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General Relativity


There notes are just to give you the very basics of the General Relativity that is needed to do cosmology. For a slightly longer and more detailed introduction to General Relativity I recommend A No-Nonsense Introduction to General Relativity by Sean Carroll. There are also some decent lectures introducting GR on Youtube if you prefer this (a 5min overview: General Relativity Explained in 7 Levels of Difficulty). This introduction to the foundatations of General Relativity by Chris Blake is also very good. We'll start with a summary of the most important things you will need to know: The fundamental quantity in General Relativity is the metric tensor $g_{\mu\nu}$. This is determined by solving Einstein's equations $$R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R = 8\pi G T_{\mu\nu}$$ where the Ricci tensor and Ricci scalar are given by $$R_{\mu\nu} = \partial_\alpha \Gamma^\alpha_{\mu\nu} - \partial_\mu \Gamma^\alpha_{\alpha\nu} + \Gamma^\alpha_{\alpha\beta}\Gamma^\beta_{\mu\nu} - \Gamma^\beta_{\mu\alpha}\Gamma^\alpha_{\beta\nu}$$ and $$R = g^{\mu\nu}R_{\mu\nu}$$ respectively and where the Christoffel symbols are given by $$\Gamma^\sigma_{\mu\nu} = \frac{1}{2}g^{\delta\sigma}(g_{\mu\delta,\nu} + g_{\delta\nu,\mu} - g_{\mu\nu,\delta})$$ $T_{\mu\nu}$ is the (total; sum over all components) energy-momentum tensor. For a perfect fluid we usually encounter in cosmology we have $$T_{\mu\nu} = (\rho + P)U^\mu U^\nu + P g^{\mu\nu}$$ where $\rho$ is the energy density, $P$ is the pressure and $U^\mu = \frac{dx^\mu}{d\lambda}$ is the 4-velocity of the fluid. In the simplest case of having a fluid comoving with the expansion of the Universe we simply have $U^\mu = (1,0,0,0)$ for which $T^0_0 = -\rho$, $T^i_j = p\delta_{ij}$ and the rest of the components are zero. The motion of particles are determined by the geodesic equation $$\frac{d^2x^\mu}{d\lambda^2} + \Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda} = 0$$ where $\lambda$ is an affine parameter along the trajectory. What it shows is that given a matter content (the $T_{\mu\nu}$) Einstein's equations tells us what the metric tensor $g_{\mu\nu}$ is and with this in hand we can compute how particles or photons will move in the space-time. If all this is known then you can skip the stuff below. If not then all this is for now just a bunch of symbols, but we'll get to explaining all of these things more thoroughly below. We only have time for a brief overview that is meant to get you to the point where you can do calculations in GR relevant for cosmology. For a deeper understanding of this you should take a full course in GR (see e.g. FYS4160 at the University of Oslo or these great lecture notes by Sean Carroll if you want to self study). A lot of the material below is based on this note by Haavard Ihle.


Newtonian gravity


Before we start with General Relativity lets talk a bit about Newtonian gravity. At the end of this note we'll get back to how Newtonian gravity emerges as a limit of General Relativity (which is good since Newtonian gravity works really well). Newtons gravitational law states that the graviational force on an point mass $m$ from a point mass $M$ is given by $$F = \frac{GmM}{r^2}$$ or in true vector form $$\vec{F} = -\frac{GmM \vec{r}}{r^3}$$ where $\vec{r}$ is the vector from $m$ to $M$. The gravitational force is conservative so its convenient to introduce what we call the gravitational potential $\Phi$ (the gravitational potential set up by $M$ is simply $\frac{GM}{r}$) for which the force is $$\vec{F} = -m\nabla\Phi$$ where $\nabla = e_i\frac{\partial}{\partial x_i}$ is the gradient operator. Given the force Newtons second law $m\ddot{x} = \vec{F}$ then tells us how the masses move. This simple formulation is only for two objects, but Newtonian gravity is linear (the super-position principle applies) so if we have a set of objects denoted by $i=1,2,\ldots,n$ then the total force on object $i$ is simply the sum $$\vec{F}_i = -\sum_{j=1,j\not= i}^n \frac{Gm_im_j \vec{r}_{ij}}{|\vec{r}_{ij}|^3}$$ where $\vec{r}_{ij} = \vec{r}_i - \vec{r}_j$ and where the sum excludes $i=j$ (there is no self-gravity). This also applies for the case where the objects are not point-masses (like the Earth) as long as we use the center of gravity of the objects as the positions above. It can also be used to study a continuous mass-distributions: we simply break the mass into tiny elements and sum up the contribution. Taking the elements to be smaller and smaller this effectivly turn the discrete sum into an integral and we end up with $$\Phi = -4\pi G\int_{\mathbb{R}^3} \frac{\rho(\vec{y})}{4\pi|\vec{x} - \vec{y}|}d^3\vec{y}$$ The connection to the point-particle formulation is seen by taking the density-field to be a sum of point masses at different locations: $\rho(\vec{y}) = \sum_{i=1}^n m_i\delta^{(3)}(\vec{y} - \vec{r}_i)$ where $\delta^{(3)}$ is the Dirac Delta Function in 3D. We can also use a formulation that is more common from electrodynamics and which will be the main formulation we will use in this course. Recall in electrodynamics we have Maxwell's equations, the first of which reads $$\nabla\cdot \vec{E} = \frac{\rho_Q}{\epsilon_0}$$ where $\vec{E}$ is the electric field, $\rho_Q$ is the charge density and $\epsilon_0$ a constant. The electric force is given by $\vec{F} = q\vec{E}$ and for two point charges we have Couloumb's law $$\vec{F} = \frac{1}{4\pi\epsilon_0} \frac{qQ\vec{r}}{r^3}$$ Notice that this is almost exactly the same as for Newtonian gravity if we identify $\frac{1}{4\pi\epsilon_0}$ with $G$, the charge density $\rho_Q$ with the mass density and $\vec{E}$ with $\nabla\Phi$ (and change a sign which comes from the fact that we need opposite signed charges to get an attractive force). This analogy is general and shows that the gravitational potential $\Phi$ is determined by an important equation called Poisson's equation $$\nabla^2 \Phi = 4\pi G\rho$$ The general solution of this equation in 3D is given by the same expression as we mentioned above $$\Phi = -4\pi G\int_{\mathbb{R}^3} \frac{\rho(\vec{y})}{4\pi|\vec{x} - \vec{y}|}d^3\vec{y}$$ The function $S(\vec{x},\vec{y}) = -\frac{1}{4\pi|\vec{x}-\vec{y}|}$ is a so-called Greens function for the Laplace operator, i.e. it has the property $$\nabla_x^2 S(\vec{x},\vec{y}) = \delta^{(3)}(\vec{x} - \vec{y})$$ Thus if we apply $\nabla_x^2$ (using a subscript $x$ to denote that the derivatives is with respect to $x$ not $y$) to the equation for $\Phi$ we arrive at the desired result $\nabla^2\Phi = 4\pi G\rho$. This differential formulation for gravity will be key to seeing how General Relativity reduces to Newtonian gravity. We will show that in the limit of weak gravity and low velocities the equation of motions following from General Relativity becomes what Newtons second law + Newtons gravitational law says it is: $\ddot{\vec{x}} = -\nabla\Phi$, where the gravitational potential is determined by Poissons equation $\nabla^2\Phi = 4\pi G\rho$. Poisson's equation will also be central when we get to talking about how perturbations evolve in our Universe. Mass-energy sets up a gravitational potential which we will find is determined by a (slightly modified to account for the expansion of the Universe) Poisson's equation.


