# Layman's explanation and understanding of Einstein's field equations

Most of us have heard of Einstein's amazing equations which describe the universe around us, yet only some of us understand what the equations are actually saying.

What are these equations actually saying, and is there a simple (relatively) way to derive them?

Here they are, from Wikipedia:

$$R_{\mu\nu}-\dfrac{1}{2}g_{\mu\nu}R+g_{\mu\nu}\Lambda=\dfrac{8\pi G}{c^4}T_{\mu\nu}$$

I have a vague notion of what a tensor is (it describes things as an array and higher orders define more complex transformations), but I don't understand what all of these tensors are doing. And why is there a $c^{4}$ in the equation!?

Einstein's equations can be loosely summarized as the main relation between matter and the geometry of spacetime. I will try to give a qualitative description what every term in the equation signifies. I will, however, have to warn potential readers that this will not be a short answer. Furthermore, I will refrain from trying to derive the equations in "elementary" manner, as I certainly don't know of any.

## Matter

On the right hand side of the equation, the most important thing is the appearance of the energy-momentum tensor $$T_{\mu\nu}$$. It encodes exactly how the matter---understood in a broad sense, i.e. any energy (or mass or momentum or pressure) carrying medium---is distributed in the universe. For understanding how to interpret the subscript indices of the $$T$$, see my explanation of the metric tensor below.

It is multiplied by some fundamental constants of nature $$\Big($$the factor $$\frac{8\pi G}{c^4}\Big)$$ but this isn't of any crucial importance: One can view them as book-keeping tools that keep track of the units of the quantities that are related by the equation. In fact, professional physicists typically take the liberty to redefine our units of measurements in order to simplify the look of our expressions by getting rid of pesky constants such as this. One particular option would be to choose "reduced Planck units", in which $$8\pi G=1$$ and $$c=1$$, so that the factor becomes $$1$$.

## Differential geometry

On the left hand side of Einstein's equations, we find a few different terms, which together describe the geometry of spacetime. General relativity is a theory which uses the mathematical framework known as (semi-)Riemannian geometry. In this branch of mathematics, one studies spaces which are in a certain sense smooth, and that are equipped with a metric. Let us first try to understand what these two things mean.

The smoothness property can be illustrated by the intuitive (and historically important!) example of a smooth (two-dimensional) surface in ordinary three-dimensional space. Imagine, for instance, the surface of an idealized football, i.e. a 2-sphere. Now, if one focuses ones attention to a very small patch of the surface (hold the ball up to your own face), it seems like the ball is pretty much flat. However, it is obviously not globally flat. Without regards for mathematical rigor, we can say that spaces that have this property of appearing locally flat are smooth in some sense. Mathematically, one calls them manifolds. Of course, a globally flat surface such as an infinite sheet of paper is the simplest example of such a space.

In Riemannian geometry (and differential geometry more generally) one studies such smooth spaces (manifolds) of arbitrary dimension. One important thing to realize is that they can be studied without imagining them to be embedded in a higher-dimensional space, i.e. without the visualization we were able to use with the football, or any other reference to what may or may not be "outside" the space itself. One says that one can study them, and their geometry, intrinsically.

## The metric

When it comes to intrinsically studying the geometry of manifolds, the main object of study is the metric (tensor). Physicists typically denote it by $$g_{\mu\nu}$$. In some sense, it endows us with a notion of distance on the manifold. Consider a two-dimensional manifold with metric, and put a "coordinate grid" on it, i.e. assign to each point a set of two numbers, $$(x,y)$$. Then, the metric can be viewed as a $$2\times 2$$ matrix with $$2^2=4$$ entries. These entries are labeled by the subscripts $$\mu,\nu$$, which can each be picked to equal $$x$$ or $$y$$. The metric can then be understood as simply an array of numbers:

