Saturday 30 March 2019

Here we meet the covariant divergence and prove a thing or two about it. (Carroll said it was 'easy'.) We also have the curved-space version of Stokes's theorem using the covariant derivative and finally the exterior derivative and commutator, where Carroll seems to have made a very peculiar typo. Sir George Gabriel Stokes, 1st Baronet was a very clever but he did not discover the theorem. He popularised it.

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


I have sent this list of eight corrections I have found to date to the author, Sean Carroll. There is one extra (p.148) which I only found by peeking at a reprint of the question.

Chapter 1
p.24 just before equation (1.68) it reads "There is also the Levi-Civita symbol a (0,4) tensor." Surely it is not a tensor.

Chapter 2

p.81 Figure 2.26 fig 2.26 shows a singularity at a point p and the text discusses a point p that is in the future of the singularity. ( part 2)

Possible error
p. 82 Equation (2.66) could be much simpler if μ'1 μ'2 ...μ'n = 01...(n-1). (It's more streamlined.)
This is trued but if the simplified for was given, the next equation would not work! (GK)

Chapter 3
p.96 equation (3.10) for the connection transformation law. The + sign should be -. (

Possible error
p.99 definition of torsion-free (connection symmetric in lower indices) given as Γλμν = Γλ(μν). Surely Γλμν = Γλνμ. would be clearer? (

Exercise 6(a) should end "Which clock ticks faster?" I picked this up from

p. 427 equation (A.11) ϕ is used to mean two different things: a map and a polar coordinate. (

The Christoffel Symbol

At last we move on to chapter three on curvature and immediately we find the Christoffel symbol Γ which is upper case γ. Elwin Bruno Christoffel was a mathematician, born in Prussia in 1829, and he studied at the University of Berlin. "He introduced fundamental concepts of differential geometry, opening the way for the development of tensor calculus, which would later provide the mathematical basis for general relativity."

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.

Sunday 24 March 2019

Transformation of Christoffel symbol

Correction to Physics Simplified

There is a derivation of the coordinate transformation for the Christoffel symbol ##\mathrm{\Gamma }## (or the connection) on Physics Simplified (here:
It is very wrong! One obvious clue is that it twists the indices giving ##{\mathrm{\Gamma }}^{\lambda {'}}_{\mu {'}\nu {'}}## in terms of ##{\mathrm{\Gamma }}^{\mu }_{\nu \lambda }##.
It is also assuming that the connection is torsion free and metric compatible. That is that
{\mathrm{\Gamma }}^{\lambda }_{\mu \nu }={\mathrm{\Gamma }}^{\lambda }_{\nu \mu } & \phantom {10000}(1) \\
\mathrm{\nabla }g_{\mu \nu }=0 & \phantom {10000}(2) \\
\end{align}These are not necessary and I will show why in my next post (I hope.)

Here is the correct derivation.

We have the (torsion free , metric compatible) equation for the connection, which can be written with or without primes\begin{align}
{\mathrm{\Gamma }}^{\lambda {'}}_{\mu {'}\nu {'}}=\frac{1}{2}g^{\lambda {'}\rho {'}}\left({\partial }_{\mu {'}}g_{\nu {'}\rho {'}}+{\partial }_{\nu {'}}g_{\rho {'}\mu {'}}-{\partial }_{\rho {'}}g_{\mu {'}\nu {'}}\right) & \phantom {10000}(3) \\
\end{align}If we take the first term in the brackets and use the tensor transformation law we get\begin{align}
{\partial }_{\mu {'}}g_{\nu {'}\rho {'}} & =\frac{\partial }{\partial x^{\mu {'}}}\left(\frac{\partial x^{\nu {'}}}{\partial x^{\nu }}\frac{\partial x^{\rho {'}}}{\partial x^{\rho }}g_{\nu \rho }\right) & \phantom {10000}(4) \\
 & =\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial }{\partial x^{\mu }}\left(\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}g_{\nu \rho }\right) & \phantom {10000}(5) \\
 & =\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}{\partial }_{\mu }g_{\nu \rho }+g_{\nu \rho }\left(\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}\frac{{\partial }^2x^{\nu }}{\partial x^{\mu }\partial x^{\nu {'}}}+\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{{\partial }^2x^{\rho }}{\partial x^{\mu }\partial x^{\rho {'}}}\right) & \phantom {10000}(6) \\
\therefore {\partial }_{\mu {'}}g_{\nu {'}\rho {'}} & =\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}{\partial }_{\mu }g_{\nu \rho }+g_{\nu \rho }\left(\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}\frac{{\partial }^2x^{\nu }}{\partial x^{\mu {'}}\partial x^{\nu {'}}}+\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{{\partial }^2x^{\rho }}{\partial x^{\mu {'}}\partial x^{\rho {'}}}\right) & \phantom {10000}(7) \\
\end{align}That was the first error in Physics Simplified. The indices ##\mu ,\nu ## in the denominator of its last term of its fourth equation were the wrong way round. That would be like swapping ##\nu ,\rho ## in the last term of (7).

