemann tensor with respect to a chart (U, x).
Recall this manifestly covariant definition
Riem(ω, Z, X, Y ) = ω(∇X ∇Y Z − ∇Y ∇X Z − ∇[X,Y ] Z) We want
Rijab .
now ∇X ∇Y Z = ∇X ((Y µ
∂ ∂ ∂ ∂ ∂ ∂ ρ µ ν β Z + Γρµν Z µ Y ν ) ρ ) = (X α α (Y µ µ Z ρ + Γρµν Z µ Y ν ) + Γραβ (Y µ µ Z α + Γα µν Z Y )X ) ∂xµ ∂x ∂x ∂x ∂x ∂xρ
For X = ∂a , Y = ∂b , Z = ∂j , then the partial derivatives of the coefficients of the input vectors become zero.
=⇒ ∇∂a ∇∂b ∂j = Now [X, Y ]i = X j
∂ (Γi ) + Γiαa Γα jb ∂xa jb
i ∂ i j ∂X Y − Y ∂xj ∂xj
For coordinate vectors, [∂i , ∂j ] = 0 ∀ i, j = 0, 1 . . . d. Thus Rijab =
∂ i ∂ i i α Γ − Γ + Γiαa Γα jb − Γαb Γja ∂xa jb ∂xb ja 21
Question :Ric(X, Y ) := Riemmamb X a Y b define (0, 2)-tensor?.
Yes, transforms as such:
EY developments. I roughly follow the spirit in Theodore Frankel’s The Geometry of Physics: An Introduction Second Ed. 2003, Chapter 9 Covariant Differentiation and Curvature, Section 9.3b. The Covariant Differential of a Vector Field. P.S. EY : 20150320 I would like a copy of the Third Edition but I don’t have the funds right now to purchase the third edition: go to my tilt crowdfunding campaign, http://ernestyalumni.tilt.com, and help with your financial support if you can or send me a message on my various channels and ernestyalumni gmail email address if you could help me get a hold of a digital or hard copy as a pro bono gift from the publisher or author. The spirit of the development is the following: “How can we express connections and curvatures in terms of forms?” -Theodore Frankel. From Lecture 7, connection ∇ on vector field Y , in the “direction” X, ∂Y i ∂ i j + Γ Y ∇ ∂k Y = jk ∂x ∂xk ∂xi Make the ansatz (approche, impostazione) that the connection ∇ acts on Y , the vector field, first: i ∂ ∂ ∂Y i ∂ ∇Y (X) = X k k + Γijk Y j X k = X k ∇ ∂k Y = (∇X Y )i i = ∇X Y i ∂x ∂x ∂x ∂xi ∂x Now from Lecture 7, Definition for Γ, dxi ∇
∂ ∂xk
∂ ∂xj
= Γijk
Make this ansatz (approche, impostazine) ∇
∂ ∂ = Γijk dxk ⊗ ∈ Ω1 (M, T M ) = T ∗ M ⊗ T M j ∂x ∂xi
where Ω1 (M, T M ) = T ∗ M ⊗ T M is the set of all T M or vector-valued 1-forms on M , with the 1-form being the following: Γijk dxk = Γij ∈ Ω1 (M )
i = 1 . . . dim(M ) j = 1 . . . dim(M )
So Γij is a dimM × dimM matrix of 1-forms (EY !!!). Thus ∇Y = (d(Y i ) + Γij Y j ) ⊗
∂ ∂xi
So the connection is a (smooth) map from T M to the set of all vector-valued 1-forms on M , Ω1 (M, T M ), and then, after “eating” a vector Y , yields the “covariant derivative”: ∇ : T M → Ω1 (M, T M ) = T ∗ M ⊗ T M ∇ : Y 7→ ∇Y ∇Y : T M → T M ∇Y (X) 7→ ∇Y (X) = ∇X (Y ) Now
∂ ∂ ∂ ∂ ∂ ∂ , f= − =0 ∂xi ∂xj ∂xi ∂xj ∂xj ∂xi (this is okay as on p ∈ (U, x); x-coordinates on same chart (U, x)) 22
EY : 20150320 My question is when is this nontrivial or nonvanishing (i.e. not equal to 0). [ea , eb ] =? for a frame (ec ) and would this be the difference between a tangent bundle T M vs. a (general) vector bundle? Wikipedia helps here. cf. wikipedia, “Connection (vector bundle)” ∇ : Γ(E) → Γ(T ∗ M ⊗ E) = Ω1 (M, E) c b ∇ea = ωab f ⊗ ec
f b ∈ T ∗ M (this is the dual basis for T M and, note, this is for the manifold, M c ∇fb ea = ωab ec ∈ E c b ωac = ωab f ∈ Ω1 (M )
is the connection 1-form, with a, c = 1 . . . dimV . EY : 20150320 This V is a vector space living on each of the fibers of E. I know that Γ(T ∗ M ⊗ E) looks like it should take values in E, but it’s meaning that it takes vector values of V . Correct me if I’m wrong: ernestyalumni at gmail and various social media. Let σ ∈ Γ(E), σ = σ a ea c a b ∇σ = (dσ c + ωab σ f ) ⊗ ec with
dσ c = =⇒ ∇X σ =
∂σ c b f b ∂x
c ∂σ ∂σ c c a b c a σ X ec = X b + ω σ ec X b b + ωab ab ∂x ∂xb
9. Lecture 9: Newtonian spacetime is curved! Axiom 1 (Newton I:). A body on which no force acts moves uniformly along a straight line Axiom 2 (Newton II:). Deviation of a body’s motion from such uniform straight motion is effected by a force, reduced by a factor of the body’s reciprocal mass. Remark: (1) 1st axiom - in order to be relevant - must be read as a measurement prescription for the geometry of space . . . (2) Since gravity universally acts on every particle, in a universe with at least two particles, gravity must not be considered a force if Newton I is supposed to remain applicable.
9.1. Laplace’s questions. Laplace ∗ 1749 †1827 Q: “Can gravity be encoded in a curvature of space, such that its effects show if particles under the influence of (no other) force we postulated to more along straight lines in this curved space?” Answer: No! Proof. gravity is a force point of view m¨ xα (t) = F α (x(t)) m¨ xα (t) = mf α (x(t)) | {z } Fα
−∂α f α = 4πGρ (Poisson) ρ mass density of matter 23
(EY : 20150330) You know this, F = Gm1 m2 /r2 x ¨α (t) − f α (x(t)) = 0 Laplace asks: Is this (¨ x(t)) of the form x ¨α (t) + Γαβγ (x(t))x˙ β (t)x˙ γ (t) = 0 Conclusion: One cannot find Γ s such that Newton’s equation takes the form of an autoparallel. 9.2. The full wisdom of Newton I. use also the information from Newton’s first law that particles (no force) move uniformly introduce the appropriate setting to talk about the difference easily uniform & straight motion
insight: in spacetime
is simply straight motion
So let’s try in spacetime: let x : R → R3
be a particle’s trajectory in space ←→ worldline (history) of the particle
X : R → R4 t 7→ (t, x1 (t), x2 (t), x3 (t)) := := (X 0 (t), X 1 (t), X 2 (t), X 3 (t))
That’s all it takes: Trivial rewritings: X˙ 0 = 1 a = 0, 1, 2, 3 =⇒
¨0 X
=0
¨ α − f α (X(t)) · X˙ 0 · X˙ 0 X
=0
(α = 1, 2, 3) =⇒
¨ a + Γa X˙ b X˙ c = 0 X bc antoparallel eqn in spacetime
Yes, choosing Γ0ab = 0 Γαβγ = 0 = Gammaα0β = Γαβ0 !
