# Can particles be created from energy?

1. Dec 17, 2006

### ShawnD

Before I even ask a question I'll say what I know just in case this confusion stems from not knowing the basics. My understanding is:
-Nuclear reactions have changes in mass even though all particles are accounted for.
-Fast moving particles approaching the speed of light have higher mass but still consist of the same basic components (1 proton becomes heavier, it does not become 2 protons).

Now for the actual question.
Thinking back to high school there was a thing called "mass defect" that accounted for the energy release in nuclear reactions. Does this concept of a mass and energy relationship apply to creating new particles, or only to changing the mass of already existing particles?

2. Dec 17, 2006

### LURCH

It does indeed work both ways. Converting mass form matter to energy is what happens in nuclear reactions, but converting from energy to matter turns out to be much more difficult. It takes place in particle colliders.

3. Dec 17, 2006

### mathman

Creating particles from energy can occur in many ways. The most common is pair production, where gamma rays (in the presence of a nucleus) convert to an electron-positron pair. At much higher energies, proton-antiproton pair formation is possible. Also right after the big bang, there was a lot of this going on (from gamma-gamma collisions).

4. Dec 21, 2006

### disregardthat

Well, it must then just be the energy stored into the smallest molecular particles and make them bind themselves, therefore appear to have a larger mass as we interpret it. The energy itself wouldn't have any mass would it? just like a wave in the water doesnt weigh anything, but the water does.

5. Dec 22, 2006

### vanesch

Staff Emeritus
Particles can be created and destroyed, period. The next thing is that total energy is conserved, and that for particles in "free initial or end states", the relation:
$$E^2 = m_0^2 c^4 + p^2 c^2$$ holds, with $$m_0$$ a constant that depends on the type of particle, and is called its "rest mass".

For a (meta-) stable bound state of particles which we want to consider as one entity, we can write a similar expression, and the m0 of that system will be a constant, which is associated with that bound state, and will correspond to the total energy (over c) of the bound system in its rest frame.

As such, we have to have that the total energy of a system is conserved, and if in a process, particles with finite m0 and/or momentum are created, other parts of the system must loose energy in order for the total amount to remain constant. There is no such thing as "pure energy", as it is always associated with a set of particles.

Now, what we usually call "mass" is often $$E/c$$, which is close to the restmass m0 of a particle or bound system if the velocity of the particle is non-relativistic, hence the confusion. Remember that m0 is just a constant that goes with a type of particle or bound system, and is not conserved in interactions. E is.

The "mass defect" of a nucleus is the difference between the sum of the different m0 of the constituting particles (neutrons and protons) when they are free, and the m0 of the bound system which is the nucleus. As such, there doesn't need to be any relation ! However, when we think of a hypothetical process where we start with an initial state of neutrons and protons (free and not moving) and ending up into the nucleus, at rest also, then the sum of their m0 will be equal to the total E of that initial state, and the m0 of the nucleus will be the total E of the end state. So this cannot be, as E is conserved. So some energy must be associated to OTHER particles, and this will usually be some gamma rays (photons) which carry away the needed energy in order for the sum of E to be the same before and after the interaction.

We have:
initial state:
k neutrons and l protons, with $$E_i = k m_n c^2 + l m_p c^2$$

final state: nucleus and gamma, $$E_f = E_{nuc} + E_{\gamma} =m_{nuc} c^2 + E_{\gamma}$$ and we must have $$E_f = E_i$$

(ok and there will be a small recoil too which I didn't take into account here, and changes $$E_{nuc} = \sqrt{m_{nuc}^2 c^4 + p_{recoil}^2 c^2}$$

So, particles are not really "created from pure energy", particles can be created (and destroyed), and there is conservation of total energy - as there is conservation of total momentum, but nobody says that particles are created from pure momentum.

Last edited: Dec 22, 2006
6. Dec 22, 2006

### ShawnD

Interesting replies. Thanks everybody