We can use (7) to get the other two terms in (3). We start with the first part, ##{\partial }_{\mu }g_{\nu \rho }##, of (7):

Second term (3), change first term of (7) with ##\mu \to \nu ,\ \ \nu \to \rho ,\ \ \rho \to \mu ## \begin{align}
T21=\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}{\partial }_{\nu }g_{\rho \mu } & \phantom {10000}(8) \\
\end{align}Third term (3), change first term of (7) with ##\mu \to \rho ,\ \ \nu \to \mu ,\ \ \rho \to \nu ## and negate
T31=-\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}{\partial }_{\rho }g_{\mu \nu } & \phantom {10000}(9) \\
\end{align}If we added up those three we get the first part of the fifth equation in Physics Simplified with ##\mu \to \lambda ##

In the above we had really substituted the primed ##\mu ,\nu ,\rho ## (which are the free indices) and then changed the dummy variables ##\mu ,\nu ,\rho##. For the next step we will have to be more careful. Following the same procedure only on the primed indices we get
T22=g_{\nu \rho }\left(\frac{\partial x^{\rho }}{\partial x^{\mu {'}}}\frac{{\partial }^2x^{\nu }}{\partial x^{\nu {'}}\partial x^{\rho {'}}}+\frac{\partial x^{\nu }}{\partial x^{\rho {'}}}\frac{{\partial }^2x^{\rho }}{\partial x^{\nu {'}}\partial x^{\mu {'}}}\right) & \phantom {10000}(10) \\
T32=-g_{\nu \rho }\left(\frac{\partial x^{\rho }}{\partial x^{\nu {'}}}\frac{{\partial }^2x^{\nu }}{\partial x^{\rho {'}}\partial x^{\mu {'}}}+\frac{\partial x^{\nu }}{\partial x^{\mu {'}}}\frac{{\partial }^2x^{\rho }}{\partial x^{\rho {'}}\partial x^{\nu {'}}}\right) & \phantom {10000}(11) \\
\end{align}We now swap the dummy variables ##\nu ,\rho ## in (11) and because of the symmetry of ##g_{\nu \rho }## we can swap them again on that and it becomes\begin{align}
T32=-g_{\nu \rho }\left(\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{{\partial }^2x^{\rho }}{\partial x^{\rho {'}}\partial x^{\mu {'}}}+\frac{\partial x^{\rho }}{\partial x^{\mu {'}}}\frac{{\partial }^2x^{\nu }}{\partial x^{\rho {'}}\partial x^{\nu {'}}}\right) & \phantom {10000}(12) \\
\end{align}exploiting the rule that partial derivatives commute we can see that the first term of (12) cancels the last term of (7). Likewise the second term of (12) cancels the first of (10). After swapping dummy variables ##\mu,\rho## again the second term of (10) is the same as its equivalent in (7). So adding up (7), (8), (9) (10) and (12) and putting them into (3) we get
{\mathrm{\Gamma }}^{\lambda {'}}_{\mu {'}\nu {'}}=\frac{1}{2}g^{\lambda {'}\rho {'}}\left(\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}\left({\partial }_{\mu }g_{\nu \rho }+{\partial }_{\nu }g_{\rho \mu }-{\partial }_{\rho }g_{\mu \nu }\right)+2g_{\nu \rho }\frac{\partial x^{\nu }}{\partial x^{\rho {'}}}\frac{{\partial }^2x^{\rho }}{\partial x^{\nu {'}}\partial x^{\mu {'}}}\right) & \phantom {10000}(13) \\
\end{align}If we now apply the tensor transformation law to the ##g^{\lambda {'}\rho {'}}## at the front of that we get
{\mathrm{\Gamma }}^{\lambda {'}}_{\mu {'}\nu {'}}=\frac{1}{2}\frac{\partial x^{\lambda {'}}}{\partial x^{\lambda }}\frac{\partial x^{\rho {'}}}{\partial x^{\rho }}g^{\lambda \rho }\left(\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}\left({\partial }_{\mu }g_{\nu \rho }+{\partial }_{\nu }g_{\rho \mu }-{\partial }_{\rho }g_{\mu \nu }\right)+2g_{\nu \rho }\frac{\partial x^{\nu }}{\partial x^{\rho{'}}}\frac{{\partial }^2x^{\rho }}{\partial x^{\nu {'}}\partial x^{\mu {'}}}\right) & \phantom {10000}(14) \\
which is similar to, but not I think the same as the seventh equation in Physics Simplified. Nevertheless simplifying:
{\mathrm{\Gamma }}^{\lambda {'}}_{\mu {'}\nu {'}}=\frac{\partial x^{\lambda {'}}}{\partial x^{\lambda }}\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{1}{2}g^{\lambda \rho }\left({\partial }_{\mu }g_{\nu \rho }+{\partial }_{\nu }g_{\rho \mu }-{\partial }_{\rho }g_{\mu \nu }\right)+\frac{\partial x^{\lambda {'}}}{\partial x^{\lambda }}{\delta }^{\lambda }_{\nu }{\delta }^{\nu }_{\rho }\frac{{\partial }^2x^{\rho }}{\partial x^{\nu {'}}\partial x^{\mu {'}}} & \phantom {10000}(15) \\
\end{align}The ##g^{\lambda \rho }\left(\dots \right)/2## part of that is of course ##{\mathrm{\Gamma }}^{\lambda }_{\mu \nu }##, so we get
{\mathrm{\Gamma }}^{\lambda {'}}_{\mu {'}\nu {'}}=\frac{\partial x^{\lambda {'}}}{\partial x^{\lambda }}\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}{\mathrm{\Gamma }}^{\lambda }_{\mu \nu }+\frac{\partial x^{\lambda {'}}}{\partial x^{\nu }}\frac{{\partial }^2x^{\nu }}{\partial x^{\nu {'}}\partial x^{\mu {'}}} & \phantom {10000}(16) \\
This is a correct transformation for the connection.

