Deriving the meaning of the Ricci Tensor


by jstrunk
Tags: deriving, meaning, ricci, tensor
jstrunk
jstrunk is offline
#1
May6-12, 06:47 PM
P: 8
I am trying to understand the meaning of the Ricci Tensor. I tried to work it out in a way that was meaningful to me based on ideas from Baez and Loveridge. Unfortumately, the forum tool wont allow me to include the URLs to those documents in this post. Anyway, I get the wrong answer. Can someone tell me where I am going wrong?


Ricci Tensor

Is given by [itex]R_{jk} = R_{jki}^i[/itex].

Describes the evolution of the size of a volume element as each point in the element flows along its
geodesic curve.

Motivation:

1. Consider a small volume element in a 3-D Riemannian space. Each point in the element is
traveling along its geodesic curve. At t=0 the element is aligned with the local coordinate basis
so that it is described by the vectors [itex]\left\{ {dx^1 \vec e_1 ,dx^2 \vec e_2 ,dx^3 \vec e_3 } \right\}[/itex]. [itex]\vec v[/itex] is the velocity of the reference point at
the corner of the element at the origin of the local coordinate basis. We will use the Geodesic
Deviation formula to find how the edges of the volume element are changing and use that
information to determine how the volume of the element is changing.

2. The change to an edge of the element is given by the Geodesic Deviation formula
[itex]
\nabla _v \nabla _v u^\alpha = R_{\beta \gamma \delta }^\alpha v^\beta v^\gamma u^\delta
[/itex]
where [itex]\vec v[/itex] is the geodesic velocity vector of a reference point and [itex]\vec u[/itex] is the geodesic vector
connecting the reference point to another point at equal times.

3. Edge [itex]\vec u = dx^1 \vec e_1[/itex] could be rewritten as [itex]\vec u = dx^1 \vec e_1 + 0\vec e_2 + 0\vec e_3 [/itex] giving geodesic deviation
$$
\eqalign{
& \nabla _v \nabla _v u^1 = R_{\beta \gamma \delta }^1 v^\beta v^\gamma u^\delta \cr
& \nabla _v \nabla _v u^1 = R_{\beta \gamma 1}^1 v^\beta v^\gamma u^1 + R_{\beta \gamma 2}^1 v^\beta v^\gamma u^2 + R_{\beta \gamma 3}^1 v^\beta v^\gamma u^3 \cr
& \nabla _v \nabla _v u^1 = \left( {R_{\beta \gamma 1}^1 v^\beta v^\gamma } \right)dx^1 + \left( {R_{\beta \gamma 2}^1 v^\beta v^\gamma } \right)0 + \left( {R_{\beta \gamma 3}^1 v^\beta v^\gamma } \right)0 \cr
& \nabla _v \nabla _v u^1 = \left( {R_{\beta \gamma 1}^1 v^\beta v^\gamma } \right)dx^1 + \left( {R_{\beta \gamma 2}^2 v^\beta v^\gamma } \right)0 + \left( {R_{\beta \gamma 3}^3 v^\beta v^\gamma } \right)0 \cr}
$$

4. The last expression contracts the Riemann Tensor on symmetric indices giving the Ricci Tensor.
The expression can be rewritten as
$$
\eqalign{
& \nabla _v \nabla _v u^1 = R_{\beta \gamma } v^\beta v^\gamma dx^1 + R_{\beta \gamma } v^\beta v^\gamma 0 + R_{\beta \gamma } v^\beta v^\gamma 0 \cr
& \nabla _v \nabla _v u^1 = R_{\beta \gamma } v^\beta v^\gamma dx^1 \cr
& \nabla _v \nabla _v dx^1 = R_{\beta \gamma } v^\beta v^\gamma dx^1 \cr}
$$
Note that there could be non-zero values for [itex]\nabla _v \nabla _v u^2 [/itex] and [itex]\nabla _v \nabla _v u^3 [/itex]but we dont care about them.
They effect the shape or orientation of the volume element but not the length of this edge.

