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A Meaning of ds^2 according to Carroll

  1. Apr 19, 2016 #1
    Hi all, I need some help- I was reading Carroll's GR book, and on pages 71-71 he discusses the metric in curved spacetime. I have a few questions regarding this section:

    (1) He says
    In our discussion of path lengths in special relativity we (somewhat handwavingly) introduced the line element as [itex] ds^2=\eta_{\mu\nu}dx^{\mu}dx^{\nu}[/itex], which was used to get the length of the path. Of course now that we know that [itex]\text{d} x^{\mu} [/itex] is really a basis dual vector, it becomes natural to use the terms "metric" and "line element" interchangeably, and write [tex] ds^2=g_{\mu\nu}\text{d}x^{\mu}\text{d}x^{\nu}[/tex]
    My question here is: isn't what he's doing right now just as handwavy as before? It's very hard to see a concrete reason justifying what he's doing, unless he's defining [itex] ds^2 [/itex] by the above expression, which I'd have no problem with. But just come out and say it man!

    (2) Why does he talk about [itex] ds^2 [/itex] in terms of basis dual vectors, instead of just differentials? What's the point of doing this? I realize that this is not just Carroll's idea; that's the way it is in GR. But in treating [itex] ds^2 [/itex] as a (0,2) tensor, instead of the square of a differential length, don't we lose the physical meaning inherent in [itex] ds^2 [/itex]? How do we go back to interpreting [itex] ds^2 [/itex] as a length so quickly when we changed its meaning so drastically?

    Thanks in advance!
     
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  3. Apr 19, 2016 #2

    PeterDonis

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    It is in the sense that he isn't giving any more justification than he did before. But that doesn't mean there isn't more rigorous justification--it only means that he's not going to that level of rigor (which requires some fairly intensive differential geometry, more than is usually done in a physics course).

    Also, the expression using coordinate dual basis vectors is correct mathematically in a way that the expression using differentials is not. See below.

    Because the metric ##g_{\mu \nu}## is an (0, 2) tensor, and to get a scalar, ##ds^2##, from an (0, 2) tensor, you need to contract it with two vectors. See below.

    It isn't. ##ds^2## is a scalar. The (0, 2) tensor is ##g_{\mu \nu}##, as noted above. So what the equation is saying is that you take the (0, 2) metric tensor and contract it with two coordinate dual basis vectors to get a scalar, ##ds^2##. You can't do that with differentials; the idea of contracting an (0, 2) tensor with differentials makes no sense.

    To be more precise, you would put back the integral signs in the equation, to obtain:

    $$
    s^2 = \int_A^B ds^2 = \int_A^B g_{\mu \nu} \text{d} x^\mu \text{d} x^\nu
    $$

    In other words, to get the squared length ##s^2## along a curve between two given events, A and B, you integrate the contraction of the metric tensor, an (0, 2) tensor, with the coordinate dual basis vectors, along the curve between those two events.
     
  4. Apr 19, 2016 #3

    Orodruin

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    I find this confusing. By definition, (0,2) tensor should be contracted with two tangent vectors to give a scalar, not with two dual vectors. Of course, with ##g_{\mu\nu}## being the components of the metric, ##g_{\mu\nu} dx^\mu \otimes dx^\nu## is the metric tensor.

    To me, the meaning of the line element does not become clear until you consider the squared length of the tangent vector of a curve ##\gamma##, ie, ##\dot\gamma = (dx^\mu/dt)\partial_\mu## and obtain ##g(\dot\gamma,\dot\gamma) = g_{\mu\nu}\dot x^\mu \dot x^\nu##, where you are dealing with actual coordinate displacements ##\dot x^\mu##.
     
