2019-02-08 16:11:20 8 Comments

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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## 3 comments

## @Salvador Villarreal 2019-02-08 17:05:36

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

Obviously,

non-relativisticis 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.

## @Cort Ammon 2019-02-08 22:35:05

The measurement accuracy required by GPS astonishes me, doubly so when I'm reminded it appears in cheap cell phones.

## @Salvador Villarreal 2019-02-09 01:46:03

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.

## @Loren Pechtel 2019-02-09 03:19:40

Are you sure of that 10^-7?? I thought it was more like 10^-12.

## @Salvador Villarreal 2019-02-09 03:37:45

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.

## @jpmc26 2019-02-09 06:06:15

@LorenPechtel Could it be that different quantities in the calculation need higher accuracy and precision?

## @Loren Pechtel 2019-02-10 03:11:50

@SalvadorVillarreal If I'm following you right that you seem to be considering the velocity of the satellite as if it was the minimum velocity that would matter.

## @Jawad 2019-02-10 08:30:42

@LorenPechtel I think the 10^-12 refers to the accuracy of the atomic clocks while SalvadorVillarreal is talking about gamma

## @Salvador Villarreal 2019-02-11 00:42:44

Sorry for delaying so much to answer. I didn't find the notes I told you about. I did the calculation for the accuracy required for GPS and @Loren_Pechtel is correct. First, note that the distance from the satellite can be expressed as $d=(t_R-t_S)c$, where $t_R$ is the time at which the device receives a signal by the satellite, $t_S$ is the time at which it's sent (encoded in the message) and $c$ is the speed of light in Earth's atmosphere. Also, we have that $t_S$ can be expressed as $t_S=t_0+t_C$, where $t_0$ is a calibration time and $t_C$ is a time calculated on the satellite.

## @Salvador Villarreal 2019-02-11 01:06:27

We also have that $t_C=\tau \gamma$ where $\tau$ is the proper time of the satellite (the time measured on the clock of the satellite). Using these, we can get the expressions for the uncertainty of every quantity: $\Delta d=c \Delta t_S$ ($c$ has no uncertainty because it's a defined constant, and $t_R$ is a datum given by the device), and since $t_0$ is constant, we have $\Delta t_S =\Delta t_C$, where $\Delta t_C$ is given by $\Delta t_C = \gamma\Delta\tau +\tau\Delta\gamma$. From there we have to substitute, I took all data from the internet so I apologise if anything is mistaken.

## @Salvador Villarreal 2019-02-11 01:07:45

The horizontal accuracy of GPS is of about $4m$. I calculated $\Delta d$ by splitting Earth's surface area among the 24 GPS satellites orbiting earth in squares. We have that the furthest points are the ones with most uncertainty $\Delta d_{horizontal}= \Delta d sin(\theta)$. So that the smallest $\Delta d$ required is $27m$. Also $\Delta\tau=10^{-10}s$ taken from the typical error of Rubium atomic clocks. Substituting everything we get that $\tau\Delta\gamma=9\times 10^{-8}$.

## @Salvador Villarreal 2019-02-11 01:11:00

As you can see, as time passes $\tau$ increases and we need higher precision on $\gamma$, if recalibration is done every day we arrive at the mentioned precision of $10^{-12}$ in $\gamma$. I suppose that actual GPS systems have algorithms that help giving a little more precision that the simple calculation I provided, but the actual value should not difference by that much. Also Loren, I apologise if my comment was misleading and only focused on the velocity, now the explanation is a little more complete, I hope it's clearer.

## @RogerJBarlow 2019-02-08 16:19:33

'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.

## @John Rennie 2019-02-08 16:38:44

When we say a particle is

non-relativisticwe 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-relativisticbehviour.