5. If we denote the signed magnitude of the first and second covariant derivatives of [itex]u^\alpha
[/itex] as [itex]\dot u^\alpha[/itex] and [itex]\ddot u^\alpha[/itex],
then we have
$$
\ddot u^1 = d\ddot x^1 = R_{\beta \gamma } v^\beta v^\gamma dx^1
$$

6. By a similar analysis we obtain the the signed magnitude of the second covariant derivative of the
all the edges of the volume element.
$$
\eqalign{
& d\ddot x^1 = R_{\beta \gamma } v^\beta v^\gamma dx^1 \cr
& d\ddot x^2 = R_{\beta \gamma } v^\beta v^\gamma dx^2 \cr
& d\ddot x^3 = R_{\beta \gamma } v^\beta v^\gamma dx^3 \cr}
$$

7. The size of an volume element and its covariant derivatives in this notation are

$$
\eqalign{
& V = \sqrt g \left( {dx^1 dx^2 dx^3 } \right) \cr
& \dot V = \sqrt g \left( {d\dot x^1 dx^2 dx^3 + dx^1 d\dot x^2 dx^3 + dx^1 dx^2 d^3 } \right) \cr
& \ddot V = \sqrt g \left[ {\left( {d\ddot x^1 dx^2 dx^3 + d\dot x^1 d\dot x^2 dx^3 + d\dot x^1 dx^2 d\dot x^3 } \right) + \left( {dx^1 d\ddot x^2 dx^3 + d\dot x^1 d\dot x^2 dx^3 + dx^1 d\dot x^2 d\dot x^3 } \right) + \left( {dx^1 dx^2 d\ddot x^3 + d\dot x^1 dx^2 d\dot x^3 + dx^1 d\dot x^2 d\dot x^3 } \right)} \right] \cr}
$$
8. The terms containing two first covariant derivatives are not caused by the curvature of space. They
are non-zero even in some coordinate systems in Euclidean space. For instance, imagine a cube
moving along the x-axis of a rectangular coordinate system. Then superimpose a spherical
coordinate system on the same space with the same origin. As the cube moves to greater x, its
thickness [itex]dx = dr[/itex] will stay the same. But the angles [itex]d\theta[/itex] and [itex]d\varphi[/itex] subtended by the cube will change.
You have two components changing times one that is isnt, even when volume isnt changing. We
subtract out these terms to get the expression for the volume change due to the curvature of space,
which we designate with subscript C.

$$
\ddot V_C = \sqrt g \left[ {d\ddot x^1 dx^2 dx^3 + dx^1 d\ddot x^2 dx^3 + dx^1 dx^2 d\ddot x^3 } \right]
$$

9. Substitute [itex]d\ddot x^\alpha = R_{\beta \gamma } v^\beta v^\gamma dx^\alpha[/itex] into
the last equation.
$$
\eqalign{
& \ddot V_C = \sqrt g \left[ {d\ddot x^1 dx^2 dx^3 + dx^1 d\ddot x^2 dx^3 + dx^1 dx^2 d\ddot x^3 } \right] \cr
& \ddot V_C = \sqrt g \left[ {\left( {R_{\beta \gamma } v^\beta v^\gamma dx^1 } \right)dx^2 dx^3 + dx^1 \left( {R_{\beta \gamma } v^\beta v^\gamma dx^2 } \right)dx^3 + dx^1 dx^2 \left( {R_{\beta \gamma } v^\beta v^\gamma dx^3 } \right)} \right] \cr
& \ddot V_C = \left( {R_{\beta \gamma } v^\beta v^\gamma } \right)\sqrt g \left[ {dx^1 dx^2 dx^3 + dx^1 dx^2 dx^3 + dx^1 dx^2 dx^3 } \right] \cr
& \ddot V_C = 3\left( {R_{\beta \gamma } v^\beta v^\gamma } \right)\sqrt g \left[ {dx^1 dx^2 dx^3 } \right] \cr}
$$

10. Divide both sides by the size of the volume element, [itex]V = \sqrt g \left( {dx^1 dx^2 dx^3 } \right)[/itex].
$$
\eqalign{
& {{\ddot V_C } \over V} = {{3\left( {R_{\beta \gamma } v^\beta v^\gamma } \right)\sqrt g \left[ {dx^1 dx^2 dx^3 } \right]} \over {\sqrt g \left( {dx^1 dx^2 dx^3 } \right)}} \cr
& {{\ddot V_C } \over V} = 3\left( {R_{\beta \gamma } v^\beta v^\gamma } \right) \cr}
$$
11. The answer is supposed to be [itex]
{{\ddot V_C } \over V} = - \left( {R_{\beta \gamma } v^\beta v^\gamma } \right)
[/itex].
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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clamtrox
clamtrox is offline
#2
May7-12, 07:50 AM
P: 937
Each component of [itex] \nabla_v \nabla_v u^{\alpha} [/itex] gives you only one component to the Ricci tensor contraction, right? So 4 does not follow yet from 3. Also your sign difference comes probably from the fact that you are contracting a funny index in Riemann tensor: usually it's defined [itex] R^{i}_{j i k} = R_{jk} [/itex]