  5. Apr 19, 2016 #4
    I'm sorry, but that directly contradicts a lot of things that I'm pretty sure are true. Among them, and this one being directly from Carroll's book:
    In fact our notation "[itex] ds^2 [/itex]" does not refer to the differential of anything, or the square of anything; it's just conventional shorthand for the metric tensor, a multilinear map from two vectors to the real numbers. Thus, we have a set of equivalent expressions for the inner product of two vectors [itex] V^{\mu} [/itex] and [itex] W^{\nu} [/itex]: [tex] g_{\mu\nu}V^{\mu}W^{\nu}=g(V,W)=ds^2(V,W) [/tex]

    You also seem to have misunderstood the equation in the beginning of my post, as Orodruin pointed out. I think the reason for misunderstanding here is that [itex] g_{\mu\nu} [/itex] isn't the metric tensor, as you interpreted it; they're the components of the metric tensor, and the [itex] \text{d}x^{\mu} \text{d}x^{\nu} [/itex] next to it are basis dual vectors; not vectors. Thus resulting in a (0,2) tensor, [itex]g[/itex], or [itex]ds^2[/itex].

    Still, thank you for trying to help- I initially got mixed up with exactly the same thing as you.

    Orodruin, thank you too for your post, that helped a lot :)
     
  6. Apr 19, 2016 #5

    PeterDonis

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    Hm, yes, good point.

    Yes, this is a more precise way of expressing the tensor. Are you interpreting the RHS of Carroll's equation ##ds^2 = g_{\mu \nu} \text{d} x^\mu \text{d} x^\nu## as just a different way of writing this tensor (i.e., leaving out the ##\otimes## symbol for brevity)? That doesn't seem right either. Or are you interpreting Carroll's equation as a hand-wavy way of writing ##ds^2 = g_{\mu \nu} \dot{x}^\mu \dot{x}^\nu##?
     
  7. Apr 19, 2016 #6
    That is indeed what Carroll meant, I think. It's unfortunate, but apparently the tensor product is often omitted.
     
  8. Apr 19, 2016 #7

    PeterDonis

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    You're right that the RHS of Carroll's equation can be interpreted as a tensor, as Orodruin says. But ##ds^2## is a scalar, no matter how you interpret the rest of the equation. See my response to Orodruin.
     
  9. Apr 19, 2016 #8

    Orodruin

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    As mentioned by the OP, I think Carrol uses it as a "short-hand" (actually, "long-hand") for the metric.
     
  10. Apr 19, 2016 #9
    Wait, so this (unfortunate) notation isn't universal in GR speak?
     
  11. Apr 19, 2016 #10

    Orodruin

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    I would not say ##ds^2## is given by ##g_{\mu\nu}\dot x^\mu \dot x^\nu## but rather add a ##dt^2## to that, where ##dt## is a differential increment of the curve parameter. In this way, you will have a direct relation between the differential increment in the parameter and the corresponding distance on the manifold.
     
  12. Apr 19, 2016 #11
    Hmm what do you mean by add a [itex] dt^2 [/itex] to that?
     
  13. Apr 19, 2016 #12
    Never mind, I'm guessing you meant that [itex] ds^2 dt^2 [/itex], when acting on two tangent vectors, would give you the classic product of the metric tensor components with differential length elements.
     
  14. Apr 19, 2016 #13

    PeterDonis

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    Yes, you're right, there needs to be a differential on the RHS. So we would have this (I am using ##\lambda## as the parameter instead of ##t## to avoid confusion with the ##t## coordinate):

    $$
    ds^2 = g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} d\lambda^2
    $$
     
  15. Apr 19, 2016 #14

    George Jones

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    I believe that there is another possible source of confusion here with respect to omitted tensor product symbols and non-diagonal metrics. I think that ##dx^\mu dx^\nu## is notation for

    $$dx^\mu dx^\nu = \frac{1}{2} \left( dx^\mu \otimes dx^\nu + dx^\nu \otimes dx^\mu \right).$$

    Then, we still have

    $$\begin{align}
    g_{\mu \nu} dx^\mu dx^\nu &= g_{\mu \nu} \frac{1}{2} \left( dx^\mu \otimes dx^\nu + dx^\nu \otimes dx^\mu \right)\\
    &= \frac{1}{2} g_{\mu \nu} dx^\mu \otimes dx^\nu + \frac{1}{2} g_{\mu \nu} dx^\nu \otimes dx^\mu\\
    &= g_{\mu \nu} dx^\mu \otimes dx^\nu ,
    \end{align}$$

    where relabeling of indices and the symmetry of ##g_{\mu \nu}## has been used.

