2011-01-02 14:30:50 8 Comments

Regarding general relativity:

- What is the physical meaning of the Christoffel symbol ($\Gamma^i_{\ jk}$)?
- What are the (preferably physical) differences between the Riemann curvature tensor ($R^i_{\ jkl}$), Ricci tensor ($R_{ij}$) and Ricci scalar ($R$)? For example why do the Einstein equations include the Ricci tensor and scalar, but not the Riemann tensor?

To be clear, by "physical meaning" I mean something like - what physical effect do these components generate? Or, they make the GR solutions deviate from Newton because of xxx factor... or something similarly physically intuitive.

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## 4 comments

## @ghostRepeater 2014-02-04 08:09:32

As for the 'physical meaning' of Christoffel symbols, there is a sense in which they don't have a physical meaning, because the information they encode is not really information about the curvature of space but about the geometry of the coordinate system you're using to describe the space.

As for an intuition about them, they encode how much the basis vector fields change for infinitesimal changes in the coordinates being used. This is why in a flat space (i.e. locally) it is always possible to make them zero: transform to a coordinate system where the basis vector fields don't change from point to point.

To know how the spacetime curves, you can look at how the metric function changes from point to point. To see this, you can look at how the basis vectors change from point to point (since the metric is completely determined by the basis vectors). This is the information the Christoffel symbol encodes.

## @Ron Maimon 2011-09-18 08:25:21

The connection has a physical significance--- it is the gravitational field. The metric is the gravitational potential.

The fact that the Christoffel symbols are not tensors does not change the fact that they are meaningful. They can be made to vanish at any one point by a coordinate transformation, but in GR, this is just saying that you can make the gravitational field vanish by choosing a freely falling coordinate frame. That's a physical statement about the gravitational field.

The transformation law for Christoffel symbols is well defined, and one way to think about the mathematical concept of the abstract connection is by identifying two different symbol descriptions when they only differ by coordinate transformation. The abstract connection does not have a value at a point, but it has holonomy values on loops.

There are no local gauge invariant observables in a generally covariant theory, so you have to make do with coordinate transforming things like the metric tensor and the connection.

## @Ben Crowell 2012-09-27 02:25:33

"There are no local gauge invariant observables in a generally covariant theory." Counterexample: the Kretschmann scalar.

## @Ron Maimon 2012-09-27 02:44:31

@BenCrowell: This is not gauge invariant, in that a gauge transformation changes it by the derivative of the Kretschmann scalar. The statement I made is correct and well known, and trivial--- it is saying that gauge transformations in GR move interior points around the manifold, so a gauge invariant function is constant.

## @Marek 2011-01-02 18:49:20

Note that there is no

physical meaningof Christoffel symbols as they are not tensors. It's always possible to choose local coordinates such that all of $\Gamma$ vanish.But their mathematical meaning is that they form a pseudotensor. Technically, if we have two covariant derivatives $\nabla_1$ and $\nabla_2$ then their difference $\Gamma := \nabla_1 - \nabla_2$ satisfies some nice mathematical properties (namely that it is an ultralocal operator) and so its acting on any object is just local and can be represented by a tensor.

For $\nabla_1$ we usually take the covariant derivative we are interested in (e.g. a metric covariant derivative with vanishing torsion induced by some metric tensor $g$). For $\nabla_2$ there are two general (and widely used) choices. One can either use coordinate covariant derivative $\partial$ (which anihilates coordinate vector ${\partial \over \partial x}$ and covector fields ${\rm d} x$ and this gives the usual expression $\nabla = \partial + \Gamma_{Christoffel}$. The other choice (which generalizes the previous one) is a covariant derivative $\bar \partial$ which annihilates some tetrad $e$ (in the previous case we had the tetrad ${\rm d} x$ which is very specific; for general tetrad there need not exist associated coordinates). This leads to the tetrad formalism and one writes $\nabla = \bar \partial + \gamma$ where $\gamma$ are Ricci rotation coefficients.

As for Riemann tensor, it is once again a tensorial representation of an ultralocal operator, namely the curvature operator $R(u, v)$. This is a black box that takes two vector fields (thought of as a direction) and returns an ultralocal operator that tells you how much the space curves along those directions. More precisely, it tells you what happens with a vector if you parallel transport it along the infinitesimal polygon $0 \to u \to u+v \to v \to [u,v] \to 0$; it can be thought of as a square except that the two fields don't need to close and this is measured by their commutator $[u, v]$. So you can express it as $R(e_a, e_b) e_c = {R_{abc}}^d e_d$ and you'll obtain the usual Riemann tensor.

Now, because of the (a)symmetry of the Riemann tensor, two inequivalent contractions are possible. One of them is the trace ${R_{abc}}^c$ and this can be trivially seen to be zero for Riemann tensor derived from Levi-Civita connection (more generally for connections that preserve volume elements). The other contraction, ${R_{abc}}^a$ gives the Ricci tensor. This will be symmetric for Levi-Civita connection (because the trace of the Riemann tensor is zero and because the torsion vanishes).

One useful (quite mathematical though) view of the Ricci tensor is as a "laplacian of the metric", $R_{ij} \sim -{1 \over 2}\Delta g_{ij}$ and by an analogy to heat flows this relates Ricci flows which are a basic tool used in the study of PoincarĂ© conjecture.