So you'd have
[tex] \ddot{V}_C = \sqrt g \left[ {\left( {R^1_{\beta \gamma 1} v^\beta v^\gamma dx^1 } \right)dx^2 dx^3 + dx^1 \left( {R^2_{\beta \gamma 2} v^\beta v^\gamma dx^2 } \right)dx^3 + dx^1 dx^2 \left( {R^3_{\beta \gamma 3} v^\beta v^\gamma dx^3 } \right)} \right] = -\sqrt{g} R_{\beta \gamma} v^{\beta} v^{\gamma} dx^1 dx^2 dx^3 [/tex]
jstrunk
jstrunk is offline
#3
May7-12, 07:13 PM
P: 8
Thanks for yout suggestion.
I am going to have to think about it, which may take a while.
I am mostly concerned about the factor of 3. The negative sign is probably just from mixing formulas from different sources that use different conventions.

clamtrox
clamtrox is offline
#4
May8-12, 02:03 AM
P: 937

Deriving the meaning of the Ricci Tensor


So starting from the formula above, you don't get a factor of 3:
[tex]
\ddot{V}_C = \sqrt{g} \left[ {\left( {R^1_{\beta \gamma 1} v^\beta v^\gamma dx^1 } \right)dx^2 dx^3 + dx^1 \left( {R^2_{\beta \gamma 2} v^\beta v^\gamma dx^2 } \right)dx^3 + dx^1 dx^2 \left( {R^3_{\beta \gamma 3} v^\beta v^\gamma dx^3 } \right)} \right] =
[/tex]
[tex]
- \sqrt{g} \left[ (R^1_{\beta 1 \gamma} v^\beta v^\gamma + R^2_{\beta 2 \gamma } v^\beta v^\gamma + R^3_{\beta 3 \gamma } v^\beta v^\gamma \right] dx^1 dx^2 dx^3 = - \sqrt{g} \left[ R^{\mu}_{\beta \mu \gamma} v^\beta v^\gamma \right] dx^1 dx^2 dx^3 = - \sqrt{g} \left[ R_{\beta \gamma} v^\beta v^\gamma \right] dx^1 dx^2 dx^3,
[/tex]
where the first equality holds due to Riemann tensor antisymmetry and for second, I just write the sum as a contraction and in third I use the usual definition of Ricci tensor.
jstrunk
jstrunk is offline
#5
May9-12, 08:50 PM
P: 8
This solves my problem. Thanks so much.
This is a question that all the Relativity books I have seen just gloss over but I thought it was important to understand.
When I clean up my arguement I will append it to this post in case it is of interest to other people.
jstrunk
jstrunk is offline
#6
May26-12, 09:51 AM
P: 8
In case anyone is interested, this is a cleaned up version of my attempt to understand what the Ricci Tensor means.

Ricci Tensor

Describes the evolution of the size of a volume element as each point in the element flows along its
geodesic curve. Specifically, it is the 2nd covariant derivative of volume per unit of volume caused
by geodesics deviation. This is my attempt to explain how this interpretation of the Ricci Tensor
arises.

1. Consider a small volume element in a 3-D Riemannian space. Each point in the element is
traveling along a geodesic curve. At t=0 the element is aligned with the local coordinate basis
and is described by the vectors [itex]\left\{ {dx^1 \vec e_1 ,dx^2 \vec e_2 ,dx^3 \vec e_3 } \right\}
[/itex]. [itex]\vec v[/itex] is the velocity of the reference point at
the corner of the element at the origin of the local coordinate basis. We will use the Geodesic
Deviation formula to find how the edges of the element are changing and use that information to
determine how the volume of the element is changing.

