# Feynman Diagrams: Mass Conservation

Can a Feynman diagram end with more matter than it began with? The answer should be no because it would break the law of conservation of mass, right?

Example: I learned that two up-quarks could interact to create a Higgs boson (via what I learned as "vector fusion"). If this new Higgs were to decay into an up-quark pair, wouldn't the new up-quark be able to interact with the old one to produce another Higgs, creating a potentially infinite amount of particles?

I'm new to this site (and physics), so if this is unclear, please comment.

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Remember, there is no law of conservation of mass. It's only energy and momentum that are conserved. Plus, what do you mean by "more matter"? If it's number of particles, that's not a problem - a collision between two particles will routinely produce 50 products. –  David Z Nov 6 '13 at 17:25

I believe that what you are suggesting would violate causality. The original up-quarks are spacelike separated from the final ones, so they couldn't possibly interact.

Your observation that this would contravene the conservation of mass is in fact a deep one. Particle interactions like the one you describe are usually considered in the context of quantum field theory (QFT). Here the theory is explicitly constructed to preserve the symmetry group of special relativity, namely the Poincare group. By Noether's theorem, such global symmetries have associated conserved quantities.

One conserved quantity of interest is the total $4$-momentum, which remains the same throughout a process. The associated symmetry is spacetime translation. The total mass is defined to be the square of the $4$-momentum, so must be preserved in any QFT interaction.

To conclude, the process you describe is impossible by symmetry considerations, either from the perspective of causality, or via the conservation of mass.

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However even in QFT, this would be impossible right? –  Arc676 Nov 5 '13 at 15:20
Yes that's just my point - it's impossible in QFT because QFT has the requisite symmetries built into it. –  Edward Hughes Nov 5 '13 at 17:26
To whoever downvoted, fancy telling me why? Cheers! –  Edward Hughes Nov 5 '13 at 21:11
I would like to discuss this with you in chat chat.stackexchange.com/rooms/71/the-h-bar –  Arc676 Nov 6 '13 at 16:30
Ah, I just realized what may be going on here: the original question was about vector boson fusion between two Z bosons emitted by up quarks. But I think your answer is addressing the case where the up and anti-up actually annihilate with each other to create the Higgs. I suspect that's where the confusion comes from. –  David Z Nov 8 '13 at 21:12

Your first port of call might the relevant wiki page, Conservation of mass. Though I admit that the page is confusing because it includes the "relativistic mass" and "rest mass" terminology, e.g. "For the special type of mass called invariant mass$\ldots$" - I suggest it is rewritten.

It was found that in classical physics, mass is conserved, which tallies with our intuition. That is, in a process $a,b,c\ldots\to \alpha, \beta, \gamma\ldots$ $$\sum_{i \in a,b,c\ldots} m_i = \sum_{i \in \alpha,\beta,\gamma\ldots} m_i.$$

With relativistic physics, however, we have a notion that "matter can be converted to energy, and vice-versa i.e., (rest) mass, $m_0$, is not conserved". "Relativistic mass", $\gamma m_0$, however, is conserved. These kinds of masses confuse many physicists and are somewhat old-fashioned. It is probably easier to understand if I write, equivalently, that "energy is conserved".

Furthermore, one can also say that "invariant mass" is conserved. Careful with this statement, though, because the invariant mass of a group of objects is not the sum of their individual invariant masses. That is, in a process $a,b,c\ldots\to \alpha, \beta, \gamma\ldots$, if invariant mass is conserved, $$\left(\sum_{i \in a,b,c\ldots} E_i\right)^2 - \left(\sum_{i \in a,b,c\ldots} p_i\right)^2= \left(\sum_{i \in \alpha, \beta, \gamma\ldots} E_i\right)^2 - \left(\sum_{i \in \alpha, \beta, \gamma\ldots} p_i\right)^2$$ where $E$ is energy and $p$ is momentum.

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