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Feynman diagram

$\textbf{Note that this diagram hasn't anything to do with the question directly.}$

After a particle and its antiparticle annihilate, their energy is converted into a force carrier particle, such as a gluon, $W$ or $Z$ force carrier particle or a photon.

When I contemplate this phenomenon, it seems curious to me that a system can have a mass, and then accelerate to the speed of of a massless particle. This curiosity stems from the special relativistic notion that it takes an infinite amount of energy to make any particle with mass accelerate to the speed of light.

The equation that I am speaking of is graphed here, and demonstrates the well recognized $\textbf {Energy-Mass against Velocity}$ asymptote of Special Relativity:

enter image description here

Besides the fact that this process adheres to the many other conservation laws which we have discovered, is there a reasonable explanation for this phenomenon? Is it just like putting a value into a function, and the universe knows to produce a photon (or another moderator)? Or, does something else take place, where the particles are exposed to energy in its purest form, so they accelerate tremendously together until they separate to form another particle pair?

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If the colours in your diagram are supposed to represent different QCD/quark/gluon colours then you should not that the colour structure changes after a single gluon emission. –  jk88 Aug 7 at 8:30
4  
All the internal lines of your diagram are not particles. They are field correlations, also called (unfortunately), "virtual particles". To these field correlations, we may associate any value of momentum and energy, including negative energy. The consequence is that any value of the squared mass, including negative values, may be associated to these field correlations. –  Trimok Aug 7 at 8:48

3 Answers 3

up vote 3 down vote accepted

In Feynman diagrams all four vectors conform with special relativity algebra, BUT the internal lines, even though they have the name "photon" are virtual, which means the mass can be different than zero, only the quantum numbers identify them as a photon, quark, electron, etc..

Feynman diagrams are an iconic representation one to one with the integrals and functions necessary to compute the reaction depicted. In this case electron positron annihilation into two quarks which then decay into two resonances. There are functions called propagators included under the integrals which correspond with the named off-mass-shell virtual "photon" etc which contain the mass of the virtual particle in the denominator, for example.

propagator

note that the square of the four momentum would cancel with the mass in the denominator generating an infinity if it were not for the i*epsilon.

Edit* after edit of question

The energy mass relationship you are quoting uses the relativistic mass, as in E=m*c^2.

relativisticmass

The relativistic mass is not the mass entering Feynman diagrams or a mass that characterizes the particles in the Feynman diagrams . That is the length of the four vector describing the particle in any reference frame and is an invariant of the Lorenz transformations, it is called the rest mass or invariant mass.

The mass of the electron is always the same , it is the relativistic mass that changes and depends on the frame of reference. When an electron and a positron annihilate there is conservation of momentum and energy and the virtual force carrier will have as mass the same as the invariant mass of the incoming pair. That is how we have detected a number of mesons, including the Z .

enter image description here

(Go to the link for a clearer image)

As the invariant mass of the incoming pair scans the region of the mass of the Z for example, the probability of interaction goes up, as we are crossing a pole in the propagator, and the Z appears at its invariant mass with its width.

So there is no problem with the mass of the intermediate boson, which is virtual except exactly on mass shell as it passes the resonance, and can be bigger or smaller than the mass in the data table as the energies of interaction, the invariant mass of the e+e- pair increases. The relativistic mass is irrelevant in the interactions of elementary particles and the use of the term has fallen out of fashion.

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Yes, that equation really is quite clever, but is there not a limit with epsilon approaching zero? (I may be thinking of a variation of the propagator). Actually, the Feynman diagram was a bit of a mistake on my part, I added as an aesthetic(which I now see was extremely misleading, I am quite sorry for wasting your time). My main purpose for asking this question was concerning the special relativistic notion that it takes an infinite amount of energy to accelerate something with mass to light speed. It really is quite an honor to get feedback from an experienced particle physicist. –  NaturalPhilosopy Aug 7 at 11:21
    
Thank you very much, truly. –  NaturalPhilosopy Aug 7 at 11:22
1  
theta and delta functions are extensively used for such limits if they are within the phase space of the variables in the integration. –  anna v Aug 7 at 12:36

it seems curious to me that a system can have a mass, and then accelerate to the speed of of a massless particle

There is no paradox here. Special relativity doesn't preserve velocity - in fact the opposite - it allows one to move between frames of reference in which the velocity of objects is relative to the observer.

