Question on cosmological redshift

I have a question on cosmological redshift which I have just learned about from Sean Carroll. After calculating it for an expanding universe he does a thought experiment to show that it is different to Doppler redshift which would be detected if two galaxies were flying away from each other in a flat (therefore not expanding) universe.

We have flat universe L on the left with two galaxies separated by distance ## s##. A photon is emitted from galaxy 1, galaxy 2 is quickly propelled to a separation of ##2s##, galaxy 2 stops and the photon arrives. Since the galaxies are now not relatively moving there would be no Doppler redshift.

On the right the galaxies are also separated by ## s## but, instead of moving a galaxy, the universe expands by a factor of 2 (it briefly gets a metric like ## ds^2=-dt^2+t^2dx^2##), then stops expanding and then the photon arrives. According to the cosmological redshift formula, the photon has a redshift.

This implies that the galaxies in universe R are still separated by ## s##, because rulers would expand along with everything else. One can also check this by drawing out and back light paths before and after expansion.

This is spooky. It also implies that in our 'expanding' universe distant galaxies are not really moving away! One also wonders how we tell that it's cosmological not Doppler redshift.

Have I got the picture roughly right? The next step will be to compare these to real values like the Hubble constant.
On PP at https://www.physicsforums.com/threads/have-i-got-the-right-picture-for-cosmological-redshift.977816/

The great Orodruin replied:

He wrote about this in January 2018 here. It's about 10 pages and the punch line is:
"This example underlines the main message of this Insight: That the assignment of properties and interpretations based on an assumed set of preferred coordinates is not necessarily coordinate invariant and we need to be careful not to impose any coordinate interpretation as absolute truth. In particular, I have seen many instances where people in popular texts make a very strong claim that cosmological redshift is fundamentally different from Doppler shift. The computations above clearly show that this is not the case, instead cosmological redshift and Doppler shift are two sides of the same coin, just viewed in different coordinates. I also have to admit to being among the set of people who did this error until I actually performed these calculations myself."

I have scanned the article (thus getting to the punch line). It deserves further study, especially I know more about the  Robertson–Walker (RW) universe which are the subject of chapter 8 section 2.

Question about geodesic on sphere

I am working from Sean Carroll's Spacetime and Geometry : An Introduction to General Relativity and have got to the geodesic equation. I wanted to test it on the surface of a sphere where I know that great circles are geodesics and is about the simplest non-trivial case I can think of.

Carroll derives the geodesic equation:
\begin{align}
\frac{d^2x^\sigma}{d\lambda^2}+\Gamma_{\mu\nu}^\sigma\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}=0&\phantom {10000}(1)\nonumber
\end{align}##\Gamma_{\mu\nu}^\sigma## is the Christoffel symbol (torsion-free and metric compatible)
\begin{align}
\Gamma_{\mu\nu}^\sigma=\frac{1}{2}g^{\sigma\rho}\left(\partial_\mu g_{\nu\rho}+\partial_\nu g_{\rho\mu}-\partial_\rho g_{\mu\nu}\right)&\phantom {10000}(2)\nonumber
\end{align}In this case the indices are 0,1 and ##x^0=\phi## the colatitude (angle from north pole), ##\ x^1=\theta## , longitude. The metric and inverse metric are
\begin{align}
g_{\mu\nu}=\left(\begin{matrix}1&0\\0&\sin^2{\phi}\\\end{matrix}\right)\ \ ,\ \ g^{\mu\nu}=\left(\begin{matrix}1&0\\0&\sin^{-2}{\phi}\\\end{matrix}\right)&\phantom {10000}(3)\nonumber
\end{align}Those should give me equations for ## \phi,\theta## parameterized by ## \lambda##.

I now expand the second term in (1) for ## \sigma=1## and using the fact that ## \Gamma## is torsion-free (##\Gamma_{\mu\nu}^\sigma=\Gamma_{\nu\mu}^\sigma##).
\begin{align}
\Gamma_{\mu\nu}^1\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}&=\Gamma_{00}^1\frac{dx^0}{d\lambda}\frac{dx^0}{d\lambda}+2\Gamma_{10}^1\frac{dx^1}{d\lambda}\frac{dx^0}{d\lambda}+\Gamma_{11}^1\frac{dx^1}{d\lambda}\frac{dx^1}{d\lambda}&\phantom {10000}(4)\nonumber\\

