## Monday, 30 September 2019

### The Riemann tensor

I'm now reading section 3.6 about the Riemann tensor which expresses the curvature of a manifold. In component form it is$$R_{\ \ \ \sigma\mu\nu}^\rho=\partial_\mu\Gamma_{\nu\sigma}^\rho-\partial_\nu\Gamma_{\mu\sigma}^\rho+\Gamma_{\mu\lambda}^\rho\Gamma_{\nu\sigma}^\lambda-\Gamma_{\nu\lambda}^\rho\Gamma_{\mu\sigma}^\lambda$$I want to check some of Carroll's assertions and challenges.

• Equation 3.111 which he said was very straightforward
• Prove that the Riemann tensor really is a tensor
• Showing that the Riemann tensor as a map is the same as its component form

I succeeded on the first and the last. The second sounds odd but what is needed is to use the transformation law for $\Gamma$, which is not a tensor, and see if the transformed equation gives the proper transformation for the Riemann tensor. Basically tons of stuff needs to cancel. The first step gives this

and the second this (the line numbers on the left refer to the parts from above)

That was 22 terms in total. The green ones were the ones that were wanted. The rest had to vanish. I only succeeded in removing four.

On the way I relearnt a few techniques and so created the tensor tricks section of Important equations. I also learnt how to take a 'second order covariant derivative'. I had failed to spot the example Carroll gave in equation 3.111. It's quite interesting. The problem is to work out $\nabla_\lambda\left(\nabla_\eta Z^\rho\right)$. If one was doing an ordinary second order derivative one would differentiate the inner part then differentiate the outer part. The reverse happens with a covariant derivative: One must first take the covariant derivative of the inner part and then calculate the covariant derivatives that are left over.
Full details at  Commentary 3.6 Riemann tensor.pdf

## Tuesday, 24 September 2019

### Question

Consider a 2-sphere with coordinates $\left(\theta,\phi\right)$ and a metric
\begin{align}
{ds}^2={d\theta}^2+\sin^2{\theta}{d\phi}^2&\phantom {10000}(1)\nonumber
\end{align}a) Show that lines of constant latitude ($\phi=\rm{constant}$) are geodesics, and that the only line of constant latitude ($\theta=\rm{constant}$) that is a geodesic is the equator  ($\theta=\pi/2$).

b) Take a vector with components $V^\mu=\left(1,0\right)$ and parallel transport it once round a circle of constant latitude. What are the components of the resulting vector, as a function of $\theta$?

### Answers

Part (a) was quite easy because I have laboured over it before. If I had known that the general solution to $$\frac{d^2V}{d\phi^2}+kV=0$$was$$V=A\cos{\left(k\phi\right)}+B\sin{\left(k\phi\right)}$$then I could have done the second part on my own. I needed some help from Prof Anthony Aguirre at UC Santa Cruz for that differential equation.

It seems that 'parallel transporting' is pretty wonky. Here's how it goes at 70° colatitude (polar angle):
Once the vector has got all round the sphere (at 70°) it's pointing in a new direction, even thought it's length never changes. (It appears to here from perspective effects.) Around equation 3.42 Carroll writes "It follows that the inner product of two parallel-transported vectors is preserved ... This means that parallel transport ... preserves the norm of vectors, the sense of orthogonality and so on.💡💡

## Sunday, 22 September 2019

### 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.

## Friday, 20 September 2019

### Question

In Euclidean three-space we can define paraboloidal coordinates $\left(u,v,\phi\right)$ via$$x=uv\cos{\phi}\ \ y=uv\sin{\phi}\ \ z=\frac{1}{2}\left(u^2-v^2\right)$$(a) Find the coordinate transformation matrix between paraboloidal and Cartesian coordinates ${\partial x^\alpha}/{\partial x^{\beta^\prime}}$ and the inverse transformation. Are there any singular points in the map?

(b) Find the basis vectors and the basis one-forms in terms of Cartesian basis vectors and forms.

(c) Find the metric and inverse metric in paraboloidal coordinates.

(d) Calculate the Christoffel symbols.

(e) Calculate the divergence $\nabla_\mu V^\mu$ and Laplacian $\nabla_\mu\nabla^\mu f$.

### Overview

I completely screwed up calculation of the metric, getting one that was very non-diagonal and would have been hard work and I had to go back to spherical polar coordinates to see where I had gone wrong, then it all worked except for a sign error in the inverse transformation matrix. It was fairly obvious which component contained the error.

