Wednesday, 13 May 2020

Reciprocals of prime numbers

On a program about Gauss (BBC In Our Time), near 48:00, they said that the reciprocal of a prime number can be a recurring decimal and the maximum length of the recurring part is one less than the number. My curiosity was aroused.

Examples are
1/7=0.142857142857142857142857142857142857...
1/59=0.01694915254237288135593220338983050847457627118644067796610169491525423728813559322033898305084745762711864406779661...
the latter was found by a VBA program which could theoretically go to 2 billion digits because that is the maximum size of an array and length of a string. However it would run out of time or memory well before it got there. Calculation the reciprocals of the numbers from 80,000-89,999 and putting them into an Excel spreadsheet found 316 reciprocals which had recurring digits one less than the number you first thought of in 7 hours 50 minutes. 89,989 is prime 1/89,989 has 89,988 recurring digits and the VBA program  took 44 seconds just to calculate that. A different programming language would be much faster.

I now have a list (in several spreadsheets) of all the reciprocals from 2 to 999,999. It is fascinating.

A 'number whose reciprocal has recurring digits one less than the number' is a bit of a mouthful, so I abbreviate it to an ##R_-## ('R minus') number. It is also useful to define a function ##R\left(n\right)## which gives the number of Recurring digits in ##n##. The VBA program calculates ##R\left(n\right)##. For ##R_-## numbers, ##R\left(n\right)=n-1## . A related function is ##U\left(n\right)## which is the number of non-recurring digits, or Unique digits, in the reciprocal. So for example we have
##R\left(7\right)=6,\ U\left(7\right)=0## and
##\frac{1}{22}=0.0454545\ldots\ \ ,\ \ R\left(22\right)=2,\ U\left(22\right)=1##

The scatter chart contains a point at ##n,R\left(n\right)## for every number from 2 to 9,999.

Given positive integers ##m,n,p## greater than 0 or 1, I have proved that
1) ##R\left(n\right)<n## for all ##n##. This is shown by the graph. There are no data points above the main diagonal line which has gradient ##~1##.

2) ##R\left(n\right)=n-1## only for prime numbers. Or only prime numbers are ##R_-## numbers.

3) ##R_-## numbers occur about 2.6 times less often than prime numbers as shown by the density of the main diagonal line.

4) For multiples of ##R_-## numbers ##n## , ##R\left(m\times n\right)## is likely to be ##R\left(n\right)##. These are shown by the less dense diagonal straight lines which have gradients 1/2, 1/3, 1/5 ....

5) ##U\left(p\times2^m\times5^n\right)=m\ or\ n## whichever is larger of ##m,n##. (We can have ##m,n=0## in this case and ##p## must not have 2 or 5 as a factor.)

6) A consequence of 5 is that two fifths of consecutive numbers have ##Q\left(n\right)=0##.

7) Another consequence of 5 is that all prime numbers, except 2 and 5, and therefore all ##R_-## numbers have no non-recurring digits.

The first two have mathematical proofs, the rest are really conjectures based on experiments with lots of numbers. The mathematical proofs are algorithmic: One thinks of how to do the calculation and that proves the result.

Here's why Reciprocals of prime numbers.pdf (11 pages)

On Gauss and me

Carl Friedrich Gauss
On the program they mention the famous story about Gauss when he was age 10 at elementary school adding up all the numbers from 1 to 100. His class was asked to do this by the teacher who, no doubt, thought she could have some quiet time while the class was busy. Gauss added the numbers up in a minute, foiling the teacher's plan. Apparently Gauss used to love telling this story.

When I was about 10 or 11 and day dreaming at school assembly, I worked out a formula for the sum of the first ##n## numbers. It is of course ##\Sigma=n(n+1)/2##. I thought I had made an fantastic mathematical and nervously went to tell Mr Fillingham, head maths teacher. He gave me a strange look and showed me the general formulas for arithmetic and geometric progressions. I was crestfallen and never repeated the story.

Gauss was given a book of logarithm tables when he was 15 and it had a list of prime numbers in the back. He said that there was poetry in the tables.

He spent so much time day dreaming about mathematics that he memorised his times tables up to very large numbers. He would have loved computers. They give you the power to play with very large sets of numbers.

Wednesday, 22 April 2020

Exercise 5.3 Inside the event horizon

Question
Consider a particle (not necessarily on a geodesic) that has fallen inside the event horizon, ##r<2GM##. Use the ordinary Schwarzschild coordinates ##\left\{t,r,\theta,\phi\right\}##.
Observable Universe

a) Show that the radial coordinate must decrease at a minimum rate given by $$
\left|\frac{dr}{d\tau}\right|\geq\sqrt{\frac{2GM}{r}-1}
$$b) Calculate the maximum lifetime for a particle along a trajectory from ##r=2GM## to ##r=0##.
c) Express this in seconds for a black hole with mass measured in solar masses.
d) Show that this maximum proper time is achieved by falling freely with ##E\rightarrow0##.
Answers
The most interesting exercise so far, especially parts b and c. If the sun were a black hole then the maximum lifetime for our particle would be $$
{\Delta\tau}_\bigodot=1.55\times\ {10}^{-5}\  \rm{s}
$$That's a pretty short time, but if the Sun were a black hole it would have ##2GM=3\ \text{km}## so our test particle would have an average speed of about ##2\ \times\ {10}^8\ \text{m s}^{-1}## which is just below the speed of light and a hundred times faster than the Parker Solar Probe launched in 2018 which should only reach 0.064% the speed of light.

However big the black hole is, the average minimum speed for the fall for the centre is constant at$$
v_\rm{AvMin}=\frac{2c}{\pi}
$$M87* the black hole at the centre of our galaxy is about ##6.5\times\ {10}^9## solar masses so we get$$
{\Delta\tau}_\rm{M87\ast}=1.55\times\ {10}^{-5}\times6.5\times\ {10}^9={10}^5\ s=28\ \rm{hours}
$$There is a short time to prepare in M87*.