only: Γα00 = −f α Question: Is this a coordinate-choice artifact? No, since Rα0β0 = − ∂x∂ β f α (only non-vanishing components) (tidal force tensor, − the Hessian of the force component) Ricci tensor =⇒ R00 = Rm0m0 = −∂α f α = 4πGρ Poisson: −∂α f α = 4πG · ρ writing: T00 = 12 s =⇒ R00 = 8πGT00 hhhh ( (( (( Einstein in 1912 ( Rab 8πGT h (= ab h h 24
Conclusion: Laplace’s idea works in spacetime Remark Γα00 = −f α Rαβγδ = 0
α, β, γ, δ = 1, 2, 3
R00 = 4πGρ Q: What about transformation behavior of LHS of x ¨a + Γabc X˙ b X˙ c | {z } (∇vX vX )a | {z }
=0
:=aa “acceleration vector”
9.3. The foundations of the geometric formulation of Newton’s axiom. new start Definition 20. A Newtonian spacetime is a quintuple (M, O, A, ∇, t) where (M, O, A) 4-dim. smooth manifold t : M → R smooth function (i) “There is an absolute space” (dt)p 6= 0
∀p ∈ M
(ii) “absolute time flows uniformly” ∇dt
= |{z}
0
everywhere
space of (0, 2)-tensor fields
∇dt is a (0, 2)-tensor field (iii) add to axioms of Newtonian spacetime ∇ = 0 torsion free Definition 21. absolute space at time τ Sτ := {p ∈ M |t(p) = τ } a dt6=0 −−−→ M = Sτ Definition 22. A vector X ∈ Tp M is called (a) future-directed if dt(X) > 0 (b) spatial if dt(X) = 0 (c) past-directed if dt(X) < 0 picture Newton I: The worldline of a particle under the influence of no force (gravity isn’t one, anyway) is a future-directed autoparallel i.e. ∇ v X vX = 0 dt(vX ) > 0 25
Newton II: ∇vX v X =
F ⇐⇒ m · a = F m
where F is a spatial vector field: dt(F ) = 0 Convention: restrict attention to atlases Astratefied whose charts (U, x) have the property x0 : U → R x1 : U → R .. .
.. .
x0 = t|U
0
=⇒
“absolute time flows uniformly”
=
0=∇
∂ ∂xa
dx0 = −Γ0ba
∇dt
a = 0, 1, 2, 3
x3 Let’s evaluate in a chart (U, x) of a stratified atlas Asheet : Newton II:
∇vX v X =
F m
in a chart. ( a 0(( (( Γ0cd ) (X b )0stratified atlas = 0 (X 0 )00 + ( ((X 0
0
0
0
0
0
γ δ α 0 0 α γ 0 (X α )00 + Γα = γδ X X + Γ 00 X X + 2Γ γ0 X X
=⇒ (X 0 )00 (λ) = 0 =⇒ X 0 (λ) = aλ + b X 0 (λ) = (x0 ◦ X)(λ)
stratified
=
Fα m
α = 1, 2, 3
constants a, b with (t ◦ X)(λ)
convention parametrize worldline by absolute time d d =a dλ dt α ¨ α + a2 Γα X˙ γ X˙ δ + a2 Γα X˙ 0 X˙ 0 + 2Γα X˙ γ X˙ 0 = F a2 X γδ 00 γ0 m α 1 F ¨ α + Γαγδ X˙ γ X˙ δ + Γα00 X˙ 0 X˙ 0 + 2Γαγ0 X˙ γ X˙ 0 = =⇒ X {z } a2 m | aα
10. Lecture 10: Metric Manifolds We establish a structure on a smooth manifold that allows one to assign vectors in each tangent space a length (and an angle between vectors in the same tangent space). From this structure, one can then define a notion of length of a curve. Then we can look at shortest curves. Requiring then that the shortest curves coincide with the straightest curves (wrt ∇) will result in ∇ being determined by the metric structure.
g
straight=short T =0 26
∇
Riem
10.1. Metrics. Definition 23. A metric g on a smooth manifold (M, O, A) is a (0, 2)-tensor field satisfying (i) symmetry g(X, Y ) = g(Y, X) ∀ X, Y vector fields (ii) non-degeneracy: the musical map “flat” [ : Γ(T M ) → Γ(T ∗ M ) X 7→ [(X)
where
[(X)(Y ) := g(X, Y )
[(X) ∈ Γ(T ∗ M ) In thought bubble: [(X) = g(X, ·) . . . is a C ∞ -isomorphism in other words, it is invertible.
Remark: ([(X))a Xa ([(X))a := gam X m
or
Thought bubble: [−1 = ] [−1 (ω)a := g am ωm 00 [−1 (ω)a := (g “−1 )am ωm =⇒ not needed. (all of this is not needed) 00
Definition 24. The (2, 0)-tensor field g “−1 with respect to a metric g is the symmetric 00
g “−1 : Γ(T ∗ M ) × Γ(T ∗ M ) → − C ∞ (M ) (ω, σ) 7→ ω([−1 (σ))
[−1 (σ) ∈ Γ(T M ))
chart: gab = gba (g −1 )am gmb = δba Example: (S 2 , O, A) chart (U, x) ϕ ∈ (0, 2π) θ ∈ (0, π) define the metric −1
gij (x
" R2 (θ, ϕ)) = 0
R ∈ R+ “the metric of the round sphere of radius R” 27
0 2 R sin2 θ
# ij
λ1 0
Aam v m = λv a 1
10.2. Signature. Linear algebra: gam v
m
a
=λ·v ?
0 ..
. λn
..
. 1 −1 ..
. −1 0 ..
.
