Do particles and anti-particles attract each other?

From the very basic understanding that they are created out of nothing mutually and collide to annihilate each other seems to indicate this happens due to an attraction.

Is it a measurable force?

Plus they are exactly alike except their opposite charge, which would indicate attraction, as this source indicates: http://www.thenakedscientists.com/forum/index.php?topic=36719.0 I'd like more information on this, if possible.

For the record, I'm speaking about them under the construct of time. If you ignore the construct of time, both particles could actually be considered the same particle, one moving backwards through "time" in a loop that occurs "once". Ignore this if you do not know what I'm talking about, or feel free to correct me on it.

  • $\begingroup$ Related: physics.stackexchange.com/q/9371/2451 and physics.stackexchange.com/q/391/2451 $\endgroup$ – Qmechanic Feb 13 '12 at 20:09
  • $\begingroup$ @Qmechanic, thank you for these links. Very informative. $\endgroup$ – Xonatron Feb 13 '12 at 20:19
  • $\begingroup$ In that forum, they are discussing antimatter as if it's a completely different thing from matter. The matter-antimatter divide is arbitrary. If you want gravitational repulsion, then... cue dark energy. $\endgroup$ – Manishearth Feb 14 '12 at 6:58

From the very basic understanding that they are created out of nothing mutually and collide to annihilate each other seems to indicate this happens due to an attraction.

Why? this just means that if two of them are nearby, they can annihilate. Remember that particles are waves, and thus are quite spread out. They don't have to be directed to collide with each other using any kind of force, they just need to be near each other.

Plus they are exactly alike except their opposite charge

Not true. Particle-antiparticle pairs have the same mass, and spin/isospin (I think), but they have opposite charge, baryon number, lepton number, strangeness, charm, bottomness, (and probably more stuff).

Neither is it mandatory for them to have opposite charge. They can both be neutral. For example, the neutron and all neutrinos have distinct antiparticles, and so does the neutral kaon (giving rise to the strange symbol $\bar{K}^0$). The neutral antikaon has a strangeness of +1, while the neutral kaon has a strangeness of -1. (Strangeness is a property rather whimsically named due to the observation that certain "strange" particles always appeared in pairs or not at all). That being said, there are particles who are their own antiparticle (pi-0 mesons, and all the neutral guage bosons-- photons, gluons, Z, Higgs, gravitons)

That being said, there are only four forces (listed in increasing order of strength)

  • Gravity.
  • Electromagnetism. Note that the force felt when pushing a wall (normal reaction force) or when two balls collide is a manifestation of this force, as what's actually happening is that the electron clouds of atoms are repelling. Many people use this "collision force" while thinking about particles, and this is wrong. Particles don't collide, they can only exchange other particles (as well as absorb/emit each other).
  • Weak nuclear force: Manifests itself in beta decays.
  • Strong nuclear force: Holds atoms together. Gives boom-boom in nuclear explosions.

For a particle-antiparticle pair, there usually will be some sort of force, and yes, it will usually be attractive. But the force can be classified into the four given above. Since gravity is weak and always acts, I'm neglecting it here:

  • Pair of neutral hadrons (neutron, kaon, etc): These have the strong force between them. There also is a bit of the weak nuclear force, though it is not necessarily attractive
  • Pair of charged hadrons (protons,pions, etc): Here the EM force as well as the strong force acts, but the strong force is attractive
  • Pair of electron-like leptons(e,$\mu$,$\tau$): Have EM attraction as well as the weak force. The weak force isn't attractive of repulsive, and it dominates. So an electron-positron pair need not attract all the time
  • Pair of neutrinos ($\nu_e$,$\nu_{\tau}$, etc): Only the weak force, need not attract
  • Pair of quarks (u,d,c,s,&c): Strong force, EM force, little bit of weak force, will attract
  • For guage bosons, it becomes a bit complicated.

So yes, we can see that a general attractive force is prevalent, but not in all cases, and not due to the same phenomenon.

Why don't they need to be close?

(Addendum from the comments below)

Quantum mechanics has a nice concept called wave particle duality. Any particle can be expressed as a wave. In fact, both are equivalent. Exactly what sort of wave is this? Its a probability wave. By this, I mean that it tracks probabilities.

I'll give an example. Lets say you have a friend, A. Now at this moment, you don't know where A is. He could be at home or at work. Alternatively, he could be somewhere else, but with lesser probability. So, you draw a 3D graph. The x and y axes correspond to location (So you can draw a map on the x-y plane), and the z axis corresponds to probability. Your graph will be a smooth surface, that looks sort of like sand dunes in a desert. You'll have "humps" or dunes at A's home and at A's workplace, as there's the maximum probability that he's there. You could have smaller humps on other places he frequents. There will be tiny, but finite probabilities, that he's elsewhere (say, a different country). Now, lets say you call him and ask him where he is. He says that he's on his way home from work. So, your graph will be reconfigured, so that it has "ridges" along all the roads he will most probably take. Now, he calls you when he reaches home. Now, since you know exactly where he is, there will be a "peak" with probability 1 at his house (assuming his house is point-size, otherwise ther'll be a tall hump). Five minutes later, you decide to redraw the graph. Now you're almost certain that he's at home, but he may have gone out. He can't go far in 5 minutes, so you draw a hump centered at his house, with slopes outside. As time progresses, this hump will gradually flatten.