Tensors


The main motivation for using tensors is to write down equations that are invariant under coordinate transformations, that is, the equations look the same in any coordinate system. The reason we want to write down such equations is that we want our physical theories to be independent of the coordinates we use to describe (or express) these theories. Hence we will here study coordinate transformations, and how objects change onder such transformations in order to learn what tensors are, and why they are so useful. So what is a tensor? Well there are many ways of defining this depending on your preferred level of mathematical rigor (take a look at the different definitions on Wikipedia). A very rough definition, just to give you some intuition, is to simply say that they are the generalization of scalars, vectors and matrices (that carry free indices). However this naive definition does not include the most important part: how a tensor changes when we do a coordinate transformation. The way it transforms is really what determines if an object is a tensor or not as we shall see below.


3D Rotations


We want to understand what tensors are, and why they are so important, but let us first start with something more familiar. Here are some equations we all know and love: $$ \begin{aligned} m \vec{a} &=\vec{F} \\ \vec{\nabla} \times \vec{E} &=-\frac{\partial \vec{B}}{\partial t} \end{aligned} $$ These equations (vector equations) are nice because, in a sense, they are "always true". We may not always agree about what $a_{x}$ is (e.g. if my $x$ -axis points in a different direction than yours), but we will both agree that (1) is true. Why is this? I claim this is because $\vec{a}$ transforms in exactly the same way as $\vec{F}$ under rotations and translations. Specifically, if: $$ \vec{x} \underbrace{\rightarrow}_{\text {"goes to" }} \overrightarrow{\tilde{x}}=\hat{R} \vec{x} + \vec{c} $$ then $$ \begin{array}{l} \vec{a} \rightarrow \overrightarrow{\tilde{a}}=\hat{R} \vec{a} \\ \vec{F} \rightarrow \overrightarrow{\tilde{F}}=\hat{R} \vec{F} \end{array} $$ ensuring that $$ m \overrightarrow{\tilde{a}}=\overrightarrow{\tilde{F}} $$ where by $\vec{x}$ "goes to" $\vec{x}$, we mean that we change coordinates from $\vec{x}$ to $\vec{x}$. $\vec{x}$ is then given as a function of the old coordinates, here just by $\vec{x}$ multiplied by a rotation matrix $\hat{R}$ and translated by $\vec{c}$. The rotation matrix $\hat{R}$ is given by e.g. (for rotations in the $x-y$ plane): $$ \hat{R}=\left(\begin{array}{ccc} \cos (\theta) & -\sin (\theta) & 0 \\ \sin (\theta) & \cos (\theta) & 0 \\ 0 & 0 & 1 \end{array}\right) $$


Einstein Summation Convention


Before we continue further it is useful to take this moment to introduce the Einstein Summation Convention. This may seem a bit unnecessary at this point, but it will make our lives much simpler when we get to more complicated tensor expressions soon. The basic idea is to write all tensors (We have not defined what a tensor is yet, but for now you can just think of matrices and vectors. We will soon enough learn exactly what a tensor is!) on index form and to not bother writing summation signs. For example: $$ x^{i} \rightarrow \tilde{x}^{i}=R_{j}^{i} x^{j} \equiv \sum_{j} R_{j}^{i} x^{j} $$ This convention works because (as we will learn) in tensor equations, indices are always summed in a very specific way. The Einstein Summation Convention can be summarized in the following rules:

  • If an index is repeated on the same side of an equation this index is summed over. I.e. the index $j$ in the equation above.
  • Always sum one upper and one lower index. If you find yourself wanting to sum over two upper or two lower indices, or find yourself wanting to sum more than two repeated indices, then something has probably gone wrong!
  • Indices that are summed over (called dummy indices) can be changed to another index symbol. I.e. in the expression $T_{j}^{i} F^{j k}, j$ can be changed to $l,$ giving us $T^{i}{ }_{l} F^{l k},$ which is an entirely equivalent expression. This often needs to be done to prevent using the same symbol for multiple repeated indices.
  • The free indices (the ones that are not summed over) have to match in each term of a tensor equation. I.e. $$ T^{i j} V_{j}=W^{i} F_{k}^{k} $$ is fine, but $$ \begin{aligned} T^{i j} V_{j} &=W^{l} F_{k}^{k} \\ T^{i j} &=W^{i} F_{k}^{k} \end{aligned} $$ are not valid tensor equations. In the first case above the free index, $i,$ on the left hand side does not show up on the right hand side, likewise the free index $l$ only shows up on the right hand side. In the second case the index $j$ on the left does not have a match on the right.
  • We use latin indices $i, j, k, l, \ldots$ to denote purely spatial indices. These take the values $1,2,3,$ denoting the three spatial dimensions. We use greek indices $\mu, \nu, \rho, \sigma, \ldots$ to denote spacetime indices. These take the values $0,1,2,3,$ where 0 denotes the timelike dimension and 1,2,3 the spatial dimensions.