0
(1, 1) tensor has eigenvalues (0, 2) has signature (p, q) (well-defined) (+ + +) (+ + −) d + 1 if p + q = dimV (+ − −) (− − −) Definition 25. A metric is called Riemannian if its signature is (+ + · · · +) Lorentzian if (+ − · · · −) 10.3. Length of a curve. Let γ be a smooth curve. Then we know its veloctiy vγ,γ(λ) at each γ(λ) ∈ M . Definition 26. On a Riemannian metric manifold M, O, A, g), the speed of a curve at γ(λ) is the number q ( g(vγ , vγ ))γ(λ) = s(λ) F. Schuller: “I feel the need for speed.” -Top Gun. (I feel the need for speed, then I feel the need for a metric) Aside: [v a ] = T1 [gab ] = L2 q p 2 [ gab v a v b ] = TL2 =
L T
Definition 27. Let γ : (0, 1) → M a smooth curve. Then the length of γ is the number Z 1 Z R 3 L[γ] := dλs(λ) = 0
1
dλ
q
(g(vγ , vγ ))γ(λ)
0
F. Schuller: “velocity is more fundamental than speed, speed is more fundamental than length” Example: reconsider the round sphere of radius R Consider its equator: π θ(λ) := (x1 ◦ γ)(λ) = 2 2 ϕ(λ) := (x ◦ γ)(λ) = 2πλ3 28
θ0 (λ) = 0 ϕ0 (λ) = 6πλ2 on the same chart gij =
" R2
# R2 sin2 θ
F.Schuller: do everything in this chart Z 1 q Z L[γ] = dλ gij (x−1 (θ(λ), ϕ(λ)))(xi ◦ γ)0 (λ)(xj ◦ γ)0 (λ) =
1
q dλ R2 · 0 + R2 sin2 (θ(λ))36π 2 λ4 =
0
0 1
Z = 6πR 0
1 dλλ2 = 6πR[ λ3 ]10 = 2πR 3
Theorem 4. γ : (0, 1) → M and σ : (0, 1) → (0, 1) smooth bijective and increasing “reparametrization” L[γ] = L[γ ◦ σ] Proof. =⇒ Tutorials
10.4. Geodesics. Definition 28. A curve γ : (0, 1) → M is called a geodesic on a Riemannian manifold (M, O, A, g) if its a stationary curve with respect to a length functional L. Thought bubble: in classical mechanics, deform the curve a little, times this deformation, to first order, it agrees with L[γ] Theorem 5. γ geodesic iff it satisfies the Euler-Lagrange equations for the Lagrangian
L :T M → R p X 7→ g(X, X) In a chart, the Euler Lagrange equations take the form: · ∂L ∂L =0 − ∂ x˙ m ∂xm F.Schuller: this is a chart dependent formulation here: L(γ i , γ˙ i ) =
q gij (γ(λ))γ˙ i (λ)γ˙ j (λ)
Euler-Lagrange equations: ∂L 1 = √ gmj (γ(λ))γ˙ j (λ) ∂ γ˙ m ... !· · ∂L 1 1 = √ gmj (γ(λ)) · γ˙ j (λ) + √ gmj (γ(λ))¨ γ j (λ) + γ˙ s (∂s gmj )γ˙ j (λ) m ∂ γ˙ ... ... Thought bubble: reparametrize g(γ, ˙ γ) ˙ = 1 (it’s a condition on my reparametrization) By a clever choice of reparametrization ( √1... )· = 0 ∂L 1 = √ ∂m gij (γ(λ))γ˙ i (λ)γ˙ j (λ) ∂γ m 2 ... putting this together as Euler-Lagrange equations: 1 gmj γ¨ j + ∂s gmj γ˙ s γ˙ j − ∂m gij γ˙ i γ˙ j = 0 2 29
Multiply on both sides (g −1 )qm 1 γ¨q + (g −1 )qm (∂i gmj − ∂m gij )γ˙ i γ˙ j = 0 2 1 γ¨q + (g −1 )qm (∂i gmj + ∂j gmi − ∂m gij )γ˙ i γ˙ j = 0 2 geodesic equation for γ in a chart.
Thought bubble:
·
∂L a+dimM ∂ξx
σ(x)
1 (g −1 )qm (∂i gmj + ∂j gmi − ∂m gij ) =: Γqij (γ(λ)) 2 ∂L =0 − ∂xi a
Definition 29. “Christoffel symbol” L.C. ∇
x
L.C.
σ(x)
Γ are the connection coefficient functions of the so-called Levi-Civita connection
We usually make this choice of ∇ if g is given. (M, O, A, g) → (M, O, A, g,
L.C.
∇)
abstract way: ∇g = 0 and T = 0 (torsion) =⇒ ∇ = L.C. ∇ Definition 30.
(a) The Riemann-Christoffel curvature is defined by Rabcd := gam Rmbcd
(b) Ricci: Rab = Rmamb Thought bubble: with a metric,
L.C.
∇
(c) (Ricci) scalar curvature: R = g ab Rab L.C.
Thought bubble:
∇
Definition 31. Einstein curvature (M, O, A, g) 1 Gab := Rab − gab R 2 00
Convention: g ab := (g “−1 )ab F. Schuller: these indices are not being pulled up, because what would you pull them up with (student) Question: Does the Einstein curvature yield new information? Answer: 1 1 d g ab Gab = Rab g ab − gab g ab R = R − δaa R = R − dimM R = (1 − )R 2 2 2 Tutorial 9: Metric manifolds. Exercise 3: Levi-Civita Connection. compatible connection ∇g = 0
Suppose torsion-free T = 0 and metric-
Question Recall T = 0 on a chart.
1 −1 cm (g ) 2
∂gma ∂gab ∂gbm + − ∂xa ∂xb ∂xm
1 = (g −1 )am 2
∂gbm ∂gmc ∂gbc + − ∂xc ∂xb ∂xm
Γcba = or Γabc
30
11. Symmetry EY : 20150321 This lecture tremendously and lucidly clarified, for me at least, what a symmetry of the Lie algebra is, and in comparing structures (M, O, A) vs. (M, O, A, ∇), clarified differences, and asking about differences is a good way to learn, the difference between L and ∇, respectively. Feeling that the round sphere (S 2 , O, A, g round ) has rotational symmetry, while the potato (S 2 , O, A, g potato ) does not. 11.1. 11.2.
Important
11.3. Flow of a complete vector field. Let (M, O, A) smooth X vector field on M Definition 32. A curve γ : I ⊆ R → M is called an integral curve of X if vγ,γ(λ) = Xγ(λ) Definition 33. A vector filed X is complete if all integral curves have I = R EY: 20150321 (i.e. domain is all of R) Ex. minute 48:30 EY : reall good explanation by F.P.Schuller; take a pt. out for an incomplete vector field. Theorem 6. compactly supported smooth vector field is complete. Definition 34. The flow of a complete vector field X is a 1-parameter family hX = R × M → M where γp : R → M is the integral curve of X with γ(0) = p Then for fixed λ ∈ R hX λ : M → M smooth picture hX λ (S) 6= S( if X 6= 0) 11.4. Lie subalgebras of the Lie algebra (Γ(T M ), [·, ·]) of vector fields. (a) Γ(T M ) = { set of all vector fields }
C ∞ (M )-module = R-vector space
=⇒ [X, Y ] ∈ Γ(T M )
[X, Y ]f := X(Y f ) − Y (Xf )
(i) [X, Y ] = −[Y, X] (ii) [λX + Z, Y ] = λ[X, Y ] + [Z, Y ] (iii) [X, [Y, Z]] + [Z, [X, Y ]] + [Y, [Z, X]] = 0 (Γ(T M ), [·, ·]) Lie algebra (b) Let X1 . . . Xs for s (many) vector fields on M , such that 31
Tutorial 11 Symmetry. Exercise 1. : True or false? (a)
• • φ∗ : T ∗ N → T ∗ M i.e. φ∗ ν(X) = ν(φ∗ X) for smooth φ : M → N , so the pullback of a covector ν ∈ T ∗ N maps to a covector in T ∗ M . • • • •
(b) (c) Exercise 2. : Pull-back and push-forward Question . Let’s check this locally j ∂y j ∂ i ∂y ∂f ) = X where ∂xi ∂y j ∂xi ∂y j j ∂f ∂y j i i ∂y ∂f dx (X) = X d(φ∗ f )(X) = d(f (φ))(X) = ∂y j ∂xi ∂xi ∂y j
φ∗ (df )(X) = (df )(φ∗ X) = (df )(X i
φ∗ X = X i
∂y j ∂ ∂xi ∂y j
So φ∗ (df ) = d(φ∗ f )
∀ p ∈ M, ∀ X ∈ X(M )
The big idea is that this is a showing of the naturality of the pullback φ∗ with d, i.e. that this commutes: φ∗
Ω1 (M )
Ω1 (N )
d
d
∞
∞
C (M )
φ∗
C (N )
Question .
