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$(1+1)$-dimensional general relativity

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The study of $(1+1)$-dimensional spacetimes is quite useful, as it is a mostly tractable toy model for many issues, though it has some shortcomings due to its simplicity. It can be used in particular in the study of string theory, where the world sheet can be considered as scalar fields on a $(1+1)$-dimensional spacetime.

1. Topology

From well-known topological theorems, the problem of classifying $2$-dimensional surfaces is much more tractable than in higher dimensions.

The classification of compact $2$-manifolds without boundaries is composed of either the sphere $S^2$, the connected sum of $g$ tori $T^2$, or the connected sum of $k$ projective planes $\mathbb{R}P^2$. As always with compact spacetimes, a compact manifold only admits a Lorentz metric if its Euler characteristic $\chi(M)$ is zero. As the connected sum of two manifolds obeys

\begin{equation} \chi(M_1 \# M_2) = \chi(M_1) + \chi(M_2) - 2 \end{equation}

and the sphere, torus and projective plane each respectively have Euler characteristic $2$, $0$ and $1$, there are only two possible compact spacetimes here : $T^2$, and $\mathbb{R}P^2 \# \mathbb{R}P^2$, which is simply the Klein bottle $K$.

For non-compact manifolds, we can borrow Richards' paper on the topic for this theorem :

For the compact case, this is equivalent to the previous definition, with the proper identifications of disk boundaries for adding another torus or projective plane. In the non-compact case, we need at least one point removed (to give us $\mathbb{R}^2$), and then this will be roughly equivalent to $\mathbb{R}^2$ minus a number of points and with a number of (possibly non-orientable) handles attached.

Of importance for string theory are the possible cobordisms. Any $(1+1)$-manifold with boundaries will have $1$-manifolds as boundaries, and there are therefore only two possible connected such manifolds : the line and the circle, therefore any boundary can be constructed as the disjoint union of a countable number of lines and circles (in string theory these would correspond to closed and open strings).

Minkowski space and Lorentz trigonometry

As per usual, a spacetime will locally have properties similar to Minkowski space, as the tangent plane will have the same structure. So let's consider some general properties of $(1+1)$-dimensional Minkowski space :

The usual definition applies. Minkowski space in two dimensions is a vector space $V$, that can be spanned from two basis vectors. Canonically we'll take them to be $\e_t$ and $\e_x$, so that any vector can be expressed as

\begin{equation} \forall v \in V,\ \exists (t, x) \in \mathbb{R}^2,\ v = t e_t + x e_x \end{equation}

It is equipped with a non-degenerate symmetric bilinear map $\eta$, such that, given the dual basis $\{ e^t, e^x \}$, it can be expressed as

\begin{equation} \eta = - e^t \otimes e^t + e^x \otimes e^x \end{equation}

and, as per usual, we can, given a basis, have some homeomorphism with$\mathbb{R}^2$ for vectors, while the metric will belong to the set of matrices $M_{2 \times 2}$, with

\begin{equation} \mathrm{\eta} = \operatorname{diag}(-1, 1) \end{equation}

Due to its low dimensionality, it's not terribly hard to compute the symmetry group here. Let's consider some linear transformation $v' = \Lambda v$, with $\Lambda \in \operatorname{GL}(2, \mathbb{R})$. In our basis, this is also a $2 \times 2$ matrix, so that we want

\begin{eqnarray} \eta(v', v') &=& \eta(\Lambda v, \Lambda v)\\ &=& v^\dagger \Lambda^\dagger \eta \Lambda v \end{eqnarray}

To preserve the norm of vectors here, we therefore need $\Lambda^\dagger \eta \Lambda = \eta$, or

\begin{eqnarray} \Lambda^\dagger \eta \Lambda &=& \begin{pmatrix} a & c \\ b & d \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix}\\ &=& \begin{pmatrix} c^2 - a^2 & cd - ab \\ cd - ab & d^2 - b^2 \end{pmatrix} \end{eqnarray} \begin{eqnarray} cd &=& ab\\ c^2 &=& a^2 - 1\\ d^2 &=& 1 + b^2 \end{eqnarray}

