heard:emelyanov

# Introduction to General Relativity (Viacheslav Emelyanov)

These lecture notes are from a four lecture course on general relativity. The information is thus very dense and probably works best for an audience which has already some background knowledge (at least calculus, SR… of course differential geometry works well… ideally fiber bundles). The lecture is a great collection of the essential features of GR.

Mostly for the speed of light $c=1$ has been used. In some important results, $c$ has been put in explicitly.

## Lecture 1: Equations of Motion in a Gravitational Field

### The Geodesic Equation

We start with the equations of motion for a freely moving particle

$$\frac{d^2 \tilde{x}^{\mu}}{ds^2} = 0$$

here $ds^2 = dt^2 - d\vec{x}^2$ is the line element invariant under Lorentz transformations. We can identify it with the eigenzeit $ds \equiv d \tau$. But nevertheless, treating space and time on equal footing, the question is how does the position/velocity/momentum… of a particle change with respect to what? Some kind of direction in spacetime? Taking an invariant intuitively seems like a good idea, but this can surely be well motivated.

The question is, will this be preserved under a general coordinate transformation? The answer is no, using

$$d\tilde{x}^{\mu} = \frac{\partial \tilde{x}^{\mu}}{\partial x^{\lambda}} dx^{\lambda}$$

we end up with

$$\boxed{ \frac{d^2 x^{\lambda}}{ds^2} + \Gamma^{\lambda}_{\mu \nu} \frac{dx^{\mu}}{ds} \frac{dx^{\nu}}{ds} = 0 }$$

which is the Geodesic equation, where the object $\Gamma^{\lambda}_{\mu \nu}$ is the affine connection and given by

$$\boxed{ \Gamma^{\lambda}_{\mu \nu} = \frac{\partial x^{\lambda}}{\partial \tilde{x}^{\rho}} \frac{\partial^2 \tilde{x}^{\rho}}{\partial x^{\mu} \partial x^{\nu}} }$$

With the definition of the metric

$$g_{\mu \nu} = \frac{\partial \tilde{x}^{\lambda}}{\partial x^{\mu}} \frac{\partial \tilde{x}^{\rho}}{\partial x^{\nu}} \eta_{\lambda \rho}$$

Sure, we assume that we can locally always choose a frame which has Minkowski metric, therefore we can obtain the general metric for any coordinate system always by the transformation from the local Minkowski coordinate system to the desired coordinates.

we can express the affine connection in the following way by use of the equivalence principle:

$$\Gamma^{\lambda}_{\mu \nu} = \frac{1}{2} g^{\lambda \rho} \left( g_{\mu \rho, \nu} + g_{\nu \rho, \mu} - g_{\mu \nu, \rho} \right)$$

In this special form, the affine connection is called Christoffel Symbol. The notation $(.)_{,\nu} \equiv \partial_{\nu} (.)$ denotes a partial derivative.

In general, the affine connection and the metric are two different objects and $\Gamma^{\lambda}_{\mu \nu}$ can not always be expressed through $g_{\mu \nu}$ (e.g. when including torsion?). Being able to express the connection through the metric is where the equivalence principle is used. Ok where does the equivalence principle enter here? Need to think about this…

### Newton's approximation

• non-relativistic: $\frac{v}{c} \ll 1 \quad \leftrightarrow \quad u^{\mu} = \frac{dx^{\mu}}{ds} \approx (1,0,0,0)$
• gravity is weak: $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu} \quad$ where $\quad |h| \ll |\eta|$

In this limit, we obtain from the geodesic equation $t=s$ from the $\lambda = 0$ equation and

$$\frac{d^2 \vec{x}^2}{dt^2} \approx - \frac{1}{2} \vec{\nabla} h_{00}$$

If we now rename $h_{00} = 2 \phi$ then we reproduce Newton's Theory with $\phi = - \frac{G M}{r}$.

