# 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. ### FOURRegistered Member

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3
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....

3. ### originHeading towards oblivionValued 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 $E = \sqrt{(mc^2)^2 + (pc)^2}$.

5. ### Fednis48Registered Senior Member

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725
What is this new equation? It might help if we knew what you were trying to prove.

7. ### rpennerFully WiredValued 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. ### rpennerFully WiredValued Senior Member

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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. ### brucepValued Senior Member

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4,098
Great post. Nice derivation.

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10. ### brucepValued Senior Member

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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
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. ### rpennerFully WiredValued Senior Member

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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. ### eramSciengineerValued 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.