$$\begin{pmatrix} g_{xx}&g_{xy}\\ g_{yx}&g_{yy}\end{pmatrix}$$

We should also say that the metric is defined such that $$g_{\mu\nu}=g_{\nu\mu}$$, i.e. it is symmetric with respect to its indices. This implies that, in our example, $$g_{xy}=g_{yx}$$. Now, consider two points that are nearby, such that the difference in coordinates between the two is $$(\mathrm{d}x,\mathrm{d}y)\;.$$ We can denote this in shorthand notation as $$\mathrm{d}l^\mu$$ where $$\mu$$ is either $$x$$ or $$y\;,$$ and $$\mathrm{d}l^x=\mathrm{d}x$$ and $$\mathrm{d}l^y=\mathrm{d}y\;.$$ Then we define the square of the distance between the two points, called $$\mathrm{d}s\;,$$ as

$$\mathrm{d}s^2= g_{xx}\mathrm{d}x^2+g_{yy} \mathrm{d}y^2 + 2 g_{xy}\mathrm{d}x \mathrm{d}y= \sum_{\mu,\nu\in\{x,y\}}g_{\mu\nu}\mathrm{d}l^\mu \mathrm{d}l^\nu$$

To get some idea of how this works in practice, let's look at an infinite two-dimensional flat space (i.e. the above-mentioned sheet of paper), with two "standard" plane coordinates $$x,y$$ defined on it by a square grid. Then, we all know from Pythagoras' theorem that

$$\mathrm{d}s^2=\mathrm{d}x^2+\mathrm{d}y^2= \sum_{\mu,\nu\in\{x,y\}}g_{\mu\nu}\mathrm{d}l^\mu \mathrm{d}l^\nu$$

This shows that, in this case, the natural metric on flat two-dimensional space is given by

$$g_{\mu\nu}=\begin{pmatrix} g_{xx}&g_{xy}\\ g_{xy}&g_{yy}\end{pmatrix}= \begin{pmatrix} 1&0\\0&1\end{pmatrix}$$

Now that we known how to "measure" distances between nearby points, we can use a typical technique from basic physics and integrate small segments to obtain the distance between points that are further removed:

$$L=\int \mathrm{d}s= \int \sqrt{\sum_{\mu,\nu\in\{x,y\}}g_{\mu\nu}\mathrm{d}l^\mu \mathrm{d}l^\nu}$$

The generalization to higher dimensions is straightforward.

## Curvature tensors

As I tried to argue in the above, the metric tensor defines the geometry of our manifold (or spacetime, in the physical case). In particular, we should be able to extract all the relevant information about the curvature of the manifold from it. This is done by constructing the Riemann (curvature) tensor $$R^{\mu}_{\ \ \ \nu\rho\sigma}$$, which is a very complicated object that may, in analogy with the array visualization of the metric, be regarded as a four-dimensional array, with each index being able to take on $$N$$ values if there are $$N$$ coordinates $$\{x^1,\dots x^N\}$$ on the manifold (i.e. if we're dealing with an $$N$$-dimensional space). It is defined purely in terms of the metric in a complicated way that is not all too important for now. This tensor holds pretty much all the information about the curvature of the manifold---and much more than us physicists are typically interested in. However, sometimes it is useful to take a good look at the Riemann tensor if one really wants to know what's going on. For instance, an everywhere vanishing Riemann tensor ($$R^\mu_{\ \ \ \nu\rho\sigma}=0$$) guarantees that the spacetime is flat. One famous case where such a thing is useful is in the Schwarzschild metric describing a black hole, which seems to be singular at the Schwarzschild radius $$r=r_s\neq 0$$. Upon inspection of the Riemann tensor, it becomes apparent that the curvature is actually finite here, so one is dealing with a coordinate singularity rather than a "real" gravitational singularity.

