# Is it possible to create matter? [duplicate]

Is it possible to create matter? In a recent discussion I had, it was suggested that with enough energy in the future, "particles" could be created.

It seems like this shouldn't be possible due to conservation but perhaps I could be wrong. Would any of you Physics masters care to elaborate?

(Note... I will understand the basics, I am by no means an expert.)

• You sure you wanna ask this? – user3459110 May 23 '14 at 6:19
• @AwalGarg haha I'm doing my best to understand – codedude May 23 '14 at 13:43
• Conservation applies to matter and energy. Converting one into the other is permitted, though making it happen is difficult and creating any significant amount of matter requires a huge amount of energy, as documented by Einstein's equation. – keshlam May 24 '14 at 0:45

The pair production process $$\gamma \to e^+ + e^-$$ where a photon creates a positron and an electron is allowed based on conservation of electric charge and number of leptons (number of particles minus number of anti-particles). However it is forbidden by relativistic kinematics: the right-hand side has a rest frame but the the left-hand side does not, so momentum cannot be conserved in the process. Pair production in the vicinity of for example an atomic nucleus, $$\gamma + Ze \to e^+ + e^- + Ze$$ is allowed since the atomic nucleus can absorb the recoil.

Naturally from conservation of energy the photon energy must exceed a minimum value $E_\text{min}$, since the particles created have rest mass. Now $E_\text{min}$ is not quite $2m_e$, twice the electron mass, since there exists a bound $e^+e^-$ state with an energy slightly lower than $2m_e$, but this correction is $\approx 7\; \text{eV}$ and $2m_e \approx 1\; \text{MeV}$.

This process can be observed in the lab with some fairly basic equipment. You need a radioactive sample that emits $\gamma$-rays above $E_\text{min}$ (preferably at much higher energies, say $\approx 5 E_\text{min}$), a piece of dense metal (like lead) and a photon detector. Pair production will take place in the lead, and you can observe a peak of photons with energies close to $m_e$. They come from the positron created annihilating with an electron, $$e^+ + e^- \to \gamma + \gamma$$ which is most likely to occur when the particles are at relative rest, giving each photon an energy of $m_e$.

(The inverse process $$\gamma + \gamma \to e^+ + e^-$$ is also allowed but it is much harder to create in the lab. You need a very, very big laser.)

• Soooooo....yes? – codedude May 23 '14 at 2:05
• – BMS May 23 '14 at 2:31
• Also, pair production in presence of strong magnetic field applies here, but I don't know if it is feasible on Earth. – auxsvr May 23 '14 at 7:22
• Inverse process probably will be recreated soon: theguardian.com/science/2014/may/18/… – user11153 May 23 '14 at 8:37

Einstein's famous equation, $e=mc^2$ states that it is possible to convert matter into energy. It can then be deduced that it also works the other way around. Otherwise, the amount of energy would increase in the universe, while the amount of matter decreased (very, very, slowly).

• Counts of the total energy in the universe include mass energy anyway. So actually what would happen is that the amount of energy would stay the same while the amount of matter decreased. I think this is a plausible end-game for an open universe: everything ends up as photons (and maybe neutrinos so there's some rest mass left? Not sure). But I'm not a cosmologist and anyway even if that is possible it doesn't mean that the conversion is one-way, just that this argument for why it isn't one way doesn't clinch things :-) – Steve Jessop May 23 '14 at 16:32

To add to Robin Ekman's answer, the answer is "yes" as stated there, but you need to be very careful these days with the word "matter" - this is a word that is becoming outmoded in physics as we understand more and more that everything is made of quantum fields. So the pair production process

$$\gamma + \gamma \to e^+ + e^-$$

is better understood simply as a change of state of quantum fields: we withdraw two photons from the photon field, whilst adding a positron and electron to the electron field.

Incidentally, the easiest way to do pair production $\gamma + \gamma \to e^+ + e^-$ in the lab is with a Tesla coil: if you're handy with a lathe and mechanical construction and are meticulous with laying down insulation, you can make a 1 million volt one in your own backyard as a friend of mine has - his is very like the design shown here. I'd reckon that, accelerating ions to a million electron volts, you will get significant $511keV$ $\gamma$ production. These $\gamma$ in turn will produce pairs: the rest energies of the electron and positron are $511keV$ each. I certainly would not recommend this: the electric hazard is huge and I can't see why the radiation hazard wouldn't be significant, so I stay away from my friend's house when he is playing with his coil.

The dichotomy between "matter" and "energy" is an outdated one: the word "matter" is disappearing in particle physics as too imprecise and now all "particles" have a property called energy which is simply one component of the momentum four vector, so that the "energy" property $E$ for a given particle:

$$E^2 \,c^2= p^2 + m^2 \,c^4\tag{1}$$

where $p^2 = \vec{p}\cdot\vec{p}$ is the squared magnitude of the everyday momentum vector $\vec{p}$. We don't speak of something being energy anymore. Some particles, like photons, have zero rest mass $m$ and they are always observed to be travelling at the speed of light with a momentum $|p| = E / c$. These are the ones that people used to talk about as being pure energy in Einstein's early days. Some particles have nonzero rest mass and can be at rest in your frame, like an electron. In a frame at rest relative to the particle in question, their energy is $E = m\,c^2$, which I'm sure you've seen before and is a special case of (1) above. This famous equation is in general, when the particle has nonzero momentum, incorrect (or, more fairly, not the full picture)!

On the other hand, people who study relativity call anything with an energy property "matter": it affects the Einstein Field Equations (EFE) in pretty much the same way whether or not it has rest mass and behaves gravitationally pretty much as a Newtonian mass of $E/c^2$ (although there are subtleties: the right hand side of the EFE is a tensor, not a simple scalar $E/c^2$, but the latter gives a good idea of what's going on).

If you look up the word Matter on Wikipedia you'll see the confusion explained in more detail. I wish henceforth that everyone would simply use the word "stuff" for anything that can be construed as a non ground state of a quantum field.

• Generally good answer, but I highly doubt that accelerating ions in a $1 MV$ field will create a significant ammount of $511 keV$ photons, let alone the $1 MeV = 2 * 511 keV$ photons you need for pair production... – Neuneck May 23 '14 at 11:07
• Indeed - while our various MV accelerators do produce radiation, most of it is from the field accelerating a few electrons, which then produce x-rays when they hit something. Gammas can come from slamming MeV range ions into the appropriate target, which is used for compositional analayis, called PIGE, particle induced gamma emission. – Jon Custer May 23 '14 at 13:56

If we're allowed to use some matter as well as energy (to create more matter than you started with) then the answer is yes.

If we're not allowed to use matter - we have to make the matter out of nothing but energy - then strictly speaking the answer is that we don't know. But we might soon.

We're pretty much certain that it is possible; it would be astonishing to discover otherwise. But we don't really know until we've done the experiment.