Vectors and Tensors in spacetime


What about vectors and tensors in spacetime? This is what we are really interested in. What is the relevant transformation for these tensors (analogous to rotations in space)? In general relativity (GR), the transformations that the theory is unchanged under, are completely general coordinate transformations. In spacetime we have a set of four coordinates. A coordinate transformation is then $$ x^{\mu} \rightarrow \tilde{x}^{\mu}=\left(\begin{array}{l} \tilde{x}^{0}\left(x^{0}, x^{1}, x^{2}, x^{3}\right) \\ \tilde{x}^{1}\left(x^{0}, x^{1}, x^{2}, x^{3}\right) \\ \tilde{x}^{2}\left(x^{0}, x^{1}, x^{2}, x^{3}\right) \\ \tilde{x}^{3}\left(x^{0}, x^{1}, x^{2}, x^{3}\right) \end{array}\right) \equiv \tilde{x}^{\mu}(x) $$ You should already be familiar with some simple coordinate transformations like the ones going from Cartesian coordinates $(x,y)$ to polar coordinates $(r,\theta)$ where the transformation is simply $x = r\cos\theta$ and $y = r\sin\theta$. What we are talking about here is exactly the same, just in spacetime. We define a spacetime vector $^{1}, v^{\mu},$ as something that transforms in the following way under a transformation of the coordinates $^{2}$ : $$ v^{\mu} \rightarrow \tilde{v}^{\mu}=\underbrace{\frac{\partial \tilde{x}^{\mu}}{\partial x^{\nu}}}_{\text {Jacobian matrix }} v^{\nu} $$ For now you will just have to take my word that this is a good definition for the transformation of a vector (but you can check yourself that e.g. the transformations in eq. 4 and 5 are special cases of this transformation rule). Note that under this definition $x^{\mu}$ does not transform like a vector! This is different from what we usually have under $3 \mathrm{D}$ rotations. There is also another kind of vector, $w_{\mu}$ with the index downstairs. This transforms slightly differently $$ w_{\mu} \rightarrow \tilde{w}_{\mu}=\underbrace{\frac{\partial x^{\nu}}{\partial \tilde{x}^{\mu}}}_{\text {inverse Jacobian }} w_{\nu} $$ The Jacobian and it's inverse combine to give the Kronecker-delta, $\delta_{\sigma}^{\mu}$ ($=1$ if the two indices are the same and $0$ otherwise), which is the tensor equivalent of the identity matrix $$ \frac{\partial \tilde{x}^{\mu}}{\partial x^{\nu}} \frac{\partial x^{\nu}}{\partial \tilde{x}^{\sigma}}=\delta_{\sigma}^{\mu} $$ which essentially follows from the chain rule. Let's look at the transformation of the combination $$ v^{\mu} w_{\mu} \rightarrow \tilde{v}^{\mu} \tilde{w}_{\mu}=\frac{\partial \tilde{x}^{\mu}}{\partial x^{\nu}} v^{\nu} \frac{\partial x^{\sigma}}{\partial \tilde{x}^{\mu}} w_{\sigma}=\frac{\partial \tilde{x}^{\mu}}{\partial x^{\nu}} \frac{\partial x^{\sigma}}{\partial \tilde{x}^{\mu}} v^{\nu} w_{\sigma}=\delta_{\nu}^{\sigma} v^{\nu} w_{\sigma}=v^{\sigma} w_{\sigma}=v^{\mu} w_{\mu} $$ 1 When we say "vector" here (and later for tensors) we usually mean vector field, $v^{\mu}(x)$. That is, a field with a vector (or tensor) at each point in spacetime. The spacetime equivalents of $3 \mathrm{D}$ vector fields like the electric or magnetic field. "For a comparison of different types of transformations, see Figure 4.

We see that the combination $v^{\mu} w_{\mu}$ is invariant under a general coordinate transformation. Such a quantity we call a scalar, in general a scalar, $\phi,$ is something that transforms like $$ \phi \rightarrow \tilde{\phi}=\phi $$ In general, a tensor is something with indices $$ T^{\mu \nu \cdots}_{\alpha \beta \cdots} $$ transforming like $$ T^{\mu \nu \cdots}_{\alpha \beta \cdots} \rightarrow \tilde{T}^{\mu \nu \cdots}_{\alpha \beta \cdots} =\frac{\partial \tilde{x}^{\mu}}{\partial x^{\rho}} \frac{\partial \tilde{x}^{\nu}}{\partial x^{\sigma}} \cdots \frac{\partial x^{\lambda}}{\partial \tilde{x}^{\alpha}} \frac{\partial x^{\delta}}{\partial \tilde{x}^{\beta}} \cdots T^{\rho \sigma \cdots}_{\lambda \delta \cdots} $$ So each upper index transforms with a Jacobian matrix, while each lower index transforms with an inverse Jacobian. So vectors with upper or lower indices are just tensors with one index.