[do the particles] accelerate tremendously together until they separate to form another particle pair?

No- they cancel out exactly to produce the mediating boson, however it's not sensible to ask what this `looks like'. You have to bare in mind that in quantum mechanics one can only observe the initial state and the final state. The observables here are the two annihilating particles/anti-particles and the created particles/anti-particles, not the mediator.

Edit after edit of question

The energy mass relationship you are quoting uses the relativistic mass, as in E=m*c^2.

relativistic mass

The relativistic mass is not the mass entering Feynman diagrams or a mass that characterizes the particles in the elementary particles table . That is the length of the four vector describing the particle in any reference frame and is an invariant of the Lorenz transformations, it is called the rest mass. The mass of the electron is always the same , it is the relativistic mass that changes and depends on the frame of reference.

When an electron and a positron annihilate there is conservation of momentum and energy and the virtual force carrier will have the mass the same as the invariant mass of the incoming pair. That is how we have detected a number of mesons, including the Z, by scanning the energy of the e+e- at rest . As the invariant mass of the incoming pair scans the region of the mass of the Z for example, the probability of interaction goes up and the Z appears at its invariant mass with its width.

So there is no problem with the mass of the intermediate boson, which is virtual except exactly on mass shell as it passes the resonance, and can be bigger or smaller than the mass in the data table. The relativistic mass is irrelevant in the interactions of elementary particles and the use of the term has fallen out of fashion.

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The paradox I was talking about had to do with the special relativistic notion that it, in theory, it takes an infinite amount of energy applied to an object (even if it's mass is minuscule) to accelerate that to light speed (the vertically asymptotic kinetic energy/rest energy function). But yes, I figured as much. I don't know why I thought someone would have some esoteric explanation of the matter. Thank you for your answer. –  NaturalPhilosopy Aug 7 at 11:01
    
(I still gave you the up arrow for the confusion that was likely my fault). –  NaturalPhilosopy Aug 7 at 11:22

From the OP's comments I have a feeling that the real issue is not the virtual particle, but an actual annihilation of a positron and electron into 2 real photons. Like here: enter image description here

And the question is, how can the system of 2 massive electrons accelerate into the speed of light, when accelerating to c requires infinite amount of energy. The answer has two parts:

1) When the you have the 2 photons, they are no longer massive particles. Photons do not accelerate. They are born moving at the speed of light. The positron and electron just cease to exist. They also don't accelerate.

2) The 2 photons are moving away from each other (in the center of mass frame of the original colliding leptons). So the center of mass of the 2-photon system is at rest in the selected frame. The center of mass did not accelerate.

Please also note that particle and antiparticle cannot annihilate into just 1 real photon. There must be at least 2 real photons due to conservation of momentum.

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I was under the impression that we were detecting single photon emissions from positronium? But yes I know what you are talking about. –  NaturalPhilosopy Aug 7 at 13:10
    
I just cannot seem to grasp that concept of them ceasing to exist you see... It doesn't make any sense. I can say it over and over again. –  NaturalPhilosopy Aug 7 at 13:13
    
but it doesn't help it make sense –  NaturalPhilosopy Aug 7 at 13:13
    
@NaturalPhilosopy Positronium decays into 2-3 photons. See here: en.wikipedia.org/wiki/Positronium Or see the article on annihilation: en.wikipedia.org/wiki/Electron%E2%80%93positron_annihilation The fact that 2 photons are produced is utilized in PET (positron emission tomography): en.wikipedia.org/wiki/… where the position of annihilation lies on the line connecting the 2 photons. –  mpv Aug 7 at 14:48
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@NaturalPhilosopy The paper says that one-photon annihilation was never observed. It is only a theoretical phenomenon that may be happening in magnetic fields in the range of 10^12 Gauss, where the extreme magnetic field provides the momentum that is normally taken by the second photon. Please note that the required field is 100 million Tesla. The strongest continuous fields on Earth were achieved 45 Tesla, strongest pulse field was 2800 Tesla, but during this experiment the equipment and the laboratory were destroyed. And it's still 5 orders of magnitude below the required induction. –  mpv Aug 7 at 15:54

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