The Kodama vector field and the gravitational red-shift

I was reading a survey article by Jaramillo and Gourgoulhon on mass and angular momentum when I come across the notion of the Kodama vector field. (Actually, I first heard of it from Mihalis Dafermos last summer; I've just completely forgotten about it until now.) I took notice of it because of a discussion yesterday in a seminar on the red/blue-shift effect. Since this post will focus mostly on the use of the Kodama vector field in analyzing gravitational red/blue-shift, I will start by an overview of the effect as well as some philosophical caveats.

Gravitational red/blue-shift

It is already well-known in folklore that a strong gravitational well causes light emitted to lengthen in wave-length. The standard, classical (not involving general relativity too heavily) explanation goes something like this: we know that the total energy is conserved in the motion of classical mechanics. We now take a hybrid quantum-mechanical/Newtonian-corpuscular view that light is carried by particles called photons, and that the intrinsic kinetic energy of the photon is $E_{\text{kinetic}} = \hbar \omega$, where $\hbar$ is Planck's constant, and $\omega$ is the angular frequency of the radiation. Suppose a photon is "produced" inside a gravitational well (say, near a very heavy star, or near a black hole), its total energy will be $E_{\text{total}} = E_{\text{kinetic}} + E_{\text{potential}}$. Now the photon escapes the gravitational well, to a point where the gravitational potential energy is higher. Then since total energy is conserved, the kinetic energy must correspondingly decrease. Looking back at the definition of the kinetic energy of a photon, this means a reduction in the frequency, which also means a increase in wavelength, since the product of frequency and wavelength is a fixed constant, the wave-speed, or the speed of light. This is why light escaping from the vicinity of a black hole will be said to be red-shifted. Analogously, a photon falling toward a black hole will gain energy and become blue-shifted.

Now we move toward general relativity. A problem immediately arises: there is no such thing as a preferred inertial frame. Whereas it makes sense in special relativity and in Galilean mechanics to compare two distant observers as "at rest with respect to each other", the Principle of General Covariance makes the same comparison nonsensical in general relativity. Geometrically, this is the problem of having curvature: parallel transports of a vector is now path dependent, so there is no one unique way to define a "parallel vector" at a distant point B of some known vector here at point A. As is well-known, the frequency of light can also be shifted by the Doppler effect (in the acoustic sector, this is the effect where the siren of an approaching ambulance sounds higher pitched than that of an ambulance driving away). We will describe the geometric interpretation of this effect later. If there is no canonical way to pick an observer, there is also no way to say definitively what the frequency of light observed at a point is.

(Actually, there is a preferred way of specifying, geometrically, a "parallel" observer. This involves parallel transporting the four-velocity of the observer along the path of propagation of the light signal. However, this specification is trivial, in the sense that by definition any such-defined observer will see the light signal at the same frequency as seen at the source.)

Let us make this notion slightly more precise. Using the geometric optics approximation, a light signal is given by a null geodesic vector field. Or, letting $(M,g)$ be a space-time, and $p\in M$ be an event, a light signal originating from point $p$ is given by a null geodesic emanating from $p$. The wave-vector of the light signal is written as $k$, a generator of the null geodesic, so in particular $\nabla_k k = 0$. An observer in the space-time is specified by an event, and his instantaneous velocity at that event. So an observer is a point $o = (p,v)$ in $TM$. We impose the physical condition that the instantaneous velocity of the observer is unit time-like $g(v,v) = -1$ and future pointing. Relative to this observer, the frequency of the light signal is given by $- g(v,k)$. (The minus sign just for normalization to a positive number. Also, two observers at the same event but with different velocities will see different frequencies as given by this formula.) Now, let $p,p'$ be two points connected by a null geodesic, and $o,o'$ two observers at those two points respectively. Then we can discuss the frequency-shift-factor as the ratio between $-g(v,k)$ and $-g(v',k)$. If the observer $o'$ were obtained by parallel transporting $o$ along the geodesic, then by definition the two frequencies will agree. Hence we have the following:

Any meaningful discussion of the red/blue-shift effect will necessarily involve a "god-given" vector field on the manifold.
Notice that the preferred vector field should be time-like (as it will represent an observer field), but it does not need to be unit: as long as the vector field is non-vanishing and time-like, we can divide by the square root of the negative of its Lorentzian length to get a unit time-like vector field to represent our observer.