### Classical Tests of GR

• Gravitational redshift
• Precession of Mercury
• Deflection of light in gravitational field

These follow from the geodesic equation already. For gravitational waves and black holes, we need full GR, the Einstein equations.

### Basic Notions of Riemannian Geometry

The guiding principle is the relativistic principle, name that all frames are equal. Therefore we need objects, which are invariant under general coordinate transformations (this means these objects keep their meaning, no matter which frame we choose). This puts us in the position to define equations, which do not depend on the frame chosen. These equations then do not depend on the observer and thus model laws of nature.

In special relativity, we had the notion of Lorentz Tensors as objects, which are invariant under Lorentz transformations. In component form:

$$\tilde{T}^{\mu_1, \ldots ,\mu_m}_{\nu_1, \ldots ,\nu_n} = \Lambda^{\mu_1}_{\lambda_1} \ldots \Lambda^{\mu_m}_{\lambda_m} \Lambda^{\rho_1}_{\nu_1} \ldots \Lambda^{\rho_n}_{\nu_n} T^{\lambda_1, \ldots ,\lambda_m}_{\rho_1, \ldots ,\rho_n}$$

e.g. $\tilde{u}^{\mu} = \Lambda^{\mu}_{\nu} u^{\nu}$

Ok the general form just looks awful. Basically the statement is, that something deserves the name 'Lorentz Tensor' if it does not obtain additional terms when a Lorentz transformation acts on it? Actually an easy example of the transformation of something that is not a Lorentz Tensor would be helpful… And how does the Lorentz transformation act on it? One transformation matrix $\Lambda^{\mu}_{\nu}$ on every index of the object I am trying to transform.

The generalization we need (here an example with 4 indices) is

$$\tilde{T}^{\mu \nu}_{\lambda \rho} (\tilde{x}) = \frac{\partial \tilde{x}^{\mu}}{\partial x^{\alpha}} \frac{\partial \tilde{x}^{\nu}}{\partial x^{\beta}} \cdot \frac{\partial x^{\gamma}}{\partial \tilde{x}^\lambda} \frac{\partial x^{\delta}}{\partial \tilde{x}^\rho} T^{\alpha \beta}_{\gamma \delta} (x(\tilde{x}))$$

Remark: these are actually components of the tensor $T(\tilde{x})$ and the components do depend on the reference frame, just the total tensor does not. Compare to a vector in three dimensions, where also the components change with the frame. The full tensor can be written like this

$$T(x) = T^{\mu \nu}_{\lambda \rho} (x) \left( \partial_{\mu} \otimes \partial_{\nu} \otimes \partial x^{\lambda} \otimes \partial x^{\rho} \right)$$

So first of all, this transformation law again looks awful where the basic idea is really simple. Again we apply a transformation to each of our indices and the transformations are just derivatives from one frame wrt the other. Lower indices get the inverse transformation. The remark on vectors is really helpful, although I would like some clarification on the whole invariance thing. For a 3D vector, the length is the invariant under rotations. The vector as an object I think cannot even draw such concepts as invariance, can it? In that sense, I need to think/read more about what actually classifies invariance for tensor objects. And how to express that in simple words.

$\Gamma^{\lambda}_{\mu \nu}$ is not a tensor. And this is good for the following reason. Locally we can choose a reference frame with flat spacetime, this means that $\Gamma = 0$. If it was a tensor, it would stay zero everywhere because the transformation is then only multiplicative.

This is interesting. How does deSitter symmetry solve this when the connection actually transforms like the adjoint representation of the group? Do we have to give up on local Minkowski frames (which I doubt is a good idea) or does it solve it at a totally different stage?

### The Covariant Derivative

We want to model laws of nature so in some sense we need a derivative structure to say things like 'this changes according to this quantity with that rate…'. Now we need to investigate how to do this with invariant objects, i.e. what we defined as tensors, because in the end we want our (differential) equation to be independent of the reference frame. Otherwise you would just change the frame (hopping into a car) and the 'law' we thought we found wouldn't hold any more.

