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 Zelaron Gaming Forum Deriving e=mc^2

 Deriving e=mc^2
Posted 2010-03-30, 12:19 AM
We begin with the definition of relativistic mass:

 $\reverse m = \frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$.

We can now define relativistic momentum as:

 $\reverse \vec{p} = \frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\vec{v}$.

Since force is defined as

 $\reverse \vec{F} = \frac{d\vec{p}}{dt}$

we can define force in the relativistic context analogously:

 $\reverse \vec{F} =\frac{d\vec{p}}{dt} =\frac{d}{dt} \left[ \frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\vec{v} \right] = \frac{d}{dt} \left[m\vec{v} \right]$.

In classical physics work is defined as

 $\reverse W = \int\limits_{r} \! \vec{F} \! \cdot \! d\vec{r}$.

We can reduce this to a problem in a single dimension for simplicity, and recall that the kinetic energy added to a body is equal to the work done on the body by an external force:

 $\reverse W = \int_{0}^{x} \! Fdx = E_{k}$.

From here we can say:

 $\reverse \int_{0}^{x} \! Fdx = \int_{0}^{x} \! \frac{dp}{dt}dx = \int_{0}^{x} \! \frac{d}{dt} \left[ mv \right] dx = \int_{0}^{t} \! \frac{d}{dt} \left[ mv \right] vdt$.

Applying the definition of relativistic mass, the last expression becomes:

[
 $\reverse \int_{0}^{t} \! \frac{d}{dt} \left[ mv \right] vdt = \int_{0}^{v} \! vd \! \left[ \frac{m_{0}v}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \right] = m_{0} \int_{0}^{v} \! \left( \frac{v}{\sqrt{1-\frac{v^{2}}{c^{2}}}} + \frac{v^{3}c}{\left( 1-\frac{v^{2}}{c^{2}} \right)^{3/2} \right) \! dv$.

Integrating through, we have

 $\reverse E_{k} = m_{0}c^{2} \left( \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} - 1 \right) = mc^{2}-m_{0}c^{2}$

The term on the left,
 $\reverse mc^{2}$
, is the total energy of the body after an external force is applied, the term on the right,
 $\reverse m_{0}c^{2}$
is the rest energy, and the final expression is the kinetic energy of the body.

The kinetic energy can be expressed more clearly, in my opinion, as

 $\reverse E_{k} = (m-m_{0})c^{2} = \Delta m c^{2}$.

And there you have it. Mass is energy!

Last edited by Sovereign; 2010-03-30 at 02:10 AM. Reason: Added spoiler tags to make font stand out.
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Demosthenes

 Posted 2010-03-30, 12:20 AM in reply to Demosthenes's post "Deriving e=mc^2" Latex isn't working entirely properly. But you should get the gist of it.
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Demosthenes

 Posted 2010-03-30, 12:24 AM in reply to Demosthenes's post starting "Latex isn't working entirely properly. ..." I have fixed it for now, but certain things aren't rendering correctly.
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Demosthenes

 Posted 2010-03-30, 08:23 AM in reply to Demosthenes's post starting "I have fixed it for now, but certain..." I'm glad I know this now. Or rather that I've read it. D3V: I need to get on zelaron more. I'm having withdrawls. Skurai: You should be used to withdrawls by now, D3V. !King_Amazon!: It's "withdrawals" you illiterate douchebags
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 !King_Amazon!

 Posted 2010-03-30, 08:28 AM in reply to Demosthenes's post "Deriving e=mc^2" Well, at least now I know HOW e = mc^2 I'd heard it used to many times in cartoons, it was in my head before teachers ever got around too it (infact, I've probably learned more from cartoons than text books), but at least now I know how it fits together.
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Skurai

 Posted 2010-03-30, 10:24 AM in reply to Skurai's post starting "Well, at least now I know HOW e = mc^2..." Well, I'm glad it helped someone, and was not an exercise in LaTeX. If there is anything I can clear up, let me know.
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Demosthenes

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