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Rieman metric tensor

  1. Mar 28, 2014 #1
    Hello All,

    Sorry if my question seems to be elementary. I am trying to find out a little bit details of the Riemann metric tensor but not too much in details. I found out the metric (g11, g12, g13, g14......). It provides information on the manifold and those parameters have the information.

    Looking into the notation: I found that g00, g01 have certain partial derivatives.

    I mean to say what actually it is? What g11 and g12 have?

    Excuse me, if I am sounding too elementary.

  2. jcsd
  3. Mar 29, 2014 #2
    No. You are at the wrong level of abstraction. Let me give you an example to illustrate your error.
    Lets say there are vectors in 2-dimensional euclidean space. They have a direction and a magnitude (defined in the usual way). Given a vector, how would you describe it? Lets agree that the coordinate system has an orthonormal basis (x & y axes are orthogonal and 1 unit in x direction has same magnitude as 1 unit in y direction). These are your common x,y Euclidean coordinates. Now someone says here is a vector V. It has coordinates (1,1) {which indicate the line from (0,0) to the end-point of the vector (hence give its direction and magnitude). Suddenly, (to be dramatic) someone else says:"What?! No, no the coordinates are (√2,0)!!" Then, someone else says "Huh? They're (0,-√2)!!!" My question to you is: can all three of them be right?" The answer is: yes. How? They each chose different directions for their x,y axes. The point is that the vector exists INDEPENDENTLY of the coordinate system being used to describe it. Its DESCRIPTION (representation) depends on the choice of coordinates (frame of reference) but it (its existence) does NOT. So, you need to understand that if you allow different coordinate systems, then the coordinates of a vector (or any line, or point, or surface, or volume, etc.) depend on the coordinate basis so comparing (1,1) and (0,√2) can not distinguish whether they are the same vector or not UNLESS we know what the coordinate frame being used for each is.
    Once you grasp that a point or a vector are expressed as coordinates, then it follows that a metric is also independent of the coordinate frame but its representation is NOT. So, unless the frame of reference is agreed upon, it is meaningless to talk about g11, etc. (because the components depend on the coordinate basis).
    In a Riemannian Vector Space, length (magnitude) has certain characteristics which other types of vector spaces may not share. The metric is a way to define those characteristics so that changes in coordinates (which do NOT change the space) keep the properties of any vector invariant (unchanged). (Any change to the vector would NOT be just a change in coordinates, rather you actually create something different, while a change to just the coordinate system and the vector's coordinates keep the vector "the same" (just its representation changes)).
    The metric is a tensor. A tensor is DEFINED by the way it behaves under differentiation (with respect to the two coordinate systems (the new one and the old one, say). In general there are two possibilities. For any set of indices, it can remain invariant when differentiated with respect to the new coordinates OR it can remain invariant when differentiated with respect to the old coordinates...(if it does neither, it is by definition NOT a tensor). For Special Relativity the indices run from 0 to 3 (usually...compared to 1 to 3 for 3-dimensional euclidean space). A metric is generally a set of one or more n x n matrices which describe the transformation from one coordinate system to another. For a vector, you only need one matrix. To transform objects more complicated than vectors (ie tensors of higher order) you need more than one matrix, meaning a matrix of matrices. Note that in the simplest case, there may be no difference between the contravariant and covariant transformation rules (differentiation wrt new or old coordinates).
    All this mumbo-jumbo aside, the best way to understand the metric tensor is to use it. You need to see it being used in the equations of relativity, and how it acts to keep the underlying physical objects (particles, waves, forces, energies, etc.) invariant under change of coordinates (observers). You will grow more comfortable with it as you apply it. It is a learning process. I don't think you can really understand it except by application. Practice.
    Last edited: Mar 29, 2014
  4. Apr 2, 2014 #3
    Hello abitslow,

    Thank you very much for the wonderful, detailed and simple answer. I have few more queries.
    Before the advent of General Relativity, Riemann has already developed the metric tensor to define curved manifolds. The higher dimension is already been extended to 4 dimensions. The attempt of physics to unify all the laws to higher dimension is the goal for theories like GUT and TOE.

    There are several attempts like adding Maxwell's equation using the Riemann metric tensor, making it 4. After that adding general relativity, Yang Mills theory. In that way the metric becomes more complicated and goes on adding -- attempting finally to include all the important equations and creating a single metric which defines all the laws of nature.

    Am I correct? If there is any misconception, please do let me know.

  5. Apr 2, 2014 #4


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    What you have described sounds to me like using higher dimensions to try to include other fields. Kaluza tried this (in the 1920's) with Electromagnetism and found that 5 dimensional (4 spatial+1time) GR is basically regular GR+E&M (plus some scalar field). Where curvature in the 5th dimension gave rise to electric and magnetic fields. However he could not figure out why, in everyday life, we don't see 4 spatial dimensions (as is obvious!). Klein, I believe, later figured out how to compactify the extra spatial dimension. Attempts at adding the weak and strong forces to this framework, so far, have been unsuccessful as far as I know.

    Yang Mills theories are gauge-symmetry theories. The gauge groups are internal degrees of freedom of the particles, and do not manifest as symmetries of the underlying spacetime (at least not in an obvious way). They are usually thought of as being in a background 4-D Minkowski spacetime, with extra gauge freedoms occuring in particle fields. The (Lie) groups of gauge symmetries are curved manifolds, but the underlying spacetime is flat.

    That is the limit of my knowledge on this area. Maybe someone more knowledgeable can enlighten you further.
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