Derivatives in spacetime


Since we have our nice spacetime coordinates, $x^{\mu},$ it seems natural to define the spacetime derivative $$ \partial_{\mu} \equiv \frac{\partial}{\partial x^{\mu}}=\left(\frac{\partial}{\partial x^{0}}, \frac{\partial}{\partial x^{1}}, \frac{\partial}{\partial x^{2}}, \frac{\partial}{\partial x^{3}}\right) $$ By defining this derivative with a lower index we have sort of hinted that this should transform like a vector with a lower index, but this needs to be shown. In order to see how the derivative transforms we need to act on different stuff, and then transform to new coordinates. The simplest thing to act on is a scalar, $\phi .$ So the question is: Does $\partial_{\mu} \phi$ transform like a vector? $$ \partial_{\mu} \phi \rightarrow \underbrace{\tilde{\partial}_{\mu}}_{\equiv \partial / \partial \tilde{x}^{\mu}} \tilde{\phi}=\underbrace{\frac{\partial x^{\nu}}{\partial \tilde{x}^{\mu}} \partial_{\nu}}_{\text {chain rule }} \phi $$ We see that indeed $\partial_{\mu} \phi$ transforms as a vector with a lower index! What about the derivative of a vector? $$ \begin{aligned} \partial_{\mu} A^{\nu} \rightarrow \tilde{\partial}_{\mu} \tilde{A}^{\nu} &=\frac{\partial x^{\alpha}}{\partial \tilde{x}^{\mu}} \partial_{\alpha}\left(\frac{\partial \tilde{x}^{\nu}}{\partial x^{\beta}} A^{\beta}\right) \\ &=\underbrace{\frac{\partial x^{\alpha}}{\partial \tilde{x}^{\mu}} \frac{\partial \tilde{x}^{\nu}}{\partial x^{\beta}} \partial_{\alpha} A^{\beta}}_{\text {regular tensor term }}+\underbrace{\frac{\partial x^{\alpha}}{\partial \tilde{x}^{\mu}} \frac{\partial^{2} \tilde{x}^{\nu}}{\partial x^{\alpha} \partial x^{\beta}} A^{\beta}}_{\text {annoying extra term! }} \end{aligned} $$ We see that $\partial_{\mu} A^{\nu}$ actually does not transform as a tensor! The first term looks exactly like we expect a tensor to transform, but we also have the second term which ruins the fun. This is pretty bad! We want to write all our physical theories as tensor equations to ensure that the physics is invariant under coordinate transformations, but we can't really write all of physics without using derivatives (you can try!). What we can do is to define a new quantity (also not a tensor), $\Gamma_{\alpha \beta}^{\mu},$ called the connection coefficients, Christoffel symbols or just Gamma's, such that $$ \Gamma_{\alpha \beta}^{\mu} \rightarrow \tilde{\Gamma}_{\alpha \beta}^{\mu}=\frac{\partial \tilde{x}^{\mu}}{\partial x^{\lambda}} \frac{\partial x^{\rho}}{\partial \tilde{x}^{\alpha}} \frac{\partial x^{\sigma}}{\partial \tilde{x}^{\beta}} \Gamma_{\rho \sigma}^{\lambda}-\frac{\partial x^{\rho}}{\partial \tilde{x}^{\alpha}} \frac{\partial x^{\sigma}}{\partial \tilde{x}^{\beta}} \frac{\partial^{2} \tilde{x}^{\mu}}{\partial x^{\rho} \partial x^{\sigma}} $$ Using this new quantity we can define a new derivative, $\nabla_{\mu},$ called the covariant derivative $$ \begin{aligned} \nabla_{\mu} \phi & \equiv \partial_{\mu} \phi \\ \nabla_{\mu} A^{\nu} & \equiv \underbrace{\overbrace{\partial_{\mu} A^{\nu}}^{\text {Not a tensor }}+\overbrace{\Gamma_{\mu \lambda}^{\nu} A^{\lambda}}^{\text {Not a tensor }}}_{\text {A tensor! }} \end{aligned} $$ The covariant derivative of a vector now transforms as a tensor $$ \begin{aligned} \nabla_{\mu} A^{\nu} \rightarrow \tilde{\nabla}_{\mu} \tilde{A}^{\nu} &=\frac{\partial x^{\alpha}}{\partial \tilde{x}^{\mu}} \frac{\partial \tilde{x}^{\nu}}{\partial x^{\beta}} \partial_{\alpha} A^{\beta}+\frac{\partial x^{\alpha}}{\partial \tilde{x}^{\mu}} \frac{\partial^{2} \tilde{x}^{\nu}}{\partial x^{\alpha} \partial x^{\beta}} A^{\beta} \\ &+\frac{\partial x^{\alpha}}{\partial \tilde{x}^{\mu}} \frac{\partial \tilde{x}^{\nu}}{\partial x^{\beta}} \Gamma_{\alpha \lambda}^{\beta} A^{\lambda}-\frac{\partial x^{\alpha}}{\partial \tilde{x}^{\mu}} \frac{\partial^{2} \tilde{x}^{\nu}}{\partial x^{\alpha} \partial x^{\beta}} A^{\beta} \\ &=\frac{\partial x^{\alpha}}{\partial \tilde{x}^{\mu}} \frac{\partial \tilde{x}^{\nu}}{\partial x^{\beta}}\left(\partial_{\alpha} A^{\beta}+\Gamma_{\alpha \lambda}^{\alpha} A^{\lambda}\right) \\ &=\frac{\partial x^{\alpha}}{\partial \tilde{x}^{\mu}} \frac{\partial \tilde{x}^{\nu}}{\partial x^{\beta}}\left(\nabla_{\alpha} A^{\beta}\right) \end{aligned} $$ Likewise, we can show that $$ \nabla_{\mu} A_{\nu} \equiv \partial_{\mu} A_{\nu}-\Gamma_{\mu \nu}^{\lambda} A_{\lambda} $$ also transforms as a tensor (note the minus sign!). The covariant derivative of a general tensor is then defined as you might expect, e.g. $$ \nabla_{\mu} T^{\alpha \beta}_{\gamma} \equiv \partial_{\mu} T_{\gamma}^{\alpha \beta}+\Gamma_{\mu \lambda}^{\alpha} T_{\gamma}^{\lambda \beta}+\Gamma_{\mu \lambda}^{\beta} T_{\gamma}^{\alpha \lambda}-\Gamma_{\mu \gamma}^{\lambda} T_{\lambda}^{\alpha \beta} $$ So the covariant derivative of a tensor comes with one $\Gamma$ -term for each index of the tensor, and the terms corresponding to an upper index come with a positive sign, while the terms corresponding to a lower index come with a negative sign. We introduce the following simplified notation for the "regular" and covariant derivatives of tensors $$ \begin{aligned} T^{\alpha \beta}_{, \mu} & \equiv \partial_{\mu} T^{\alpha \beta} \\ T^{\alpha \beta}_{; \mu} & \equiv \nabla_{\mu} T^{\alpha \beta} \end{aligned} $$ Great, we have now found a derivative that we can use on tensors and still end up with tensors. However, we still have no way to actually calculate $\Gamma_{\alpha \beta}^{\mu} .$ It will turn out that the $\Gamma$ 's carry information about the curvature of spacetime (as well as information about the coordinates used), so we will have to talk about curvature before we can learn how to actually calculate the $\Gamma$'s. For the case where we have no curvature (good old flat space) then thee $\Gamma$'s are identical to zero and the covariant derivative reduces to the good old normal derivative.


The metric tensor



    Figure 1: The metric tensor defines the spacetime interval, $ds$, between events $x^\mu$ and $x^\mu + dx^\mu$.