We can do the same for a black hole with the mass of the Universe: The observable Universe contains ordinary matter equivalent to ##{10}^{23}## solar masses. So$$
{\Delta\tau}_{Universe}=1.55\times\ {10}^{-5}\times{10}^{23}\approx{10}^{18}s=300\ \text{billion years}
$$The estimated 'age' of the Universe is 14 billion years, so there is plenty of time inside the big black hole - we have hardly started the journey.

See proof and calculations at Ex 5.3 Inside the event horizon.pdf (4 pages). Also contains speculations on what happens to a photon and links to other answers.
My document on Constants and conversion factors also came in very handy.

Thursday, 16 April 2020

Eddington and Finkelstein take us into a black hole


In the first part of section 5.2 (two posts ago) it seemed to be impossible to get inside the Schwarzschild radius. In the second part we look at other coordinate systems and find the Eddington-Finkelstein coordinates which show us how. The metric then takes a different form (which says something about Birkhoff's theorem) and it does not have an infinity at the Schwarzschild radius (##r=2GM##). 

The properties of the function ##1-2GM/r## (which causes the trouble) frequently amaze. It keeps eating itself up which is very satisfactory.

Eddington-Finkelstein coordinates eliminate the time coordinate ##t## and introduce
$$v=t+r+2GM\ln{\left(\frac{r}{2GM}-1\right)}$$
so the relationship between ##v## and ##t## is complicated.

The image shows a light cone at various distances out from the centre at ##r=0## . The ingoing light beam always heads for the centre the outgoing beam can get away when ##r>2GM## but flips towards the centre once it originates at a distance less than the Schwarzschild radius (aka the event horizon). The closer the starting point is to the centre, the less room there is for manoeuvre. 

I am slightly dubious about the direction of the ingoing side of a light cone inside the Schwarzschild radius.

See why and check my maths: Commentary 5.6#2 Schwarzschild Black Holes.pdf (6 pages)

Friday, 10 April 2020

2 years on


Two years on and I am about half way through Spacetime and Geometry : An Introduction to General Relativity – by Sean M Carroll. It's probably the best value for money book I have read (er... studied) in my life. When I started I could barely remember how to differentiate. Now I can use the chain rule almost without thinking and the tensor things which I had never met before are a doddle. I am now on the Eddington-Finkelstein metric and discovering how to get into a black hole (which seems to be impossible in the obvious ##t,r,\theta,\phi## coordinates) and why you never get out. I can't praise the book and Sean Carroll highly enough!

Schwarzschild Black Holes


We're just looking at the first two pages of section 5.2 here where Carroll shows closing up light cones as one approaches an event horizon and then a beacon on a radial geodesic which purports to show that clock ticks on the beacon appear to get slower and slower according to a stationary observer who maintains a safe distance. 

I slightly improve on the former and show the world lines of in- and out-going photons starting at a given radius. I then try to reproduce the latter, first with an invented geodesic, and then with a properly calculated one. The invented geodesic (shown above) produced the result Carroll suggests but the 'properly calculated ones' did not. The first method produced correct geodesics, but not the desired ones. I believe that the second method failed due to my mathematical inexperience. Possibly there is no exact solution.

The diagram is like Carroll's Fig 5.8. It shows the world lines of a photon and a beacon falling directly towards the centre of the black hole. There is an observer hovering above them at a safe distance. Using the Schwarzschild metric the photon never seems to cross the event horizon at ##r=R_S##! The beacon, also falling directly in, sends signals (flashes of light) back out at intervals ##\Delta\tau_1##. They arrive at the observer separated by longer and longer times. The beacon also appears to take forever to get to the event horizon!

The maths didn't really work. Witness my struggles at Commentary 5.6 Schwarzschild Black Holes.pdf (8 pages)

Monday, 30 March 2020

Gravitational redshift

The second piece of evidence for general relativity we examine is gravitational redshift in section 5.5. That's when the wavelength (or frequency) of light changes as it moves to stronger or weaker parts of a gravitational field.

Apparently Pound and Rebka were the first to measure it using gamma rays going up 72 feet (that's 22m in new money). They did it in the Jefferson laboratory (pictured)  at Harvard in 1959. That's about 40 years after Einstein predicted it. The change in the wavelength was 2 parts in a thousand trillion (##2## in ##10^{15}##). They measured it by wiggling the source of the gamma rays about in a speaker cone and seeing when the Doppler shift cancelled the gravitational shift!

The calculations are quite simple (and I got the right answer without cheating!) but I really needed to understand a few other things which sent me right back to chapters 3 and then 1. It all concerns the energy-momentum vector for a massive particle and then a massless e.g. gamma ray) particle and how the energy and therefore frequency fits into that. Light dawned.

The basic calculations: Commentary 5.5 Gravitational Redshift.pdf (2 pages)
On four-momentum and energy: Commentary 3.4 Particle energy.pdf (6 pages)

Monday, 23 March 2020

Precession of perihelion of Mercury

In 5.5, we continue our adventures from the previous section 5.4 where we calculated a few things from the Killing vectors and geodesics of the Schwarzschild metric. I learnt a good lesson about Killing vectors then that went over my head previously.

There are several amazing things in this section one group being mathematical dexterity and the other being the feats of astronomers. The latter have measured the precession of the perihelion of Mercury at ##{44}^{\prime\prime}  \text{per century}##. ##{44}^{\prime\prime}## is 44 seconds of an arc or 44 times 1/3600 degrees. That's about the angle that a soccer ball would subtend if it was one kilometer away and if you could see it. How did they do that?

