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Relativistic terminology

  1. Jan 19, 2008 #1
    I find in the literature the following equation in which the OX(O'X') components of the same tardyon are present (u(x) and u'(x)) in I and in I' respectively.
    1=[1-V^2/c^2)]/[(1-V/u(x))(1+V/u'(x)]
    How would you call it, relativistic identity, reference frame independent...?
    Thanks in advance.
     
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  3. Jan 19, 2008 #2

    Fredrik

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    First of all, I think "tardyon" is a useless concept. I would just call it a "particle", or an "object". I might clarify by saying that it's a "massive particle", or a "particle moving at velocity u<c", but I think that isn't usually necessary.

    As for the rest, I don't understand what you're asking. What does that equation say? What's u(x) and u'(x)? Velocities as functions of position?? Does the prime mean that it's a different frame? Then what are they doing in the same equation? Should there be a prime on the x too?
     
  4. Jan 19, 2008 #3
    English is not my first language. Learning from Wikipedia I found out that


    "A tardyon or bradyon is a particle that travels slower than light. Therefore this is a synonym with a massive particle (i.e. a particle having a non-zero mass). This includes all known particles (except luxons). The term "tardyon" is constructed to contrast with "tachyon", which refers to hypothetical particles that travel faster than light." but that is of second importance.
    u(x) and u'(x) represent the OX(O'X') components of the speed of the same particlle moving at velocity u<c , V representing the rellative speed of the involved reference frames.
    In order to make my question more transparent please tell me what is the name of
    x^2-c^2t^2=x'^2-c^2t'*2 ?
    Is it a relativistic identity, does it relate two covariant algebraic combinations of physical quantities measured in I and I' respectively...?
    Thanks.
     
  5. Jan 19, 2008 #4

    Dale

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    I think it is perfectly fine to use the word "tardyon". It certainly is faster to type than "particle moving at velocity u<c" or even "massive particle".

    That said, I don't know of a specific name for the equation either. It is just an expression of the invariance of the spacetime interval.
     
    Last edited: Jan 19, 2008
  6. Jan 19, 2008 #5

    Fredrik

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    Maybe the term you're looking for is just "Lorentz invariant". The left-hand side is Lorentz invariant, because it "looks the same" expressed in the coordinates of another inertial frame, as we can see by looking at the right-hand side. The right hand-side is of course also Lorentz invariant for the same reason.

    So both sides of the equation are Lorentz invariant. I don't know a word that describes what the equation is though. I don't think there is one. An equation like

    [tex]E^2=\vec{p}+m^2[/tex]

    (where c=1) is Lorentz invariant because it looks the same in all inertial frames. (m is coordinate independent and [itex]\vec{p}^2[/itex] changes in a way that exactly compensates for the change in E).

    The equation in your first post seems to contain quantities expressed in two different coordinate systems. I still don't understand it. Why u(x) and u'(x) instead of just u and u'? Are OX and O'X' what you call the frames? Then why did you also call them I and I'? How can u and u' appear in the same equation? (If I have already answered your question, I guess these things don't matter much at this point).

    And I didn't mean to imply that it's wrong to use the word "tardyon". It is fine to use it. I just don't find the concept useful, and I don't think many authors use it. Yes it's faster to type than "particle moving at velocity u<c", but is it faster for the reader? Compare reading "particle moving at velocity u<c" and immediately knowing what it means, with reading "tardyon" followed by a visit to Wikipedia...
     
    Last edited: Jan 20, 2008
  7. Jan 20, 2008 #6
    I have never heard of a specific name too. Only "interval invariance".
     
  8. Jan 20, 2008 #7

    pam

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    I call it "incorrect". What if u=0?
     
  9. Jan 20, 2008 #8
    Are you referring to the quantity that appears on both sides of this equation, expressed in reference to different frames? That is the spacetime interval, as has been pointed out above.
     
  10. Jan 20, 2008 #9

    Fredrik

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    I don't think he's asking about space-time intervals or anything specific like that. He's asking about a term that describes a whole class of mathematical expressions and/or equations. I'm still not sure what that class is. My best guess is still that the term he's looking for is "Lorentz invariant", but it's hard to tell since the equation in post #1 doesn't seem to make sense.
     
  11. Jan 20, 2008 #10
    Thanks. I present it again in a changed way

    1=[1-VV/cc]uu'/[(u'-V)(u+V) (1)
    where u and u' represent the OX(O'X') components of the same tardyon and so related by the addition law of parallel speeds. For say u'=0, u=-V and so the result is 0/0!
    I have obtained (1) by considering the same Doppler Effect experiment from I and I' without using the addition law of parallel speeds. Solved for u or u' (1) gives them.
    Regards
     
  12. Jan 21, 2008 #11

    Fredrik

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    Your equation becomes u'=u+V in the limit c goes to infinity, so based on that and on what you said, I'm guessing that it's just a strange looking way to express the velocity addition law (assuming that it isn't just wrong). It's clear that "Lorentz invariant" isn't at all what you're looking for. Are you looking for a name that describes all the equations that can be derived using only the properties of Minkowski space? I don't think there is one, except I guess "relativistic" or "special relativistic" to be more precise.
     
  13. Jan 21, 2008 #12
    You guessed my question and thanks for your answer.
    I think the equation is correct, il leads solved for u or for u' to the transformation of speeds. The fact that when c goes to infinity it leads to the classical addition law of speeds could be an argument for the fact that in Galileo's relativity the clocks are synchronized using a signal propagating at infinite speed, a natural conclusion for the fact that he was not able to detect a finite speed of it. (Dt=0, Dx diferent from zero).
    Regards
     
  14. Jan 22, 2008 #13

    pam

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    Multiply (u+V)=u'(1+u.V) by (u'-V)=u(1+u'.V).
     
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