So what makes a choice of a vector field a good one? Going back to the Newtonian picture, one might ask that the vector field represent a family of stationary observers relative to each other. In the case where we have a stationary space-time (one endowed with a continuous time-translation symmetry), one such preferred vector field will simply be the stationary Killing vector field. Denoting the Killing vector field by $X$, we observe the following \[ \nabla_k[ g(X,k) ] = (\nabla_k X^\flat)(k) + X^\flat (\nabla_k k) = 0 \] where in the middle expression, the second term vanishes by virtue of $k$ being geodesic, and the first term vanishes because the Killing equation implies $\nabla X^\flat$ is antisymmetric. (This is the well known result that the inner-product of a Killing vector with a geodesic vector field is a conserved quantity constant along a geodesic.) So given two points $p,p'$ on the null geodesic, and picking the observers $o = (p,\tilde{X}|_p ), o' = (p' ,\tilde{X} |_{p'})$ where $\tilde{X} = X / \sqrt{- g(X,X)}$, the frequency shift factor will be \begin{equation} \label{eqn2} \frac{ - g(\tilde{X},k) |_{p'} }{ -g(\tilde{X},k) |_p } = \frac{ g(X,k)|_{p'}}{g(X,k)|_p} \frac{\sqrt{-g(X,X)}|_p}{\sqrt{-g(X,X)}|_{p'}} = \sqrt{\frac{g(X,X)|_p}{g(X,X)|_{p'}}} \end{equation}

Now take the example of the Schwarzschild solution, and set $X$ to be the standard time-translation. $g(X,X) = - (1 - \frac{2M}{r})$ where $M$ is the mass of the black hole. A little bit of rearrangement shows that the frequency shift of a signal going from $p$ to $p'$ is \begin{equation} \sqrt{ \frac{ r(p') [r(p) - 2M] }{ r(p)[r(p') - 2M] }} \end{equation} which shows that the frequency decreases if $r(p') > r(p)$ and vice versa, confirming the intuition that light escaping from near the black hole decreases in frequency, and light traveling toward the black hole gains in frequency. An interesting effect from this is that at the moment that the light reaches $r = 2M$, the frequency is infinitely shifted to the blue. This confirms the heuristic notion that $r = 2M$ is a horizon, and there there is a infinite potential well which is impossible to escape from. In reality, however, this is a coordinate effect! Physically the wave-vector $k$ does not "blow up" as it cross the boundary $r = 2M$ (in fact, since it is parallel transported, it will not blow up unless we run into a bona fide space-time singularity). It is merely that our measure of stationarity, the vector field $X$, becomes no longer time-like at the juncture. If one picks another distinguished field of observers such that they remain time-like at the event horizon, then any shifts will in general not be infinite (but then one may actually have relative motion of the observers which will contribute a non-negligible factor in the shift). (Now, if we propagate a signal along the horizon, the above formula yields a ratio of zeroes, which is ill defined. In this case however, there is a way of talking about the gravitation red-shift through the notion of surface gravity.

Before moving on, let us just write down the formula for the shift formula relative to a family of observers given by the vector field $v$ (here we assume that $v$ is future-pointing and time-like, but not necessarily normalized). Let $\gamma$ be a null geodesic, affinely parametrized by $s$ where $\partial_s = k$ the geodesic vector field. The shift between $\gamma(a)$ and $\gamma(b)$ is then \begin{equation}\label{eqn3} \exp \left( \log \left[\frac{g(v,k)}{g(v,v)}\right]^{\gamma(b)}_{\gamma(a)}\right) = \frac{g(v,v)|_{\gamma(a)}}{g(v,v)|_{\gamma(b)}}\exp\left( \int_a^b \frac{g(\nabla_kv,k)}{g(v,k)} \circ \gamma(s) ds \right). \end{equation}

The Kodama vector field

In view of Condition 1, to discuss any notion of red-shift it is imperative that we have a distinguished time-like vector field $v$. In the case of spherical symmetry, such a vector field is readily available.

The Kodama vector field, first observed by Hideo Kodama, is a "canonical" geometric vector field in spherical symmetry. Recall that under the assumption of spherical symmetry, a space-time $(M,g)$ can be reduced to the quotient manifold $(Q,h)$, whose boundary corresponds to fixed points of the symmetry. On $Q$, and hence on $M$, a natural function $ r$ is defined to be the area-radius function: we set $4\pi r^2(q)$ for $q\in Q$ to equal the area of of the symmetry orbit of $q$ in $M$ under the metric induced by $g$. The Kodama vector field $K$ is defined on $Q$ to be the Hodge-dual relative to the metric $h$ of the gradient $\nabla r$ (with the orientation of the volume form chosen in such a way that $K$ is future-pointing time-like if $\nabla r$ is space-like). In a local oriented frame with $\epsilon_{ab}$ the Levi-Civita symbol standing in for the volume element of $(Q,h)$, we can write $K^a = \epsilon^{ab}\nabla_br$. By an abuse of notation we will also denote by $K$ the natural lift of the Kodama vector field to $M$ (so that it will be always orthogonal to the vector fields generating the Lie symmetries). Observe that by definition, $g(K, \nabla r) = 0$, so whenever $\nabla r$ is non-null and non-vanishing, $\{K,\nabla r\}$ defines an orthogonal, but not normalised, frame.