∂ )) ∂xb ∞ Let g ∈ C (N ) b ∂ ∂x ∂ ∂ φ∗ g= φ(p) = gφx−1 x(p) = (gyy −1 φx−1 )(x) = ∂xb g ∂xb ∂xb ∂ ∂g b ∂y a ∂y a ∂g −1 −1 = (gy (yφx (x(p)))) = = b b ∂x ∂y ∂x x ∂xb ∂y a (φ∗ )ab := (dy a )(φ∗ (
y
Then
φ∗
∂ ∂xb
=
and so (φ∗ )ab =
∂y a ∂ ∂xb ∂y a ∂y a ∂xb
Question . Exercise 3. :Lie derivative-the pedestrian way
While it is true that ∀ p ∈ S 2 , for x(p) = (θ, ϕ), and (yix−1 )(θ, ϕ) = (y 1 , y 2 , y 3 ) ∈ R3 and that, at this point p, (y ) /a + (y 2 )2 /b2 + (y 3 )2 /c3 = 1, this doesn’t imply (EY: 20150321 I think) that, globally, it’s an ellipsoid (yet). In the familiar charts given, Question . 1 2
2
32
spherical chart (U, x) ∈ A and (R3 , y = idR3 ) ∈ B it looks like an ellipsoid, but change to another choice of charts, and it could look something very different. Question .
Equip (R3 , Ost , B) with the Euclidean metric g, and pullback g. Note that the pullback of the inclusion from R3 onto S 2 for the Euclidean metric is the following: a ∂ ∂ ∂ ∂x ∂ ∂xb ∂ ∂xa ∂xb ∂ ∗ , = g i , i = g , = g i g ∗ ∗ ab ∂θi ∂θj ∂θi ∂θj ∂θi ∂xa ∂θj ∂xb ∂θi ∂θj With gab = δab , the usual Euclidean metric, this becomes the following: ellipsoid gij =
∂xa ∂xa ∂θi ∂θj
At this point, one should get smart (we are in the 21st century) and use some sort of CAS (Computer Algebra System). I like Sage Math (version 6.4 as of 20150322). I also like the Sage Manifolds package for Sage Math. I like Sage Math for the following reasons: • Open source, so its open and freely available to anyone, which fits into my principle of making online education open and freely available to anyone, anytime • Sage Math structures everything in terms of Category Theory and Categories and Morphisms naturally correspond to Classes and Class methods or functions in Object-Oriented Programming in Python and theyve written it that way and I like Sage Manifolds for roughly the same reasons, as manifolds are fit into a category theory framework thats written into the Python code. e.g. sage: S2 = Manifold(2, ’S^2’, r’\mathbb{S}^2’, start_index=1) ; print S2 sage: print S2 2-dimensional manifold ’S^2’ sage: type(S2)
With code (Ive provided for convenience; you can make your own as I wrote it based upon to example of S 2 on the sagemanifolds documentation website page), load it and do the following: cf. https://github.com/ernestyalumni/diffgeo-by-sagemnfd/blob/master/S2.sage http://sagemanifolds.obspm.fr/examples.html sage: load("S2.sage") sage: U_ep = S2.open_subset(’U_{ep}’) sage: eps. = U_ep.chart() sage: a = var(a) sage: b = var(b) sage: c = var("c") sage: inclus = S2.diff_mapping(R3, {(eps, cart): [ a*cos(phi)*sin(the), b*sin(phi)*sin(the),c*cos(the) ]} , name="inc",latex_name=r’\mathcal{i}’) sage: inclus.pullback(h).display() inc_*(h) = (c^2*sin(the)^2 + (a^2*cos(phi)^2 + b^2*sin(phi)^2)*cos(the)^2) dthe*dthe - (a^2 - b^2)*cos(phi)*cos(the)*sin(phi)*sin(the) dthe*dphi - (a^2 - b^2)*cos(phi)*cos(the)*sin(phi)*sin(the) dphi*dthe + (b^2*cos(phi)^2 + a^2*sin(phi)^2)*sin(the)^2 dphi*dphi sage: inclus.pullback(h)[2,2].expr() (b^2*cos(phi)^2 + a^2*sin(phi)^2)*sin(the)^2
A new open subset Uep was declared in S 2 , a new chart (Uep , (θ, φ)) was declared, the constants, a, b, c, were declared, and the inclusion map given in the problem y ◦ i ◦ x−1 : (θ, φ) 7→ (a cos φ sin θ, b sin φ sin θ, c cos θ) Then the pullback of the inclusion map i was done on the Euclidean metric h, defined earlier in the file S2.sage 33
. Then one can access the components of this metric and do, for example, simplify_full(),full_simplify(), reduce_trig() on the expression. In Python, I could easily do this, and give an answer quick in LaTeX: sage: for i in range(1,3): ....: for j in range(1,3): ....: print inclus.pullback(h)[i,j].expr() ....: latex(inclus.pullback(h)[i,j].expr() ) ....: c^2*sin(the)^2 + (a^2*cos(phi)^2 + b^2*sin(phi)^2)*cos(the)^2 (EY: I’ll suppress the LaTeX output but this sage math function gives you LaTeX code) and so 2 2 2 2 i∗ g = c2 sin (the) + a2 cos (φ) + b2 sin (φ) cos (the) dθ ⊗ dθ+ −2 a2 − b2 cos (φ) cos (the) sin (φ) sin (the) dθ ⊗ dφ+ 2 2 2 + b2 cos (φ) + a2 sin (φ) sin (the) dφ ⊗ dφ Question . sage: polar_vees = eps.frame() sage: X_1 = - sin(phi) * polar_vees[1] - cot( the ) * cos(phi) * polar_vees[2] sage: X_2 = cos( phi ) * polar_vees[1] - cot( the ) * sin( phi) * polar_vees[2] sage: X_3 = polar_vees[2] sage: X_2.lie_der(X_1).display() (cos(the)^2 - 1)/sin(the)^2 d/dphi sage: X_3.lie_der(X_1).display() cos(phi) d/dthe - cos(the)*sin(phi)/sin(the) d/dphi sage: X_3.lie_der(X_2).display() sin(phi) d/dthe + cos(phi)*cos(the)/sin(the) d/dphi
Indeed, one can check on a scalar field feps ∈ C ∞ (S 2 ): sage: f_eps = S2.scalar_field({eps: function(’f’, the, phi ) }, name=’f’ ) sage: (X_1( X_2(f_eps)) - X_2(X_1(f_eps) ) ).display() U_{ep} --> R (the, phi) |--> -D[1](f)(the, phi) sage: X_2.lie_der(X_1) == -X_3 True sage: X_3.lie_der(X_1) == X_2 True sage: X_3.lie_der(X_2) == -X_1 True
=⇒ [Xi , Xj ] = −ijk Xk So spanR {X1 , X2 , X3 } equipped with [ , ] constitute a Lie subalgebra on S 2 (It’s closed under [ , ] 12. Integration 12.1. 34
12.2. 12.3. Volume forms. Definition 35. On a smooth manifold (M, O, A) a (0, dimM )-tensor field Ω is called a volume form if (a) Ω vanishes nowhere (i.e. Ω 6= 0 ∀ p ∈ M ) (b) totally antisymmetric X ...) Y , . . . , |{z} Y . . . ) = −Ω(. . . , |{z} Ω(. . . , |{z} X , . . . , |{z} ith
ith
jth
jth
In a chart: Ωi1 ...id = Ω[i1 ...id ] Example (M, O, A, g) metric manifold construct volume form Ω from g In any chart: (U, x)
Ωi1 ...id :=
q
det(gij (x))i1 ...id
where Levi-Civita symbol i1 ...id is defined as 123...d = +1 1...d = [i1 ...id ] Proof. (well-defined) Check: What happens under a change of charts q Ω(y)i1 ...id = det(g(y)ij )i1 ...id = s ∂xm ∂xn ∂y m1 ∂y md = det(gmn (x) i ) . . . [m1 ...md ] = ∂y ∂y j ∂xi1 ∂xid q q ∂x ∂y ∂x det = |detgij (x)| det i1 ...id = detgij (x)i1 ...id sgn det ∂y ∂x ∂y EY : 20150323 Consider the following: id i1 Ω(y)(Y(1) . . . Y(d) ) = Ω(y)i1 ...id Y(1) . . . Y(d) = q id i1 = det(gij (y))i1 ...id Y(1) . . . Y(d) = s ∂xm ∂xn ∂y i1 ∂y id = det(gmn (x)) i i1 ...id m1 . . . m X m1 . . . X md = j ∂y ∂y ∂x ∂x d s ∂xm ∂xn ∂y = det(gmn (x)) i det m1 ...md X m1 . . . X md = j ∂y ∂y ∂x p ∂x ∂y = det(gmn (x)) det det m1 ...md X m1 . . . X md = ∂y ∂x p ∂x ∂x m1 md = det(gmn (x))m1 ...md sgn det X ...X = sgn(det )Ωm1 ...md (x)X m1 . . . X md ∂y ∂y 35
If det
∂y ∂x
> 0, Ω(y)(Y(1) . . . Y(d) ) = Ω(x)(X(1) . . . X(d) )
This works also if Levi-Civita symbol i1 ...id doesn’t change at all under a change of charts. (around 42:43 https://youtu. be/2XpnbvPy-Zg)
Alright, let’s require, restrict the smooth atlas A to a subatlas (A↑ still an atlas) A↑ ⊆ A s.t. ∀ (U, x), (V, y) have chart transition maps y ◦ x−1 x ◦ y −1 s.t. det
∂y ∂x
>0
↑
such A called an oriented atlas
(M, O, A, g) =⇒ (M, O, A↑ , g) Note: associated bundles.
