In summary, the introduction gives a brief summary of the theory of geodesic congruences. The theory is extensively covered in many textbooks, and the following is a summary of what was covered. Geodesics are curves in a vector space that are tangent to curves of constant affine parameter lambda. The vector field φ is tangent to curves of constant lambda and is interpreted as a deviation vector between neighbouring geodesics. In some neighbourhood of the family, (s, lambda, x2, x3) is a coordinate chart that satisfies φ(x2, x3) = 0. By the equality of mixed partial derivatives, the commutator of φ and x2x3 is zero, which
ergospherical
Introduction
The theory of geodesic congruences is extensively covered in many textbooks (see References); what follows in the introduction is a brief summary. Consider a 1-parameter family of timelike geodesics ##\gamma_s(\lambda)##, where ##s## labels each geodesic in the family whilst ##\lambda## is an affine parameter along each ##\gamma_s##. Then the vector field ##\xi \equiv \partial / \partial s## is tangent to curves of constant ##\lambda## and is interpreted as a deviation vector between neighbouring geodesics.
In some neighbourhood of the family, ##(s,\lambda, x^2, x^3)## is a coordinate chart satisfying ##\xi = \partial/\partial s## and ##u = \partial/\partial \lambda##. By the equality of mixed partial derivatives, the commutator of ##u## and ##\xi## is zero (i.e. ##\xi## is Lie transported along ##u##),\begin{align*}
0 = [u, \xi]^a = (L_{u} \xi)^a = \xi^b \nabla_b u^a – u^b \nabla_b \xi^a
\end{align*}which implies that ##\dfrac{D\xi^a}{d\lambda} = u^b...

Abhishek11235, vanhees71, fresh_42 and 1 other person
Hi,

congratulation for the job done. I would like to point out some topic already discussed in PF so far.

I noted you use both Latin and Greek indices even if not together in the same formula ! So for example in the first part of the Introduction you use Latin indices (i.e. Abstract Index Notation) while in FRW section Greek ones (Ricci calculus notation).

My understanding, as discussed so far and in line with the first insight's reference (H. Reall, Part 3 General Relativity section 1.6), is that the second one (Greek indices) actually involves objects's components in a given basis (namely a given basis for the vector space and its associated dual-vector basis).

So, from a general point of view, a Greek indices equation valid in a particular/specific basis is not true in other bases.

That said, I believe the reason since you employed Greek indices in the rest of the insight is that you were assuming specific coordinate charts for each of the spacetimes discussed.

vanhees71
Yeah as far as possible I tried to use Latin indices (or no indices) for coordinate independent expressions and Greek indices when evaluating the components in a particular basis. (That's the convention of Wald and Reall.)

vanhees71
To be onest, I'm often in trouble with expression like that in FRW section, namely ##\delta_t^{\mu}##. Here ##\mu## as Greek index has the role of "which component" whereas ##t## is the name of a fixed given component (coordinate name).

Do you think there is a way to get rid of that (and similar) ambiguity ? Thank you.

Can you clarify what is confusing you? In the coordinates ##(t,r,\theta, \phi)##, the vector ##u = \partial/\partial t## has a ##t## component of ##1## and the rest of the components 0, i.e. ##u^{\mu} = \delta^{\mu}_t##. Recall that the symbol ##\delta^{\mu}_{\nu}## is ##1## if ##\mu = \nu## and 0 if ##\mu \neq \nu##.

Alternatively, you can write ##u^{\mu} = dx^{\mu}(u) = dx^{\mu} \left( \dfrac{\partial}{\partial t} \right) = \dfrac{\partial x^{\mu}}{\partial t} = \delta^{\mu}_t##.

ergospherical said:
Can you clarify what is confusing you? In the coordinates ##(t,r,\theta, \phi)##, the vector ##u = \partial_t## has a ##t## component of ##1## and the rest of the components 0, i.e. ##u^{\mu} = \delta^{\mu}_t##.

Recall that the symbol ##\delta^{\mu}_{\nu}## is ##1## if ##\mu = \nu## and 0 if ##\mu \neq \nu##.
No no it makes sense. IMO the point to be highlighted is that the letter ##t##, actually, is not Greek so in this case there is no problem.

My point is broader in the sense that many times mixing Greek and Latin index names turns out to be confusing.

Ohhh, haha, okay. I wouldn't lose sleep over that.

Another common convention is for latin letters to be 3-tensors, and greek letters 4-tensors. It always helps to have a section making explicit any conventions that are not universal. An unobtrusive way to do it is to have an appendix on this, with one sentence in the intro referring to the appendix for conventions.

Last edited:
vanhees71, cianfa72 and ergospherical
Yes, and also all the sign conventions about the curvature and Ricci tensor are also nice to have ;-).

## 1. What is a geodesic congruence?

A geodesic congruence is a set of geodesics, or the shortest paths between points in a curved space, that all have the same tangent vector at each point. This means that the geodesics in the congruence are all parallel to each other and have the same direction of motion.

## 2. How does a geodesic congruence behave in FRW spacetime?

In FRW (Friedmann-Robertson-Walker) spacetime, a geodesic congruence will follow the expansion of the universe. This means that the geodesics will become increasingly separated from each other as the universe expands. In a closed universe, the geodesics will eventually converge and cross each other, while in an open universe, they will continue to diverge.

## 3. What happens to a geodesic congruence in a Schwarzschild spacetime?

In a Schwarzschild spacetime, which describes the curvature of space around a non-rotating massive object, a geodesic congruence will behave differently depending on its distance from the object. Close to the object, the geodesics will be attracted towards it, while further away, they will be repelled. This is due to the strong gravitational field of the object.

## 4. Can a geodesic congruence exist in a Kerr spacetime?

Yes, a geodesic congruence can exist in a Kerr spacetime, which describes the curvature of space around a rotating massive object. However, the geodesics will not be parallel to each other, as they will be affected by the rotation of the object. This means that the geodesics will have a spiraling motion around the object.

## 5. How do geodesic congruences relate to the curvature of spacetime?

Geodesic congruences are closely related to the curvature of spacetime, as they follow the shortest paths in a curved space. The behavior of a geodesic congruence can give us information about the curvature of the space it is moving through. In a highly curved spacetime, the geodesics may deviate significantly from straight lines, while in a flat spacetime, they will remain straight and parallel.

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