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!)

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)

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)

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)

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)

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.

More at Commentary 4.1 Physics in curved spacetime.pdf (5 pages)

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. But in 2020 I could not prove C. In January 2021 I found and proved the Voss-Weyl formula and C becomes easy.

The proofs are in Commentary 4.1 Laplacian.pdf on the first three pages. The other four record my 2020 struggles.

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)

Thanks to Physics forums with whose I proved C for a diagonal metric but they did not know about Voss-Weyl.