So far in my explanation of General Relativity, I’ve discussed the metric \(g_{ab}\), from which one can calculate the curvature tensor \(R^{a}_{bcd}\) by way of the connection \(\Gamma^{a}_{bc}\).

In practical astrophysical contexts:

• The metric is related to the gravitational potential at a point, i.e. how much “potential energy” a unit mass will have sitting in the gravitational field.  But I haven’t said anything about energy yet, so you’re entitled to ignore this remark…
• The connection (which involves a derivative of the metric) tells you the gravitational force at a point, i..e the amount by which freely-falling objects will accelerate in a given coordinate system.
• Finally, the curvature (which involves a derivative of the connection) tells you the tidal forces at a point, i.e. a difference in the force acting on a nearby object.  Yes, the ocean tides happen because the moon’s gravitational field has nonzero curvature at the Earth’s location.  That’s why it’s called that.

So far this is just the kinematics of general relativity—that is, what kind of entities are involved, and the basic outline of their behavior.  For example, If I wanted to tell you the kinematics for basic Newtonian mechanics (what you learn in high school physics), I’d say that A) there are a bunch of objects which have masses and positions (and orientations if you want things to get complicated…), B) the position of an object can change with time, but its mass is “conserved” and therefore doesn’t, and C) if you want to work out the “force” of an object, you can do so using \(F = ma\).

OK, so I’ve told you all about Newtonian Mechanics, and now you can go use it to solve problems, right?  No, of course not!  You can recite “the time-derivative of the position is the velocity, the time-derivative of the velocity is the acceleration, and the acceleration equals the force over the mass” over and over again, but it’s totally useless until I tell you what the forces actually are!  Without that, you can’t make any predictions at all about what the objects are doing.

Unless you count boring predictions like “the object will be somewhere”, you need to know something else.  This something else is called the dynamics, which means the rules for how things actually change with time.  (For example, if I told you that any two objects with mass \(m_1\) and \(m_2\) at a distance \(r\) are gravitationally attracted towards each other’s positions, with a force that is proportional to \(F = Gm_1m_2/r^2\), and if you know the initial positions and velocities, then you can work out their orbits!  At least, you can if you’re clever at math, like Newton was.)

So we need to write down an equation which says how things can change with time.  We call this the equations of motion.  Ever since Newton wrote down \(F = m{\ddot x}\) (each dot being a time derivative, so that his archnemesis Leibnitz would have written \(F = m (d^2x/dt^2)\) to say the same thing) we’ve realized that these equations typically involve taking two derivatives.  So we shouldn’t be surprised that the equation of motion for general relativity involves the curvature tensor \(R^{a}_{bcd}\), since it’s a double derivative of the metric, which is the basic field of General Relativity.

To write down the equations of motion, we need to massage the curvature tensor a little bit.  If you’ve forgotten the ground rules for tensors, click on the link.  We start with the the Riemann curvature tensor \(R^{a}_{bcd}\).  Since each of the letters is a spacetime vector index with four possible values, it looks like this has \(4 \times 4 \times 4 \times 4 = 256\) components.  Fortunately there are a lot of symmetries and constraints, so there’s actually only 20 independent components per spacetime point.  We can define the Ricci tensor \(R_{ab}\) by contracting the top index with the middle index on the bottom, like so:$$R_{ab} = R^c_{acb};$$Recall that the Einstein summation convention says that if you ever see the same letter as both a subscript and as a superscript, you’ve got to add up all of the four possible ways for them to be the same (i.e. both 0, both 1, both 2, or both 3).  Since the Ricci tensor is symmetric (\(R_{ab} = R_{ba}\)), it only represents 10 out of the 20 curvature components.  If this is not enough simplification for you, we can go further by contracting again using the inverse metric:$$R = R_{ab} g^{ab}.$$\(R\) is called the Ricci scalar, because it has just one component.

Whew!  Without further ado, here’s the equation of motion for General Relativity, called “the Einstein equation” after you know who: $$R_{ab} – \tfrac{1}{2} g_{ab} R = 8\pi G T_{ab}.$$Compact, beautiful, and probably completely incomprehensible since I haven’t explained all of the symbols yet!

