I've been sick the past week, and also busy with some other things, so I haven't spent much time on this recently. I'm going to write down what I understand and what I don't, mainly just to get my thoughts in order.(adsbygoogle = window.adsbygoogle || []).push({});

I've been reading Wald. The relevant pages are 217 and 46. Page 46 is where he defines the "orthogonal displacement vector". It seems impossible to understand any of this without a thorough understanding of that, so I'll start there. Consider a one-parameter family of timelike curves [itex]\{\gamma_s\}[/itex], parametrized by proper time. I'll write the tangent vector of [itex]t\mapsto\gamma_s(t)[/itex] at t as [itex]\dot\gamma_s(t)[/itex]. I'll write the curve [itex]s\mapsto\gamma_s(t)[/itex] as [itex]\gamma(t)[/itex], and its tangent vector at s as [tex]\dot{\gamma(t)}(s)[/tex]. (I hope this notation isn't too confusing). We define two vector fields T and X by

[tex]T(\gamma_s(t))=\dot\gamma_s(t)[/tex]

[tex]X(\gamma_s(t))=\dot{\gamma(t)}(s)[/tex]

There's an extremely important step in all of this that Wald mentions without proof: It's possible to reparametrize each [itex]\gamma_s[/itex], without destroying the "parametrized by proper time" property, so that X is orthogonal to T at all points that the congruence passes through. If someone wants to see the details, I can type them up.

Why is X a measure of the "distance" to nearby curves in the congruence? The best answer I've been able to come up with uses a coordinate system x:

[tex]x^i(\gamma_{s+r}(t))=x^i(\gamma_s(t))+r\frac{d}{dr}\bigg|_0 x^i(\gamma_{s+r}(t))+\mathcal O(r^2)=\cdots=x^i(\gamma_s(t))+rX^i(\gamma_s(t))+\mathcal O(r^2)[/tex]

[tex]X^i(\gamma_s(t))=\lim_{r\rightarrow 0}\frac{x^i(\gamma_{r+s}(t))+x^i(\gamma_s(t))}{r}[/tex]

So the ith component of X in the coordinate system x is the limit of "coordinate change"/"parameter change" in the direction of X as the parameter change goes to zero.

This result is all we need from page 46. Jump to page 217. I will continue to call the timelike vector field T and the displacement vector X. I will also continue to consider a one-parameter family of curves, because the notation is already complicated enough to be annoying. We have [itex]\mathcal L_T X=0[/itex], because that Lie derivative is equal to the commutator [T,X], which is zero because T and X are the basis vectors associated with the coordinate system [itex]\gamma_s(t)\mapsto(s,t)[/itex].

A crucial step that I haven't done yet: Prove that this really is a coordinate system, given that T is a smooth vector field with integral curves [itex]\gamma_s[/itex], reparametrized to make X orthogonal to T.

It's not hard to show that [T,X]=0 implies [itex]\nabla_X T=\nabla_T X[/itex]. Wald's definition of the tensor B is equivalent to

[tex]B(Y,Z)=g(\nabla_Z T,Y)[/tex]

for all Y,Z. We can then define [tex]B^\sharp=B^i{}_j\partial_i\otimes dx^j[/tex], and use [itex]\nabla_X T=\nabla_T X[/itex] to show that [tex]B^\sharp(\cdot,X)=\nabla_T X[/tex]. This is why [tex]B^a{}_b X^b=B^\sharp(\cdot,X)[/tex] "measures the failure of X to be parallel transported". Note that to get this result, we used[itex]\nabla_X T=\nabla_T X[/itex], which is implied by [X,T]=0.

After that, I don't understand much. I understand that [itex]h^a{}_b=\delta^a_b+T^aT_b[/itex] is a projection operator, but I don't understand why we're looking for a projection operator at all. When I look at the definitions of the tensors that are defined at the end, I see that they don't involve the orthogonal displacement vector X in any way. I have no idea how they can have anything to do with things like rigidity without it. Now I'm also wondering if it was a waste of time to work through the details of the orthogonal displacement vector. It doesn't seem to have anything to do with anything at the end. I really don't get any of this.

I'll give it another shot within the next few days, but if I don't start to get it soon, I'm just going to abandon it. This is taking too much time.

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# My attempt to understand congruences, rigidity, etc.

Loading...

Similar Threads - attempt understand congruences | Date |
---|---|

I Attempt at Disproving Velocity Curving Space-Time | Feb 17, 2017 |

Attempting to learn tensor calculus | Feb 1, 2014 |

Scenario in an attempt to understand relativity | Mar 19, 2011 |

Attempting to Understand Alcubierre | May 10, 2006 |

Twinship, an attempt to understand SR | Apr 11, 2004 |

**Physics Forums - The Fusion of Science and Community**