On the other hand, recall that $\nabla r$ is actually pointing in the same direction as the mean curvature vector of the symmetry orbits. And thus its causal character defines the region as either trapped, anti-trapped, or regular. In the regular region, then, $K$ is a distinguished time-like future pointing vector field.

First let us describe some properties of this vector field by recalling some elementary computations in spherical symmetry. The spherical symmetry splitting of $(M,g)$ implies that the metric takes the warped product decomposition \begin{equation} g = h + r^2 g_{\mathbb{S}}~,\quad M = Q\times_{r} \mathbb{S}^2 \end{equation} where $h$ is the natural lift of the metric of the quotient space $Q$ and $g_{\mathbb{S}}$ is the standard metric on the two-sphere, acting on the fibre. A standard computation in semi-Riemannian geometry yields that, for $X,Y\in TQ$, $V,W\in T\mathbb{S}^2$, we have that1 \begin{align} {}^MRic(X,Y) &= {}^QRic(X,Y) - \frac{2}{r}\nabla^2r(X,Y) \newline {}^MRic(X,V) &= 0 \newline {}^MRic(V,W) &= {}^{\mathbb{S}^2}Ric(V,W) - g(V,W) \left( \frac{\triangle r}{r} + \frac{g(\nabla r, \nabla r)}{r^2}\right) \end{align}

Now, for since both parts of the warped product are two dimensional, the Ricci curvatures are simply given by the metric times one-half the scalar curvature. So \begin{align} \label{eqn4a} {}^MG(X,Y) &= - \frac{2}{r}\nabla^2r(X,Y) - \frac{2}{r}\left(\frac{m}{r^2} - \triangle r \right)g(X,Y) \newline \label{eqn4b} {}^MG(V,W) &= \frac{1}{r}\triangle r g(V,W) - \frac12 {}^QR\cdot g(V,W) \end{align} where $G = Ric - \frac12 R g$ is the Einstein tensor and $m = \frac{r}{2}(1 - g(\nabla r, \nabla r))$ is the Hawking mass function. Restricting the first equation to $Q$ (since $r$ is constant on symmetry spheres), and re-arranging the terms, we have \begin{equation} \label{eqn5} \nabla^2 r = \frac{m}{r^2}h - 4\pi r ( \tilde{T} - (\mathop{tr}\tilde{T}) h) \end{equation} where $\tilde{T}$ is the restriction of the stress-energy tensor to $Q$, and we used Einstein's equation $G = 8\pi T$. Dotting both-sides by $\nabla r$, we also find after simple algebraic manipulations that \begin{equation}\label{eqn6} d m = 4\pi r^2 \ast_h(\tilde{T}\circ K) \end{equation}

Now, let us first consider the space-time divergence $\mathop{div} K = c(\nabla K)$ ($c(\cdot)$ is the contraction operation). By the definition of the warped product, the horizontal connection along $Q$ is the same as the connection on $M$. To examine the vertical derivatives, we note that $K$ is orthogonal to the symmetry spheres, so the horizontal projection of a horizontal derivative of $K$ is given by $g(\nabla_VK,W) = - g(K,II(V,W))$ where $II$ is the second fundamental form. Taking the horizontal trace then gives that the horizontal part of the divergence is given by the inner product of $K$ with the mean curvature vector $H$ of the symmetry spheres, which is zero by construction. Therefore the space-time divergence $\mathop{div} K$ gives a scalar function identical to the $ Q$-divergence of $K$. But on $Q$, $\mathop{div} K = \delta K^\flat = * d * * dr = 0$, so we have that

$K$ has vanishing $Q$- and space-time divergences.

Now, let us compute the deformation tensor ${}^K\pi = \mathcal{L}_K g = sym(\nabla K^\flat)$. Now \[ \mathcal{L}_K g = \mathcal{L}_K h + \mathcal{L}_K ( r^2 g_{\mathbb{S}}). \] Since $K$ is orthogonal and "symmetric" to the symmetry spheres, we have that $\mathcal{L}_K g_{\mathbb{S}} = 0$. In addition $\mathcal{L}_K r^2 = 2 r K(r) = 0$ by construction. So the deformation tensor in question is merely the symmetric, horizontal part of $\nabla K^\flat$. Now we arrive at the amazing property of the Kodama vector field. Let ${}^KJ = T\circ K$ by the energy current associated with the vector field $K$. A standard calculation shows that $\mathop{div} {}^KJ = g(T, {}^K\pi)$. Now as $K$ is divergence free, its deformation tensor is trace-free. The equations $\eqref{eqn4a}$ and $\eqref{eqn4b}$ then implies that \[ \mathop{div} {}^KJ = - \frac{1}{4\pi r} h( \nabla^2 r, sym(\nabla K^\flat)) = - \frac{1}{4\pi r} h(\nabla^2 r, \nabla * \nabla r) = 0. \] So we conclude that