    There is a special tensor, called the metric tensor $g_{\mu\nu}$. This tensor is symmetric, $g_{\mu \nu}=g_{\nu \mu}$, and defines the infinitesimal spacetime interval, $d s$ (see Figure 1), between two events (points), $x^{\mu}$ and $x^{\mu}+d x^{\mu}$ given by $$ d s^{2}=g_{\mu \nu}(x) d x^{\mu} d x^{\nu} $$ The metric tensor tells us how to compute distances and norms of vectors in our spacetime and is the fundamental quantity in General Relativity. Let us look at some simple examples:

    • The spatial interval on a 2 -D sphere (with coordinates $(\theta, \phi)$ ) is given by $$ \begin{aligned} d s^{2} &=g_{\theta \theta} d \theta^{2}+g_{\theta \phi} d \theta d \phi+g_{\phi \theta} d \phi d \theta+g_{\phi \phi} d \phi^{2} \\ &=d \theta^{2}+\sin ^{2} \theta d \phi^{2} \end{aligned} $$ I.e. $$ g_{\mu \nu}=\left(\begin{array}{cc} 1 & 0 \\ 0 & \sin ^{2} \theta \end{array}\right) $$ This tells us how to compute the distance - along the surface of the sphere - between two points separated by spherical angles $d\theta$ and $d\phi$ on the unit sphere (this curve will be an arc; a part of circle).
    • In spacetime, the first example is the completely flat spacetime, called Minkowski space, which is the spacetime of special relativity $$ d s^{2}=-d t^{2}+\delta_{i j} d x^{i} d x^{j} $$ or $$ g_{\mu \nu}=\eta_{\mu \nu} \equiv\left(\begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) $$
    • As another example, take the (spatially) flat FLRW metric (the metric that describes the geometry of our Universe as a whole) $$ d s^{2}=-d t^{2}+a^{2}(t) \delta_{i j} d x^{i} d x^{j} $$ implying a metric $$ g_{\mu \nu}=\left(\begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & a^{2} & 0 & 0 \\ 0 & 0 & a^{2} & 0 \\ 0 & 0 & 0 & a^{2} \end{array}\right) $$ and the inverse metric (written just as $g$ with upper indices) $$ g^{\mu \nu} \equiv\left(g^{-1}\right)^{\mu \nu}=\left(\begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 / a^{2} & 0 & 0 \\ 0 & 0 & 1 / a^{2} & 0 \\ 0 & 0 & 0 & 1 / a^{2} \end{array}\right) $$ Multiplying the meric and the inverse metric together gives the identity matrix and the equivalent statement of that in tensor language is $$g_{\mu \nu} g^{\nu \beta} = \delta^{\beta}_{\nu}$$ Writing out the matrices this statement is $$ g_{\mu \nu} g^{\nu \beta} = \left[ \left(\begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & a^{2} & 0 & 0 \\ 0 & 0 & a^{2} & 0 \\ 0 & 0 & 0 & a^{2} \end{array}\right) \left(\begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 / a^{2} & 0 & 0 \\ 0 & 0 & 1 / a^{2} & 0 \\ 0 & 0 & 0 & 1 / a^{2} \end{array}\right) \right]_{\text{The element in row }\mu \text{ column }\beta} = \left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)_{\text{The element in row }\mu \text{ column }\beta} \equiv \delta_\mu^\beta $$

    The metric tensor carries all the information about the curvature of spacetime, and it is the dynamical variable that is "solved for" in GR using the Einstein equation (which we will get to). This also means that the Christoffel symbols are determined by the metric $^{3}$ $$ \Gamma_{\alpha \beta}^{\mu}=\frac{1}{2} g^{\mu \nu}\left(g_{\alpha \nu, \beta}+g_{\beta \nu, \alpha}-g_{\alpha \beta, \nu}\right) $$ Another useful thing we can do with the metric tensor is to move indices up and down $$ \begin{aligned} V_{\mu} & \equiv g_{\mu \nu} V^{\mu} \\ T_{\lambda}^{\mu} & \equiv g^{\mu \nu} T_{\lambda \nu} \end{aligned} $$ and so on. Since $g$ is a tensor, the new tensors we get by moving indices up or down will also transform as we would expect based on the new positions of the indices. Sometimes we wish to take a tensor with a given number of indices and make another tensor from it that has fewer indices. We also want to make sure to do this in a way such that the resulting object still transforms like a tensor. One way to do this is to set an upper index equal to a lower index, and sum over all values. E.g. $$ S_{\mu} \equiv S_{\mu \lambda}^{\lambda} $$ This is called contracting two indices, and it works out because upper and lower indices transform with the inverse transformations to each other. What if we want to contract two lower or two upper indices? We can't sum over these (remember that we are only allowed to sum one upper and one lower index!), however, we can use the metric (or inverse metric). E.g. $$ T_{\lambda} \equiv g_{\mu \nu} T_{\lambda}^{\mu \nu} $$ Note that we can not make a tensor with only one index less than the original tensor using contraction, we always need to contract two indices at a time.

    ${ }^{3}$ Here we have assumed $\nabla_{\mu} g_{\alpha \beta}=0$ (often called metric compatibility) and that the antisymmetric part of the connection coefficients, called the torsion, vanish, $\Gamma^{\mu}_{\alpha \beta}-\Gamma^{\mu}_{\beta \alpha}=0$. The first of these assumptions is simply a common and useful convention, while the second is an important, but also very common, physical assumption.


    Parallel transport and the Riemann curvature tensor



      Figure 2: Parallel transport of a vector around a loop on a sphere. Even though the vector always stays parallel to itself, it has changed directions when it returns to it's original position.

