Background
Following last time, we are almost ready to write down the Einstein equations. Before doing any math, let's understand what we're trying to do. Minkowski realized that Einstein's special relativity was best understood by combining space and time into 4-dimensional spacetime, with Lorentzian metric
\[ ds^2 = -dt^2 + dx^2 + dy^2 + dz^2. \]
The spacetime approach works wonderfully and even explains the Lorentz invariance of Maxwell's equations (indeed, it was Maxwell's equations that motivated Einstein to postulate his principle of relativity). However, (for reasons that I may discuss later) gravity is not a "force" but rather the geometry of spacetime itself.
By mass-energy equivalence (which is one of the most basic consequences of relativity), the gravitational field, whatever it is, must couple to the stress-energy tensor \(T_{ij}\). I won't get into details, but the stress-energy tensor is a familiar object from physics that roughly tells you what the energy-momentum density/flux is in each direction at every point in spacetime. If the matter is completely static, then it is ok to think of this as measuring the mass density, but for nonstatic matter it also takes things like pressure into account.
Now, as I said above, the gravitational field is just the geometry of spacetime, which is measured by the metric tensor \(g_{ij}\). Mass-energy equivalence says that it must couple to the stress-energy tensor \(T_{ij}\). The simplest field equation then would be
\[ G_{ij} = c T_{ij} \]
where \(G_{ij}\) is some tensor built out of \(g_{ij}\) and its derivatives, and \(c\) is some constant. The equations of Newtonian gravity are 2nd order in the gravitational field, so if we want these equations to reduce to Newton's in the appropriate limit, \(G_{ij}\) should only depend on the metric and its first two derivatives. Now there is an obvious 2nd rank tensor satisfying these constraints: \(R_{ij}\), the Ricci tensor. However, this turns out to be completely wrong (except in the vacuum).
Any reasonable matter will satisfy local energy-momentum conservation,
\[ \nabla_j T^{ij} = 0. \]
It turns out that the Ricci tensor does not satisfy this condition in general. So to look for the right tensor \(G_{ij}\), we turn to the Bianchi identity.
The Bianchi Identity
As discussed in the previous post, the curvature of a connection is the endomorphism-valued 2-form
\[ F = d\Omega - \Omega \wedge \Omega \]
where \(\Omega\) is the matrix of 1-forms telling us how to take the covariant derivative of a frame, i.e.
\[ \nabla_i e_j = \Omega_{ij} \otimes e_j. \]
Since a connection can be extended to all tensor powers in a natural way, we can consider the covariant derivative of the curvature \(F\) (thought of as a section of the appropriate bundle). Quick calcluation:
\begin{align}
\nabla F &= \nabla(d\Omega - \Omega \wedge \Omega) \\
&= d^2 \Omega - d\Omega \wedge \Omega + \Omega \wedge d\Omega \\
& \ \ + d\Omega \wedge \Omega - \Omega \wedge \Omega \wedge \Omega \\
& \ \ - \Omega \wedge d\Omega + \Omega \wedge \Omega \wedge \Omega \\
&= 0.
\end{align}
Thus the endomorphism valued 3-form \(\nabla F\) is identically 0. Writing \(F\) in components as \(R_{ijkl}\), this is equivalent to
\[ R_{ijkl|m} + R_{ijlm|k} + R_{ijmk|l} = 0. \]
Now let's contract:
\begin{align}
0 &= g^{ik} g^{jl} R_{ijkl|m} + g^{ik} g^{jl} R_{ijlm|k} + g^{ik} g^{jl} R_{ijmk|l}\\
&= g^{ik} R_{ik|m} - g^{ik}R_{im|k} - g^{jl} R_{jm|l} \\
&= \nabla_m S - 2 \nabla^k R_{mk}
\end{align}
So we see that the tensor
\[ G_{ij} = R_{ij} - \frac{S}{2} g_{ij} \]
is divergence free. This yields the Einstein field equations:
\[ R_{ij} - \frac{S}{2} g_{ij} = c T_{ij}. \]
Actually, there is another obvious divergence free tensor: \(g_{ij}\) itself. So a more general form is
\[ G_{ij} + \Lambda g_{ij} = c T_{ij} \]
where \(\Lambda\) is a constant called the cosmological constant.