&=\Gamma_{00}^1\left(\frac{d\phi}{d\lambda}\right)^2+2\Gamma_{10}^1\frac{d\phi}{d\lambda}\frac{d\theta}{d\lambda}+\Gamma_{11}^1\left(\frac{d\theta}{d\lambda}\right)^2&\phantom {10000}(5)\nonumber
\end{align}using the metric and (2) and taking each ## \Gamma## in turn, many terms vanish because the metric components are constant or 0
\begin{align}
\Gamma_{00}^1&=\frac{1}{2}g^{1\rho}\left(\partial_0g_{0\rho}+\partial_0g_{\rho0}-\partial_\rho g_{00}\right)=g^{1\rho}\partial_0g_{0\rho}&\phantom {10000}(6)\nonumber\\

&=g^{10}\partial_0g_{00}+g^{11}\partial_0g_{01}=0&\phantom {10000}(7)\nonumber\\
2\Gamma_{10}^1&=g^{1\rho}\left(\partial_1g_{0\rho}+\partial_0g_{\rho1}-\partial_\rho g_{10}\right)=g^{1\rho}\partial_1g_{0\rho}+g^{1\rho}\partial_0g_{\rho1}&\phantom {10000}(8)\nonumber\\

&=g^{10}\partial_1g_{00}+g^{11}\partial_1g_{01}+g^{10}\partial_0g_{01}+g^{11}\partial_0g_{11}&\phantom {10000}(9)\nonumber\\

&=\frac{1}{\sin^2{\phi}}\frac{\partial}{\partial\phi}\left(\sin^2{\phi}\right)=\frac{2\sin{\phi}\cos{\phi}}{\sin^2{\phi}}=2\cot{\phi}&\phantom {10000}(10)\nonumber\\
\Gamma_{11}^1&=\frac{1}{2}g^{1\rho}\left(\partial_1g_{1\rho}+\partial_1g_{\rho1}-\partial_\rho g_{11}\right)=g^{1\rho}\partial_1g_{\rho1}-\frac{1}{2}g^{1\rho}\partial_\rho g_{11}&\phantom {10000}(11)\nonumber\\

&=g^{10}\partial_1g_{01}+g^{11}\partial_1g_{11}-\frac{1}{2}g^{10}\partial_0g_{11}-\frac{1}{2}g^{11}\partial_1g_{11}&\phantom {10000}(12)\nonumber\\

&=\frac{1}{2}\frac{1}{\sin^2{\phi}}\frac{\partial}{\partial\theta}\left(\sin^2{\phi}\right)=0&\phantom {10000}(13)\nonumber
\end{align}Putting the expressions for ## \Gamma## back into (5) we get
\begin{align}
\Gamma_{\mu\nu}^1\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}&=2\cot{\phi}\frac{d\phi}{d\lambda}\frac{d\theta}{d\lambda}&\phantom {10000}(14)\nonumber
\end{align}Putting that back into (1) we have
\begin{align}
\frac{d^2\theta}{d\lambda^2}+2\cot{\phi}\frac{d\phi}{d\lambda}\frac{d\theta}{d\lambda}=0&\phantom {10000}(15)\nonumber
\end{align}I can use the same method to get a solution for ## \sigma=0##, but that's not relevant yet. Put together they show that lines of longitude are always geodesics as is the equator. So far so good… but

A great circle equation (which we know is a geodesic) can be written as
\begin{align}
\theta=\lambda\ \ ,\ \ \phi=\tan^{-1}{\left(\frac{C}{A\cos{\lambda}+B\sin{\lambda}}\right)}&\phantom {10000}(16)\nonumber
\end{align}where ## A,B,C## are constants depending on the start and end points. I calculated this and checked it against the slightly ambiguous Wolfram maths article at http://mathworld.wolfram.com/GreatCircle.html.