The ordinary partial derivative of a tensor is not a tensor: $\partial_{\lambda} T^{\mu}_{\nu} =$ not a tensor, even though $T^{\mu}_{\nu}$ is a tensor.

We need a derivative, that conserves tensor properties, a covariant derivative. e.g.

This covariant derivative is the ordinary derivative plus one affine connection for every index of the tensor. The affine connection gets contracted with that index and if it is a lower index on the tensor, it obtains a minus sign. This is still somehow awkward, I am sure there is an easier and more intuitive way to explain this.

\begin{align} \nabla_{\lambda} T^{\mu}_{\nu} & = \partial_{\lambda} T^{\mu}_{\nu} + \Gamma^{\mu}_{\color{green}{\rho} \lambda} T^{\color{green}{\rho}}_{\nu} \color{green}{-} \Gamma^{\color{green}{\rho}}_{\nu \lambda} T^{\mu}_{\color{green}{\rho}} \nonumber\\ \nabla_{\lambda} S & = \partial_{\lambda} S \end{align}

contracted indices and the minus sign in green for easier spotability.

Remark: In GR we have $\nabla_{\lambda} g_{\mu \nu} = 0$ which follows directly from the Christoffel Symbol. In extended GR, when the affine connection can not be represented as the Christoffel Symbol we can get $\nabla_{\lambda} g_{\mu \nu} = N_{\lambda \mu \nu}$ with a non-metricity tensor $N_{\lambda \mu \nu}$.

### The Riemann Tensor

Since $\nabla_{\lambda} g_{\mu \nu} = 0$, we try the second derivative. It turns out, there is only one unique tensor which can be constructed from the metric using second derivatives, and this is the Riemann Tensor.

$$R^{\mu}_{\nu \lambda \rho} = \partial_{\lambda} \Gamma_{\nu \rho}^{\mu} - \partial_{\rho} \Gamma_{\nu \lambda}^{\mu} + \Gamma_{\lambda \color{green}{\kappa}}^{\mu} \Gamma_{\nu \rho}^{\color{green}{\kappa}} - \Gamma_{\rho \color{green}{\kappa}}^{\mu} \Gamma_{\nu \lambda}^{\color{green}{\kappa}}$$

Contracting two indices, we can define the Ricci Tensor (by symmetry properties of the Riemann Tensor, this is also unique)

$$R_{\mu \nu} = R^{\color{green}{\kappa}}_{\mu \color{green}{\kappa} \nu}$$

and contracting the Ricci Tensor with the metric, we find the Ricci Scalar. (I'm sure this thing is also unique)

$$R = g^{\mu \nu} R_{\mu \nu}$$

Properties of the Riemann Tensor
• Symmetry: $R_{\color{red}{\mu \nu} \color{green}{\lambda \rho}} = R_{\color{green}{\lambda \rho} \color{red}{\mu \nu}}$
• Antisymmetry: $R_{\color{red}{\mu} \color{green}{\nu} \lambda \rho} = - R_{\color{green}{\nu} \color{red}{\mu} \lambda \rho}$ and same for the last two indices
• Cyclicity: $R^{\mu}_{\nu \lambda \rho} + R^{\mu}_{\lambda \rho \nu} + R^{\mu}_{\rho \nu \lambda} = 0$
• Bianchi Identity: $R^{\mu}_{\nu \lambda \rho; \kappa} + R^{\mu}_{\nu \rho \kappa; \lambda} + R^{\mu}_{\nu \kappa \lambda; \rho} = 0$

Where the semicolon $(.)_{;\nu} \equiv d_{\nu} (.)$ is the total derivative.

Cyclicity looks like the Jacobian Identity, is there something to that? The interpretation of the Bianchi Identity, that a boundary is necessarily a closed surface, is there some insight to be gained by thinking about this here?

