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Derivation of the momentum-energy relation

  1. Feb 23, 2005 #1
    In deriving the relation [tex] E^2=(pc)^2+(mc^2)^2 [/tex], I have used the relations [tex] p=\gamma mu [/tex] and [tex]E=\gamma m_0c^2[/tex]. I am currently stuck until a certain step and would appreciate it if someone could show its derivation, thanks alot...

    : )
     
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  3. Feb 23, 2005 #2

    dextercioby

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    Okay,here goes:

    [tex] p^{2}=\gamma^{2}m^{2}v^{2}=\frac{c^{2}m^{2}v^{2}}{c^{2}-v^{2}} [/tex]

    Add what u need,but only after multiplying by [itex]c^{2}[/itex]:

    [tex] p^{2}c^{2}+m^{2}c^{4}=m^{2}c^{4}(\frac{v^{2}}{c^{2}-v^{2}}+1)=
    m^{2}c^{4}\frac{c^{2}}{c^{2}-v^{2}}=(\gamma m c^{2})^{2}=E^{2} [/tex]

    ,where i made use of Einstein's formula...

    Daniel.



    [tex] p
     
  4. Feb 23, 2005 #3

    robphy

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    Just to encourage more use of the rapidity [tex]\theta[/tex] (and one's trigonometric intuition):

    [tex]u=c\tanh\theta[/tex] and [tex]\gamma=\cosh\theta[/tex] (So, [tex]\gamma u=c\sinh\theta[/tex])

    So, start with [tex]p=m_0c\sinh\theta[/tex] and [tex]E=m_0c^2\cosh\theta[/tex].

    Since [tex]
    \begin{align*}
    \cosh^2\theta - \sinh^2\theta&\equiv1\\
    \left(\frac{E}{m_0 c^2}\right)^2 - \left( \frac{p}{m_0c}\right)^2&=1\\
    E^2 - (pc)^2&=(m_0c^2)^2\\
    \end{align*}
    [/tex]
     
  5. Feb 23, 2005 #4

    dextercioby

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    Though not taught in HS,your method is elegant.:smile:Mine is simply lacking in inspiration.:yuck:

    Daniel.
     
  6. Feb 23, 2005 #5
    For those of you who have not yet followed the derivation for the mass-energy relation please see
    http://www.geocities.com/physics_world/sr/mass_energy_equiv.htm

    Pete
     
  7. Feb 23, 2005 #6

    selfAdjoint

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    IMHO, there is no reason why hyperbolic trig functions should not be taught in HS trig (in 1950 we got them in "College Math" which I took Junior year in HS). Nor is there that rapidity should not be taught in intro to relativity for math-enabled students.
     
  8. Feb 23, 2005 #7

    dextercioby

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    I didn't mean hyperbolic trig functions,but special relativity & SLT using them...Maybe they're taught somewhere,but i would find that level (of teaching physics) too high...:wink:

    Daniel.
     
  9. Feb 23, 2005 #8
    hey, thanks for the help ! and i'll definitely check out that trig derivation once I've started on hyperbolic functions though...

    : )
     
  10. Feb 23, 2005 #9

    DB

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    Umm, not to bother you guys but, how does
    [tex] p^{2}=\gamma^{2}m^{2}v^{2}=\frac{c^{2}m^{2}v^{2}}{ c^{2}-v^{2}} [/tex]?
    What I get is:

    [tex](\frac{1}{\sqrt{1-v^2/c^2}})^2=\frac{1}{1-v^2/c^2} *m^2v^2[/tex]

    [tex]\frac{m^2v^2}{1-v^2/c^2}[/tex]

    Now I'm stuck....
     
  11. Feb 23, 2005 #10

    robphy

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    multiply by [tex]\frac{c^2}{c^2}[/tex]
     
  12. Feb 24, 2005 #11

    DB

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    oo, ya now I got it, thanks. Is that what Dexter meant when he said:
    "Add what u need,but only after multiplying by [itex]c^2[/itex]:"?
     
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