# Theory of relativity/General relativity/Introduction

A great many attempts have been made to explain the basic concepts of general relativity to non-experts. These can range from museum exhibits that roll a ball around on a curved surface, to treatments that are mathematically quite daunting. This article will attempt to explain it at the level of undergraduate, or ambitious high-school, mathematics and physics.
The general relativity formulation of gravity states that gravity arises from the curvature of spacetime, and, in analogy with the classical notion that massive objects create a gravitational field attracting other objects, matter causes spacetime to curve. In the words of physicist John Wheeler, "Space tells matter how to move, matter tells space how to curve."

## Coordinate Systems and Spacetime Diagrams

Figure 1 shows a "spacetime diagram", with my house, my neighbour's house, and my neighbour walking from his house to mine.
    This is the same kind of diagram that is used in explanations of special relativity. The "spacetime" is sometimes called "Minkowski space". Spacetime is actually four-dimensional, but we can only show two dimensions, so we leave out y and z. The single x spatial coordinate is good enough for our purposes, so the diagram has x going from left to right, and t (time) going upward. For the purposes of this explanation, don't worry about the considerations of special relativity such as the speed of light, the Lorentz transform, or light cones. None of that is important just now.
The diagram shows the calibration, in space (that is, x) and time. These measurements are made with respect to my (stationary) frame of reference. My house is at x=0, and my neighbour's house is at x=5 (all units are arbitrary). My neighbour walks at 1 unit per minute.
The diagram shows some "events"—my house, now; my house, 5 minutes from now; and my neighbour's house now and 5 minutes from now. The green diagonal line depicts my neighbour walking from his house to mine, arriving 5 minutes from now. That line is called his world line. The red line going straight up in my house is my own world line (I'm sitting at home.)
A car is driving down the street, from left to right. Figure 2 shows the same four events and two world lines, but with different calibration—the car's own coordinate system. The car is driving at 1 unit per minute, but in the opposite direction from my neighbour's walking. The event of my neighbour's arrival at my house is now at x=-5. It's way behind the car, though the car was directly adjacent to my house at t=0. The car passes my neighbour (that is, my neighbour is observed to be at x=0 in this coordinate system) at t=2.5.
Because the car's frame of reference is in motion, the calibration lines in figure 2 are not perpendicular. The formerly vertical lines are now slanted. But there is something very important to notice about both this coordinate system and the previous one: They are both flat. The flatness comes from the fact that the calibration lines are straight and parallel. The boxes created by the lines are parallelograms. But note that the lines don't have to be perpendicular, and the boxes don't have to be rectangles. Straight parallel lines and parallelograms are all that is required.
These two flat coordinate systems have a very important physical property: Neither I, sitting at home, nor a passenger in the car, experience any "fictitious forces" (this concept is described below.) That is, people in the car don't feel any recoil from acceleration, or centrifugal force, or Coriolis force. These frames of reference are said to be inertial. This leads to an important principle of geometrical physics:
   * Inertial frames of reference have flat coordinate systems. Flat coordinate systems lead to an absence of fictitious forces.
Now consider figure 3. The coordinate system is once again that of the car, but the car is accelerating, starting at a standstill in front of my house at t=0. Its world line is curved. Once again, it crosses paths with my neighbour. This case is very different from the other two. The calibration lines are curved, and the boxes that they create are not parallelograms. This coordinate system is curved. Another thing to notice is that people in the car will feel a fictitious force—a "recoil" force agains the back of the seat. This frame of reference is not inertial.
   * Accelerating frames of reference have curved coordinate systems. Curved coordinate systems lead to fictitious forces.
The notion of flat and curved coordinate systems is actually quite familiar from analytic geometry. Figures 4 and 5 show two common coordinate systems (or "ways of drawing the calibration lines on graph paper".) Figure 4 shows Cartesian coordinates, which are flat, and figure 5 shows polar co-ordinates, which are curved.
* More Advanced Mathematical Treatment
* While we can often visualize curved and flat coordinate systems in terms of pictures, the mathematically correct way is with the "metric", or "metric tensor". This is a formula for measuring the distance between two points in space. (Actually, it only measures the distance between two points that are infinitesimally close to each other.) The measurement is in terms of the two points' coordinates—different coordinate systems mean different coordinates for any given point, so the distance formula will be different.
* In ordinary Cartesian coordinates, the Pythagorean theorem applies, so the distance is given by:
* $s^{2}=(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}\,$
* Or, since we are only interested in the distance between points that are infinitesimally close, we write it in terms of "differentials":
* $ds^{2}=dx^{2}+dy^{2}\,$
* For polar coordinates, the formula is:
* $ds^{2}=dr^{2}+r^{2}d\theta ^{2}\,$
* These expressions are sometimes known as "quadratic forms" or as "symmetric second-rank covariant tensors".
* These metrics are often written as 2x2 (or whatever the dimensionality is) matrices. For Cartesian coordinates the matrix is ${\begin{bmatrix}1&0\\0&1\\\end{bmatrix}}$
* For polar coordinates it is ${\begin{bmatrix}1&0\\0&r^{2}\\\end{bmatrix}}$
* For the spacetime coordinate system of Figure 1, it could be ${\begin{bmatrix}1&0\\0&1\\\end{bmatrix}}$, depending on our choice of physical calibration factors, including the scale factor between space and time. In true special relativity we would choose these very carefully, using the Lorentz metric, and with the speed of light as the scale factor. But we promised not to get into that yet. The important point is that all the coefficients are constants.
* For the "oblique" coordinate system of Figure 2, it is ${\begin{bmatrix}1&1\\1&2\\\end{bmatrix}}$, or
* $ds^{2}=dx^{2}+2\ dx\ dt+2\ dt^{2}\,$
* Once again, all the coefficients are constants.
* For the coordinate system of the accelerating car, in Figure 3, it is ${\begin{bmatrix}1&t/2\\t/2&1+t^{2}/4\\\end{bmatrix}}$, or
* $ds^{2}=dx^{2}+t\ dx\ dt+(1+t^{2}/4)\ dt^{2}\,$
* The noteworthy fact here is that the coefficients depend on time. They are not constants.
* The general principle is that:
* A coordinate system is flat if its metric coefficients are constant.

