Tuesday, 14 September 2021

Question

Take the mathematician's view that tangent vectors and directional derivatives are the same thing, $u\equiv\partial_u$. Let $u,v$ be two vector fields and define their commutator in the manner familiar from quantum mechanics [not familiar to me!]$$\left[u,v\right]\equiv\left[\partial_u,\partial_v\right]\equiv\partial_u\partial_v-\partial_v\partial_u$$(a) Derive the following expression for $\left[u,v\right]$ valid in any coordinate basis$$\left[u,v\right]=\left(u^\beta v_{\ \ ,\beta}^\alpha-v^\beta u_{\ \ ,\beta}^\alpha\right)e_\alpha$$Thus despite that it looks like a second-order differential operator, $\left[u,v\right]$ is actually of first order - i.e. it is a tangent vector.
(b) For any basis $\left\{e_\alpha\right\}$ one defines the "commutation coefficients" $c_{\beta\gamma}^{\ \ \ \ \ \alpha}$ and $c_{\beta\gamma\alpha}$ by $$\left[e_\beta,e_\gamma\right]\equiv c_{\beta\gamma}^{\ \ \ \ \ \alpha}e_\alpha;\ \ c_{\beta\gamma\alpha}=g_{\alpha\mu}c_{\beta\gamma}^{\ \ \ \ \ \mu}$$Show that $c_{\beta\gamma}^{\ \ \ \ \ \alpha}=c_{\beta\gamma\alpha}=0$ got any coordinate basis.
(c) Calculate $c_{\hat{\beta}\hat{\gamma}}^{\ \ \ \ \ \hat{\alpha}}$ for the spherical noncoordinate basis of exercise 8.1.

This was straight forward. But I learned that Hermann Schwarz outshone Euler,  Lagrange, Cauchy who were all very clever. It was Schwarz who proved that partial derivatives commute. The others didn't get it quite right apparently. So I repeated his proof and now I know when partial derivatives commute and when they don't. Hurrah.
On this exercise: 8.2 Exercise Commutators.pdf (3 pages)
On Schwarz's theorem: 8.2 Schwarz's theorem.pdf (4 pages)

Thursday, 9 September 2021

MTW Ex 8.1 Practice with tensor algebra

Question

Let $t,x,y,z$ be Lorentz coordinates in flat spacetime, and let $$r=\left(x^2+y^2+z^2\right)^\frac{1}{2}\ ,\ \theta=\arccos{\left(\frac{z}{r}\right)}\ ,\ \phi=\arctan{\left(\frac{y}{x}\right)}$$be the corresponding spherical coordinates. Then$$e_0=\frac{\partial\mathcal{P}}{\partial t}\ ,\ e_r=\frac{\partial\mathcal{P}}{\partial r}\ ,\ e_\theta=\frac{\partial\mathcal{P}}{\partial\theta}\ ,\ e_\phi=\frac{\partial\mathcal{P}}{\partial\phi}$$is a coordinate basis, and $$e_{\hat{0}}=\frac{\partial\mathcal{P}}{\partial t}\ ,\ e_{\hat{r}}=\frac{\partial\mathcal{P}}{\partial r}\ ,\ e_{\hat{\theta}}=\frac{1}{r}\frac{\partial\mathcal{P}}{\partial\theta}\ ,\ e_{\hat{\phi}}=\frac{1}{r\sin{\theta}}\frac{\partial\mathcal{P}}{\partial\phi}$$is a non-coordinate basis.
(a) Draw a picture of $e_\theta,e_\phi,e_{\hat{\theta}},e_{\hat{\phi}}$ at several different points on a sphere of constant $t,r$.
(b) What are the one-form bases dual to these tangent-vector bases?
(c) What is the transformation matrix linking the original Lorentz frame to the spherical coordinate frame $\left\{e_a\right\}$?
(d) Use this transformation matrix to calculate the metric components $g_{\alpha\beta}$ in the spherical coordinate basis and invert the result to get $g^{\alpha\beta}$.
(e) Show that the non-coordinate basis $\left\{e_{\hat{a}}\right\}$ is orthonormal everywhere; i.e. that $g_{\hat{\alpha}\hat{\beta}}=\eta_{\alpha\beta}$; i.e. that $$g=-\omega^{\hat{0}}\otimes\omega^{\hat{0}}+\omega^{\hat{r}}\otimes\omega^{\hat{r}}+\omega^{\hat{\theta}}\otimes\omega^{\hat{\theta}}+\omega^{\hat{\phi}}\otimes\omega^{\hat{\phi}}$$(f) Write the gradient of a function $f$ in terms of the spherical coordinate and noncoordinate bases.
(g) What are the components of the Levi-Civita tensor in the spherical coordinate and noncoordinate bases?