The 8 and the \(\pi\) are the same numbers which you learned about in school.  \(G\) is Newton’s constant, which I sneakily introduced earlier in this post.  Note that the \(8\pi\) isn’t really just there for backwards compatibility with Newton’s force law.  If Einstein’s equation had been discovered first, we would have left out the \(8\pi\) from it, and then we would have written the force law as \(F = G m_1 m_2/ 8\pi r^2\).  But as it is, Newton got his \(G\) before Einstein did, so we’re stuck with it.

But the really important symbol here is \(T_{ab}\).  This is the energy-momentum tensor, or (because why should anything have only one name!) the stress-energy tensor.  It’s a \(4 \times 4\) symmetric matrix which tells you how the energy and momentum of matter (stuff) are flowing through a given point.  Now if you are a true Israelite in whom there is no guile, you should be asking: “What on earth (or in the heavens) are energy and momentum!  You haven’t explained that yet!”  No I haven’t.  For now, let’s just say it’s a property of matter, but we will get to it in a later post.

The combination of curvatures \(R_{ab} – \tfrac{1}{2} g_{ab} R\) which appears on the left-hand-side is also known as the Einstein tensor.  It has the same 10 components as the Ricci tensor \(R_{ab}\); they’re just repackaged a bit differently.  So the Einstein equation is actually 10 equations.

So, if you know what the matter is doing, you can figure out something about the geometry of matter.  At least, you can figure out the 10 of the components of the curvature which correspond to the Ricci tensor \(R_{ab}\).  Since the full Riemann tensor \(R_{abcd}\) has 20 components, there are 10 components left which are undetermined.  The remaining 10 components are called the Weyl tensor, and can be nonzero even in regions in which there is no matter.  That’s why there can be tidal forces outside of the surface of the sun or moon, even though there isn’t any solar or lunar matter there.  It’s the Weyl tensor which does that.  Also, as I wrote in Geometry is a Field:

There can also be distortions of the spacetime geometry which exist independently of matter.  These gravity waves are to gravity what light is to electromagnetism, ripples in the field which travel through empty space, and can be emitted and absorbed.  The propagation of these waves is also determined by the Einstein equation.  Since gravity comes from massive objects, gravity waves are emitted when extremely large masses oscillate, for example when two neutron stars orbit each other.  We know gravity waves are there, but we haven’t detected them directly.  However, we hope to detect them soon with the LIGO experiment.

It’s also the Weyl tensor which allows for gravity waves.

Clever readers may notice that I never wrote down what the Weyl tensor actually is.  There’s a clever formula where you start with \(R^a_{bcd}\), and then cleverly suck out all of the information about \(R_{ab}\), and end up with the Weyl tensor \(C^a_{bcd}\).  But it’s a bit complicated, so don’t ask.  The important thing is even when all of the components of \(R_{ab}\) are zero, \(R^a_{bcd}\) doesn’t have to be zero.

When we say that the Einstein equation is the “equation of motion” for General Relativity, we mean that you can use it to work out how the metric changes with time.   So, if you know the metric everywhere at some “time” which we will call $t = 0$ (think of this as being like the position of the gravitational field), and if you also know its first derivative \(\dot{g}_{ab}\) (think of this as being like the velocity), and if you know what the matter is doing, then the Einstein equation (which is like the force law) lets you work out the second derivative \(\ddot{g}_{ab}\).  By continuing to apply the Einstein equation, you can work out the value of the metric for all time!

Well, not quite.  Remember that coordinates don’t matter!  This means that we can’t actually hope to totally determine the metric, since if we start with a metric which obeys the Einstein equation, and distort it by changing the coordinate system, we get an equally good solution to Einstein’s equation.  So what we should really say, is that if you know the metric and its first derivative at \(t=0\) (and you know how matter behaves so you can figure out \(T_{ab}\)), then you can determine the fields at \(t > 0\) or \(t < 0\) up to coordinate transformations.

So we can actually only need to figure out \(\ddot{g}_{ab}\) up to coordinate transformations.  There are 10 components of  \(g_{ab}\), but there are also 4 spacetime coordinates \((t,\,x,\,y,\,z)\) whose values can be freely determined.  As a result, we actually only need to use 10 – 4 = 6 of the Einstein equations in order to figure out how the metric changes with time.

The remaining 4 equations are called constraints, because they don’t involve second derivatives of the metric.  Instead, they restrict which values of \((g_{ab}(x,y,z),\,\dot{g}_{ab}(x,y,z))\) you are allowed to start with.  These constraints are one of the most subtle features of General Relativity, because they ensure that the total energy and momentum of an object (like the sun) are encoded in the gravitational field coming out from it.  However, since I haven’t yet explained what energy and momentum are, I should probably say something about that first, before going into this.

