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For example, nucleons in nucleus are in motion with kinetic energies of 10 MeV. Their rest energies are about 1000 MeV. Kinetic energy of nucleons is small compared to rest energy. They are hence considered non-relativistic.

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When we say a particle is non-relativistic we mean the Lorentz factor $\gamma$ is close to one, where $\gamma$ is given by:

$$ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} $$

So saying $\gamma$ is close to one means that the velocity $v$ must be much less than $c$.

With a bit of algebra we can show that the kinetic energy of a particle is given by:

$$ T =(\gamma - 1)mc^2 $$

And the rest mass energy is the usual $mc^2$, so if we take the ratio of the kinetic energy to the rest mass energy we get:

$$ \frac{T}{E} = \frac{(\gamma - 1)mc^2}{mc^2} = \gamma - 1 $$

And if this ratio is small that means $\gamma \approx 1$, which was our original criterion for non-relativistic behviour.

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I would like to add something to the already great answers posted.

Obviously, non-relativistic is a qualitative term, you can translate it to "relativistic effects are so small that they're negligible in this problem".

In the particular case you're talking about, and as was pointed out by Roger JBarlow and John Rennie, you can calculate the Lorentz factor to be $\gamma=1.01$. This means you are going to have measurement errors on the order of $10^{-2}$. In some fields this may be acceptable (it would be beyond amazing in fluid mechanics) , but I recall a great professor I had on relativity (he works in numerical relativity, and is one of the leading figures on the field, at least in my country) who said "If the errors are on the order of $10^{-4}$, the results are basically useless". This is further illustrated by the fact that accurate GPS measurements rely on accurate calculation of relativistic effects which are (if I recall correctly) on the order of $10^{-12}$, and would otherwise give errors of kilometers.

The bottom line is that the question "is this particle non-relativistic?" Is basically the same as asking "is $\gamma$ close enough to 1 so that I can just assume it's 1?". This will change depending on the problem under consideration.

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    $\begingroup$ The measurement accuracy required by GPS astonishes me, doubly so when I'm reminded it appears in cheap cell phones. $\endgroup$ – Cort Ammon Feb 8 at 22:35
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    $\begingroup$ It is pretty amazing. And now that you bring it up, there are some amusing details to the story: When the GPS was first going to be introduced, the engineers in charge of the project were not going to take relativistic effects into account. It seems like one physicist warned the they should and they reluctantly agreed. They had the two versions, just to check how much it really affected... After a few months they checked back, just to realise that the error was already of a few km. The errors add up over time because accurate GPS requires a good knowledge of the positions of the satellites. $\endgroup$ – Salvador Villarreal Feb 9 at 1:46
  • $\begingroup$ Are you sure of that 10^-7?? I thought it was more like 10^-12. $\endgroup$ – Loren Pechtel Feb 9 at 3:19
  • $\begingroup$ I need to check my notes from back then, but a quick calculation of $\gamma$ (taking the velocity of 14000 km/h found on the web) gives 1.00000648. The difference is on the order of $10^{-6}$. Maybe the calculation involved a factor of $\sqrt{\gamma}$, but I honestly don't remeber. I will check and change the answer if necessary. $\endgroup$ – Salvador Villarreal Feb 9 at 3:37
  • $\begingroup$ @LorenPechtel Could it be that different quantities in the calculation need higher accuracy and precision? $\endgroup$ – jpmc26 Feb 9 at 6:06
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'Non-relativistic' means $v\ll c$.

That is effectively the same as $\gamma \approx 1$ as $\gamma={1 \over \sqrt{1-v^2/c^2}}$.

But also $\gamma={E_{tot}\over E_{rest}} \equiv 1+{E_{kin} \over E_{rest}}$

So if the kinetic energy is small compared to the rest mass, $\gamma$ is only slightly bigger than 1, and $v/c$ is small. And one is justified in ignoring relatistic effects.

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