      In flat spacetime parallel transport, that is, moving a vector from one point to another along a curve while keeping it constant, seems simple enough, we just move it without changing it. If we are using Cartesian coordinates we can just keep the components of the vector constant. Written in mathematics, moving a vector along a curve $x^{\mu}(\lambda)$ while keeping the components constant, is ensured by requiring that the derivative with respect to $\lambda$ vanishes, $$ \frac{d}{d \lambda} V^{\mu}=\frac{d x^{\nu}}{d \lambda} \frac{\partial}{\partial x^{\nu}} V^{\mu}=0 $$ or just $V^{\mu}\left(x\left(\lambda_{1}\right)\right)=V^{\mu}\left(x\left(\lambda_{0}\right)\right),$ where $\lambda$ is a parameter monotoncally increasing along the curve. If we are in curved spacetime (or curved space), this is not so trivial. This can be seen, for example, by moving a vector around on the surface of a sphere while keeping it parallel to itself. If you start at the north pole with a vector pointing towards the equator, and you move it along the surface to the equator, then along the equator some angle $\theta$ and then back up to the north pole in a straight line, all while keeping the direction of the vector parallel to itself, you will see that the vector has changed! It now points in a different direction (in fact, the direction differs by the angle $\theta,$ that you moved along the equator.). See Figure 2. In a sense this is bad. We cannot uniquely compare (add, subtract, take dot product etc) vectors defined at different points in spacetime. We could of course parallel transport one vector to the other, but the vector you end up with would depend on the path you took between the points, so the combination would be entirely arbitrary. On the other hand, we now have a method to detect curvature! If someone hands you a complicated metric, and you want to know if the spacetime corresponding to the metric is curved, or if it is secretly just flat spacetime and your friend was using some weird coordinates, you can just check this by moving a vector around in this spacetime using parallel transport, and see if the vector changes. If the vector changes, the space has curvature, and if the vector stays the same, it means that the space is (probably) flat $^{4}$. To generalize the equation for parallel transport (Eq. 50 ) to a case where we don't use Cartesian coordinates we just replace the spacetime derivative $\partial / \partial x^{\nu}$ by the covariant derivative $\nabla_{\nu} .$ This way the equation becomes a valid tensor equation. $$ \frac{d x^{\nu}}{d \lambda} \nabla_{\nu} V^{\mu}=0 $$ We argued that parallel transporting a vector around a curve and comparing to the original vector was a way for us to detect curvature. Let's apply this to a general infinitesimal curve, and see what we get. If we start with a vector $V^{\mu}$ and parallel transport it first along an infinitesimal vector $\delta A^{\alpha},$ then along an infinitesimal vector $\delta B^{\beta},$ then backwards along $\delta A^{\alpha}$ and finally backwards along $\delta B^{\beta},$ so we are back where we started (see Figure 3). The change in the vector $V^{\mu}$ from moving around this infinitesimal loop using parallel transport should then somehow contain information about the curvature of the spacetime. In addition we expect the change to be proportional to $\delta A^{\alpha}$ and $\delta B^{\beta},$ as well as $V^{\mu}$ itself. We then get: $$ V_{\text {after }}^{\mu}-V_{\text {before }}^{\mu}=R_{\nu \alpha \beta}^{\mu} V^{\nu} \delta A^{\alpha} \delta B^{\beta} $$ where $R_{\nu \alpha \beta}^{\mu}$ is called the Riemann curvature tensor, or simply the Riemann tensor, and it tells us how spacetime is curved. From Eq. 51 we have all the information we need to calculate $R_{\nu \alpha \beta}^{\mu}$ explicitly. It turns out, after a somewhat tedious derivation, that the Riemann tensor is given by the commutator ($[a,b] \equiv ab-ba$) of two covariant derivatives $$ R_{\nu \alpha \beta}^{\mu} V^{\nu}=\left[\nabla_{\alpha}, \nabla_{\beta}\right] V^{\mu} $$

      ${ }^{4}$ Of course there are sometimes paths you can take in spacetime where your vector would happen to end up with the same components (for example if you move the vector in Fig. 2 around an angle $\theta=2 \pi$ at the equator, before you move it back up to the north pole.)

      After another tedious derivation you can show that this reduces to $$ R_{\nu \alpha \beta}^{\mu}=\Gamma_{\nu \beta, \alpha}^{\mu}-\Gamma_{\nu \alpha, \beta}^{\mu}+\Gamma_{\alpha \lambda}^{\mu} \Gamma_{\beta \nu}^{\lambda}-\Gamma_{\beta \lambda}^{\mu} \Gamma_{\alpha \nu}^{\lambda} $$ The Riemann tensor is then the ultimate test for flatness. If all the components of the Riemann tensor are zero, then the spacetime is flat.


        Figure 3: Moving a vector, $V^\mu$, around loop along two infinitesimal vectors, $\delta A^\alpha$ and $\delta B^\alpha$, and seeing how it changes due to the curvature, allows us to define the Riemann curvature tensor.


        The Ricci tensor, the Ricci scalar and the Einstein equation


        We can contract the first and third indices of the Riemann tensor in order to obtain the Ricci tensor $$ R_{\mu \nu} \equiv R_{\mu \lambda \nu}^{\lambda} $$ Contracting the two indices if the Ricci tensor gives us the Ricci scalar $$ R \equiv g^{\mu \nu} R_{\mu \nu} $$ Together these tensors make up the Einstein tensor, $G_{\mu \nu},$ (sometimes called $E_{\mu \nu}$) which is the left hand side of the Einstein equation $$ G_{\mu \nu} \equiv R_{\mu \nu}-\frac{1}{2} g_{\mu \nu} R = 8 \pi G T_{\mu \nu} $$ where $T_{\mu \nu}$ is the Energy-momentum tensor. The Einstein equation relates the curvature of the universe, represented by $G_{\mu \nu},$ to the matter and energy content of the universe, represented by $T_{\mu \nu} .$ So any distribution of matter and energy will create curvature in spacetime around it, in much the same way as any distribution of charge will create an electromagnetic field. Let me note briefly that $G_{\mu \nu}=0$ does not imply that spacetime is flat, it just means that there is no energy or momentum locally. In the same way that the electromagnetic field does not vanish outside a charge distribution, the curvature will also extend outside a distribution of energy. Curvature is only zero when all the components of the Riemann tensor vanishes $\left(R_{\nu \alpha \beta}^{\mu}=0\right)$ In addition, there are gravitational wave solutions of the Einstein equations, which are solutions of the sourceless Einstein equation $\left(G_{\mu \nu}=0\right),$ analogous to the radiative solutions of the sourceless Maxwell equations $\left(\partial_{\mu} F^{\mu \nu}=0\right)$


        Minimal coupling and the Geodesic equation


        We have described the equations governing the curvature of spacetime, the Einstein equation, but what about the rest of physics? How do we do physics in curved spacetime? There is a simple prescription we can use to generalize physics to curved spacetime, called the minimal-coupling principle:

        • Take a law of physics valid in special relativity (SR).
        • Write it in tensorial form by making the replacements $\eta_{\mu \nu} \rightarrow g_{\mu \nu}$ and $\partial_{\mu} \rightarrow \nabla_{\mu}$

        Let us illustrate with some examples:

        • Consider the conservation of energy and momentum. In SR this is equivalent to the four divergence of the energy momentum tensor vanishing: $$ \partial_{\mu} T^{\mu \nu}=0 $$ In GR this equation becomes: $$ \nabla_{\mu} T^{\mu \nu}=0 $$ Simple! In GR this equation is a consequence of the Einstein tensor satisfying the equation $\nabla^\mu G_{\mu\nu} = 0$ as can be checked by direct computation. Historically Einstein initially proposed different field equations like $R_{\mu\nu} = T_{\mu\nu}$ which does not have the property that it implies $\nabla^\mu T_{\mu\nu} = 0$ so this would have then had to be included a seperate assumption, but ultimately he found the right combination for which this key property is a direct conseqence of his field equation.
        • As a more interesting example consider Newton's second law, which, in the absence of forces is given by $$ \frac{d^{2} x^{\mu}}{d \lambda^{2}}=0 $$ This is not a tensor equation. The velocity, $\frac{d x^{\mu}}{d \lambda},$ is a vector, but the acceleration involves the derivative of a vector, which, as you may remember, does not transform as a tensor $$ \frac{d^{2} x^{\mu}}{d \lambda^{2}}=\frac{d x^{\alpha}}{d \lambda} \partial_{\alpha} \frac{d x^{\mu}}{d \lambda} $$ where we used the chain rule. Replacing the spacetime derivative by a covariant derivative gives us the general relativistic version Newtons second law (in the absence of forces - If you have forces present, they can be added to the right hand side once they are written on covariant form) $$ \frac{d x^{\alpha}}{d \lambda} \nabla_{\alpha} \frac{d x^{\mu}}{d \lambda}=\frac{d^{2} x^{\mu}}{d \lambda^{2}}+\Gamma_{\alpha \beta}^{\mu} \frac{d x^{\alpha}}{d \lambda} \frac{d x^{\beta}}{d \lambda}=0 $$ This is called the geodesic equation (more about this below), and is the equation of motion for a particle moving in a curved spacetime, and will be used a lot by us during this course. Looking at the left hand side of Eq. 62 we see that this can also be interpreted as the parallel transport equation (Eq. 51 ) for a curve where the velocity vector of the curve, $\frac{d x^{\mu}}{d \lambda},$ is parallelly transported along the curve itself. Such a curve will be in a sense as "straight as possible". What this means is that if you, at any point along the curve approximate spacetime as flat, and make a nice cartesian coordinate system, the curve will look straight in these coordinates. Such a "locally straight" curve is called a geodesic. So particles in curved spacetime move (in the absence of forces) along geodesics.

        We mentioned the geodesic equation above, lets motivate it a bit better. A fundamental principle in General Relativity is the Equivalence principle: The local effects of motion in a curved spacetime (gravitation) are indistinguishable from those of an accelerated observer in flat spacetime, without exception. Gravity is a manifistation of spacetime curvature. If we focus our attention on a small enough region of spacetime, that region of can be considered to have no curvature and hence no gravity. Although we cannot transform away the gravitational field globally, we can get closer and closer to an ideal inertial reference frame if we make the laboratory become smaller and smaller in spacetime volume. In such a freely falling frame occupying a small region of spacetime, the laws of physics are the laws of special relativity. Hence all special relativity equations can be expected to work in this small segment of spacetime. We can always choose a frame (coordinates) for which gravity vanishes locally. This is called a freely falling reference frame. In such a frame the equation of motion is simply what we would expect from special relativity $$\frac{d^2X^\mu}{d\lambda^2} = 0$$ i.e. there is no force. What happens in other coordinate systems? In the freely falling reference frame we have $g_{\mu\nu} = \eta_{\mu\nu}$ and changing to a coordinate system $x^\mu = x^\mu(X^\alpha)$ then since the line-element is invariant we have $$g_{\mu\nu}dx^\mu dx^\nu = \eta_{\mu\nu}dX^\mu dX^\nu\implies g_{\mu\nu} = \eta_{\rho\delta} \frac{dX^\rho}{dx^\mu} \frac{dX^\delta}{dx^\nu}$$ which can be seen by using that $dX^\delta = \frac{dX^\delta}{dx^\beta}dx^\beta$. Performing the transformation to the new reference frame using $$\frac{dX^\mu}{d\lambda} = \frac{dX^\mu}{dx^\nu} \frac{dx^\nu}{d\lambda}$$ $$\frac{d^2X^\mu}{d\lambda^2} = \frac{d^2X^\mu}{dx^\alpha dx^\nu} \frac{dx^\alpha}{d\lambda} \frac{dx^\nu}{d\lambda} + \frac{dX^\mu}{dx^\nu} \frac{d^2x^\nu}{d\lambda^2}$$ $$\implies 0 = \frac{dx^\rho}{dX^\mu}\frac{d^2X^\mu}{d\lambda^2} = \frac{d^2x^\rho}{d\lambda^2} + \frac{dx^\rho}{dX^\mu}\frac{d^2X^\mu}{dx^\alpha dx^\nu} \frac{dx^\alpha}{d\lambda} \frac{dx^\nu}{d\lambda}$$ we find $$\frac{d^2x^\mu}{d\lambda^2} = -\Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}$$ where $$\Gamma^\mu_{\nu\delta} = \frac{\partial x^\mu}{\partial X^\rho}\frac{\partial^2 X^\rho}{\partial x^\nu \partial x^\delta}$$ This form has a similar looking form as Newtons second law where the right hand side plays the role of the usual gravitational force (per unit mass)! The gravitational force is a fictious force. This equation is called the geodesic equation and the Christoffel symbols are more conveniently expressed in terms of the metric as $$\Gamma^\sigma_{\mu\nu} = \frac{1}{2}g^{\delta\sigma}(g_{\mu\delta,\nu} + g_{\delta\nu,\mu} - g_{\mu\nu,\delta})$$ Another way to derive this equation is using the principle of least action (see the link for a derivation).