Note also: det
∂y b ∂xa
∂y b ∂xa
= det(∂a (y b x−1 ))
is an endomorphism on vector space V . ϕ : V → V detϕ
g is a (0, 2) tensor field, not endomorphism (not independent of choice of basis)
p
independent of choice of basis
|det(gij (y))|
Definition 36. Ω be a volume form on (M, O, A↑ ) and consider chart (U, x)
Definition 37. ω(X) := Ωi1 ...id i1 ...id same way 12...d = +1 [... ] one can show
ω(y) = det
∂x ∂y
ω(x)
scalar density
12.4. Integration on one chart domain U . Definition 38. Z (12.1)
(U,y)
Z
f: = U
dd βω(y) (y −1 (β))f(y) (β)
y(U )
Proof. : Check that it’s (well-defined), how it changes under change of charts Z Z Z Z ∂y ∂x (U,y) −1 f: = dd βω(y) (y −1 (β))f(y) (β) = = dd α det f (α)ω (x (α)det = (x) (x) ∂x ∂y (U,y) U y(U ) x(U ) Z = dd αω(x) (x−1 (x))f(x) (α) x(U )
36
On an oriented metric manifold (M, O, A↑ , g) Z Z f := U
dd α
x(U )
q |
det(gij (x))(x−1 (α)) f(x) (α) {z } √
g
12.5. Integration on the entire manifold. 13. Lecture 13: Relativistic spacetime Recall, from Lecture 9, the definition of Newtonian spacetime ∇ torsion free t ∈ C ∞ (M ) (M, O, A, ∇, t)
dt 6= 0 ∇dt = 0
(uniform time)
and the definition of relativistic spacetime (before Lecture ) ∇ torsion-free (M, O, A↑ , ∇, g, T )
g Lorentzian metric(+ − −−) T time-orientation
13.1. Time orientation. Definition 39. (M, O, A↑ , g) a Lorentzian manifold. Then a time-orientation is given by a vector field T that (i) does not vanish anywhere (ii) g(T, T ) > 0 Newtonian vs. relativistic Newtonian X was called future-directed if dt(X) > 0 ∀ p ∈ M , take half plane, half space of Tp M also stratified atlas so make planes of constant t straight relativistic half cone ∀ p, q ∈ M , half-cone ⊆ Tp M
This definition of spacetime Question I see how the cone structure arises from the new metric. I don’t understand however, how the T , the time orientation, comes in
Answer ( (M, O, A, g) g ← − + − −−) requiring g(X, X) > 0, select cones T chooses which cone
This definition of spacetime has been made to enable the following physical postulates: 37
(P1) The worldline γ of a massive particle satisfies (i) gγ(λ) (vγ,γ(lambda) , vγ,γ(λ) ) > 0 (ii) gγ(λ) (T, vγ,γ(λ) ) > 0 (P2) Worldlines of massless particles satisfy (i) gγ(λ) (vγ,γ(λ) , vγ,γ(λ) ) = 0 (ii) gγ(λ) (T, vγ,γ(λ) ) > 0 picture: spacetime: Answer (to a question) T is a smooth vector field, T determines future vs. past, “general relativity: we have such a time orientation; smoothness makes it less arbitrary than it seems” -FSchuller, Claim: 9/10 of a metric are determined by the cone spacetime determined by distribution, only one-tenth error 13.2. Observers. (M, O, A↑ , ∇, g, T ) Definition 40. An observer is a worldline γ with g(vγ , vγ ) > 0 g(T, vγ ) > 0 together with a choice of basis vγ,γ(λ) ≡ e0 (λ), e1 (λ), e2 (λ), e3 (λ)
of each Tγ(λ) M where the observer worldline passes, if g(ea (λ), eb (λ)) = ηab
1 =
−1 −1
−1
precise: observer = smooth curve in the frame bundle LM over M 13.2.1. Two physical postulates. (P3) A clock carried by a specific observer (γ, e) will measure a time Z λ1 q τ := dλ gγ(λ) (vγ,γ(λ) , vγ,γ(λ) ) λ0
between the two “events” γ(λ0 )
“start the clock”
γ(λ1 )
“stop the clock”
and Compare with Newtonian spacetime: t(p) = 7 Thought bubble: proper time/eigentime τ M = R4 Application/Example.
