## Thursday, 28 January 2021

### Photon sphere

 M87* By Event Horizon Telescope
I come across the photon sphere (or last photon orbit) which is another radius around a Schwarzschild black hole at a distance $3R_s/2$. It is the closest distance for a stable orbit and light would orbit there in an exact circle. I would like to calculate it using Carroll conventions. $R_s=2GM$ is the Schwarzschild radius.

There is a Wikipedia article on it. The proof it gives has some peculiarities. I followed it avoiding those. It is one page.
Here: Commentary 5.4 Photon Sphere.pdf

The famous picture is of M87* the black hole in our neighbour galaxy Messier 87. The black central sphere has a radius of $2.6R_s$ so the photon sphere is well inside that black bit and this is not a good illustration of a photon sphere!

## Friday, 22 January 2021

### Tensor Tricks

 Cat gets Carroll
The file Commentary Tensor Tricks.pdf contains some useful equations for tensor manipulation which I have collected.

The very first on tensor rank comes from early in the book (page 21) where we are told that: a tensor $T$ of type (or rank) $\left(k,l\right)$ is a multilinear map from a collection of dual vectors and vectors to $\mathbf{R}$:
$$T:\left(T_p^\ast\times\cdots\times T_p^\ast\right)_{k\ times}\times\left(T_p\times\cdots\times T_p\right)_{l\ times}\rightarrow\mathbf{R}$$
That is Carroll's 1.56 and I am pretty sure he has that the wrong way round. It should say: a tensor $T$ of type (or rank) $\left(k,l\right)$ is a multilinear map from a collection of $k$ vectors and $l$ dual vectors to  $\mathbf{R}$:$$T:\left(T_p\times\cdots\times T_p\right)_{k\ times}\times\left(T_p^\ast\times\cdots\times T_p^\ast\right)_{l\ times}\rightarrow\mathbf{R}$$But it turns out that you don't really need to know what $k$ and $l$ are separately in General Relativity (because we always have a metric). You only need to know the total rank $k+l$!

#### Contents

• What tensor rank?
• Multi-dimensional Chain Rule
• Partial derivative gives Kronecker delta: Coordinates, Vectors, Tensors
• Partial derivatives commute
• Metric is always symmetric
• Contracting with metric lowers / raises index
• You can lower or raise indices on a tensor equation
• Swap indices with metric or any similar tensor
• Inverse of a matrix
• The determinant of the inverse is reciprocal of the determinant
• Determinant of a tensor in terms of Levi-Civita symbol
• Inverse tensor
• A relationship for the derivative of the determinant
• Fully contracted symmetric × antisymmetric tensor vanishes
• Symmetrising a tensor equation
• Two formulas involving four-velocity
• Second formula
• The projection tensor on four-velocity
• Contra / co-variant tensor transformation matrices
• Tensor contractions using matrices

## Tuesday, 19 January 2021

### Inverses, determinants, Levi-Civita symbol and Laplacian

I have been wandering among inverses of matrices and tensors, determinants, Levi-Civita symbol and the Laplacian and found a wonderful formula for an inverse tensor (4). This leads to the differential equation connecting the determinant of a tensor, the determinant of its inverse, its inverse and itself at (5) and (6). The former can be used to prove the equivalence of Wikipedia's (7) Carroll's (8) formulas for the Laplacian which was where this all started.

I have presented the chain of reasoning as a set of questions and answers:

## Questions

Consider the following:
For a square matrix $A$ and its inverse $A^{-1}$
\begin{align}
\end{align}The adjugate is the transpose of the cofactor matrix. The cofactor matrix is the matrix whose $i,j$th element is the $\left(-1\right)^{i+j}$ times the determinant of the matrix formed by removing row $i$ column $j$ from $A$.
a) Prove that equation (1) is true. Easy.
b) Rewrite it in component form. Very easy.
Carroll's 2.66 for the determinant of a matrix $A$ with components $A_{\ \ \ \ \ {\mu\prime}_1}^{\mu_1}$, was
\begin{align}{\widetilde{\epsilon}}_{{\mu\prime}_1{\mu\prime}_2\ldots{\mu\prime}_n}\det{\left(A\right)}={\widetilde{\epsilon}}_{\mu_1\mu_2\ldots\mu_n}A_{\ \ \ \ \ {\mu\prime}_1}^{\mu_1}A_{\ \ \ \ \ {\mu\prime}_2}^{\mu_2}\ldots A_{\ \ \ \ \ {\mu\prime}_n}^{\mu_n}&\phantom {10000}(2)\nonumber
\end{align}Vanhees71's PF post uses the first simplification of (2) in four dimensions and then says, slightly rearranged, that components of $g^{\mu_11}$ are
\begin{align}
gg^{\mu_11}=\left(-1\right)^{\mu_1}\epsilon^{\mu_1\mu_2\mu_3\mu_4}g_{\mu_22}g_{\mu_33}g_{\mu_44}&\phantom {10000}(3)\nonumber
\end{align}I find a more general formula
\begin{align}
\left|A\right|{\bar{A}}^{\alpha\mu_\alpha}=\epsilon^{\mu_1\mu_2\mu_3\ldots\mu_n}\prod_{\beta\neq\alpha} A_{\mu_\beta\beta}&\phantom {10000}(4)\nonumber
\end{align}Vanhees71 used (3) to get
\begin{align}
\partial_\nu g=gg^{\rho\sigma}\partial_\nu g_{\rho\sigma}&\phantom {10000}(5)\nonumber
\end{align}c) Are (3) and (4) (in 4 dimensions) the same? Easy.
d) Use (2) to prove that the correct one really is correct. Hard.
e) Use the correct one to prove (5). Middling.
f) Prove the generalization of (5):
\begin{align}
\left|\bar{A}\right|\partial_\nu\left|A\right|={\bar{A}}^{\rho\sigma}\partial_\nu A_{\sigma\rho}&\phantom {10000}(6)\nonumber
\end{align}There is a Wikipedia formula in "arbitrary curvilinear coordinates in N dimensions" for the Laplacian:
\begin{align}
\nabla^2=\frac{1}{\sqrt{\left|g\right|}}\frac{\partial}{\partial x^i}\left(\sqrt{\left|g\right|}g^{ij}\frac{\partial}{\partial x^j}\right)&\phantom {10000}(7)\nonumber
\end{align}And Carroll's formula (from exercise 3.4):
\begin{align}
\nabla^2=\nabla_\mu\nabla^\mu=g^{\mu\nu}\nabla_\mu\nabla_\nu&\phantom {10000}(8)\nonumber
\end{align}g) Use (5) to prove that (7) and (8) are the same. Middling to hard.