Suppose we have a field \(\Phi\) in a curved spacetime, and we want to know how fast it is changing as you move in some direction in space or time.  Because there is more than one possible direction to move in, we have to select a vector \(\delta x^a\) which tells us which direction in the coordinate space \(x^a\) to move in (remember, \(x^a\) stands for a list of all 4 spacetime coordinates.)  Then we can calculate it by taking a partial derivative.  If your calculus is rusty, the partial derivative \(\partial_a\) is defined by: $$ \delta x^a\, \partial_a \Phi = \lim_{\epsilon \to 0} \frac{\Phi(x^a + \epsilon\,\delta x^a) – \Phi(x^a)}{\epsilon}.$$In other words, we compare the value of \(\Phi\) at two different points (\(x^a\) and \(x^a + \epsilon\,\delta x^a\)).  As \(\epsilon\) gets smaller, these two points get closer and closer together, so the values of \(\Phi\) typically get more and more similar, but because we divide by \(\epsilon\) we end up with a nonzero answer in the limit.  I’ve written \(\partial_a\) instead of \((\partial / \partial x^a)\) because I’m lazy.

That was the formula for the partial derivative in a particular direction \(\delta x^a\) (which is itself a list of 4 numbers).  If we want to have a list of all 4 possible partial derivatives at each point, we can just write \(\partial_a \Phi\) without the \(\delta x^a\).  This is the partial derivative covector, where a covector is a thing which eats a vector (like \(v^a\)) and spits out a number.  That’s almost the same thing as a vector, but not quite, which is why its index is downstairs instead of upstairs.  (You can convert between covectors and vectors by using the metric, e.g. \(\partial_b \Phi = g_{ab} \partial^a \Phi\), where as usual we sum over all 4 possible values of the index.)

Now, \(\Phi\) was a scalar field, meaning that it didn’t have any indices attached to it.  What if we tried to do the same trick with some vector field \(v^a\) (or a covector \(v_a\))?  Well, nothing stops us from taking the partial derivative of a vector in the exact way: $$ \delta x^a\, \partial_a v^b = \lim_{\epsilon \to 0} \frac{v^b(x^a + \epsilon\,\delta x^a) – v^b(x^a)}{\epsilon}.$$Unfortunately, this turns out to be a stupid thing to do.  The problem is that (before we take the limit) it involves comparing two vectors at different points.  But in a curved spacetime, it doesn’t make sense to talk about the same direction at different points, because coordinates are arbitrary.  There’s no particular sense in comparing the “t” component of a vector at a point \(x_1\) with the “t” component of a vector at another point \(x_2\), because the definition of “t” is arbitrary.  If you change the coordinate system at \(x_2\) but not \(x_1\) you’ll get confused.

In a curved spacetime, you can only compare vectors at different points if you select a specific path to go between the two points.  You can then drag (or if you prefer, parallel transport) the vector along this path, but if you choose a different path you might get a different answer.

Well here, because the points are really close, there’s an obvious path to pick.  Since spacetime looks flat when you zoom up really close, you can just parallel transport along the very short straight line connecting the two points.  This allows you to relate the coordinate system at the starting point \(x_1\) to the destination point \(x_2\).  Thus, when we take the derivative, we want to compare \(v^a(x_1)\) not to the same coordinate component of \(v^a(x_2)\), but to the parallel translated component of the vector.  When we do this, we get the covariant derivative, defined as follows: \(\nabla_a\):$$\nabla_a v^b = \partial_a v^b + \Gamma^{b}_{ac} v^c.$$Well, that’s not very useful until I tell you what capital gamma means.  It’s called the Christoffel symbol or the connection, and it tells us how to parallel transport vectors by an infinitesimal amount.  Basically if you take a vector pointing in the \(c\) direction and drag it a little bit in the \(a\) direction, then \(\Gamma^{b}_{ac}\) says how much your vector ends up shifting in the \(b\) direction, relative to your system of coordinates.  It turns out that the bottom two indices are symmetric: \(\Gamma^{b}_{ac} = \Gamma^{b}_{ca}\).

