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## General Relativity for Dummies

Well, the paper by Baez and Bunn is on-line and you can grab it in about three seconds. You are clearly not going to understand tensor calculus this week, so I really urge you to stop struggling with that until you are better prepared, and to try Baez and Bunn instead http://www.arxiv.org/abs/gr-qc/0103044 (see also http://math.ucr.edu/home/baez/gr/gr.html and maybe other suggestions in http://math.ucr.edu/home/baez/RelWWW/HTML/tutorial.html) After you read it once through, we can probably help you render stuff like $dV/V$ for AP high school students.

Hmmm... another idea might be to use high school trig and bit of elementary calculus to compute the tidal force in the Schwarzschild vacuum, which you can compare with the Newtonian tidal force. They agree, which ought to be surprising! Then you can talk a bit about how this can be understood. See http://en.wikipedia.org/w/index.php?...oldid=28781955
and see if you can figure out the derivation I have in mind. Don't try to understand the Newtonian tidal tensor, obviously--- look for the two arguments I gave for the components. One is an AP calculus level argument for why the radial tidal force is a tensile stress scaling like $2 m/r^3$. The other is a high school geometry argument (with appeal to the small angle approximation, which says $\sin(\alpha) \approx \alpha$ for small angles) for why the orthogonal tidal force is a compressive stress scaling like $-m/r^3$.

Another idea: you can explain in words and a bit of high school geometry type math how to interpret Carter-Penrose conformal diagrams, which are a convenient way to exhibit the global structure of the the full Schwarzschild vacuum solution (and other solutions). You can compare with the diagram for a black hole formed by gravitational collapse, and then you can discuss a thought experiment in which a hollow spherical shell collapses. The shell could be made of matter, e.g. dust particles, but it's more fun to consider a collapsing shell made of radially infalling EM radiation. You can sketch the event horizon in the conformal diagram and point out that this shows that you can be inside a black hole before you know it, i.e. in the case of a collapsing shell of EM radiation (the field energy of the EM field contributes to the curvature of spacetime, as per Einstein's field equation), the shell is approaching at the speed of light, so you don't know its comin g until it passes your location. After it passes, you find you are falling toward what a concentration of mass-energy (the shrinking shell of radiation). If you are very unlucky, you are in fact inside the horizon of a newly formed black hole. In the conformal diagram you can see that the EH has expanded past your location even BEFORE the shell arrives. The beautifully illustrates the global nature of horizons. This is very exciting and subtle idea which can however be explained using diagrams.

(For a pedantic citation, I offer a picture in the monograph by Frolov and Novikov cited here http://math.ucr.edu/home/baez/RelWWW....html#advanced)

Any talk on theory X should end with mention of some outstanding open problems concerning theory X. I'd suggest saying something like this:

 Quote by (no-one; this is a suggested comment for your talk) You probably know that finding a quantum theory of gravitation is a very hard open problem. But even in classical gtr many important problems are still unsolved. For example, the solution space of the Einstein field equation is not yet very well understood in many respects. The EFE is a kind of generalized wave equation, but it is nonlinear. One can obtain much insight into the solution space of linear wave equations by methods such as Green's functions, but these integral transform methods mostly break down for nonlinear equations. So for example, showing that all vacuum solutions which are "nearby" flat spacetime are in some sense "almost flat" was fabulously difficult, and still has hardly been generalized at all. We'd like to have a similar result describing nonlinear perturbations of the Kerr vacuum used to describe the exterior field of a rotating black hole, for example.
Again, noone is discouraging you from studying linear algebra, tensor calculus, exterior calculus, differential equations, the theory of manifolds, and other background for gtr. To the contrary, this is such wonderful stuff that it is well worth learning and I wish everyone were as eager as you are to do just that! What we are saying, though, is that something well worth doing is worth doing well, but you can't possibly master the math, much less the physics behind said math, in a few days, given where you are starting from.

Giving a talk is a wonderful experience and if all goes well can be very rewarding. But I think you owe it to your classmates to do the best job you can, which requires having clear and achievable goals for what points you want to get across. I don't know what those could be--- you can decide what you most want to convey after reading Baez and Bunn.