Now, the geometrical meaning of the Ricci tensor is that it measures the deformation of the volume element in normal geodesic coordinates. These are coordinates that you can obtain around any point if you parametrize the neighborhood by geodesic flows. So the Ricci tensor measures how geodesics tend to get denser or sparser around a point in a given direction. Think about how sphere with positive curvature has less volume because its geodesics converge (they are the big circles on the sphere) than a hyperbolic space with negative curvature where geodesics diverge (there are infinitely many straight lines parallel to a given line). In particular, Ricci-flat manifolds (which are the solutions of vacuum Einstein's equations with zero cosmological constant) behave in this regard like the usual Euclidean space. You need to generalize this to Einstein manifolds (which are vacuum solutions with non-zero cosmological constant) to obtain analogues of sphere and hyperbolic space (namely, deSitter and anti-deSitter space).

There is a lot more to be said on these topics but I hope this will be helpful at least a little to you.

## @Jeremy 2011-01-03 01:59:45

this is a very important point, and should be voted up! The connection has NO PHYSICAL significance. Although I wouldn't say because it is not a tensor, but rather because it is only dependent on coordinates, and is not invariant.

## @Jerry Schirmer 2011-01-03 02:22:54

@Jeremy: If we're being pedantic, then <b>no</b> non-scalar is invariant--things that carry indices certainly change under a change of coordinates--they are <b>co</b>variant, not invariant.

## @Andrew Steane 2019-07-09 12:15:15

"no physical significance" is wrong: step on board a roller coaster or turntable and you will experience the physical significance right away. Even sitting in a chair you experience the physical significance of $\Gamma^a_{bc}$---or do all those high-school mechanics lessons with forces such as $m {\bf g}$ and pressures such as $m g h$ have "no physical significance"?

## @Jerry Schirmer 2011-01-02 15:18:52

The simplest way to explain the Christoffel symbol is to look at them in flat space. Normally, the laplacian of a scalar in three flat dimensions is:

$$\nabla^{a}\nabla_{a}\phi = \frac{\partial^{2}\phi}{\partial x^{2}}+\frac{\partial^{2}\phi}{\partial y^{2}}+\frac{\partial^{2}\phi}{\partial z^{2}}$$

But, that isn't the case if I switch from the $(x,y,z)$ coordinate system to cylindrical coordinates $(r,\theta,z)$. Now, the laplacian becomes:

$$\nabla^{a}\nabla_{a}\phi=\frac{\partial^{2}\phi}{\partial r^{2}}+\frac{1}{r^{2}}\left(\frac{\partial^{2}\phi}{\partial \theta^{2}}\right)+\frac{\partial^{2}\phi}{\partial z^{2}}-\frac{1}{r}\left(\frac{\partial\phi}{\partial r}\right)$$

The most important thing to note is the last term above--you now have not only second derivatives of $\phi$, but you also now have a term involving a

firstderivative of $\phi$. This is precisely what a Christoffel symbol does. In general, the Laplacian operator is:$$\nabla_{a}\nabla^{a}\phi = g^{ab}\partial_{a}\partial_{b}\phi - g^{ab}\Gamma_{ab}{}^{c}\partial_{c}\phi$$

In the case of cylindrical coordinates, what the extra term does is encode the fact that the coordinate system isn't homogenous into the derivative operator--surfaces at constant $r$ are much larger far from the origin than they are close to the origin. In the case of a curved space(time), what the Christoffel symbols do is explain the inhomogenities/curvature/whatever of the space(time) itself.

As far as the curvature tensors--they are contractions of each other. The Riemann tensor is simply an anticommutator of derivative operators--$R_{abc}{}^{d}\omega_{d} \equiv \nabla_{a}\nabla_{b}\omega_{c} - \nabla_{b}\nabla_{a} \omega_{c}$. It measures how parallel translation of a vector/one-form differs if you go in direction 1 and then direction 2 or in the opposite order. The Riemann tensor is an unwieldy thing to work with, however, having four indices. It turns out that it is antisymmetric on the first two and last two indices, however, so there is in fact only a single contraction (contraction=multiply by the metric tensor and sum over all indices) one can make on it, $g^{ab}R_{acbd}=R_{cd}$, and this defines the Ricci tensor. The Ricci scalar is just a further contraction of this, $R=g^{ab}R_{ab}$.

Now, due to Special Relativity, Einstein already knew that matter had to be represented by a two-index tensor that combined the pressures, currents, and densities of the matter distribution. This matter distribution, if physically meaningful, should also satisfy a continuity equation: $\nabla_{a}T^{ab}=0$, which basically says that matter is neither created nor destroyed in the distribution, and that the time rate of change in a current is the gradient of pressure. When Einstein was writing his field equations down, he wanted some quantity created from the metric tensor that also satisfied this (call it $G^{ab}$) to set equal to $T^{ab}$. But this means that $\nabla_{a}G^{ab} =0$. It turns out that there is only one such combination of terms involving first and second derivatives of the metric tensor: $R_{ab} - \frac{1}{2}Rg_{ab} + \Lambda g_{ab}$, where $\Lambda$ is an arbitrary constant. So, this is what Einstein picked for his field equation.

Now, $R_{ab}$ has the same number of indicies as the stress-energy tensor. So, a hand-wavey way of looking at what $R_{ab}$ means is to say that it tells you the "part of the curvature" that derives from the presence of matter. Where does this leave the remaining components of $R_{abc}{}^{d}$ on which $R_{ab}$ does not depend? Well, the simplest way (not COMPLETELY correct, but simplest) is to call these the parts of the curvature derived from the dynamics of the gravitational field itself--an empty spacetime containing only gravitational radiation, for example, will satisfy $R_{ab}=0$ but will also have $R_{abc}{}^{d}\neq 0$. Same for a spacetime containing only a black hole. These extra components of $R_{abc}{}^{d}$ give you the information about the gravitational dynamics of the spacetime, independent of what matter the spacetime contains.

This is getting long, so I'll leave this at that.