Proton-proton collision and virtual photons

With that in mind, let's look at your calculation. Using Newton's Second Law, we can calculate the change in momentum of the protons during the collision. This is equal to the momentum transferred by the virtual photon, so we can set up the equation:Change in momentum = Momentum transferred by one virtual photonmΔv = ΔpWhere m is the mass of the proton and Δv is the change in velocity of the proton.Solving for Δv, we get:Δv = Δp/mPlugging in the
  • #1
TheTourist
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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
 
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  • #2
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
 

FAQ: Proton-proton collision and virtual photons

1. What is a proton-proton collision?

A proton-proton collision is a type of high-energy particle collision that occurs in particle accelerators, such as the Large Hadron Collider. In this process, two protons are accelerated to extremely high speeds and then collide with each other, resulting in the creation of new particles and energy. These collisions allow scientists to study the fundamental building blocks of matter and the laws of physics.

2. What are virtual photons?

Virtual photons are particles that are not directly observable, but are used to explain interactions between charged particles. They are considered "virtual" because they exist for a very short amount of time and cannot be directly detected. In the context of proton-proton collisions, virtual photons mediate the electromagnetic force between the colliding protons.

3. How are virtual photons involved in proton-proton collisions?

In a proton-proton collision, the two protons approach each other and interact through the exchange of virtual photons. These virtual photons carry energy and momentum, allowing the protons to interact and produce new particles. The number and energy of the virtual photons involved in the collision determine the characteristics of the resulting particles.

4. What can we learn from studying proton-proton collisions and virtual photons?

By studying proton-proton collisions and virtual photons, scientists can gain a deeper understanding of the fundamental laws of physics and the structure of matter. This research can also help us better understand the origins of the universe and the behavior of particles in extreme conditions. Additionally, the technologies developed for studying these collisions have practical applications, such as in medical imaging and cancer treatment.

5. Are there any potential dangers associated with proton-proton collisions and virtual photons?

While proton-proton collisions and virtual photons used in these collisions are highly energetic, they do not pose any significant danger to the public. Particle accelerators are designed with numerous safety measures in place to ensure the protection of workers and the environment. Any potential risks are carefully assessed and monitored by scientists and regulatory bodies before any experiments are conducted.

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