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Well for one thing, it would have to be. I can't tell you exactly why it happens to be *that* number--that's just a fundamental law of the universe--but it definitely has to be c^2. You're thinking strictly in terms of numbers, but the *UNITS* are important.
The units of energy are Joules, or (kilograms x meters^2)/(seconds^2). The mass is kilograms and the speed of light is meters/second. Squaring it gives you not only the proper units but the proper magnitude of the number as well. If you raised it to 1.999 you'd have kg(m^1.999)/(s^1.99). That simply isn't energy. Just like kilograms are not a unit of electrical current.
One way that you can know that this value is correct is by thinking about the nuclear binding energy. Let's say that you have an unstable nucleus that undergoes nuclear fission in a reaction. A big radioactive uranium nucleus becomes two smaller nuclei--often barium and krypton. You also release a lot of energy. If you found the mass of the barium and krytpon nuclei, you'd find that it was less than the mass of the original uranium nucleus. Where'd that mass go? It was converted into energy. If you measured the total energy released from the fission, it would be equal to the missing mass multiplied by the speed of light squared.
Conversely, in nuclear fusion, you combine two hydrogen nuclei and produce a helium nucleus, releasing alot of energy. It turns out the helium nucleus has less mass than the two hydrogens that combined to produce it. Again, the missing mass x the speed of light squared tells you the energy that was released.
That's also a nice way of putting it. However, lets say you wanted to express this equation in atomic mass units instead of kilograms. The equation then becomes E= m (in amus) x 931.5MeV (mega electron volts). Not so pretty now, is it?
The units of energy are Joules, or (kilograms x meters^2)/(seconds^2). The mass is kilograms and the speed of light is meters/second. Squaring it gives you not only the proper units but the proper magnitude of the number as well. If you raised it to 1.999 you'd have kg(m^1.999)/(s^1.99). That simply isn't energy. Just like kilograms are not a unit of electrical current.
One way that you can know that this value is correct is by thinking about the nuclear binding energy. Let's say that you have an unstable nucleus that undergoes nuclear fission in a reaction. A big radioactive uranium nucleus becomes two smaller nuclei--often barium and krypton. You also release a lot of energy. If you found the mass of the barium and krytpon nuclei, you'd find that it was less than the mass of the original uranium nucleus. Where'd that mass go? It was converted into energy. If you measured the total energy released from the fission, it would be equal to the missing mass multiplied by the speed of light squared.
Conversely, in nuclear fusion, you combine two hydrogen nuclei and produce a helium nucleus, releasing alot of energy. It turns out the helium nucleus has less mass than the two hydrogens that combined to produce it. Again, the missing mass x the speed of light squared tells you the energy that was released.
That's also a nice way of putting it. However, lets say you wanted to express this equation in atomic mass units instead of kilograms. The equation then becomes E= m (in amus) x 931.5MeV (mega electron volts). Not so pretty now, is it?
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The square in E=mc^2 is derived through algebra; the same sort of algebra as Pythagora's theorem, which certainly says a^2+b^2=c^2, and not c^1.999.
Read "Why does E=mc^2" by Brian Cox for a complete explanation. It's a good read.
Nevertheless your point is an interesting one. I think the answer would have to be that integer values play an important, deep role in mathematical physics.
By the way, the convention is to write E=mc^2 with lower case m and c - otherwise it means something slightly different.
:-)
Read "Why does E=mc^2" by Brian Cox for a complete explanation. It's a good read.
Nevertheless your point is an interesting one. I think the answer would have to be that integer values play an important, deep role in mathematical physics.
By the way, the convention is to write E=mc^2 with lower case m and c - otherwise it means something slightly different.
:-)
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It's supposed to represent the speed of light in a vacuum.
The "squared" part isn't what he calculated, it's the speed of light multiplying on itself that he figured.
The "squared" part isn't what he calculated, it's the speed of light multiplying on itself that he figured.