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Metric notation

  1. Aug 26, 2010 #1
    The Wikipedia article Metric tensor (general relativity) has the following equation for the metric tensor in an arbitrary chart, g =

    [tex]g_{\mu\nu} \, \mathrm{d}x^\mu \otimes \mathrm{d}x^\nu[/tex]

    It then says, "If we define the symmetric tensor product by juxtaposition, we can write the metric in the form"

    [tex]g=g_{\mu\nu} \, \mathrm{d}x^\mu \, \mathrm{d}x^\nu[/tex]

    and

    [tex]\mathrm{d}s^2=g_{\mu\nu} \, \mathrm{d}x^\mu \, \mathrm{d}x^\nu[/tex]

    and, "In general relativity, the terms metric and line element are often used interchangeably."

    Given that this quantity, g = ds2, if it has any units, would have units of length or time squared, would "area element" be a more apt name? Sometimes a superscript 2 is attached to a first order tensor such as a cotangent vector, w, to mean g-1(w,w), or, in matrix terms, if we represent its components as a row, w wT, a scalar. In this case, it looks as though the superscript 2 corresponds rather to wT w, a 2x2 matrix. A tensor product, rather than a contraction. Is that right?

    Regarding the expression "if we define the symmetric tensor product by juxtaposition", would another way of putting this be: here juxtaposition will mean the tensor product symmetrised,

    [tex]\mathrm{d}x^\mu \, \mathrm{d}x^\nu \equiv \mathrm{d}x^{(\mu} \, \mathrm{d}x^{\nu)} = \frac{1}{2}(\mathrm{d}x^\mu \otimes \mathrm{d}x^\nu + \mathrm{d}x^\nu \otimes \mathrm{d}x^\mu)[/tex]

    which, in this case,

    [tex]=\mathrm{d}x^\mu \otimes \mathrm{d}x^\nu[/tex]

    (if it's okay, in this instance, to use indices denoting which 1-form in the same way as indices are used when they denote components)?

    This notation seems completely at odds with the use of juxtaposition to denote the geometric product, since, in that case, I think

    [tex]\mathrm{d}x^\mu \, \mathrm{d}x^\nu = \mathrm{d}x^\mu \wedge \mathrm{d}x^\nu[/tex]

    i.e. something antisymmetric. Is there a less ambiguous notation for what the Wikipedia article want to say; would it be clearer to use the regular tensor product symbol as they did in their first equation and just state that this tensor is symmetric?
     
    Last edited: Aug 26, 2010
  2. jcsd
  3. Aug 26, 2010 #2

    bcrowell

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    In Cartesian coordinates, the metric of the Euclidean plane is [itex]ds^2=dx^2+dy^2[/itex], which is simply the Pythagorean theorem. The Pythagorean theorem basically tells you the length of the hypotenuse, ds, not an area. Another good example is the form of the metric of the Euclidean plane in oblique Cartesian coordinates: http://www.lightandmatter.com/html_books/genrel/ch03/ch03.html#Section3.4 [Broken] , subsection 3.4.1, example 7. The metric is [itex]ds^2=dx^2+dy^2+\cos\phi dx dy[/itex], where [itex]\phi[/itex] is the angle between the x and y axes. If the metric represented area, then this expression should vanish for [itex]\phi=0[/itex], but it doesn't.

    In Euclidean space, we could simply talk about [itex]ds=\sqrt{ds^2}[/itex], and that would make it clearer that we were talking about a length, not an area. The reason we don't do this in relativity is that ds2 can have either sign, so ds would sometimes be imaginary, and, more importantly, wouldn't be an analytic function. The whole structure of GR is based on the assumption that certain functions are smooth, e.g., any coordinate transformation is allowed, provided that it's smooth.
     
    Last edited by a moderator: May 4, 2017
  4. Aug 26, 2010 #3

    jcsd

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    The metric defines the scalar product, so ds2 is the square norm in a tangent space. The norm is usually thought of as the 'length' of a vector.
     