Similarly, if you want to define the covariant derivative of a covector, you just have to attach the indices a little bit differently:$$\nabla_a v_b = \partial_a v_b – \Gamma^{c}_{ab} v_c.$$The minus sign comes in because covectors are the opposite of vectors, so they need to do behave oppositely under a coordinate change.  Or, if you have a complicated tensor with multiple upstairs or downstairs indices, you have to have a separate correction term involving \(\Gamma\) for each of the indices.  How tedious!  But, in the case of a scalar field \(\Phi\), we get off scot free: the covariant and partial derivative are just the same.

If your spacetime is flat and you use Minkowski coordinates, then \(\Gamma = 0\).  But even in flat spacetime you can have \(\Gamma \ne 0\) if you use a weird coordinate system, like polar coordinates.

All of this is a little bit circular so far, since I haven’t actually told you how to calculate \(\Gamma^{b}_{ac}\) yet.  It’s just some thing with the right number of indices to do what it does.  In fact, you could choose to think of the connection \(\Gamma^{b}_{ac}\) as a fundamental field in its own right, in which case there would be no need to define it in terms of anything else.  But that is NOT what people normally do in general relativity.  Instead they define the connection in terms of the metric \(g_{ab}\), because it turns out there is a slick way to do it.

We want to find a way to use the metric to compare things at two different points.  In other words, the metric is a sort of standard measuring stick we want to use to see how other things change.  But obviously the metric cannot change relative to itself.  (If you define a yard as the length of a yardstick, then other things can change in size, but the stick will always be 1 yard by definition.)  Therefore, the covariant derivative of the metric itself is zero: \(\nabla_c g_{ab} = 0\).  But if we write out the correction terms we get: $$\nabla_c g_{ab} = \partial_c g_{ab} – \Gamma^{d}_{bc} g_{ad} – \Gamma^{d}_{ac} g_{bd} = 0.$$We can use this equation to solve for \(\Gamma\) in terms of the metric.  To do this, we just switch around the roles of the \(a\), \(b\), and \(c\) indices to get $$\partial_a g_{bc} – \Gamma^{d}_{ac} g_{bd} – \Gamma^{d}_{ab} g_{cd} = 0.$$and$$\partial_b g_{ac} – \Gamma^{d}_{ab} g_{cd} – \Gamma^{d}_{bc} g_{ad} = 0.$$By adding up two of these equations and subtracting the other, and dividing by two, one can prove that$$\Gamma^{d}_{ab} g_{dc} = \frac{1}{2}(\partial_a g_{bc} + \partial_b g_{ac} – \partial_c g_{ab}).$$We can then define \(\Gamma^{d}_{ab}\) directly as $$\Gamma^{d}_{ab} = \frac{1}{2} g^{cd}(\partial_a g_{bc} + \partial_b g_{ac} – \partial_c g_{ab}).$$To do that, we had to introduce something called the inverse metric \(g^{ab}\).  You get this by writing the metric \(g_{ab}\) out as a matrix and inverting it.  (Technically we write \(g_{ab} g^{bc} = \delta^c_a\) where \(\delta^c_a\) is a very boring tensor which is always 1 if \(a\) and \(c\) are the same index, and 0 if they are different.)

So then, the connection (which allows us to transport vectors from place to place) can be written in terms of the first derivative of the metric.  We’ll need to take a second derivative of the metric to get the curvature \(R^{a}_{bcd}\), but that will be the subject of another post.

One of the things I’ve been trying to do on this blog is to explain Einstein’s theories of relativity.  Here are my previous posts on this subject:

Time as the Fourth Dimension?
The Ten Symmetries of Spacetime
Fields
Geometry is a Field
Coordinates don’t matter
All points look the same

The first two have to do with Einstein’s first theory of Special Relativity, in which spacetime is taken to be fixed.  The third post describes what fields are.  The last four describe his second theory, General Relativity, in which the geometry of spacetime is itself a field: the gravitational field.

The gravitational field is also called the metric \(g_{ab}\), where \(a\) and \(b\) are indices selected from the four spacetime dimensions \((0, 1, 2, 3)\).  However, the exact form of the metric depends on your choice of coordinates, and it is always possible to pick coordinates so that the metric at any point looks just like the metric of Special Relativity.  For this reason, in order to discuss curved spacetimes (those that differ from the flat spacetime of Special Relativity, e.g. the gravitational field of the Sun or the Earth) we need to compare the metric at different points.

