# Will 1 gram of matter moving at relativistic speeds completely annihilate a larger quantity of stationary antimatter?

This is a question about the relativistic mass concept which I am having trouble understanding, mainly because of the scenario below.

Simple scenario:

Suppose 1 gram of matter is accelerated to 99% the speed of light. At this speed, the relativistic mass is 7 times greater than the rest mass. If this matter collides with a stationary quantity of 7 grams of antimatter, will the two masses annihilate completely with each other? Or will the matter just annihilate 1 grams worth of antimatter?

If the latter is true then what exactly am I overlooking about the relativistic mass concept that makes the former incorrect?

• Think about annihilation in terms of conserved quantum numbers, the most important one of which is charge. Relativistic motion does not change the charge of a body, so for every charged particle in your matter you need a charged particle of opposite charge in the antimatter. – CuriousOne May 2 '15 at 22:08
• Yet another reason why we shouldn't be teaching "relativistic mass". – dmckee --- ex-moderator kitten May 2 '15 at 22:36
• if you have 1 gram of matter accelerated to 0.99c and 1 gram of stationary anti matter then in the center of mass frame you will have the same "relativistic mass" of both. That's symmetric and that's what you need for complete annihilation. Not 7 grams – borilla May 2 '15 at 22:47
• From the matter's point of view, the large quantity of antimatter is the one moving at relativistic speeds. – BlueRaja - Danny Pflughoeft May 3 '15 at 0:56
• The same thing that happens to the rest of the energy: it gets radiated out as photons and neutrino-antineutrino pairs and maybe some other stuff. – DanielLC May 3 '15 at 4:45

A sophisticated, yet easy way to see that this the answer must be "No." is to recall that velocity is relative — that there is no absolute notion of velocity.

You said the matter was moving and the antimatter still, but that point of view (AKA frame of reference) is not privileged in any way. An observer at rest with respect to the matter has just as much right to conclude that the anti-matter is in motion as you have to conclude that the matter is moving.

So you can't rely on a velocity dependent notion of mass to work out the consequences of the scenario.

The modern approach to relativity is to define the (only!) mass of a particle or system as the square of its energy-momentum four vector (with appropriate factors of $c$): $$m = \frac{\sqrt{E^2 - (pc)^2}}{c^2} \,.$$

The thing that you you've been taught to call "relativistic mass", $\gamma m$, is (to within two factors of $c$) described as the "total energy" of the particle or system.

• "there is no absolute notion of velocity" ... : I'm considering that the Solar System was found to be travelling at V=369 km/s in relation to the CMB i.e. the distant stars. V is the SS absolute velocity in official references. – Helder Velez May 4 '15 at 13:22
• We've been over this ground before, but for the benefit of those that come later: identifiable is not the same as privileged. Local co-moving coordinates are identifiable, but they are not absolute. And there is nothing "official" about you statement. – dmckee --- ex-moderator kitten May 4 '15 at 14:14
• This answer doesn't really explain why that notion of mass (rest mass) is the relevant one. Also, there is no absolute notion of velocity, but there is an absolute notion of relative velocity between the two particles, isn't there? – Bzazz May 5 '15 at 23:14
• @Bzazz On the contrary it does explain it. No "relative" notion of mass will make the same prediction in all frames, but there can only be one outcome of the experiment so the proper theory has to offer that same prediction from the point of view of all observer. That is, it must use an invariant notion of mass, and that means a Lorentz scalar. – dmckee --- ex-moderator kitten Apr 12 '17 at 21:24

The particle-antiparticle annihilation is on a per-particle basis. One electron annihilates on positron. One up quark annihilates one anti-up quark. One down quark annihilates one anti-down quark. Moving at relativistic speeds doesn't change the number of particles.

For that matter, you could annihilate an electron with an anti-muon, since an electron and a muon have the same quantum numbers, even though a muon is heavier.

Also, you could just as well say the matter is still and the antimatter is moving, so by the same logic you'd have 49 times as much antimatter.

• An electron cannot annihilate an anti-muon since they do not posses exactly the same quantum numbers. They have different lepton numbers. – PhotonBoom May 2 '15 at 22:52
• They have different lepton family numbers, but those are only approximately conserved. The total lepton number, which must be conserved, is the same in each. But you do seem to be correct in that they wouldn't annihilate. Or at least they'd leave an electron neutrino and muon antineutrino. – DanielLC May 3 '15 at 4:53

Each particle only annihilates its exact antiparticle. Electrons annihilate positrons. A blue up quark annihilates an anti-blue anti-up quark. A muon annihilates an anti-muon. The thing about anti-matter is that it postulates an exact opposite of every particular particle type (except for things like photons that are their own antiparticles). It's about having identically opposite particles, same mass (and spin), opposite everything else.

What you are overlooking about the relativistic mass concept is that it is a horrible idea that will almost always get you into serious trouble. Relativistic mass is just energy measured in units of mass, and if you use it like mass you will get the wrong acceleration for tangentially applied forces and almost any other physical situation.

When considering relativistic speeds, the notion of "particle & anti-particle" somewhat blurs. The correct treatment of a relativistic free electron for example is given by the Dirac Equation, which relates Dirac Spinors. A spinor is something like a 4-vector, describing the wave function of our electron.

In it's rest frame, two of the spinor's components will represent an electron state, while the other two represent an anti-electron (positron) state.

Now comes the twist: if you change the frame of reference, you have to act on the spinor with a Lorentzian-boost. This will in general affect all components of the spinor, rendering the positron-state-related components of the spinor non-zero. That means, that you cannot really distinguish between electron and positron if the particle moves at relativistic speeds w.r.t. you.

This is not meant as an answer to your question (it is not easily answerable), but rather as comment.

The answer to the main question is no. One gram of matter (electrons), will annihilate exactly one gram of antimatter (positrons), regardless of the speed with which they approach each other. If they collide with a speed other than zero, the energy due to the motion will also be added to the energy produced by the annihilation.