  5. Aug 26, 2010 #4
    Okay, if I've understood correctly: although the metric has units of length squared in the context of relativity, so as to keep it real and differentiable, its purpose is still calculating length, and the metric isn't the thing to integrate if we want to calculate an area.

    How can g = ds2, if g is tensor of valence (0,2) and ds2 a tensor of valence (0,0), i.e. a scalar? Perhaps there's been some change in the meaning of the notation between the two equations. Could it be that the first represents the metric as function, g(__,__), a valence (0,2)-tensor, while the second represents the value of this fuction after some arbitrary vector has been operated on in both slots, g(v,v), the value being a scalar?

    Is it meaningful to ask whether this symbol should be read as (ds)2 or d(s2)?
     
  6. Aug 26, 2010 #5

    jcsd

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    Yep, plus 'spacetime distance' is a pretty abstract concept, why bother taking the root?





    the metric tensor field defines the scalar product in each tangent space.

    At a given event in spacetime, the metric tensor field will define a metric tensor that will give the scalar product in that tangent space i.e. a function g(_,_) that takes two tangent vectors as it's argument. g(ds,ds) = ds2

    ds is a tangent vector and ds2 is it's square norm. to find ds2 ypou are integrating the square norms of tangent vectors along some spacetime curve to find it's 'spacetime length'.
     
  7. Aug 26, 2010 #6

    DrGreg

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    There is a bit of abuse of notation. g is a (0,2)-tensor, [itex]g_{\mu\nu}[/itex] denotes the components of the tensor (or the tensor itself in abstract index notation), and the equation

    [tex](ds)^2 = g_{\mu\nu}\,dx^\mu\,dx^\nu[/tex]​

    is, strictly speaking, a definition of s in terms of g. Nevertheless, it is conventional to use that last equation, with the components written out explicitly, to specify the metric of a particular spacetime in a particular coordinate system. It's easier to write

    [tex]ds^2 = x^2\,dt^2 - dx^2 - dy^2 - dz^2[/tex]​

    for example, instead of writing the components of g as a 4×4 matrix.

    In my opinion it's misleading and wrong to write g = ds2.
     
    Last edited: Aug 26, 2010
  8. Aug 26, 2010 #7

    Fredrik

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    Suppose that [itex]x:U\rightarrow \mathbb R^n[/itex] is a coordinate system of a manifold M, and that [itex]C:[a,b]\rightarrow M[/itex] is a curve in M such that [itex]C([a,b])\subset U[/itex]. The function [itex]B=x\circ C:[a,b]\rightarrow \mathbb R^n[/itex] is a curve in [itex]\mathbb R^n[/itex]. I will write the tangent vector of C at C(t) as [tex]\dot C(t)[/tex]. Its components are

    [tex]\dot C^i(t)=\dot C(t)x^i=(x^i\circ C)'(t)=B^i'(t)[/tex]

    The differential of [itex]B^i[/itex] is defined as a real-valued function of two variables:

    [tex]dB^i(t,dt)=B^i'(t)dt[/tex]

    The metric g at C(t) acting on two copies of the tangent vector of the curve at that point:

    [tex]g_{C(t)}(\dot C(t),\dot C(t))=g_{ij}(C(t))B^i'(t)B^j'(t)[/tex]

    When [itex]M=\mathbb R^2[/itex] and the metric is Euclidean, the right-hand side is

    [tex]=\sum_i (B^i'(t))^2=\frac{1}{dt^2}\sum_i (dB^i(t,dt))^2[/tex]

    and it makes sense to write the sum as [itex]dx^2+dy^2[/itex]. So it seems that this would be an appropriate definition of [itex]ds^2[/itex]:

    [tex]ds^2(t,dt)=g_{C(t)}(\dot C(t),\dot C(t))dt^2[/tex]

    Edit: We could also define ds instead of ds2:

    [tex]ds(t,dt)=\sqrt{\pm g_{C(t)}(\dot C(t),\dot C(t))}dt[/tex]

    where the sign is chosen to make the square root positive, and for a given t, we have to use the same sign for all values of dt. (The function [itex]dt\mapsto ds(t,dt)[/itex] is only defined on an interval around 0 on which [tex]g_{C(t)}(\dot C(t),\dot C(t))[/tex] doesn't change sign or reach 0).
     
