Guys... i think i need a help with my Science Investigatory Project about E=mc2

Discussion in 'Physics & Math' started by FOUR, Nov 12, 2015.

  1. FOUR Registered Member

    Messages:
    3
    Guys please help me with my research...
    i need a proof/example that can prove my new equation about the E=mc2 + pc..
    I don't know if there's something wrong with my equation but i feel like it's complete though....
    Im gonna present this equation in front of the judges in our school... PLEASE HELP ME
     
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  3. origin Heading towards oblivion Valued Senior Member

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    11,573
    I am afraid you are going to have a difficult time trying to prove \(E = mc^2 + pc\).
    It would be much easier to prove \(E = \sqrt{(mc^2)^2 + (pc)^2}\).
     
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  5. Fednis48 Registered Senior Member

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    What is this new equation? It might help if we knew what you were trying to prove.
     
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  7. rpenner Fully Wired Valued Senior Member

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    4,833
    \( E= \sqrt{ (mc^2)^2 + (pc)^2 } \\ E \vec{v} = c^2 \vec{p} \)
    hold for both massive and massless particles that carry momentum and energy in situations where gravity can be neglected..
     
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  8. rpenner Fully Wired Valued Senior Member

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    4,833
    Thus we have the following derived relationships when \(m > 0\):

    \(E = mc^2 \sqrt{1 + \left( \frac{p}{mc} \right)^2 } \)
    \(\vec{v} = \frac{ \vec{p} }{ m \sqrt{1 + \left( \frac{p}{mc} \right)^2 } }\)

    From the latter we get
    \(v^2 m^2 + p^2 \frac{v^2}{c^2} = p^2 \\ v^2 m^2 = p^2 ( 1 - \frac{v^2}{c^2} ) \\ p = \frac{m v}{\sqrt{ 1 - \frac{v^2}{c^2} }}\)
    Thus
    \(E = mc^2 \sqrt{1 + \left( \frac{p}{mc} \right)^2 } \\ = mc^2 \sqrt{1 + \left( \frac{mv}{mc} \frac{1}{\sqrt{ 1 - \frac{v^2}{c^2} }}\right)^2 } \\ = mc^2 \sqrt{1 + \frac{v^2}{c^2} \frac{1}{ 1 - \frac{v^2}{c^2} } } \\ = mc^2 \sqrt{ \frac{ 1 - \frac{v^2}{c^2}}{ 1 - \frac{v^2}{c^2}} + \frac{ \frac{v^2}{c^2} }{ 1 - \frac{v^2}{c^2} } } \\ = mc^2 \sqrt{ \frac{ 1 }{ 1 - \frac{v^2}{c^2} } } \)

    Thus we introduce the simplifying notation (when \(m > 0\) ):
    \( \gamma = \sqrt{ \frac{ 1 }{ 1 - \frac{v^2}{c^2} } } = \frac{ 1 }{\sqrt{1 - \frac{v^2}{c^2} } } = \sqrt{1 + \left( \frac{p}{mc} \right)^2 } \\ E = \gamma m c^2 \\ \vec{p} = \gamma m \vec{v} \)

    If we introduce \(\rho\) such that \(\gamma = \cosh \, \rho\) then:
    \( E = mc^2 \cosh \, \rho \\ p = m c \sinh \, \rho \\ v = c \tanh \, \rho\)
    which is a nice result using hyperbolic analogues of trigonometric functions.

    It also has application in the 1-dimensional version of the Einstein law for composition of velocities:
    \( \rho_w = \rho_u + \rho_v \\ \Rightarrow w = c \tanh \left( \tanh^{-1} \frac{u}{c} + \tanh^{-1} \frac{v}{c} \right) = \frac{( u + v)c^2 }{c^2+u v} = \frac{u + v}{1 + \frac{u v }{c^2}}\)
     
    Last edited: Nov 13, 2015
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  9. brucep Valued Senior Member

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    4,098
    Great post. Nice derivation.
     
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  10. brucep Valued Senior Member

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    4,098
    What you're looking for is the relativistic energy equation. Origin wrote it correctly and rpenner derived it. This is another way to derive it. Choose Chapter one Speeding
    http://www.eftaylor.com/download.html#general_relativity
    It starts at 4 The Principle of Extremal Aging which is a form of Noethers Theorem the principle of least action. You should read starting at 1 to familiarize yourself with the metric. For school it would be cool to be able to explain the derivation of the equation.
     
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  11. rpenner Fully Wired Valued Senior Member

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    4,833
    Using Taylor series we can show when \(p << mc\) that
    \( \gamma = \sqrt{1 + \left( \frac{p}{mc} \right)^2 } \approx 1 + \frac{1}{2} \left( \frac{p}{mc} \right)^2 - \frac{1}{8} \left( \frac{p}{mc} \right)^4 + \frac{1}{16} \left( \frac{p}{mc} \right)^6 - \frac{5}{128}\left( \frac{p}{mc} \right)^8 + \dots \\ E \approx mc^2 + \frac{p^2}{2 m} - \frac{p^4}{8 m^3 c^2} + \dots \)
    Which means the Newtonian formula for kinetic energy \(\frac{p^2}{2 m}\) is accurate when \(p << mc\).
    If \(p = \frac{1}{10} m c\) then Newtonian mechanics predicts the kinetic energy to be \(\frac{mc^2}{200}\) while Einstein predicts it to be \( \left( \sqrt{1 + \frac{1}{100}} - 1 \right) mc^2 \) which is different by only about a quarter of a percent.

    Likewise we can show when \(v<<c\) that
    \( \gamma = \frac{ 1 }{\sqrt{1 - \frac{v^2}{c^2} } } \approx 1 + \frac{1}{2} \frac{v^2}{c^2} + \frac{3}{8} \frac{v^4}{c^4} + \frac{5}{16} \frac{v^6}{c^6} + \frac{35}{128} \frac{v^8}{c^8} + \dots \\ E \approx mc^2 + \frac{m v^2}{2} + \frac{3 m v^4}{8 c^2} + dots \\ p \approx mv + \frac{m v^3}{2 c^2} + \frac{3 m v^5}{8 c^4} + dots \)
    So Newtonian formulas approximate the low-velocity behavior of the relativistic formulas. That's why Newton's physics had centuries of success.
     
  12. eram Sciengineer Valued Senior Member

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    1,877
    Have you ever watched the College Humor video "If Google was a Guy?"



    If we had "If Wolfram Alpha was a Guy?", rpenner would be perfect.

    Please Register or Log in to view the hidden image!

     

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