Saturday, 12 October 2019

Symmetries and Killing vectors

Sean Carroll, my guide and nemesis
I'm now reading section 3.8 on symmetries and Killing vectors. It's not too hard to follow but there are a few stumbling blocks.

After equation 3.161 for the geodesic in terms of 4 momentum ##p^\lambda\nabla_\lambda p^\mu=0##  Carroll says that by metric compatibility we are free to lower the index ## \mu##. Metric compatibility means that ##\nabla_\rho g_{\mu\nu}=\nabla_\rho g^{\mu\nu}=0## so I tried to show that given that, ##\nabla_\lambda p^\mu=\nabla_\lambda p_\mu##. Here was my first attempt:
Lower the index with the metric, use the Leibnitz rule, use metric compatibility$$
\nabla_\lambda p^\mu=\nabla_\lambda g^{\mu\nu}p_\nu=p_\nu\nabla_\lambda g^{\mu\nu}+g^{\mu\nu}\nabla_\lambda p_\nu=0+\nabla_\lambda p^\mu
$$Then I tried painfully expanding ##\nabla_\lambda g^{\mu\nu}p_\nu## and got the same result. So then I asked on Physics Forums: Why does metric compatibility imply  ##\nabla_\lambda p^\mu=\nabla_\lambda p_\mu##? I got my wrist slapped by martinbn who pointed out that  ##\nabla_\lambda p^\mu=\nabla_\lambda p_\mu## made no sense because there are different types of tensors on each side of the equation. (The ## \mu## is up on one side and down on the other). I was embarrassed😡. 

What Carroll is really saying is that metric compatibility means that$$
\nabla_\lambda p^\mu=0\Rightarrow\nabla_\lambda p_\mu=0
$$which is quite different and easy to show:$$
\nabla_\lambda p^\mu=0
$$$$
\Rightarrow g^{\mu\nu}\nabla_\lambda p_\nu=0
$$$$
\Rightarrow g_{\rho\mu}g^{\mu\nu}\nabla_\lambda p_\nu=0
$$$$
\Rightarrow\delta_\rho^\nu\nabla_\lambda p_\nu=0
$$$$
\Rightarrow\nabla_\lambda p_\rho=0
$$I posted something very like those steps and there was silence which usually means they are correct. The first step uses, ##\nabla_\lambda p^\mu=g^{\mu\nu}\nabla_\lambda p_\nu##, which can be done in several ways
1) ##\nabla_\lambda p^\mu## is a tensor so you can lower (or raise) an index with the metric as usual.
2) ##\nabla_\lambda p^\mu=\nabla_\lambda\left(g^{\mu\nu}p_\nu\right)=p_\nu\nabla_\lambda g^{\mu\nu}+g^{\mu\nu}\nabla_\lambda p_\nu=g^{\mu\nu}\nabla_\lambda p_\nu## as in (1) use the Leibnitz rue and metric compatibility
3) ##\nabla_\lambda p^\mu=\nabla_\lambda\left(g^{\mu\nu}p_\nu\right)=g^{\mu\nu}\nabla_\lambda p_\nu## using Carroll's third rule for covariant derivatives: That they commutes with contractions.

The Leibnitz rule was the second rule of covariant derivatives and I discussed all four in Commentary 3.2 Christoffel Symbol. The third caused angst and another question on PF. I now think that the third rule is just saying that because the covariant derivative is a tensor you can raise and lower indexes on it. 3 and 1 above are really the same. I have suitably amended Commentary 3.2 Christoffel Symbol.

Sometimes I hate Carroll! 

There was also another post on the thread ahead of the first two which referred to a similar question on Stack Exchange. MathematicalPhysicist was asked to show that $$
U^\alpha\nabla_\alpha V^\beta=W^\beta\Rightarrow U^\alpha\nabla_\alpha W_\beta=W_\beta
$$The proof for this is very similar to the above:$$
U^\alpha\nabla_\alpha V^\beta=W^\beta
$$$$
\Rightarrow U^\alpha g^{\beta\gamma}\nabla_\alpha V_\gamma=g^{\beta\gamma}V_\gamma
$$$$
\Rightarrow U^\alpha g_{\mu\beta}g^{\beta\gamma}\nabla_\alpha V_\gamma=g_{\mu\beta}g^{\beta\gamma}V_\gamma
$$$$
\Rightarrow U^\alpha\delta_\mu^\gamma\nabla_\alpha V_\gamma=\delta_\mu^\gamma V_\gamma
$$$$
\Rightarrow U^\alpha\nabla_\alpha V_\mu=V_\mu
$$Once again there are three ways to do the first step. Metric compatibility is not essential.

See Commentary 3.8 Symmetries and Killing vectors.pdf first two pages. Then I run into another problem with Killing.

Thursday, 10 October 2019

Exercise 3.06 Metric outside Earth

Question

A good approximation to the metric outside the surface of the Earth is provided by$$
{ds}^2=-\left(1+2\Phi\right){dt}^2+\left(1-2\Phi\right){dr}^2+r^2\left({d\theta}^2+\sin^2{\theta}{d\phi}^2\right)
$$where$$
\Phi=-\frac{GM}{r}
$$may be thought of as the familiar Newtonian gravitational potential. Here ## G## is Newton's constant and ## M## is the mass of the Earth. For this problem ## \Phi## may be assumed to be small.
a) Imagine a clock on the surface of the Earth at a distance ##R_1## from the Earth's center, and another clock on a tall building at distance ##R_2## from the Earth's center. Calculate the time elapsed on each clock as a function of the coordinate time ## t##. Which clock moves ticks faster?
b) Solve for a geodesic corresponding to a circular orbit around the equator of the Earth (##\theta=\pi/2## ). What is ## d\phi/dt##?
c) How much proper time elapses while a satellite at radius  ##R_1## (skimming along the surface of the Earth, neglecting air resistance) completes one orbit? You can work to  first order in ## \Phi## if you like. Plug in the actual numbers for the radius of the Earth and so on (don't forget to restore the speed of light) to get an answer in seconds. How does this number compare to the proper time elapsed on the clock stationary on the surface?

Answers

Karl Schwarzschild
Image credit
Apparently the metric given is the Newtonian metric. I learnt how to reinstate the speed of light ## c## and so do real calculations. I was brutally reminded that ##{ds}^2=-{d\tau}^2## where ## \tau## is the proper time and that the proper time is what appears on ideal clocks. I also found the Schwarzschild radius - the radius of the event horizon of a black hole with a given mass. Some time ago I had found two sample solutions which had been left lying around on the internet by careless professors. They have now gone but luckily I had downloaded them before. I found that they were both littered with errors! The worst was probably UCSB's who popped an extra question and got totally the wrong answer for gravitational time dilation on a GPS satellite. Lucky they weren't involved in the construction!  Finally I had a look to see if I  could do anything with elliptical orbits but then decided I had spent enough time on this exercise.

Full answer and bonus questions at Ex 3.06 Metric outside Earth.pdf (16 pages)

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

Exercise 3.05 2-sphere: geodesics and parallel transport

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

Exercise 3.04 Paraboloidal coordinates

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!