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Minkowski Space Metric

  1. Oct 14, 2014 #1
    I've never seen a satisfactory explanation of the metrics used in a calculation of distance in Minkowski space. In Euclidean space, the distance is:
    ds^2 = dx^2 + dy^2 + dz^2
    But in Minkowski space, the distance is
    ds^2 = (dt * c)^2 - dx^2 - dy^2 - dz^2
    Why are the signs reversed? This implies that space (or time depending on your convention) is imaginary.
     
  2. jcsd
  3. Oct 14, 2014 #2

    Dale

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    That is one way to look at it, but it faded into disuse quite some time ago. Now, the usual approach is not to consider the time coordinate to be imaginary, but to consider the minus sign to be in the metric. So (in units where c=1):

    ##ds^2 = g_{\mu\nu} dx^{\mu} dx^{\nu} = -dt^2 + dx^2 + dy^2 + dz^2##

    This can, as you suggested, be achieved by ##dx = (i~dt,dx,dy,dz)## and

    ##g = \left(
    \begin{array}{cccc}
    1 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1 \\
    \end{array}
    \right)##

    But it can also be achieved by ##dx = (dt,dx,dy,dz)## and

    ##g = \left(
    \begin{array}{cccc}
    -1 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1 \\
    \end{array}
    \right)##

    The usual modern approach is the latter
     
  4. Oct 14, 2014 #3

    Fredrik

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    The ##dx^\mu## notation is from differential geometry. In the context of SR, we can talk about matrices instead. The Euclidean inner product (i.e. the dot product) on the space of 4×1 matrices is given by ##\langle x,y\rangle=x^Ty##. If you insist on using this formula in SR, you have to make some components of x and y imaginary. A nicer way is to modify the definition to ##\langle x,y\rangle =x^Tg y##, where g is defined in DaleSpam's post.
     
  5. Oct 14, 2014 #4

    Nugatory

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    The different sign on the ##t## coordinate means that the Minkowski metric describes a space-time in which the distance between points on the line corresponding to the path of a light beam is zero. Experiments confirm that this model accurately describes the universe that we live in, so that's the model that we use. Thus, your "Why?" question comes down to "Why is the universe built this way and not some other way?" - and science isn't going to give you a satisfactory answer to that question.

    As DaleSpam points out above, the modern style of moving the sign difference into the metric tensor reduces the embarrassing appearance of "imaginary" (better to say "complex" instead) numbers in the formulas. The older style, in which sooner or later you find yourself treating ##ict## (with ##i=\sqrt{-1}##) as a coordinate, was used mostly because it made the Lorentz transformations look like the already familiar problem of rotating the coordinate axes in Euclidean space. That helped people who were familiar with the mathematical underpinnings of classical mechanics make the jump to special relativity (it's worth noting that Goldstein introduces relativistic mechanics this way) but it's something you'll have to unlearn when you move on to general relativity.
     
  6. Oct 14, 2014 #5

    Chalnoth

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    Also, whether to put the minus sign on the time coordinate or the spatial coordinates is a matter of convention, and both conventions are in wide use.
     
  7. Oct 14, 2014 #6

    ghwellsjr

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    Since you have the option of using either convention, when you are actually doing a calculation for the distance between two events, if it comes out imaginary in one convention, you can switch to the other convention. Then you can think of the distance as being either an actual spatial distance or an actual time interval, depending on which convention evaluates to a positive value by taking its squareroot. In the first case, it can be measured with a ruler at rest in an Inertial Reference Frame where the two events occur at the same time or in the second case, it can be measured with an inertial clock that is present at both events. Another commonly used term for this distance is the Spacetime Interval, which in the first case is called "Spacelike" and in the second case is called "Timelike". If the Spacetime Interval evaluates to zero, that means that it cannot be measured either with a ruler or with a clock and it is called "Null".
     
  8. Oct 14, 2014 #7

    Chalnoth

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    I think we can do a bit better than that.