Apparently we are following d'Inverno 1992, not Einstein, and we start with a differential equation for the orbit in terms of radial distance ##r## and affine parameter ##\lambda## and turn it into a simpler differential equation in terms of ##x\propto1/r## and azimuth ##\phi##. A crafty differentiation then makes the equation further collapse into the equation of an ellipse with a small perturbation, as witnessed by astronomers. Using more trigonometrical tricks we then solve the GR perturbation part and find a term in that which must correspond to the precession. A further trig trick gets us to "the equation for an ellipse with an angular period that is not quite ##2\pi##" and from that we extract the precession of the perihelion of Mercury. How did anybody think up all that?

The image shows the path of a Mercury that precesses about a million times faster than our own Mercury. I plotted it to check the  "the equation for an ellipse with an angular period that is not quite ##2\pi##". It is surprisingly accurate for such a large precession.

Read my attempts to follow Carroll at
Commentary 5.5 Precession of perihelia.pdf (7 pages)
Commentary 5.4 Geodesics of Schwarzschild.pdf (8 pages)


Monday, 9 March 2020

The Schwarzschild metric

Event horizon

In this section 5.1 Carroll rediscovers the Schwarzschild metric and I discover a little fib of Carroll's which leads us to an event horizon, which he doesn't even mention. What a cheek!

It might be interesting to some that Schwarzschild was German and in German Schwarz means black and Schild means shield so his name means 'black shield'. Like Rotschild (red shield) the great bankers. Schwarzschild published his metric in the same year that Einstein published his theory of general relativity. Quick work. He died the same year. Perhaps the effort killed him, but I bet he was pleased.

FYI  the Schwarzschild metric  is$$

{ds}^2=-\left(1-\frac{2GM}{r}\right){dt}^2+\left(1-\frac{2GM}{r}\right)^{-1}{dr}^2+r^2{d\Omega}^2
$$where ##{d\Omega}^2## is the metric on a two sphere$$
{d\Omega}^2={d\theta}^2+\sin^2{\theta}{d\phi}^2
$$and, of course, the spatial coordinates are spherical polar, in all they are ##\left\{t,r,\theta,\phi\right\}##.

The Schwarzschild radius is given by$$

R_S=2GM
$$That is also the radius of the event horizon of a black hole of mass ##M## so we are able to calculate how compressed the Sun would have to be to be a black hole: radius 3 km.

Read all about it: Commentary 5.1 The Schwarzschild metric.pdf (6 pages including the Riemann tensor)

Lagrange Formulation of General Relativity

Section 4.3 on the Lagrange formulation of General Relativity was pretty tough. I had to start by reading and understanding section 1.10 on classical field theories of which I knew nothing. That took about a month. Back on this section  I got very confused about small variations, which are vital for this branch of calculus. I was misled (even lied to) by JG on math.stackexchange and then helped by Physics Forums. I had to collect all my new knowledge about variations in a separate commentary. Carroll threw several interesting  challenges at me. We end with a new definition of the energy momentum tensor.

Read my thoughts: Commentary 4.3 Lagrange Formulation of GR.pdf (7 pages)
And nuggets on variations: Commentary Variations of objects as in calculus of variations.pdf (3 pages)

Thursday, 5 March 2020

Planck dimensions

At equations 4.91-4.95 Carroll gives the Planck's set of four dimensioned quantities: Planck's mass, length, time and energy. I wanted to compare them with actual things. Planck first noticed these way back in 1899.
\begin{align}

m_p=\sqrt{\frac{\hbar c}{G}}&=2.18\times{10}^{-8}\rm{kg}&\rm{{10}^{7}\ E. coli}\phantom {100000000000000000000}&\phantom {10000}(1)\nonumber\\

l_p=\sqrt{\frac{\hbar G}{c^3}}&=1.63\times{10}^{-35}\rm{m}&\rm{Radius\ of\ proton\ ={10}^{-15}\ m}\phantom {10000}&\phantom {10000}(2)\nonumber\\

t_p=\sqrt{\frac{\hbar G}{c^5}}&=5.39\times{10}^{-44}\rm{s}&\rm{Cosmic\ inflation\ ends\ at\ {10}^{-32}s}\phantom {10000}&\phantom {10000}(3)\nonumber\\

E_p=\sqrt{\frac{\hbar c^5}{G}}&=1.95\times{10}^9\rm{J}&\rm{Sun\ emits{\ 10}^{26}\ Js^{-1}.  \text{ A-bomb}\rm={10}^{12}}\ J&\phantom {10000}(4)\nonumber\\

&=1.22\times{10}^{19}\rm{GeV}&

&\phantom {10000}\nonumber

\end{align}
Max Planck 1858-1947
He then says "Most likely, quantum gravity does not become important until we consider particle masses greater than ##m_p##, or times shorter than ##t_p##, or lengths smaller than ##l_p##, or energies greater than ##E_p##; at lower scales classical GR should suffice. Since these are all far removed from observable phenomena, constructing a consistent theory of quantum gravity is more an issue of principle than of practice."

Whilst it is unimaginable that we will see things shorter than ##t_p## or smaller than ##l_p##, particle masses ('point masses') greater than ##m_p## are commonplace in GR and energies greater than ##E_p## are happening all the time. Can anybody help me make sense of the ##m_p,E_p## parts? And why are those 'greater than' and the others 'less than'?

I found out when I asked on physics forums. Read it at Commentary 4.5 Planck dimensions.pdf (2 pages).

Tuesday, 25 February 2020

Einstein's equation

Einstein age 18. Credit.
We are now on section 4.2 which has some very shady approximations. However we do get to Einstein's equation for general relativity if we tolerate that. The equation is$$
R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=8\pi GT_{\mu\nu}
$$where ##R_{\mu\nu},R## are the Ricci tensor and scalar which tell us about the curvature of spacetime, ##g_{\mu\nu}## is the metric, ##G## is Newton's constant and ##T_{\mu\nu}## is the energy-momentum tensor. So the equation tells us how the curvature of spacetime reacts to the presence of energy-momentum (which includes mass). Newton is not forgotten altogether😊.

