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Why does $E=mc^2$ give results in Joules ? Why didn't Einstein need to create a separate unit and how come using two standard units (speed of light in km.s and mass in grams) give another result in Joules ?

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$c$ is the speed of light and therefore has units $m/s$. Hence, $mc^2$ has units $kg \cdot m^2 / s^2$ which is equivalent to a Joules. –  Lagerbaer Apr 22 '11 at 15:02
Even if you don't feel like writing out the definition of J in terms of the base units, you can tell that $mc^2$ will have the same units of energy, by thinking about good old Newtonian kinetic energy. In Newtonian physics, kinetic energy is ${1\over 2}mv^2$. When we have a mass in kg and a speed in m/s, this quantity comes out in J. That means that $mc^2$ must come out in J too. After all, $v$ has the same units as $c$, and the $1\over 2$ has no units. –  Ted Bunn Apr 22 '11 at 15:33
Actually this raises an interesting question (for me at least) of why dimensional analysis always works. I mean, I completely believe that it does, I'm not doubting it; but I can't prove it, other than to say that it seems obvious to me that it should work. –  Colin K Apr 22 '11 at 17:27
@Colin: Dimensional analysis doesn't always work--you still have to combine variable meaningfully--xkcd.com/687 –  Jerry Schirmer Apr 22 '11 at 19:09
Agree with @Colin :-) Anyway, Colin, I'm not sure exactly what you're getting at by saying you're not sure why dimensional analysis always works. If you can elaborate, that might be a good thing to ask as a separate question. (Or you could take it up in the chat room) –  David Z Apr 22 '11 at 22:32

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It's an energy. The SI unit for energy is a Joule. If the theory is going to predict an energy-mass correspondance, then it better give the energy in units of energy. If the units of $mc^{2}$ didn't work out, for the equation to make any sense, you'd have to include some constant $\alpha$ so that $E=\alpha mc^{2}$ would have units of energy on both sides.

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Because of the definition of Joules in terms of the base kg/m/s units.

If you used lbs feet/second and calories you would need a conversion factor, because calories are defined in terms of heating water rather than mechanics, but the equation still works.

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