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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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    Does 12500J fall inside the value of "probably somewhat larger than 0"? – BowlOfRed Nov 08 '20 at 22:44
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    @BowlOfRed I don't think any calculator would round down such a large number – jng224 Nov 08 '20 at 22:47
  • c^2 is 90000000000 m^2/s^2. 12500J seems pretty small compared to that. I think you need to figure out by how much you're rounding. – BowlOfRed Nov 08 '20 at 22:48
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    Your calculator doesn’t have sufficient precision to compute $\gamma$ when $v$ is only 50 meters per second. There is nothing wrong with the formula. If you use, say, Mathematica and do a high-precision calculation, you’ll see that it works. – G. Smith Nov 08 '20 at 23:18
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    Rounding error. – my2cts Nov 09 '20 at 15:52
  • Here: https://physics.stackexchange.com/questions/263986/kinetic-energy-relativistic-correction?rq=1 you can find the first few terms of the analytic approximation of $\gamma-1$ in powers of $\frac{v^2}{c^2}$. You can see that the classical term is the leading term, followed by a correction smaller by a factor proportional to $\frac{v^2}{c^2}$. Numerical calculations require a full control on the used precision. – GiorgioP-DoomsdayClockIsAt-90 Nov 14 '20 at 12:43
  • You got a correct answer to a different question under the assumption $\beta^2=v^2/c^2≈0$: from it, $v≈0$ and $E_K$ is Newtonian, $E_K=mv^2/2≈0$. $\beta$ is the $v$ in units of $c$, a tiny number. A tiny number $\sim10^{-n}$ squared is, using a rigorous math term, a teeny-tiny number, $\sim10^{-2n}$. Your assumption is valid for the total SR mass-energy of motion in your other question, $E=E_0+E_K=[m_0c^2]+[(\gamma-1)m_0c^2]=\gamma m_0c^2$; in the $v\ll c$ regime $E_0$ dominates, $E_K=12,\text{kJ}\lll E_0=(10,\text{kg})\cdot c^2$; $E_K≈0$ is a negligible correction. – kkm -still wary of SE promises Feb 05 '23 at 23:26

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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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    You can’t get 14 significant digits of output (12500.000000002) from 3 significant digits of input (1.06). – G. Smith Nov 08 '20 at 23:24
  • The hyperlink 'is' uses the whole formula so no precision is lost. I truncated the result for clarity, but it's a fair point. – MarcosMFlores Nov 08 '20 at 23:28
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    Please fix your MathJax. An equation isn’t rendering. – G. Smith Nov 08 '20 at 23:38
  • Thanks! Should be fixed now. – MarcosMFlores Nov 08 '20 at 23:49
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    The input to WolframAlpha is incorrect... don’t space out the digits. Use https://www.wolframalpha.com/input/?i=%281%2FSqrt%281+-+50%5E2%2F%28299792458%29%5E2%29+-+1%29 or alternatively https://www.wolframalpha.com/input/?i=Cosh%28arctanh%2850%2F299792458%29%29-1 – robphy Nov 09 '20 at 05:18
  • @robphy Thanks for the heads-up. I've corrected the links. – MarcosMFlores Nov 09 '20 at 05:57
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    You’re missing a $-1$ in your substitution for $\gamma$ in $\gamma-1$ (just after the WolframAlpha link). – robphy Nov 09 '20 at 07:07