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Minkowski's 4 dimensional world

  1. Jul 5, 2006 #1

    i'm reading Einstein's popular book, On the Special and General Relativity,
    i'm on Appendix two, Minkowski's Four-Dimensional Space ("World")


    however, i ran into such problem here. Einstein says

    "We can characterise the Lorentz transformation still more simply if we introduce the imaginary square root of -1 times ct in place of t, as time-variable."

    i know that the goal is to square this imaginary square root of -1 times ct in order to get -c^2t^2 which is in the previous simple derivation of Lorentz Transformation:
    x'^2 + y'^2 + z'^2 - c2^t'^2 = x^2 + y^2 + z^2 - c^2t^2

    but this the imaginary square root of -1 times ct is the distance travelled by light after time t, not time t (time travelled) itself, isn't it? if anyone knows this please tell me.
  2. jcsd
  3. Jul 5, 2006 #2


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    Welcome to these Forums therapeuter!!

    A good question to start a disucssion. :smile:

    Einstein notwithstanding, MTW do a good article on this subject: "Farewell to ict" Gravitation Box 2.1, page 51.

    The problems with it are:

    1. Vectors (contravariant) and one-forms (covariant) are confused.

    2. Thus it hides the character of the geometric object being dealt with.

    3. The essentially very different rotations in Minkowski and Euclidean space are confused.

    4. Thus it hides the nature of the parameter in transformations; is it cyclical or does it asymptotically tend to infinity?

    5. It hides the completely different metric structure of ++++ and -+++ geometry.

    In ++++ Euclidean geometry a zero interval between two events implies they are the same event, in -+++ Minkowskian geometry it implies they both lie on a null geodesic, one event may be a SN Ia explosion at the far side of the universe and the other the observation of that explosion on Earth billions of years later.

    6. The causal structure of the universe , limited to all events in the past light cone of a particular event Xa that influence Xa, is broken.

    7. Finally, and as a consequence of the above, no-one has been able to discover a way to make an imaginary ict coordinate work in the general curved space-time manifold.

    Here, I am afraid, we have to part company with Einstein. :frown:

    Note also: Stephen Hawking re-introduced ict in his 'theory of everything' to explain what happens to time 'at the north pole', (the BB - "the only boundary condition is there is no boundary condition".) Perhaps that is why he couldn't get that to work either. :rolleyes:

    Last edited: Jul 5, 2006
  4. Jul 5, 2006 #3
    LOL Garth. Excellent.
  5. Jul 5, 2006 #4


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    In spite of all the problems with it that Garth noted and the universal practice of modern physicists, it is worth noting that every pre-WWII physicist and textbook that I've ever seen used ict. It would be an interesting historical study to find out the locus of change.
  6. Jul 5, 2006 #5
    What do you mean when you say "here we have to part company with einstein"?
  7. Jul 5, 2006 #6


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    Isn't it clear? Einstein's use of ict, as found in his book, does not have the theoretical strength to support Einstein's own curved spacetime in General Relativity consistently, but the modern way of expressing Minkowski geometry does.
  8. Jul 7, 2006 #7
    hi Garth, thank you for such detailed explanation....

    however, that's not the question i asked....

    i was asking, since the - c2^t'^2 in the simple derivation of Lorentz trans. is the minus of the square of the distance travelled by light (r = ct),


    then the ict here is the distance travelled by light, right? then why is it that einstein said ict is a substitute for time coordinate?

    and if it's distance travelled by light, then how can it make the 4 dimensional continuum analogous with euclidean? for the analogous euclidean equation would simply be:

    x^2 + y^2 = z^2,
    so x^2 + y^2 - z^2 = 0

  9. Jul 7, 2006 #8
    The Euclidean way to measure distances between two points is to square the differences in all the components, add them, and then square root it. So:

    [tex]ds^2 = dx^2+dy^2+dz^2.[/tex]

    We can happily extend this to more dimensions. Say we go for four:

    [tex]ds^2 = dx^2+dy^2+dz^2+d\alpha ^2.[/tex]

    Now, as it turns out, in Special Relativity, the only consistent way to measure separations in space time between events is by the following rule

    [tex]ds^2=dx^2 + dy^2+dz^2-c^2dt^2,[/tex]

    where all the letters have their usual meanings and units.

    So Einstein said, if we make our fourth Euclidean co-ordinate (here written as [itex]\alpha[/itex]) equal to [itex]ict[/itex], we'll get (measuring things the Euclidean way):

    ds^2 &= dx^2+dy^2+dz^2+d(ict)^2 \\
    &= dx^2 + dy^2+dz^2-c^2dt^2,

    which gives us the correct way of measuring things.

    The motivation behind doing that was so that we could stick to the usual Euclidean distance measuring system (i.e. the Euclidean metric), and not have to change to the new Minkowskian way (i.e. the Minkowski metric). But as it turns out (as Garth has pointed out) the Minkowskian way is actually better for many reasons, and we just have to get used to the fact that the Euclidean metric is not the real metric of SR.

    EDIT: it appears I've ignored a large amount of the OP's original post. So I'll try and answer his/her question below...

    So the [itex]t[/itex] that appears in the [itex]ict[/itex] is simply the separation in time between two spacetime events. The factor [itex]i[/itex] appears to make the metric Euclidean, and the factor [itex]c[/itex] appears as a conversion factor.

