Proton-proton collision and virtual photons

[TheTourist]

a) Two protons, approaching each other with 3keV of energy each, collide head on. Calculate how close they get to each other.

b) Estimate the momentum transferred by one virtual photon at the closest approach.

c) Hence, estimate roughly how many virtual particles are exchanged during the collision.

For parts a) and b) I used Coulombs law for energy I calculated that they get to about 4x10-13m apart, and hence to estimate the momentum for one virtual photon I used The Uncertainty Principle (momentum) to be about 2x10-22kgms-2.
It is with part c) that I am struggling. I used Newtons second law combined with Coulombs law to calculate the change in momentum of the protons, and found (by taking the ratio) that there are 1/137 virtual photons exchanged. Ordinarily I would have shrugged this off immediately as wrong, however this is the fine structure constant (alpha), and it seemed too great a coincidence. The probability of each virtual photon transition is 1/SQRT(137), so for an emission of a photon and absorbance of photon, this would be 1/137, and hence one virtual photon is exchanged! Is this right? Does it even make sense?! The main thing that is making me doubt is the fact that change in momentum for one photon is much greater than change in momentum of the protons.

Thanks

 

[mark726]

for your post! Let's go through each part separately to make sure we have a clear understanding of the calculations and results.

a) Using Coulomb's law, we can calculate the potential energy of the two protons at a distance of 4x10^-13m apart. This is equal to the kinetic energy of the protons (3keV each), so we can set up the equation:

Potential energy = Kinetic energy
k (q1q2)/r = 2(3keV)
Where k is the Coulomb constant, q1 and q2 are the charges of the protons, and r is the distance between them.

Solving for r, we get:
r = (2kq1q2)/2(3keV)
Plugging in the values for k, q1, and q2, we get:
r = (2x8.99x10^9 Nm^2/C^2)(1.6x10^-19 C)^2/6x10^3 eV
r = 4.65x10^-13m

So your calculation for the closest distance they get to each other is correct!

b) To estimate the momentum transferred by one virtual photon, we can use the Uncertainty Principle, which states that the product of the uncertainty in position and the uncertainty in momentum must be greater than or equal to Planck's constant divided by 4π. In this case, the uncertainty in position is the distance between the protons (4.65x10^-13m) and the uncertainty in momentum is given by the kinetic energy of the protons (3keV). So we have:

ΔxΔp ≥ h/4π
(4.65x10^-13m)(3x10^3 eV) ≥ h/4π
Δp = (6.63x10^-34 Js)/(4πx4.65x10^-13m)
Δp = 2.2x10^-22 kgm/s

So your estimation for the momentum transferred by one virtual photon is also correct!

c) Now for the tricky part. Let's start by clarifying what we mean by "virtual particles" in this context. In particle physics, virtual particles are particles that are not directly observable, but are used to explain interactions between particles. They are a mathematical tool, rather than actual physical particles. So in this