Saturday, February 25, 2012
Monday, February 6, 2012
Geometry of Curved Spacetime 4
Today I had to try to explain connections and curvature in local frames (as opposed to coordinates), and I really feel that Wald's treatment of this is just awful (this is one of the few complaints I have with an otherwise classic textbook). It is particularly baffling since the treatment in Misner, Thorne, and Wheeler is just perfect. What follows is the modern math (as opposed to physics) point of view. This is more abstract than any introductory GR (or even Riemannian geometry) text I've seen, but in this case the abstraction absolutely clarifies and simplifies things.
Let \(M\) be a smooth manifold and suppose \(E\) is a smooth vector bundle over \(M\). A connection on \(E\) is a map \nabla taking sections of \(E\) to sections of \(T^\ast M \otimes E\), \(\mathbb{R}\)-linear and satisfying the Leibniz rule
\[ \nabla(f\sigma) = df \otimes \sigma + f \nabla \sigma. \]
Now consider the sheaf of \(E\)-valued \(p\)-forms on \(M\). Call it \(\Omega^p(E)\). Then we can extend the connection to a map
\[ \nabla: \Omega^p(E) \to \Omega^{p+1}(E) \]
via the Leibniz rule:
\[ \nabla(\eta \otimes \sigma) = d\eta \otimes \sigma + (-1)^p \eta \wedge \nabla \sigma. \]
Let us define the curvature \(F\) associated to a connection \(\nabla\) by the composition
\[ F = \nabla^2: \Omega^p(E) \to \Omega^{p+2}(E). \]
Claim \(F\) is \(C^\infty\)-linear, i.e. it is tensorial.
Proof
\begin{align}
\nabla(\nabla(f \sigma)) &= \nabla( df \otimes \sigma + f \nabla \sigma) \\\
&= d^2 f \otimes \sigma - df \wedge \nabla \sigma + df \wedge \nabla \sigma + f \nabla^2 \sigma \\\
&= f \nabla^2 \sigma.
\end{align}
So far we have not made any additional choices (beyond \(\nabla\)). In order to actually compute something locally, we have to make some choices. Let \(\hat{e}_a\) be a frame, i.e. a local basis of sections of \(E\). Then \(\nabla \hat{e}_a\) is an \(E\)-valued 1-form, hence it can be expressed as a sum
\[ \nabla \hat{e}_a = \sum_{b} \omega_a^b \otimes \hat{e}_b \]
where the coefficients \(\omega_a^b\) are 1-forms, often called the connection 1-forms. Let \(\Omega\) denote the matrix of 1-forms whose entries are exactly \(\omega_a^b\).
Claim Let \(\sigma = \sigma^a \hat{e}_a\). Then we have
\[ \nabla \sigma = d\sigma + \Omega \sigma. \]
Proof The coefficients \(\sigma^a\) are functions (i.e. scalars), so \(\nabla \sigma^a = d\sigma^a\). Using the Leibniz rule we have
\begin{align}
\nabla(\sigma^a \hat{e}_a) &= (\nabla \sigma^a) \hat{e}_a + \sigma^a \nabla \hat{e}_a \\\
&= d\sigma^a \hat{e}_a + \sigma^a \omega_a^b \hat{e}_b \\\
&= d\sigma^a \hat{e}_a + \omega_c^a \sigma^c \hat{e}_a \\\
&= (d\sigma + \Omega \sigma)^a \hat{e}_a.
\end{align}
Claim The curvature satisfies \(F = d\Omega - \Omega \wedge \Omega\).
Proof Just apply the above formula twice using Leibniz.