As $\sinh(u)$ is a bijection on $\mathbb{R}$, we can always consider $c$ to be $c = \sinh(u)$ for some $u$, for any value of $c$. Therefore, we have

\begin{equation} a^2 = c^2 + 1 = \sinh^2(u) + 1= \cosh^2(u) \end{equation}

Therefore, $a = \pm \cosh(u)$. The same reasoning can be used with $b = \sinh(v)$, in which case we have $d = \pm \cosh(v)$. And from $ab = cd$, we have that $u$ must be equal to $v$, and $a$ and $d$ must be of the same sign. This can be checked by considering all possible sign combinations and recasting it as $ab - cd = 0$, which gives us equalities of the type $\sinh(u \pm v) = 0$, only obeyed by $u = v$ for the minus case, and by $u = v = 0$ otherwise, which is simply the identity matrix. Therefore, any Lorentz transform can be constructed from the discrete transformation $-I$ and the one-parameter transformation

\begin{equation} \Lambda(u) = \begin{pmatrix} \cosh(u) & \sinh(u) \\ \sinh(u) & \cosh(u) \end{pmatrix} \end{equation}

The

Tensor quantities

The two dimensional case being the lowest dimension in which we can actually define a spacetime, things are somewhat simplified and we can write out tensor quantities quite naturally.

The determinant of the metric tensor is, easily enough, with coordinates $(t,x)$,

\begin{equation} \det(g) = g_{tt} g_{xx} - g_{xt}^2 \end{equation}

The inverse metric is

\begin{equation} g^{-1} = \frac{1}{\det(g)} \begin{pmatrix} g_{xx} & -g_{xt} \\ -g_{xt} & g_{tt} \end{pmatrix} \end{equation}

There are only six Christoffel symbols to consider, which are

\begin{eqnarray} {\Gamma^t}_{tt} &=& g^{tt} g_{tt,t} + g^{tx} (2g_{tx,t} - g_{tt,x})\\ {\Gamma^t}_{xt} &=& g^{tt}g_{tt,x} + g^{tx} (g_{xx,t} + g_{xt,x} - g_{})\\ {\Gamma^t}_{xx} &=& g^{\sigma\tau} (g_{} + g_{} - g_{})\\ {\Gamma^x}_{tt} &=& g^{\sigma\tau} (g_{} + g_{} - g_{})\\ {\Gamma^x}_{xt} &=& g^{\sigma\tau} (g_{} + g_{} - g_{})\\ {\Gamma^x}_{xx} &=& g^{\sigma\tau} (g_{} + g_{} - g_{})\\ \end{eqnarray}

The Riemann tensor, due to its symmetries, is greatly simplified. As there are only two possible indices, there are very few combination of indices that will not be zero due to the antisymmetry of may of them. In fact, thanks to the usual formula for the number of independant components of the Riemann tensor,

\begin{equation} C_n = \frac{1}{12} n^2 (n^2-1) \end{equation}

we can see that in two dimensions, there are only one. The Riemann tensor can thus be expressed as

\begin{equation} R_{abcd} = C \varepsilon_{ab} \varepsilon_{cd} = C (g_{ad}g_{bc} - g_{ac} g_{bd}) \end{equation}

And the only possible non-zero components are the ones of different $a$ and $b$ as well as $c$ and $d$, or, in full, $(txtx)$, $(txxt)$, $(xttx)$ and $(xtxt)$. As these are all connected, it would be of some interest to compute a single one :

\begin{eqnarray} {R^t}_{xtx} &=& \end{eqnarray}

The Gauss-Bonnet theorem

An important fact about $(1+1)$-dimensional spacetimes stems from the Gauss-Bonnet theorem. Very roughly, the Einstein-Hilbert action is a constant :

\begin{equation} S[g] = \int_{M} R d\mu[g] = \text{cte} \end{equation}

meaning that $\delta S = 0$ for any value of $g$, therefore any metric is admissible as a solution to the Einstein equations. This is much tougher to prove as is usually implied, although a fairly simple proof stems from the fact that the number of components of the Riemann tensor is

\begin{equation} C_n = \frac{1}{12} n^2 (n^2-1) \end{equation}

or simply $1$ in $2$ dimensions. The Riemann tensor can therefore simply be expressed, due to its symmetries, by

\begin{equation} R_{abcd} = C \varepsilon_{ab} \varepsilon_{cd} = C (g_{ad}g_{bc} - g_{ac} g_{bd}) \end{equation}

and therefore

\begin{eqnarray} R_{bd} &=& R_{abcd} g^{ac} = -C g_{bd}\\ R &=& -2C \end{eqnarray}