Don't forget to cite your sources
 Very well then, I'll look over this Baez & Bunn article and Baez's "general relativity tutorial" looks amusing. Do you recommend any place to start within your own list of tutorials?
 "Let me describe the Ricci scalar, R, in 2d. This is positive at a given point if the surface looks locally like a sphere or ellipsoid there, and negative if it looks like a hyperboloid - or "saddle". If the R is positive at a point, the angles of a small triangle there made out of geodesics add up to a bit more than 180 degrees. If R is negative, they add up to a bit less." - http://math.ucr.edu/home/baez/gr/oz1.html This is mind blowing.
 Recognitions: Science Advisor Staff Emeritus I'm afraid you're rather missing the point on the vectors. The vectors being talked about are what one would call "4-vectors" in special relativity, i.e for example $$(u^0, u^1, u^2, u^3 ) = (t,x,y,z)$$ The square-norm of a 4-vector, what I have been informally calling length^2, is the actually the square of the invariant Lorentz interval. GR is built on special relativity. The geometry of space time in GR s constructed from the geometry of the Lorentz interval. With a metric tensor of $$\left| \begin{array}{cccc} -c^2&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{array} \right|$$ the standard Minkowski tensor for a flat SR space-time, using a more-or-less standard sign convention, the square-norm of a contravariant vector with components (t,x,y,z) would be -c^2 t^2 + x^2 + y^2 + z^2 or in the usual notation. the square norm of $u^i$ would be $$g_{ij}x^i x^j$$ You can thus see that the square-norm of a unit timelike vector (1,0,0,0) is -c^2, while the square-norm of a unit spacelike vector, for example (0,1,0,0) or (0,0,1,0) or even (0,0,3/5,4/5) is +1. You'll find much more about 4-vectors in Space-time physics. You will notice a few small differences in approach, which will perhaps make things easier. For instance, you'll probably say "hello" to ict. Note that you can grab the first few chapters of an older edition of space-time physics online from Taylor's website. There is very little sense in struggling with tensors until you learn the basics of 4-vectors. It sounds like the SR treatment you had didn't cover them :-(.

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 Quote by Alabran Very well then, I'll look over this Baez & Bunn article and Baez's "general relativity tutorial" looks amusing. Do you recommend any place to start within your own list of tutorials?
Yes, the Baez & Bunn article

Recognitions:
 Quote by Alabran "Let me describe the Ricci scalar, R, in 2d. This is positive at a given point if the surface looks locally like a sphere or ellipsoid there, and negative if it looks like a hyperboloid - or "saddle". If the R is positive at a point, the angles of a small triangle there made out of geodesics add up to a bit more than 180 degrees. If R is negative, they add up to a bit less." - http://math.ucr.edu/home/baez/gr/oz1.html This is mind blowing.
Yup, and you can explain it with a good diagram too. That's what we are all trying to say-- look for cool stuff you can explain well with diagrams and perhaps a few lines of math your classmates are already comfortable with, like trig and a bit of differential calculus.

BTW, the fact about angles has nothing to do with gtr per se; this is one of the most basic results in the theory of surfaces, which was created by Gauss c. 1830 and later picked up by Riemann to create what we now call Riemannian geometry. Riemann himself speculated that curved manifolds could be useful for describing physics, but died before he could pursue this. This fact about angles is in fact integral for one definition of the Gaussian curvature, which in gtr appears as sectional curvatures (the components of the Riemann curvature tensor taken wrt a frame field).

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Staff Emeritus
 Quote by Alabran "Let me describe the Ricci scalar, R, in 2d. This is positive at a given point if the surface looks locally like a sphere or ellipsoid there, and negative if it looks like a hyperboloid - or "saddle". If the R is positive at a point, the angles of a small triangle there made out of geodesics add up to a bit more than 180 degrees. If R is negative, they add up to a bit less." - http://math.ucr.edu/home/baez/gr/oz1.html This is mind blowing.
And you can draw triangles on a sphere to illustrate this, using geodesics (great circles) for straight lines.

It's easy enough to draw a triangle with three right angles on a sphere. Try doing that on a plane :-).