Connection 1-forms from Christoffel symbols. Suppose now that we are in the Riemannian setting and we already know the Christoffel symbols in some coordinates. Then we can express our frame \(\hat{e}_a\) in terms of coordinate vector fields, i.e.
\[ \hat{e}_a = \hat{e}_a^i \frac{\partial}{\partial x^i} \]
Then we have that
\[ \nabla_j \hat{e}_a^i = \frac{\partial \hat{e}_a^i}{\partial x^j} + \Gamma^i_{jk} \hat{e}_a^k \]
So, as a vector-valued 1-form, we have
\[ \nabla \hat{e} = \frac{\partial \hat{e}_a^i}{\partial x^j} dx^j \otimes \frac{\partial}{\partial x^i}
+ \Gamma^i_{jk} \hat{e}_a^k dx^j \otimes \frac{\partial}{\partial x^i}. \]
Juggling things a bit using the metric, we find
\[ \nabla \hat{e}_a = \frac{\partial \hat{e}_a^i}{\partial x^j} \hat{e}^b_i dx^j \otimes \hat{e}_b
+ \Gamma^i_{jk} \hat{e}_a^k \hat{e}_i^b dx^j \otimes \hat{e}_b. \]
So the connection 1-forms are given by
\[ \omega_a^b = \frac{\partial \hat{e}_a^i}{\partial x^j} \hat{e}^b_i dx^j
+ \Gamma^i_{jk} \hat{e}_a^k \hat{e}_i^b dx^j. \]
To come later (if I ever get around to it): some explicit computations.
Let \(M\) be a smooth manifold and suppose \(E\) is a smooth vector bundle over \(M\). A connection on \(E\) is a map \nabla taking sections of \(E\) to sections of \(T^\ast M \otimes E\), \(\mathbb{R}\)-linear and satisfying the Leibniz rule
\[ \nabla(f\sigma) = df \otimes \sigma + f \nabla \sigma. \]
Now consider the sheaf of \(E\)-valued \(p\)-forms on \(M\). Call it \(\Omega^p(E)\). Then we can extend the connection to a map
\[ \nabla: \Omega^p(E) \to \Omega^{p+1}(E) \]
via the Leibniz rule:
\[ \nabla(\eta \otimes \sigma) = d\eta \otimes \sigma + (-1)^p \eta \wedge \nabla \sigma. \]
Let us define the curvature \(F\) associated to a connection \(\nabla\) by the composition
\[ F = \nabla^2: \Omega^p(E) \to \Omega^{p+2}(E). \]
Claim \(F\) is \(C^\infty\)-linear, i.e. it is tensorial.
Proof
\begin{align}
\nabla(\nabla(f \sigma)) &= \nabla( df \otimes \sigma + f \nabla \sigma) \\\
&= d^2 f \otimes \sigma - df \wedge \nabla \sigma + df \wedge \nabla \sigma + f \nabla^2 \sigma \\\
&= f \nabla^2 \sigma.
\end{align}
So far we have not made any additional choices (beyond \(\nabla\)). In order to actually compute something locally, we have to make some choices. Let \(\hat{e}_a\) be a frame, i.e. a local basis of sections of \(E\). Then \(\nabla \hat{e}_a\) is an \(E\)-valued 1-form, hence it can be expressed as a sum
\[ \nabla \hat{e}_a = \sum_{b} \omega_a^b \otimes \hat{e}_b \]
where the coefficients \(\omega_a^b\) are 1-forms, often called the connection 1-forms. Let \(\Omega\) denote the matrix of 1-forms whose entries are exactly \(\omega_a^b\).
