Fine-Structure Const, inverse square-law, bonding

[Widdekind]

Imagine an electron e^- approaching a proton p⁺, at a distance [math]\Delta X[/math] away. The EM force, between those charged particles, is said to be mediated, by virtual photons, "borrowed into existence on a Heisenberg loan", according to the HUP. Now, naively, that EM force would plausibly be:

 

[math]F = \frac{dp}{dt} \approx \frac{\Delta p}{\Delta t} \approx \frac{\hbar}{\Delta X} \frac{c}{\Delta X} \rightarrow \frac{\hbar c}{r^2}[/math]

Such symbols seem to "say", that the virtual photons exchanged, between a pair of charged particles:

 

• have a characteristic wavelength, comparable to the particle separation distance, i.e. [math]\lambda \approx \Delta X[/math]
  
• are exchanged at a rate, comparable to the (inverse) light-crossing time, i.e. [math]c \Delta t \approx \Delta X[/math]
  

Such a combination of effects could account, for an "inverse-square law", for EM, i.e. as charged particles approach each other, they exchange more energetic virtual photons, more frequently, i.e. "stronger x faster = inverse-square-law". However, the actual strength, of the EM interaction, between pairs of charged particles, is much weaker, than the above "theoretical Quantum maximum":

 

[math]F_{actual} = \frac{e^2}{4 \pi \epsilon_0 r^2} = \alpha \frac{\hbar c}{r^2}[/math]

where [math]\alpha = 1/137[/math] is the FSC, i.e. the EM interaction coupling strength. Thus, this simple picture qualitatively accounts, for the "Quantum weakness" of EM, being only [math]\approx 0.007[/math] of the theoretical maximum interaction strength.

 

 

QUESTION:

 

Does this analysis imply, that as an electron approaches a proton, so that their separation distance decreases, then the virtual photons "flitting back-and-forth between them" increase in ("borrowed") energy -- until if-and-when the particles approach closely enough, so that those virtual photons become energetic enough, to "carry away" all of the energy needed, i.e. "13.7eV + KE₀", for the e^- + p⁺ to bond, into H ? Do the virtual photons then become "promoted", "actualizing" into real photons, and "carrying off" the correct amount of energy, to leave the electron "perfectly bound" onto the proton ?

 

Note, that whilst the "strength" of a real photon is [math]E = hc/\lambda[/math], the "strengths" of "Heisenberg-borrowed" virtual photons are only [math]E \approx \alpha \hbar c / \lambda[/math], i.e. virtual photons are "enfeebled" by a factor of [math]2 \pi \alpha \approx 860[/math]. Thereby, whilst a real "promoted" photon, of energy E=14eV, has a wavelength near 900 A, virtual "ghost" photons only carry that much energy, when the charges are separated by a distance of about 1 A. Were it not, for the "enfeeblement", of the virtual photons, mediating the EM interaction, electrons would be "pushed around" by 14eV virtual photons, whilst still thousands of atomic radii away, from the protons.

[questionposter]

The inverse square law exists because of the properties of spheres, and as virtual particles get emitted from a source, they spread out more according to the surface of a sphere which I guess makes sense because they are emitted from a sphere-like object. Classical distance apparently exists in QM.

Edited January 7, 2012 by questionposter

[derek w]

Question.you say"flitting back-and-forth between them" increase in ("borrowed") energy.

 

Where is the energy being borrowed from?You must have conservation of energy?

[questionposter]

Question.you say"flitting back-and-forth between them" increase in ("borrowed") energy.

 

Where is the energy being borrowed from?You must have conservation of energy?

 

Maybe it's imaginary, in which case your not actually taking real energy, and that would make sense because when it interacts with an object it could multiply by some kind of complex conjugate to form real results and real consequences.

Edited January 7, 2012 by questionposter

[derek w]

You could say that the energy is borrowed from the electron & proton,the field of virtual particles would be part and parcel of the electron & proton(travelling with them),the virtual particles being created and annihilated constantly.

 

 

The force of attraction being created by the virtual particle field not the electron & proton?

[Widdekind]

virtual photons have mass ?

 

Consider two 'particles', separated by a distance [math]r[/math], that are interacting through some force, via massive force-carrying bosons, which propagate at a speed [math]v < c[/math], so that [math]r = v \Delta t = c \Delta t \beta[/math]. Recall, for massive particles [math]p = \gamma m_0 v[/math], i.e. [math]c p = \gamma E_0 \beta = E \beta[/math]. Thus:

 

[math]F \equiv \frac{\Delta p}{\Delta t} \approx \frac{\Delta E \beta}{c \Delta t} = \frac{\hbar c}{\left( c \Delta t \right)^2} \beta = \frac{\hbar c}{r^2} \beta^3[/math]

Now

 

[math]\Delta E = \gamma E_0[/math]

 

[math]\approx \frac{\hbar}{\Delta t} = \frac{\hbar c}{r}\beta[/math]

Let

 

[math]r_0 \equiv \frac{\hbar c}{E_0}[/math]

 

[math]x \equiv \frac{r}{r_0}[/math]

Then

 

[math]x = \frac{\beta}{\gamma} = \beta \sqrt{1-\beta^2}[/math]

So [math]\beta(x)[/math] is a solution of [math]\beta^4 - \beta^2 + x^2 = 0[/math], i.e.