2. The change to an edge of the element is given by the Geodesic Deviation formula
[itex]\nabla _v \nabla _v u^\alpha = R_{\beta \gamma \delta }^\alpha v^\beta v^\gamma u^\delta[/itex]
where [itex]\vec v[/itex] is the geodesic velocity vector of a reference point, [itex]\vec u[/itex] is the geodesic vector connecting
the reference point to another point at equal times and [itex]\nabla _v u^\alpha = v^\beta u_{;\beta }^\alpha[/itex] is the covariant derivative of
[itex]\vec u[/itex] along [itex]\vec v[/itex].

3. Edge [itex]\vec u = dx^1 \vec e_1[/itex] could be rewritten as [itex]\vec u = dx^1 \vec e_1 + 0\vec e_2 + 0\vec e_3 [/itex] giving geodesic deviation
$$
\eqalign{
& \nabla _v \nabla _v u^\alpha = R_{\beta \gamma \delta }^\alpha v^\beta v^\gamma u^\delta \cr
& \nabla _v \nabla _v u^1 = R_{\beta \gamma 1}^1 v^\beta v^\gamma u^1 + R_{\beta \gamma 2}^1 v^\beta v^\gamma u^2 + R_{\beta \gamma 3}^1 v^\beta v^\gamma u^3 \cr
& \nabla _v \nabla _v dx^1 = R_{\beta \gamma 1}^1 v^\beta v^\gamma dx^1 + R_{\beta \gamma 2}^1 v^\beta v^\gamma 0 + R_{\beta \gamma 3}^1 v^\beta v^\gamma 0 \cr
& \nabla _v \nabla _v dx^1 = R_{\beta \gamma 1}^1 v^\beta v^\gamma dx^1 \cr}
$$

Note that there could be non-zero values for [itex]\nabla _v \nabla _v u^2 [/itex] and [itex]
\nabla _v \nabla _v u^3 [/itex] but we dont care about them.
They effect the shape or orientation of the volume element but not the length of this edge.

4. If we denote [itex]\nabla _v dx^\alpha[/itex] as [itex]d\dot x^\alpha[/itex] and [itex]
\nabla _v \nabla _v dx^\alpha[/itex] as [itex]d\ddot x^\alpha[/itex], then we have
$$
d\ddot x^1 = R_{\beta \gamma 1}^1 v^\beta v^\gamma dx^1
$$

5. By a similar analysis we obtain the lengths of the other two edges of the volume element, so
$$
\eqalign{
& d\ddot x^1 = R_{\beta \gamma 1}^1 v^\beta v^\gamma dx^1 \cr
& d\ddot x^2 = R_{\beta \gamma 2}^2 v^\beta v^\gamma dx^2 \cr
& d\ddot x^3 = R_{\beta \gamma 3}^3 v^\beta v^\gamma dx^3 \cr}
$$
This can be summarized as
$$
d\ddot x^\alpha = R_{\beta \gamma \alpha }^\alpha v^\beta v^\gamma dx^\alpha
$$ (no summation on alpha).

6. The size of an volume element and its covariant derivatives in this notation are
$$
\eqalign{
& V = \sqrt g \left( {dx^1 dx^2 dx^3 } \right) \cr
& \dot V = \sqrt g \left( {d\dot x^1 dx^2 dx^3 + dx^1 d\dot x^2 dx^3 + dx^1 dx^2 dx^3 } \right) \cr
& \ddot V = \sqrt g \left[ \matrix{
\left( {d\ddot x^1 dx^2 dx^3 + d\dot x^1 d\dot x^2 dx^3 + d\dot x^1 dx^2 d\dot x^3 } \right) + \left( {dx^1 d\ddot x^2 dx^3 + d\dot x^1 d\dot x^2 dx^3 + dx^1 d\dot x^2 d\dot x^3 } \right) \hfill \cr
+ \left( {dx^1 dx^2 d\ddot x^3 + d\dot x^1 dx^2 d\dot x^3 + dx^1 d\dot x^2 d\dot x^3 } \right) \hfill \cr} \right] \cr}
$$

