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Basis vectors of Minkowski space

  1. Oct 2, 2015 #1
    Hi, I'm doing a first course in GR and have just found out that

    [itex]\eta_{ab} = g(\vec{e}_{a}, \vec{e}_{b}) = \vec{e}_{a} \cdot \vec{e}_{b}[/itex]

    where g is a tensor, here taking the basis vectors of the space as arguments. I haven't seen this written explicitly anywhere but does this mean that

    [itex]\vec{e}_{0} = (i, 0, 0, 0)[/itex]?

    Isn't this strange? And does it have any particular significance or is it just an oddity? Apologies if this is asked before or just plain wrong.

    EDIT: I should say that I know this preserves the time component in the spacetime metric, but it still seems weird to think of imaginary basis components that describe a physical space. Or is it wrong to think of spacetime as anything other than purely mathematical?
     
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  3. Oct 2, 2015 #2

    Orodruin

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    The normal thing when specifying components is to use real components for all vectors. In addition, you cannot make any statement about the components of a vector without referring to a particular basis, ie, just giving a list of components does not mean anything without the knowledge about which basis they are written in.

    The expression you have written defines the inner product, i.e., we define ##e_0\cdot e_0 = g(e_0,e_0)##, not in terms of any components.
     
  4. Oct 3, 2015 #3
    Hi, according to Schultz, the expression defines the components of the tensor g( , ). But the Minkowski metric has

    [itex]\eta_{00} = -1[/itex]

    surely this means that

    [itex]\vec{e}_{0} = (i, 0, 0, 0)[/itex] ?
     
  5. Oct 3, 2015 #4

    Orodruin

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    No, the inner product is ##\vec v \cdot \vec w= v^iw^j g_{ij}##.
     
  6. Oct 3, 2015 #5
    hmm... I'm not following you. Schulz in "A first course in general relativity" says the following:

    The components in a frame O of a tensor of type (0 N) are the the values of the function when its arguments are the basis vectors [itex]\{\vec{e}_{a}\}[/itex] of the frame O.

    By function he means the function that defines the tensor object's mapping of vectors into the reals. In this case, the metric tensor g is a type (0 2) tensor and its components can be defined as follows:

    [itex]g(\vec{e}_{a}, \vec{e}_{b}) = \vec{e}_{a} \cdot \vec{e}_{b} = \eta_{ab}[/itex]

    η is then the array of components of the tensor g which we know as the metric matrix. However in order to generate this array, it seems to me that one of the basis vectors must contain an imaginary term. I will accept that this isn't true (because I'v not found this anywhere), but why not?
     
  7. Oct 3, 2015 #6

    vanhees71

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    Minkowski space is a real affine space. Only in older textbooks you find a convention, where one uses imaginary time arguments and writes the Minkowski product like a Euclidean product. Fortunately this very cumbersome idea is outdated, and one uses a manifestly real vector space. The Minkowski product is not positive definite but has the signature (3,1) (in the convention of your textbook, called the east-coast convention). One also talks about a pseudo-scalar product. For a pseudo-Cartesian basis ##(e_0,e_1,e_2,e_3)## you have thus for the representing matrix
    $$\eta_{\mu \nu} = e_{\mu} \cdot e_{\nu}=\mathrm{diag}(-1,1,1,1).$$
     
  8. Oct 3, 2015 #7

    DrGreg

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    sunrah, you need to understand that the dot symbol and the dot product notation are not being used the way you may be familiar with in 3D Euclidean space. It has been redefined so that[tex]
    \begin{pmatrix}A^0\\A^1\\A^2\\A^3 \end{pmatrix}
    \cdot
    \begin{pmatrix}B^0\\B^1\\B^2\\B^3 \end{pmatrix}
    = -A^0B^0 + A^1B^1 + A^2B^2 + A^3B^3
    [/tex]by definition. See Schutz 2nd edition p. 45.
     
  9. Oct 3, 2015 #8
    Thanks, I thought that in the explanation I quoted (p. 62, 1st edition) he was going to into more detail about how this new definition comes about, e.g. it has its origins in the tensor that defines the metric which introduces the minus sign, and that the components of this tensor are defined by the basis vectors of the space. But I'll accept it really is just a redefinition for now, I'm still very much at the beginning of this subject.
     
  10. Oct 3, 2015 #9
    thanks, it's nice to see that everything is infinitely more complicated than you think :-)
     
  11. Oct 3, 2015 #10

    vanhees71

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    No, it's much simpler than you think. You must just carefully learn linear algebra and quadratic forms for real vector spaces. Because you should first read a good intro into special relativity than to jump straight to general relativity. A very good introduction can be found in Landau-Lifshitz vol. II or the Feynman Lectures.
     
  12. Oct 4, 2015 #11
    So for the metric tensor g

    [itex]g(\vec{e}_{a},\vec{e}_{b}) = g_{\mu\nu}\tilde{w}^{\mu\nu}(\vec{e}_{a},\vec{e}_{b})[/itex]

    where wμν are the bases of g. It can be shown that

    [itex]g_{\mu\nu}\tilde{w}^{\mu\nu}(\vec{e}_{a},\vec{e}_{b}) = g_{\mu\nu}\tilde{d}x^{\mu}\otimes\tilde{d}x^{\nu} [/itex]

    So is gμν in fact not the metric but the array holding the coefficients of the metric? This contains the minus one in the g00 term. And is the outer product of bases what is in fact determined by the geometry of the space we are working, e.g. for flat space it would be a matrix made up of dt2, dx2, dxdy etc. terms. The metric is then the whole of the last expression and the spacetime interval is in fact the metric of spacetime and in 3D euclidean space, the gμν is the identity matrix?
     
    Last edited: Oct 4, 2015
  13. Oct 5, 2015 #12

    vanhees71

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    Yes, and it's not a metric in the mathematical sense but a pseudo metric, because its non-degenerate but not positive definite. Physicists nevertheless usually talk about "the metric" ;-).
     
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