Claim Let \(\sigma = \sigma^a \hat{e}_a\). Then we have
\[ \nabla \sigma = d\sigma + \Omega \sigma. \]
Proof The coefficients \(\sigma^a\) are functions (i.e. scalars), so \(\nabla \sigma^a = d\sigma^a\). Using the Leibniz rule we have
\begin{align}
\nabla(\sigma^a \hat{e}_a) &= (\nabla \sigma^a) \hat{e}_a + \sigma^a \nabla \hat{e}_a \\\
&= d\sigma^a \hat{e}_a + \sigma^a \omega_a^b \hat{e}_b \\\
&= d\sigma^a \hat{e}_a + \omega_c^a \sigma^c \hat{e}_a \\\
&= (d\sigma + \Omega \sigma)^a \hat{e}_a.
\end{align}
Claim The curvature satisfies \(F = d\Omega - \Omega \wedge \Omega\).
Proof Just apply the above formula twice using Leibniz.
Connection 1-forms from Christoffel symbols. Suppose now that we are in the Riemannian setting and we already know the Christoffel symbols in some coordinates. Then we can express our frame \(\hat{e}_a\) in terms of coordinate vector fields, i.e.
\[ \hat{e}_a = \hat{e}_a^i \frac{\partial}{\partial x^i} \]
Then we have that
\[ \nabla_j \hat{e}_a^i = \frac{\partial \hat{e}_a^i}{\partial x^j} + \Gamma^i_{jk} \hat{e}_a^k \]
So, as a vector-valued 1-form, we have
\[ \nabla \hat{e} = \frac{\partial \hat{e}_a^i}{\partial x^j} dx^j \otimes \frac{\partial}{\partial x^i}
+ \Gamma^i_{jk} \hat{e}_a^k dx^j \otimes \frac{\partial}{\partial x^i}. \]
Juggling things a bit using the metric, we find
\[ \nabla \hat{e}_a = \frac{\partial \hat{e}_a^i}{\partial x^j} \hat{e}^b_i dx^j \otimes \hat{e}_b
+ \Gamma^i_{jk} \hat{e}_a^k \hat{e}_i^b dx^j \otimes \hat{e}_b. \]
So the connection 1-forms are given by
\[ \omega_a^b = \frac{\partial \hat{e}_a^i}{\partial x^j} \hat{e}^b_i dx^j
+ \Gamma^i_{jk} \hat{e}_a^k \hat{e}_i^b dx^j. \]
To come later (if I ever get around to it): some explicit computations.
Saturday, February 4, 2012
Geometry of Curved Spacetime 3
Today, some numerology. The Riemann curvature tensor is a tensor \(R_{abcd}\) satisfying the identities:
1. \(R_{abcd} = -R_{bacd}.\)
2. \(R_{abcd} = R_{cdba}. \)
3. \(R_{abcd} + R_{acdb} + R_{adbc} = 0. \) (First Bianchi)
4. \(R_{abcd|e} + R_{acec|d} + R_{abde|c} = 0. \) (Second Bianchi)
By 1, the number of independent \(ab\) indices is \(N = n(n-1)/2\), and similarly for \(cd\). By 2, the number of independent pairs of indices is \(N(N+1)/2\). Now the cyclic constraint 3 can be written as
\[ R_{[abcd]} = 0, \]
and thus constitutes \({n \choose 4}\) equations. So the number of independent components is
\begin{align}
N(N+1)/2 - {n \choose 4} &= \frac{n(n-1)((n(n-1)/2+1)}{4} - \frac{n(n-1)(n-2)(n-3)}{24} \\
&= \frac{(n^2-n)(n^2-n+2)}{8} - \frac{(n^2-n)(n^2-5n+6}{24} \\
&= \frac{n^4-2n^3+3n^2+2n}{8} - \frac{n^4-6n^3+11n^2-6n}{24} \\
&= \frac{2n^4-2n^2}{24} \\
&= \frac{n^4-n^2}{12} \\
&= \frac{n^2(n^2-1)}{12}
\end{align}
Now consider the Weyl tensor \(C_{abcd}\) which is defined as the completely trace-free part of the Rienmann tensor. The trace is determined by the Ricci tensor \(R_{ab}\) which as \(n(n+1)/2\) indepdendent components, so the Weyl tensor has
\[ \frac{n^2(n^2-1)}{12} - \frac{n^2-n}{2} = \frac{n^4-7n^2+6n}{12} \]
independent components. Now, for \(n = 1\) we see that \(R_{abcd}\) has no independent components, i.e. it vanishes identically. In \(n=2\), it has only 1 independent component, and so the scalar curvature determines everything. In \(n=3\), it has 6 independent components. Note that in this case, the Weyl tensor has no independent components, i.e. it is identically 0. So we see that in \(n = 2, 3\) every Riemannian manifold is conformally flat. So things only start to get really interesting in \(n=4\), where the Riemann tensor has 20 independent components, and the Weyl tensor has 10.