By taking certain "parts of" the Riemann tensor, we can discard some of the information it contains in return for having to only deal with a simpler object, the Ricci tensor:

$$R_{\nu\sigma}:=\sum_{\mu\in \{x^1,\dots x^N\}} R^\mu_{\ \ \ \nu\mu\sigma}$$

This is one of the tensors that appears in the Einstein field equations. the second term of the equations features the Ricci scalar $$R$$, which is defined by once again contracting (a fancy word for "summing over all possible index values of some indices") the Ricci tensor, this time with the inverse metric $$g^{\mu\nu}$$ which can be constructed from the usual metric by the equation

$$\sum_{\nu\in\{x^1,\dots,x^N\}}g^{\mu\nu}g_{\nu\rho}= 1\ \text{if }\mu=\rho\ \text{and }0\ \text{otherwise}$$

As promised, the Ricci scalar is the contraction of the Ricci tensor and inverse metric:

$$R:=\sum_{\mu,\nu\in\{x^1,\dots x^N\}}g^{\mu\nu}R_{\mu\nu}$$

Of course, the Ricci scalar once again contains less information than the Ricci tensor, but it's even easier to handle. Simply multiplying it by $$g_{\mu\nu}$$ once again results in a two-dimensional array, just like $$R_{\mu\nu}$$ and $$T_{\mu\nu}$$ are. The particular combination of curvature tensors that appears in the Einstein field equations is known as the Einstein tensor

$$G_{\mu\nu}:=R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}$$

## The cosmological constant

There is one term that we have left out so far: The cosmological constant term $$\Lambda g_{\mu\nu}$$. As the name suggests, $$\Lambda$$ is simply a constant which multiplies the metric. This term is sometimes put on the other side of the equation, as $$\Lambda$$ can be seen as some kind of "energy content" of the universe, which may be more appropriately grouped with the rest of the matter that is codified by $$T_{\mu\nu}$$.

The cosmological constant is mainly of interest because it provides a possible explanation for the (in)famous dark energy that seems to account for certain important cosmological observations. Whether or not the cosmological constant is really non-zero in our universe is an open issue, as is explaining the value observations suggest for it (the so-called cosmological constant problem a.k.a. "the worst prediction of theoretical physics ever made", one of my personal interests).

PS. As pointed out in the comments, if you enjoyed this you may also enjoy reading this question and the answers to it, which address that other important equation of general relativity, which describes the motion of "test particles" in curved spacetimes.

Einstein's equation relates the matter content (right side of the equation) to the geometry (the left side) of the system. It can be summed up with "mass creates geometry, and geometry acts like mass".

For more detail, let's consider what a tensor is. A two-index tensor (which is what we have in Einstein's equation), can be thought of as a map which takes one vector into another vector. For example, the stress-energy tensor takes a position vector and returns a momentum vector (mathematically, $p_{\nu}=T_{\nu\mu}x^{\mu}$, and I'm mixing up vectors and co-vectors all over the place to simplify the discussion). The interpretation is that the right side of Einstein's equation tells us the momentum which is passing through a surface defined by the position vector.

The left side can be interpreted in this manner as well. The Ricci curvature $R_{\mu\nu}$ takes a position vector and returns a vector telling us how much the curvature is changing through the surface defined by $\vec{x}$. The second and third terms, both having factors of the metric $g_{\mu\nu}$, tell us how much distance measurements are changed when traveling along the vector. There are two contributions to this change in distance - the scalar curvature $R$ and the $\Lambda$. If $R_{\mu\nu}$ is "curvature in a single direction", than $R$ is the "total curvature". $\Lambda$ is a constant which tells us how much innate energy empty space has, making all distances get larger for $\Lambda >0$.

So, reading the equation right to left, "Einstein's equation tells us that momentum (moving mass) causes both curvature and a change in how distances are measured." Reading left to right, "Einstein's equation tells us that curvature and changing distance acts just like moving mass."

• Wow - great simplified explanation. Sep 12, 2016 at 13:26
• @levitopher Why it's called "stress" in stress-energy? Jan 7, 2017 at 9:33