It is definitely not the same as the one given Physics Simplified which was\begin{align}
{\mathrm{\Gamma }}^{\lambda {'}}_{\mu {'}\nu {'}}=\frac{\partial x^{\mu {'}}}{\partial x^{\mu }}\frac{\partial x^{\lambda }}{\partial x^{\lambda {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}{\mathrm{\Gamma }}^{\mu }_{\nu \lambda }+\frac{\partial x^{\mu {'}}}{\partial x^{\mu }}\frac{{\partial }^2x^{\mu }}{\partial x^{\lambda {'}}\partial x^{\nu {'}}} & \phantom {10000}(17) \\
Pdf and docx files at Commentary 3.2 Correction to Physics Simplified.pdf

Saturday 23 March 2019

Exercise 3.01 Consequences of metric compatibility

Christoffel Symbol
At last I am on to chapter 3 on curvature. It has taken me three weeks to do the first four pages and exercise 3.01 was done on the way. I learned or relearned many things and had two interesting discussions on Physics forums (here and here). The question asked us to prove that the covariant derivative of the inverse metric and and that the covariant derivative of the Levi-Civita tensor both vanished if the covariant derivative of the metric vanishes and the Christoffel Symbol is symmetric in its lower indices. These conditions are known as metric compatibility and torsion freedom(?) respectively. One would say 'a connection on a manifold is torsion free and metric compatible'.

In maths that is


Verify the consequences of metric compatibility: If
{\mathrm{\nabla }}_{\sigma}g_{\mu \nu }=0 & \phantom {10000}(1) \\
\end{align}then (a)\begin{align}
{\mathrm{\nabla }}_{\sigma}g^{\mu \nu }=0 & \phantom {10000}(2) \\
\end{align}and (b)\begin{align}
{\mathrm{\nabla }}_{\lambda}{\varepsilon }_{\mu \nu \sigma \rho }=0 & \phantom {10000}(3) \\
\end{align} I am not sure if we are assuming ##{\mathrm{\Gamma }}^{\tau }_{\lambda \mu }={\mathrm{\Gamma }}^{\tau }_{ \mu \lambda }## or not.


