I am reading Spacetime and Geometry : An Introduction to General Relativity – by Sean M Carroll. The blog contains answers to his exercises, commentaries, questions and more.
Friday, 16 April 2021
Luminosity distance
Friday, 19 February 2021
Cosmological vs Doppler redshift
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| Milne Universe. Flat and expanding. |
- more about Taylor series which I find peculiar.
- that coordinates are orthogonal if the metric is diagonal and I now almost understand the notation ##e_\tau=\partial_t##.
- about proper distance and simultaneity conventions, which I had never heard of before.
- I learnt about the varying speed of light!
Friday, 22 January 2021
Tensor Tricks
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| Cat gets Carroll |
Contents
- What tensor rank?
- Multi-dimensional Chain Rule
- Partial derivative gives Kronecker delta: Coordinates, Vectors, Tensors
- Partial derivatives commute
- Metric is always symmetric
- Contracting with metric lowers / raises index
- You can lower or raise indices on a tensor equation
- Swap indices with metric or any similar tensor
- Inverse of a matrix
- The determinant of the inverse is reciprocal of the determinant
- Determinant of a tensor in terms of Levi-Civita symbol
- Inverse tensor
- A relationship for the derivative of the determinant
- Fully contracted symmetric × antisymmetric tensor vanishes
- Symmetrising a tensor equation
- Two formulas involving four-velocity
- Second formula
- The projection tensor on four-velocity
- Contra / co-variant tensor transformation matrices
- Tensor contractions using matrices
The rest of the fab four are
Monday, 21 December 2020
Maxwell's equations have something missing
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| James C. Maxwell is peeved |
\nabla\times\mathbf{B}-\partial_t\mathbf{E}=\mathbf{J}$$
Friday, 4 December 2020
Robertson-Walker metrics
{ds}^2=-{dt}^2+R^2\left(t\right)\left[\frac{{d\bar{r}}^2}{1-k{\bar{r}}^2}+{\bar{r}}^2{d\Omega}^2\right]=-{dt}^2+a^2\left(t\right)\left[\frac{{\rm dr}^2}{1-\kappa r^2}+r^2{d\Omega}^2\right]
$$The second version is Carroll's preferred form - 'flouting' conventional wisdom.
Four people found the equations in various teams. They are Alexander Friedmann (Russian), Georges Lemaître (Belgian), Howard Robertson (USian) and Arthur Walker (British) and the metrics are often named after one or some or all. Carroll favours the English speakers.
\Gamma_{01}^1=\Gamma_{02}^2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Gamma_{03}^3=\frac{\dot{a}}{a}
$$there should be an ##=## sign between ##\Gamma_{02}^2## and ##\Gamma_{03}^3##!
Wednesday, 19 August 2020
Einstein-Rosen bridges: Wormholes in Schwarzschild spacetime
Thursday, 13 August 2020
Big Bang!
Now we want to do a conformal diagram for an expanding universe. The metric equation is$$
{ds}^2=-{dt}^2+t^{2q}\left({dr}^2+r^2{d\Omega}^2\right)
$$and ##0<q<1\ ,0<t<\infty\ ,\ 0\le r<\infty##. It should be pretty easy because we did most of the heavy lifting when we did the conformal diagram for flat spacetime. However I think Carroll made another mistake!