From (16) clearly
\begin{align}
\frac{d^2\theta}{d\lambda^2}=0&\phantom {10000}(17)\nonumber
\end{align}so the other term in (15) should always be zero. It is
\begin{align}
2\cot{\phi}\frac{d\phi}{d\lambda}\frac{d\theta}{d\lambda}=-2\frac{\left(B\cos{\lambda}-A\sin{\lambda}\right)\left(A\cos{\lambda}+B\sin{\lambda}\right)}{C^2+\left(A\cos{\lambda}+B\sin{\lambda}\right)^2}&\phantom {10000}(18)\nonumber
\end{align}which is not always zero! So something is wrong.
It's either
(15) which I have calculated in another way to check it or
(16) which I have also plotted and gives good results or
my differentiation of ##\frac{d\phi}{d\lambda}## which I checked with Symbolab: https://www.symbolab.com/solver/ordinary-differential-equation-calculator/?or=dym&query=%5Cfrac%7Bd%7D%7Bdx%7D(%5Carctan(%5Cfrac%7BC%7D%7BD%5Ccos(x)%2BB%5Csin(x)%7D))

Help! What has gone wrong?
On Physics Forums at https://www.physicsforums.com/threads/question-about-geodesic-on-sphere.975370/

Question on partial derivative

I am working on the geodesic equation for the surface of a sphere. While doing so I come across the partial derivative
\begin{align}
\frac{\partial}{\partial\theta}\left(\sin^2{\phi}\right)=0?&\phantom {10000}(1)\nonumber
\end{align}where ## \phi, \theta## are colatitude(angle from north pole) and longitude. I could say that the expression must vanish because ## \phi,\theta## are orthogonal. But I know thate ## \phi## is a function ## \theta##  so I could say, using the chain rule twice, that the expression is
\begin{align}
\frac{\partial}{\partial\theta}\left(\sin^2{\phi}\right)=2\sin{\phi}\cos{\phi}\frac{d\phi}{d\theta}?&\phantom {10000}(2)\nonumber
\end{align}The answer might be very obvious and simple. I think the former is correct but would like some confirmation please.

The detailed reasons for the question are given below. They include my fears about the metric.

The geodesic equation is
\begin{align}
\frac{d^2x^\sigma}{d\lambda^2}+\Gamma_{\mu\nu}^\sigma\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}=0&\phantom {10000}(3)\nonumber
\end{align}where ## \Gamma## is the Christoffel symbol (torsion-free and metric compatible), given by
\begin{align}
\Gamma_{\mu\nu}^\sigma=\frac{1}{2}g^{\sigma\rho}\left(\partial_\mu g_{\nu\rho}+\partial_\nu g_{\rho\mu}-\partial_\rho g_{\mu\nu}\right)&\phantom {10000}(4)\nonumber
\end{align}The coordinates are  ##x^0=\phi## and ##\ x^1=\theta## and the metric and inverse metric are
\begin{align}
g_{\mu\nu}=\left(\begin{matrix}1&0\\0&\sin^2{\phi}\\\end{matrix}\right)\ \ ,\ \ g^{\mu\nu}=\left(\begin{matrix}1&0\\0&\sin^{-2}{\phi}\\\end{matrix}\right)&\phantom {10000}(5)\nonumber
\end{align}## \lambda## will parameterise a line which gives the required geodesic. First I derived what these equations mean and I get two second order differential equations involving ## \phi,\theta,\lambda##. I believe these are
\begin{align}
\frac{d^2\phi}{d\lambda^2}-\sin{\phi}\cos{\phi}\left(\frac{d\theta}{d\lambda}\right)^2=0&\phantom {10000}(6)\nonumber
\end{align}and
\begin{align}
\frac{d^2\theta}{d\lambda^2}+2\cot{\phi}\frac{d\phi}{d\lambda}\frac{d\theta}{d\lambda}=0&\phantom {10000}(7)\nonumber
\end{align}There were two ways to work these equations out. Assuming (2) (that ##{\partial f(\phi)}/{\partial\theta}\neq0##) each method gave a different answer for equation (7) unless I made a mistake, which is quite likely.

I have not yet tried to solve these equations, but I know that a geodesic on a sphere is a great circle so I can work out its equation and it can be written
\begin{align}
\theta=\lambda\ \ ,\ \ \phi=\tan^{-1}{\left(\frac{C}{A\cos{\lambda}+B\sin{\lambda}}\right)}&\phantom {10000}(8)\nonumber
\end{align}where ## A,B,C## are constants depending on the start and end points (in quite a complicated way.) It should be simple to check that (8) satisfies (6) and (7). Clearly ## \phi## depends only on ## \theta## and the start and end points, as one would expect. This leads to the original question.

I checked my equation (8) by using my wonderful 3-D plotter. The geodesics look perfect.