This exercise seems mainly to be concerned with putting the rules in Box 8.4 into practice. We meet a noncoordinate (anholonomic) basis. One of the features of these is that coordinates cannot be used to describe positions but you can use components to describe other tensors. So$$p^{\hat{\mu}}=\left(a,b,c,d\right)\equiv ae^{\hat{0}}+be^{\hat{r}}+ce^{\hat{\theta}}+de^{\hat{\phi}}$$is at best meaningless but$$g_{\hat{\alpha}\hat{\beta}}=\left(\begin{matrix}a&0&0&0\\0&b&0&0\\0&0&c&0\\0&0&0&d\\\end{matrix}\right)$$is valid and is the answer to (d) when $a=-1,b=c=d=1$ .

I found an excellent new way of  creating animated gifs. Which I used for the answer to (a) below. The diagrams for $e_{\hat{\theta}},e_{\hat{\phi}}$ are a good indication of (e).

 Coordinate bases going round equator Coordinate bases going over pole Non-coordinate bases going round equator Non-coordinate bases going round equator

Monday, 30 August 2021

Question

The two-dimensional metric for a flat sheet of paper in polar coordinates is $\left(r,\theta\right)$ is$${ds}^2={dr}^2+r^2{d\phi}^2$$or in modern notation$$\mathbf{g}=\mathbf{d}r\otimes\mathbf{d}r+r^2\mathbf{d}\phi\otimes\mathbf{d}\phi$$Presumably the coordinates are $\left(r,\phi\right)$ not $\left(r,\theta\right)$.
(a) Calculate the connection coefficients using 8.24.
(b) Write down the geodesic equation in $\left(r,\phi\right)$ coordinates.
(c) Solve these equations for $r\left(\lambda\right)$ and $\phi\left(\lambda\right)$ and show that the solution is a uniformly parameterized straight line. ($x\equiv r\cos{\phi}=a\lambda+b$ for some $a$ and $b$, $y\equiv r\sin{\phi}=j\lambda+k$ for some $j$ and $k$).
(d) Verify that the noncoordinate basis $\mathbf{e}_{\hat{r}}\equiv\mathbf{e}_r=\frac{\partial\mathcal{P}}{\partial r},\ \mathbf{e}_{\hat{\phi}}\equiv r^{-1}\mathbf{e}_\phi=r^{-1}\frac{\partial\mathcal{P}}{\partial\phi},\ \ \mathbf{\omega}^r=\mathbf{d}r,\ \mathbf{\omega}^{\hat{\phi}}=r\mathbf{d}\phi$ is orthonormal, and that $\left<\mathbf{\omega}^\alpha,\mathbf{e}_{\hat{\beta}}\right>=\delta_{\ \ \hat{\beta}}^{\hat{\alpha}}$. Then calculate the connection coefficients of this basis from a knowledge [part (a)] of the connection of the coordinate basis.

I think 1) there are hats missing from omega indices and 2) 'modern notation' might not be very modern. I find it surprising that such an old, respected book has so many misprints.

a,b,c were straightforward. (d) contained the surprises. The $\left(\hat{r},\hat{\phi}\right)$ system (as we might call it) was orthonormal. The $\left(r,\phi\right)$ system was not, it was only orthogonal. The connection coefficients of the $\left(\hat{r},\hat{\phi}\right)$ system are not all symmetric in the lower two indices: $\Gamma_{\hat{\phi}\hat{r}}^{\hat{\phi}}=0\neq\Gamma_{\hat{r}\hat{\phi}}^{\hat{\phi}}=\frac{1}{r}$ which we prove. The method of calculating the coefficients is a great exercise in the piercing counter $\left<,\right>$.