We introduce the coordinate ##\eta## with ##{dt}^2=t^{2q}{d\eta}^2## and we get a metric$$
{ds}^2=\left[\left(1-q\right)\eta\right]^{2q/\left(1-q\right)}\left(-{d\eta}^2+{dr}^2+r^2{d\Omega}^2\right)
$$The part on the right is the same as the flat metric with ##t\rightarrow\eta## so we can use all the work we did before to transform that into$$
{ds}^2=\omega^{-2}\left[-{dT}^2+{dR}^2+\sin^2{R}{d\Omega}^2\right]
$$with$$
\omega^{-2}=\left(\frac{\left[\left(1-q\right)\eta\right]^{q/\left(1-q\right)}}{\left(\cos{T}+\cos{R}\right)}\right)^2
$$and a bit of work on that gives $$
\omega=\left[\left(1-q\right)\sin{T}\right]^{q/\left(q-1\right)}\left(\cos{T}+\cos{R}\right)^{1/\left(1-q\right)}
$$But Carroll says that$$
\omega=\left(\frac{\cos{T}+\cos{R}}{2\sin{T}}\right)^{2q}\left(\cos{T}+\cos{R}\right)
$$I'm pretty sure that Carroll is wrong, even though his formula is more attractive. I also worked out how he went wrong. Carroll writes "The precise form of the conformal factor is actually not of primary importance" (because you throw it away for the diagram). Perhaps that's why he did not check it very carefully.
And here's the diagram
At the singularity very near ##t=0## space can apparently be as big as you like. Never fear: ##r## might be big but ##t^{2q}## will be very small, so distances are very small too.
Read all the details at
Commentary App H Conformal Diagram Expanding Universe.pdf (6 pages including a diversion on values of ##q##)
Saturday, 8 August 2020
Conformal Diagrams
{ds}^2=-{dt}^2+{dr}^2+r^2\left({d\theta}^2+\sin^2{\theta}{d\phi}^2\right)
$$
\bar{t}=\arctan{t}\ \ ,\ \ \bar{r}=\arctan{r}
$$which certainly pack spacetime into the range$$
-\frac{\pi}{2}<\bar{t}<\frac{\pi}{2}\ ,\ 0\le\bar{r}<\frac{\pi}{2}
$$as you will see below if you press the button. Carroll says it might be fun to draw the light cones on that, so I made a movie:
Tuesday, 25 February 2020
Einstein's equation
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| Einstein age 18. Credit. |
R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=8\pi GT_{\mu\nu}
$$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.
Friday, 17 January 2020
Physics in curved spacetime
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| Fools straight line |
\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}
$$which says ##h^{\mu\nu}=0##. Oops! But it turns out it turns out that the sign on ##h^{\mu\nu}## is immaterial in this section. There is a full analysis in the pdf.
g_{00}=-1-2\Phi
$$which is also what we were given in Exercise 3.6.
Saturday, 12 October 2019
Symmetries and Killing vectors
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| Sean Carroll, my guide and nemesis |
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:$$
U^\alpha\nabla_\alpha V^\beta=W^\beta
$$$$
\Rightarrow U^\alpha g^{\beta\gamma}\nabla_\alpha V_\gamma=g^{\beta\gamma}W_\gamma
$$$$
\Rightarrow U^\alpha g_{\mu\beta}g^{\beta\gamma}\nabla_\alpha V_\gamma=g_{\mu\beta}g^{\beta\gamma}W_\gamma
$$$$
\Rightarrow U^\alpha\delta_\mu^\gamma\nabla_\alpha V_\gamma=\delta_\mu^\gamma W_\gamma
$$$$
\Rightarrow U^\alpha\nabla_\alpha V_\mu=W_\mu
$$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.