O = Ost A 3 (R4 , idR4 ) g : g(x)ij = ηij
;
i T(x) = (1, 0, 0, 0)i
=⇒ Γi(x) jk = 0 everywhere 38
ab
=⇒ (M, O, A↑ , g, T, ∇) =⇒ spacetime is flat
Riemm = 0
This situation is called special relativity. Consider two observers: γ : (0, 1) → M i γ(x) = (λ, 0, 0, 0)i
δ : (0, 1) → M (λ, αλ, 0, 0)i i α ∈ (0, 1) :δ(x) = (λ, (1 − λ)α, 0, 0)i
λ≤ λ>
1 2 1 2
let’s calculate: Z
1
τγ :=
q
0 Z 1/2
τδ :=
j i γ ˙ (x) g(x)ij γ˙ (x)
dλ
p
1 − α2 +
0
Z
1
dλ1 = 1
= 0
Z
1
p
12 − (−α)2 =
1/2
Z
1
p
1 − α2 =
p
1 − α2
0
Note: piecewise integration Taking the clock postulate (P3) seriously, one better come up with a realistic clock design that supports the postulate. idea. 2 little mirrors (P4) Postulate Let (γ, e) be an observer, and δ be a massive particle worldline that is parametrized s.t. g(vγ , vγ ) = 1 (for parametrization/normalization convenience) Suppose the observer and the particle meet somewhere (in spacetime) δ(τ2 ) = p = γ(τ1 ) This observer measures the 3-velocity (spatial velocity) of this particle as vδ : α (vδ,δ(τ2 ) )eα
(13.1)
α = 1, 2, 3
where 0 , 1 , 2 , 3 is the unique dual basis of e0 , e1 , e2 , e3 EY:20150407 There might be a major correction to Eq. (13.1) from the Tutorial 14 : Relativistic spacetime, matter, and Gravitation, see the second exercise, Exercise 2, third question: (13.2)
v :=
α (vδ ) eα 0 (vδ )
Consequence: An observer (γ, e) will extract quantities measurable in his laboratory from objective spacetime quantities always like that. Ex: F Faraday (0, 2)-tensor of electromagnetism:
F (ea , eb ) = Fab
0 −E 1 = −E2 −E3
observer frame ea , eb 39
E1 0 −B3 B2
E2 B3 0 −B1
E3 −B2 B1 0
Eα := F (e0 , eα ) B γ := F (eα , eρ )αβγ where 123 = +1 totally antisymmetric
13.3. Role of the Lorentz transformations. Lorentz transformations emerge as follows: Let (γ, e) and (e γ , ee) be observers with γ(τ1 ) = γ e(τ2 ) (for simplicity γ(0) = γ e(0) Now e0 , . . . , e 1
at τ = 0
and ee0 , . . . , ee1
at τ = 0
both bases for the same Tγ(0) M Thus: eea = Λba eb
Λ ∈ GL(4)
Now:
ηab = g(e ea , eeb ) = g(Λma em , Λnb en ) = = Λma Λnb g(em , en ) | {z } ηmn
i.e. Λ ∈ O(1, 3) Result: Lorentz transformations relate the frames of any two observers at the same point. “e xµ − Λµν xν ” is utter nonsense
Tutorial. I didn’t see a tutorial video for this lecture, but I saw that the Tutorial sheet number 14 had the relevant topics. Go there.
14. Lecture 14: Matter two types of matter point matter field matter point matter massive point particle more of a phenomenological importance field matter electromagnetic field more fundamental from the GR point of view both classical matter types 40
14.1. Point matter. Our postulates (P1) and (P2) already constrain the possible particle worldlines. But what is their precise law of motion, possibly in the presence of “forces”, (a) without external forces Z Smassive [γ] := m
dλ
q
gγ(λ) (vγ,γ(λ) , vγ,γ(λ) )
with: gγ(λ) (Tγ(λ) , vγ,γ(λ) ) > 0 dynamical law Euler-Lagrange equation similarly Z Smassless [γ, µ] =
dλµg(vγ,γ(λ) , vγ,γ(λ) )
δµ
g(vγ,γ(λ) , vγ,γ(λ) ) = 0
δγ
e.o.m.
Reason for describing equations of motion by actions is that composite systems have an action that is the sum of the actions of the parts of that system, possibly including “interaction terms.” Example. S[γ] + S[δ] + Sint [γ, δ] (b) presence of external forces or rather presence of fields to which a particle “couples” Example Z S[γ; A] =
dλm
q gγ(λ) (vγ,γ(λ) , vγ,γ(λ) ) + qA(vγ,γ(λ) )
where A is a covector field on M . A fixed (e.g. the electromagnetic potential) m Consider Euler-Lagrange eqns. Lint = qA(x) γ˙ (x)
˙ ! ∂Lint ∂Lint a m(∇vγ vγ )a + −qF am γ˙ m m γ− m = 0 =⇒ m(∇vγ vγ ) = | {z } ∂ ˙ (x) ∂γ(x) Lorentz force on a charged particle in an electromagnetic field | {z } ∗
˙ ∂L ∂ m γ=q· (A(x)m ) · γ˙ (x) ∂ ˙m ∂xm
∂L = qA(x)a , ∂ γ˙ a
∂L ∂ =q· (A(x)m )γ˙ m ∂γ a ∂xa ∗=q
∂Aa ∂Am − m ∂x ∂xa
m γ˙ (x)
m = q · F(x)am γ˙ (x)
F ← Faraday
Z S[γ] =
(m
q g(vγ , vγ ) + qA(vγ ))dλ 41
14.2. Field matter. Definition 41. Classical (non-quantum) field matter is any tensor field on spacetime where equations of motion derive from an action. Example: SMaxwell [A] =
1 4
√ d4 x −gFab Fcd g ac g bd
Z M
A (0, 1)-tensor field = thought cloud: for simplicity one chart covers all of M √ − for −g (+ − −−)
Fab := 2∂[a Ab] = 2(∇[a A)b] Euler-Lagrange equations for fields 0=
∂L ∂ − ∂Am ∂xs
∂L ∂∂s Am
+
∂2L ∂ ∂ s t ∂x ∂x ∂∂t ∂s Am
Example . . . (∇
∂ ∂xm
F )ma = j a
inhomogeneous Maxwell thought bubble j = qvγ
∂[a Fb] − () homogeneous Maxwell Other example well-liked by textbooks Z SKlein-Gordon [φ] :=
√ d4 x −g[g ab (∂a φ)(∂b φ) − m2 φ2 ]
M
φ (0, 0)-tensor field 14.3. Energy-momentum tensor of matter fields. At some point, we want to write down an action for the metric tensor field itself. But then, this action Sgrav [g] will be added to any Smatter [A, φ, . . . ] in order to describe the total system.
Stotal [g, A] = Sgrav [g] + SMaxwell [A, g]
δA δgab
:=⇒ Maxwell’s equations :
1 Gab + (−2T ab ) = 0 16πG
G Newton’s constant
Gab = 8πGN T ab Definition 42. Smatter [Φ, g] is a matter action, the so-called energy-momentum tensor is −2 ∂Lmatter ∂Lmatter ab T := √ − ∂s + ... −g ∂gab ∂∂s gab 42
− of
−2 √ g
is Schr¨ odinger minus (EY : 20150408 F.Schuller’s joke? but wise)
choose all sign conventions s.t. T (0 , 0 ) > 0 Example: For SMaxwell : 1 Tab = Fam Fbn g mn − Fmn F mn gab 4 Tab ≡ TMaxwellab T (e0 , e0 ) = E 2 + B 2 T (e0 , eα ) = (E × B)α Fact: One often does not specify the fundamental action for some matter, but one is rather satisfied to assume certain properties / forms of Tab Example Cosmology: (homogeneous & isotropic) perfect fluid
of pressure p and density ρ modelled by T ab = (ρ + p)ua ub − pg ab radiative fluid What is a fluid of photons: ab TMaxwell gab = 0 !
ab observe: Tp.f. gab = 0 = (ρ + p)ua ub gab − p g ab gab | {z } 4
↔ρp 04p = 0 ρ = 3p p=
1 3ρ
Reconvene at 3 pm? (EY : 20150409 I sent a Facebook (FB) message to the International Winter School on Gravity and Light: there was no missing video; it continues on Lecture 15 immediately) Tutorial 14: Relativistic Spacetime, Matter and Gravitation. Exercise 2: Lorentz force law. Question electromagnetic potential.