If you have a saddle, I suppose you could draw triangles on a saddle-surface to illustrate the case where R is negative, but a sphere is easier to deal with.
 Recognitions: Science Advisor You can also make a cardboard model of geodesics on surface with negative and positive curvature concentrated at points, to show that it isn't neccessary that spacetime be curved where you are to have measureable effects "in the large". This actually edges toward discussing the Aharonov-Bohm effect http://en.wikipedia.org/wiki/Aharonov-Bohm_effect I am thinking of a model of certain "Sturmian tilings" (these are generalizations of Penrose tilings, due to de Bruijn) which can be obtained by looking from the right angle and an array of stacked cubes in E^3. There are three types of vertices: ones where three squares meet (angular deficit $\pi/2$; positive curvature concentration) and ones where six squares meet (angular deficit $-\pi$; i.e. an angular excess; negative curvature concentration). (I should stress that we are discussing two dimensional Riemannian manifolds here. Spacetime models are four-dimensional Lorentzian manifolds, which are significantly different both mathematically (noncompact symmetry group, geodesic paths maximize proper time under small endpoint-fixing perturbations, etc.) and conceptually (timelike geodesics can't "turn around in time", etc.) from Riemannian manifolds. Again, you should avoid trying to actually learn much about manifolds for purposes of preparing for your talk.) Alabran, one of the great results of Gauss in what we now call Riemannian geometry is the Gauss-Bonnet theorem, which states that in such models of a surface which is topologically spherical, the angular deficits must add up to $4 \pi$, the area of a unit sphere. Try this with a cube: it has eight corners, each having deficit $\pi/2$, since three squares meet at each corner. Now put a small cube on one face. That adds four new vertices with angular excess of $\pi/2$ and four new verticles with angular deficit of $\pi/2$, so the sum is unchanged. With a lot of work, you can add a square pyramid (say) to another face and verify that the sum is still unchanged. Want a proof of the Gauss-Bonnet theorem? Good for you! Well, one nifty approach is based upon Sperner's lemma, which is of independent interest. Ask after you've given your talk! I once had the pleasure of pointing out to Penrose that you can interpret a rhombic Penrose tiling as a tiling by squares thought of as an abstract manifold with curvature concentrated at the vertices. Naturally it has net curvature near zero over any large disk. Even better, you can declare that each square has metric tensor $ds^2 = -dt^2 + dx^2$! Then, the null geodesics are extremely intricate and interesting! Superficially they resemble something studied by Conway but they are not the same.
 Any chance you could switch topics to special relativity? You might have a prayer there.

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 Quote by Thrice Any chance you could switch topics to special relativity? You might have a prayer there.
Ditto the thought, but that was the first thing at least two posters suggested. He's already rejected the notion of changing topics, however.
 Thank you all for your help, this has been very illuminating. While we're all familiar with gaussian surfaces through it's use in finding electric and magnetic flux, I doubt any of us understood it at this level. I just returned from dinner in consideration on how to make a 270 degree triangle. The concept of drawing triangles on the surface of a sphere, treating it as a plane, as Pervect mentioned is revolutionary in my mind. I didn't really understand this "The Parable of The Apple" when I tried to get into the MTW Gravitation book, I think I do now. Never-the-less, I think I'll have to perform it myself. "You can also make a cardboard model of geodesics on surface with negative and positive curvature concentrated at points, to show that it isn't neccessary that spacetime be curved where you are to have measureable effects "in the large". This actually edges toward discussing the Aharonov-Bohm effect http://en.wikipedia.org/wiki/Aharonov-Bohm_effect" I'm familiar with this effect, it seems to be quite famous as far quantum mechanics go. I am thinking of a model of certain "Sturmian tilings" (these are generalizations of Penrose tilings, due to de Bruijn) which can be obtained by looking from the right angle and an array of stacked cubes in E^3. There are three types of vertices: ones where three squares meet (angular deficit ; positive curvature concentration) and ones where six squares meet (angular deficit ; i.e. an angular excess; negative curvature concentration). I'm a little curious about how to go about doing this. This is the "cardboard model" of geodesics you mentioned. By $$E^3$$ do you mean stacked cubes along the direction $$Ek$$? Or something else. If it's feasable to make such a model for class, I'd consider doing it, though I don't think I could construct one from this description. Also, (assuming a "Unit Sphere" to be a sphere of radius r=1) wouldn't the area of a Unit Sphere be $$(4/3) \pi$$ using the equation for the volume of a sphere: $$(4/3) \pi r^3$$ ? I don't quite understand why where 3 squares meet there is an "angular deficit" or what that implies. You mean there are $$\pi/2$$ less radians within a corner of a cube than within a quarter of a sphere? [Add] I just noticed you said "area" and you probably did not mean volume. Area of a sphere is, of course, 4pir^2.