Wednesday, 24 April 2019
Wolfram Mathworld great circle equation error
According to www.mathworld.wolfram.com/GreatCircle.html (19) the geodesic equation on a sphere (great circle) is given below. It is derived from a somewhat specialised equation for a geodesic on a surface (http://mathworld.wolfram.com/Geodesic.html (30)), which itself is derived by considering a minimised line integral. Wolfram's (19) is given as
\begin{align}
a{\mathrm{cos} u\ }{\mathrm{s}\mathrm{i}\mathrm{n} v\ }{\mathrm{sin} c_2\ }+a{\mathrm{sin} u\ }{\mathrm{s}\mathrm{i}\mathrm{n} v\ }{\mathrm{cos} c_2\ }-\frac{a{\mathrm{c}\mathrm{o}\mathrm{s} v\ }}{\sqrt{{\left(\frac{a}{c_1}\right)}^2-1}}=0 & \phantom {10000}(1) \\
\end{align}where ##a## is the radius of the sphere, ##c_1,c_2## are constants of integration, ##u,v## are respectively longitude and latitude. In the next equation it recasts that in Cartesian coordinates as\begin{align}
x{\mathrm{sin} c_2\ }+y{\mathrm{cos} c_2\ }-\frac{z}{\sqrt{{\left(\frac{a}{c_1}\right)}^2-1}}=0 & \phantom {10000}(2) \\
\end{align}"which shows that the geodesic giving the shortest path between two points on the surface of the equation lies on a plane that passes through the two points in question and also through center of the sphere." (2) is indeed the equation of a plane which contains the origin, but it also implies that\begin{align}
x=a{\mathrm{cos} u\ }{\mathrm{s}\mathrm{i}\mathrm{n} v\ }\ \ ,\ y=a{\mathrm{sin} u\ }{\mathrm{s}\mathrm{i}\mathrm{n} v\ }\ \ ,\ z=\ a{\mathrm{c}\mathrm{o}\mathrm{s} v\ } & \phantom {10000}(3) \\
\end{align}This is very wrong. It would be correct if ##v## was the colatitude (angle measured from the pole). The colatitude is normally called ##\phi ## and ##\phi ={\pi }/{2}-v##, as they say in their #7. Alternatively one can swap all ##{\mathrm{sin} v\ },{\mathrm{cos} v\ }##. I guessed that the equation is therefore\begin{align}
a{\mathrm{cos} u\ }{\mathrm{cos} v\ }{\mathrm{sin} c_2\ }+a{\mathrm{sin} u\ }{\mathrm{cos} v\ }{\mathrm{cos} c_2\ }-\frac{a{\mathrm{sin} v\ }}{\sqrt{{\left(\frac{a}{c_1}\right)}^2-1}}=0 & \phantom {10000}(4) \\
\end{align}This is correct as I have proved by other means. (Here). Numerically it can be shown with my great 3-D graph plotter (here and here). The incorrect equation is obviously not a great circle, whereas the correct one looks plausible;
The schematic shows great circles between cities. The right hand one shows the London-Peking great circle according Wolfram Mathworld. Perhaps this is what happened to the British Airways pilot who flew from London to Edinburgh instead of Düsseldorf in a month ago. (On the BBC here)
I have not been able to trace the source of the error in the Wolfram Mathworld proof. It may go as far back as their (7) where they might have intended to introduce ##\phi ##.
This error on Wolfram Mathworld caused me a lot of grief!
And another small error
There is also a typo on http://mathworld.wolfram.com/Geodesic.html between equations (11) and (12). It reads "Starting with equation (##\mathrm{\Diamond }##)" which should be "Starting with equation (5)"Saturday, 30 March 2019
Here's the very impressive Stokes's theorem, which applies to the diagram
$$\int^{\ }_{\mathrm{\Sigma }}{{\mathrm{\nabla }}_{\mu }V^{\mu }\sqrt{\left|g\right|}}d^nx=\int^{\ }_{\mathrm{\partial }\mathrm{\Sigma }}{n_{\mu }V^{\mu }\sqrt{\left|\gamma \right|}}d^{n-1}x
$$
At Carroll's (3.36) he says "if ##\mathrm{\nabla }## is the Christoffel symbol, ##{\omega }_{\mu }## is a one-form, and ##X^{\mu }## and ##Y^{\mu }## are vector fields, we can write
$${\left(\mathrm{d}\omega \right)}_{\mu \nu }=2{\partial }_{[\mu }{\omega }_{\nu ]}=2{\mathrm{\nabla }}_{[\mu }{\omega }_{\nu ]}
$$The phrase "if ##\mathrm{\nabla }## is the Christoffel symbol" is bizarre and it is easy to prove the equation without it, assuming the Christoffel connection is torsion-free (##{\mathrm{\Gamma }}^{\lambda }_{\mu \nu }={\mathrm{\Gamma }}^{\lambda }_{\nu \mu }##). I think our author meant "if the connection is torsion-free".