15. Lecture 15: Einstein gravity Recall that in Newtonian spacetime, we were able to reformulate the Poisson law ∆φ = 4πGN ρ in terms of the Newtonian spacetime curvature as
R00 = 4πGN ρ R00 with respect to ∇Newton GN = Newtonian gravitational constant 43
This prompted Einstein to postulate < 1915 that the relativistic field equations for the Lorentzian metric g of (relativistic) spacetime Rab = 8πGN Tab However, this equation suffers from a problem LHS: (∇a R)ab 6= 0 generically RHS: (∇a T )ab = 0 thought bubble: = formulated from an action Einstein tried to argue this problem away. Nevertheless, the equations cannot be upheld. 15.1. Hilbert. Hilbert was a specialist for variational principles. To find the appropriate left hand side of the gravitational field equations, Hibert suggested to start from an action Z √ −gRab g ab SHilbert [g] = M
thought bubble = “simplest action” aim: varying this w.r.t. metric gab will result in some tensor Gab = 0 15.2. Variation of SHilbert . √ √ √ [δ −gg ab Rab + −gδg ab Rab + −gg ab δRab ] | {z } | {z } | {z } M
Z
!
0 = |{z} δ SHilbert [g] = gi
1
2
3
mn
√ −(detg)g δgmn 1√ √ and 1 : δ −g = = −gg mn δgmn 2 −g 2 thought bubble δdet(g) = det(g)g mn δgmn e.g. from det(g) = exp trln g ad 2: g ab gbc = δca =⇒ (δg ab )gbc + g ab (δgbc ) = 0 =⇒ δg ab = −g am g bn δgmn ad 3: ∆Rab
= |{z}
δ∂b Γmam − δ∂m Γmab + ΓΓ − ΓΓ =
normal coords at point
= ∂b δΓmam − ∂m δΓmab = = ∇b (δΓ)mam − ∇m (δΓ)mab √ √ =⇒ −gg ab δRab = −g “if you formulate the variation properly, you’ll see the variation δ commute with ∂b ” EY : 20150408 I think one uses the integration at the bounds, integration by parts trick ei Γi(x) jk − Γ (x) jk are the components of a (1, 2)-tensor. Notation: (∇b A)ig =: Aij;b 44
√ =⇒ −gg ab δRab √ √ √ √ −g(g ab δΓmam );b − −g(g ab δΓmab );m = −gAb;b − −gB m,m = |{z}
∇g=0
Question: Why is the difference of coefficients a tensor? Answer: Γi(y) jk =
∂y i ∂xm ∂xq m Γ ∂xm ∂y j ∂y k (x)
,nq
+
∂y i ∂ 2 xm ∂xm ∂y j ∂y k
Collecting terms, one obtains Z
!
0 = δSHilbert =
[ M
√ √ √ 1√ −gg mn δgmn g ab Rab − −gg am g bn δgmn Rab + ( −gAa ) ,a − ( −gB b ) ,b ] 2 | {z } | {z } surface
Z =
√
−gδ
M
gmn |{z}
surface term
1 1 [ g mn R − Rmn ] =⇒ Gmn = Rmn − g mn R 2 2
arbitrary variation
Hence Hilbert, from this “mathematical” argument, concluded that one may take 1 Rab − gab R = 8πGN Tab 2 Einstein equations Z SE−H [g] =
√
−gR
M
15.3. 3. Solution of the ∇a T ab = 0 issue. One can show (→ Tutorials) that the Einstein curvature 1 Gab = Rab − gab R 2 satisfy the so-called contracted differential Bianchi identity (∇a G)ab = 0 15.4. Variants of the field equations. (a) a simple rewriting: 1 Rab − gab R = 8πGN Tab = Tab 2 GN =
1 8π
Contract on both sides g ab 1 Rab − gab R = Tab ||g ab 2 R − 2R = T := Tab g ab =⇒ R = −T 1 =⇒ Rab + gab T = Tab 2 1 ⇐⇒ Rab = (Tab − T gab ) =: Tbab 2 Rab = Tbab 45
(b) √
Z SE−H [g] :=
−g(R + 2Λ)
M
thought bubble: Λ cosmological constant History: 1915: Λ < 0 (Einstein) in order to get a non-expanding universe >1915: Λ = 0 Hubble today Λ > 0 to account for an accelerated expansion Λ 6= 0 can be interpreted as a contribution − 12 Λg to the energy-momentum “dark energy” Question: surface terms scalar? Answer: for a careful treatment of the surface terms which we discarded, see, e.g. E. Poisson, “A relativist’s toolkit” C.U.P. “excellent book” Question: What is a constant on a manifold? R√ R√ Answer: −gΛ = Λ −g1 [back to dark energy] [Weinberg, QCD, calculated] idea: 1 could arise as the vacuum energy of the standard model fields Λcalculated = 10120 × Λobs “worst prediction of physics” Tutorials: check that • Schwarzscheld metric (1916) • FRW metric • pp-wave metric • Reisner-Nordstrom =⇒ are solutions to Einstein’s equations in high school m¨ x + mω 2 x2 = 0 x(t) = cos (ωt) ET: [elementary tutorials] study motion of particles & observers in Schwarzscheld S.T. Satellite: Marcus C. Werner Gravitational lensing odd number of pictures Morse theory (EY:20150408 Morse Theory !!!) Domenico Giulini Hamiltonian form Canonical Formulations 46
Key to Quantum Gravity
Tutorial 13 Schwarzschild Spacetime EY : 20150408 I’m not sure which tutorial follows which lecture at this point. The tutorial video is excellent itself. Here, I want to encourage the use of CAS to do calculations. There are many out there. Again, I’m partial to the Sage Manifolds package for Sage Math which are both open-source and based on Python. I’ll use that here. Exercise 1. Geodesics in a Schwarzschild spacetime Question Write down the Lagrangian.
Load “Schwarzschild.sage” in Sage Math, which will always be available freely here https://github.com/ernestyalumni/ diffgeo-by-sagemnfd/blob/master/Schwarzschild.sage: sage: load("Schwarzschild.sage") 4-dimensional manifold ’M’ open subset ’U_sph’ of the 4-dimensional manifold ’M’ Levi-Civita connection ’nabla_g’ associated with the Lorentzian metric ’g’ on the 4-dimensional manifold ’M’
and so on. Look at the code and I had defined the Lagrangian to be L . To get out the coefficients of L of the components of the tangent vectors to the curve, i.e. t0 , r0 , θ0 , φ0 , denoted tp,rp,thp,php in my .sage file, do the following: sage: L.expr().coefficients(tp)[1][0].factor().full_simplify() (2*G_N*M_0 - r)/r sage: L.expr().coefficients(rp)[1][0].factor().full_simplify() -r/(2*G_N*M_0 - r) sage: L.expr().coefficients(php)[1][0].factor().full_simplify() r^2 sage: L.expr().coefficients(thp)[1][0].factor().full_simplify() r^2*sin(th)^2 Question There are 4 Euler-Lagrange equations for this Lagrangian. Derive the one with respect to the function t(λ)!.