## Lecture 2: Gravitational Field Equations

So far we investigated how objects behave under the influence of gravity. Now, how does the presence of objects change the gravitational field?

### The Energy Momentum Tensor

We start from the Newtonian approach, where we have the gravitational potential $\phi(x)$ given by the matter density $\rho(x)$ through the Poisson equation:

$$\Delta \phi(x) = 4 \pi G \rho(x)$$

Now we need a general invariant version of this. The solution is that the matter density enters the $T^{00}$ component of the Energy Momentum Tensor. For a nonrelativistic, pointlike body we can motivate the limit $\rho(x) \rightarrow T^{\mu \nu} = \delta^{\mu}_0 \delta^{\nu}_0 M \delta(\vec{x})$. The spatial components of the Energy Momentum Tensor are suppressed by the low velocity of the body.

This is more like a justification than a motivation. A better motivation would be good, but I cant think of any arguments right now. The poisson equation seems like a good starting point, but going from a scalar to a tensor surely needs some good motivation

We define the action of the gravitational field and its variation with respect to the metric

$$\delta_g S = - \frac{1}{2} \int d^4x \sqrt{-g} T^{\mu \nu}(x) \cdot \delta g_{\mu \nu}(x)$$

what? where does it come from, how does the action look like? why do we vary wrt the metric? … Things I should clarify. The volume elements $d^4 x$ is not invariant, but the 'scalar volume element' with the factor $\sqrt{-g}$ is. $g$ is here the determinant of the metric. For flat spacetime this is $-1$, thus the square root gives just the identity.

“Field equations are local, they don't feel topology.”

This whole section is very rushed, I should add and emphasis properly motivated starting points, comment on where we want to go with this, by what means we then get there and what the result tells us.

### Maxwell Theory

Ok now we want to construct the energy momentum tensor with an example. Maxwell theory is seems like a good idea? Well, the motivation for this section strongly depends on a solid explanation and motivation of the last section. So this needs some work.

$$S_M = - \frac{1}{4} \int d^4x \sqrt{-g} F_{\mu \nu} F^{\mu \nu}$$

some algebraic geometry notation remarks made
Which uses the one-form $A(x) = A_{\mu}(x) dx^{\mu}$. Also $dA(x) = \frac{1}{2} F_{\mu \nu} dx^{\mu} \wedge dx^{\nu}$ with the wedge-product which is antisymmetric. And the nilpotent operator $d^2 = 0$.

I need to read up about this formalism some more, Baez book does it.

The definition of the field strength tensor (in flat spacetime) is

$$F_{\mu \nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$$

we generalize this by the use of covariant derivatives

$$F_{\mu \nu} = \nabla_{\mu} A_{\nu} - \nabla_{\nu} A_{\mu} = \partial_{\mu} A_{\nu} - \Gamma_{\mu \nu}^{\rho} A_{\rho} - \partial_{\nu} A_{\mu} + \Gamma_{\nu \mu}^{\rho} A_{\rho}$$

because of the symmetry of the Christoffel Symbol in the lower indices, the terms cancel and the general invariant field strength tensor is the same as the flat spacetime field strength tensor. Thus, $F_{\mu \nu}$ has a trivial dependence on the metric.

Since the raising and lowering of indices is performed by the metric, the Maxwell Lagrangian contains the metric implicitly in the contraction of the two field strength tensors. Written explicitly, this is

$$S_M = - \frac{1}{4} \int d^4x \sqrt{-g} F_{\mu \nu} F_{\lambda \rho} g^{\lambda \mu} g^{\rho \nu}$$

If we now vary the Maxwell action wrt the metric, we obtain

$$\frac{\delta S_M}{\delta g_{\mu \nu}} = \int d^4x \frac{1}{4} \sqrt{-g} \left( \frac{1}{2} g^{\mu \nu} F_{\lambda \rho} F^{\lambda \rho} - 2 F^{\mu \lambda} F^{\nu}_{\lambda} \right)$$

