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Can a physicist look at this please and tell me if i have derived this correctly?

  1. Feb 26, 2009 #1
    1. The problem statement, all variables and given/known data

    I need a physicist to look over these derivations and help me see if there are any mistakes. Thank you in advance, it is much appreciated.

    2. Relevant equations

    - reltivistic math derivations

    3. The attempt at a solution

    The First Part

    [tex]E^2-(pc)^2=(Mc^2)^2[/tex] where the expression [tex](Mc^2)^2[/tex] is by definition, the squared mathematical precision of an ''invariant mass'', hence, [tex]Mc^4[/tex].

    [tex]\rightarrow Mv(\frac{E}{M})=Mc^2(v)[/tex]

    allow [tex]v=c[/tex] then this simplifies to [tex]Mv^3=Mc^2[/tex] (Just to show that these are relativistic equivalances without the need of gamma function. This now leads me to calculate:


    The Second Part

    my equation, albiet as simple as it is, will show its importance throughout the metric work:

    [1] [tex]M(1+M)=2M[/tex] if

    [2] [tex]-(\frac{E}{c})^2+mv^2=p^2[/tex]

    then combine by division of [1] and [2] equations, allowing the relativistic proof:

    [tex]\frac{\eta^{\mu v}p_{\mu}p_{\mu}}{2m}=\frac{p^2}{2m}=\frac{p^2}{2}\frac{E}{c^2}[/tex]

    which then follows

    [tex]p^2=\eta^{\mu v}p_{mu}p_{v}=-(\frac{E}{c^2})^2+p^2[/tex]

    I need to go the now, i will finish this later, but sooner than later.
    Last edited by a moderator: Feb 26, 2009
  2. jcsd
  3. Feb 26, 2009 #2


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    Science Advisor
    Homework Helper

    That is true

    That is true.

    No you need a mathematician. Because those are also the only equations I could find that don't have (mathematical) errors in them.
  4. Feb 26, 2009 #3
    That is true, friend.
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