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Why C in E=MC2?


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#1 browndn

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Posted 2 March 2008 - 09:48 PM

please explain it. all i kno is that energy and matter[mass] are manifestations of the same thing and C = speed of light but what does it have to do with anything?
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#2 Reaper

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Posted 2 March 2008 - 09:55 PM

From what I remember, it has something to do with the fact that you can't accelerate anything past the speed of light.
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#3 Klaynos

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Posted 2 March 2008 - 10:21 PM

It appears in the equations as c is constant in all rest frames...

The derivation for E=mc2 (note it's a lower case c) is a bit long for me to remember and post here but it's perfectly possible to work through without a massive knowledge of maths.
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#4 Reaper

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Posted 2 March 2008 - 10:28 PM

The derivation for E=mc2 (note it's a lower case c) is a bit long for me to remember and post here but it's perfectly possible to work through without a massive knowledge of maths.



Here is a derivation of the equation: http://www.adamauton.../warp/emc2.html

Let me know if you spot a mistake, it seems alright to me.
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#5 NeonBlack

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Posted 3 March 2008 - 08:24 AM

The fact that you can't go faster than c is the result of, not the reason for E=mc2. You do not have to assume that nothing can go faster than c, only that all physics (including the speed of light) is the same in all inertial frames. This is what (in my opinion) makes special relativity so beautiful.

I only briefly skimmed the link Lockheed posted is nice that it's based on a physical situation, but I believe this mathematical derivation is much more straight-forward. It can get a little messy, but it's not very long at all. Anyone with university calculus should be able to handle it.

If you're learning about relativity, you probably know \gamma=\frac{1}{\sqrt{1-v^2/c^2}}
The relativistic momentum is p=mv\gamma
(If you're not satisfied beginning here, you can derive the expression for momentum with the velocity addition and by assuming that momentum must be conserved).
Newton's 2nd law says: F=\frac{dp}{dt}=m\frac{dv}{dt}\gamma^3
The regular old definition of work: W=\int F dx
Now here we do a little chain-rule trick to rewrite this as
W=m \int \gamma^3 \frac{dx}{dt} dv
And since \frac{dx}{dt}=v
W=\int \gamma^3 v dv = mc^2 \gamma
Now just let v=0 and bam E=mc^2
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#6 Severian

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Posted 3 March 2008 - 01:36 PM

please explain it. all i kno is that energy and matter[mass] are manifestations of the same thing and C = speed of light but what does it have to do with anything?


I think the reason for your confusion is probably that the "speed of light" is a very bad name for 'c'. c is a limiting velocity caused by the non-linear nature of the relation between velocity and momentum. Light happens to travel at this speed (i.e. as fast as possible) simply because it has no mass - any other particle with no mass would also travel at c too.

The actual value c=3x108ms-1 is just a matter of convention in the units we choose, i.e. the definition of metres and seconds. If we measured time in uints of seconds and distance in units of light-seconds, then c = 1 light-second s-1.
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#7 thedarkshade

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Posted 3 March 2008 - 07:35 PM

The derivation for E=mc2 (note it's a lower case c) is a bit long for me to remember and post here but it's perfectly possible to work through without a massive knowledge of maths.

Fair enough! And it tells the huge amount of energy stored in matter!
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