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Lorentz basis

  1. Feb 18, 2009 #1
    I found an interesting statement in the following exercise 3.2 a), but I'm confused:


    Why is there the restriction to [tex] Y_{0}^{0} > 1 [/tex] and [tex] det( Y_{\mu}^{\lambda} ) = 1 [/tex] in the definition of a Lorentz basis? In the literature you can also find the definition withouth the restriction to proper Lorentz-Transformations. So my question is if it is always possible to construct a basis with this restrictions and why?

    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Feb 18, 2009 #2
    I think it's based on the following: The set of proper Lorentz transformations have the property [itex]\textrm{det}(\Lambda_\mu^\nu) = 1[/itex] and [itex]\Lambda_0^0[/itex]. When acting on the basis states [itex]Y_\mu[/itex] we end up with new basis states. The proper Lorentztransformations leaves the restrictions you mention intact. So if we start out with a Lorentz basis, and we only consider proper Lorentz transformations, then we will always deal with bases that have this property.

    Also, I think that with such a choice of bases we also have a "natural" time direction and handedness of the coordinate system. Proper Lorentztransformations leave these properties intact.
  4. Feb 23, 2009 #3
    I've a question about Relativistic Kinetic Energy. I understand the Proof but I see an infinite solution on the hyperphysics website when I put in the speed of light as the velocity v.

    To reproduce

    Enter mass m=1, 10^1
    Enter velocity 2.998 10^8 which is C Metres Per Second and in the Javascript as C

    I get K.E.(relativistic) Infinite solution.

    Shouldn't the solution be E=MC^2? or 10 * (2.998 10^8)^2 where V=C?

    Is it a bug in the program or a bug in my reasoning?
  5. Feb 26, 2009 #4
    Right so, I get it now. We're not really dealing with the Relative Kinetic Energy of the moving object. We dealing with the work done on it. Thanks for your lack of response.
  6. Feb 26, 2009 #5


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    You probably would have gotten a response
    if you started your own thread and
    not added to an existing thread that is not directly related to your question.

    To answer your question,
    who said that you could set v=c in that equation for the relativistic kinetic energy?
    Mathematically, [with a fixed value for m_0 anc c] you get infinity... as reported by the program.
    (The relativistic generalization of the so-called work-energy theorem still applies here:
    the [relativistic] net work done on it is equal to the change in its [relativistic] kinetic energy.)
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