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Minkowski Metric and the Sign of the Fourth Dimension

  1. Jul 24, 2012 #1
    Why is the unit vector for time in Minkowski space i.e. the fourth dimension unit vector always opposite in sign to the three other unit vectors?

    The standard signature for Minkowski spacetime is either (-,+,+,+) or (+,-,-,-).

    Is there some particular reason or advantage for making time opposite to the spatial dimensions?
  2. jcsd
  3. Jul 24, 2012 #2
    A mixed signature space has certain geometric properties different from ordinary, Euclidean space. One of them is the presence of directions that don't change regardless of "rotations" that mix up space and time. These "rotations" are Lorentz transformations, and these directions that are unchanged are the paths of light rays. Minkowski space is the only flat space with this property* being in line with what we know about relativity.

    *You can have a (0,+,+,+) space where the timelike vector is invariant, but you can show this corresponds to Galilean invariance, which we know not to be present in the physical world.
  4. Jul 24, 2012 #3
    Right okay that is a grand answer to my question, thank you. I am familiar with the Poincare group that contains the isometries of Minkowski spacetime. Essentially the answer you gave me was, "The negative time unit vector is required to fulfill the requirements of the 10-dimensional Poincare group, consisting of a translation through time, a transition through any 3 directions of space, a rotation around any of the three spatial axes, and a boost in any of the three directions."

    If the time unit vector was not negative the the group of isometries of the spacetime of metric (+,+,+,+) would only be 7-dimensional, lacking in the rotation around any of the spatial axes.

    Is this correct?
  5. Jul 24, 2012 #4

    Ben Niehoff

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    No, the group of isometries in (++++) signature is still ten-dimensional (6 rotations in the 6 orthogonal planes, and 4 translations).

    The reason spacetime has the Minkowski signature (-+++) is because of the experimental fact that light has the same speed in every reference frame. Lorentz transformations can rotate among the 3 spatial directions, or they can mix time and space (via 3 boosts), but they always leave the lightcone intact.
  6. Jul 24, 2012 #5
    Hm, I don't think so. Even a 4d Euclidean space would have 4 translational and 6 rotational degrees of freedom. The point I was trying to make was that the Minkowski signature is chosen because, in general terms, out of the three possibilities for a flat space--Euclidean, Galilean, and Minkowskian--only the latter has the correct notion for the principle of relativity. The others can be ruled out experimentally.

    If you're asking from a pure math standpoint, then I'm not sure I understand the question.
  7. Jul 24, 2012 #6
    Okay so there must be a mathematical way of showing that for a (+,+,+,+) signature the speed of light is not the same in every reference frame, and that for a (-,+,+,+) or it's equivalent the speed of light has a constant speed in every frame?

    And if that can be shown, then it could then be shown that the constant that the speed of light travels at is equivalent to the speed a massless particle travels?
  8. Jul 24, 2012 #7
    The key is to show that, under generalized rotations, a Euclidean space has no invariant vectors in the plane of the rotation, whereas a Minkowski space does.
  9. Jul 24, 2012 #8
    Okay so is what you're saying that the definition of a vector is that it is invariant under translation and rotation, and that in Euclidean space, vectors are not invariant under rotation and that in Minkowski space they are?
  10. Jul 24, 2012 #9
    Definition of a vector, no. Obviously not every vector is invariant under rotations. But for there to be lightlike paths, there must be some vectors that are invariant under rotations, even when that vector is in the plane of rotation.
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