φ_{*} ∂_{x} = (cos θ) ∂_{x} + (sin θ)∂_{y}
φ_{*} ∂_{y} = -(sin θ) ∂_{x} + (cos θ) &part_{y}
The way in which Faraday made use of his ideas of lines of force in co-ordinating the phenomena of magneto-electric induction shew him to have been in reality a mathematician of a very high order — one from whom the mathematicians of the future may derive valuable and fertile methods. For the advance of the exact sciences depends upon the discovery and development of appropriate and exact ideas, by means of which we may form a mental representation of the facts, sufficiently general, on the one hand, to stand for any particular case, and sufficiently exact, on the other, to warrant the deductions we may draw from them by the application of mathematical reasoning. From the straight line of Euclid to the lines of force of Faraday this has been the character of the ideas by which science has been advanced, and by the free use of dynamical as well as geometrical ideas we may hope for a further advance. The use of mathematical calculations is to compare the results of the application of these ideas with our measurements of the quantities concerned in our experiments.... We are probably ignorant even of the name of the science which will be developed out of the materials we are now collecting.... — James Clerk Maxwell
Here's a list of known errors that appear in both the first and second version:
J=D_{k}(F_{ij}) \otimes *(dx^k\wedge *(dx^i \wedge dx^j))
should be
J=D_{k}(F_{ij}) \otimes *(dx^k\wedge *(dx^i \wedge dx^j)) + F_{ij} \otimes *(d*(dx^i \wedge dx^j)).
I need to look at the book to see what's going on, but the main result is still true: you can gauge transform a solution of the Yang-Mills equations and get a new solution.
Your convention for the Riemann tensor as compared with other authors: R_(Baez)^a_{bcd} = R_(Penrose)_{bcd}^a = -R_(Wald)_{bcd}^a = -R_(Ashtekar)_{bcd}^a The major difference in your notation is that the "a" index above is in the first "column", whereas all other authors put it in the last column. I think this difference in index ordering on the Riemann tensor causes some confusion when you introduce the "imitation Riemann tensor" in the chapter on the Palatini action. On page 408, you have: R^c_{ab}^d = F_{ab}^{IJ} (e_I^c) (e_J^d) (where I have omitted the tilde over the R, and replaced the Greek indices with lower case Latin indices due to the typographical limits of email.) I'm pretty sure the "c" and "d" on the right hand side should be reversed. I know this is just a definition and so cannot be right or wrong per se, but (1) if the imitation Riemann is truly to be analogous to the Riemann, the "I" index above on F should really correspond to the "d" index on R, and the "J" index on F should really correspond to the "c" index on R when using the tetrad to "transform" internal indices to space-time indices. (2) In any case, I concluded that it was impossible to do exercise 35 without modifying the definition of the imitation Riemann in the manner I described. While I'm at it, I think there's a missing factor of 1/2 in the equation atop p. 411. More precisely, I believe there should be a factor of 1/2 in front of the second term on the second line. I think you forgot it when expanding out the antisymmetrization brackets.
John Huerta wrote:
I think I may have spotted another erratum! This has to do with what Jonathan Engle was writing to you about: On page 408, you have: R^c_{ab}^d = F_{ab}^{IJ} (e_I^c) (e_J^d) (where I have omitted the tilde over the R, and replaced the Greek indices with lower case Latin indices due to the typographical limits of email.) I'm pretty sure the "c" and "d" on the right hand side should be reversed. I know this is just a definition and so cannot be right or wrong per se. This has consequences, not just for Exercise 35, but also for the main text. Specifically, transforming δS from the form with the coframe field and internal curvature, shown in the last line of page 409, to the form with the imitation Einstein tensor shown in the first equation of page 410, one gets the wrong sign (at least for the first term, but it must work the same for the second), unless you make the correction Jonathan Engle suggests!