Part (a) was quite simple but I struggled with the part (b) until 23 March and had to give up. Along the way I had lots of practice at index manipulation, I reacquainted myself with Cramer's rule for solving simultaneous equations, proved (b) on the surface of a sphere, found the 'dynamite' version of Carroll's streamlined matrix determinant equation (2.66) and added some equation shortcut keys to my keyboard. The time was not wasted.

We make frequent use here of  the fact that ##g_{\mu \nu }g^{\mu \rho }\mathrm{=}{\delta}^{\rho }_{\nu }## and the indexing effect of the Kronecker delta: ##{\delta}^{\lambda }_{\beta }\mathrm{\Gamma }^{\mu }_{\sigma \lambda }={\mathrm{\Gamma }}^{\mu }_{\sigma \beta }## because we are summing over ##\lambda ## and the only non-zero term is when ##\beta =\lambda ##. In this case ##\mathrm{\Gamma }## can be replaced by any symbol or tensor of any rank.

Here is the full effort Ex 3.01 Consequences of metric compatibility.pdf (7 pages of which 4 might be worth looking at).

The comment left by JSBach1801 solved this problem very easily as I fully realised in Jan 2021. Thanks! That answer is right at the end of the pdf.

Thursday 14 March 2019

Second order partial derivatives vanish?

At the end of a long proof I came across something in tensor calculus that seems too good to be true. And if something seems too good to be true .....

The something is that a second order partial derivative vanishes if one of the parts in the denominator is in the same reference frame as the numerator. That is for example
\frac{{\partial }^2x^{\mu }}{\partial x^{\rho }\partial x^{{\mu }^{'}}}=\frac{{\partial }^2x^{\mu }}{\partial x^{{\mu }^{'}}\partial x^{\rho }}=0 & \phantom {10000}(1) \\
\end{align}The equality of the first two parts follows because partial derivatives commute. We have
\frac{{\partial }^2x^{\mu }}{\partial x^{{\mu }^{'}}\partial x^{\rho }}=\frac{\partial }{\partial x^{{\mu }^{'}}}\left(\frac{\partial x^{\mu }}{\partial x^{\rho }}\right) & \phantom {10000}(2) \\
\frac{\partial x^{\mu }}{\partial x^{\rho }}={\delta }^{\mu }_{\rho } & \phantom {10000}(3) \\
\end{align}where ##{\delta }^{\mu }_{\rho }## is the Kronecker delta which is a constant. (3) seems very reasonable because when ##\mu \neq \rho ##, ##\partial x^{\mu }## and ##\partial x^{\rho }## are orthogonal so ##{\partial x^{\mu }}/{\partial x^{\rho }}## vanishes and when ##\mu =\rho ##, ##{\partial x^{\mu }}/{\partial x^{\rho }}=1##.

So (2) is the derivative of a constant which always vanishes. QED.

Have I made a very stupid mistake? Or am I stating something everybody knows?

On PF at
The good stevendaryl pointed out my error and gave a little example that proved$$
\frac{{\partial }^2x^{\mu }}{\partial x^{\rho }\partial x^{{\mu }^{'}}}≠\frac{{\partial }^2x^{\mu }}{\partial x^{{\mu }^{'}}\partial x^{\rho }}
$$Partial derivatives only commute if they are in the the same coordinate system, so this IS true$$
\frac{{\partial }^2x^{\mu }}{\partial x^{\rho }\partial x^{{\nu }}}=\frac{{\partial }^2x^{\mu }}{\partial x^{{\nu }}\partial x^{\rho }}$$Say we had$$
f\left(x,y\right)=x^2{\mathrm{sin} y\ }
\frac{\partial }{\partial x}\left(\frac{\partial }{\partial y}f\right)=\frac{\partial }{\partial x}\left(x^2{\mathrm{cos} y\ }\right)=2x{\mathrm{cos} y\ }
\frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}f\right)=\frac{\partial }{\partial y}\left(2x{\mathrm{sin} y\ }\right)=2x{\mathrm{cos} y\ }$$Say we had
$$x^{\mu }=z$$then $$\frac{\partial }{\partial x}\left(\frac{\partial }{\partial y}z\right)=0$$and
$$\frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}z\right)=0$$Say we had$$x^{\mu }=y
\frac{\partial }{\partial x}\left(\frac{\partial }{\partial y}y\right)=\frac{\partial }{\partial x}\left(1\right)=0$$and$$
\frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}y\right)=\frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}y\right)=0$$