Read more at Commentary 3.2 Properties of covariant derivative.pdf (7 pages)
Monday, 25 March 2019
Corrections
This is trued but if the simplified for was given, the next equation would not work! (GK)
The Christoffel Symbol
| Christoffel |
I followed equations (3.5)-(3.10) carefully because I fell into the same tramp as I had before on one step and found an error of a sign in Carroll's (3.10) which is the important equation for the transformation of the connection. This is fairly obvious because it comes from (3.9) and a + term has gone to the other side of the equation without changing sign. It was also confirmed by notes I found at Physics 171 and the proof about Carroll's (3.26), see below. I struggled with that proof (and part 2 of exercise 1) for too long. Eventually I found it after I found another erroneous proof on another website which nevertheless gave me a great new indexing trick (note d). I have corrected the error here.
I still do not understand Carroll's third rule for covariant derivatives that they commute with contractions but he never seems to use it. Its meaning provoked a discussion on physics forums which did not help me. In another discussion my false assumption about commuting partial derivatives was exposed.
The three most important equations here are$$
{\mathrm{\Gamma }}^{\nu '}_{\mu '\lambda '}=\frac{\partial x^{\mu }}{\partial x^{\mu '}}\frac{\partial x^{\lambda }}{\partial x^{\lambda '}}\frac{\partial x^{\nu '}}{\partial x^{\nu }}{\mathrm{\Gamma }}^{\nu }_{\mu \lambda }-\frac{\partial x^{\mu }}{\partial x^{\mu '}}\frac{\partial x^{\lambda }}{\partial x^{\lambda '}}\frac{{\partial }^2x^{\nu '}}{\partial x^{\mu }\partial x^{\lambda }}
$$ and $$
{\mathrm{\Gamma }}^{\nu '}_{\mu '\lambda '}=\frac{\partial x^{\mu }}{\partial x^{\mu '}}\frac{\partial x^{\lambda }}{\partial x^{\lambda '}}\frac{\partial x^{\nu '}}{\partial x^{\nu }}{\mathrm{\Gamma }}^{\nu }_{\mu \lambda }+\frac{\partial x^{\nu '}}{\partial x^{\lambda }}\frac{{\partial }^2x^{\lambda }}{\partial x^{\mu '}\partial x^{\lambda '}}
$$ which are alternatives for the transformation of a connection. The first one was Carroll's (3.10) (corrected.)
The third is Carroll's (3.27). He writes it is "one of the most important equations in this subject; commit it to memory." It is for a torsion-free (##{\mathrm{\Gamma }}^{\lambda }_{\mu \nu }={\mathrm{\Gamma }}^{\lambda }_{\nu \mu }##) metric-compatible (##{\mathrm{\nabla }}_{\rho }g_{\mu \nu }=0##) connection and is$$
{\mathrm{\Gamma }}^{\sigma }_{\mu \nu }=\frac{1}{2}g^{\sigma \rho }\left({\partial }_{\mu }g_{\nu \rho }+{\partial }_{\nu }g_{\rho \mu }-{\partial }_{\rho }g_{\mu \nu }\right)
$$For some reason Carroll writes ##{\mathrm{\Gamma }}^{\lambda }_{\mu \nu }={\mathrm{\Gamma }}^{\lambda }_{\nu \mu }## as ##{\mathrm{\Gamma }}^{\lambda }_{\mu \nu }={\mathrm{\Gamma }}^{\lambda }_{(\mu \nu )}## which is the same but more complicated. The brackets are the symmetrisation operator.
Thursday, 28 February 2019
Important Equations for General Relativity
They are in Commentary Important Equations.pdf along with references and some notes.
