# The norm of the vector sum of two 4-momentum vectors before and after pair production

Two photons traveling along the x-axis (in a lab frame of reference) of different frequencies are about to collide. Their 4-momentum vectors are (h$\nu_1$/c , h$\nu_1$/c , 0, 0) and (h$\nu_2$/c , -h$\nu_2$/c , 0 , 0). Each are null vectors. The norm squared of the vector sum of the two vectors yields 4$h^2\nu_1\nu_2/c^2$ using the (+---) metric. If the sum of the energies of the two photons is greater than 1.022 MeV and pair production results then the norm squared of the vector sum of the 4-momenta of the positron and the electron after the interaction in a frame of reference where their total momentum is 0, would yield 4$m_e^2c^2$. Since the morm of the 4-momentum is lorentz invariant and a conserved quantity, this would lead to a useful result like calculating the minimum energy needed for a gamma photon interacting with a CMB photon to lead to pair production.

I understand mathematically that it is possible to get a vector whose morm is not 0 from the addition of two null vectors. However, I have some trouble understanding from a physics perspective how two photons about to collide would have a non-zero rest mass, at least in some lab frame of reference before the collision.

I am guessing that the interaction is a process, and not simply vector addition at a point in time.

• The squared norm of the vector sum of 4-momenta is at at least $4m_e^2c^4$ but can have any value greater than that too. Commented Jun 12, 2015 at 17:48

The mass of a system is just $c^{-2}\sqrt{E^2-c^2\vec p^2}$ where $E$ is the total energy and $\vec p$ is the total momentum, it is never the sum of the masses of the parts.

So why did we ever think it was? The formula $c^{-2}\sqrt{E^2-c^2\vec p^2}$ is very similar to the formula $\sqrt{x^2+y^2+ z^2}$ for geometric length and it is really like a different kinda of length for a different kind of geometry. Which means the following geometrical fact still holds.

The length of a sum of a vectors is approximately the sum of the lengths when the vectors point in almost the same direction.

When masses are moving slowly relative to each other then their energy momentum vectors $(E,c\vec p)$ point in almost the same directions so the mass (length) of the sum of those vectors is approximately the sum of the masses (sum of the lengths).

But that geometric fact tells you nothing about a situation where vectors aren't pointing in the same direction. And you expect it to fail and fail badly because when things point in wildly different directions the sum of the lengths is totally different than the length of the sum.

If you accept that mass is the length of the energy momentum vector (it is the energy divided by $c^2$ in the frame where the momentum is zero) then it shouldn't surprise you. In fact if you just replaced the word mass with "energy divided by $c^2$ in the frame where the momentum is zero" most sentences would make total sense, the only real problem is when there is no frame but then we say no mass. No frame no mass. So take those both and you are set.

Expecting mass to be something else will hurt you. Thinking mass is something else will hurt you. But realizing it is a kind of length of a vector is fine. And using "energy divided by $c^2$ in the frame where the momentum is zero" really tells you that when you wanted to use mass you probably really should have been using energy and the physical reason something happens.

Having different masses just tells you that you have a different balance between energy and momentum. For zero mass you have equal amount of both, for positive mass you have more energy than momentum and the value of mass tells you how much more. But it isn't as simple as adding them, you get $E^2=(c\vec p)^2+(mc^2)^2$ instead.

The mass of a system (length of total energy-momentum) does not equal sum of the masses of parts. Don't expect it to. Your experience dealing with energy-momentum vectors that point in almost the same direction simply didn't prepare you for situations where they do not.

• I see. The invariant mass of a system includes contributions from the kinetic energy of each particle even if their respective rest-masses are 0. To be in the center of mass frame (total momentum 0) of the two photons before collision, would you need a frame that is moving as such that the doppler shift would result in equal frequencies for each of the photons? Commented Jun 12, 2015 at 18:39
• @mv311 You are correct. Commented Jun 12, 2015 at 20:21