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In this question Can hydrogen atom exchange induce attractional forces between $e^-e^-$? one answer showed the range of exchange particle but not how to calculate. how do we calculate/know what are range of exchange particle? Do we calculate or observe?

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We can calculate it . To explain this, let us consider nuclear beta decay

$n \rightarrow p + e^- + \nu_e$

where the exchange particle is the $W$ boson. This particle was calculated (and observed*) to have a mass of about $80$ $GeV/c^2$. We calculate its range using the uncertainty relation

$\Delta E \Delta t \geq \frac{h}{2}$

Now assuming that the uncertainty in energy $E \approx mc^2$ is comparable to the energy of the $W$ boson we can rearrange the above relation

$$\Delta t \approx \frac{h}{mc^2}$$

to give us the distance

$$D \approx c\Delta t = \frac{h}{mc}$$

and given

$mc^2 \approx 80$ $GeV/c^2 = 1.28 \times 10^{-8} J$

we get

$$mc = 4.27 \times 10^{-17} \, J s/m$$

and using

$$h= 6.62 \times 10^{-34} Js $$

giving us the range of the $W$ to be

$$D = 1.55 \times 10^{-17} m$$

This is an order of magnitude 10 times the diameter of a proton.

*I noticed you have "virtual particles" as one of your tags and in the link you provided, but here the W boson is real (observable) but has a very small lifetime due to its instability (when considering energies above its rest mass). It can also arise as an unobservable virtual exchange particle during certain scattering processes).

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  • $\begingroup$ "is comparable to the energy of the W boson" the energies available in a neutron decay is very small. To be real, it has to be on mass, which is not possible at the mass of the neutron. I think the argument is wrong. $\endgroup$ – anna v Oct 27 '20 at 5:04
  • $\begingroup$ "is comparable to the energy of the W boson" the energies available in a neutron decay is very small. To be real, it has to be on mass, which is not possible at the mass of the neutron. I think the argument is wrong.if we accept your last comment of real mass.. The calculation here hyperphysics.phy-astr.gsu.edu/hbase/Forces/exchg.html . $\endgroup$ – anna v Oct 27 '20 at 5:30
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    $\begingroup$ The mass/energy above is that for the W boson. Not a neutron. $\endgroup$ – joseph h Oct 27 '20 at 5:57
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For clarity:

If a force involves the exchange of a particle, that particle has to "get back home before it is missed" in the sense that it must fit within the constraints of the uncertainty principle. A particle of mass m and rest energy E=mc2 can be exchanged if it does not go outside the bounds of the uncertainty principle in the form

et

A particle which can exist only within the constraints of the uncertainty principle is called a "virtual particle", and the time in the expression above represents the maximum lifetime of the virtual exchange particle. Since this exchange particle cannot exceed the speed limit of the universe, it cannot travel further than c times that lifetime. The maximum range of the force would then be on the order of

et2

Note that this expression implies that a zero mass for the exchange particle implies a force of infinite range. The rest masses of the exchange particles for the electromagnetic force and gravity, the photon and the graviton, are taken to be zero and those forces are presumed to be infinite in range.

The same as the answer by Dr jh .

It is improtant to note that Heisenberg uncertainty arguments give a large envelope within which the specific interactions are limited. The potentials though are very important. In the case of zero mass, the gluon has zero mass but the range of strong interactions is very small, due to the form of the strong potential .

The paper that started the discussion does not use this Heisenberg uncertainty envelope way of calculating ranges. See the complexity in a recent paper here in getting an effective potential.

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  • $\begingroup$ Very good Anna. $\endgroup$ – joseph h Oct 27 '20 at 6:07

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