# $E=mc^2$ resembles non-relativistic kinetic energy formula $E_K = \frac{1}{2}mv^2$?

The simplest equation expressing mass–energy equivalence is the famous $E=mc^2$ where $c$ represents the speed of light. Compare this with $E_K = \frac{1}{2}mv^2$.

Since $E=mc^2$ can be applied to rest mass ($m_0$) and rest energy ($E_0$) to show their proportionality as $E_0=m_0c^2$, I ask whether this resemblance is just a coincidence created by the need of any equation to be homogeneous for units, or are the two equations fundamentally related?

I know about Mass-Energy Equivalency but I could not find the answer I'm looking for there.

• I don't know $E=mv^2$. There is only $E=\frac{1}{2}mv^2$. – peterh May 12 '14 at 11:59
• why is $E=mc^2$ infamous? – Antonio Ragagnin May 12 '14 at 12:06
• @AntonioRagagnin $E=mc^2$ is infamous because laypeople believe $E=mc^2$ made the nuclear bomb possible. Even when one knows their physics, a connection between the two can't be denied... – hb20007 May 12 '14 at 12:40

In short, no, it is not a coincidence, they are related. Namely, you may derive the kinetic energy as the first order approximation to the relativistic energy.

We have,

$$E_0 = mc^2$$

as you say correctly. Then

$$E = \gamma m c^2 = \left( 1 - \frac{v^2}{c^2}\right)^{-\frac{1}{2}} m c^2$$

or using a binomial expansion

$$E \simeq \left( 1 + \frac{1}{2} \frac{v^2}{c^2} + \dots \right) m c^2 \simeq mc^2 + \frac{1}{2} mv^2$$

So subtacting the rest energy $E_0$ we get

$$E_k = E - E_0 \simeq \left( mc^2 + \frac{1}{2} mv^2 \right) - \left( mc^2 \right) = \frac{1}{2} m v^2$$

Note that we can of course only use this expansion when $v \ll c$. This makes sense, because that is exactly the case in Newtonian mechanics, which is where we use the more familiar kinetic energy formula.

• Two simultaneous identical answers in two days, @Flint72! But this time it was I who included a link! +1 anyway :-) – rob May 12 '14 at 12:06
• @rob: Ya, I was meaning to link to this page on the derivation in wikipedia, though it's a little messy. – Flint72 May 12 '14 at 12:08
• Oh, you are the same 2 guys from physics.stackexchange.com/q/112028 – hb20007 May 12 '14 at 12:29
• Ahh... two great answer posted in the exact same minute:S. I'm picking this one as 'best answer' because the derivation is a bit more straightforward. – hb20007 May 12 '14 at 12:54

Every relationship between mass and energy will contain two factors of velocity, for dimensional consistency.

In special relativity we have the more exact relationship $$E^2 - p^2c^2 = m^2c^4$$ where the momentum $p$ is $$p = \frac{mv}{\sqrt{1-v^2/c^2}}.$$ You can do a little algebra to show that the total energy is always $$E = \frac{ mc^2}{\sqrt{1-v^2/c^2}}.$$

In mathematics we have a tool called the "binomial theorem," which tells us that $$(1+\epsilon)^n = 1 + n\epsilon + \frac{n(n-1)}{2!}\epsilon^2 + \cdots$$ This expression turns out to hold even if the power $n$ is not a positive integer! If $\epsilon\ll1$, we also have the luxury of being able to throw away the higher powers. For small speeds, then, the Einstein equation becomes $$E = \frac{ mc^2}{\sqrt{1-v^2/c^2}} = mc^2 + mc^2 \cdot \frac{-1}{2}\frac{-v^2}{c^2} + mc^2 \cdot\mathcal O\left( (v/c)^4 \right)$$ which is clearly the rest energy, the classical kinetic energy, and a relativistic correction that becomes large when $v\approx c$.

• Great minds, eh? – Flint72 May 12 '14 at 12:06

## protected by Qmechanic♦Aug 9 '14 at 17:41

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