According to Einstein’s equation $$E=mc^2$$ Matter can be converted into Energy. An example of this is a nuclear reaction. What happens to the matter in the process? Do the atoms/subatomic particles just vanish? Any insights into this process are appreciated.

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    $\begingroup$ Energy is always conserved. Rest energy can be converted to other types of energy. Did you do much research on this? $\endgroup$
    – user234190
    Commented Jan 15, 2020 at 4:47
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    $\begingroup$ If mass is converted to energy it can be either photons or kinetic energy of other mass, which is also heat. $\endgroup$
    – Yukterez
    Commented Jan 15, 2020 at 4:53
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    $\begingroup$ The equation does not say that matter can be converted into energy. I doesn't even say that mass can be converted into energy. What it says is that mass and energy are the same thing. Energy has inertia. It takes a somewhat stronger force to accelerate a body when it's warm than when it's cold. $\endgroup$
    – md2perpe
    Commented Jan 15, 2020 at 20:06
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    $\begingroup$ When matter is converted into energy, it is converted into energy. $\endgroup$
    – Aron
    Commented Jan 16, 2020 at 4:09
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    $\begingroup$ I think this is a good question: The OP is wondering what happens to the actual fundamental particles when mass is converted to energy. $\endgroup$ Commented Jan 16, 2020 at 7:03

6 Answers 6


Much of mass is just binding energy, so in a chemical reaction the electrons rearrange themselves and energy is released and the total mass of the molecules goes down (in an exothermic reaction, for example).

The same is true for common nuclear reactions like spontaneous fission of uranium, with the caveat that some important nuclear reactions do involve the changing of fundamental particles, e.g., beta decay:

$$ n \rightarrow p + e^- + \bar\nu_e $$

Here the mass of the RHS is less than that of LHS, so the electron and antineutrino are energetic (in the neutron rest frame). The energy comes from the difference in the neutron (936.6 MeV) and (938.3 MeV) proton masses, which is about 1.3 MeV.

That 1.3 MeV is not binding energy, rather it is the quark mass difference coming from:

$$ d \rightarrow u + e^- + \bar\nu_e $$

where the up (down) quark mass is $2.3\pm 0.7 \pm 0.5$ MeV ($4.8\pm 0.5 \pm 0.3$ MeV). Note that most of the nucleon mass is once again binding energy.

So what happens to mass when the down decays into an up? Nothing. Mass is not stuff, mass is a quadratic coupling to the Higgs boson that leaves finite energy at zero momentum:

$$ E = \sqrt{(pc)^2 + (mc^2)^2} \rightarrow_{|p=0}=mc^2$$

or in other terms free particle waves (divide by $\hbar$):

$$ \omega = \sqrt{(kc)^2 + (mc^2/\hbar)^2}\rightarrow_{|p=0}=mc^2/\hbar$$

it is just finite finite frequency at zero wavenumber.

How does it work that matter is not stuff and is just a quanta in the quantum field? While we are generally comfortable with the photon being a quanta in the electromagnetic field which can appear and disappear at will (provided at least energy and momentum are conserved), this is because we view the EM field as a fundamental object. Moreover, the photon is its own antiparticle, and is neutral, so the conservation of charge and other quantum number does not always come to mind. They are also boson, so we don't think of them as "stuff" because they can be in the same quantum state.

Well so is the quark field also fundamental, and its quanta are up and down quarks (and more), and in the case of beta decay, the $W$ boson changes the quark's flavor (and charge, and other quantum numbers) such that at rest, the initial quanta ($d$) has a stronger coupling to the Higgs than the final quanta ($u$), which we see as more mass in the initial state and hence more kinetic energy in the final state.

Of course, you may still see the quarks as "stuff" and consider the Higgs coupling as some sort of binding energy leading to mass. That's OK.

So let's look at electron-positron annihilation:

$$ e^+ + e^- \rightarrow 2\gamma $$

Here, matter (2 leptons) just disappear and turn into 2 gamma rays. The initial state really acts like stuff: they are fermions, they cannot be in the same quantum state, they have mass, they are have charge, and so on. Enough electrons make a lightning bolt: that is very real.