    Last edited: Aug 26, 2010
  9. Aug 26, 2010 #8

    Fredrik

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    Hm...

    [tex]dx^\mu\otimes dx^\nu(u,v)=u^\mu v^\nu\neq u^\nu v^\mu=dx^\nu\otimes dx^\mu(u,v)[/tex]

    Lee uses that symmetric product to get [itex]g=g_{ij}dx^i dx^j[/itex] to hold. I suspect that's the source the Wikipedia article used.
     
    Last edited: Aug 26, 2010
  10. Aug 26, 2010 #9

    Fredrik

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    I think it's more of a mnemonic for a definition than a full definition. The equation doesn't even make it clear that it's a statement about curves. The full definition is that the proper time (proper length) of a timelike (spacelike) curve [itex]C:[a,b]\rightarrow M[/itex] is

    [tex]\int_a^b\sqrt{|g_{C(t)}(\dot C(t),\dot C(t))|}dt[/tex]
     
    Last edited: Aug 26, 2010
  11. Aug 26, 2010 #10
    Thanks everyone for your answers, which I'm currently pondering...

    So

    [tex]\mathrm{d}s^2 = \mathrm{d}s \cdot \mathrm{d}s = g_{ij} \, \mathrm{d}x^i\otimes \mathrm{d}x^j((\mathrm{d}s)^k \, \partial_k,(\mathrm{d}s)^m \partial_m)[/tex]

    [tex]= g_{ij} \, \mathrm{d}x^i ((\mathrm{d}s)^k \, \partial_k) \otimes \mathrm{d}x^j((\mathrm{d}s)^m \partial_m)[/tex]

    [tex]= g_{ij} \, (\mathrm{d}s)^i \, (\mathrm{d}s)^j \enspace ?[/tex]

    I've begun reading this book by David Bachman about differential forms. On p. 18, he talks about dx and dy (in relation to R2) as 1-form fields that act as coordinate functions for tangent spaces:

    [tex]\mathrm{d}x^i(\textbf{v})=v^i[/tex]

    and I gather the set of 1-forms [itex]\left \{ \mathrm{d}x^i \right \}[/itex] of the cotangent space at each point makes a convenient basis for the cotangent space there, called a coordinate basis, which is dual to the coordinate basis [itex]\partial_i[/itex] of tangent vectors. I've also been reading about the idea of exterior derivatives, a kind of derivative (wrt position) of k-form fields, the value of this derivative being a (k+1)-form field. Presumably the 1-form coordinate basis [itex]\left \{ \mathrm{d}x^i \right \}[/itex] consists of exterior derivatives of the coordinate curves [itex]\left \{ x^i \right \}[/itex].

    From the notation then, I'd have expected ds to be the exterior derivative of a form field, perhaps a scalar field, s. But here you say it's a tangent vector. Is the notation just inconsistent with the exterior derivative notation, or is the ds that we input into g related to the exterior derivative of some function via the inverse of g, thus: g("ds","ds") = g(g-1(ds,__),g-1(ds,__))? I must be wrong about that though, as I guess the metric couldn't be defined circularly in terms of itself.