The equation can also be written as $$
R_{\mu\nu}=8\pi G\left(T_{\mu\nu}-\frac{1}{2}Tg_{\mu\nu}\right)
$$where ##T=g_{\mu\nu}T_{\mu\nu}## and in empty space where ##T_{\mu\nu}=0## that gives us$$
R_{\mu\nu}=0
$$The equation is a field equation for the metric and the Newtonian gravity field equation is Poisson's equation$$
\nabla^2\Phi=4\pi G\rho
$$where ##\Phi## is the gravitational potential and ##\rho## the mass density.

The section starts by plausibly guessing that GR field equation must be of the form $$
R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=\kappa T_{\mu\nu}
$$where ##\kappa## is a constant we must find. The GR field equation must be the same as Poisson's equation in almost-flat spacetime. So we use a small perturbation ##h_{\mu\nu}## on the flat metric: $$
g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}
$$and discarding second order terms in ##h_{\mu\nu}## eventually work out that ##\kappa=8\pi G## to bring the two equations into line.

However! Carroll's 4.38 is wrong. It says ##T_{00}=\rho## and in fact ##T_{00}=\rho\left(1-h_{00}\right)## and  if 4.38 were right then 4.39 would be wrong, but in fact it is right. Carroll is sort of having it both ways and we only get ##\ \kappa\approx8\pi G## at best. It contains first order terms in ##h_00##. Hopefully the next section using the Lagrangian formulation will do better!

See Commentary 4.2 Einsteins equation.pdf (5 pages) for the details.

Sunday, 23 February 2020

Thursday, 20 February 2020

Exercise 1.12 Energy momentum tensors of two field theories

EM waves by Lomonosov Moscow State University
Question
Consider the two field theories we explicitly discussed, Maxwell's electromagnetism (let ##J^\mu=0##) and the scalar field theory defined by (1.148)

(a) Express the components of the energy-momentum tensors of each theory in three-vector notation, using divergence, gradient, curl, electric and magnetic fields, and an overdot to denote time derivatives.

(b) Using the equations of motion, verify (in any notation you like) that the energy-momentum tensors are conserved.

The energy momentum tensors in question were
\begin{align}
T_{scalar}^{\mu\nu}=\eta^{\mu\lambda}\eta^{\nu\sigma}\partial_\lambda\phi\partial_\sigma\phi-\eta^{\mu\nu}\left(\frac{1}{2}\eta^{\lambda\sigma}\partial_\lambda\phi\partial_\sigma\phi+V\left(\phi\right)\right)&\phantom {10000}(1)\nonumber
\end{align}and
\begin{align}
T_{EM}^{\mu\nu}=F^{\mu\lambda}F_{\ \ \ \lambda}^\nu-\frac{1}{4}\eta^{\mu\nu}F^{\lambda\sigma}F_{\lambda\sigma}&\phantom {10000}(2)\nonumber
\end{align}Answer
We had to answer a and b for each tensor so that makes four questions really.

a scalar) The first one was quite easy but we had to use the outer product ##\bigotimes## and the Kronecker delta ##\delta^{ij}## too. (Assuming ##\nabla\phi\bigotimes\nabla\phi## means what I think it means.)

a TM) The second was quite difficult and I found a new tool Microsoft Mathematics to multiply horrid matrices which come from the likes of ##F^{\mu\lambda}F_{\ \ \ \lambda}^\nu## in (2).

In both the b cases we need to show that the tensors are conserved we just have to show that
\begin{align}
\partial_\mu T^{\mu\nu}=0&\phantom {10000}(3)\nonumber
\end{align}which is four equations one for each ##\nu##, the four coordinates.

b scalar) I was able to do that entirely by tensor and index manipulation which was cool. The time and space coordinates worked the same.

b EM) This was a horrible mixture of tensor and index manipulation and then decomposition into electric and magnetic fields. I had to do the ##t,x## coordinates separately and assumed that  ##y,z## are like ##x##. It was nasty.

Read the full answer here: Ex 1.12 Energy momentum tensors of two field theories.pdf 

Wednesday, 19 February 2020

Microsoft Mathematics

I was recently doing a problem which involved calculating the electromagnetic energy momentum tensor$$
T^{\mu\nu}=F^{\mu\lambda}F_{\ \ \ \lambda}^\nu-\frac{1}{4}\eta^{\mu\nu}F^{\lambda\sigma}F_{\lambda\sigma}
$$In a previous problem I had already calculate$$
F_{\mu\nu}=\left[\begin{matrix}0&-E_1&-E_2&-E_3\\E_1&0&B_3&-B_2\\E_2&-B_3&0&B_1\\E_3&B_2&-B_1&0\\\end{matrix}\right],\ F_{\ \ \ \lambda}^\nu=\eta^{\nu\rho}F_{\rho\lambda}=\left[\begin{matrix}0&E_1&E_2&E_3\\E_1&0&B_3&-B_2\\E_2&-B_3&0&B_1\\E_3&B_2&-B_1&0\\\end{matrix}\right]
$$
and$$
F^{\mu\nu}=F_{\ \ \ \lambda}^\mu\eta^{\lambda\nu}=\left[\begin{matrix}0&E_1&E_2&E_3\\-E_1&0&B_3&-B_2\\-E_2&-B_3&0&B_1\\{-E}_3&B_2&-B_1&0\\\end{matrix}\right]
$$I was using ##c=1## and metric signature ##-++++##
So I had to do a bit of unpleasant matrix multiplication. I found Microsoft Mathematics useful. It successfully tells me that $$
F^{\mu\lambda}F_{\ \ \ \lambda}^\nu=\left[\begin{matrix}{E_1}^2+{E_2}^2+{E_3}^2&B_3E_2-B_2E_3&B_1E_3-B_3E_1&B_2E_1-B_1E_2\\B_3E_2-B_2E_3&{B_2}^2+{B_3}^2-{E_1}^2&-B_1B_2-E_1E_2&-B_1B_3-E_1E_3\\B_1E_3-B_3E_1&-B_1B_2-E_1E_2&{B_1}^2+{B_3}^2-{E_2}^2&-B_2B_3-E_2E_3\\B_2E_1-B_1E_2&-B_1B_3-E_1E_3&-B_2B_3-E_2E_3&{B_1}^2+{B_2}^2-{E_3}^2\\\end{matrix}\right]
$$
(without working out that ##{E_1}^2+{E_2}^2+{E_3}^2=E##) and was good at the other bits. The above matrix will easily paste back into a MS-Word equation. Sadly one cannot paste matrices from Word into it.