But the electron and positron are both quanta in the electron-field. That is it. They have opposite charge and lepton-number (electron number), so they can annihilate without violating any conservation laws. The initial state mass is just energy at zero momentum, it is not something more "real" or fundamental than the electron field itself. The rest energy is now available for the 511 KeV gamma rays, which is just 2 quanta in the EM field (and it is charge that couples the two fields).

So in summary: all matter is made from quanta in quantum fields, and the fields are the fundamental objects. Mass is just Higgs coupling. If particle and antiparticle meet, they can annihilate, in which case the mass disappears (or not, there can be massive particle in the final state).

Since things like baryon number and electron number are conserved, the basic particles they make (atoms), in the absence of antimatter, appear stable and look like "stuff". But fundamentally: they are neither.

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    $\begingroup$ @OrangeDog Definitely, and then as a macroscopic superfluid they act in incredibly non-intuitive ways. For me, the fact that all He4's are not just identical, but indistinguishable, only makes sense if they are a coherent collection of field quanta. $\endgroup$
    – JEB
    Commented Jan 15, 2020 at 17:10
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    $\begingroup$ Lead-208 might be a better example, as (AFAIK) nobody's managed to make a superfluid of it. $\endgroup$
    – OrangeDog
    Commented Jan 15, 2020 at 17:13
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    $\begingroup$ I would also think that the gamma photons after the electron-positron annihilation exert the same gravitational force on their surroundings as the original "particles". The overall mass of the system does not change at all (how could it?). That is part of the meaning of Einstein's mass/energy equivalence: They are one. $\endgroup$ Commented Jan 16, 2020 at 2:13
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    $\begingroup$ @lcv mass is conserved in a chemical reaction, but it gets unbound from the reagents and dissipated as various forms of energy. For each 20 grams of oxygen-hydrogen mixture that you burn into water, the water is 11 nanograms lighter than the original mixture. Those 11 nanograms get redistributed to the surrounding environment, primarily through kinetic energy and electromagnetic radiation. $\endgroup$ Commented Jan 16, 2020 at 13:03
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    $\begingroup$ @lcv baryon number != mass $\endgroup$
    – OrangeDog
    Commented Jan 16, 2020 at 15:45

First off, I'd like to point out, because the term comes up often, in strict usage there is no such thing as "pure energy". Energy is not a stuff, as in particles, but a number that is associated with stuff, a quantity, though that quantity can indeed be thought of as acting, given its conservation property, like a sort of intangible "stuff" that you can put some of here or there or move around in different ways. Of course, I suppose, there's some philosophical wrangling you can do there - after all, "particles" could be considered "just numbers", too, in a sense, if you just stick to a crude instrumentalist understanding. So perhaps I should say "energy is not particles", as opposed to "not a 'stuff'".

Instead of talking about "matter" turning to "energy", it is more proper to talk about mass as a form of energy, so there is no question of "conversion": mass is already an energy. What happens is we have conversion of that energy from one form to another form, and in doing so, other things have to change, and that may mean particles being rearranged or one kind of particles turn into another kind of particles.

In this case, since you talk of "pure energy" and I am well in touch with how people talk of things colloquially, I presume what you're thinking of is annihilation - such as between a particle and its antiparticle. Well, what happens there is the second kind of process: the old particles disappear and simultaneously (at least, with the usual caveat that we cannot watch quantum processes, but the maths for what we can run of the inbetween tell us that if anything there can be called the "annihilation event", "simultaneous" ain't too bad in dealing with it) new particles appear. If it's an electron and a positron, you get two photons. If it's a proton with antiproton, you get (ultimately) some photons and also neutrinos. The latter actually have some mass, but the former do not. All of them have energy, though, and the energy is conserved throughout.

What happens is that in these cases, the rest (or "mass") energy of the particles, proportional to their mass, is converted into mass and also kinetic energy of other particles. In the electron-positron case, the final energy is purely kinetic, as in a sense all of photons' energy is kinetic energy, while in the proton-antiproton case, the neutrinos have a little bit of rest mass energy, but the overwhelming majority of the energy is still kinetic in this case as well.