## Geodesics and Fictitious Forces

A geodesic is a line that is "objectively straight". What we mean by that is that a line's straightness transcends different coordinate systems. Different coordinate systems may have different equations for a line, but whether a line is a geodesic is independent of that.
We have drawn the figures above (and those below) so that geodesics are straight lines on your computer screen. The various colored world lines of Figures 1, 2, and 3 are all geodesics except for the blue line of Figure 3, for the accelerating car.
In a flat coordinate system, the equation of a geodesic is a linear equation. For example, the equation of the green line in figure 2 is $x=5-t$.
The equation of the blue line in Figure 3 is $x=0$, but that doesn't make it a geodesic, because the coordinate system is not flat.
In a curved coordinate system, the formula for a geodesic is much more complicated. It will be explained below that, in the presence of gravitational fields, the space itself is curved. This means that all coordinate systems are curved, which makes the calculation of geodesics very complicated. This is one of the reasons that general relativity is so difficult.
Everyone except the passengers in the accelerating car (blue line, Figure 3), are undergoing inertial motion, and feel no fictitious forces:
       Observers following geodesics in spacetime are undergoing inertial motion, and feel no fictitious forces.
An observer following a geodesic is said to be in "free fall". Planets are in free fall around the Sun. People falling down elevator shafts are also in free fall. (We consider the various observers in Figures 1 and 2 to be in free fall, because we are ignoring gravity for now. They feel no fictitious forces other than gravity.)
How about observers whose world lines are not geodesics? They are being accelerated by some external force, which is what makes their world line deviate from straightness. If they are "at rest", or moving uniformly, in their own coordinate system, the force making their world line curve may not seem real, but their recoil against that force will seem real. That recoil is a fictitious force.
For example, passengers in the car of Figure 3 feel a recoil force against the back of the car seat. People in various rotating amusement park rides feel a "centrifugal force", another fictitious force, seeming to push them outward. An observer standing on the ground can see that the centrifugal force is fictitious—what is really happening is that the ride mechanism is pushing them into circular motion. There is another fictitious force associated with rotation: the Coriolis force.