7. The terms containing two first covariant derivatives are not caused by the curvature of space. They
are non-zero even in some coordinate systems in Euclidean space. For instance, imagine a cube
moving along the x-axis of a rectangular coordinate system. Then superimpose a spherical
coordinate system on the same space with the same origin. As the cube moves to greater x, its
thickness [itex]dx = dr[/itex] will stay the same. But the angles [itex]d\theta[/itex] and [itex]
d\varphi[/itex] subtended by the cube will change.
You have two components changing times one that is isnt. This type of component change is just
an artifact of the fact that the basis is changing and doesnt reflect an actual change in volume. We
subtract out these terms to get the expression for the real volume change due to the geodesic
deviation which we designate with subscript G.

$$
\ddot V_G = \sqrt g \left[ {d\ddot x^1 dx^2 dx^3 + dx^1 d\ddot x^2 dx^3 + dx^1 dx^2 d\ddot x^3 } \right]
$$

Note: I am not entirely convinced by this argument, which I based on one given by Loveridge. If
someone has a better rationale for getting rid of these terms, I would like to hear it.

8. Substitute [itex]d\ddot x^\alpha = R_{\beta \gamma \alpha }^\alpha v^\beta v^\gamma dx^\alpha[/itex] (no summation on alpa) into the last equation.
$$
\eqalign{
& \ddot V_G = \sqrt g \left[ {d\ddot x^1 dx^2 dx^3 + dx^1 d\ddot x^2 dx^3 + dx^1 dx^2 d\ddot x^3 } \right] \cr
& \ddot V_G = \sqrt g \left[ {\left( {R_{\beta \gamma 1}^1 v^\beta v^\gamma dx^1 } \right)dx^2 dx^3 + dx^1 \left( {R_{\beta \gamma 2}^2 v^\beta v^\gamma dx^2 } \right)dx^3 + dx^1 dx^2 \left( {R_{\beta \gamma 3}^3 v^\beta v^\gamma dx^3 } \right)} \right] \cr
& \ddot V_G = \left( {R_{\beta \gamma 1}^1 + R_{\beta \gamma 2}^2 + R_{\beta \gamma 3}^3 } \right)v^\beta v^\gamma \sqrt g \left[ {dx^1 dx^2 dx^3 } \right] \cr
& \ddot V_G = \left( { - R_{\beta \gamma } v^\beta v^\gamma } \right)\sqrt g \left[ {dx^1 dx^2 dx^3 } \right] \cr}
$$

9. Divide both sides by the size of the volume element, [/itex]V = \sqrt g \left( {dx^1 dx^2 dx^3 } \right)[/itex].
$$
\eqalign{
& {{\ddot V_G } \over V} = {{\left( { - R_{\beta \gamma } v^\beta v^\gamma } \right)\sqrt g \left[ {dx^1 dx^2 dx^3 } \right]} \over {\sqrt g \left( {dx^1 dx^2 dx^3 } \right)}} \cr
& {{\ddot V_G } \over V} = - R_{\beta \gamma } v^\beta v^\gamma \cr}
$$.

10. We conclude that the Ricci Tensor describes the 2nd covariant derivative of volume per unit of
volume caused by geodesics deviation.
clamtrox
clamtrox is offline
#7
May30-12, 03:03 AM
P: 937
Quote Quote by jstrunk View Post
7. The terms containing two first covariant derivatives are not caused by the curvature of space. They
are non-zero even in some coordinate systems in Euclidean space. For instance, imagine a cube
moving along the x-axis of a rectangular coordinate system. Then superimpose a spherical
coordinate system on the same space with the same origin. As the cube moves to greater x, its
thickness [itex]dx = dr[/itex] will stay the same. But the angles [itex]d\theta[/itex] and [itex]
d\varphi[/itex] subtended by the cube will change.
You have two components changing times one that is isnt. This type of component change is just
an artifact of the fact that the basis is changing and doesnt reflect an actual change in volume. We
subtract out these terms to get the expression for the real volume change due to the geodesic
deviation which we designate with subscript G.

$$
\ddot V_G = \sqrt g \left[ {d\ddot x^1 dx^2 dx^3 + dx^1 d\ddot x^2 dx^3 + dx^1 dx^2 d\ddot x^3 } \right]
$$

Note: I am not entirely convinced by this argument, which I based on one given by Loveridge. If
someone has a better rationale for getting rid of these terms, I would like to hear it.
This is just a guess... But what happens if you calculated [itex] \frac{\ddot{V}}{V}-\frac{\dot{V^2}}{V^2} [/itex]?


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