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    $\begingroup$ Unfortunately the title of my question was just energy and not pure energy. I believe it was edited to say pure energy. $\endgroup$ Commented Jan 15, 2020 at 14:13
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    $\begingroup$ "mass is already an energy". This is what confuses a lot of people, but it's right there in the equation. There is an equality sign in E=mc², not an arrow as there is in chemical reactions. $\endgroup$
    – Suppen
    Commented Jan 16, 2020 at 7:56
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    $\begingroup$ @Suppen does V = kT mean that "temperature is a volume"? $\endgroup$
    – OrangeDog
    Commented Jan 16, 2020 at 15:25
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    $\begingroup$ @OrangeDog : $k_B T$ is not a volume. It is an energy. Temperature is not an energy, either, but in the very specific case of an ideal gas, it is directly proportional to the kinetic energy of its (idealized) molecules. The energy $k_B T$ is on the order of (but not exactly!) the average such for a single molecule. It also tends to be a good characteristic energy dimension for thermodynamic systems in general, but one must not make the error of equating temperature straightforwardly with energy, as it is not. $\endgroup$ Commented Jan 17, 2020 at 5:29
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    $\begingroup$ I know. Tell @Suppen $\endgroup$
    – OrangeDog
    Commented Jan 17, 2020 at 8:17

Einstein, in his famous but rarely-read paper "Does inertia of a body depend upon its energy content?", didn't write that matter (or, better, mass) can be converted into energy. The way he stated its result was: $$ \Delta E = \Delta m \cdot c^2, $$ i.e. in the rest frame of any system, variations of mass and variations of energy (of the system) are proportional. Such a result is unconditionally valid, and then it is applicable also to the case where the initial or final mass in $\Delta m$ is zero.

However the relation does not say anything about the mechanism or even if a process with initial or final massless state is actually possible. That is not the task of a conservation principle like this.

Actually, the first application of Einstein's relation to nuclear physics, made by Lise Meitner to estimate the energy obtained in one of the first fission processes obtained in lab, was not related to any annihilation process, but it connected the mass defect between the initial nucleus and the fission fragments to the kinetic energy of the fragments in the center of mass frame. Once again, Einstein's relation does not say anything about which fission process is possible. It is just establishing a connection between mass and energy variations.

  • $\begingroup$ and certainly nothing is said about the efficiency of that relation ... we "know" that supermassive black holes reach an efficiency of about 43% (that is the plasma-beams and gamma- and x-rays + visible light being ejected from time to time ...) while the remaining ~57% more or less end up adding to the mass of the black hole (the actual "eating" process ) $\endgroup$
    – eagle275
    Commented Jan 15, 2020 at 16:38

To start with the $m$ in $E=mc^2$ is the relativistic mass, and in particle physics this is out of use it causes confusion. One uses four vectors , where the "length" of the four vector is the invariant mass, uniquely identified with elementary particles, and with systems of elementary particles.

Four vectors are good in keeping track of energy, as one of the components of the four vector is the energy.

$$\overrightarrow{P} = \left[\begin{matrix}E\\ p_xc\\ p_yc\\ p_zc\end{matrix}\right] = \left[\begin{matrix}E\\ \overrightarrow{p}c\end{matrix}\right]$$

It is vector algebra and allows rigorous calculations.

The elementary particles in the table of of particle physics have fixed masses seen in the table and their four vector always has the length of that mass:

$$\sqrt{P\cdot {P}} = \sqrt{E^2 - (pc)^2} = m_0c^2$$

What happens to the matter in the process? Do the atoms/subatomic particles just vanish

All matter is composed out of elementary particles, in bound systems where there are a large number of four vectors to add to get to the invariant mass of the composite system. Look at how complicated a proton is


That is why it has a large mass of almost 1000 Mev whereas the constituent quarks and antiquarks have masses of mev and the gluons zero mass. The added four vectors give the mass of the proton ( in a complicated to calculate way)

There are quantum number conservation rules that matter composed out of elementary particles has to obey. Charge, baryon number lepton number etc have to be conserved. So one way of turning mass to energy is by annihilation, particle hits antiparticle , quantum numbers add up to zero (by antiparticle definition)and then other pairs of particles and radiation can appear ,taking away kinetic energy. Proton antiproton annihilating even if they have small momenta, create a large number of pions, of course following quantum number conservation.

