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Grasping the Properties of Minkowski Space

  1. Jan 6, 2016 #1
    I'm trying to get an intuitive feel for Minkowski space in the context of Special Relativity. I should mention that I have not studied (but hope to) the mathematics of topology, manifolds, curved spaced etc., but I'm loosely familiar with some of the basic concepts.

    I understand that spacetime can be described using a 4-dimensional space with the metric signature (-,+,+,+), or equally (+,-,-,-). From what I have read Minkowski space is flat like Euclidean space (I believe the term is that that have the same topological structure?). Where then, lie the differences between Euclidean space and Minkowski space, apart from the metric?

    Is the geometry of Minkowski space non-Euclidean? I would think it is due to its different metric structure (different notion of distance). What is this geometry?

    The Lorentz transformations, i.e. rotations of one reference frame into another, involve the hyperbolic trigonometric functions. What does this mean for Minkowski space? Is it a hyperbolic space, or does it just have "some hyperbolic property"? What is this property exactly, that sets it apart from Euclidean space? I get the feeling here that circles are somehow replaced with hyperbolas in some Minkowskian context, but I can't put my finger on it.

    Do all these properties arise simply by having a Lorentzian metric?

    Some web searches brought up the following links. While they are helpful, they still leave me without a satisfactory grasp of the meat in Minkowski space.


  2. jcsd
  3. Jan 6, 2016 #2


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    Although it's possible to draw some connections between hyperbolic geometry and the geometry of Minkowski space, the connection is not that close. This is explained well in Michael Brown's comment on physics.SE. The math.SE answer IMO vastly exaggerates the connection.

    You don't need to know any of those things to understand the geometry of Minkowski space. Spacetime in SR has the trivial topology, and it's flat, so topology and curvature are basically irrelevant.

    Topology is not the same as curvature. An example of a space with a different topology would be Euclidean space with a hole taken out of it.

    The metric is the only fundamental difference.

    Yes, it's non-euclidean. The geometry of Minkowski space is the geometry of Minkowski space. It isn't equivalent to some other geometry.

    The Lorentz transformations can be expressed in terms of hyperbolic trig functions, but they don't have to be, and you seem to have been misled by the fact that they can. It doesn't indicate that the geometry is hyperbolic geometry or anything.

    Probably the best way to clear this up would be to read up on the Erlangen program: https://en.wikipedia.org/wiki/Erlangen_program . Euclidean geometry is the geometry that keeps a circle invariant when you rotate. The geometry of 1+1-dimensional Minkowski space is the geometry that keeps the light cone invariant under a boost.
  4. Jan 7, 2016 #3
    Thanks for your reply. It seems that I've misunderstood the nature of Minkowski space indeed. Regarding the Lorentz transformations - what other forms can they take? I'm only familiar with their representation as 4x4 square matrices that include elements which are hyperbolic trigonometric expressions. Are you saying that the appearance of these hyperbolic functions is just coincidental, in a sense?
  5. Jan 7, 2016 #4

    Mister T

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    Note that if ##\tanh{\theta}=v## as is the usual case, then ##\sinh{\theta}=\gamma v## and ##\cosh{\theta}=\gamma## where ##\gamma \equiv (1-v^2)^{-\frac{1}{2}}##.

    Make these substitutions in the matrix and you have the transformation in a form that contains no hyperbolic trig functions.
  6. Jan 7, 2016 #5


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    You can make an analogous substitution in a Euclidean rotation matrix in order to express everything as a slope instead of circular trig functions.

    However, in both cases, there is an angle measure which additive ... unlike the velocity or slope.
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