The basic idea is this: Suppose we are sitting at one particular point \(X\) of spacetime, and we imagine we have an infinitesimal arrow whose base sits at the point \(X\) and which points (ever so infinitesimally) away from \(X\) in some direction.  We call such a thing a vector and write it as \(v^a\), where \(a = (0, 1, 2, 3)\).  This is actually a set of four numbers: \(v^0\) would be a number saying how far \(v\) points in the 0-direction, \(v^1\) would be a number saying how far \(v\) points in the 1-direction, and so on.

Now, suppose we take this vector on a vecation tour through spacetime.  Perhaps it would be simpler just to think about space for a minute.  Imagine vectors lying on a 2-dimensional sphere.  (Warning: when an ordinary person uses the word sphere, they usually mean to include the interior, but when a math or physics person says sphere, they only mean the surface!)  Imagine for example, that the vector \(v^a\) lies on the surface of a Earth, and pretend that the vertical up-down direction doesn’t exist.  So on a random point on the Earth’s surface, \(v^a\) can point north, south, east, west, or in-between, but it can’t tilt up or down.

Imagine that \(v^a\) starts on the North Pole, pointing towards Hawaii.  Suppose we slide the vector down in the direction it is pointing, until it hits the Equator.  The vector now points South.  Now let’s drag the vector 1/4 of the way around the Equator, without rotating it.  The vector still points South.  Finally we drag it back to the North Pole.  It has now been rotated 90 degrees from its original position!  This is true even though on each step of the journey, we were careful not to rotate it as it travelled.  (Dragging vectors without rotating them is called parallel transport, by the way.)

This happens because the geometry of the surface of the Earth is intrinsically curved, i.e. the geometry of the sphere is not flat.  This must be distinguished from extrinsic curvature, which refers to something of lower dimension being bent within a higher-dimensional space.  The surface of the Earth is extrinsically curved in our ordinary 3-dimensional space, but that’s not what were talking about.  We’re talking here about the geometry of the sphere, quite apart from whether it is or is not embedded in some higher dimensional space.

The distinction is quite important in the case of our 4-dimensional spacetime, because (as far as we know) it is not embedded in any kind of higher-dimensional space.  When we say that spacetime is curved, we do not mean that spacetime is sitting in some kind of  “hyperspace” in which it is bent into a funny shape.  We mean that if you move vectors around in spacetime in a loop (i.e. a path that starts and ends at the same spacetime point \(X\)), it may come back rotated compared to its original position.  That is what curvature means.

I should note that it’s okay to drag these vectors either forwards or backwards in time, or along spacelike directions.  That’s because these are imaginary vectors serving as a visualization aid to probe the geometry of the spacetime.  They aren’t tangible physical objects which have to travel slower than light, and towards the future.

Now, ever since Archimedes, math people have liked to study things by breaking them up into tiny infinitesimal pieces.  So we want to think about what happens if you drag a vector around an infinitesimal loop.  To define this, we imagine we have a vector \(v^a\), which we drag around in a tiny parallelogram whose sides are given by vectors \(x^a\) and \(y^a\) pointing away from \(X\).  When we drag \(v^a\) along this parallelogram, if there is curvature it can come back rotated as an ever-so-slightly different vector \(w^a\).  To keep track of this we write: $$w^a = v^a + \epsilon^2 R^a_{bcd}\,v^b x^c y^d,$$ where \(R^a_{bcd}\) is a gadget with four indices known as the “Riemann curvature tensor”, which keeps track of the amount of curvature at any given point \(X\).  \(\epsilon\) is an infinitesimally tiny parameter which keeps track of how big the sides of the unit parallelogram are, for vectors of some unit length.

What’s a tensor?  The metric \(g_{ab}\), vectors such as \(v^a\), and \(R^a_{bcd}\) are all examples of tensors.  Tensors are similar to vectors, except that they are allowed to point in any number of directions.  A tensor is a kind of field which is allowed to depend on two things: 1) which point you are at in space and time, and 2) zero or more spacetime indices written as subscripts or superscripts \(a, b, c …\).  These indicate the total number of vectors it takes as inputs or outputs.  Note that the Riemann curvature has 3 of its indices downstairs and 1 upstairs.  That notation tells us that it eats 3 vectors as inputs and spits out 1 vector as an output.