        Newtonian limit of General Relativity


        Now that we have introduced General Relativity lets see how Newtonian gravity arises as an approximation of it (the Newtonian limit). Lets start from the case where we have no gravity, there is no sources and the metric can be taken to be the Minkowski metric. What happens if we introduce some matter into the system? This can be studied by starting from the flat space for which the metric takes the form of the Minkowski metric $$ds^2 = -dt^2 + (dx^2 + dy^2 + dz^2)$$ and adding in some energy sources in the energy momentum tensor by putting $T^\mu_\nu = {\rm diag}(-\rho,0,0,0)$. The presence of the sources will mean the metric is no longer completely flat, but it should be close to flat. We therfore expect the metric to take the form $$ds^2 = -dt^2(1+2\Phi) + (1-2\Phi)(dx^2 + dy^2 + dz^2)$$ where $\Phi$ is some small perturbation (using this symbol as we will later see that this term will correspond to the usual gravitational potential in Newtonian theory). We also cheated a bit (knowing what the answer will be). For a proper derivation you would take a more general expansion of the metric and we will do this when dealing with perturbations of the metric in cosmology later in this course. Plugging this metric into the Einstein equation and discarding all second order terms (i.e. we assume $\Phi \ll 1$ so that any term that contains two or more products of $\Phi$ like $\Phi^2$ are ignored) we end up with $$R_{00} - \frac{1}{2}Rg_{00} = 2\nabla^2 \Phi $$ which together with $8\pi G T_{00} = 8\pi G \rho$ in the Einstein equation gives us $$\nabla^2 \Phi = 4\pi G \rho$$ and we see that $\Phi$ is determined by the Poisson equation we encountered when talking about the differential formulation of Newtonian gravity. Now lets look at the geodesic equation. Assuming we have small velocities so that $v/c \ll 1$ then can ignore the $\frac{dx^i}{d\tau}$ terms as they are much smaller than $\frac{dx^0}{d\tau} \simeq 1$ and the only term that sizeably contributes is $\Gamma^i_{00} \frac{dx^0}{d\tau}\frac{dx^0}{d\tau} = \Gamma^i_{00}$ so the geodesic equation becomes $$\frac{d^2x^i}{d\tau^2} = - \Gamma^i_{00} = -\nabla^i \Phi$$ which is on the form of Newtons second law where $-\nabla\Phi$ plays the role of the gravitational acceleration. These two equations is therefore exactly the same as in the differential formulation of Newtonian gravity and shows that General Relativity does indeed reduce to Newtonian gravity as long as the sources we add are not too massive (we assumed $\Phi \ll 1$; this breaks down for black holes for example where we would have $\Phi \to 1/2$ at the event horizon), the velocities involved are non-relativistic (we assumed $v/c \ll 1$) and the matter involved is that of 'normal' matter (we assumed the EM tensor only has a sizable density term). This derivation also answers another question: why the prefactor $8\pi G$ to $T_{\mu\nu}$ in the Einstein equations and not something else? Because this is the factor needed to get the correct Newtonian limit. The terms we have ignored are indeed small in the solar-system (velocities are much less than the speed of light and the gravitational potential of the sun is $\sim 10^{-6}$, the Earth $\sim 10^{-10}$ and so on), but that does not mean these terms are not important. They do imply some very interesting observable effects in the solar-system like the perihelion precession of Mercury for example which has provided strong experimental evidence for general relativity. Apart from these thing the Newtonian description is still very accurate. Where it is not good is when we deal with very massive objects like black holes and neutron stars and also for describing how the Universe as a whole evolves. Here the full power of General Relativity is needed to make progress. This connection, that the perturbations in the metric is closely related to the usual Newtonian potential (which is true in the frames we will work in this course) will be useful later in the course when we look at perturbations to the cosmological FRLW metric and will allow us to use our Newtonian intuition to understand how the metric perturbations will evolve in our Universe.


        Summary


        We are now done with a short introduction to GR. Let us summarize the most important points:

        • Tensors are great, because they change in specific and predictable ways under coordinate transformations, assuring that tensor equations look the same in any coordinate system. This is crucial for physics, because we don't want the physical laws to depend on what coordinate system we are using.
        • In order to take derivatives of tensors (and for the result to transform as a tensor) we need to use the covariant derivative $\nabla_{\mu}$ instead of the regular spacetime derivative $\partial_{\mu}$. The covariant derivative of a tensor introduces a term with a connection coefficient (or Christoffel symbol) $\Gamma_{\alpha \beta}^{\mu}$ for each index of the tensor e.g. $$ \nabla_{\mu} T_{\gamma}^{\alpha \beta}=T_{\gamma ; \mu}^{\alpha \beta}=T^{\alpha \beta}_{\gamma, \mu}+\Gamma_{\mu \lambda}^{\alpha} T_{\gamma}^{\lambda \beta}+\Gamma_{\mu \lambda}^{\beta} T_{\gamma}^{\alpha \lambda}-\Gamma_{\mu \gamma}^{\lambda} T_{\lambda}^{\alpha \beta} $$
        • The metric tensor $g_{\mu \nu}$ tells you how to measure distances in a given spacetime $\left(d s^{2}=\right.$ $\left.g_{\mu \nu}(x) d x^{\mu} d x^{\nu}\right) .$ The metric tensor, and it's inverse $g^{\mu \nu},$ can also be used to move indices of tensors up or down, and contract indices together e.g. $$ \begin{aligned} V_{\mu} & \equiv g_{\mu \nu} V^{\nu} \\ T_{\lambda}^{\mu} & \equiv g^{\mu \nu} T_{\lambda \nu} \\ S_{\lambda} & \equiv g_{\mu \nu} S_{\lambda}^{\mu \nu} \end{aligned} $$
        • The connection coefficients (this is not a tensor!) can be calculated from the first derivatives of the metric $$ \Gamma_{\alpha \beta}^{\mu}=\frac{1}{2} g^{\mu \nu}\left(g_{\alpha \nu, \beta}+g_{\beta \nu, \alpha}-g_{\alpha \beta, \nu}\right) $$
        • The Riemann curvature tensor is the ultimate measure of the curvature of a spacetime $$ R_{\nu \alpha \beta}^{\mu}=\Gamma_{\nu \beta, \alpha}^{\mu}-\Gamma_{\nu \alpha, \beta}^{\mu}+\Gamma_{\alpha \lambda}^{\mu} \Gamma_{\beta \nu}^{\lambda}-\Gamma_{\beta \lambda}^{\mu} \Gamma_{\alpha \nu}^{\lambda} $$ Spactime is flat if and only if all the components of the Riemann curvature tensor are zero.
        • From the Riemann curvature tensor we can get the Ricci tensor and Ricci scalar $$ \begin{aligned} R_{\mu \nu} & \equiv R_{\mu \lambda \nu}^{\lambda} \\ R & \equiv g^{\mu \nu} R_{\mu \nu} \end{aligned} $$ Together these tensors make up the Einstein tensor, $G_{\mu \nu},$ which is the left hand side of the Einstein equation $$ G_{\mu \nu} \equiv R_{\mu \nu}-\frac{1}{2} g_{\mu \nu} R = 8 \pi G T_{\mu \nu} $$ $T_{\mu \nu}$ is the energy momentum tensor arising from all the content (i.e. matter and energy) of the spacetime. The Einstein equation is the equation of motion for the metric $g_{\mu \nu},$ and describes how matter and energy affects the curvature of spacetime.
        • Particles in spacetime move (in the absence of forces) along "locally straight" curves, called geodesics, governed by the geodesic equation $$ \frac{d^{2} x^{\mu}}{d \lambda^{2}}+\Gamma_{\alpha \beta}^{\mu} \frac{d x^{\alpha}}{d \lambda} \frac{d x^{\beta}}{d \lambda}=0 $$


          Figure 4: Comparison of different types of transformations, 3D-rotations, Lorentz transformations, and general coordinate transformations, and how vector transformations and invariant dot products are defined in each case.