I also searched the internet for the equation and found one on Wolfram MathWorld (http://www.mathworld.wolfram.com/GreatCircle.html). It was slightly different from mine and cause me great pain. It is incorrect! Their colatitudes and latitudes are mixed up. I have told them, but there is no response. I also plotted their solution and it gave a rather long route to Peking. Whatever the case ## \phi## depends on ## \theta## and the start and end points.

(8) tells us that
\begin{align}
\frac{d^2\theta}{d\lambda^2}=0&\phantom {10000}(9)\nonumber
\end{align}Which does not sit well with (7). If I had the metric at (5) 'the wrong way round', it would be
\begin{align}
g_{\mu\nu}=\left(\begin{matrix}\sin^2{\phi}&0\\0&1\\\end{matrix}\right)\ \ ,\ \ g^{\mu\nu}=\left(\begin{matrix}\sin^{-2}{\phi}&0\\0&1\\\end{matrix}\right)&\phantom {10000}(10)\nonumber
\end{align}this would have given (9) at (7). However I want to cross one bridge at a time …

Posted on PF at https://www.physicsforums.com/threads/question-on-partial-derivative.975058/

The first three answers satisfy me that$$
\frac{\partial}{\partial\theta}\left(\sin^2{\phi}\right)=0
$$Equation (8) is just the equation of some line on the sphere, in this case a geodesic.

Second order partial derivatives vanish?

At the end of a long proof I came across something in tensor calculus that seems too good to be true. And if something seems too good to be true .....

The something is that a second order partial derivative vanishes if one of the parts in the denominator is in the same reference frame as the numerator. That is for example
\begin{align}
\frac{{\partial }^2x^{\mu }}{\partial x^{\rho }\partial x^{{\mu }^{'}}}=\frac{{\partial }^2x^{\mu }}{\partial x^{{\mu }^{'}}\partial x^{\rho }}=0 & \phantom {10000}(1) \\
\end{align}The equality of the first two parts follows because partial derivatives commute. We have
\begin{align}
\frac{{\partial }^2x^{\mu }}{\partial x^{{\mu }^{'}}\partial x^{\rho }}=\frac{\partial }{\partial x^{{\mu }^{'}}}\left(\frac{\partial x^{\mu }}{\partial x^{\rho }}\right) & \phantom {10000}(2) \\
\end{align}but\begin{align}
\frac{\partial x^{\mu }}{\partial x^{\rho }}={\delta }^{\mu }_{\rho } & \phantom {10000}(3) \\
\end{align}where ##{\delta }^{\mu }_{\rho }## is the Kronecker delta which is a constant. (3) seems very reasonable because when ##\mu \neq \rho ##, ##\partial x^{\mu }## and ##\partial x^{\rho }## are orthogonal so ##{\partial x^{\mu }}/{\partial x^{\rho }}## vanishes and when ##\mu =\rho ##, ##{\partial x^{\mu }}/{\partial x^{\rho }}=1##.

So (2) is the derivative of a constant which always vanishes. QED.

Have I made a very stupid mistake? Or am I stating something everybody knows?

On PF at https://www.physicsforums.com/threads/second-order-partial-derivatives-vanish.968079/
The good stevendaryl pointed out my error and gave a little example that proved$$
\frac{{\partial }^2x^{\mu }}{\partial x^{\rho }\partial x^{{\mu }^{'}}}≠\frac{{\partial }^2x^{\mu }}{\partial x^{{\mu }^{'}}\partial x^{\rho }}
$$Partial derivatives only commute if they are in the the same coordinate system, so this IS true$$
\frac{{\partial }^2x^{\mu }}{\partial x^{\rho }\partial x^{{\nu }}}=\frac{{\partial }^2x^{\mu }}{\partial x^{{\nu }}\partial x^{\rho }}$$Say we had$$
f\left(x,y\right)=x^2{\mathrm{sin} y\ }
$$then$$
\frac{\partial }{\partial x}\left(\frac{\partial }{\partial y}f\right)=\frac{\partial }{\partial x}\left(x^2{\mathrm{cos} y\ }\right)=2x{\mathrm{cos} y\ }
$$and$$
\frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}f\right)=\frac{\partial }{\partial y}\left(2x{\mathrm{sin} y\ }\right)=2x{\mathrm{cos} y\ }$$Say we had
$$x^{\mu }=z$$then $$\frac{\partial }{\partial x}\left(\frac{\partial }{\partial y}z\right)=0$$and
$$\frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}z\right)=0$$Say we had$$x^{\mu }=y
$$then$$
\frac{\partial }{\partial x}\left(\frac{\partial }{\partial y}y\right)=\frac{\partial }{\partial x}\left(1\right)=0$$and$$
\frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}y\right)=\frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}y\right)=0$$