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 Quote by Alabran While we're all familiar with gaussian surfaces through it's use in finding electric and magnetic flux, I doubt any of us understood it at this level.
Anyone with graduate training in math or physics knows the Gauss-Bonnet theorem, even good undergraduate students know it, but yes, this would surely be new to most high school students.

 Quote by Alabran I just returned from dinner in consideration on how to make a 270 degree triangle. The concept of drawing triangles on the surface of a sphere, treating it as a plane, as Pervect mentioned is revolutionary in my mind.
This was indeed revolutionary in 1830 or so, when the idea was independently invented by Gauss, Lobachevski, and Bolyai. But by 1870 there were so many noneuclidean geometries that it was becoming difficult to keep track of them all! After your talk, see recent This Weeks Finds by John Baez, in which he has been discussing Kleinian geometry, one of the best ways to organize homogeneous geometries like the hyperbolic plane, the sphere (or a "round" metric version of RP^2, a certain nonorientable boundaryless surface, the "quotient" of the sphere obtained by identifying antipodal points), plus the less structured notions of geometry such as affine or conformal geometry (the former naturally belongs on the plane, the latter on the toplogical sphere).

 Quote by Alabran I didn't really understand this "The Parable of The Apple" when I tried to get into the MTW Gravitation book, I think I do now. Never-the-less, I think I'll have to perform it myself.
After you've studied Taylor and Wheeler, since for spacetime you need the Lorentzian version whereas we are talking about the Riemannian version.

Quote by Alabran
 Quote by Chris Hillman "You can also make a cardboard model of geodesics on surface with negative and positive curvature concentrated at points, to show that it isn't neccessary that spacetime be curved where you are to have measureable effects "in the large". ... I am thinking of a model of certain "Sturmian tilings" (these are generalizations of Penrose tilings, due to de Bruijn) which can be obtained by looking from the right angle and an array of stacked cubes in E^3.
I'm a little curious about how to go about doing this. This is the "cardboard model" of geodesics you mentioned. By $$E^3$$ do you mean stacked cubes along the direction $$Ek$$? Or something else. If it's feasable to make such a model for class, I'd consider doing it, though I don't think I could construct one from this description.
Well, Quasitiler with D=3 is what you want, but that seems to be broken.

Well, for a surface made of square faces which is naturally embedded in five dimensional euclidean space (!) try http://www2.spsu.edu/math/tile/aperi...se/josleys.htm Can you see three rhombs meeting at a common vertex as three faces of a cube? This particular picture won't "pop into" E^3 but rather into E^5, but this surface made of squares in E^5 does remain close to a hyperplane. When you have given your talk we can talk about decompositions of rotations in E^n; the hyperplane is an eigenspace (linear algebra) of the rotation which cycles the five basis vectors and that is why Penrose tilings---but not most other Sturmian tilings from D=5--- appear so symmetrical! This example is more complicated, so we have vertices where 7,6,5 4, or 3 squares meet instead of just 3 or 6. Can you spot some?

 Quote by Alabran Also, (assuming a "Unit Sphere" to be a sphere of radius r=1)
Correct.

 Quote by Alabran wouldn't the area of a Unit Sphere be $$4}/3\pi$$ using the equation for the volume of a sphere: $$4/3(\pi)r^2$$ ?
No, but you can figure this out by differentiating $4/3 \pi r^3$ wrt r!

 Quote by Alabran I don't quite understand why where 3 squares meet there is an "angular deficit" or what that implies.
In the plane, four squares meet at a vertex. I said that in the surface I have in mind, sometimes three squares meet at a vertex (missing one right angle, so a deficit of one right ange) and sometimes six (two extra right angles, so an excess of two right angles). I couldn't find a three-dimensional picture on-line, so try to figure it out from the five-dimensional one.