1. \(R_{abcd} = -R_{bacd}.\)
2. \(R_{abcd} = R_{cdba}. \)
3. \(R_{abcd} + R_{acdb} + R_{adbc} = 0. \) (First Bianchi)
4. \(R_{abcd|e} + R_{acec|d} + R_{abde|c} = 0. \) (Second Bianchi)
By 1, the number of independent \(ab\) indices is \(N = n(n-1)/2\), and similarly for \(cd\). By 2, the number of independent pairs of indices is \(N(N+1)/2\). Now the cyclic constraint 3 can be written as
\[ R_{[abcd]} = 0, \]
and thus constitutes \({n \choose 4}\) equations. So the number of independent components is
\begin{align}
N(N+1)/2 - {n \choose 4} &= \frac{n(n-1)((n(n-1)/2+1)}{4} - \frac{n(n-1)(n-2)(n-3)}{24} \\
&= \frac{(n^2-n)(n^2-n+2)}{8} - \frac{(n^2-n)(n^2-5n+6}{24} \\
&= \frac{n^4-2n^3+3n^2+2n}{8} - \frac{n^4-6n^3+11n^2-6n}{24} \\
&= \frac{2n^4-2n^2}{24} \\
&= \frac{n^4-n^2}{12} \\
&= \frac{n^2(n^2-1)}{12}
\end{align}
Now consider the Weyl tensor \(C_{abcd}\) which is defined as the completely trace-free part of the Rienmann tensor. The trace is determined by the Ricci tensor \(R_{ab}\) which as \(n(n+1)/2\) indepdendent components, so the Weyl tensor has
\[ \frac{n^2(n^2-1)}{12} - \frac{n^2-n}{2} = \frac{n^4-7n^2+6n}{12} \]
independent components. Now, for \(n = 1\) we see that \(R_{abcd}\) has no independent components, i.e. it vanishes identically. In \(n=2\), it has only 1 independent component, and so the scalar curvature determines everything. In \(n=3\), it has 6 independent components. Note that in this case, the Weyl tensor has no independent components, i.e. it is identically 0. So we see that in \(n = 2, 3\) every Riemannian manifold is conformally flat. So things only start to get really interesting in \(n=4\), where the Riemann tensor has 20 independent components, and the Weyl tensor has 10.
Path Integrals 3: Recovering the Spectrum from Asymoptotics
In my previous posts on path integrals, I described (rather tersely) how the path integral, suitably defined and interpreted, can be used to compute the Schwartz kernel of the operators \(e^{iHt}\) (Lorentzian signature) and \(e^{-Ht}\) (Euclidean signature).