Question on covariant derivatives

I am reading I am reading Spacetime and Geometry : An Introduction to General Relativity -- by Sean M Carroll and have arrived at chapter 3 where he introduces the covariant derivative ##{\mathrm{\nabla }}_{\mu }##. He makes demands on this which are \begin{align}
\mathrm{1.\ Linearity:}\mathrm{\ }\mathrm{\nabla }\left(T+S\right)=\mathrm{\nabla }T+\mathrm{\nabla }S & \phantom {10000}(1) \\
\mathrm{2.\ Leibniz\ rule:}\mathrm{\nabla }\left(T\ \otimes \ \ S\right)=\left(\mathrm{\nabla }T\right)\ \ \otimes \ \ S+T\ \otimes \ \ \left(\mathrm{\nabla }S\right) & \phantom {10000}(2) \\
{\mathrm{3.\ Commutes\ with\ contractions:}\mathrm{\nabla }}_{\mu }\left(T^{\lambda }_{\ \ \ \lambda \rho }\right)={\left(\mathrm{\nabla }T\right)}^{\mathrm{\ \ \ }\lambda}_{\mu \ \ \lambda \rho } & \phantom {10000}(3) \\
{\mathrm{4.\ Reduces\ to\ partial\ derivative\ on\ scalars:}\mathrm{\nabla }}_{\mu }\phi ={\partial }_{\mu }\phi  & \phantom {10000}(4) \\
\end{align}1,2 and 4 seem reasonable but I cannot understand 3 and he does not seem to use it, even though he implies that he does.

The LHS of (3) seems straight forward\begin{align}
{\mathrm{\nabla }}_{\mu }\left(T^{\lambda }_{\ \ \ \lambda \rho }\right) & ={\partial }_{\mu }T^{\lambda }_{\ \ \ \lambda \rho }+{\mathrm{\Gamma }}^{\lambda }_{\mu \kappa }T^{\kappa }_{\ \ \ \lambda \rho }-{\mathrm{\Gamma }}^{\kappa }_{\mu \lambda }T^{\lambda }_{\ \ \ \kappa \rho }-{\mathrm{\Gamma }}^{\kappa }_{\mu \rho }T^{\lambda }_{\ \ \ \lambda k} & \phantom {10000}(5) \\
 & ={\partial }_{\mu }T^{\lambda }_{\ \ \ \lambda \rho }-{\mathrm{\Gamma }}^{\kappa }_{\mu \rho }T^{\lambda }_{\ \ \ \lambda k} & \phantom {10000}(6) \\
\end{align}Which is very like the rule for the covariant derivative of a (0,1) tensor.

I understand that the ##\mathrm{\nabla }T## in (1) and (2) means ##{\mathrm{\nabla }}_{\sigma}T## where ##T## is some tensor. So the RHS of (3) appears to be ##{\left({\mathrm{\nabla }}_{\sigma}T\right)}^{\mathrm{\ \ \ }\lambda}_{\mu \ \ \lambda \rho }## which leaves too many indices on the RHS. Otherwise the RHS is some kind of derivative with one contra- and three co-variant indices. What is that?

Help!
Posted on Physics Forums at https://www.physicsforums.com/threads/question-on-covariant-derivatives.967509/

Exercise 2.08 Exterior derivative and modified Leibniz rule

This exercise started in section 2.9 on which I wrote a seven page commentary before accepting Carroll's challenge to prove the modified Leibniz rule for the exterior derivative.

Carroll introduces differential forms: "A differential p-form is a completely antisymmetric (0,p) tensor. Thus scalars are automatically 0-forms and dual vectors (downstairs index) are one-forms. We also have the 4-form ##{\epsilon }_{\mu \nu \rho \sigma }##, the Levi-Civita tensor." I had a slight problem with 0- and 1-forms but this was resolved on Physics Forums.