Subscripted variables can easily be entered as E_x becomes ##E_x##. Greek letters are allowed too. Superscripts are always exponents. It also does differentiation, integration and quite a bit more which I have not explored.

It has some drawbacks on my computer at least.

As mentioned I was unable to copy/paste matrices from MS-Equations into MS-Mathematics, so all the data entry must be done in the latter. There is an addin which might do the trick. But it did not like my version of Word (Office 365). The main install program, crashed when it came to the addin part.

The 'keyboard' is tiny and the seven choices of colours are demented. The image on the right is about the right size but it is less legible than the real thing.

Links to resources
Microsoft mathematics: https://www.microsoft.com/en-us/download/details.aspx?id=15702
Addin:  https://www.microsoft.com/en-US/download/details.aspx?id=36777

Sunday, 16 February 2020

Classical field theory

Sir William Rowan Hamilton
I really want to do section 4.3 the Lagrangian formulation of general relativity but first I am going back to section 1.10 on classical field theory which I skipped  because it involved the mysterious Lagrangian. I also needed to know about the fascinating Taylor expansions (Commentary 1.10 Taylor and Maclaurin series) and its extension for a function of two variables (which I guessed to begin with). My investigations on the Laplace operator or Laplacian (##\nabla^2## see Commentary 4.1 Laplacian) also came in handy.

Considering the number of things I did not know it was a good idea to postpone reading this section until now.

The section describes how we use Hamilton's least action principle and the Lagrangian which becomes the Lagrange density to get field equations.
It then has examples of how to use the procedure to get
  • a scalar field theory, 
  • a field theory for a harmonic oscillator which might be related to the relativistic equation of motion of an electron and 
  • an Electro Magnetic field theory which (with some assumptions) gives Maxwell's equations.
The key thing is to guess the Lagrange density in each case .

Read about my chapter 1 "Endkampf" here: Commentary 1.10 Classical Field Theory.pdf (9 pages)

Thursday, 13 February 2020

Exercise 1.10 Transformations of Electro Magnetic Field Tensor

Question

Using the tensor transformation law applied to ##F_{\mu\nu}##, show how the electric magnetic field 3-vectors ##E## and ##B## transform under 
a) a rotation about the ##y##-axis,
b) a boost along the ##z##-axis.

Answer

Electric and magnetic fields increase and contra-rotate perpendicular to direction of relative motion.
Boost is relative speed in ##z##-direction as a fraction of ##c## speed of light.
Carroll has already given us the Electro Magnetic Field Tensor is $$
F_{\mu\nu}=\left[\begin{matrix}0&-E_1&-E_2&-E_3\\E_1&0&B_3&-B_2\\E_2&-B_3&0&B_1\\E_3&B_2&-B_1&0\\\end{matrix}\right]=-F_{\nu\mu}
$$and the Lorentz rotation transformation matrix $$
\Lambda_{\ \ \nu}^{\mu^\prime}=\left[\begin{matrix}1&0&0&0\\0&\cos{\theta}&0&\sin{\theta}\\0&0&1&0\\0&-\sin{\theta}&0&\cos{\theta}\\\end{matrix}\right]
$$and the Lorentz transformation under a boost ##v## along the ##z##-axis $$
\Lambda_{\ \ \mu}^{\mu^\prime}=\left[\begin{matrix}\cosh{\phi}&0&0&-\sinh{\phi}\\0&1&0&0\\0&0&1&0\\-\sinh{\phi}&0&0&\cosh{\phi}\\\end{matrix}\right]
$$Where the boost parameter ##\phi=\tanh^{-1}{v}##.

There are many ways to solve this problem and I learnt many lessons.

Method 1

The Cartesian rotation matrix by geometry.
The simplest way to do the first question (a) is just to use the Cartesian rotation matrix on ##E## and ##B##. For any vector ##X## that gives $$
\left(\begin{matrix}{X^\prime}_1\\{X^\prime}_2\\{X^\prime}_3\\\end{matrix}\right)=\left(\begin{matrix}X_1\cos{\theta}+X_3\sin{\theta}\\X_2\\-X_1\sin{\theta}+X_3\cos{\theta}\\\end{matrix}\right)
$$Put ##E## then ##B## into that and back into the equation for ##F_{\mu\nu}## and you get the answer. But that would be cheating and will not work for (b). On the other hand, it is useful to check the answer.

Method 2

I found a crib-sheet that helped me here: One really should apply ##\Lambda## to ##F## but as it stands ##\Lambda_{\ \ \nu}^{\mu^\prime}## will only work on ##F^{\mu\nu}##  not on ##F_{\mu\nu}##. So one needs to get ##\Lambda_{\mu^\prime}^{\ \ \ \ \nu}## to operate on##\ F_{\mu\nu}## or ##F^{\mu\nu}## to be operated on by ##\Lambda_{\ \ \nu}^{\mu^\prime}##. We have$$
\ F^{\mu\nu}=\eta^{\nu\sigma}\eta^{\mu\rho}F_{\rho\sigma}
$$then we would get$$
F^{\mu^\prime\nu^\prime}=\Lambda_{\ \ \mu}^{\mu^\prime}\Lambda_{\ \ \nu}^{\nu^\prime}F^{\mu\nu}
$$and one can find the transformed ##E,B## by comparing ##F^{\mu^\prime\nu^\prime}## and ##F^{\mu\nu}##.
The crib-sheet recommended to just do the sums that are implicit in the repeated ##\mu,\nu##  indices in second equation. It is definitely easier to do two separate matrix multiplications! One has to be careful with where the indices are. I learnt about that as well.