Therefore, the Energy Momentum Tensor for Maxwell Theory is

$$\boxed{ T^{\mu \nu} = F^{\mu \lambda} F^{\nu}_{\lambda} -\frac{1}{4} g^{\mu \nu} F_{\lambda \rho} F^{\lambda \rho} }$$

Note that the conservation of this energy momentum tensor is fulfilled:

\begin{align} \textrm{special relativity: } & \partial_{\mu} T^{\mu \nu} = 0 \nonumber\\ \textrm{general relativity: } & \nabla_{\mu} T^{\mu \nu} = 0 \end{align}

the proof uses the equations of motion and covariance (the action is a scalar) to prove this. This actually is a nice and quick derivation of the Maxwell energy momentum tensor. Id just need to add some motivation and comments to clear up the steps

### Einstein Equations

Now we motivated the generally covariant form of the rhs of the Poisson equation what about the lhs?

$$? \quad \leftarrow \quad \Delta \phi = 4 \pi G \rho(x) \quad \rightarrow \quad \sim T^{\mu \nu}$$

Because on the rhs we have $\sim T^{\mu \nu}$ we should also have a rank 2 tensor of some sort, call it $\Pi^{\mu \nu}$. Since $T^{\mu \nu}$ is symmetric, $\Pi^{\mu \nu}$ should also be symmetric. And it should only depend on the metric (the geometry).

This is at least reasonable, we put all matter dependence in the energy momentum tensor and can examine the vacuum solutions where it is zero and only geometry remains

Since the Riemann tensor is the object which describes our geometry and we need a rank 2 tensor, this leaves the Ricci tensor as a good candidate. Now we also know, that the energy momentum tensor is generally conserved, i.e. $\nabla_{\mu} T^{\mu \nu} = 0$, therefore the lhs also needs to fulfill this requirement. We can also add the Ricci scalar with a metric for the tensor structure and fix their relative factor to make the combination generally conserved,

$$G^{\mu \nu} = R^{\mu \nu} - \frac{1}{2} g^{\mu \nu} R$$

this we call the Einstein Tensor.

Now we arrived at $G^{\mu \nu} = \kappa T^{\mu \nu}$ and what is left is to fix the proportionality constant $\kappa$. We do this by looking at the $G^{00} = \kappa T^{00}$ equation and work in the limit of weak gravity again. In this way we can recover the Poisson equation as the weak gravity limit which fixes $\kappa$.

details
Look at the $\mu = 0, \nu = 0$ equation.

$$R^{00} - \frac{1}{2} g^{00} R = \kappa T^{00}$$

We use the trace of the Einstein tensor to get rid of the Ricci scalar:

$$\textrm{Tr} G = g_{\mu \nu} G^{\mu \nu} = R - \frac{1}{2} 4 R = \kappa g_{\mu \nu} T^{\mu \nu}$$

This leads to $R = - \kappa T^{\mu}_{\mu}$. Inserting this into the $00$ equation, we get

$$R^{00} = \kappa (T^{00} - \frac{1}{2} g^{00} T^{\mu}_{\mu})$$

We expand in weak gravity $g^{\mu \nu} \approx \eta^{\mu \nu} + h^{\mu \nu}$ where at leading order, we can use $\eta^{00}$ instead of $g^{00}$ here. The Ricci tensor is for $|h_{\mu \nu}| \ll 1$

$$R_{\mu \nu} = \partial_{\lambda} \Gamma^{\lambda}_{\mu \nu} - \partial_{\nu} \Gamma^{\lambda}_{\mu \lambda} + \mathcal{O}(h^2)$$

thus

$$R_{00} \approx \partial_{i} \Gamma^{i}_{00} \approx \frac{1}{2} \eta^{ij} \partial_i \partial_j (-h_{00}) = \frac{1}{2} \Delta h_{00}$$