sage: L.expr().diff(t) 0 This confirms that
∂L ∂t
=0
d ∂L For dλ ∂t0 , then one needs to consider this particular workaround for Sage Math (computer technicality). One takes derivatives with respect to declared variables (declared with var) and then substitute in functions that are dependent upon λ, and then take the derivative with respect to the parameter λ. This does that: sage: L.expr().diff( thp ).factor().subs( r == gamma1 ).subs( thp == gamma3.diff( tau ) ).subs( th == gamma3 ).diff(tau)\ ....: .factor() 2*(2*cos(gamma3(tau))*gamma1(tau)*D[0](gamma3)(tau)^2 + 2*sin(gamma3(tau))*D[0](gamma1)(tau)*D[0](gamma3)(tau) + gamma1(tau)*sin(gamma3(tau))*D[0, 0](gamma3)(tau))*gamma1(tau)*sin(gamma3(tau))
Question Show that the Lie derivative of g with respect to the vector fields Kt := 47
∂ . ∂t
The first line defines the vector field by accessing the frame defined on a chart with spherical coordinates and getting the time vector. The second line is the Lie derivative of g with respect to this vector field. sage: K_t = espher[0] sage: g.lie_der(K_t).display() # 0, as desired 0 EY : 20150410 My question is this: ∀ X ∈ Γ(T M ) i.e. X is a vector field on M , or, specifically, a section of the tangent bundle, then does LX g = 0 instantly mean that X is a symmetry for (M, g)? LX g is interpreted geometrically as how g changes along the flow generated by X, and if it equals 0, then g doesn’t change. 16. 17. 18. 19. 20. 21. 22. Lecture 22: Black Holes Only depends on Lectures 1-15, so does lecture on “Wednesday” Schwarzschild solution also vacuum solution (from tutorial EY : oh no, must do tutorial) Study the Schwarzschild as a vacuum solution of the Einstein equation: m = GN M where M is the “mass” 2m 1 g = 1− dt ⊗ dt − dr ⊗ dr − r2 (dθ ⊗ dθ + sin2 θdϕ ⊗ dϕ r 1 − 2m r
t
in the so-called Schwarzschild coordinates
(−∞, ∞)
r
θ
(0, ∞)
(0, π)
ϕ (0, 2π)
What staring at this metric for a while, two questions naturally pose themselves: (i) What exactly happens r = 2m?
t (−∞, ∞)
r (0, 2m) ∪ (2m, ∞)
θ (0, π)
ϕ (0, 2π)
(ii) Is there anything (in the real world) beyond t → −∞? t → +∞ idea: Map of Linz, blown up Insight into these two issues is afforded by stopping to stare. 48
Look at geodesic of g, instead. 22.1. Radial null geodesics. null - g(vγ , vγ ) = 0 Consider null geodesic in “Schd” "
Z S[γ] =
dλ
2m 1− r
# −1 2m 2 2 2 ˙2 2 t − 1− r˙ − r (θ + sin θϕ˙ ) r ˙2
with [. . . ] = 0 and one has, in particular, the t-eqn. of motion: 1−
2m r
. t˙ = 0
=⇒ 2m ˙ t = k = const. 1− r Consider radial null geodesics ! θ = const. ϕ = const. From 2 and 2 =⇒ r˙ 2 = k 2 ↔ r˙ = ±k =⇒ r(λ) = ±k · λ Hence, we may consider e t(r) := t(±kλ) Case A: ⊕ e de t k r t˙ = = = 2m dr r˙ r − 2m 1− r k =⇒ e t+ (r) = r + 2m ln |r − 2m| (outgoing null geodesics) Case b. ± (Circle around −, consider −):
e t− (r) = −r − 2m ln |r − 2m| (ingoing null geodesics) Picture 22.2. Eddington-Finkelstein. Brilliantly simple idea: change (on the domain of the Schwarzschild coordinates) to different coordinates, s.t. in those new coordinates, ingoing null geodesics appear as straight lines, of slope −1 This is achieved by
t¯(t, r, θ, ϕ) := t + 2m ln |r − 2m| Recall: ingoing null geodesic has e t(r) = −(r + 2m ln |r − 2m|) 49
(Schdcoords)
⇐⇒ t¯ − 2m ln |r − 2m| = −r − 2m ln |r − 2m| + const. ∴ t¯ = −r + const. (Picture) outgoing null geodesics
t¯ = r + 4m ln |r − 2m| + const. Consider the new chart (V, g) while (U, x) was the Schd chart.
U |{z}
[
{ horizon } = V
Schd
“chart image of the horizon” Now calculate the Schd metric g w.r.t. Eddington-Finkelstein coords.
t¯(t, r, θ, ϕ) = t + 2m ln |r − 2m| r¯(t, r, θ, ϕ) = r ¯ r, θ, ϕ) = θ θ(t, ϕ(t, ¯ r, θ, ϕ) = ϕ EY : 20150422 I would suggest that after seeing this, one would calculate the metric by your favorite CAS. I like the Sage Manifolds package for Sage Math. Schwarzschild BH.sage on github Schwarzschild BH.sage on Patreon Schwarzschild BH.sage on Google Drive
sage : load ( ‘ ‘ Schw a r z s c h i l d _ B H . sage ’ ’) 4 - dimensional manifold ’M ’ / Applications / Sage -6.6. app / Contents / Resources / sage / local / lib / python2 .7/ site - packages / sage / geometry / manifolds / utilities . py :283: See http :// trac . sagemath . org /11912 for details . expr = expr . sim p l i f y _ r a d i c a l () Levi - Civita connection ’ nabla_g ’ associated with the Lorentzian metric ’g ’ on the 4 - dimensional manifold ’M ’ Launched png viewer for Graphics object consisting of 4 graphics primitives
Then calculate the Schwarzschild metric g but in Eddington-Finkelstein coordinates. Keep in mind to calculate the set of coordinates that uses t¯, not e t: sage : gI . display () gI = (2* m - r )/ r dt * dt - r /(2* m - r ) dr * dr + r ^2 dth * dth + r ^2* sin ( th )^2 dph * dph sage : gI . display ( X_EF_I_null . frame ()) gI = (2* m - r )/ r dtbar * dtbar + 2* m / r dtbar * dr + 2* m / r dr * dtbar + (2* m + r )/ r dr * dr + r ^2 dth * dth + r ^2* sin ( th )^2 dph * dph
50
Part 2. Special Relativity Physics Stackexchange question (from Wesley) and answer (from David Bar Moshe had an excellent question and explanation for special relativity.
http://physics.stackexchange.com/questions/12221/what-does-a-frame-of-reference-mean-in-terms-of-manifold Recall that for charts (U, x) ∈ AM , of smooth atlas AM of smooth M . Now x : U ⊂ M → Rn (U 0 , x0 ) ∈ AM
x 0 : U 0 ⊂ M → Rn
If U ∩ U 0 6= ∅, by def. of smooth atlas AM , x0 ◦ x : Rn → Rn are diffeomorphisms. x ◦ x0 : Rn → Rn For notation, A, B ∈ Ω1 (M ) A ∧ B ∈ Ω2 (M ) A ∧ B = Ai dxi ∧ Bj dxj = Ai Bj dxi ∧ dxj A, B ∈ Tp M or T M A ∧ B ∈ ∧2 (T M ), A ∧ B = Ai ei ∧ B j ei = Ai B j ei ∧ ej (A × B)i = eijk Aj B k or A × B = Aj B k ijk ei if dimM = 3, ∗(A ∧ B) =
√ g il jm lmn dxn (3−2)! Ai Bj g g
=
√
gg il Ai g jm Bj lmn dxn
References [1] Eric Poisson, A Relativist’s Toolkit: The Mathematics of Black-Hole Mechanics, Cambridge University Press, 2004. ISBN 0 521 83091 5 EY’s references: [2] Klaus Jnich (Author), S. Levy (Translator), Topology (Undergraduate Texts in Mathematics), Springer; 1st ed. 1984. 2nd Corr. printing 1994 edition (January 1, 1995). ISBN-13: 978-0387908922 [3] John M. Lee. Introduction to Smooth Manifolds. 2010. Springer. ISBN-10: 0-387-95495-3 (hardcover), ISBN-13: 978-0387-95448-6
E-mail address:
[email protected] URL: http://ernestyalumni.wordpress.com ernestyalumni.tilt.com
51
