Thread: E=mc^2 is the same thing as E=1/2mv^2?
The thread name needs work, but I love the ideas since it connects things I “heard of” in high school, but I didn’t learn it until after grad school, which felt like a travesty.
Prereqs: high school math + binomial theorem
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The thread name needs work, but I love the ideas since it connects things I “heard of” in high school, but I didn’t learn it until after grad school, which felt like a travesty.
Prereqs: high school math + binomial theorem
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So even starting middle school you might have heard stuff like E = mc^2, which was supposed to be famous. But in high school physics you learn things like E = mgh or E = 1/2mv^2. What’s up with that? Why doesn’t the former generalize anything in the latter?
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Some of you are nodding. “Relativity,” sagely. That’s great. My brain stopped there and never went deeper until later on.
But yes, relativity; the actual equation is
E = mc^2 / \\sqrt(1 - v^2/c^2),
which when velocity is small basically gives E = mc^2. Now...
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But yes, relativity; the actual equation is
E = mc^2 / \\sqrt(1 - v^2/c^2),
which when velocity is small basically gives E = mc^2. Now...
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Here is where we can get some small mileage - we know v^2/c^2 is some small value x. What is 1/\\sqrt(1 - x) for small x? We can rewrite this as (1-x)^{-1/2}.
Here, the world (seemingly) splits into two. If you follow the Demon of Analysis, you should use Taylor expansion.
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Here, the world (seemingly) splits into two. If you follow the Demon of Analysis, you should use Taylor expansion.
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But I follow the Angel of Algebra, so I see the binomial theorem
(1-x)^k = 1 - (k choose 1) x + (k choose 2) x^2 - ...
Now it is weird to do (-1/2 choose 1), but usually (k choose 1) = k, right? So we squint our eyes (sorry that’s how math is done), and
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(1-x)^k = 1 - (k choose 1) x + (k choose 2) x^2 - ...
Now it is weird to do (-1/2 choose 1), but usually (k choose 1) = k, right? So we squint our eyes (sorry that’s how math is done), and
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(1-x)^{-1/2} ~ 1 + 1/2 x^2 + (smaller x)
So E = mc^2/(1-v^2/c^2)^{1/2} ~ mc^2 + 1/2 mv^2 (!) + (smol)
(and yes, whether you follow the Devil or the Angel, you get here since physics is Rome. Tribalism was so 2020)
How to interpret this?
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So E = mc^2/(1-v^2/c^2)^{1/2} ~ mc^2 + 1/2 mv^2 (!) + (smol)
(and yes, whether you follow the Devil or the Angel, you get here since physics is Rome. Tribalism was so 2020)
How to interpret this?
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The dominant term is that (when we sum over all particles in a system) the sum of the mc^2’s remain. Since c^2 is a constant, we get *conservation of mass*: the sum of the masses pretty much stays constant.
This also means mass isn’t *really* conserved! (esp. for high v’s)
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This also means mass isn’t *really* conserved! (esp. for high v’s)
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The cool part is that the second order term is also pretty much conserved since in practice the m’s don’t change that much! This gives *conservation of kinetic energy,* which is the 1/2 mv^2 stuff we learned in physics class!
So we somehow got 2 classic “laws” at once!
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So we somehow got 2 classic “laws” at once!
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It also explains the dodgy reason why conservation of kinetic energy is often “violated;” often there’s some other fast moving particles (with high temperature) that weren’t accounted for in your problem involving 2 colliding balls — your teacher said something about heat.
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