Thursday 21 November 2019

Success at killing vectors

With help from Desmos
I had more success at killing vectors in the second part of section 3.8. We were asked to prove that a linear combination of Killing vectors with constant coefficients is still a Killing vector and also show that the commutator of two Killing vector fields is a Killing vector field. The first was indeed trivial as Carroll promised. The second was not. The problem is to show that, when ## A,B## are Killing vectors (##\nabla_\nu A_\lambda+\nabla_\lambda A_\nu=0## and ##\nabla_\nu B_\lambda+\nabla_\lambda B_\nu=0##) that, $$
$$The first problem was that I only knew (from way back) that the commutator can be expressed as $$
\left[A,B\right]^\lambda=A^\rho\partial_\rho B^\lambda-B^\rho\partial_\rho A^\lambda
$$which is not much use in this case. But it was easy to get to these two$$
\left[A,B\right]^\lambda=A^\rho\nabla_\rho B^\lambda-B^\rho\nabla_\rho A^\lambda
\left[A,B\right]_\sigma=A^\rho\nabla_\rho B_\sigma-B^\rho\nabla_\rho A_\sigma
$$which are much better. I also needed a covariant form of the Riemann tensor which is$$
R_{\ \ \sigma\nu\mu}^\tau X_\tau=\left[\nabla_\mu,\nabla_\nu\right]X_\sigma
$$They all earned a place in Important Equations for General Relativity.

It was another 13 equations to get to the desired conclusion with evil side-tracks and errors on the way. I have not included those in the answer.

The answer does include much index juggling, including use of antisymmetries to really throw the Riemann tensor indices about. That was the final breakthrough. It's at
Commentary 3.8 Symmetries and Killing vectors.pdf (pages 4-6).

Friday 15 November 2019

Exercise 3.12 Derivatives of Killing vectors


Show that any Killing vector ##K^\mu## satisfies the relations mentioned in the text:
\nabla_\mu\nabla_\sigma K^\rho=R_{\ \ \ \sigma\mu\nu}^\rho K^\nu&\phantom {10000}(1)\nonumber\\
K^\lambda\nabla_\lambda R=0&\phantom {10000}(2)\nonumber


Wilhelm Karl Joseph Killing 1847-1923
The text gives a few clues about how to get from (1) to (2). What we know about Killing vectors is that they satisfy Killing's equation
\nabla_{(\mu}K_{\nu)}\equiv\frac{1}{2}\left(\nabla_\mu K_\nu+\nabla_\nu K_\mu\right)=0&\phantom {10000}(3)\nonumber\\
\Rightarrow\nabla_\mu K_\nu=-\nabla_\nu K_\mu&\phantom {10000}(4)\nonumber
\end{align}##\nabla_\mu K_\nu## is antisymmetric.

The Killing vectors in (1) and (2) are given in contravariant form.
The Riemann tensor is antisymmetric* in its last two indices like (7) and has other interesting symmetries, not all independent, e.g. (6) comes form (7) and (8):
&R_{\rho\sigma\mu\nu}=g_{\tau\rho}R_{\ \ \ \sigma\nu\mu}^\tau&\phantom {10000}(5)\nonumber\\
\text{Swap first two }~~~~~~~~~
&R_{\rho\sigma\mu\nu}=-R_{\sigma\rho\mu\nu}&\phantom {10000}(6)\nonumber\\
\text{Swap last two  }~~~~~~~~~
&R_{\rho\sigma\mu\nu}=-R_{\rho\sigma\nu\mu}&\phantom {10000}(7)\nonumber\\
\text{Swap first and last pair  }~~~~~~~~~
&R_{\rho\sigma\mu\nu}=R_{\mu\nu\rho\sigma}&\phantom {10000}(8)\nonumber\\
\text{Rotate last three  }~~~~~~~~~
&R_{\rho\sigma\mu\nu}+R_{\rho\mu\nu\sigma}+R_{\rho\nu\sigma\mu}=0&\phantom {10000}(9)\nonumber\\
\text{Permute last three  }~~~~~~~~~
&R_{\rho\left\lceil\sigma\mu\nu\right\rceil}=0&\phantom {10000}(10)\nonumber
\end{align}(6) implies
g_{\tau\rho}R_{\ \ \ \sigma\mu\nu}^\tau=-g_{\tau\sigma}R_{\ \ \ \rho\mu\nu}^\tau&\phantom {10000}(11)\nonumber\\
\Rightarrow g^{\lambda\rho}g_{\tau\rho}R_{\ \ \ \sigma\mu\nu}^\tau=-g^{\lambda\rho}g_{\tau\sigma}R_{\ \ \ \rho\mu\nu}^\tau&\phantom {10000}(12)\nonumber\\
\Rightarrow R_{\ \ \ \sigma\mu\nu}^\lambda=-R_{\ \sigma\ \ \rho\mu\nu}^{\ \ \ \ \lambda}&\phantom {10000}(13)\nonumber
\end{align}We surmise that a similar manoeuvre on (7) would get us (14) which was the assertion * above and that (8) would get us (15)
R_{\ \ \ \sigma\nu\mu}^\tau=-R_{\ \ \ \sigma\mu\nu}^\tau&\phantom {10000}(14)\nonumber\\
R_{\ \ \ \sigma\nu\mu}^\tau=R_{\nu\mu\ \ \ \sigma}^{\ \ \ \ \ \ \tau}&\phantom {10000}(15)\nonumber
\end{align}These might will not be useful, although it might will be better to work with fully covariant Riemann tensors.

