# Conservation of Mass during Electron - Positron Annihilation

An electron has a mass of $m_e = \; 9.1094 × 10^{-31} kg$. A positron has the same mass. So during electron positron low-energy annihilation, won't the law of conservation of mass be violated?

Note: I am a high school student interested in physics and read about electron positron annihilation in a book. So if there is a flaw in a concept behind the question, please do point it out.

• No, no, why? Intermediate particles would be formed in that annihilation. May 25, 2017 at 13:17
• @WrichikBasu But radiations are produced as a result. SO where does the mass disappear? May 25, 2017 at 13:19
• @WrichikBasu I am talking about low energy collision May 25, 2017 at 13:19
• Low energy - that I'm not quite sure of. I was talking of high energy ones, where an electron and a positrion annihilate to produce a muon and it's antiparticle $\mu ^+$. Maybe others will help you. May 25, 2017 at 13:22

won't the law of conservation of mass be violated?

One has to understand that presently there is classical physics, to which belong Newtonian physics , thermodynamics and some other branches of mathematically modeling nature, and quantum physics, which models the data coming from the microcosm of particles : molecules atoms nuclei and elementary particles. Special relativity is much more important in the quantum mechanical framework. For ordinary velocities it reduces to Newtonian mechanics.

The law of conservation of mass belongs to classical macroscopic descriptions of nature. There mass is conserved.

The electron positron annihilation belongs to the quantum mechanical and special relativity framework.

In quantum mechanics and special relativity mass is not conserved; only energy and momentum are retained from the newtonian conservation laws. Mass is the "length" of the four vector describing a system of particles, called invariant mass, which is not the sum of individual masses. Invariant mass of single particle characterizes it uniquely.

Two 511keV photons are produced when the electron and positron annihilate. What is conserved in this case is the total energy of the system.

The rest mass energy of an electron or positron is 511keV (use $e = mc^2$) so their masses are converted to EM radiation. The angular and linear momentum of the system is also conserved as the two photons are at 180 degrees to each other.

edited to remove incorrect statement.

• I don't know of a quantity called "mass-energy"; the only place I'd expect to see this term is in "mass-energy equivalence". Do you mean "invariant mass"? May 25, 2017 at 13:58
• Mass-energy is surely not conserved in eg e e to photon photon May 25, 2017 at 13:58
• Energy-momemtum is conservedl May 25, 2017 at 13:58
• Sorry! Total energy is conserved. Linear and angular momentum are also conserved among other things. The way I remembered it was the mass is converted to energy (mass energy equivalence). My wording wasn't very careful, my apologies. May 25, 2017 at 14:03

Mass is not a conserved quantity in the sub-molecular particle realm (so, in everyday life it's not conserved, but it almost is, and that's good enough for most household chemists).

Energy is a conserved quantity as is momentum, and that conservation includes the possibility that mass either appears or disappears in the system. For particle reactions (like electron-positron annihilation, or its reverse, pair-production) we use a calculated quantity which is invariant under Lorentz transformations: $$E^2-\left(\vec{p}\cdot\vec{p}\right)c^2,$$ where $E$ is the total energy of the interacting system including mas energies, and $\vec{p}$ is the total momentum of the system.

Let's consider your low-energy system of an electron and positron in a reference frame where the electron is at rest and the positron has kinetic energy $\left(\gamma-1\right)mc^2$ and momentum $\gamma mv$. We shouldn't use Newtonian forms because we are including mass-energy and producing photons. The quantity $\gamma$ is just a shorthand symbolic way of writing the quantity $$\gamma=\left(1-\frac{v^2}{c^2}\right)^{(-1/2)}=\left(1-\beta^2\right)^{(-1/2)},\text{ where }\beta=\frac{v}{c}.$$ We can also write the momentum as $p=\gamma m\beta c.$

Then we have $$E=2mc^2+\left(\gamma-1\right)mc^2=mc^2\left(\gamma+1\right).$$

For our conserved quantity we get $$m^2c^4\left(\gamma^2+2\gamma+1\right)-\left(\gamma^2m^2\beta^2c^2\right)c^2.$$ With some simple algebra one can show that $\gamma^2\beta^2=\gamma^2-1,$ so we can write $$m^2c^4\left(\gamma^2+2\gamma+1\right)-m^2c^4\left(\gamma^2-1\right)=2m^2c^4\left(\gamma+1\right).$$

Now, if we consider a system of two photons with energies $E_1$ and $E_2$ and corresponding momenta magnitudes $p_1=E_1/c$ and $p_2=E_2/c$, we can write our invariant quantity, keeping in mind that momenta are vector quantities, $$\left(E_1+E_2\right)^2-\left(\vec{p}_1+\vec{p}_2\right)\cdot\left(\vec{p}_1+\vec{p}_2\right)c^2$$ $$E_1^2+E_2^2+2E_1E_2-\left(p_1^2c^2+p_2^2c^2+2p_1p_2c^2\cos\theta_{12}\right),$$ where $\theta_{12}$ is the angle between the photon momenta. Because each $pc$ term is a photon energy $E$, the invariant quantity of these two photons (which is also the square of the invariant mass or square of the length of the momentum-energy 4-vector) becomes $$2E_1E_2\left(1-\cos\theta_{12}\right).$$

Now we can equate these two quantities to see if the system of the positron and electron can produce two photons without leftover individual mass: $$2E_1E_2\left(1-\cos\theta_{12}\right)=2m^2c^4\left(\gamma+1\right).$$

We could certain chose numbers to make this equation work, so the next thing is to see what the experiment provides. An experiment with a positron emitter such as $^{~22}$Na shows two equal energy photons ($E_1=E_2$) emitted in directly opposite directions ($\cos\theta_{12}=\cos\pi=-1$) with energies equal to the mass-energy of an electron (511 keV): $$4E_1^2=2m^2c^4(\gamma+1)$$ which tells us that $\gamma=1$, or that positron is also at rest when it interacts with the electron.