$K$ defines a conserved current.
Looking at $\eqref{eqn6}$, which we can re-write as \[ d m = 4\pi \ast_g(T\circ K)( r^2 dA^\sharp ) \] we see that the Hawking or Misner-Sharp mass is the dual potential for the Kodama current.

Perhaps a remark is in order here: The proof of Lemma 3 is not the usual conserved current derivation from the energy conditions. In the usual energy condition derivation, for a vector field $X$, $\mathop{div}(T\circ X) = T\circ \mathcal{L}_Xg$ via the matter evolution equation $\mathop{div}T = 0$. And thus the current $T\circ X$ is divergence-free whenever $X$ is a Killing vector. However, for the Kodama vector field $K$, $\mathcal{L}_Kg$ is in general not vanishing. The fact that the corresponding current is divergence free is a orthogonality condition between the stress energy tensor $T$ and the deformation tensor $\mathcal{L}_Kg$ brought about by the rigid structure of the spherical symmetry.

Suppose now our space-time which admits already a spherical symmetry, also admits a translation symmetry $X$. There is a unique direction, orthogonal to the spherical symmetry, in which the area-radius $r$ is constant; the Kodama field points in this direction by definition, and any translation symmetry also since a symmetry must preserve the geometry. In general the two differ by a scalar factor. In Schwarzschild, for example, the two are identical (up to a sign-change due to orientation).

Heuristically, then, the Kodama vector field, at least in the regular region of the space-time, should be taken to be the vector-field that defines stationary observers in spherical symmetry. It has several nice properties: (a) it agrees with the time-translation symmetry in Schwarzschild (b) its flow preserves the area radius (c) it generates a conserved current (d) and lastly, in the asymptotically flat situation, $K$ approaches a time translation at spatial infinity (so its corresponding family of observers approaches "stationary" in the Newtonian/Euclidean sense in the asymptotic region).

Frequency shift associated to the Kodama field

Now we assume we are in a spherically symmetric space-time $(M,g)$ satisfying Einstein's equations. Applying $K$ to $\eqref{eqn3}$, we see that the frequency shift factor is \[ \frac{1 - 2m/r |_{\gamma(a)}}{1 - 2m/r |_{\gamma(b)}} \exp\left( \int_a^b \frac{g(\nabla_kK,k)}{g(K,k)} \circ \gamma(s) ds \right) \] Compared to the Schwarzschild case we have an additional shift factor coming from the integral. Now let us assume that $k$, the wave-vector, is tangent to $Q$ for ease of computation. A simple computation in null frame shows that \[ \frac{g(\nabla_kK,k)}{g(K,k)} = \frac{\nabla^2r(k,k)}{k(r)} \] Applying $\eqref{eqn5}$, and observing that $k$ is null, we have that $\nabla^2r(k,k) = - 4\pi r T(k,k) \leq 0$ if we assume the dominant or null energy condition on the matter fields. This shows that the integral term contributes a net blue-shift for incoming radiation ($k(r) < 0$) and a net red-shift for outgoing ones. For non-vacuum space-times, however, the quantity $(1 - 2m/r)$ does not necessarily have any monotonicity properties. One can see this by computing $\nabla_k (1 - 2m/r) = \nabla_k( g(\nabla r,\nabla r) ) = 2 \nabla^2r (\nabla r, k)$ using $\eqref{eqn5}$. This is no different from the Newtonian situation: if one remains outside a gravitating body, then moving towards the body decreases the gravitational potential. But if one is inside an extended gravitating body (say a cloud of dust), then moving "inwards" will, on the one hand, move you closer to the center of mass while on the other, reduce the net number of particles asserting a central pull on you. (An example is the gravitational field inside a ball of uniformly distributed of matter. The gravitational potential has an absolute minimum on the surface of the ball.) Yet just like the Schwarzschild case, if gravitational collapse occurs and a trapped surface forms, then the blue-shift factor approaches infinity as a wave travels toward the trapped surface.

  1. These can be found by a direct computation in local coordinates adapted to the warped product; see Chapter 7 of O'Neill, Semi-Riemannian Geometry. ^
Willie WY Wong
Associate Professor

My research interests include partial differential equations, geometric analysis, fluid dynamics, and general relativity.