In any equation involving tensors, each index letter is repeated, either 1) once in each term of the equation, always upstairs or always downstairs, or 2) twice in the same term of the equation, once upstairs and once downstairs.  In case (1) we interpret the tensor equation as being true for any possible choice of index, as long as it is the same for all terms, on both sides of the equals sign.  In case (2), we consider all 4 possible choices for the index and add them together (the Einstein summation convention).  These rules prevent us from doing nonsensical things like e.g. trying to add scalars and vectors together.

Tensors are not themselves coordinate-invariant, but when you change your system of coordinates, the value of the tensors changes in a particularly simple way.  This makes them useful when trying to describe physics in a coordinate-invariant way.  So long as you follow the rules in the previous paragraph, a tensor equation is a coordinate-invariant idea, i.e. if it is true in one coordinate system it is true in all of them.  That’s because if you change your coordinates, both the left-hand-side and the right-hand-side of the equation change in the same way, so it doesn’t matter.

The last thing I need to say here is that the Riemann curvature tensor \(R^a_{bcd}\) is not a new field additional to the metric tensor \(g_{ab}\).  If you know what the metric is, you can work out the Riemann tensor.  At any given point, \(R^a_{bcd}\) depends on the metric \(g_{ab}\), its first derivative \((\partial / \partial x^c) g_{ab}\) and its second derivative \((\partial / \partial x^d) (\partial / \partial x^c) g_{ab}\).  But the formula looks complicated, so I’ll spare you for the time being, until I can think of a simple way to justify it.

I’ve told you so far that the gravitational field is encoded in a \(4 \times 4\) matrix known as the metric.  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.  The 4 diagonal components \((g_{00}, g_{11}, g_{22}, g_{33})\) tell you how to measure length-squared along the four coordinate axes.  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.  The remaining 6 off-diagonal terms keep track of the spatial angle between the coordinate axes.  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: $$\cos(\theta) = \frac{g_{12}} {\sqrt{g_{11} g_{22}}}$$

However, I’ve also said that the metric depends on the choice of coordinates, which is arbitrary.  We can use this freedom to choose a set of coordinates where the metric looks particularly simple at any given point.   We can start by choosing our four coordinate axes to be at right-angles to each other.  This gets rid of all those funky 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.  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.  Since we all know that’s not the way the real world works, we’ll ignore this possibility.

We can also rescale the tick marks along any coordinate axis.  This allows us to multiply each diagonal component of the metric by a positive real number.  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 \(-1\).  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.  This choice is called the signature of the spacetime.

Now if you remember my very first post on spacetime geometry, \(+\) directions in the metric correspond to spatial dimensions, while the funny \(-\) sign is what makes for a time dimension.  But the real world has one time dimension, everywhere.  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.  So that means that the correct signature for spacetime has \((-, +, +, +)\) along the diagonal, which is called Lorentzian (a.k.a. Minkowskian) signature.  (If we had wanted to describe a timeless four-dimensional space, we would instead select the Riemannian (a.k.a. Euclidean) signature \((+, +, +, +)\).)  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.  So after all this fidgeting around, we end up with a somewhat profound conclusion: in General Relativity, every point of spacetime looks the same as every other point.

This is related to what Einstein called the Equivalence Principle, which says that at short enough distances, the effects of acceleration are indistinguishable from being in a gravitational field.  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.  In other words, you can always choose a coordinate system in which there is no gravitational force at any given point.

(Lewis Carroll actually described this principle several decades before Einstein in Sylvie and Bruno, which includes a description of a tea party taking place in a freely-falling house.  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.  I like this book better than his more famous classics, but don’t read it unless you can withstand LD20 of Victorian sentimentality about fairy children.  Also, Carroll didn’t go on to discover a revolutionary theory of gravity based on this principle.)

It might seem now like everything has become too simple.  If the metric looks the same at every single point, then why did we even bother with it?  Where’s the information in the gravitational field?  Well, it’s true that for any one point, there’s a coordinate system where the metric looks just like \(\eta_{ab}\).  But there’s no coordinate system for which the metric looks like \(\eta_{ab}\) everywhere at once.  (Unless there’s no gravitational field anywhere, in which case Special Relativity is true).  If you make the metric look simple in one place, it has to look complicated somewhere else.

So in order to describe the gravitational field properly, we have to find a way to compare the metric at different points.  We can do this using something called parallel transport.  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.  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.  If so, we say that the spacetime contains curvature.  If the spacetime contains curvature, this is a fact about the gravitational field which is invariant, i.e. objectively true.  You can’t eliminate it just by changing your coordinates.