Mathematics
- Definition of \partial\mu (Carroll 1.54)
- (Anti)symmetrisation operator (Carroll 1.79)
- Vector as derivative (Carroll 2.16)
- Commutator (Carroll 2.20/2.23)
- Tensor transformation equation (Carroll 2.30)
- Basis vectors (Physics Forums)
- Covariant derivative / Christoffel symbol (Carroll section 3.2)
- Torsion Tensor (C Eq 3.22)
- The Christoffel connection Γ (C Eq 3.27)
- The geodesic equation (C Eq 3.44)
- Directional covariant derivative (C Eq 3.38)
- The parallel transport equation (C Eq 3.39, 3.40)
- Riemann tensor (c Eq 3.112/3.113)
- Bianchi identity (c Eq 3.140)
- Ricci tensor and scalar, Weyl tensor (c Eq 3.144-3.147)
- Einstein tensor and 'contravariant' derivative (c Eq 3.1452,2)
- Killing's equation, Killing vectors (c Eq3.174)
- Geodesic deviation equation (c Eq3.208)
Tensor tricks
- What tensor rank?
- Multi-dimensional Chain Rule
- Partial derivative of components gives Kronecker delta
- Coordinates
- Vectors (C1.152)
- Tensors
- Partial derivatives commute
- Metric is always symmetric (C section 2.5)
- Contracting with metric lowers / raises index
- You can lower or raise indices on a tensor equation
- Swap indices with metric or any similar tensor
- Inverses and determinants
- Inverse of a matrix
- The determinant of the inverse is reciprocal of the determinant
- Determinant of a tensor in terms of Levi-Civita symbol (C Eq 2.66)
- Inverse tensor
- A relationship for the derivative of the determinant
- Fully contracted symmetric × antisymmetric tensor vanishes
- Symmetrising a tensor equation
- Two formulas involving four-velocity
- Second formula
- The projection tensor on four-velocity (C Eq 1.21)
Physics
- Electro Magnetic Field Tensor (C Eq 1.69)
- Maxwell's equations (C Eq 1.96-1.98)
- Energy Momentum tensor for a perfect fluid (C Eq 3.93 and 1.114)
- Energy Momentum tensor for dust in SR (C Eq 1.110)
- Energy-momentum tensor from action for matter (C Eq 4.75)
- Energy-momentum conservation equation (C Eq 3.92 & 4.8)
- Einstein's equation x 3 for general relativity (C Eq2.44-4.46)
- Friedmann equations (C Eq 8.67)
More Maths
- Differential and Integration on Web
- Pullback / Pushforward operators (Carroll A.9, A.10)
- Levi-Civita symbol and tensor (Carroll section 2.8)
- p-forms (Carroll section 2.9)
- Exterior derivative (Carroll 2.76)
- Wedge product (Carroll 2.73)
- Hodge star operator (Carroll 2.82)
- Stokes's theorem (C Eq 3.35)
- Euler-Lagrange Equation
The rest of the fab four are
Carroll's (1.68) where the Levi-Civita symbol is defined says "the Levi-Civita symbol is a ##(0,4)## tensor. It is NOT a tensor as he reminds us elsewhere.
Thursday, 10 January 2019
Commentary 2.8 Tensor Densities
We are told that the Levi-Civita symbol, which is not a tensor, is defined as$${\widetilde{\epsilon }}_{{\mu }_1{\mu }_2\dots {\mu }_n}=\left\{ \begin{array}{ll}
+1 & \mathrm{if}\mathrm{\ }{\mu }_1{\mu }_2\dots {\mu }_n\mathrm{\ is\ an\ even\ permitation\ of}\ 01..(n-1)\ \\
-1 & \mathrm{if\ }{\mu }_1{\mu }_2\dots {\mu }_n\mathrm{\ is\ an\ odd\ permitation\ of}\ 01..\left(n-1\right) \\
0 & \mathrm{otherwise} \end{array}
\right.$$and (Carroll's (2.66)) that given any ##n\times n## matrix ##M^{\mu }_{\ \ \ \mu '\ }##, the determinant ##\left|M\right|## obeys $${\widetilde{\epsilon }}_{{\mu '}_1{\mu '}_2\dots {\mu '}_n}\left|M\right|={\widetilde{\epsilon }}_{{\mu }_1{\mu }_2\dots {\mu }_n}M^{{\mu }_1}_{\ \ \ \ \ {\mu '}_1}M^{{\mu }_2}_{\ \ \ \ \ {\mu '}_2}\dots M^{{\mu }_n}_{\ \ \ \ \ {\mu '}_n}$$We are invited to check this for 2×2 and 3×3 matrices which we do and do discover some beauty comparing traditional methods for calculating the determinant of a matrix using cofactors or using the Levi-Civita symbol.