    Now, suppose that the metric written explicitly has a term like ##f dt d \phi##. Thinking that there is only a single omitted tensor product, it is tempting to write

    $$f dt d \phi = f dt \otimes d \phi,$$

    so that ##g_{t \phi} = f##.

    Actually,

    $$
    f dt d \phi = f \frac{1}{2} \left( dt \otimes d\phi + d\phi \otimes dt \right)\\,
    $$

    so that ##g_{t \phi} = f/2##.

    This pitfall does not appear in diagonal metrics like Schwarzschild like, since the non-zero terms have ##\mu = \nu##, and then

    $$ \frac{1}{2} \left( dx^\mu \otimes dx^\nu + dx^\nu \otimes dx^\mu \right) = dx^\mu \otimes dx^\nu .$$
     
  16. Apr 20, 2016 #15

    vanhees71

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    [corrected because of mistake found by PeterDonis in#17]

    This shows, how important it is to distinguish between tensors (INvariant objects) and its components with respect to a basis and the corresponding dual basis of the (tangent-)vector space (COvariant objects).

    A 2nd"=rank tensor ##g## is invariant and maps two elements of the the vector space to a scalar, and this is an invariant number. Now you introduce bases. The most simple ones are holnomous bases. Having generalized coordinates ##q^{\mu}## for the space-time manifold you take at any point the tangent vectors to the coordinate lines, i.e., those lines, where only one of the ##q^{\mu}## is varied. The (holonomous) basis vectors defined as the tangent vectors on these coordinate lines are denoted by ##\partial_{\mu}##, and dual-basis vectors of the dual space to this basis are denoted by ##\mathrm{d} q^{\mu}##.

    A 2nd-rank tensor (here the pseudometric tensor) can be decomposed as
    $$g=g_{\mu \nu} \mathrm{d} q^{\mu} \otimes \mathrm{d} q^{\nu}.$$
    The components are the mapping of the basis vectors ##\partial_{\mu}##, i.e.,
    $$g_{\mu \nu}=g(\partial_{\mu},\partial_{\nu}).$$
    Now it is also simple to get the various transformation properties. Take another map with coordiantes ##q^{\prime \mu}##. Then you have (the mnemotechnically very easy to remember) transformation properties for the holonomous bases and co-bases,
    $$\mathrm{d} q^{\prime \mu}=\frac{\partial q^{\prime \mu}}{\partial q^{\nu}} \mathrm{d} q^{\nu}, \quad \partial_{\mu}'=\frac{\partial q^{\nu}}{\partial q^{\prime \mu}} \partial_{\nu},$$
    i.e., the transformations are contragredient to each other. Now it's easy to see, how the components transform
    $$g_{\mu \nu}'=g(\partial_{\mu}',\partial_{\nu}')=g \left (\frac{\partial q^{\rho}}{\partial q^{\prime \mu}} \partial_{\rho} ,
    \frac{\partial q^{\sigma}}{\partial q^{\prime \nu}} \partial_{\sigma} \right)=
    \frac{\partial q^{\rho}}{\partial q^{\prime \mu}} \frac{\partial q^{\sigma}}{\partial q^{\prime \nu}} g(\partial_{\rho},\partial_{\sigma}) = \frac{\partial q^{\rho}}{\partial q^{\prime \mu}} \frac{\partial q^{\sigma}}{\partial q^{\prime \nu}} g_{\rho \sigma}.$$
    It's also easy to see that
    $$g=g_{\mu \nu} \mathrm{d} q^{\mu} \otimes \mathrm{d} q^{\nu} = g_{\mu \nu}' \mathrm{d} q^{\prime \mu} \mathrm{d} q^{\prime \nu}$$
    etc.

    To understand ##\mathrm{d} s^2## think about a curve parametrized with a parameter ##\lambda##. Then
    $$\left (\frac{\mathrm{d} s}{\mathrm{d} \lambda} \right)^2=\dot{s}^2 =g (\dot{q}^{\mu} \partial_{\mu},\dot{q}^{\nu} \partial_{\nu}) = g_{\mu \nu} \dot{q}^{\mu} \dot{q}^{\nu}$$
    is the pseudo-length (pseudo, because the metric is not positive definite in GR but of signature (1,3) or (3,1) depending on whether you use west- or east-coast convention). The dot means derivative with respect to ##\lambda##.
     