Wednesday 6 March 2019

Question on covariant derivatives

I am reading I am reading Spacetime and Geometry : An Introduction to General Relativity -- by Sean M Carroll and have arrived at chapter 3 where he introduces the covariant derivative ##{\mathrm{\nabla }}_{\mu }##. He makes demands on this which are \begin{align}
\mathrm{1.\ Linearity:}\mathrm{\ }\mathrm{\nabla }\left(T+S\right)=\mathrm{\nabla }T+\mathrm{\nabla }S & \phantom {10000}(1) \\
\mathrm{2.\ Leibniz\ rule:}\mathrm{\nabla }\left(T\ \otimes \ \ S\right)=\left(\mathrm{\nabla }T\right)\ \ \otimes \ \ S+T\ \otimes \ \ \left(\mathrm{\nabla }S\right) & \phantom {10000}(2) \\
{\mathrm{3.\ Commutes\ with\ contractions:}\mathrm{\nabla }}_{\mu }\left(T^{\lambda }_{\ \ \ \lambda \rho }\right)={\left(\mathrm{\nabla }T\right)}^{\mathrm{\ \ \ }\lambda}_{\mu \ \ \lambda \rho } & \phantom {10000}(3) \\
{\mathrm{4.\ Reduces\ to\ partial\ derivative\ on\ scalars:}\mathrm{\nabla }}_{\mu }\phi ={\partial }_{\mu }\phi  & \phantom {10000}(4) \\
\end{align}1,2 and 4 seem reasonable but I cannot understand 3 and he does not seem to use it, even though he implies that he does.

The LHS of (3) seems straight forward\begin{align}
{\mathrm{\nabla }}_{\mu }\left(T^{\lambda }_{\ \ \ \lambda \rho }\right) & ={\partial }_{\mu }T^{\lambda }_{\ \ \ \lambda \rho }+{\mathrm{\Gamma }}^{\lambda }_{\mu \kappa }T^{\kappa }_{\ \ \ \lambda \rho }-{\mathrm{\Gamma }}^{\kappa }_{\mu \lambda }T^{\lambda }_{\ \ \ \kappa \rho }-{\mathrm{\Gamma }}^{\kappa }_{\mu \rho }T^{\lambda }_{\ \ \ \lambda k} & \phantom {10000}(5) \\
 & ={\partial }_{\mu }T^{\lambda }_{\ \ \ \lambda \rho }-{\mathrm{\Gamma }}^{\kappa }_{\mu \rho }T^{\lambda }_{\ \ \ \lambda k} & \phantom {10000}(6) \\
\end{align}Which is very like the rule for the covariant derivative of a (0,1) tensor.

I understand that the ##\mathrm{\nabla }T## in (1) and (2) means ##{\mathrm{\nabla }}_{\sigma}T## where ##T## is some tensor. So the RHS of (3) appears to be ##{\left({\mathrm{\nabla }}_{\sigma}T\right)}^{\mathrm{\ \ \ }\lambda}_{\mu \ \ \lambda \rho }## which leaves too many indices on the RHS. Otherwise the RHS is some kind of derivative with one contra- and three co-variant indices. What is that?

Posted on Physics Forums at

Saturday 2 March 2019

Exercise 2.10 Maxwells equations in 2-dimensional spacetime


Consider Maxwell's equations, d= 0 , d*= *J in 2-dimensional spacetime. Explain why one of the two equations can be discarded. Show that the electro-magnetic field can be expressed in terms of a scalar field. Write out the field equations for this scalar field in component form.


"2-dimensional spacetime"? Is that one dimension of space and another of time or two space and one time? We assume the former and see what happens.

This question seemed too easy, especially after my labours on Maxwell's equation with three-forms. In one time and one space dimension, the first of Maxwell's equations becomes 0 = 0 and I think that the second produced the scalar field which I wrote down in almost one page: Ex 2.10 Maxwells equations in 2-dimensional spacetime.pdf.