Shouldn’t people teach more accurate versions of $E=mc^2$?

Question

I know they teach $E^2=(mc^2)^2+(pc)^2$ but in my opinion this equation is ugly. There is an equation that is even more accurate and, in my opinion, looks better. So why not teach it?

First off, this is something I did for fun, but my question, while not completely serious, does stand.

I like working with equations and a while ago I decided to turn my attention to the famous equation $E=mc^2$ (if you don’t know this equation then you should do some research). When I started working with it I knew of the quote unquote “true” equation $E^2=(mc^2)^2+(pc)^2$. I thought this equation looked ugly but I didn’t know how to fix that so I turned to just making it more accurate.

You see, there is something known as relativistic momentum. $p=\gamma mv$ you may consider this to be the more accurate version of momentum, so I decided to plug it into the “true” equation I mentioned earlier.

First the definitions. $$E^2=(mc^2)^2+(pc)^2$$ $$p=\gamma mv$$ $$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ Then the algebra. $$E=\sqrt{(mc^2)^2+(pc)^2}$$ $$\sqrt{(mc^2)^2+(\gamma mvc)^2}$$ $$\sqrt{(mc^2)^2+\bigg(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}mvc\bigg)^2}$$ $$\sqrt{m^2 c^4 +\bigg(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\bigg)^2 m^2 v^2 c^2}$$ $$\sqrt{m^2\bigg(c^4 +\frac{v^2 c^2}{1-\frac{v^2}{c^2}}\bigg)}$$ $$\sqrt{m^2\bigg(c^4 +\frac{v^2 c^2}{\frac{c^2}{c^2}-\frac{v^2}{c^2}}\bigg)}$$ $$\sqrt{m^2\bigg(c^4 +\frac{v^2 c^2}{(\frac{c^2 - v^2}{c^2})}\bigg)}$$ $$\sqrt{m^2\bigg(c^4 +\frac{v^2 c^4}{c^2 - v^2}\bigg)}$$ $$\sqrt{m^2 c^4\bigg(1 +\frac{v^2}{c^2 - v^2}\bigg)}$$ $$\sqrt{m^2 c^4\bigg(\frac{c^2 - v^2}{c^2 - v^2} +\frac{v^2}{c^2 - v^2}\bigg)}$$ $$\sqrt{m^2 c^4\bigg(\frac{c^2 - v^2 +v^2}{c^2 - v^2}\bigg)}$$ $$\sqrt{m^2 c^6\bigg(\frac{1}{c^2 - v^2}\bigg)}$$ $$mc^3\sqrt{\frac{1}{c^2 - v^2}}$$ $$E=\frac{mc^3}{\sqrt{c^2 - v^2}}$$ And there it is $E=mc^3$ sort of. Now it could be that I made this equation, but I think it looks better than the “true” equation. So that brings up a question. If it is more accurate and, in my opinion, looks better, why isn’t it taught in school? One could make the argument that with the methods used in the algebra, $v$ cannot equal $c$ but we are talking about an object with mass so this would be true anyway.

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EDIT: I’m not saying $E=mc^3$ should be taught instead of $E=mc^2$. I am instead asking why is $E^2=(mc^2)^2+(pc)^2$ taught instead of $E=\frac{mc^3}{\sqrt{c^2 - v^2}}$?

Answer 1

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Answer 2

There’s a notion of relativistic mass $m=m_0 \gamma$ so that $E=m_0 \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}c^2=m_0 \gamma c^2= m c^2$ is in fact correct.

Answer 3

How can be $c^2-v^2=1 $ ? wehn $c=300000 \frac{km}{sec}$ and $v$ is our velocity . Usually $$v<<c$$ and it lead again the relation to $$E=\frac{mc^3}{\sqrt{c^2 - v^2}} \sim \frac{mc^3}{\sqrt{c^2 }}=mc^2 $$ The fastest things on earth until now is The Lockheed SR-71 Blackbird is the current record-holder for a crewed airbreathing jet aircraft. It's about $3500 \ \frac {km}h$ if you approx it to $3600 \ \frac{km}{h}=600 \ \frac{m}{sec}$ and $$v=600 \frac{m}{sec} << c\sim 30000000 \frac{m}{sec}$$