Method 3

Just before I finished I proved that the transformation ##\ \Lambda_{\mu^\prime}^{\ \ \ \ \nu}## to operate on the covariant ##F_{\mu\nu}## is simply the inverse of ##\Lambda_{\ \ \nu}^{\mu^\prime}## the original, useless, transformation. The inverse of ##\Lambda_{\ \ \nu}^{\mu^\prime}## can be found by making  ##\theta\rightarrow-\theta## or ##\phi\rightarrow-\phi## - reversing the angle or boost. So the answer can be obtained with many fewer calculations. Duh!

Full answer at Ex 1.10 Transformations of Electro Magnetic Field Tensor.pdf (11 pages!)

Monday, 3 February 2020

Taylor and Maclaurin series

In section 1.10 Carroll Taylor-expands a Lagrangian. So I had a look at Taylor series. There is a very good article on Wikipedia about them.




In particular I looked at the Taylor series of 1/(1-x), polynomials especially 3rd order, sines and 'other functions' which states that they often converge to their Taylor series expansion which is a technique often used in physics.


I was able to proved that a 3rd order polynomial is its Taylor series, that is:$$
k_0+k_1x+k_2x^2+k_3x^3=\sum_{i=0}^{i=3}\left(\left(\sum_{n=i}^{n=3}{\frac{n!}{\left(n-i\right)!}k_na^{n-i}}\right)\left(\sum_{j=0}^{j=i}\frac{x^{i-j}\left(-a\right)^j}{j!\left(i-j\right)!}\right)\right)
$$The bit on the right is the Taylor series (after a bit of work) and there are many terms like  ##a^ix^j## which amazingly cancel each other out. I was not able to prove that any polynomial is its Taylor series as Wikipedia states.

The Taylor series for the sine function is $$
x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots
$$which is Madhava's sine series, the well known formula for ##\sin{x}##. It was discovered in the west by Isaac Newton (1670) and Wilhelm Leibniz (1676). Taylor prospered in the 1700's so perhaps he found it independently. But since Newton and Leibniz independently invented calculus and argued about who was first, they probably new quite a bit about Taylor and McLaurin series. However they were all preceded by Madhava of Sangamagrama (c. 1350 – c. 1425), the founder of the Kerala school of astronomy and mathematics, who proved it as recorded in 1530 in the Indian text the Yuktibhāṣā. The picture shows the sine Taylor series for terms up to ##{x^15}/{15!}##. Amazingly accurate!
Read it all in Commentary 1.10 Taylor and Maclaurin series.pdf (4 pages)

Thursday, 30 January 2020

Exercise SI.01 Simple Lagrangians

Line according to formula For values
##-25\leq t\leq 25, a,b,c,d##. 
Question
Derive Newton's first law in two dimensions using Hamilton's principle and the Euler-Lagrange equation. First do it in Euclidean ##x,y## coordinates then in polar ##r,\theta## coordinates.

Do it by filling in the gaps in the Wikipedia article on the subject as quoted below.

1) Prove that ##m\ddot{x}=0## follows from the Euclidean Lagrangian as at (1), (2)
2) Prove that the two Euler–Lagrange at (4),(5) produce the two equations given.
3) Prove that the solutions to those two equations are indeed (6) and (7)
4) Plot (6), (7) for values of ##a,b,c,d## showing that it is a straight line and that ##a## is the velocity, ##c## is the distance of the closest approach to the origin, and ##d## is the angle of motion. What is ##b##?
Extracts from the article:
In two Euclidean dimensions in the absence of a potential, the Lagrangian is simply equal to the kinetic energy
\begin{align}
L=\frac{1}{2}mv^2=\frac{1}{2}m\left({\dot{x}}^2+{\dot{y}}^2\right)&\phantom {10000}(1)\nonumber
\end{align}in orthonormal ##(x,y)## coordinates, where the dot represents differentiation with respect to the curve parameter (usually the time, ##t##). Therefore, upon application of the Euler–Lagrange equations, [to the ##x## coordinate]
\begin{align}
\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{x}}\right)-\frac{\partial L}{\partial x}=0\ \ \Rightarrow\ \ m\ddot{x}=0&\phantom {10000}(2)\nonumber
\end{align}[The Lagrangian and the solution are the same for the ##y## coordinate.]

In polar coordinates ##(r,\theta)## the kinetic energy and hence the Lagrangian becomes
\begin{align}
L=\frac{1}{2}m\left({\dot{r}}^2+r^2{\dot{\theta}}^2\right)&\phantom {10000}(3)\nonumber
\end{align}[where ##x=r\cos{\theta}\ \ ,\ y=r\sin{\theta}##.]
The radial ##r## and ##\theta## components of the Euler–Lagrange equations become, respectively
\begin{align}
\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{r}}\right)-\frac{\partial L}{\partial r}=0\ \Rightarrow\ \ \ddot{r}-r{\dot{\theta}}^2=0&\phantom {10000}(4)\nonumber\\
\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{\theta}}\right)-\frac{\partial L}{\partial\theta}=0\ \Rightarrow\ddot{\theta}+\frac{2\dot{r}\dot{\theta}}{r}=0&\phantom {10000}(5)\nonumber
\end{align}The solution to those two equations is given by
\begin{align}
r=\sqrt{\left(at+b\right)^2+c^2}&\phantom {10000}(6)\nonumber\\
\theta=\tan^{-1}{\left(\frac{at+b}{c}\right)}+d&\phantom {10000}(7)\nonumber
\end{align}for a set of constants ##a,\ b,\ c,\ d##  determined by initial conditions. Thus, indeed, the solution is a straight line given in polar coordinates: ##a## is the velocity, ##c## is the distance of the closest approach to the origin, and ##d## is the angle of motion." It does not reveal what ##b## is.
Answer 
I struggled with this and question 2 for over a day until I worked out what was really happening and came to the very important observation in the pdf. This enabled me to move on in section 1.10.