## Gravity as a Fictitious Force

Finally, we come to gravity itself.
       Gravity is a fictitious force, just like all the others. It arises from particles not following geodesics.
Figure 6 shows a spacetime diagram of the coordinate system of a person falling down an elevator shaft. His world line is shown in red. The left-to-right axis, which was the "x" axis in the earlier figures, is now the "z" axis, that is, height. The coordinate system travels down the shaft with him, and he thinks he is at rest in it. He does not feel any fictitious forces, including gravity. The coordinate system is flat and inertial.
Figure 7 shows the same person in the coordinate system of "stationary" people standing on the top floor of the building. The world line of the falling person is still red, and that of the stationary people is blue. The stationary coordinate system is curved; the stationary people are not following geodesics. The floor is pushing up against them, making them deviate from geodesics. Their recoil against this is the fictitious force that they call "gravity". The curved coordinate lines are the lines of "stationary" constant height. Think of them as being like the floors of the building. Note that the unfortunate victim is passing downward through them.
The aspect of gravity that indicates that it is a fictitious force is that it shares an important property with the standard fictitious forces: it is exactly proportional to the object's mass. For the standard fictitious forces, it is easy to see why this is so—they are actually the result of accelerations. By Newton's law $F=ma$, the force is proportional to the mass for a given acceleration.
This connection is the basis for Einstein's equivalence principle—one can't tell the difference between gravity and a traditional fictitious force. The thought experiment making this plausible was the "elevator experiment". A person in a closed elevator can't tell whether it is stationary in a building on Earth, or is out in space being dragged, at 9.8 meters per second squared, by a cable attached to a rocket.
For ordinary fictitious forces, we can readily explain things by transforming away from the curved coordinate system to some flat coordinate system. For example, we can "transform away" the centrifugal force of a rotating amusement park ride by viewing the action from a stationary observer on the ground.
For gravity, the equivalent transformation is to a coordinate system that is in free fall, such as that of the person falling down an elevator shaft. This leads to the following not-very-satisfactory explanation of the Earth's gravity: The gravity observed on the Earth's surface arises because the Earth is flat, and there are rocket engines on the underside, accelerating everything upward at 9.8 meters per second squared. This would explain everything, at least locally, but it has one fatal flaw: The Earth is round, with gravity going inward everywhere. So the surface of the Earth would have to be accelerating outward in all directions. This would require that the Earth continually get bigger, which it does not do.