As I found out after much effort, it is also important to remember the definition of the Riemann tensor when the connection is torsion free. It measures the difference between taking covariant derivatives of a vector going the two opposite ways round a path (Carroll 3.112)
\left[\nabla_\rho,\nabla_\sigma\right]X^\mu=R_{\ \ \ \nu\sigma\rho}^\mu X^\nu&\phantom {10000}(16)\nonumber
\end{align}That equation is pretty similar to (1) and then there's (4).

The second part is very simple if you remember that if a Killing vector exists then it is always possible to find a coordinate system where it is one of the basis vectors and the metric will be independent of the coordinate for that basis vector. Carroll stated this after the Killing equation at his 3.174.

I was unable to answer the questions but was guided by a solution I stumbled across on  Semantic Scholar by Professor Alan Guth.

More on my struggles and links to Guth solution at
Ex 3.12 Derivatives of Killing vectors.pdf (4 pages)

Wednesday 6 November 2019

Tensor Calculus with Word VBA macros

Presentation of Word VBA macros for helping with tensor equations!
One of the things you often to do when working in with tensor equations is to expand things like Riemann tensors, covariant derivatives and Christoffel symbols. The expansions are shown below at (1) to (6). They all involve shuffling indices and introducing dummy indices which are used in 'contractions'. These are summations over the dummy index so more like an expansion than a contraction.

The expansions are fiddly and after a while my eyes start to pop out. You can see that they pretty much all turn into Christoffel symbols, the ##\Gamma_{\mu\nu}^\sigma## thing. That in turn can be expressed in terms of the metric and inverse metric: ##g_{\mu\nu}\ ,\ g^{\mu\nu}##. So if you know the metrics you can write out all the Christoffel symbols in terms of coordinates - proper equations. There are only slightly less than ##n^3## of these in ## n## dimensions. There's a reduction to only ##n^2\left(n+1\right)/2## coefficients because ##\Gamma_{\mu\nu}^\sigma=\Gamma_{\mu\nu}^\sigma## (that's called being 'torsion free', more like drudgery free). Even when it's six on a the surface of a sphere, that process finally pops my eyes right out of their sockets and it's all too easy to make a mistake at any stage of the process. And it's  40 Christoffel symbols in the four dimensions of General Relativity.

When I came to Exercise 3.12 on 14 October I had to expand a Riemann tensor again. So I wrote some code to do it. That was quite hard and I decided to do covariant derivatives and Christoffel symbols while I was at it. Then I suffered from mission creep and decided to fully expand those ##n^2\left(n+1\right)/2## coefficients of the Christoffel symbol. There are lots of terms like ##g^{\sigma\lambda}\partial_\mu g_{\lambda\nu}## which very often vanish, it was quite hard to work out which they were and use the information. I had to learn about the structure of MS equations from scratch. That may be the subject of another post. Three weeks slipped by and now I now have this all-singing toolbox:


LxL: Insert inline equation
x: Insert equation on new line
x|(n): Insert numbered equation in new table row
x|x|(n): Inserted two part numbered equation row

Grid: Operations for enhancing tabular equations

Expand: Expand equation containing Riemann, Christoffel and/or covariant derivative
Renumber: Renumber all equations and references
To Web: Prepare document for Web/MathJax in new window

Pick up: Pick up definitions from equations

Write Metrics: Write out (& calculate) picked up metrics
Write ##\Gamma##s: Write out full expansion of Riemann symbols using metrics and coordinates.

The box at the bottom is for information and error messages, when it might beep.

Riemann tensor

R_{\ \ \ \sigma\mu\nu}^\rho=\partial_\mu\Gamma_{\nu\sigma}^\rho-\partial_\nu\Gamma_{\mu\sigma}^\rho+\Gamma_{\mu\lambda}^\rho\Gamma_{\nu\sigma}^\lambda-\Gamma_{\nu\lambda}^\rho\Gamma_{\mu\sigma}^\lambda&\phantom {10000}(1)\nonumber
\end{align}Covariant derivative
(0,2) tensor
\nabla_\mu V^{\nu\rho}=\partial_\mu V^{\nu\rho}+\Gamma_{\mu\lambda}^\nu V^{\lambda\rho}+\Gamma_{\mu\lambda}^\rho V^{\nu\lambda}&\phantom {10000}(2)\nonumber
\nabla_\mu V^\nu=\partial_\mu V^\nu+\Gamma_{\mu\lambda}^\nu V^\lambda&\phantom {10000}(3)\nonumber
\end{align}one form
\nabla_\mu\omega_\nu=\partial_\mu\omega_\nu-\Gamma_{\mu\nu}^\lambda\omega_\lambda&\phantom {10000}(4)\nonumber
\end{align}Other tensor
\nabla_\mu U_{\ \ \lambda\ \ \kappa}^{\nu\ \rho}=\partial_\mu U_{\ \ \lambda\ \ \kappa}^{\nu\ \rho}+\Gamma_{\mu\sigma}^\nu U_{\ \ \lambda\ \ \kappa}^{\sigma\ \rho}+\Gamma_{\mu\sigma}^\rho U_{\ \ \lambda\ \ \kappa}^{\nu\ \sigma}-\Gamma_{\mu\lambda}^\sigma U_{\ \ \sigma\ \ \kappa}^{\nu\ \rho}-\Gamma_{\mu\kappa}^\sigma U_{\ \ \lambda\ \ \sigma}^{\nu\ \rho}&\phantom {10000}(5)\nonumber
\end{align}Christoffel symbol
\Gamma_{\mu\nu}^\sigma=\frac{1}{2}g^{\sigma\lambda}\left(\partial_\mu g_{\lambda\nu}+\partial_\nu g_{\mu\lambda}-\partial_\lambda g_{\mu\nu}\right)&\phantom {10000}(6)\nonumber
\end{align} To read more about the macros click Read More.
Macros available at Archive2019-11-06.
Pdf file here: Tensor Calculus.pdf.