    So

    [tex]dx^2+dy^2 = \left ( \frac{\mathrm{d} x}{\mathrm{d} t} \frac{u}{u} \right )^2 + \left ( \frac{\mathrm{d} y}{\mathrm{d} t} \frac{v}{v} \right )^2[/tex]

    where u and v are real numbers, particular values of dt? Or does u = v? I guess it doesn't matter if we're dividing each by itself. But then, since these values seem to have been put into the equation only to be divided back out of it, wouldn't it be simpler to just write

    [tex]\left ( \frac{\mathrm{d} x}{\mathrm{d} t} \right )^2 + \left ( \frac{\mathrm{d} y}{\mathrm{d} t} \right )^2[/tex]

    or

    [tex]\dot{x}^2+\dot{y}^2[/tex]

    if that's all it means?
     
  12. Aug 26, 2010 #11

    atyy

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    This guy distinguishes uses roman and italic "d" symbols to distinguish between rigorous and non-rigourous notation! You may find his remark on roman "d" on p3 and his remark on notation on p29 interesting."Thus, we will write dx2 and dt dx when the rigorous but cumbersome notation dx⊗dx, 1/2 (dt ⊗ dx + dx ⊗ dt) does not yield any computational advantages. In those cases, we write “dx” with an italic “d” since we are not actually invoking the exterior differential d." http://homepages.physik.uni-muenchen.de/~Winitzki/T7/GR_course.pdf [Broken]
     
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  13. Aug 26, 2010 #12

    jcsd

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    Rasalhague, I think just step a back and ask yourself what a metric function actually is (though ignore the fact that a Lorentzian metric function only nods it's head the axioms of a metric function). How does a metric function relate to the norm? How does the norm relate to the scalar product? How does the scalar product relate to the metric tensor? How can you use the metric function on a tangent space to define a notion of length of a curve in a manifold?

    Also remember, slightly confusingly a metric function and a metric tensor are not the same. The metric tensor is called the metric tensor because it defines a metric function, ratehr than because it is one.

    I have a theory that the notation and the terminology of this subject is delibrately designed to confuse people! Mostly me!
     
  14. Aug 27, 2010 #13
    Thanks for the link, atyy. That looks really useful.

    jcsd, I guess "metric function" is synonymous with "distance function", an example being the integral Fredrik gave in #9. And metric function, or even just metric, would be a perfectly reasonable name for it, consistent with the name metric space, if people didn't have to talk, in the same context, about another function, a tensor field, also called "the metric".

    The similarity of the ds2 notation to the Pythagorean formula made me think of it, at first, as another way of writing delta s squared = delta t squared etc. I thought when people said "the metric" they sometimes meant this Pythagoras-like approximate distance function, and sometimes the metric tensor field. But reading that Wikipedia article in the light of Bachman's introduction to differential forms suggested that ds2 might actually be just another notation for the metric tensor field, and the link atyy posted makes this (refreshingly) explicit. That said, I wonder if there's an element of "notation hijacking" going on here, as described in this article promoting the idea of infinitesimals. Bachman rejects the infinitesimal interpretation, but these people seem to be saying their approach is not incompatible with the idea of differential forms (and Roger Penrose makes use of both ideas in The Road to Reality); it's a shame their section 4.3 "Differentials in higher dimensions" is so short.
     
    Last edited: Aug 27, 2010
  15. Aug 27, 2010 #14

    Fredrik

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    I can't even read this. It hurts my eyes too much to see a tangent vector called "ds". It's like calling an integer "7/13". :smile:

    I don't understand what you're doing here. I was just using the definition of a differential in the last step, because it's the only way to separate the dx from the dt in dx/dt.

    Yes, it would. But I wasn't just trying to calculate [tex]g_{C(t)}(\dot C(t),\dot C(t))[/tex]. I was trying to find a rigorous version of the non-rigorous [itex]ds^2=dx^2+dy^2[/itex]. The definition of "ds" that I suggested after those calculations seems to accomplish that, but it also seems pretty useless. It's probably better to do what DrGreg suggested in #6, which is to think of [itex]ds^2=dx^2+dy^2[/itex] as nothing more than a sloppy way to specify the components of the metric in a coordinate system. As I said in #9, you can also think of it as a mnemonic for the definitions of proper length/time of a curve.
     