Suppose that we understand the spectrum of \(H\) completely (nb: for a given system described by \(H\), this is the goal). For example, suppose we know that the spectrum of \(H\) consists of discrete eigenvalues \(E_n, n = 0, \cdots\) with corresponding eigenvectors \(|n\rangle\),
\[ H|n\rangle = E_n|n\rangle. \]
(For simplicity, I assume there is no continuous spectrum and that the eigenvalues are nondegenerate.) Then we have
\[ e^{-iHt} = \sum_n e^{-i E_n t} \langle n|n\rangle \]
and
\[ e^{-Ht} = \sum_n e^{-E_n t} \langle n|n\rangle \]
Now the second expression turns out to be very useful. Assume the eigenvalues are ordered so that
\[ E_0 < E_1 < \cdots \]
Then we can write
\[ e^{-Ht} = e^{-E_0 t} |0\rangle + \sum_{n \geq 1} e^{-(E_n-E_0)t}|n\rangle \]
Now suppose that \(v\) is some vector which is close to the ground state, in the sense that
\[ \langle v|0\rangle \neq 0 \]
(This is obviously a generic condition, so if we just pick \(v\) randomly we can expect this to be true.) Then we can consider
\[ e^{-Ht} v = e^{-E_0 t} v_0 |0\rangle + \sum_{n \geq 1} e^{-(E_n-E_0)t} v_n|n\rangle \]
Now for \(n \geq 1\), \(E_n-E_0\) is strictly positive, and so for large \(t\) all of the higher terms are exponentially damped. So, we have the asymptotic
\[ e^{-Ht} v \sim e^{-E_0 t }v_0|0\rangle \]
Next comes the really interesting part. Multiply on the right by a position-representation eigenbra \(\langle x|\):
\[ \langle x|e^{-Ht} v \sim e^{-E_0 t} v_0 \langle x|0\rangle \]
Now \(v_0\) is an irrelevant constant, so we might as well take it to be 1 (rescale \(v\) as necessary). The expression \(\langle x|0\rangle\) is exactly the ground state wavefunction in the position representation! Call it \(\psi_0(x)\). So to conclude: the large-t asymptotic of the expression \(\langle x|e^{-Ht}v\) is (up to an overall constant) given by \(e^{-E_0 t} \psi_0(x)\), hence we can recover both the ground state energy and the ground state wavefunction. But the value of this expression is exactly given by the Euclidean path integral. So we have a correspondence:
Asymptotics of Euclidean path integral \(\leftarrow\rightarrow\) The spectrum of \(H\).
Coming next: instantons.
Path Integrals 2: Euclidean Path Integrals and Heat Kernels
Consider the heat equation
\[ \frac{\partial \psi}{\partial t} = -\hat{H} \psi \]
We find similarly
\[ \langle x_N|U_t|x_0\rangle = \int \exp \sum_{j=0}^{N-1} ik_j(x_j - x_{j+1}) -\Delta t H(x_j, k_j) dx dk. \]
Now take \(H(x,k) = k^2/2m + V(x)\) and complete the square:
\[ ik_j(x_j - x_{j+1}) - \Delta t k_j^2/2m - \Delta tV(x_j) + ik_{j+1}(x_{j+1} - x_{j+2}) - k_{j+1}^2/2m - V(x_{j+1}) \]
\begin{align}
ik_j a_j - \Delta t k_j^2/2m &= -(-2mi k_j a_j / \Delta t + k_j^2) \Delta t/2m \\
&= -(k_j^2 -2mi k_j a_j/\Delta t -m^2 a_j^2/(\Delta t)^2 + m^2 a_j^2/(\Delta t)^2) \Delta t/2m \\
&= -(k_j -mia_j/\Delta t)^2 \Delta t/2m -ma_j^2/2\Delta t
\end{align}
Combining things together, we have that the heat kernel is given by
\[ \langle y|e^{-t \hat{H}}|x\rangle = \int e^{-S_{\textrm{euc}}} \mathcal{D}x \]
That is, Schwartz kernel of time evolution operator is given by the oscialltory Lorentzian signature path integral, whereas the heat kernel is given by the exponentially decaying path integral (better chance of being well-defined). Most importantly, the heat kernel contains most of the essential information about the spectrum of \(\hat{H}\), which is really all we need in order to understand the dynamics.
See ABC of Instantons. (I never understood the title of Nekrasov's "ABCD of Instantons" until I found this classic).
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