Therefore, $C = -R/2$ and $R_{\mu\nu} = R/2 g_{\mu\nu}$, so that no matter the metric,

\begin{equation} R_{\mu\nu} - \frac{R}{2} g_{\mu\nu} = 0 \end{equation}

Any metric will therefore obey the EFE as long as the stress-energy tensor is identically zero, while no metric will if it does not. With basic general relativity, there is no strong link between the metric and the matter content. To do a more proper proof, the one usually hinted at in most GR papers, we'll need to prove the Gauss-Bonnet theorem in those circumstances, which is fortunately done in [1] and [2]. Here's the proof redone here.

First, we need to define some geometric elements for our manifold, which is not usually done for Lorentzian manifolds. In particular we'll need to define angles, which are tricky here (we strictly speaking don't need to define angles but it will be nice to have the full version of the Gauss-Bonnet theorem for spacetimes).

As we are in two dimensions, for any unit timelike vector $T$, we can define a unique unit spacelike vector $S$ such that both vectors are orthogonal and together form a positively oriented basis at that point. In other words,

\begin{eqnarray} g(T,T) &=& -1\\ g(S,T) &=& 0\\ g(S,S) &=& 1\\ T \wedge S &=& 1 \end{eqnarray}

We define at a point the basis $\{ E, E^\perp \}$, such that $E$ and $E^\perp$ are a timelike and spacelike vector obeying such properties. Such a basis is called an allowable basis. Let $X$ and $Y$ be two unit timelike vectors with the same time orientation, and components $(x_1, x_2)$ and $(y_1, y_2)$ with respect to some allowable basis. The angle between two such vectors is then the number $u$ such that

\begin{equation} \begin{pmatrix} \cosh(u) & \sinh(u) \\ \sinh(u) & \cosh(u) \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} y_1\\ y_2 \end{pmatrix} \end{equation}

This number is independent of the orthonormal basis used. We'll note $(X,Y) = U$, and we have

\begin{eqnarray} (Y,X) &=& -(X,Y) (-X, -Y) &=& (X, Y) \end{eqnarray}

No boost will switch the time-orientation of a vector, so that with this definition, $(X, -Y)$ or $(-X, Y)$ have no angles.

Weyl transform and conformal symmetry

Lorentz surfaces

Lorentz geometry

A few important examples

Due to its simplicity, the solution space for spacetimes in $(1+1)$ dimensions is much "smaller" in some sense. While still very much infinite, the notable solutions tend to be on a much shorter list, as a lot of possible spacetimes are simply not possible. In particular :

• Any globally hyperbolic spacetime either has the topology $\mathbb{R} \times \mathbb{R}$ or $\mathbb{R} \times S^1$
• Although one can attach a handle to our spacetime, there can be no wormhole on every spacelike hypersurface, as the topology would simply be the same (although there is a wormhole toy model that can be used).
• Any ultrastatic spacetime, being of the form $-dt^2 + f(x) dx^2$, can be recast as (locally) Minkowski space, by a coordinate change for $x$.
• Any static spacetime $-f(x) dt^2 + g(x) dx^2$

Minkowski space

The Minkowski torus

Misner space

Non-time orientable Minkowski cylinder

de Sitter space

The Clifton-Pohl torus

The Clifton-Pohl torus is an interesting example as it is the most well-known case of a geodesically incomplete compact spacetime. While for Riemannian spaces compact manifolds are always geodesically complete, this is not necessarily the case for Lorentzian ones.

The trouser spacetime

This is the simplest "classic" Lorentz cobordism in $(1+1)$ dimensions (by which I mean one where everything is compact), which is the cobordism between $S^1$ and $S^1 \sqcup S^1$ (or vice versa). Topologically it is equivalent to $(S^1 \times \mathbb{R}) \setminus D^2$ (or a sphere minus three disks, equivalently). Due to well-known theorems on cobordisms, any Lorentz metric on such a spacetime will either fail to be causal or time-oriented.