To begin with I got this: The first term in (2) is
\begin{align}
\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{x}}\right)=\frac{1}{2}m\frac{d}{dt}\left(\frac{\partial{\dot{x}}^2}{\partial\dot{x}}\right)=m\frac{d\dot{x}}{dt}=m\ddot{x}&\phantom {10000}(8)\nonumber
\end{align}and the second term is
\begin{align}
\frac{\partial L}{\partial x}=\frac{1}{2}m\frac{\partial}{\partial x}\left({\dot{x}}^2\right)=m\dot{x}\frac{\partial}{\partial x}\left(\dot{x}\right)=m\frac{dx}{dt}\frac{\partial}{\partial x}\left(\frac{dx}{dt}\right)=m\frac{d}{dt}\left(\frac{dx}{dt}\right)=m\ddot{x}&\phantom {10000}(9)\nonumber
\end{align}put those back into (2) and we get
\begin{align}
m\ddot{x}-m\ddot{x}=0&\phantom {10000}(10)\nonumber
\end{align}Which does not prove that ##m\ddot{x}=0##!!

To see all what went wrong and the vital observation, look in Ex SI.01 Simple Lagrangians.pdf (5 pages)

Monday, 27 January 2020

Exercise 1.08 un-conserved dust energy momentum tensor

Very bad dust by George E. Marsh
Question

If the energy momentum tensor ##\partial_\nu T^{\mu\nu}=Q^\mu## what physically does the spatial vector ##Q^i## represent? Use the dust energy momentum tensor to make your case.



Answer
We had at 1.110 that the dust energy momentum tensor was$$
T_{dust}^{\mu\nu}=p^\mu N^\nu=mnU^\mu U^\nu=\rho U^\mu U^\nu
$$##p_i## or ##p^i## is the pressure (not momentum) in the ##x^i## direction, that is force per unit area. I'm not sure if the index should be up or down.

##n## is the number density as measured in the dust's rest frame, ##n=## particles of dust per unit volume in rest frame.

##N^\nu=nU^\nu## is the number-flux four-vector,  ##N^0## is the number density of particles in any frame, ##N^i## is the flux of particles in the ##x^i## direction. So if there is no flux, that's the rest frame, ##N^\mu=\left(n,0,0,0\right)##

##m## is the mass of each dust particle (in the dust's rest frame)) which we assume to be the same.

Moreover the dust particles are all moving with the same four-velocity ##U^\mu## - I think.

##\partial_\mu T^{\mu\nu}=0## was the conservation equation for ##T^{\mu\nu}## so if ##\partial_\mu T^{\mu\nu}=Q^\mu## then clearly ##T^{\mu\nu}## is not being conserved and it is ##Q^\mu## that is disturbing the equilibrium. That answer is correct but rather feeble. I did a bit better with help from Valter Moretti on physics.stackexchange and learnt about the theorem of divergence and that ##T^{i0}=T^{0i}## components of this energy momentum tensor are roughly momentum.
More at Ex 1.08 Dust Energy  Momentum tensor.pdf (2 pages and a bit)

Friday, 24 January 2020

The energy-momentum tensor in SR

Fluid flow by Thierry Dugnolle
Many moons ago I had skipped the end of section 9 and all of section 10 in chapter 1. It is time to return to them and flat space time starting with the end of section 9 it looks at the energy-momentum tensor for a perfect fluid. I had written at the top of the page "? don't follow much of this from 1.116". Now I do. It shows how to get from that tensor to two well known classical (non-relativistic) equations in the non-relativistic limit: a continuity equation and the Euler equation from fluid dynamics. This gives us faith that the tensor is correct - even though I had not heard of either of these equations.

Along the way we find some tricks with four-velocity ##U^\mu={dx^\mu} / {d\tau}##, meet the projection tensor ##P_{\ \ \ \ \nu}^\sigma## which projects a vector orthogonal to another do some dimensional analysis and find out what energy density really is (I missed this in the book, or Carroll forgot to mention it). I had to consult the great Physics Forums on that and got liked for a facetious comment about measuring it in £sd 😀- I'm sure that oil companies must do it: kerosene is more valuable per kg than bunker fuel.

All the details here Commentary 1.9 Energy Momentum Tensor.pdf (6 pages)

Friday, 17 January 2020

Physics in curved spacetime

I've now started chapter 4 on Gravitation just in time for 2020. Very exciting!
Fools straight line
Carroll first states two formulas of Newtonian Gravity his 4.1 and 4.2$$
\mathbf{a}=-\nabla\Phi
$$where ##\mathbf{a}## is the acceleration of a body in a gravitational potential ##\Phi##. And Poisson's differential equation for the potential in terms of the matter density ##\rho## and Newton's gravitational constant ##G##:$$
\nabla^2\Phi=4\pi G\rho
$$I had a long pause thinking about the various formulas for the Laplacian ##\nabla^2## here.
How to these tie up with the old-fashioned laws? Newton's law of gravity is normally stated as$$
F=G\frac{m_1m_2}{r^2}
$$which combined with Newton's second law ##F=m\mathbf{a}## gives us the acceleration of a mass in the presence of another as$$
\mathbf{a}=G\frac{M}{r^2}
$$In exercise 3.6 we were given 'the familiar Newtonian gravitational potential'$$
\Phi=-\frac{GM}{r}
$$A bit of rough reasoning shows these are equivalent.