    It is really nothing to do with the distance that light (i.e. EM waves) travels in that time t. It is more to do with the fact that nothing (not just light) could travel faster than c, and all massless things travel at c.
    Last edited: Jul 7, 2006
  10. Jul 8, 2006 #9


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    I and others have explained why ict is not the substitute for the time coordinate, the answer to your OP question is simply no:

    ict is not the distance travelled by light - that distance is given simply by ct.

  11. Jul 8, 2006 #10
    Last edited: Jul 9, 2006
  12. Jul 9, 2006 #11


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    That is so very true. The metric tensor of Euclidean space is
    [tex]\left ( \begin {array} {cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array} \right )[/tex]
    and the metric tensor for Minkoski spacetime is
    [tex]\left ( \begin {array} {cccc} -1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array} \right )[/tex]
    (or the first element can be 1 and the others -1).
    Last edited: Jul 9, 2006
  13. Jul 9, 2006 #12
    Just to add to sA's point, I hope it's obvious that the first component would be [itex]ct[/itex] and the rest are [itex]x,y,z.[/itex]
  14. Jul 10, 2006 #13
    but einstein's statement that lorentz transformation corresponds to a rotation of the coordinates in 4 dimensional space time is still valid right?

    (i also guess that means that motion results in the rotation of your coordinate system: is it right?)

    You have to go through the modern linear algebra/differential geometry version of rotation but yes. The Poincare Group is SO(1,3), the orthogonal transformations with unit determinant. Turns out the entries in the matrices can be either hyperbolic functions like sinh and cosh, for boosts, or circular functions like sin and cos for rotations. In Euclidean space you would have its twin SO(4) the four dimensional rotations, with all circular functions. What you lose is Einstein's pretty, but specious derivation of just replacing t with it in the Euler definitions to turn hyperbolic functions into circular functions and Lorentz boosts into rotations.
    Last edited by a moderator: Jul 10, 2006
  15. Jul 10, 2006 #14
    It should be noted here that a space is neither Euclidean nor pseudo-Riemanian until a metric is defined. A metric is something that defines the inner product of two vectors. In SR the metric is pseudo-Riemanian, in particular it is Minkowskian.

    What "pseudo-Riemanian/Riemanian" means is a whole different apple.:biggrin:

  16. Jul 11, 2006 #15
    this seems to mean that euclidean space is a special case of minkowskian space... is it right? (like circle is a special case of ellipsis, when the eccentricity is zero)
  17. Jul 11, 2006 #16
    No that is not correct.

    In Euclidean 4 dimensional space the distance between two points is defined as:

    [tex]ds^2 = dx^2+dy^2+dz^2+d\alpha ^2[/tex]

    In 4 dimensional Minkowski space it is defined as:

    [tex]ds^2=dx^2 + dy^2+dz^2-c^2dt^2[/tex]

    Or (and I prefer this):

    [tex]ds^2=c^2dt^2 - dx^2 - dy^2 - dz^2[/tex]

    The superset of Euclidean and Minkowski space is a 4 dimensional complex space.
    Last edited: Jul 11, 2006
  18. Jul 13, 2006 #17
    Thanks mejen, what i still don't understand is, what is the difference between

    [tex]ds^2=dx^2 + dy^2+dz^2-c^2dt^2[/tex]


    [tex]ds^2=c^2dt^2 - dx^2 - dy^2 - dz^2[/tex]

    i mean, if the distance comes up 0.75 using the first equation, it will be -0.75 with the second. is that of no consequence? i mean, like, measuring something forward or backward, no matter?

    but then, as i was reading this:


    "Whenever the displacement in time (cdt) is greater than the displacement in space, the square of the interval is positive. Consider a real body. In the rest frame of the body there is no space displacement, there is only time displacement. The square of the interval is positive." but this is if you use the second equation. using the first you'd have to say, "Whenever the displacement in time (c dt) is greater than the displacement in space, the square of the interval is negative." does it not matter?
  19. Jul 13, 2006 #18


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    The difference is called the "signature convention".

    The first is often referred to as +++-, with the "time coordinate" x4.

    The second is +---, with the "time coordinate" x0.

    A third convention that is used (seen, e.g., in selfAdjoint's post above) is -+++, with the "time coordinate" x0.

    The choice is often one of convenience.

    In SR/GR, it makes no physical difference which you use... as long as you are consistent in the application of your definitions and conventions. [It is sometimes a pain to compare the results from different conventions.]

    (Similarly, geometric optics has various sign conventions. http://en.wikipedia.org/wiki/Sign_convention)

    [Apparently, there is at least one exception to the above:
    "Where the sign of the metric makes a difference."
    Steven Carlip and C├ęcile DeWitt-Morette
    ...but this is beyond the scope of this discussion.]
  20. Jul 14, 2006 #19
    excellent clarification, robphy. infinite thanx.

    as for mejen, where are you from?
  21. Jul 23, 2006 #20
    hi i want to see if i got this right.

    i read that the interval between two events connected by light is zero, since the displacement in space, dx, is equal to the time displacement, cdt (x^2 + y^2 + z^2 - c^2t^2 = 0). does this not only explain why time stops for the object that reaches velocity c, but also mean that nothing can go faster than c because spacetime distance can't be negative?
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