Setting ##{\mu '}_1{\mu '}_2\dots {\mu '}_n=01\dots (n-1)## we get the even simpler$$\left|M\right|={\widetilde{\epsilon }}_{{\mu }_1{\mu }_2\dots {\mu }_n}M^{{\mu }_1}_{\ \ \ \ \ 0}M^{{\mu }_2}_1\dots M^{{\mu }_n}_{\ \ \ \ \ (n-1)}$$Other combinations of ##{\mu '}_1{\mu '}_2\dots {\mu '}_n## either give ##0=0## or the same as ##01\dots (n-1)## or other cofactor expansions of the determinant.
However if the equation had been simplified the next equation (2.67) would not work and (2.67) is the punch line.
We also prove that the determinant of the metric under a coordinate transformation is given by
$$g\left(x^{\mu '}\right)={\left|\frac{\partial x^{\mu '}}{\partial x^{\mu }}\right|}^{-2}g\left(x^{\mu }\right)$$ See the details at
Saturday, 29 December 2018
Commentary 2.7 Causality
Part 1 Causality jargon
This section was introducing a ton of jargon and, as ever, Carroll confused me with his brevity and power sentences. Early on we had an achronal hypersurface which is one where no two points are connected by a timelike curve. Carroll gives any edgeless spacelike hypersurface in Minkowski space as an example. I was having a bit of trouble imagining an 'edgeless spacelike hypersurface in Minkowski space' when I found Fig 1. That made it obvious. The HYPERSURFACE OF THE PRESENT therein is achronal.![]() |
| Fig 1 |
timelike
|
inside the light cone - massive
particle
|
null (aka lightlike)
|
on the light cone - photon
|
spacelike
|
outside the light cone
|
On a flat (x,t) spacetime diagram any line is a hypersuface, if it's edgeless it continues forever and it is spacelike if its gradient m is always limited: -1 < m < 1 - it is more parallel to the space axis than the time axis.
![]() |
| Fig 2 Various types of hypersurface. Both spacelike surfaces are achronal. |
Thinking about an achronal hypersurface S, Carroll defines one + four (or eight) new terms.
Causal curve
|
One which is timelike or null
everywhere. Two are shown.
|
|
Causal future of S
|
J+(S)
|
Set of points that can be
reached from S by following a future directed causal curve
|
Chronological future of S
|
I+(S)
|
Set of points that can be
reached from S by following a future directed timelike curve
|
Future domain of dependence of
S
|
D+(S)
|
Set of all points that p such
that every past moving inextendible* causal curve through p intersects S.
Points predictable from S (see below).
|
Future Cauchy horizon of S
|
H+(S)
|
Boundary of D+(S).
Limit of predictable points (see below).
|
* inextendible means the curve goes on forever.
We'll now concentrate on the spacelike hypersurfaces S and T, which are achronal, in fig 3.
![]() |
| Fig 3 |
Before all the definitions, Carroll had mysteriously said that "We look at the problem of evolving matter fields …".
Light dawned: The evolution from events (the initial conditions) in S can only completely specify future events in D+(S) its future domain of dependence. Events beyond its future Cauchy horizon cannot be predicted from the initial conditions.
