I’ve told you so far that the gravitational field<\/a> is encoded in a $$4 \\times 4$$ matrix known as the metric.\u00a0 Here it is, displayed as a nice table: $$! g_{ab} = \\left( \\begin{array}{cccc} g_{00} & g_{01} & g_{02} & g_{03}\\\\ g_{01} & g_{11} & g_{12} & g_{13} \\\\ g_{02} & g_{12} & g_{22} & g_{23} \\\\ g_{03} & g_{13} & g_{23} & g_{33} \\end{array} \\right)$$There’s 10 components because the matrix is symmetric when reflected diagonally.\u00a0 The 4 diagonal<\/em> components $$(g_{00}, g_{11}, g_{22}, g_{33})$$ tell you how to measure length-squared along the four coordinate axes.\u00a0 For example, the length along the $$1$$-axis is given by $$!\\Delta s = \\sqrt{g_{11}} \\Delta x^1,$$ where $$\\Delta x^1$$ is the coordinate difference in the $$1$$-direction.\u00a0 The remaining 6 off-diagonal<\/em> terms keep track of the spatial angle between the coordinate axes.\u00a0 If you know enough Trigonometry, you can figure out that the angle $$\\theta$$ between e.g. the $$1$$-axis and the $$2$$-axis is given by this formula: $$!\\cos(\\theta) = \\frac{g_{12}} {\\sqrt{g_{11} g_{22}}}$$<\/p>\n However, I’ve also said that the metric depends on the choice of coordinates<\/a>, which is arbitrary.\u00a0 We can use this freedom to choose a set of coordinates where the metric looks particularly simple at any given point.\u00a0\u00a0 We can start by choosing our four coordinate axes to be at right-angles to each other.\u00a0 This gets rid of all those funky <\/em>off-diagonal components of the metric, which involve two different directions: $$! g_{ab} = \\left( \\begin{array}{cccc} g_{00} & 0 & 0 & 0\\\\ 0 & g_{11} & 0 & 0 \\\\ 0 & 0 & g_{22} & 0 \\\\ 0 &0 & 0 & g_{33} \\end{array} \\right)$$If any of the four remaining numbers happen to be 0, we say that the metric is degenerate<\/em>.\u00a0 This would correspond to a weird geometry in which you can move in one of the directions for free without it affecting your total distance travelled.\u00a0 Since we all know that’s not the way the real world works, we’ll ignore this possibility.<\/p>\n We can also rescale the tick marks along any coordinate axis.\u00a0 This allows us to multiply each diagonal component of the metric by a positive <\/em>real number.\u00a0 So if say $$g_{22}$$ is positive, we can choose coordinates where it’s $$+1$$, and if it’s negative, we can choose coordinates where it’s $$-1$$.\u00a0 This gives us:$$! g_{ab} = \\left( \\begin{array}{cccc} \\pm 1 & 0 & 0 & 0\\\\ 0 & \\pm 1 & 0 & 0 \\\\ 0 & 0 & \\pm 1 & 0 \\\\ 0 &0 & 0 & \\pm 1 \\end{array} \\right)$$Since it also doesn’t matter what order we list the four coordinate directions, all that matters is the total number of $$+$$’s and $$-$$’s.\u00a0 This choice is called the signature <\/em>of the spacetime.<\/p>\n Now if you remember my very first post<\/a> on spacetime geometry, $$+$$ directions in the metric correspond to spatial dimensions<\/em>, while the funny $$-$$ sign is what makes for a time dimension<\/em>.\u00a0 But the real world has one time dimension, everywhere.\u00a0 No matter how far you travel, you’ll never find a place (so far as we know) where there isn’t any time direction, or where there are extra time dimensions.\u00a0 So that means that the correct signature for spacetime has $$(-, +, +, +)$$ along the diagonal, which is called Lorentzian<\/em> (a.k.a. Minkowskian<\/em>) signature.\u00a0 (If we had wanted to describe a timeless four-dimensional space, we would instead select the Riemannian <\/em>(a.k.a. Euclidean<\/em>) signature $$(+, +, +, +)$$.)\u00a0 We conclude that for any point of spacetime, you can always choose a set of coordinates such that the metric takes a special form that we’ll call $$\\eta_{ab}$$: $$! g_{ab} = \\left( \\begin{array}{cccc}-1 & 0 & 0 & 0\\\\ 0 & +1 & 0 & 0 \\\\ 0 & 0 & +1 & 0 \\\\ 0 &0 & 0 & +1 \\end{array} \\right) = \\eta_{ab}.$$In other words, if you zoom in on any point, you recover Special Relativity<\/a>.\u00a0\u00a0So after all this fidgeting around, we end up with a somewhat profound conclusion: <\/strong><\/em>i<\/strong>n General Relativity, every point of spacetime looks the same as every other<\/strong> point<\/strong>.<\/p>\n This is related to what Einstein called the Equivalence Principle<\/a>, which says that at short enough distances, the effects of acceleration are indistinguishable from being in a gravitational field.\u00a0 We all know from personal experience that riding in an elevator can make us weigh more or less, and from TV that astronomers in the Space Shuttle are weightless when they’re in free fall.\u00a0 In other words, you can always choose a coordinate system in which there is no gravitational force at any given point.<\/p>\n (Lewis Carroll actually described this principle several decades before Einstein in Sylvie and Bruno<\/a>, which includes a description of a tea party taking place in a freely-falling house.\u00a0 Then he describes what happens if the house is being pulled down with a rope faster than gravity would accelerate it, and explains how you could have a normal tea party as long as you have it upside-down.\u00a0 I like this book better than his more famous classics, but don’t read it unless you can withstand LD20<\/a> of Victorian sentimentality about fairy children.\u00a0 Also, Carroll didn’t go on to discover a revolutionary theory of gravity based on this principle.)<\/p>\n It might seem now like everything has become too<\/em> simple.\u00a0 If the metric looks the same at every single point, then why did we even bother with it?\u00a0 Where’s the information in the gravitational field?\u00a0 Well, it’s true that for any one <\/em>point, there’s a coordinate system where the metric looks just like $$\\eta_{ab}$$.\u00a0 But there’s no coordinate system for which the metric looks like\u00a0$$\\eta_{ab}$$ everywhere at once.\u00a0 <\/em>(Unless there’s no gravitational field anywhere, in which case Special Relativity<\/a> is true).\u00a0 If you make the metric look simple in one place, it has to look complicated somewhere else.<\/p>\n So in order to describe the gravitational field properly, we have to find a way to compare<\/em> the metric at different points.\u00a0 We can do this using something called\u00a0parallel transport<\/em>.\u00a0 I’ll give more details later, but basically it tells us how an object moves in a gravitational field when we carry it along a path through spacetime.\u00a0 When we carry the object around a tiny loop so that it returns to its original position, we might find that it comes back rotated compared to its original orientation.\u00a0 If so, we say that the spacetime contains\u00a0curvature<\/em>.\u00a0 If the spacetime contains curvature, this is a fact about the gravitational field which is invariant, i.e. objectively true.\u00a0 You can’t eliminate it just by changing your coordinates.<\/p>\n","protected":false},"excerpt":{"rendered":" I’ve told you so far that the gravitational field is encoded in a $$4 \\times 4$$ matrix known as the metric.\u00a0 Here it is, displayed as a nice table: $$! g_{ab} = \\left( \\begin{array}{cccc} g_{00} & g_{01} & g_{02} & … Continue reading