First we meet the wedge product:

Given a##\ p##-form ##A## and a ##q##-form ##B## we can form a ##(p+q)##-form known as the wedge product ##A\wedge B## by taking the antisymmetrised tensor product:
$${\left(A\wedge B\right)}_{{\mu }_1\dots {\mu }_{p+q}}=\frac{\left(p+q\right)!}{p!q!}A_{[{\mu }_1\dots {\mu }_p}B_{{\mu }_{p+1}\dots {\mu }_{p+q}]}$$
I prove that the wedge product of an n-dimensional 2-form and 1-form is completely antisymmetric in any number of dimensions n ##\mathrm{\ge}## 2 and therefore a 3-form.

Then we meet the exterior derivative

$${\left(\mathrm{d}A\right)}_{{\mu }_1\dots {\mu }_{p+1}}=\left(p+1\right){\partial }_{[{\mu }_1}A_{{\mu }_2\dots {\mu }_{p+1}]}$$They both involve the ghastly total antisymmetrisation operation [] on indices. It is defined back in his equation (1.80) as

$$T_{[{\mu }_1\dots {\mu }_n]}=\frac{1}{n!}\left(T_{{\mu }_1\dots {\mu }_n}\pm \mathrm{sum\ over\ permutations\ of}\ {\mu }_1\dots {\mu }_n\right)$$

This led on to Exercise 2.08

Question

Verify (2.78): For an exterior derivative of a product of a p-form ω and a q-form η, we have the modified Leibnitz rule:

$$\mathrm{d}\left(\omega \wedge \eta \right)=\left(\mathrm{d}\omega \right)\wedge \eta +{\left(-1\right)}^p\omega \wedge \left(\mathrm{d}\eta \right)$$

Answer 

Here we have the ghastly total antisymmetrisation operation [] again and nested in itself. I had to invent a new notation

$$\sum_{\mp \mathrm{\circlearrowleft }}{A_{{\mu }_1\dots {\mu }_n}}\equiv \left(A_{{\mu }_1\dots {\mu }_n}\pm \mathrm{sum\ over\ permuta}\mathrm{tions\ of}\ {\mu }_1\dots {\mu }_n\ where\ we\ use\ -\ for\ odd\ permutations\ and\ +\ for\ even.\right)$$

because writing the stuff about the permutations every time would be stupid and does not fit on a line. I expanded each term in the question equation and reached expressions like

$$\frac{{\left(-1\right)}^{p\left(q+1\right)}}{\left(q+1\right)!p!q!}\sum_{\mp \mathrm{\circlearrowleft }}{\left(\sum_{\mp \mathrm{\circlearrowleft }}{{\mathrm{\partial }}_{{\mu }_1}{\eta }_{{\mu }_2\dots {\mu }_{q+1}}}\right){\omega }_{{\mu }_{q+2}\dots {\mu }_{p+q+1}}}$$

where you can see the nested expansions explicitly. Each term had a different variant of the nesting so the nesting had to be removed and I proved, for example and avoiding too many subscripts, that

$$\sum_{\mp \mathrm{\circlearrowleft }}{\left(\sum_{\mp \mathrm{\circlearrowleft }}{{\mathrm{\partial }}_a{\eta }_{c_1\dots c_q}}\right){\omega }_{b_1\dots b_p}}\mathrm{=}\left(q+1\right)!{\left(-1\right)}^{q(p+q)}\sum_{\mp \mathrm{\circlearrowleft }}{{\mathrm{\partial }}_a{\omega }_{b_1\dots b_p}{\eta }_{c_1\dots c_q}}$$

factorials cancelled beautifully but I was left with

$${\mathrm{d}\left(\omega \wedge \eta \right)}_{\ }=\left(\mathrm{d}\omega \right)\wedge \eta ={\left(-1\right)}^{\left(q+p\right)}\omega \wedge \left(\mathrm{d}\eta \right)$$

which is not the same as the modified Leibnitz rule, in other words, junk.

I also show that the equation does not work for two 1-forms. In fact I get$$2\mathrm{d}\left(\omega \wedge \eta \right)=\left(\mathrm{d}\omega \right)\wedge \eta +{\left(-1\right)}^p\omega \wedge \left(\mathrm{d}\eta \right)$$which agrees with my wrong answer. I am within spitting distance of the correct proof!  I hope to come back to this problem.

Read the full account
Commentary 2.9 Differential forms.pdf (7 pages)
Ex 2.08 Exterior derivative and modified Leibnitz rule.pdf (8 pages - I have long thought that he was Leibnitz. That is acceptable, but Leibniz is better.)