There were some more terms
Cauchy surface
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Closed achronal surface Σ whose
domain of dependence D+(Σ) is the entire manifold
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Globally hyperbolic
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A space time that has a Cauchy
surface
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Partial Cauchy surface
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? Cauchy surface whose domain
of dependence D+(Σ) is not the entire manifold
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Closed time like curve
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See below
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From information on a Cauchy surface on we can predict what happens throughout the entire manifold / entire universe / all spacetime.
Part 2. Cylindrical spacetime
We then have a simple example: Consider a two-dimensional geometry with coordinates ##\{t , x\}##, such that points with coordinates ##(t , x)## and ##(t , x+1)## are identified. The topology is thus ##\boldsymbol{\mathrm{R}}\times S^1##. We take the metric to be$${ds}^2=-{\mathrm{cos} \left(\lambda \right)\ }{\mathrm{d}t}^2-{\mathrm{sin} \left(\lambda \right)\ }\left[\mathrm{d}t\mathrm{d}x+\mathrm{d}x\mathrm{d}t\right]+{\mathrm{cos} \left(\lambda \right)\ }{\mathrm{d}x}^2$$where$$\lambda ={{\mathrm{cot}}^{-1} t\ }$$which goes from ##\lambda =0## (##t = - \infty ##) to ##\lambda = \pi## (##t = \infty ##).
##\lambda ={{\mathrm{cot}}^{-1} t\ }## is the same as ##\lambda ={\mathrm{tan}}^{-1} ( 1 / t )## and so
$$t=\ 1 /{\mathrm{tan} \lambda \ }$$That's a problem. As ##\lambda \to 0, t \to \infty ## not ##-\infty ##. So we really want
$$\lambda =-{{\mathrm{cot}}^{-1} t\ }$$To find the light cone we want a null vector ##V^{\mu }##, at various times. We can get this from the metric and Desmos plotted various light cones from 0 to ##\pi## as shown below. Details are in the pdf. I had arrived at the same diagram as Carroll!
Fig. 4. Shows the light cones in red in the distant past (λ = 0) to distant future (λ = π). In our diagram we are identifying points with coordinates (t,x) and (t, x+~100), so that we can better see the strange light cones for large t or λ≈π .
Our light cones rotate the same way as Carroll's Fig 2.25.
Carroll says "When t > 0, x becomes the timelike coordinate." (Because x, not t, is in the light cone. Moreover light and particles can only move in the positive x direction and they can move in + and - t directions.) We can now draw two causal curves from a point p as shown. One reaches the surface S, the other does not. Therefore p is outside the future Cauchy horizon of S. This applies to any point p with t > 0. As he says "There is thus necessarily a Cauchy horizon at t=0." Surely it's worse than that. There is a 'global' Cauchy horizon at t = 0. Perhaps that is what he meant.
In plainer language: "Nothing at t > 0 is predictable by things at t < 0".
I don't see why we had to have the cylindrical coordinate system. The closed causal curve guarantees that the causal curve from p is inextendible, but we could have had a curve that waved around forever keeping its t coordinate always >0.
Part 3 A singularity
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| Fig 5 |
Fig 5 shows a singularity at s and an achronal surface Σ that extends indefinitely in the plus and minus x directions. The point p cannot be in D+(Σ), future domain of dependence of Σ, because there are causal curves from p that end at s. Therefore there is a future Cauchy horizon at H+(Σ) as shown. H+(Σ) also extends indefinitely in the plus and minus x directions.
I am not sure if the right branch of the past light cone from p should escape the influence of the singularity, but it does not matter for this argument.
Part 4. A Diversion
Friday, 28 September 2018
Commentary on Appendix A: Mapping S2 and R3
Why am I here?
The problem part 1
Sticking the sphere into
This was not so simple and I struggled with the substitution for days. (1) is easy to show algebraically or geometrically. Getting to the metric in polar coordinates involves various differentiation rules: (trig functions and the product rule), but then they need to be applied to infinitesimals (dx, dθ etc) rather than proper derivatives (dx/dy etc). That aspect was very novel to me.



