At his 4.4 Carroll states that the next equation gives the path of a particle subject to no forces$$
\frac{d^2x^i}{d\lambda^2}=0
$$If we solve it in polar coordinates for ##r,\theta## instead of ##x,y## Carroll says we get a circle and he cheekily suggests that we might think free moving particles follow that path. But the solution is $$
r=m\theta+k
$$where ##m,k## are constants. We can plot that and, obviously if ##m=0## we get a circle of radius ##k## but if ##m\neq0## we get other more interesting lines which are equally wrong. See above. Another error by Carroll, but only minor 😏. The next one is a corker.

Then we examine the equations in a near Newtonian environment and equation 4.13 ##g^{\mu\nu}=\eta^{\mu\nu}-h^{\mu\nu}## is wrong. The actual equation is obviously$$
g^{\mu\nu}=\eta^{\mu\nu}+h^{\mu\nu}
$$Properly 4.13 might be
$$
g^{\mu\nu}=\eta^{\mu\nu}-h_{\rho\sigma}\eta^{\mu\sigma}\eta^{\nu\rho}
$$which is true to first order and gives ##h^{00}=-h_{00}## which is used in the next section. If  one accepts the approximation that ##\eta## can be used to raise and lower indices on any object of order ##h## then that also gives us$$
g^{\mu\nu}=\eta^{\mu\nu}-h^{\mu\nu}
$$which says ##h^{\mu\nu}=0##. Oops! But it turns out it turns out that the sign on ##h^{\mu\nu}## is immaterial in this section. There is a full analysis in the pdf.

We soon arrive at the conclusion that in the near Newtonian environment we have the time, time component of the metric is $$
g_{00}=-1-2\Phi
$$which is also what we were given in Exercise 3.6.

Thursday, 9 January 2020

The Laplacian

Laplace 1749-1827
The second equation in chapter 4 was Poisson's differential equation for the gravitational potential ##\Phi##:
\begin{align}
\nabla^2\Phi=4\pi G\rho&\phantom {10000}(1)\nonumber
\end{align}What does ##\nabla^2## mean? A quick look on the internet reveals that it is the Laplace operator or Laplacian sometimes written ##\Delta## (capital delta) or possibly ##∆## (which Microsoft describes as increment). They come out different in latex \Delta and ∆ respectively. I should use the former.

I have found various formulations for the Laplacian and I want to check that they are all really the same. Two are from Wikipedia and the third is from Carroll. They are:
A Wikipedia formula in ##n## dimensions:
\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}(2)\nonumber
\end{align}A Wikipedia formula in "in 3 general curvilinear coordinates ##(x^1,x^2,x^3)##":
\begin{align}
\nabla^2=g^{\mu\nu}\left(\frac{\partial^2}{\partial x^\mu\partial x^\nu}-\Gamma_{\mu\nu}^\lambda\frac{\partial}{\partial x^\lambda}\right)&\phantom {10000}(3)\nonumber
\end{align}And Carroll's formula (from exercise 3.4) which I did not understand at the time because I had not spotted the second step:
\begin{align}
\nabla^2=\nabla_\mu\nabla^\mu=g^{\mu\nu}\nabla_\mu\nabla_\nu&\phantom {10000}(4)\nonumber
\end{align}The Wikipedia also gives a formula for the Laplacian in spherical polar coordinates:
\begin{align}
\nabla^2f&=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial f}{\partial r}\right)+\frac{1}{r^2\sin{\theta}}\frac{\partial}{\partial\theta}\left(\sin{\theta}\frac{\partial f}{\partial\theta}\right)+\frac{1}{r^2\sin^2{\theta}}\frac{\partial^2f}{\partial\phi^2}&\phantom {10000}(5)\nonumber\\

&=\frac{1}{r}\frac{\partial^2}{\partial r^2}\left(rf\right)+\frac{1}{r^2\sin{\theta}}\frac{\partial}{\partial\theta}\left(\sin{\theta}\frac{\partial f}{\partial\theta}\right)+\frac{1}{r^2\sin^2{\theta}}\frac{\partial^2f}{\partial\phi^2}&\phantom {10000}(6)\nonumber
\end{align}where ##\phi##  represents the azimuthal angle and ##\theta## the zenith angle or co-latitude. So the metric will be
\begin{align}
g_{\mu\nu}=\left(\begin{matrix}1&0&0\\0&r^2&0\\0&0&r^2\sin^2{\theta}\\\end{matrix}\right)&\phantom {10000}(7)\nonumber
\end{align}I assumed that the coordinates are ordered ##r,\theta,\phi## although Wikipedia does not say that.

I want to prove that
A) (2), (3) and (4) both give (5) or (6) the Laplacian in spherical polar coordinates.
B) (4) is equivalent to (3) the general 3-dimensional expression.
C) (4) is equivalent to (2) the general 𝑛-dimensional expression.
A was quite easy. B follows immediately from the formula for the covariant derivative. I could not prove C. After some help from Physics forums I proved it for a diagonal metric apparently it is true for a non diagonal metric but I could not penetrate the hints.

View my struggles at Commentary 4.1 Laplacian.pdf. A and B are on the first two pages and the other seven are where it all starts to go wrong.

I spent far too long on this but
1) Revisited using the Levi-Civita symbol for expanding determinant (47)
2) Met the log, exponent, trace of a matrix (63) and for diagonal matrices (67)
3) Used diagonalizing matrices (65)
4) Used the product operator ##\prod\ ##for the first time at (27) and again at (66)
5) Proved that ##\ln{\left(\det{\left(A\right)}\right)}=tr{\left(\ln{\left(A\right)}\right)}## (66)
6) Found a formula for the log of a derivative (73)
7)  Found a formula for the log of a matrix (76)