${\hat{e}}_r,{\hat{e}}_\phi$ form a noncoordinate basis because, if we use them, the same point can have different coordinates. We show that is true. $e_r,e_\phi$ do not suffer from this problem.

Additionally we calculate the commutators (or Lie derivatives) $\left[e_r,e_\phi\right]=\left[\partial_r,\partial_\phi\right]$ and $\left[{\hat{e}}_r,{\hat{e}}_\phi\right]$. The first vanishes the second does not. This is a proof that the first is a coordinate (holonomic) basis and the second a noncoordinate (anholonomic) basis where you can't use coordinates. So you can't say $\left[{\hat{e}}_r,{\hat{e}}_\phi\right]=\left[\partial_{\hat{r}},\partial_{\hat{\phi}}\right]$. even though you can use the indices as in $\Gamma_{\hat{\phi}\hat{r}}^{\hat{\phi}}=0$.

Tiptoe through noncoordinate minefield at 8.5 Exercise Plane polar coordinates.pdf  (14 pages)

Friday, 30 July 2021

Pythagoras's and Newton's formulas for 𝝅

This fun video by Veritasium gives the story of Pythagoras and Newton and $\pi$ but it doesn't go into details. I wanted to know more.

Pythagoras calculated $\pi$ by placing regular n-agons just inside and outside a unit circle and calculating their perimeters. Some examples are shown below.
Apparently he stopped at a 96-agon. I reckon the result he got was$$48\sqrt{2-2\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{4}\sqrt{2+\sqrt3}}}}<\pi<\frac{96\sqrt{2-2\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{4}\sqrt{2+\sqrt3}}}}}{\sqrt{2+2\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{4}\sqrt{2+\sqrt3}}}}}$$It was accurate to four significant digits and not surprisingly he stopped! He didn't even have decimal numbers.

The second last person to use this method in anger was Ludolph van Ceulen in about 1600 who did a $2^{62}$-agon and still only got 35 decimal places. Apparently it took him 25 years. Nutter.

Then along came Newton (about 1700) and he used calculus (which he had just invented) and the binomial theorem to get$$\pi\approx-\frac{3\sqrt3}{2}+6-\left(\frac{1}{4}+\frac{3}{320}+\frac{45}{53,760}+\frac{1,575}{22,855,680}\right)$$That's accurate to five significant digits and a good deal easier to calculate. The further you go the better it gets:
Here's how they did it Newtons formula for pi.pdf (10 pages).

Friday, 16 July 2021

Kinetic Energy ≠ ½mv²

While studying $E=mc^2$ I keep noticing that kinetic energy is not $\frac{1}{2}mv^2$ as we were always taught in school. The relativistic formula is, in various forms, $$K_{rel}=\frac{c^3}{\sqrt{c^2-v^2}}-c^2=c^2\left(\frac{1}{\sqrt{1-\beta^2}}-1\right)=c^2\left(\frac{1}{2}\beta^2+\frac{3}{8}\beta^4+\ldots\right)$$where $\beta=v/c$.

On May 2nd 2021 the Parker Solar Probe achieved a velocity of 532,000 km/h which is about $0.0005c$. That gives
\begin{align}\frac{K_{rel}}{c^2}=1.25000023\times{10}^{-7}\ ,\ \frac{K_{cl}}{c^2}=1.25\times{10}^{-7}&\phantom {10000}(8)\nonumber\end{align}Not much difference. The mass of the probe is about 600kg so the difference is $2.3\times{10}^3\ Joules$ which might keep a phone going for an hour.

The difference is more noticeable at much higher velocities. At $0.1c$ the difference is about 1%. It goes wild as you approach the speed of light.
A bit more here: 2.2 Kinetic energy.pdf (1 page)

Wednesday, 14 July 2021

Why does E=mc²?