Spinors

Spinors are a tad tricky to do in $(1+1)$ dimensions due to the fact that the usual simple definition of the spin group as the double cover of the Lorentz group does not work too well here : the connected component of the Lorentz group is simply $\mathbb{R}$ here, which is its own cover. Therefore we'll have to muddle with Clifford algebras for a bit here.

As we can recall, the Clifford algebra is constructed from the vector space of multivectors,

\begin{equation} \bigwedge V = \bigoplus_{p = 0}^{n} \bigwedge^p V \end{equation}

which is, in two dimensions, the direct sum of $0$-forms, $1$-forms and $2$-forms, or

\begin{equation} A = f_0 + f_1 dx + f_2 dt + f_3 dx \wedge dt \end{equation}

and the Clifford product must obey, for $v \in V$, $v^2 = g(v,v)$, and we also have the equality $uv = g(u,v) + u \wedge v$. If we take the bilinear product to be Minkowski space, this leads to $dxdt = dx \wedge dt$. Thanks to this and the associativity of the Clifford product, this leads to

\begin{eqnarray} AB &=& (f_0 + f_1 dx + f_2 dt + f_3 dx \wedge dt) (g_0 + g_1 dx + g_2 dt + g_3 dx \wedge dt)\\ &=& f_0 g_0 + f_0 g_1 dx + f_0 g_2 dt + f_0 g_3 dx \wedge dt + f_1 g_0 dx + f_1 g_1 + g_2 f_1 dx \wedge dt + g_3 f_1 dt + f_2 g_0 dt - g_1 f_2 dx \wedge dt - f_2 g_2 + g_3 f_2 dx + g_0f_3 dx dt - g_1 f_3 dt - g_2 f_3 dx + g_3 f_3 \end{eqnarray}

Supermanifolds

Field theory in $(1+1)$ dimensions

One big attraction of $(1+1)$ dimensional spacetimes are that quite a lot of field theories are tractable on them, even non-linear ones such as the sine-Gordon model, by using such things as the inverse scattering transform.

Klein-Gordon

The Klein-Gordon scalar field obeys the usual equation of motion in a curved spacetime :

\begin{equation} \Box \varphi = 0 \end{equation}

which can be expressed as

\begin{eqnarray} \Box \varphi &=& \frac{1}{\sqrt{g}} \partial_\mu (g^{\mu\nu} \sqrt{g} \partial_\nu \varphi)\\ &=& g^{\mu\nu}_{,\mu} \partial_\nu \varphi + g^{\mu\nu} \partial_\mu \partial_\nu \varphi + \frac{\sqrt{g}_{,\mu}}{\sqrt{g}} g^{\mu\nu} \partial_\nu \varphi \end{eqnarray}

The usual equation for Minkowski space is then of course the wave equation.

\begin{equation} \partial_x^2 \varphi - \partial_t^2 \varphi = 0 \end{equation}

Using the null coordinates $u, v = x \pm t$, with metric $ds^2 = -dudv$, this is

\begin{equation} \partial_u \partial_v \varphi = 0 \end{equation}

Hence $\partial_v \varphi$ is independant of $u$ and is simply a function of $v$,

\begin{equation} \partial_v \varphi = f(v) \end{equation}

We can then simply integrate over $v$, in which case $f$ will become its antiderivative $F$ and we'll get a constant (in $v$) of integration, that may depend on $u$.

\begin{equation} \varphi = F(v) + G(u) \end{equation}

switching coordinates back, this is the usual solution of the wave equation

\begin{equation} \varphi = F(x + t) + G(x - t) \end{equation}

Dirac field

Maxwell field

Sine-Gordon

\begin{equation} \mathcal{L} = g^{\mu\nu} \nabla_\mu \varphi \nabla_\nu \varphi + m^2 \varphi^2 + g \cos(\varphi) \end{equation}

Gross-Neuveu model

Sigma model

Bibliography

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Last updated : 2019-07-04 08:33:44