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A Riemannian Manifolds and Local Cartesian Coordinates

  1. Mar 13, 2016 #1
    Hello! Good morning to all forum members!
    I am studying general relativity through the wonderful book: "General Relativity: An Introduction for Physicists" by M.P. Hobson (Cambridge University Press) (2006). My question is about Riemannian manifolds and local cartesian coordinates (Chapter 02 - Section 2.11 - Page 43). Section starts explaining that for a general Riemannian manifold (considering only strictly Riemannian manifolds), it is not possible to perform a coordinate transformation from an arbitrary coordinate system to a desired coordinate system that will take the line element into a Euclidean form. Section goes on to explain that, however, it is possible to make a coordinate transformation such that in the neighbourhood of some especified point P the line element takes the Euclidean form. The next step is to demonstrate that fact by applying the expansion in Taylor series about the point P. The general transformation rule for the metric functions is called to perform the Taylor expansion. My question is this demonstration. In particular I am not able to understand the expression provided for expansion in Taylor series, especially terms that have derivatives of higher orders. I tried to justify this fact by considering an open ball centered on the point P, where the higher order derivatives arises from points in the neighbourhood of the point P. To be honestly I am not convinced that I am on the right track. Could someone explain me this demonstration, in particular the expansion in Taylor series. I am having doubts on writing Taylor expansion about the point P.
    I am really grateful for any response.
    PS.: Unfortunately, I still can not write equations using Latex on our forum, which would facilitate the understanding of my doubt.
  2. jcsd
  3. Mar 13, 2016 #2
    To help understand my doubt I wrote expansion in Taylor series that is disturbing me.

    Attached Files:

  4. Mar 13, 2016 #3


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    If you have any function [itex]F(x,y,z,t)[/itex] that is sufficiently well-behaved at [itex]x_0, y_0, z_0, t_0[/itex], then you can write it as a power series:

    [itex]F(x,y,z,t) = F(x_0, y_0, z_0, t_0) + \sum_j \frac{\partial F}{\partial x^j}|_{P_0} \delta x^j + \sum_{jk} \frac{1}{2} \frac{\partial^2 F}{\partial x^j \partial x^k}|_{P_0}\delta x^j \delta x^k+ \sum_{jkl} \frac{1}{6} \frac{\partial^3 F}{\partial x^j \partial x^k \partial x^l}|_{P_0} \delta x^j \delta x^k \delta x^l + ...[/itex]

    where [itex]x^0 = t, x^1 = x, x^2 = y, x^3 = z[/itex] and [itex]\delta x^i = (x^i - x^i_0)[/itex] and [itex]P_0[/itex] is the point [itex]x_0, y_0, z_0, t_0[/itex], and where [itex]|_{P_0}[/itex] means "evaluate at point [itex]P_0[/itex]".

    This is just the generalization of Taylor series to the case of multiple variables.
  5. Mar 13, 2016 #4
    Hello! I am really grateful for your answer!
    I could not fully understand the meaning of the term ##\delta x^{i}=\left ( x^{i}-x_{0}^{i} \right )##. The general transformation rule for the metric functions is:
    ##g_{ab}^{'}=\frac{\partial x^{c}}{\partial x^{'a}}\frac{\partial x^{d}}{\partial x^{'b}}g_{cd}##
    The desired condition (in the neighbourhood of ##P##) is given by:
    ##g_{ab}^{'}=\delta _{ab}+O\left [ \left ( x^{'}-x_{p}^{'} \right )^2 \right ]##
    However, at the moment, the objective is to perform the expansion in Taylor series about the point ##P##. There needs to exists a relation between the arbitrary coordinate system (##x^{a}##) and the desired coordinate system (##x^{'a}##) given by ##x^{a}\left ( x^{'} \right )##And my question lies at this point. The term ##\delta x^{i}=\left ( x^{i}-x_{0}^{i} \right )## means points of an open ball centered on the point ##P##? I could not connect the mathematical equation, expansion in Taylor series, with the physical meaning of a relation connecting the two sets of coordinates. Again, I am really grateful for any response :smile:.
    PS.: I finally succeeded in learning how to insert equations in Latex!
  6. Mar 15, 2016 #5
    Hello! Good morning for all forum members!
    I believe that one of the points for which I am having trouble is the difference between the meaning of Taylor expansion at a point ##P## and about a point ##P##. I believe that there is a relationship with the radius of convergence in the second case. Could anyone help me with this question and provide a brief explanation of how the expansion in Taylor series is related to the general transformation rule for the metric functions? Again, I am really grateful for any response :smile:.
  7. Mar 15, 2016 #6


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    I would have to see an excerpt from the book to know what they are trying to do, but there is no particular relationship. You have an original coordinate system, [itex]x^i[/itex] with corresponding metric [itex]g_{ij}[/itex]. You're transforming to a new coordinate system, [itex]y^\alpha[/itex] with corresponding metric [itex]g_{\alpha \beta}[/itex]. The relationship between them is:

    [itex]g_{\alpha \beta} = \sum_{ij} \frac{\partial x^i}{\partial y^\alpha} \frac{\partial x^j}{\partial y^\beta}g_{ij}[/itex] or just [itex]\frac{\partial x^i}{\partial y^\alpha} \frac{\partial x^j}{\partial y^\beta}g_{ij}[/itex] with the implied summation convention.

    So if you are trying to find a locally Euclidean coordinate system, then you want at the chosen point [itex]\mathcal{P}[/itex] to have:

    [itex]g_{\alpha \beta} = \delta_{\alpha \beta}[/itex]
    [itex]\frac{\partial g_{\alpha \beta}}{\partial y^\mu} = 0[/itex]
    [itex]\frac{\partial^2 g_{\alpha \beta}}{\partial y^\mu \partial y^\nu} = 0[/itex]

    So you can just use the product rule:
    [itex]\frac{\partial g_{\alpha \beta}}{\partial y^\mu} = \frac{\partial }{\partial y^\mu} (\frac{\partial x^i}{\partial y^\alpha} \frac{\partial x^j}{\partial y^\beta} g_{ij})[/itex]
    [itex] =\frac{\partial^2 x^i}{\partial y^\alpha \partial y^\mu} \frac{\partial x^j}{\partial y^\beta} g_{ij} + \frac{\partial x^i}{\partial y^\alpha} \frac{\partial^2 x^j}{\partial y^\beta \partial y^\mu} g_{ij} + \frac{\partial x^i}{\partial y^\alpha} \frac{\partial x^j}{\partial y^\beta} \frac{\partial g_{ij}}{\partial y^\mu}[/itex]

    The second derivative will have 6 terms.
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