Trying to follow the second worked example in Box 2.2 of Misner, Thorne, Wheeler I realised that I did not know how Einstein proved $E=mc^2$ and, when I checked in my books, both MWT and Carroll gloss it over! Using Einstein's 1905 paper, an internet video and the relativistic Doppler shift formulas given by Wikipedia I now do. Here's my version based on those.
We take two scenarios
1) observe the object, in its rest frame, emitting two pulses of light with energy $E/2$ in opposite directions so that it does not change velocity and then we take off in a rocket with velocity $v$ so the object has some kinetic energy $K_1$

2) measure its kinetic energy $K_2$ from a frame moving with velocity $v$ then it emits the light which will have energy doppler shifted according to $$E_r=E\left(1+\frac{v^2}{2c^2}\right)$$After the emission we are now in the final state of scenario 1 and the object must have kinetic energy $K_1$ and total energy $K_1-E$.

In both scenarios the object lost energy by radiation and gained kinetic energy (because the observer started moving) and we end up at the same place. The total energy must be the same in either scenario therefore$$K_1-E=K_2-E\left(1+\frac{v^2}{2c^2}\right)\Rightarrow\frac{Ev^2}{2c^2}=K_2-K_1$$We know how to calculate the kinetic energy ($mv^2/2$) and the velocities of the kinetic energies are the same so the masses must have changed. So$$\frac{Ev^2}{2c^2}=m_2\frac{v^2}{2}-m_1\frac{v^2}{2}\Rightarrow E=\Delta Mc^2$$where $\Delta M$ is the change in the mass of the object. Furthermore we conclude that if one could get all the energy out of an object of mass $m$ that energy would be
$$E=mc^2$$For more details including the derivation of $E_r$ see the very short 2.2 Box E=mc2.pdf.

Coming soon. More on that Box 2.2.

Friday, 2 July 2021

2.5 Differential forms

 Vector $U$ (solid) and1-form $\widetilde{U}$ (dashed)
In section 2.5 of Misner, Thorne, Wheeler 'differential forms' or '1-forms' are introduced and there is soon a fairly elaborate recipe for constructing 1-forms $\widetilde{U}$ from a vector $U$. I reproduce it below. I tried out the recipe on some of the sample vectors shown and it seemed to work. A one form $\widetilde{U}$ is shown as a set of planes, whose spacing is defined, and which have a positive sense, or direction, as shown by a dashed arrow (which is not a vector). It would seem that you could equally well show the 1-form by a dashed arrow perpendicular (in the Euclidean sense) to the planes whose direction and length would show the positive sense and the spacing of the planes of $\widetilde{U}$. You can. And I implemented it in a spreadsheet and made the amusing gif on the right. Then I saw the joke.

The recipe

Figure 2.7. Several vectors, $A,B,C,D,E$, and corresponding 1-forms $\widetilde{A},\widetilde{B},\widetilde{C},\widetilde{D},\widetilde{E}$. The process of drawing $\widetilde{U}$ corresponding to a given vector $U$ is quite simple. 1) Orient the surfaces of $\widetilde{U}$ orthogonal to the vector $U$ . (Why? Because any vector $V$ that is perpendicular to $U$ must pierce no surfaces of $\widetilde{U}$ ($0=U\bullet V=U,V$) and must therefore lie in a surface of $\widetilde{U}$.) 2) Space the surfaces of  $\widetilde{U}$ so the number of surfaces pierced by some arbitrary vector $Y$ (e.g., $Y=U$) is equal to $Y\bullet U$.

Note that in the figure the surfaces of $\widetilde{B}$ are, indeed, orthogonal to $B$; those of $\widetilde{C}$ are, indeed, orthogonal to $C$, etc. If they do not look so, that is because the reader is attributing Euclidean geometry, not Lorentz geometry, to the spacetime diagram. He should recall, for example, that because $C$ is a null vector, it is orthogonal to itself ($C\bullet C=0$), so it must itself lie in a surface of the 1-form  $\widetilde{C}$. Confused readers may review spacetime diagrams in a more elementary text, e.g., Taylor and Wheeler (1966)."

Read it here, including punchline: 2.5 Differential forms.pdf. (6 pages)