    For example, one way to understand why we formulate the dimensions in this way, with the time coordinate taking the opposite sign of the spatial coordinates, is that when we write down distances in this way, the shortest path something will take between two events is the path it actually takes. This isn't terribly useful in just Minkowski space, as this just means that things travel in straight lines. But in General Relativity this can be used to compute the paths of orbits, or of light rays being deflected by a gravitational field, or anything else you might care to estimate the path of. Everything takes the shortest distance when you use the metric, which has different signs between the spatial and time components, as the measure of distance. With the added constraints that light always takes a path which has space-time distance equal to zero and objects with mass cannot travel space-like distances (with the metric convention RCopernicus used, s^2 must be greater than zero).
     
  9. Oct 14, 2014 #8

    pervect

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    The name "distance" may be confusing you. In special relativity, distance is observer dependent, due to Lorentz contraction. What is independent of the observer, and hence an invariant, is the Lorentz interval.

    We sometimes, especially in analogies, refer to space-like Lorentz intervals as distances, or call them "proper distances". But the Lorentz interval is still a separate concept, it's distinguishing feature is that it's the same for all observers, and the formula (with it's minus sign) calculates this quantity that is the same for all observers. Without the minus sign, this quantity we calculate would not be the same for all observers, and hence would not be as of much interest.

    We'll get back to the similarities of the Lorentz interval with Euclidean distance later, but for now it's important to recognize that they are different ghings, before we point out their underlying similarity.

    Note that the Lorentz interval being equal to zero is equivalent to a lightlike separation between a pair of points, and vica versa. So, the Lorentz interval being zero is equivalent to saying that the geometry of space-time is such that a light like separation between points is independent of the observer. If this sounds like it's on the right track, good! If it seems a bit vague, read on.

    The more formal justification of the Lorentz interval follows from the Lorentz transform itself. You can verify mathematically that a consequence of the Lorentz transform is that it leaves the Lorentz interval unchanged. The Lorentz transformations don't leave distances unchanged, nor do they leave times unchanged. The only scalar quantity that the Lorentz transforms leave unchanged is the Lorentz interval.

    There are several ways of motivating the Lorentz transforms, you can use Einstein's original approach, or my favorite, the k-calculus approach due to Bondi. But the point is that after you start out with the axioms of relativity, that the speed of light is the same for all obserervers, plus whatever auxillary assumptions your particular approach to special realtivity needs (isotropy is a common one). At the end, you wind up with the Lorentz transform. I can't really get more specific than that in a short post, I will just suggest that if you don't understand how the Lorentz transformations came about, and my explanation is too brief, that there is a lot of literature out there you can read to fill in the gaps. After you've derived this transform, you notice an interesting property it has - it leaves this quantity that we call the Lorentz interval unchanged.

    If you compare this to Euclidean geometry, the invariance of the Lorentz interval under Lorentz transformations is similar to the invariance of distance under rotations in Euclidean space. So the Lorentz interval is a bit like the concept that distance used to be in Euclidean space, beuase it's independent of the observer.

    So you have this useful analogy, between the transforms induced by changes in velocity (called Lorentz boosts), and rotations in standard Euclidean space. They both leave something underlying unchanged. In the case of Euclidean space, this important thing that is unchanged by rotation is Euclidean distance. In the case of Minkowskii space, this important thing that is unchanged by a boost is the Lorentz interval.
     
  10. Oct 14, 2014 #9

    Dale

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    Yes, both are in wide use. My personal preference is to use ##ds^2## for the (-+++) convention and ##d\tau^2## for the (+---) convention.
     
  11. Oct 14, 2014 #10

    ghwellsjr

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    NOTE: the Lorentz interval that pervect was talking about in his post is the same as the Spacetime Interval that I was talking about in my post.

    And remember, between any pair of arbitrary events, it is either a pure spatial distance, or a pure time interval, or neither, which is why it is called "null". Only a flash of light can be present at both events in the null case and for that reason it is also called "lightlike".
     
