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1

Our everyday world, the world we live and die in is made out of matter, by definition. Our everyday world has been modeled successfully with the classical theories of Mechanics, Electrodynamics and Thermodynamics. The forces of gravitation and electromagnetism are the ones observable easily , again in the everyday environments. All these theories were ...


1

Electric charge is not special. Every charge is replaced by its opposite under the C conjugation. For instance, electric charge goes from positive to negative and vice versa. Color charge goes from blue to antiblue and vice versa. Mutatis mutandis with green/red color charge. Every charge is sent to its anti-charge.


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This depends entirely on the kinematics. Imagine I'm throwing an electron at a positron. If the two interact, then the resulting photons (two or more, although the more you add, the less likely the process is of occurring) will have energies entirely dependent on how fast you threw them together. If you throw them together hard enough, you can create heavier ...


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Theoretically we can, I we had enough. But making and maintaining antimatter is hard. Also, if you aren't in a vacuum, you will just see bright light and energy.


2

This is a bubble chamber antineutron event This picture, taken in the Berkeley 30-inch propane bubble chamber in 1958 Antiprotons enter from the top with momentum 684MeV/c . At the arrow one of the antiprotons in the beam disappears, shown with purple on the right. Then a vertex appears out of nothing where an annihilation event is obseved into ...


1

A beam of anti-protons are fired into a charge-exchanger. If an anti-neutron is produced, S1 and S2 will not detect anything, but the final detector will. ([Picture courtesy of John Rennie])2 [Here is a book on the matter.]3


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If q is the momentum for positron, then the propagator for it is still $i\frac{\not{q}+m}{\not{q}^2 +m^2}$.


2

For reference, the fermion propagator is $$ \left\langle 0 \right| T\psi(x)\overline\psi(y) \left|0\right \rangle= S(x-y) = \int \frac{d^4k}{(2\pi)^4} \frac{i}{\not k-m}e^{-ik\cdot(x-y)}$$ Depending on the time ordering, this describes a particle moving from $y$ to $x$, or an antiparticle moving from $x$ to $y$. Now, consider a one-loop diagram in which ...


0

... is this what is responsible for Radioactive Decay? Sort of ... radioactive decay occurs when the final state (result of the decay) has lower energy than the initial nucleus and the barrier is occasionally surmounted. This barrier penetration and the virtual pair at the black hole horizon are both quantum processes, so they have a little physics in ...


2

You are correct, it is redundant. You could also say they have the same spin and parity.


10

No, the virtual photon is not a particle, since a virtual particle is what one calls the internal lines in a Feynman diagram, and there are no asymptotic particle states associated to these lines, so a virtual particle is not a particle in the usual (or any other rigorous) sense. Therefore, the question is non-sensical because it is not clear what an ...


1

Explain why the mass of a tree cannot be converted directly into energy. That's a tricky one, because it could turn out that it is possible to turn matter alone into energy. Floris hinted at this with radioactive decay, but there are potentially other methods such as melting hadrons in a quark-gluon plasma (QGP), see for example this report. The interesting ...


6

When a radioactive element decays, part of its mass is converted to energy - no obvious need for antimatter anywhere. Instead, the energy is released because the binding energy of the sum of the fragments might be higher than that of the parent nucleus. However, to fully convert matter to energy you do need the antiparticle. Otherwise, you run into ...


4

Energy is never created nor destroyed, and to say "X is converted into energy" is just meaningless. We don't convert things distinct from energy into energy, all we ever do is convert one form of energy into another. The badly posed question from your book probably intends to ask why we cannot convert the mass energy that any chunk of matter contains as per ...


0

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.


1

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 ...


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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 ...


33

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 ...


13

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 ...


0

I'm going to give this a shot, cause I think it's a fun question, but you probobly should work it out a bit more. Virtual particles, or particle/anti-particle creation looks like 1 electron and 1 anti electron (for example) existing briefly then cancelling each other out. If it happens in the void of space, they interact with each other and - poof, gone. ...


0

At least for the radioactive decay that most people are familiar with, and that you're probably talking about, electroweak force is the cause.



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