  16. Aug 27, 2010 #15
    Heh, heh, that was the effect I was aiming at ;-) My apologies to your eyes!

    I was just trying to bring out what seemed to be a redundancy, or needless complication, in the notation/definition. Supposing that the variable dt takes on real values, the value of the differential dBi was defined, in terms of the derivative of Bi, as the derivative multiplied by a number. But to define ds2 in terms of dBi you had to then divide dBi by that number. I called the number u or v to make it look less mysterious.

    Okay, good. The link atyy posted takes a similar approach, calling [itex]ds[/itex] a "traditional (or jargon) notation for the metric" (i.e. for [itex]g[/itex]), rather than trying to justify it. It says it's going to "sometimes use the physicists' metric notation for brevity", writing [itex]\mathrm{d}t^2[/itex] for [itex]\mathrm{d}t \otimes \mathrm{d}t[/itex], but "but we will denote the metric by [itex]g[/itex] instead of the inconsistent [itex]ds^2[/itex]."

    Thanks for all your help!
     
  17. Sep 18, 2010 #16
    Here (quoted below) is a similar view to that expressed in the text atyy linked to, except that Sean Carroll allows another distinct meaning to "ds" besides "jargon notation for a metric tensor field". Sean Carroll says the "ds" notation is "perfectly consistent" with what he calls "the informal notion" of an infinitesimal displacement.

    Is it also perfectly consistent with the formal notion of an infinitesimal displacement (in nonstandard analysis), and what is the nature of this consistency?

    Is this doubling up of meanings in one notation, "ds2" = either "(square of) infinitesimal displacement" or "metric tensor field", analogous to the two- or threefold meaning given to a symbol such as [itex]T_{\alpha\beta}[/itex]--i.e. a tensor or tensor field (in abstract index notation), the set of components of a tensor or tensor field (in some basis), one unspecified individual component of a tensor or tensor field (in some basis). I mean, are the two definitions of "ds2" actually different things, just as these uses of indices all refer to different notions which it is important to distinguish between, because those differences can have consequences. Or is there some deeper sense in which a 1-form is the same thing as an infinitesimal displacement (in nonstandard analysis, or whatever system is used to give the rigor to justify the notion), and it's only a historical accident or an accident of the way our brains are wired that they've come to be conceptualised as different things?

    In other words, are the notations "ds2" = "g", and "dx" (infinitesimal) and "dx" (1-form) deceptively consistent (as the index notation might be said to be), or do they point to an underlying sameness behind the variety of names and ways of thinking about them.

    Apologies for the vagueness--and indeed repetitiveness--of this question! (And thanks to Fredrik and others who’ve done their best to enlighten me on similar matters in the past. One day I’ll get it...)

    http://preposterousuniverse.com/grnotes/

    Chapter 2, pp. 17-19:

    (Figure on p. 19 shows [itex]d \theta[/itex] and [itex]\sin(\theta) \, d\phi[/itex] as the two shorter sides of a right triangle whose longest side is labelled [itex]ds[/itex].)

    (Lecture 8, p. 33, of these other notes uses a similar approach. They call ds2 "the interval" and seem to conceptualise it as something distinct from, but related to, the metric tensor g.)
     
    Last edited: Sep 19, 2010
  18. Sep 28, 2010 #17
    The squared value does represent area in the Pythagorean theorem though, doesn't it? This is the idea behind the proofs of it that I've seen.

    If [itex]\phi=0[/itex] (i.e. the x and y axes coincide!), I suppose we could make the oblique equation meaningful by saying there's only one axis and zero displacement in the other (nonexistent) direction, so that it reduces to, for example, ds^2 = dx^2 --> ds = dx.

    Why is it important to avoid a square root in this expression, but not in Example 8 of the same section where the square root of the determinant of the metric tensor is used to calculate area?
     
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