I Carroll GR: Tangent Space & Partial Derivatives

Click For Summary
The discussion centers on the mathematical framework presented by Carroll regarding the differentiation of functions on an n-manifold M. The key point is the relationship between the derivative of a function f: M → R and its expression as a composite function f ∘ γ, where γ is a curve parameterized by λ. Participants express confusion over the assumptions made about coordinates and dimensions, particularly in cases where a single coordinate system may not suffice. There is acknowledgment that Carroll's approach implicitly defines f as a function of λ through the curve γ, which may not be immediately clear. Overall, the conversation highlights the challenges of understanding the notation and concepts in general relativity, especially regarding the mapping of functions across different dimensional spaces.
chartery
Messages
42
Reaction score
5
TL;DR
Difficulty understanding first equality in equation 2.9 (p 43) of Carroll's lecture notes.
He draws an n-manifold M, a coordinate chart φ : M → Rn, a curve γ : R → M, and a function f : M → R, and wants to specify ##\frac d {d\lambda}## in terms of ##\partial_\mu##.
##\lambda## is the parameter along ##\gamma##, and ##x^\mu## the co-ordinates in ##\text{R}^n##.

His first equality is ##\frac d {d\lambda}\text{f}## = ##\frac d {d\lambda}##(##\text {f} \circ \gamma##).
It is not clear to me how he can equate the derivative of a map from M to R, with that of a composite map from R to R.

(Feel free to indicate that the question shows my knowledge is inadequate for this level of study! :rolleyes:)
 
Physics news on Phys.org
He really means the function ##f## seen as a function of ##\lambda##, so ##f\circ \gamma## so the first step is more of a definition of what he means.
 
  • Like
Likes chartery, Ibix and malawi_glenn
Concrete (well, partly concrete) example: you have a function ##f(x,y)## defined on a 2d plane and a path ##\gamma## on that plane whose points have coordinates ##x_\gamma(\lambda)## and ##y_\gamma(\lambda)##, so the curve is parameterised by ##\lambda##. To calculate ##\frac d{d\lambda}f(x,y)## you just sub in the curve - ##\frac d{d\lambda}f(x_\gamma(\lambda),y_\gamma(\lambda))##, and now you've got a function of ##\lambda## that you can differentiate. OK so far?

The above maths, though, makes assumptions about coordinates that you may not wish to make (e.g. sometimes you can't make a single coordinate system cover all the space and the path might move between patches), and also assumes a dimension of 2. So Carrol is observing without explicit dependence on coordinates or dimensionality that to differentiate ##f## with respect to ##\lambda## he first needs to express ##f## as ##f(\lambda)##, and that's what ##f\circ\gamma## does.
 
Last edited:
  • Like
Likes chartery
Many thanks for helpful replies. To be sure I understand, he is defining his (plain!) f here to be f(##\gamma##(##\lambda##)) ?
 
  • Like
Likes Ibix
chartery said:
Many thanks for helpful replies. To be sure I understand, he is defining his (plain!) f here to be f(##\gamma##(##\lambda##)) ?
Yes, you will often see this implicit assumption for brevity of notation.
 
  • Like
Likes PeroK and Ibix
Obvious with hindsight, though I think the diagram preceding, and the lack of any reference to f as a composite sent me on the detour. Thanks again.
 
Carroll's notes were the first GR text I read. I think that idea of functions as maps from one space of one dimensionality to another, and chaining those maps together in order to export structure from one space into another, is one of the things I found harder to integrate into my thinking. Not because it's particularly difficult, but because it's a very different way of looking at something very familiar, and Carroll perhaps doesn't spend quite enough time (for me, anyway) on introducing it before using it with cheerful abandon.
 
  • Like
Likes vanhees71
@Ibix I'm with you on the cheerful, not to say complete, abandon :-)
 
  • Like
Likes Ibix