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Transformation equations presented in a different way

  1. Feb 6, 2007 #1
    Consider a particle that moves with speed u relative to the inertial reference frame I and with speed u' relative to the inertial reference frame I'. Let g(u), g(u') and g(V) be the orresponding gamma factors (V the relative speed of I and I'). m(0) stands for its rest mass, E(0) for its rest energy, p and E for its momentum and enegy measured by observers from I. It is obvious that
    p=g(u')g(V)m(0)(V+u')=g(u')g(V')E(0)(V+u')cc
    E=g(u')g(V)E(0)(1+Vu'/cc)
    I consider that such a presentation presents some (pedagogical) advantages showing clearly what observers from the I frame measure in the case when u'=0 and when u' and V are both equal to zero.
    Even if I know that the concept of relativistic mass is persona non grata on the Forum I would also suggest for the relativistic mass
    m=g(u')g(V)E(0)(1+Vu'/cc)=g(u')g(V)m(0)(1+Vu'/cc).
    The oppinion of those who teach or learn special relativity theory is highly appreciated of course in the spirit of
    sine ira et studio
     
  2. jcsd
  3. Feb 7, 2007 #2

    Meir Achuz

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    Never use "it is obvious" in pedagogy. That is like the old mathematics professor joke.
     
  4. Feb 7, 2007 #3

    Don't you think that [tex]E=\gamma(V)*E(0)[/tex] is much cleaner?
     
    Last edited: Feb 7, 2007
  5. Feb 7, 2007 #4

    Hootenanny

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    As an aside, persona non grata means 'an unwelcome person' and is not usually used in reference to a concept. Absit invidia :wink:
     
  6. Feb 7, 2007 #5
    transformation equation

    Thanks. I think that there is a big difference between posting on the Forum, where the participants can easily transform the usual transformations in those I propose and presenting them in all its steps.
    I know very much joks with teachers of physics and mathematics. Which of them do you mean?
     
  7. Feb 7, 2007 #6
    transformation equations

    Thanks. Yes it is but I think that the Forum could offer a simple equation editor.
     
  8. Feb 7, 2007 #7

    cristo

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    The forum does! LaTex is very easy to learn, and it is easy to use on this forum too; simply put [tex] [ /tex] tags (without the space) around the equations.
     
  9. Feb 7, 2007 #8
    latina ginta est Regina

    Thanks. My first language is close to Latin. I thought that physicists are able to extrapolate from persona non grata to relativistic mass which is there non grata. I end with
    absit invidia which is shorter and more adequate then sine ira et studio I used so far.
     
  10. Feb 7, 2007 #9
    [tex]\gamma(u')\gamma(V)(1+Vu'/c^2)=\gamma(u)[/tex] , so the above reduces the the much cleaner, well known :

    [tex]m(u)=\gamma(u)*m(0)[/tex]
     
    Last edited: Feb 7, 2007
  11. Feb 7, 2007 #10

    robphy

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    The identity (rearranged in a minor way)
    [tex]\gamma(u')\gamma(V)(1+u'V/cc)=\gamma(u)[/tex]
    is more recognizable in terms of rapidities:
    [tex]
    \begin{align*}
    \gamma(u')\gamma(V)(1+u'V/cc)
    &=
    \cosh{\theta'}\cosh{\phi}(1+c\tanh{\theta'}\ c\tanh{\phi}/cc)\\
    &=
    \cosh{\theta'}\cosh{\phi}+\sinh{\theta'}\sinh{\phi}\\
    &=
    \cosh{(\theta'+\phi)}\\
    &=
    \cosh{\theta}\\
    &=
    \gamma(u)
    \end{align*}
    [/tex]
    where u[itex]=c\tanh{\theta}[/itex] is the spatial-velocity obtained by "spatial-velocity-composition of u' and V".

    In addition, in terms of rapidities, one can immediately transcribe the calculation into a spacetime diagram, which provides a hopefully more intuitive interpretation of what is happening physically [and mathematically].

    So, it's not clear to me if anything is gained in the proposed formula, except maybe for a particular type of problem.
     
    Last edited: Feb 7, 2007
  12. Feb 7, 2007 #11
    transformations

    Thanks. My intention is to present the transformation equations in such a way that theirs right sides contain only a proper physical quantity and velocities reducing the long discussions related to the concept of relativistic mass. That is the direction in which I hope our discussions will evolve.
    Reading my lines please take into account that English is not my first language.
     
  13. Feb 18, 2007 #12
    Thank you for having brought the formula to a more transparent shape. Consider the concept of proper mass m(0) and multiply both sides of with it. It leads to[tex]m(0)gamma(u0=m(0)gamma(u')gamma(V)(1+u'V/cc[/tex]An exercised eye will recognise in the left side of the equation the expression of the relativistic mass in I in the left side its expresion as a function of phyhsical quantities measured in I. Do you consider that such a presentation is time saving, transparent and convincing for the fact that conservation laws are not compulsory in the derivation. I do not convinced that the equation will appear correctly in myh message.
     
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