    Last edited: Apr 20, 2016
  17. Apr 20, 2016 #16

    lavinia

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    Infinitesimally a metric is an inner product on each tangent space. It takes pairs of vectors and computes a scalar. Since it acts on pairs of vectors at each tangent space it is a dual 2 tensor. Thus, with respect to any tangent frame field ##s_{i}## it can be expressed as ##g_{ij}s^{*}_{i}s^{*}_{j}## where ##s^{*}_{i}## is the dual basis. If the frame field is a coordinate frame then the dual basis is ##dx^{μ}##, the differentials of the coordinate functions. (However, if the frame field is not a coordinate field then the dual basis is not a set of differentials.)

    If the metric is positive definite then the integral of its square root along a curve gives the length of the curve. Because of this the square root is called the line element. In Relativity, the metric is not positive definite but ##ds^2## is still a sum of products of dual tangent vectors, ##g_{uv}dx^{u}dx^{v}##. This gives a rigorous definition of ##ds^2## in the case of a non-positive definite metric. Without seeing that the ##dx^{u}## are dual tangent vectors, one is only writing an expression by analogy with a positive definite metric. This is what Carroll meant by hand waving.

    In both the case of a positive definite and a semi-definite metric, one can compute its value in terms of infinitesimal increments in the coordinate functions.
     
    Last edited: Apr 20, 2016
  18. Apr 20, 2016 #17

    PeterDonis

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    This is backwards. The quantities ##\text{d} q^\mu## are covectors, and the quantities ##\partial_\mu## are tangent vectors. The placement of the indexes can be confusing, but consider: the quantity ##g## is a (0, 2) tensor; the expression ##g = g_{\mu \nu} \text{d} q^\mu \otimes \text{d} q^\nu## is saying that this (0, 2) tensor is a particular linear combination of tensor products of two (0, 1) tensors, i.e., covectors. If those quantities were vectors, that equation would make no sense; you can't make a (0, 2) tensor from linear combinations of (1, 0) tensors.

    As for the quantities ##\partial_\mu##, they are directional derivatives along the coordinate basis directions, and there is a one-to-one-correspondence between directional derivatives and tangent vectors, so they can also be considered as tangent vectors. The equation ##g_{\mu \nu} = g \left( \partial_\mu, \partial_\nu \right)## then just says that ##g_{\mu \nu}## is a scalar--a number--obtained by contracting the (0, 2) tensor ##g## with a pair of (1, 0) tensors, i.e., vectors. If those quantities were covectors, that equation would make no sense; you can't contract a (0, 2) tensor with two (0, 1) tensors.
     
  19. Apr 20, 2016 #18

    PeterDonis

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    This doesn't look right. ##ds^2## is supposed to be a scalar, so it can't be the sum of products of dual vectors; that would be an (0, 2) tensor. The expression ##g_{uv} dx^u dx^v## can be interpreted as a linear combination of products of coordinate differentials, with the coefficients being the metric coefficients; that would give a scalar. Unless you are using ##ds^2## just as an alternate (and confusing) notation for the tensor ##g##. But I don't think that was Carroll's intention; he appears to refer to ##ds^2## as a "line element", i.e., a scalar denoting the arc length along an infinitesimal segment of a path.
     
  20. Apr 20, 2016 #19

    vanhees71

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    Ok,I mixed this up again. Indeed the tangent vectors on a curve given by ##q^{\mu}(\lambda)## in coordinate space are
    $$t=\dot{q}^{\mu} \partial_{\mu}.$$
    Thus indeed, the ##\partial_{\mu}## are the tangent vectors on the coordinate lines and ##\mathrm{d} q^{\mu}## the corresponding co-vectors (or dual vectors), i.e., linear mappings defined by ##\mathrm{d} q^{\mu}(\partial_{\nu})={\delta^{\mu}}_{\nu}##.

    I'll correct my previous posting.
     
  21. Apr 20, 2016 #20

    lavinia

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    Yes it is correct. "products" means when evaluated on tangent vectors.
     
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