BTW, a phenomenon discovered by Conway, "empires" is a lovely demonstration of how surprising global order can arise from local rules. This is the same "local to global" issue I keep mentioning. See http://www2.spsu.edu/math/tile/aperi...jackempire.gif for an empire in another variant of Penrose tilings (combinatorially equivalent but drawn with nonrhomboidal tiles). Whenever you see the blue tiles, you must also see the green tiles, in those precise positions relative to the blue tiles, but the other tiles are not determined by the blue tiles. A rather fascinating idea which I and others have bandied about is the notion that this suggests a geometric approach to mathematical logic. After all, we know that in mathematics combining lemmas can sometimes enforce a surprising conclusion.

An example much more important for science/statistics as a whole is the Szemeredi lemma. This ties together combinatorics, number theory, ergodic theory, Ramsey theory, hard analysis, and many wonderful things. It basically says that complete combinatorial disorder is impossible. That's far more profound than noneuclidean geometry, one could argue! http://www.arxiv.org/abs/math/0702396

Having mentioned number theory, I should point out that Sturmian tilings are in a sense merely a way of realizing geometrically certain phenomena in simultaneous rational approximation, a classical topic which goes back to Lagrange.

If you ever get a chance to give another talk, you should give a talk on random graphs. Because this is such a wonderful and amazing subject, yet so accessible, even at high school level.

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 Quote by Alabran I just made a 270 degree triangle on a basketball. Exciting!

 Quote by Chris Hillman Well, Quasitiler with D=3 is what you want, but that seems to be broken. Well, for a surface made of square faces which is naturally embedded in five dimensional euclidean space (!) try http://www2.spsu.edu/math/tile/aperi...se/josleys.htm Can you see three rhombs meeting at a common vertex as three faces of a cube? This particular picture won't "pop into" E^3 but rather into E^5, but this surface made of squares in E^5 does remain close to a hyperplane. When you have given your talk we can talk about decompositions of rotations in E^n; the hyperplane is an eigenspace (linear algebra) of the rotation which cycles the five basis vectors and that is why Penrose tilings---but not most other Sturmian tilings from D=5--- appear so symmetrical! This example is more complicated, so we have vertices where 7,6,5 4, or 3 squares meet instead of just 3 or 6. Can you spot some?
Oh, I understand this. As funny as this sounds, I noticed this in second grade, though I never thought of it as being a representation of extra-dimensional space, I was very interested in three-dimensional illustration at the time and, familiar with the foreshortening effects of perspective, noticed that the seperate tiles when placed together the right way created an illusion of depth. I still don't really see it "popping into" five dimensions as much as the three of the cube. Is it because of the five rotational lines of symmetry (i.e. a cube within a hypotetical tile creation with 7+ lines of symmetry would have as many dimensions for that reason.)

 Quote by Chris Hillman No, but you can figure this out by differentiating $4/3 \pi r^3$ wrt r!
I see my mistake now. I was thinking about volume...

After thinking about it, I just realized that the volume is the integral of the surface area... this is astounding. I thought to follow this into other dimensions, but it doesn't seem to work. $$\frac{d(4/3 (\pi) r^2)}{d(r)} = 4\pi r^2$$ and $$\frac{d((\pi) r^2)}{dr} = 2\pi r$$ though there doesn't seem to be a differential relation between a sphere and a circle.

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Right, a (p,q)-Sturmian tiling space $S(w)$ arises from a certain periodic tiling $O(W)$ in $E^{p+q}$, by taking a slice through the O(W) along a p-plane parallel to W in $E^{p+q}$. The result is a parallelotope tiling in $E^p$ which are in general aperiodic but almost periodic. For some choices of W it turns out that some directions are actually periodic, however. Furthermore, as the notation suggests there is a duality between (p,q) and (q,p) tilings which interchanges combinatorially interesting features. In particular, the classification of singular tilings which can occur in $S(W)$ is dual to the classification of the periodicities which appear in $S(W^\perp)$. The space of Penrose tiling is a (2,3)-Sturmian tiling space, and the dual tiling space is interesting.