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This is a follow up to my previous question Why is the kinetic energy for non-relativistic velocities not described by $KE=mc^2$?

After trying to use the actual formula for relativistic kinetic energy $$KE=(\gamma - 1)mc^2$$ where $\gamma = \frac 1 {\sqrt{1-\frac{v^2}{c^2}}}$, I tried to do some test calculations to compare it to the non-relativistic kinetic energy formula $$KE=\frac 1 2 mv^2$$ Let's do an example where $m=10kg$ and $v=50 \frac m s$, which is still pretty fast but well below relativistic speeds. So both formulas should give the same result.

For non-relativistic kinetic energy, one gets $KE=12500J$. For relativistic KE however, one gets $KE=0$ (Note: The actual value is probably somewhat larger than $0$ but has been rounded).

Although this contradicts our exception that both formulas should give the same result, it is somewhat expectable: Since $v \ll c$, $\frac{v^2}{c^2} \approx 0$, thus $\gamma \approx 1$ and $KE \approx (1-1)mc^2=0$.

So what is the takeoff here? I thought that the non-relativistic KE is a simplification of relativistic KE, so the latter should also be able to give accurate results for non-relativistic velocities. But obviously, it doesn't. Why?

jng224
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1 Answers1

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One has be to careful interpreting $v\ll c$ as equivalent to $\frac{v}{c}\to 0$. To derive the non-relativistic kinetic energy, we make use of the binomial approximation \begin{equation} (1 + x)^{\alpha} \simeq 1 + \alpha x + \mathcal{O}(x^2) \quad {\rm for} \quad x\ll 1 . \end{equation}

Therefore, for $v\ll c$ or $\frac{v}{c}\ll 1$ then \begin{equation} \gamma = \left(1 - \frac{v^2}{c^2}\right)^{-1/2}\simeq 1 + \frac{1}{2}\frac{v^2}{c^2} + \cdots . \end{equation}

Using this result, \begin{equation} {\rm KE} =(\gamma - 1)mc^2 \simeq mc^2\left[\left(1 + \frac{1}{2}\frac{v^2}{c^2}\right) - 1\right] = \frac{1}{2}mv^2 \end{equation} as expected. Now, for your numerical computation you are correct that \begin{equation} {\rm KE}_{\rm classical} = \frac{1}{2}(10\ {\rm kg})(50\ {\rm m/s})^2 = 12500\ {\rm J} . \end{equation}

However, $\gamma - 1$ while close to zero is actually,

\begin{equation} \gamma - 1= \left(1 - \frac{v^2}{c^2}\right)^{-1/2} - 1 = \left(1 - \frac{(50\ \text{m/s})^2}{(299 792 458\ \text{m/s})^2}\right)^{-1/2} - 1 \simeq 1.39\times 10^{-14} . \end{equation}

With this, the relativistic kinetic energy is \begin{equation} {\rm KE} = (\gamma - 1)mc^2 \simeq 1.39\times 10^{-14}\cdot (10\ {\rm kg})\cdot c^2 = 12500.0000000003\ \text{J} . \end{equation} This number is very close to the classical answer, as expected. The takeaway here is that approximations like $v\ll c$ should not be treated as limits and calculators will probably round for very small velocities.

Hope this helps!

MarcosMFlores
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