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Metric Tensor of a line element

  1. Jul 15, 2015 #1
    When we define line element of Minkowski space, we also define the metric tensor of the equation. What actually is the function of the tensor with the line element.
     
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  3. Jul 15, 2015 #2

    vanhees71

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    A tensor is a multi-linear form, i.e., a function that maps ##n## vectors ##\vec{v}_j##, ##j \in \{1,2,3,\ldots,n \}## to real (I assume a real vector space here as is the case for Minkowski space) numbers, which is linear in each argument, i.e., for any numbers ##\lambda_1## and ##\lambda_2## you have
    $$T(\lambda_1 \vec{x}+\lambda_2 \vec{y},\vec{x}_2,\ldots,\vec{x}_n)=\lambda_1 T(\vec{x},\vec{x}_2,\ldots,\vec{x}_n)+\lambda_2 T(\vec{y},\vec{x}_2,\ldots,\vec{x}_n)$$
    and analogous rules for all the other arguments. Such a multi-linear form is called a tensor of ##n^{\text{th}}## rank.

    In the case of Minkowski space you have a four-dimensional real vector space with a pseudometric, i.e., one has a nondegenerate symmetric bilinear form, i.e., a tensor of second rank. Usually it's written as ##x \cdot y##, where ##x## and ##y## are four-vectors (without an arrow; vectors with an arrow usually means the three spatial components in a given basis). The bilinear form is assumed to be symmetric, i.e., the so defined pseudoscalar product is commutative: ##x \cdot y=y \cdot x##.

    Given a basis ##e_{\mu}## you have ##x=x^{\mu} e_{\mu}## and ##y=y^{\mu} e_{\mu}##, where the Einstein summation convention is used, i.e., whenever there is a pair of indices (one subscript one superscript) one has to sum over it. The indices usually run from 0 to 3. Now obviously due to the linear properties oft the pseudometric tensor it is sufficient to know the numbers
    $$g_{\mu \nu} = e_{\mu} \cdot e_{\nu}$$
    because then you have
    $$x \dot y=x^{\mu} e_{\mu} \cdot y^{\nu} e_{\nu}=x^{\mu} y^{\mu} e_{\mu} \cdot e_{\nu}=x^{\mu} y^{\nu} g_{\mu \nu}.$$
    The bilinear form is called non-degenerate if the ##4 \times 4##-matrix ##(g_{\mu \nu})## is invertible. There's also a theorem telling that this matrix has always the same number of positive and negative eigenvalues, no matter how you choose the basis. Minkowski space is defined as a space, where the components of the pseudometric form a matrix that has one positive and three negative eigenvector (or the other way around, depending on which sign convention you use; I'm using the socalled "west-coast convention").

    Further one can show that there are always pseudoorthonormal bases, i.e., one can choose 4 vectors ##e_{\mu}## such that
    $$(g_{\mu \nu}=(\eta_{\mu \nu})=\mathrm{diag}(1,-1,-1,-1).$$

    Since ##g_{\mu \nu}## is invertible, it makes sense to define vector components with lower indices by
    $$x_{\mu}=g_{\mu \nu} x^{\nu}.$$
    Defining ##(g^{\mu \nu})=(g_{\mu \nu})^{-1}##, you get the inverse of this as
    $$x^{\mu}=g^{\mu \nu} x_{\nu}.$$
    Note that due to the symmetry of the pseudometric the matrices ##(g_{\mu \nu})## and ##g^{\mu \nu}## are symmetric.

    Now if you have a trajectory in Minkowski space, a world line, given by a parametrization ##x=x(\lambda)## or for the components with respect to some fixed basis ##x^{\mu} =x^{\mu}(\lambda)##, the derivative ##\mathrm{d}x/\mathrm{d}\lambda=\dot{x}## are tangent vectors along this world line, and you can define the invariant (scalar) quantity
    $$\mathrm{d} s^2=\mathrm{d} \lambda^2 \dot{x} \cdot \dot{x}=\mathrm{d} \lambda^2 g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}.$$

    If the world line describes the motion of a massive particle, then its tangent vectors are all time-like, i.e., ##\mathrm{d} s^2>0## at all points of the trajectory, and then it makes sense to define the proper time of the particle by
    $$c \tau=\int_{\lambda_0}^{\lambda} \mathrm{d} \lambda ' \sqrt{\dot{x} \cdot \dot{x}}.$$
    For a particle at rest in a given pseudoorthonormal coordinate system (Galilean basis), one has ##\vec{x}=\text{const}##, where ##\vec{x}## are the spatial components ##(x^1,x^2,x^3)## with respect to this Galilean basis, you have
    $$c \tau=\int_{\lambda_0}^{\lambda} \mathrm{d} \lambda' \dot{x}^0=x^0(\lambda)-x^0(\lambda_0),$$
    i.e., up to a conventional factor ##c## (speed of light in vacuum) the proper time of a massive particle is the time measured by a clock comoving with the particle.
     
  4. Jul 15, 2015 #3

    Nugatory

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    The expression for the squared length of an infinitesimal line element (##ds^2=...##) is the result of applying the metric tensor, so you can read off the components of the metric tensor in a given coordinate system from the line element.

    For example, we know that the line element in two-dimensional Euclidean space using Cartesian coordinates is ##ds^2=dx^2+dy^2##. We compare that with the general expression for the length of an infinitesimal line element in any two-dimensional space ##ds^2=g_{ij}x^ix^j=g_{00}dx_0^2+g_{01}dx_0dx_1+g_{10}dx_1dx_0+g_{11}dx_1^2## and we see that the components of the Euclidian-space metric tensor in these coordinates are ##g_{00}=g_{11}=1## and the rest are zero.

    Usually writing down the line element is the most natural and compact way of specifying the components of the metric tensor. This works everywhere, not just in Minkowski space, so you'll see people writing ##ds^2=....## a lot.

    And once you have the metric tensor, you know pretty much all there is to know about the space: curvature, geodesic equation, Christoffel symbols, Riemann, Ricci, Weyl tensors.... They're all defined in terms of the metric tensor.
     
    Last edited: Jul 15, 2015
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