There is also an answer here from University of California, Santa Barbara (UCSB).
It starts from ##\mathrm{d}\left(\omega \wedge \eta \right)## and uses the ordinary Leibnitz rule to split that in two. It then manipulates the two parts to convert them into the RHS of the question equation. There are some rather 'hand-waving' steps such as the first line of text "We drop the inner set of antisymmetrization brackets in the second line because the terms are antisymmetrized in all of their indices". It works in the opposite direction to what I did.

Question on canonical for of the metric

I posted a question on the canonical for of the metric to physics forums here. It is about equation (2.46) in the book.  he writes:
Quote: A useful characterisation of the metric is obtained by putting g_μν into its canonical form. In this form the metric components become

$$ g_{\mu\nu} = \rm{diag} (-1, -1,...-1,+1,+1, ... +1,0,0, ... ,0) $$

where "diag" means a diagonal matrix with the given elements. So Carroll's expression seems to imply a diagonal matrix with with a minimum size 9x9 and one extra row and column wherever one of his .'s takes a value. That's too vague!

There were two good answers and I would rewrite (2.46) as

$$ g_{\mu\nu} = \rm{diag} (-1, 0,1...-1, 0,1 ... -1, 0,1 ... -1, 0,1) $$

i.e. each diagonal element can be -1,0 or 1. 

Sadly Orodruin wrote "That would have been wrong and very unclear." He did not elucidate.
I then wrote:

Does this work?As a matrix, a non-degenerate metric in canonical form is diagonal with ±1 in each component.

Orodruin:
Technically, that would be a pseudo-metric unless you have +1 in all diagonals, but in physics we just call it metric anyway.
martinbn:
This

$$ g_{\mu\nu} =  \begin{pmatrix} -1 & 0 & 0 & 0 \\   0 & 1 & 0 & 0 \\   0 & 0 & 1 & 0 \\    0 & 0 & 0 & 1 \end{pmatrix} $$

and this

$$ g_{\mu\nu} = \rm{diag} (-1, 1, 1,1) $$ 

are the same.

I was tempted to reply Are we discussing how many angels can fit on a pinhead? But religion is not allowed.

So the canonical form of the R² metric is

$$ \begin{pmatrix} 1 & 0 \\   0 & 1  \\    \end{pmatrix} $$

and the canonical form of the Minkowski metric (which is Lorentzian) is

$$ \begin{pmatrix} -1 & 0 & 0 & 0 \\   0 & 1 & 0 & 0 \\   0 & 0 & 1 & 0 \\    0 & 0 & 0 & 1 \end{pmatrix} $$

Question on co- / contra-variant coordinates

I am studying co- and contra- variant vectors and I found the video at https://www.youtube.com/watch?v=8vBfTyBPu-4 very useful. It discusses the slanted coordinate system where the X, Y axes are at an angle of α. One can get the components of v either by dropping perpendiculars to the axes (v_i) or by dropping a line parallel to the other axis (vⁱ). These give correct results for the norm v_i vⁱ and the dot product v_i wⁱ. (I have not shown w). So the vⁱ are called contravariant and the v_i are called covariant. According to the video Dirac thought this was a great example.

But both v_i and vⁱ  contra-vary with a change of scale of the basis vectors. This contradicts some definitions of contravariant and covariant components, e.g. this one on Wikipedia. These definitions say that covariant components co-vary with a change of scale.

Is there a simple resolution to this apparent contradiction?

Submitted as comment on video and
on Physics Forums at https://www.physicsforums.com/threads/covariant-coordinates-dont-co-vary.959888/

Orodruin answered on PF seven hours later with "The covariant and contravariant components belong to different basis vectors." Cryo came up with something a bit more complicated and I followed Orodruin. So here's how it goes:

To get v with the covariant coordinates we must have

v = v_xe_x + v_ye_y      (1)

I have drawn them in on the diagram and e_x is clearly smaller than e^x, which I have conveniently drawn so that v_x = 1. We must also have

 v_xe_x = v^xe^x       (2)

and similarly

v_ye_y = v^ye^y       (3)

which give us

v = v^xe^x + v^ye^y       (4)

which is as it should be.

(3) gives us

v_y = e^y(e_y)^-1v^y       (5)

where (e_y)^-1 is the inverse of e_y: or e_y(e_y)^-1 = 1.

(5) shows us that v_y does indeed vary with e^y with a strange constant of proportionality (e_y)^-1v^y.