We can again substitute $h_{00} = 2 \phi$ and get

$$R_{00} = \Delta \phi$$

To get the rhs, we recall $T^{00} (x) = \rho(x)$

is this motivated somewhere? the section on the energy momentum tensor should really have some more comments on it and motivate its physical meaning well. Maybe there is even a quick derivation in terms of arguments possible. Also the trace is not mentioned here but is needed

Inserting this, we get from the $00$ equation

$$\Delta \phi = \frac{\kappa}{2} \rho(x)$$

and comparing with the Poisson equation gives us $\kappa = 8 \pi G$.

This gives us the Einstein Field Equations

$$\boxed{ R_{\mu \nu} - \frac{1}{2} g^{\mu \nu} R = \frac{8 \pi G}{c^4} T^{\mu \nu} }$$

which describes the gravitational interaction as geometry of spacetime resulting from a given energy-momentum distribution. Once the geometry is known, one can go back to the geodesic equation and describe the motion of particles in this spacetime. This of course assumes particles/objects which are sufficiently light to not have a significant impact on the geometry themselves.

Remarks:

• The Einstein Equations can be derived from the variational principle from the Einstein-Hilbert Action

$$S_{EH} = - \frac{1}{16 \pi} \int d^4x \sqrt{-g} R$$

• Varying $\frac{\delta S_{EH}}{\delta g_{\mu \nu}}$ generates the geometric part (the Einstein Tensor) of the Einstein equations, which are then sourceless Einstein equations. We can add a matter term to also generate an energy momentum tensor.

$$S_{tot} = S_{EH} + S_{matter}$$

• Pressure (as part of the energy momentum tensor) also influences the geometry of spacetime
• we can add a $-\Lambda g^{\mu \nu}$ term to the Einstein tensor, where $\lambda \sim x^{-4}$. Observations from far away objects comply with a nonzero but small $\Lambda$.

Lambda is the gravitational constant and describes a kind of an inflation of spacetime: every observer sees every (stationary) point accelerate away from him. This is afaik the observation of redshifts in distant galaxies, everything seems to drift away from us and the further away, the faster it drifts.

• The Dark Energy intepretation of $\Lambda$ comes from putting $\Lambda g^{\mu \nu}$ to the other side of the Einstein equations and absorbing it into the energy momentum tensor. Then it can be interpreted as an exotic contribution to $T^{\mu \nu}$ which has negative pressure:

A perfect fluid has an energy momentum tensor of the form $$T^{\mu \nu} = (\rho + p) u^{\mu} u^{\nu} - p g^{\mu \nu}$$

if now (for some absurd reason) $\rho = -p$, the pressure mimicks the cosmological constant.

This whole dark energy interpretation sounds increadibly stupid to me. The realization of a perfect fluid, where the local density always complies with the momentum sounds just silly to me. Especially since dark energy in order to explain the effect would need to make up like 70% of our universe. And not only that, as far as I know, simply resorting to de Sitter space instead of Minkowski space as the asymptotical space in the weak gravity limit works well and describes the scenario we see without introducing new weird states of energy that makes up most of our universe yet we have never seen any of it directly.

### Summary

We started by transforming the equations of motion in a general coordinate frame, which describes the motion of a particle in a given geometry. We assumed the general equivalence principle which means that we can always locally choose a frame (right?), in which spacetime is Minkowski. Then, trying to find a differential equation which describes the effect of matter on the geometry of spacetime, we tried to find objects, which keep their meaning under general coordinate transformations - which are generally invariant. From the Poisson equation (the Newtonian limit) we motivated the Einstein Field Equations which describe how the geometry of spacetime is affected by the presence of matter. I should clarify the invariant/covariant notions. Also the term 'generally', maybe 'diffeomorphism' is less misleading?

## Lecture 4: Gravitational Waves

heard/emelyanov.txt · Last modified: 2017/12/06 09:45 (external edit)