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Solve t²-x²

  1. Nov 7, 2003 #1

    At the beginning of my first thread I'd like to introduce myself

    I'm Alexander and I live in Bavaria .... because of this my English sometimes is really bad ... But I hope you all can understand me...
    And my hobbies are: Basketball, Latin (or in general: languages), Uncycling ...
    If somebody of you needs help with Latin - or German, of course - I will perhaps be able to help him/her :smile:

    But let's talk about my problem:

    I'm reading a book about the relativity theory (Sorry ... I don't know what's the correct English word - in German it's "Relativitätstheorie" ) . The author says that everything (i.e. the time and the space) is distorted only to keep difference t²-x² constant. But I don't understand why ...
    The book says that t = g-faktor and things like that ....
    I don't understand this.

    I would be very very very very very .... grateful for help and I hope that you have understood the sense of my question at least (because of my mistakes of course )

    Thank you and good night ...
  2. jcsd
  3. Nov 7, 2003 #2
    Re: t²-x²???

    Consider the emission of a spherical light wave as observed in an inertial reference frame (with coordinates {t,x,y,z}). At a time t after emission, the light front forms a sphere of radius ct, so that x2 + y2 + z2 = (ct)2, which can be rewritten as (choosing units in which c=1),

    t2 - x2 - y2 - z2 = 0

    Now, consider a different inertial frame, with coordinates {t',x',y',z'}, sharing the same origin as the original frame. According to the speed of light postulate, light will be emitted at speed c in this frame as well, so:

    t'2 - x'2 - y'2 - z'2 = 0

    Thus, we see that the quantity t2-x2-y2-z2 is a constant (equal to zero) in all frames, for the case that the point (t,x,y,z) can be reached from the origin by a light ray. (i.e., a "lightlike" or "null" separation).

    I have not proven that this quantity is the same ("invariant") for all separations (not just lightlike ones), however. That is a little more complicated. But you can find a very nice derivation of it in Schutz's book A First Course in General Relativity, if you know some matrix algebra.

    This quantity, the "spacetime interval", is quite interesting: if you write it with a sign difference as s2 = x2 + y2 + z2 - t2, it looks very much like the Pythagorean theorem; we interpret the quantity s as the "spacetime distance" between the origin and a point. It turns out that the physical theory of special relativity is completely equivalent to the geometry of Euclid in four dimensions, except with this slightly altered Pythagorean theorem. The minus sign in the equation is what makes time different from space in relativity.
  4. Nov 7, 2003 #3
    Thank you very much for your explanations ... :smile:

    Now I understand ... So I must imagine of a "ball", whose middle is the origin. And the term t²-x² is actual "accidental".

    Thank you for this clou ... but I think I begin with German books ... .... and I must learn the basics of maths and physics in the beginning, I think :frown: (I've been learning since 2 years physics at school, so we haven't even been tractating the terminus "energy" )

    By the way: What's matrix algebra? (I know ...Sometimes I'm asking dumb questions .... )
  5. Nov 7, 2003 #4
    I'm not sure what you mean by calling it "accidental" ...

    Here is another interesting fact about spacetime distance: in Euclidean geometry, you can define a "rotation" to be a (linear) transformation that keeps distances the same: if you rotate your coordinate system, the distances between points remain the same in the old and new coordinates.

    In relativity, a Lorentz transformation is a coordinate transformation between inertial reference frames in spacetime. It turns out that a Lorentz transformation is very similar to a rotation, except that it preserves the spacetime distance, not the space distance.

    A matrix represents a linear system of equations (or a linear transformation of vectors). For instance, the expressions

    5x + 6y
    2x + 4y

    can be written as a "matrix equation"

    [5 6] [x]
    [2 4] [y]


    [5 6]
    [2 4]

    is a "matrix", and


    is a "vector" (which is a kind of matrix).
  6. Nov 7, 2003 #5
    Your book sounds bad. Nothing is "distorted" in relativity theory. Everything preserves the unvarying value of lightspeed just as light would travel through space uninterrupted.

    I think it is a mistake for an elementary book to teach relativity theory without the letter 'c' for lightspeed. Physicists like it set to value '1' for computational purposes, but they already understand what it means. When 'c' is set to value 1 (light-seconds per second), then one is apt to forget that there is a speed and to assume space2 and time2 are being combined. The second term of the expression is really a space term, (ct)2, which is the square of the product of a speed and a time, and that yields a space term squared. My second objection is that complex computations with 'c' set to 1 leave the responsibility of reinserting the actual value wherever needed at the end of the computation. This requires much experience.

    Sind Sie Münchener?
    Last edited: Nov 7, 2003
  7. Nov 8, 2003 #6
    O.K. ... you mustn't listen to each words of mine becuase I don't know how to express all those things in English .... but I hope I will get better ....

    No ... When you open the Atlas, you will find the village I live in there, where the Danube crosses the Bavarian frontier (near Passau)

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