Back to the drawing board

Orodruin did not like that at his #5 and then gave me a tip at #7 that the dual (or covariant) bases are given by ê^a = ∇x^a. I will use ê for basis vectors in future, following Carroll. From the formula I could calculate the covariant basis vectors in terms of the Cartesian basis vectors because I already had the inverse metric of the covariant system. This was calculated from the transformations of the systems to and from Cartesian.

The answer came:

and draw them properly (x = 1, y = 2)

From (NF12) it is clear that  ê^x is π/2 - α below the Cartesian X axis. Therefore the angle between ê^y and ê^x is a right angle, and the projected line from v^y  is parallel to ê^x as we should have guessed and Orodruin intimated. Likewise ê^y is parallel to the projected line from v_x and always on the Cartesian Y axis. In addition |êⁱ | > |ê_i |.

Getting back to the original question, what happens to v_x if we double the contravariant basis ê_x to ê'_x? We could leave  v_x, ê^x, v_y, ê^y alone and they would still give v. We could double ê^x and halve v_x. Neither option would co-vary. Both would give incorrect values for |v|. We need to find the metric of the primed system so we can just calculate the primed covariant coordinates and bases. We also cannot use the previous technique (calculating from the transformations of the systems to and from Cartesian), because we don't know how to transform primed covariant coordinates to Cartesian - we don't know what the primed coordinates are.

Therefore we use the pullback operator which we met two posts ago in Commentary on Appendix A: Mapping S2 and R3.

If we have our contravariant primed bases, with scale factors a,b so

ê'_x = aê_x  ,  ê'_y = bê_y 

which give

v'^x  = v^x  ⁄ a   ,  v'^y =v^y  ⁄ b 

Turning all the handles we end up with

v'_x = a²v'^x  + abv'^ycos⁡α

v'_y =b²v'^y + abv'^xcos⁡α

Since v' ⁱ  decrease as a, b and v'_i  increase as a²  ,b² ,ab, we can safely say that v'_i  increase as ê_i . They co-vary. It also looks as if they will 'compensate' for v' ⁱ , giving correct |v| and v.w. We would just have to follow the trail.

The full details of all the above are contained in sections 5b and Note F of  Commentary 1.1 Tensors matrices and indexes.pdf

What is a commutator?

Wikipedia says: "Conceptually, the Lie bracket [X, Y] is the derivative of Y along the flow generated by X."

A Lie bracket is called a commutator in the book.

Question: Reductio ad absurdum on commutators?

I am studying Spacetime and Geometry : An Introduction to General Relativity by Sean M Carroll and have a question about commutators of vector fields. A vector field on a manifold can be thought of as differential operator which transforms smooth functions to smooth functions on the manifold. For a vector field X and a function f(xi) we write
X(f) = g, where g is another function. We then define the commutator of two fields X and Y as

[X,Y](f) = X(Y(f)) - Y(X(f)

In the exercise I am working on (2.05), we are asked to find two vector fields whose commutator does not vanish. An important step is to show that if the commutator vanishes for one function f, it vanishes for all functions. This is implied by the question but not proven.

I proved it this way using 'Reductio ad absurdum'. Is this correct?

Our starting point is f  ≠ 0 and [X,Y](f) = 0. We have another function g ≠ 0 and [X,Y](g) ≠ 0.
We already know that commutators are linear (from the previous exercise), so

[X,Y](f + g) = [X,Y](f) + [X,Y](g)

or

[X,Y](f + g) = [X,Y](g)

Therefore f = 0, which breaks our starting assumption, with which there must be some error. The only non trivial possibility is that [X,Y](g) = 0. QED?
The question is at
https://www.physicsforums.com/threads/spacetime-and-geometry-vanishing-commutators.954767/
I then searched for Commutator and found this
Commutator of two vector fields:
https://www.physicsforums.com/threads/commutator-of-two-vector-fields.950661/
One answer contained
"[X,Y]  describes how far the endpoints of a rectangle vary if you go along X followed by Y or the other way around. Commuting vector fields mean the two path end at the same point;"

Perhaps I should have put my question in  the physics / Special and General Relativity forum.

Question for Physics Forums: Help with integration conventions

I am reading Spacetime and Geometry by Sean Carroll. In section 1.10 on classical field theory, he uses this formula (1.132)

The curly L is a Lagrange density. S is an action, Φ is a vector potential.

Could the integral also be written as follows?

Posted on physics forums.