  12. Oct 15, 2014 #11
    As others already explained, it's just a matter of convention. The way it was written the first time (I think) is probably easier to understand:

    "[..] the invariants of the Lorentz group.
    We know that the substitutions of this group [..] are linear substitutions which do not affect the quadratic form
    x2 + y2 + z2 - t2.
    - https://en.wikisource.org/wiki/Translation:On_the_Dynamics_of_the_Electron_(July)#.C2.A7_9._.E2.80.94_Hypotheses_on_gravitation
     
  13. Oct 15, 2014 #12
    Hello Dale Spam,

    As you have written, I would like to know how the values of (-1,0,0,0) come from? Are they parameters?
     
  14. Oct 15, 2014 #13
    Hello Nugatory,

    Thank you for this wonderful, lucid answer. I would clarify (-ct), is because that t time is considered imaginary, hence i=root(sqrt-1)). Is that so?
     
  15. Oct 15, 2014 #14
    This is off topic, I'm swimming in deep waters here and maybe this is obvious, but as a note I thought relativity was the embodiment of having a universal speed limit. And that the problem with "why this way" starts out of that observation (e.g. "why locality and hence causality").
     
  16. Oct 15, 2014 #15

    Nugatory

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    The essential basis of relativity is not the universal speed limit, it is the invariance of the speed of light; the universal speed limit (and much else) follows from light-speed invariance.

    But with said...
    You've just moved the "Why?" question around. Why do we live in a universe that has a universal speed limit instead of one that does not? There's a fine and consistent mathematical model for describing a universe in which the speed of light is not invariant and there is no universal speed limit; it's called classical mechanics and there's nothing wrong with it except that observation tells us that it's not the way the universe works.
     
  17. Oct 15, 2014 #16
    I'm afraid your answer doesn't make much sense. I can claim that ds^2 = dx^2 - dy^2 describes a shorter path than ds^2 = dx^2 + dy^2, but I have no justification for arbitrarily flipping the sign on one of my dimensions. So why is Minkowski able to get away with it with time? Is not time orthogonal in every way to space?
     
  18. Oct 15, 2014 #17

    Dale

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    The flipped sign is what sets up the causal structure of spacetime. It separates spacetime and four-vectors into timelike spacelike and lightlike regions.

    Clocks measure timelike intervals and rods measure spacelike intervals. If you gave them both the same sign then you would have a theory where you could measure time with rods and simply turn towards the past as easily as turning left.
     
  19. Oct 15, 2014 #18

    Dale

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    They come directly from the line element:
    ##ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}=-dt^2+dx^2+dy^2+dz^2##

    There are no cross terms, so all of the off diagonal entries are 0. Then the diagonal entries are the corresponding coefficients: (-1,1,1,1)
     
    Last edited: Oct 16, 2014
  20. Oct 15, 2014 #19

    pervect

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    Respoinding to Chalnoth:

    Interesting,why single it out for special attention then?

    At the risk of being repetitive, I'll summarize my original longer post, which I'm concerned may have just gotten lost in the mass of replies.

    While one can pretend that the sign flip in the expression for ds^2 did just "fall out of the air", and explore the consequence of saying "this magic quantity, which we won't tell you where it came from, is the same for all inertial observers" in detail and comparing the predictions made by this claim, I think it is more in the spirit of the original question to ask, historically, where did this quantity come from.

    The process is a bit long, as I outlined in a longer post - one starts with special relativity and derives the Lorentz transform, then one looks at special quantities that are unchanged by the Lorentz transform and singles them out for special interest, eventually winding up with a geometrical interpretation of the special quantities that have this special property, which is called invariance.

    So the quantity in question has a bit of a history, and if one wants to know where it came from, one needs to study the history.

    If the real underlying question is "why study relativity at all", it might be good to remember that the end goal of science is to make predictions that agree with experiment, and focus on the experimental results. Starting out with preconceived notions of "how Nature ought to work" winds up in frustration at best, and at worst ends up with one sticking to the (incorrect) preconceived notions because one is happier with them than one is with the sometimes messy results that are actually measured.
     
  21. Oct 15, 2014 #20

    DrGreg

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