# Why do we use only nonrelativistic equations in nuclear physics?

What is limit between relativistic and non-relativistic equations? Which conditions do we have to use one of these?

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I'm pretty sure we use a lot of relativistic equations in nuclear physics. Especially the famous one with the mass-energy equivalency... – Lagerbaer Feb 13 '12 at 19:17

Nuclear physics, and nuclear engineering in particular, has no need to consider relativistic effects in probably the majority of calculations simply because the energies are not high enough.

To get to the root of the problem, I'll look at nuclear processes, which are state changes, fissions, and various emissions of nuclei. The energetics of these reactions is dictated by the mass difference between the reactants and the products. For nuclear engineering (I can't speak for the whole of "nuclear physics"), the total mass number $Z$ is conserved, so the difference that causes a mass deficit is found in binding energy alone. Even though the reaction energy comes from the binding energy, the emitted particles, with the exception of electrons, have at least a full proton of neutron mass associated with them. These are large rest masses compared to the mass deficits that come from binding energy.

Here is the familiar nuclear binding energy graph:

These values are per nucleon (meaning per neutron and proton) and they reach the neighborhood of $8\text{ MeV}$, but that's a relative value between Iron and Hydrogen and no practical direct transition between those exists. Generally we're talking about transitions between two different states, and if we look at $U^{235}$ versus the middle range, or around 150, then that's only on the order of $1\text{ MeV}$. This is vaguely consistent with the typical number of $200\text{ MeV}$ released in a single fission event. Take $1\text{ MeV} \times 235 \approx 200\text{ MeV}$, as a back-of-the envelope calculation.

The relevant question for whether or not a particle is relativistic is how the rest mass energy compares to the kinetic energy. When $200\text{ MeV}$ is released in a single reaction, that energy is turned into kinetic energy. What are the rest masses of the emitted particles? Here is a neutron and electron:

$$M_n \approx 900 \frac{\text{MeV}}{c^2}$$ $$M_e = 0.511 \frac{\text{MeV}}{c^2}$$

You can see that the entire energy of a fission reaction would not be enough to accelerate a neutron to relativistic energies. In practice, no single particle gets anywhere near that amount of kinetic energy (particularly the lighter ones). A neutron from a fission has in the neighborhood of $6 MeV$ kinetic energy. How does this make it relativistic? The common formula to use is:

$$E^2 = (pc)^2 + (m_0 c^2)^2$$

This is why we will measure mass in funny units that use electron volts. It's a way to put the kinetic contribution and rest mass contribution in comparable units. When I say that a fission neutron has $6\text{ MeV}$ energy, I mean that $E-m_0 c^2=6\text{ MeV}$ and when I say that the rest mass is about 900 MeV I mean that $m_0 c^2 \approx 900\text{ MeV}$. Obviously, 900 is much more than 6 in the context of a Pythagorean triple, and the $(pc)^2$ term in the above equation is the smallest of the 3 terms by far.

What about electrons? Electrons from nuclear reactions are very often relativistic. Consider the decay of a neutron. In the decay, a neutrino, electron, and proton are ejected. The electron carries off significant energy in the form of kinetic energy which compares to the $0.511\text{ MeV}$ of rest mass. If Newtonian mechanics are used in that calculation, you will find the electron to have a speed greater than the speed of light. It is possible to use relativistic momentum accounting for the electron momentum and Newtonian mechanics for the proton because the proton is still very heavy compared to the energetics of the reaction. The combination makes for some interesting algebra.

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There are several ways to typeset units in raw LaTeX (see the comments on How to correctly show units with a vector) that are better than just naively writing them in mathmode. I've applied one to your equations. – dmckee Feb 13 '12 at 19:55

It is useful in beam-based experimental nuclear physics (as opposed to the nuclear power context that Zassounotsukushi discusses) to use energies up to a few GeV. At those energies electrons are highly relativistic, and nucleons are fast enough to that one has to treat them relativistically, but heavy nuclei are generally still Newtonian.

Jefferson lab, for instance, does a fair bit of plain old nuclear physics as well as the non-perturbative particle physics that CEBAF was designed for. CEBAF---the Continuous Electron Beam Accelerator Facility---is the big accelerator there and runs at energies up to 6 GeV (with a 12 GeV upgrade in progress). There is also a high-power field effect laser.

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