## Curvature of Spacetime

The problem with the not-very-satisfactory explanation of gravity given above was that, while we can transform the curved coordinate system of a person standing on the ground to the coordinate system of a person falling down an elevator shaft locally, we can't do it everywhere. People falling down elevator shafts on opposite sides of the Earth will not have consistent coordinate systems. The problem is that spacetime itself is curved.
Up until now, we have been dealing with curved and flat coordinate systems on spaces (the technical term is "manifold") that are themselves flat. We put coordinate systems on them as though we were drawing lines on a flat piece of graph paper. Now we have to deal with curved manifolds.
The curvature of a 2-dimensional manifold embedded in 3-dimensional space can be visualized by looking at it. It's just what you think it means. A typical curved 2-dimensional manifold is the surface of a sphere. We can place a coordinate system on it that is nearly flat over a tiny region. For example, ordinary latitude/longitude coordinates will work near the equator. But this coordinate system doesn't work everywhere. As one moves away from the equator, it measures distances incorrectly, and at the North and South poles it turns into polar coordinates, which are clearly not flat. (One could use a different coordinate system that looks flat at the North pole, but it wouldn't be flat at the equator.) This coordinate system that is flat at the equator is analogous to the coordinate system of the person falling down the elevator shaft—it can't be extended around the world.
Whether a manifold is curved or not is determined by the Riemann tensor, also called the curvature tensor. This is a mathematical object that has an "objective" or "intrinsic" existence, independent of coordinate system. (Tensors are defined to have this property. Doing this correctly is what makes tensor calculus so difficult.) All of the spaces we have discussed so far, even those with curved coordinate systems, are flat. In fact, they are 2-dimensional Euclidean spaces.
Riemann's tensor on the 2-dimensional surface of a sphere is nonzero.
* More Advanced Mathematical Treatment
* Riemann's tensor can be calculated, by a some long formulas that we won't go into here, from the various partial derivatives of the metric coefficients. If those coefficients are constant (the coordinate system is flat), all of the derivatives are zero, and Riemann's tensor is zero. Conversely, if the manifold is curved (Riemann's tensor is nonzero), there are no flat coordinate systems.
* For the curved coordinate system of Figure 3, the derivatives are not zero, but they all cancel in the calculation of Riemann's tensor.
* For the surface of a sphere, Riemann's tensor can be shown to be nonzero, so there are no flat coordinate systems.
* The number of nontrivial components of Riemann's tensor is
* ${\frac {N^{2}\ (N^{2}-1)}{12}}$
* For a 2-dimensional surface, this is 1, so there is just one number determining the curvature. This is called the Gaussian curvature, usually denoted by K. For 4-dimensional spacetime, there are 20 components.
For the case of a 2-dimensional surface, the curvature is determined by a single number K, the Gaussian curvature. Figure 8 shows some surfaces with negative, zero, and positive Gaussian curvature.
The figure on the left, with negative Gaussian curvature, is often referred to as being (locally) "saddle-shaped". The figure in the middle has zero curvature. Only "intrinsic" curvature counts; rolling up a sheet of paper does not.
The fact that we can't make a flat coordinate system globally around a non-microscopic region of the Earth's surface tells us that in the vicinity of the Earth, spacetime itself is curved.
It is important to note that the curvature of 4-dimensional spacetime is vastly more complicated than the Gaussian curvature of a surface. It has 20 components instead of 1. Attempts to visualize it as a (perhaps saddle-shaped) surface are not correct; they merely suggest the meaning of curvature. The actual analysis of the curvature, and how it gives rise to gravity, requires careful analysis of tensor calculus.
    This point needs to be emphasized: Visualizations of gravitational curvature in terms of an ordinary curved surface, with a ball rolling across it and being deflected, such as one sees in science museum exhibits, do not do justice to the actual curvature of general relativity. They simulate a classical gravity well, such as exists in the Newtonian formulation of gravity.
The curvature of spacetime is manifested in the metric tensor, which has 10 non-trivial parameters, from which one derives the curvature itself in Riemann's tensor (20 parameters), from which one further derives Ricci's tensor and Einstein's tensor (10 parameters each.)

## The Schwarzschild Solution

Under general relativity, the metric tensor describing the curved spacetime in the vicinity of the Earth (or the Sun, or any spherically symmetric gravitating body) is the celebrated Schwarzschild solution. Exact closed-form solutions to the gravity equations are notoriously difficult to obtain, and the Schwarzschild solution is one of very few such solutions. (The Kerr solution, which takes the body's rotation into account, is another.) In the Schwarzschild solution, the strength of an object's gravity is determined by a parameter, with the dimensions of length, called the Schwarzschild radius. When the equations are solved, the instantaneous downward acceleration of a particle initially at rest (for example, the instant someone jumps into the elevator shaft) is
       ${\frac {d^{2}r}{dt^{2}}}=-{\frac {c^{2}\,a\ (r-a)}{2\,r^{3}}}$
where $a$ is the Schwarzschild radius. Classically, the acceleration would be
       ${\frac {d^{2}r}{dt^{2}}}=-{\frac {GM}{r^{2}}}$
where $G$ is Newton's constant of gravitation, and $M$ is the mass of the gravitating body.
Setting these equal, under the assumption that $a\,$ is very small in comparison to $r\,$, we get
       $a={\frac {2\,GM}{c^{2}}}$
For the Earth, $a$ is about 1 centimeter. For the Sun it is about 3 kilometers.
But, as will be seen below, this equation only applies in the absence of gravitating matter, that is, outside of the Earth. The Earth would only have a Schwarzschild radius if its mass were all concentrated inside a sphere of 1 centimeter. If that were the case, it would be a black hole, with an event horizon at the Schwarzschild radius.
* More Advanced Mathematical Treatment
* The metric tensor, under plain special relativity with no curvature, in Cartesian coordinates, is:
* ${\begin{bmatrix}-c^{2}&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}}$
* In polar coordinates, it is:
* ${\begin{bmatrix}-c^{2}&0&0&0\\0&1&0&0\\0&0&r^{2}&0\\0&0&0&r^{2}\ \sin ^{2}\,\theta \end{bmatrix}}$
* The curved Schwarzschild metric tensor, in polar coordinates, is:
* ${\begin{bmatrix}-{\frac {c^{2}\ (r-a)}{r}}&0&0&0\\0&{\frac {r}{(r-a)}}&0&0\\0&0&r^{2}&0\\0&0&0&r^{2}\ \sin ^{2}\,\theta \end{bmatrix}}$
* Since the Schwarzschild solution is curved, it has Riemann's tensor not equal to zero. But Einstein's tensor is zero:
* $G_{\mu \nu }=0\,$
* The gravitational field, being very nearly an inverse square field under Newtonian mechanics, is quite similar to the classical electrostatic field. That field is described by Maxwell's equations, specifically the equation $\nabla \cdot E=\rho /\epsilon$,
* where $\rho \,$ is the density of electric charge. In the absence of charge, we have $\nabla \cdot E=0$, giving an inverse-square electric field.
* The quantity $\nabla \cdot E$ in electrostatics is roughly analogous to $G_{\mu \nu }\,$ in general relativity. Just as the inverse-square electric field is the solution to $\nabla \cdot E=0$, the (very nearly) inverse square Schwarzschild gravitational field is the solution to $G_{\mu \nu }=0\,$.
The Schwarzschild solution is just one simple case (gravity in the vacuum surrounding a spherically symmetric massive object) of general relativity, but it provides many of the observed phenomena:
* bending of light (both around the Sun during an eclipse and around distant galaxies with gravitational lensing)
* the anomalous precession of the perihelion of Mercury
* gravitational redshift
* gravitational time dilation
* gravitational delay (Shapiro effect) of spacecraft signals past planets