In nuclear physics because of the binding energy curve of the periodic table of elements ,

binding energy

It is possible to get energy by fission or by fusion of particular elements, which has led to the atomic bomb and the hydrogen bomb. The four vectors representing a nucleus can break up into four vectors of other nuclei and release energy in the form of radiation or kinetic energy of the new nuclei.

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    $\begingroup$ Without specifying what frame reference one is using it is impossible to say if $m$ is the relativistic mass or the invariant mass (what you indicate as $m_0$). I assumed, as is nowdays costumary, that the reference frame is the center of mass frame and that $m$ is the invariant mass. Moreover, Einstein obtained the relation working in the center of mass system, and that refeence frame is the mosto obvious if one has to transform binding energies into mass. $\endgroup$ Commented Jan 15, 2020 at 13:51
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    $\begingroup$ @GiorgioP there is a very specific formula for relativistic mass dependent on γ en.wikipedia.org/wiki/… , $m_{rel}/m_0$=γ$ , I wanted to make it clear for readers accessing the question. $\endgroup$
    – anna v
    Commented Jan 15, 2020 at 16:02
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    $\begingroup$ This is why you should probably use momentum, and discard the concept of "relativistic mass". $\endgroup$
    – OrangeDog
    Commented Jan 16, 2020 at 15:29

You are basically asking about mass energy equivalence, and nuclear reactions. And you would like to know what happens to the (I assume rest) mass and matter in the process.

The mass–energy formula also serves to convert units of mass to units of energy (and vice versa), no matter what system of measurement units is used. However, use of this formula in such circumstances has led to the false idea that mass has been "converted" to energy. This may be particularly the case when the energy (and mass) removed from the system is associated with the binding energy of the system. In such cases, the binding energy is observed as a "mass defect" or deficit in the new system. The difference between the rest mass of a bound system and of the unbound parts is the binding energy of the system, if this energy has been removed after binding. For example, a water molecule weighs a little less than two free hydrogen atoms and an oxygen atom.


There are basically two types of reactions that are very interesting in this case to understand how the process works:

  1. fission

The mass of the atomic nucleus us less then the sum of the masses of the free constituents, protons and neutrons, this missing energy is the mass defect.

It is very important to understand that the constituents during fission remain, it is just the mass that changes because you need to add energy to the bound system to make it unbound again (separate the constituents).

The very meaning of the mass energy equivalence is in this case that the mass of the bound system is built up by constituents, and the binding energy modifies this.

As you go to lower QM level, at the level of quarks inside the proton and neutron, it is more easy to see that 99% of the mass of the proton or neutron is basically the binding energies of quarks and gluons, and only 1% is the mass of the quarks. This is where the mass energy equivalence becomes so important to understand. This is QM.

  1. electron positron annihilation

This process happens when an electron and a positron collide and at low energies, the electron and the proton cease to exist (in their original form), and the true nature of the underlying QM world is revealed, the total energies of the electron positron pair is converted into photons. Thus, what you exactly ask about what happens to the matter (electron positron) in the process is that they get transformed into a different type of form of energy, the very quanta of energy, photon.

This is where the true nature of QM is revealed and contrary to popular belief, the mass energy equivalence is beautifully represented by this process too.

The most probable is the creation of two or more photons. Conservation of energy and linear momentum forbid the creation of only one photon. (An exception to this rule can occur for tightly bound atomic electrons.[1]) In the most common case, two photons are created, each with energy equal to the rest energy of the electron or positron (0.511 MeV).[2]



Not to take away from other more detailed answers, I think there's something that might be more to your point.

Matter is not converted into energy. Mass is converted into energy.

This can be observed in, for example, nuclear fusion of hidrogen, resulting in a helium nucleus. With two protons (of, let's say, mass=1) you'd expect the resulting nucleus to have mass=2.

But it doesn't. The resulting mass is smaller than 2. That difference in mass is the energy that was released in the process.

Incidentally, you can get energy out of fusion up until you make an iron nucleus. Anything heavier than that will actually absorb energy (which is then transformed into mass, following the same principle). The consequence is that you can get energy out of fission of those heavier elements (which is how current nuclear power plants work).


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