The way that a covariant tensor ends up in the wrong coordinates after using the general tensor transformation law (e.g. at (81)) is very odd and needs further investigation. I think it's just inevitable and would often need transformation back into the correct coordinate system.

I learned a bit more about basis vectors, but this exercise mainly seemed to be teacher-torture with plenty of exercise on differentiation. The Laplacian $\nabla_\mu\nabla^\mu f$ is not in the index of this book. I hope it is discussed somewhere. I spent about half the time developing drawings of the basis vectors and one-forms but sadly the animation broke down. It was a surprise that the coordinate transformation matrix and its inverse not only take tensors back and forth between coordinate systems but are also inverses in the matrix sense. I suppose I should have known.

When calculating the Christoffel symbols, my first shot was 77% correct. I suppose if this was an exam question that would be quite good, except that I have taken too much time!

### Answer

See: Ex 3.04 Paraboloidal coordinates.pdf (15 pages, 147 equations)

### Exercise 3.02 Spherical gradient divergence curl as covariant derivatives

 Top of last page in German version of Jackson

### Question

You are familiar with the operations of gradient ($\nabla\phi$), divergence ($\nabla\bullet\mathbf{V}$) and curl ($\nabla\times\mathbf{V}$) in ordinary vector analysis in three-dimensional Euclidean space. Using covariant derivatives, derive formulae for these operations in spherical polar coordinates $\left\{r,\theta,\phi\right\}$ defined by
\begin{align}
x&=r\sin{\theta}\cos{\phi}&\phantom {10000}(1)\nonumber\\
y&=r\sin{\theta}\sin{\phi}&\phantom {10000}(2)\nonumber\\
z&=r\cos{\theta}&\phantom {10000}(3)\nonumber
\end{align}Compare your results to those in Jackson (1999) or an equivalent text. Are they identical? Should they be? (JD Jackson, Classical Electrodynamics, Wiley 1999)

### Answer

One of the most difficult things about this exercise was finding a copy of Jackson (1999). It is a 'classic' text, so, unlike most English mathematical text books in the Berlin library, the only copies available were translation into German. Luckily it wasn't too hard to find the required results. They were on the very last page. The heading is 'Description of vector operations in various coordinate systems'. Simple really!