## Nonzero Gravitating Mass

We have seen how the curvature of spacetime tells matter how to move. Now we will examine how matter tells space how to curve.
The "gravitational field", which we might denote by a vector $V\,$, arises from the curvature of spacetime. Two tensor fields—the Ricci tensor $R_{\mu \nu }\,$ and the closely related Einstein tensor $G_{\mu \nu }\,$ govern this. The divergence of $V\,$ is just the 00 component of Ricci's tensor:
       $\nabla \cdot V=R_{00}$
Just as the statement that an electric field is spherically symmetric and has zero divergence leads to an inverse square law, the statement that a coordinate system is spherically symmetric and has $R_{\mu \nu }=0\,$ leads to an inverse square law (in the classical approximation) for the "gravitational field". (The Schwarzschild solution was just the exact solution to $R_{\mu \nu }=0\,$.)
Now the actual law for electrostatics is Coulomb's law:
       $E={\frac {Q}{4\pi \epsilon r^{2}}}$
where $Q\,$ is the charge and $\epsilon \,$ is the physical constant governing the force. This is equivalent to Maxwell's first equation:
       $\nabla \cdot E={\frac {\rho }{\epsilon }}$
where $\rho \,$ is the charge density.
Similarly, the equation for classical gravity is:
       $V={\frac {GM}{r^{2}}}$
where $M\,$ is the gravitating object's mass and $G\,$ is Newton's constant of gravitation. This is equivalent to:
       $\nabla \cdot V=4\pi G\rho$
where $\rho \,$ is the matter density.
Therefore we have:
       $R_{00}=4\pi G\rho \,$
(We had $R_{00}=0\,$ before because we were calculating the field in a vacuum.)
The actual tensor of interest is Einstein's tensor $G_{\mu \nu }\,$, which has $G_{00}=2R_{00}\,$ in this simple case, so:
       $G_{00}=8\pi G\rho \,$
Specifically:
       $G_{\mu \nu }={\begin{bmatrix}8\pi G\rho &0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}}$
The other components relate to the motion of the gravitating matter—when the matter moves, things become even more complicated.
The stress-energy-momentum tensor $T_{\mu \nu }\,$, describes the configuration of matter and other things (e.g. electrical energy) that create gravity. For motionless matter, we have:
       $T_{\mu \nu }={\begin{bmatrix}\rho &0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}}$
So the equation of gravity, Einstein's equation, is:
       $G_{\mu \nu }=8\pi G\ T_{\mu \nu }\,$