 

[math]\beta^2 = \frac{1 \pm \sqrt{1 - 4 x^2}}{2}[/math]

By analogy, to the case of mass-less force-carrying bosons, we take the positive solution, so that [math]\beta \rightarrow 1[/math] as [math]x \rightarrow 0[/math], i.e. the force intensifies, and the force-carriers become increasingly energetic, with decreasing distance, between the interacting 'particles'. Also, massive force-carriers are limited to a maximum range, [math]x = 1/2[/math], i.e. [math]r = r_0/2[/math], beyond which they cannot reach, before their "borrowed" energy must be "repaid". At that maximum range, force-carrying bosons propagate at minimum speed, [math]\beta^2 = 1/2[/math], i.e. [math]\approx 0.7 c[/math]. Er go, forceful interactions are always relativistic, i.e. [math]v \sim c[/math].

 

Thus

 

[math]F \approx \frac{\hbar c}{r_0^2} \frac{\beta^3}{x^2}[/math]

 

[math]\beta^2 = \frac{1}{2} + \frac{\sqrt{1 - 4x^2}}{2}[/math]

 

[math]\therefore F \approx \frac{F_0}{x^2} \times \left( \frac{1}{2} + \frac{\sqrt{1 - 4x^2}}{2} \right)^{\frac{3}{2}}[/math]

So [math]F \rightarrow 1/r^2[/math] as [math]r \rightarrow 0[/math], where [math]\beta \rightarrow 1[/math]. Er go, forces mediated by massive bosons exhibit an inverse-square-law behavior, at short ranges, with a "cut-off" at longer ranges, i.e. are qualitatively "Yukawa-like". Perhaps, as the energy [math]E[/math], at which 'particle' collisions are conducted, increases, the "effective mass" of massive force-carriers decreases, i.e. [math]E_{0,eff} \approx E_0 - E[/math] ? If so, then the maximum effective range of interaction [math]r_{0,eff}[/math], approximately separating the "short range" region wherein [math]F \approx 1/r^2 [/math], from the "beyond range" region wherein [math]F \approx 0[/math], increases with collision energy [math]E[/math]

 

Note that forces are never stronger than [math]F \propto 1/r^2[/math], but, at higher energies, "you get more of that potential".

 

 

---

Where is the energy being borrowed from?You must have conservation of energy?

 

No, our universe offers "venture capital" to "aspiring entrepreneurs", to make a business analogy. For, an energy "loan", of amount [math]\Delta E[/math], may be "borrowed", from the background, for a time [math]\Delta t \approx \hbar \ \Delta E[/math], according to the Heisenberg Uncertainty Principle (HUP), before the "loan" must be "repaid".

 

Only by such "lending" can low-energy 'particles', in our cold cosmos, at present epoch [math]\left( T \approx 3K \right) [/math], "finance" the generation, of "expensive" Weak Force bosons, whose rest-mass-energies are 80-90 GeV.

 

The force of attraction being created by the virtual particle field not the electron & proton?

 

I understand, that Fundamental Forces, are conveyed, between fermions f_1,2, by force-carrying bosons B, which are generated by a transmitting 'particle' f₁; and later absorbed by a receiving 'particle' f₂

 

[math]f_1 \begin{array}{c}

B \\

\longrightarrow \\ \end{array} f_2[/math]

Perhaps the "force of attraction" is "generated" by particles, and "conveyed" by force-carriers ? In analogy, a transmitting radio dish, sends a signal, which is received by another radio dish, some-where-and-when-else. Or, fundamental 'particles' are constantly "chatting" at each other, via "walkee-talkees" or "cell phones", where the first particle "talks", then a signal is sent, which "plays out" to the other particle.

Edited January 12, 2012 by Widdekind

[Widdekind]

virtual photons have mass ?

 

...

 

Thus

 

[math]F \approx \frac{\hbar c}{r_0^2} \frac{\beta^3}{x^2}[/math]

 

[math]\beta^2 = \frac{1}{2} - \frac{\sqrt{1 - 4x^2}}{2}[/math]

 

[math]\therefore F \approx \frac{F_0}{x^2} \times \left( \frac{1}{2} - \frac{\sqrt{1 - 4x^2}}{2} \right)^{\frac{3}{2}}[/math]

 

If you take the negative root solution, then [math]\beta \rightarrow 0[/math] as [math]r \rightarrow 0[/math], i.e. the massive force carriers are "weaker", and the force "softens", at shorter ranges. Plotting that F(x) solution reveals, that the force is zero at zero range, and increases in strength linearly, up to the maximum "cut off" range, whereat the force strengthens "steeply". Now, qualitatively, that behavior resembles that of the Strong "color" force, which is weak ('asymptotic freedom') at short range, but increases sharply near its maximum range. And, qualitatively, taking the negative over the positive root, sounds similar, to the difference between "non-Abelian gauge fields", e.g. Strong force, vs. "Abelian gauge fields", e.g. EM. If so, then the Strong force could be modeled, via massive gluons. For what reason(s) do scientists say, that gluons are mass-less, e.g. "have they weighed one" ? Naively, if the maximum range of the Strong force is [math]r_0 \approx 3 fm[/math], then the implied gluon mass would be [math]m_g \approx 100 MeV[/math].