So what have I described here? It's a wavefunction (technically the modulus squared of a wavefunction), or the "wave" nature of a particle. The wavefunction can reconfigure and also "collapse" to a "peak", depending on what data you receive.

Now, everything has a wavefunction. You, me, a house, and particles. You and me have a very restricted wavefunction (due to tiny wavelength, but let's not go into that), and we rarely (read:never) have to take wave nature into account at normal scales. But, for particles, wave nature becomes an integral part of their behavior.

In the next paragraph I am simplifying some stuff with the wavefunctions, and neglecting part of their nature just so my job becomes easier

Back to the problem. Now, our particle and antiparticle are both waves. They have a small hump, but can be quite spread out. Now, these waves come near each other. Remember, the value of the wave (actually the square of its modulus, as a wavefunction is a complex number) gives the probability at which we can find a particle at a point. If the wavefunctions are$\Psi_1(x,y),\Psi_2(x,y)$, the probability of finding both particles at the same point will be $\Psi_1(x,y)\times\Psi_2(x,y)$ (normal probability rules). Now, you have a whole bunch of points where both wavefunctions exist (infinite actually, and technically both wavefunctions are spread out all over the universe, but I'm neglecting that). Adding these probabilities $$\sum\limits_{\forall\space(x,y)}\Psi_1(x,y)\times\Psi_2(x,y)$$, you will get some finite significant probability, even though the individual probabilities are infinitesimal.

So, even if part of the two wavefunctions overlap, there is some nontrivial probability that they annihilate each other. Like I said, the wavefunctions actually cover all of space, but if we neglect those parts (they're extremely small), then the wave "size" is still pretty large. So a particle/antiparticle pair need not be too close to annihilate.

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    $\begingroup$ This is a great answer. And shows the level of depth at which physics questions can be answered. I didn't expect all this but appreciated it. The biggest part I did not know was perhaps the most simple, how the pair do not need to be close to annihilate. Feel free to dive into that deeper. $\endgroup$ – Xonatron Feb 14 '12 at 13:16
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    $\begingroup$ @MatthewDoucette Added a much deeper explanation. I've kept it in layman's terms, and not touched a few irrelevant things, but it should be enough. Don't worry about the upvote. We answer questions for the satisfaction of teaching/helping someone, as well as the experience and insights we get ourselves. We don't do it for the rep. $\endgroup$ – Manishearth Feb 14 '12 at 13:45
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    $\begingroup$ If you're interested in Quantum mechanics, you might want to ask another question asking for an explanation of QM in layman's terms. I'll try to look for one myself, but I don't know a good Internet source of the top of my head. And don't get discouraged if QM makes no sense: "It is my task to convince you not to turn away because you don't understand it. You see my physics students don't understand it. ... That is because I don't understand it. Nobody does."-Richard Feynman (Pioneer of Quantum Electrodynamics) $\endgroup$ – Manishearth Feb 14 '12 at 13:50
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    $\begingroup$ Here's a bit on wave-particle duality that contains most of the stuff I left out: halexandria.org/dward161.htm $\endgroup$ – Manishearth Feb 14 '12 at 13:50
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    $\begingroup$ +1 This is a wonderful answer because it answers the question thoroughly and has bursts of semi-random knowledge in it. I particularly like the explanation of the wave function using the friend analogy. The only small objection I have is that it encourages the misconception that the wave function is a probability distribution, while that's not the case. But you've mentioned that later on when you say the value of the wave (actually its modulus, as a wavefunction is a complex number) gives the probability (though it's really the modulus squared that gives the probability). $\endgroup$ – Wouter May 1 '13 at 9:27

In particle physics, the microcosm version of everyday physics, we have four "known" forces.

1) gravitational : an attractive force, depending on the mass.

2) electromagnetic

these two have also macroscopic manifestations.

3) weak which appears in decays of particles

4) strong which is the source of nuclear forces and binds the quarks into nucleons

All these forces are involved when two elementary particles meet and what happens when they interact depends on the energy they have and the quantum numbers they carry. There are specific calculations that can be made about the probable outcomes of the interaction.

Lets take an electron and a positron, antiparticles of each other. The electric force between them is attractive; Particularly if the energy is too small, they annihilate into two photons.If given enough energy they still annihilate, and one of the possible outcomes ,as shown in the first diagram is into a quark antiquark pair through an intermediary photon. Many and various such diagrams can be and are made. Qualifying the interaction as attractive is the least of its qualities.

When an antiproton meets a proton, the electric force is attractive but once within nuclear distances the strong force appears and dominates by quark annihilations. Again it has little meaning to call it "an attractive force". The energy of the interaction and the quantum numbers of the decay products are what dominate the description of the interaction.

In general, "force" in particle physics means the possible exchanged particle in the Feynman diagrams that describe the interaction. Attractive or repulsive is not a good qualifier. For example, at the LHC, one is scattering protons on protons and is getting the whole zoo of the standard model particles as output. What meaning is there in the qualifier "repulsive" which the macroscopic forces would give between two positive charges?

  • $\begingroup$ thank you for this answer too. I chose the other answer only because it was more detailed, but basically you have addressed the same issues with my question! $\endgroup$ – Xonatron Feb 14 '12 at 13:17

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