Analyse Fonctionnelle left a comment on 7 March 2021. Here it is again with $signs round latex expressions, so it looks nicer. I also made some replies in italics. Hello, I wonder if you are still working on the exercises of Carroll's book. Continue the good work, I think it's very helpful. Thank you! I'm kind of running out of exercises, so there are not many new ones now. First of all, all the Christoffell symbols are correct, but a few things, 1. When working with Christoffell symbols, it's best not to contract, lower or raise indices because they are not tensors. So when you calculate the covariant form of the divergence, you should do$g^{ij}\nabla_i V_j$first, instead of carrying the contraction of$i$into the calculation. And then do the covariant derivative of$V_j$(there'll be a minus sign in front of the$\Gamma$'s). Quite right! 2. The covariant form of curl should be$\epsilon^{ijk}\nabla_j V_k \partial_i$and the whole thing divided by the square root of the determinant of the metric. The way you wrote in the pdf will give you a number, not a vector. And the square root of$det(g)$is because$\epsilon$is not a tensor but a tensor density. Sounds plausible. 3. As you said at the end of the pdf, you'll need to convert everything to an orthonormal basis in order to get the form in Jackson's book. Basically, the new basis is$\hat{\partial_i} = \partial_i/\sqrt{g_{ii}}$(no summation) and correspondingly, the new component is$\hat{A^i} = A^i \sqrt{g_{ii}}$(no summation). Sounds plausible. And there will be places where you need to change back and forth between$V_i$and$V^i$. It's not straightforward in any means. Thank you for putting in the time and effort to try to work out the exercises. I really appreciate it. I'm sure many other people do too. Thanks very much for the thoughtful feedback. ## Wednesday, 18 September 2019 ### Exercise 3.04 (with Spherical coordinates) This is a model for doing the real exercise 3.4 which was on paraboloidal coordinates$\left(u,v,\phi\right)$. I did it because I know, or can find, the answers and so I can check my methods. There is very little information on paraboloidal coordinates that I can find😖. I was particularly hazy on how to calculate the basis vectors. Now I know! It appears that basis one-forms point in the same direction as basis vectors and the$r$basis one-form is the same as the vector. I probably should think about that harder. I also had quite a lot of problems with signs on terms such as$\sqrt{x^2+y^2}$. Inattentive use of https://www.derivative-calculator.net/ gave wrong answers. Hopefully I will very soon complete the nasty real question. ### Question In Euclidean three-space we can define spherical coordinates$\left(r,\theta,\phi\right)via \begin{align} x=r\sin{\theta}\cos{\phi}&\phantom {10000}(1)\nonumber\\ y=r\sin{\theta}\sin{\phi}&\phantom {10000}(2)\nonumber\\ z=r\cos{\theta}&\phantom {10000}(3)\nonumber \end{align}Note: we are using the 'physicist' convention as shown on the video. \theta$is the colatitude (polar) and$ \phi$the longitude (azimuthal). (a) Find the coordinate transformation matrix between paraboloidal and Cartesian coordinates$\frac{\partial x^\alpha}{\partial x^{\beta^\prime}}$and the inverse transformation. Are there any singular points in the map? (b) Find the basis vectors and the basis one-forms in terms of Cartesian basis vectors and forms. (c) Find the metric and inverse metric in paraboloidal coordinates. (d) Calculate the Christoffel symbols. (e) Calculate the divergence$\nabla_\mu V^\mu$and Laplacian$\nabla_\mu\nabla^\mu f. Full details at Ex 3.04 Spherical coordinates.pdf. 7 pages including link to Excel file for generating the video. ## Wednesday, 11 September 2019 ### Finding the Christoffel symbols and going round in circles In section 3.3 Carroll proves that the geodesic equation gives a line that parallel transports vectors between two points. It's 'embarrassing simple'. He also claims that this line is the shortest distance between the two points. 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}(1)\nonumber \end{align}and the Christoffel symbols are 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}(2)\nonumber \end{align}He then uses calculus of variations to find the shortest distance and, low and behold, it produces the geodesic equation. So now we know that the geodesic equation which started by just parallel transporting also gives a straight line. But he then says that it is actually easier to find the Christoffel symbols by using calculus of variations and then disentangling them from the result than by using (2) and he tells us that we will see this in section 3.5. In contrast Orodruin had said that it is better to find the geodesic equation by writing down the Euler-Lagrange equations for the shortest path. I don't know whether I'm coming or going! It is twisted. Now I've been through the pain of the first part of section 3.5: finding the Christoffel symbols. I followed Carroll carefully and it involves: • doing the variation, • integration by parts, • eliminating boundary terms • and then identifying what's left with the geodesic equation and the Christoffel symbols therein. Everything except the last part is almost a re-enactment of the proof of the Euler-Lagrange equation. The last part seems like work of the devil. It seems much easier to find the Christoffel symbols directly from (2) below as we shall see. Since the metric is often diagonal almost all the terms in (2) vanish. It is also very important to remember that for any function f$where$ \mu\neq\nu\begin{align} \partial_\mu\left(f\left(x^\nu\right)\right)=\frac{\partial}{\partial x^\mu}\left(f\left(x^\nu\right)\right)=0&\phantom {10000}(3)\nonumber \end{align}as I forgot to my shame many moons ago. (The rule applies if other coordinates are part of the function as long as none of them are thex^\mu## coordinate.) So even more terms vanish from (2).

Full details in here

The geodesic equation can also be proved by finding the shortest distance (proper time in GR) between two points (events).
Commentary 3.3 Shortest path by calculus of variations.pdf (3 pages)

Calculating Christoffel symbols by both methods for a simple expanding universe
Commentary 3.5 Expanding Universe Revisited #1.pdf

## Monday, 9 September 2019

### Variational Calculus and the Euler-Lagrange equation

 Euler and Lagrange
Euler (1707-1783) was one of the most brilliant mathematicians of all time and he and Lagrange (1736-1813), a student of his and another great, invented variational calculus and the Euler-Lagrange equation.

Carroll keeps using variational calculus and I think I understand it now. Along the way I found a proof of the Euler-Lagrange equation in Youtube by the Faculty of Khan. It does it in under eight minutes!  That must be a world record. It took me about three hours to understand everything and I wrote it out in this little article at Commentary 3.3 Calculus of Variations and Euler-LaGrange equation.pdf. The proof takes about four pages and then I used it find the equation of a straight line on the plane. I then tried to use it to find a geodesic on a sphere as recommended by Orodruin. I failed. This all happens in section 3.3 and again at the start of section 3.5 'Expanding Universe Revisited' which uses calculus of variations to find geodesic equations and get Christoffel symbols from them. I have more to say on the advantages or not of this technique.

Seven months later videofountain answered my comment on the YouTube video and and pointed out a small error in Commentary 3.3. Now Corrected 😊