c) No, d) The correct one is (4). The rest of the answers are in the first six pages of
The remaining six pages contain examples to test the answers and check progress.

## Wednesday, 6 January 2021

### The Voss-Weyl formula

 Hermann Weyl
The Voss-Weyl formula is for the contraction of the covariant derivative of a vector field:$$\nabla_\mu F^\mu=\frac{1}{\sqrt{\left|g\right|}}\frac{\partial}{\partial x^\mu}\left(\sqrt{\left|g\right|}F^\mu\right)$$It immediately solved an outstanding problem from a year ago that I had with the Laplacian. While proving it I find three other interesting formulas:
Jacobi's formula is$$\partial_kQ={\widetilde{Q}}^{ji}\partial_kQ_{ij}$$where $Q_{ij}$ is a tensor, $Q$ is its determinant and what might be new is ${\widetilde{Q}}^{ij}$ which is the cofactor of $Q_{ij}$ and thus as a matrix, ${\widetilde{Q}}^{ji}$ is the transpose of the cofactors of $Q_{ij}$. The cofactor of a component $Q_{ij}$ is the determinant of the matrix made by removing row $i$, column $j$ then multiplying by $\left(-1\right)^{i+j}$. The important thing is that the transpose of the cofactors is called the adjugate, so ${\widetilde{Q}}^{ji}$ is the adjugate and the inverse of a matrix is the adjugate divided by its determinant. We write that:$$Q^{ij}=\frac{1}{Q}{\widetilde{Q}}^{ji}$$Finally on the way to proving the Voss-Weyl formula we find the marvellously symmetrical $$Q^{-1}\partial_kQ=Q^{ij}\partial_kQ_{ij}$$where $Q^{-1}$ is the determinant of the inverse of a rank 2 tensor $Q_{ij}$ which has determinant $Q$. It is also well known that $Q^{-1}=\frac{1}{Q}$.

I must thank David A. Clarke of Saint Mary’s University, Halifax NS, Canada for his guidance.

## Monday, 4 January 2021

### Hubble bubble toil and trouble

 EMPEHI
... is  the name of a 1964 hit by English pop group Manfred Mann, a misquote from Shakespeare's Macbeth but for us it is the continued mystery of the Hubble parameter $H$ which we meet at the end of section 8.3 and was discovered by the great astronomer Hubble and predicted by the Friedmann equations.

Carroll wrote his book in 2003 and gives the Hubble constant, $H_0$, (which is the current value of the Hubble parameter) as $70\pm10\ \rm{km/sec/Mpc}$. Since then many more measurements of it have been made and there seem to be two values 67 and 73 depending on the measuring technique used and it is not known why. This mystery is called the Hubble tension.

We also meet the Hubble time $t_H=1/H_0=13.6\ \rm{billion\ years}$ which is the age of the Universe and the Hubble distance $\ d_H=c/H_0=14\ \rm{billion\ lightyears}$ which is the size of the visible Universe. Obviously.

We also meet the deceleration parameter $q$ , the density parameter $\Omega$ and the critical density $\rho_{crit}$. Recent measurements of the cosmic background anisotropy lead us to believe that $\Omega$ is very close to unity which would mean the Universe is flat and infinite. Nut as Wikipedia says, it might merely be that the universe is much larger than the part we see. Similarly, the fact that Earth is approximately flat at the scale of the Netherlands does not imply that the Earth is flat: it only implies that it is much larger than the Netherlands.

See my notes  at Commentary 8.3 Hubble and more.pdf (4 pages)