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Using four-vectors in derivations and proofs

  1. Mar 17, 2005 #1

    JFo

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    I'm taking a first course in modern physics and were currently discussing special relativity. My professor keeps using four-vectors in derivations and proofs, and requires us to use them, but he hasn't developed the theory behind them; that is he's only showed us how to manipulate them. The text were using doesn't use them. Does anyone know of a book that gives a clear exposition for them?
     
  2. jcsd
  3. Mar 17, 2005 #2
    You say you already have an algebraic understanding, and that is most of what is there. It sounds like you are interested in geometry. I recommend the books on spacetime and special relativity by John Wheeler.
     
  4. Mar 17, 2005 #3

    JFo

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    hmmm,, Thanks Crosson, I'll have to check that out.
    I basically know the components and the "weird dot product" operation, But I don't understand why the first component is ct and not just t. I also don't understand the Lorentz interval (If I'm referring to the right thing) and what it represents physically. Also when taking the derivative to get the velocity or acceleration, as well as deriving the momentum seem fuzzy. I guess overall I don't see what they represent, and what they're used for. I guess an analogy would be like learning a definite integral, but not being shown that it is a limit of sums.
     
  5. Mar 17, 2005 #4

    pervect

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    Well, I'd say that the first thing to do is to understand the Lorentz interval, that's the most fundamental thing. Taylor & Wheeler's book should help a lot.

    But for a really quick explanation, the Lorentz interval of any light beam is zero. To see this, the distance a light beam moves is sqrt(x^2 + y^2 + z^2), by the pythagorean theorem, and the time it takes to move that distance is the distance /c.

    Writing t = sqrt(x^2+y^2+z^2)/c, we can manipulate it this through some algebra and say that

    (ct)^2 = x^2 + y^2 + z^2, thus x^2 + y^2 + z^2 - (c*t)^2 = 0 for any light beam.

    So, saying that the Lorentz interval is constant for any observer implies that the speed of light is constant for any observer, because the Lorentz interval of any light beam is always zero.

    It turns out that the Lorentz interval is constant for any observer in cases where the Lorentz interval is non-zero as well as in cases where it is zero. My view is that this is just something that you have to learn and accept, but perhaps the additional discussion in a textbook will help you to do that. Note that time and distance are not constant for all observers - only the Lorentz interval does that job, so that makes it very important.
     
  6. Mar 17, 2005 #5

    jtbell

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    Staff: Mentor

    There's a relativity forum here, by the way... just scroll down the forum listing a bit. I'll just tackle one of your questions here.

    It makes the units consistent among all four components of the time-position four-vector, and it makes the space-time interval ("distance") between two events invariant under a Lorentz transformation.
     
  7. Mar 17, 2005 #6

    JFo

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    Thank you all for your responses, that helps... I didn't see the special relativity forum but I've spotted it now,, thanks.
     
  8. Mar 18, 2005 #7

    JFo

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