1주: 물질과 힘 그리고 측정(Matter and forces, measuring and counting)
W1.0 환영(Welcome)
W1.1 물질(Matter)
W1.2 힘(Forces)
W1.2a 자연단위(Natural units)
W1.2b 특수 상대론과 4-벡터(Special relativity and four-vector)
W1.2c 가상입자(Virtual Particles)
W1.3 확률과 단면(Probability and cross section)
W1.3a 광자 빔의 감쇄(Attenuation of a photon beam)
W1.4 러더포드 실험(Rutherford experiment)
W1.4a 러더포드 단면(Rutherford cross section)
W1.4b 산란율 계산(Counting rate Rutherford)
W1.5 양자 산란(Quantum scattering)/동영상/영문자막/슬라이드
In this first module, we're in the process of introducing objects studied by particle physics, namely matter, forces and space-time. And in this context we must obviously discuss scattering processes.
In this fifth video, we will show how to approach the scattering processes between particles in a quantum way. The goals for you are;
- to identify the conceptual differences between the classical approach that we've used up to now, and the quantum evolution of a system.
이제까지 입자의 산란에 대해 고전적인 방식(운동에너지와 쿨롱 력)으로 접근해 봤다. 이제 양자학(quantum physics)으로 시각을 옮겨보기로 한다.
- to know how to draw a Feynman diagram for simple scattering processes and explain its ingredients.
산란의 과정을 쉽게 설명하는 파인먼 도(Feynman diagram)에 대해 배우고 (입자의 세계를) 어떻게 설명하는지 살펴보자.
In video 1.4 Mercedes has calculated the differential cross section for Coulomb scattering off a static target, which is reproduced here. By lucky coincidence, the classical result is still valid in a relativistic quantum context.
The reasons are the following.
그 이유는 다음과 같다.
- The result of quantum theory contains no factor of ħ. This means that the artificial limit ħ going to zero, ħ⟶0, which normally takes us back to the classical result will not change the answer that we obtained.
양자론으로 얻은 결과에도 ħ 인자는 포함되지 않는다. 이는 극한 ħ⟶0 일때 고전적 접근 법으로 얻은 결과와 같다는 뚯이다.
- The classical result is also valid in the relativistic regime already. This fact is not surprising, Maxwell's equations are valid for relativistic velocities. After all, they also they describe electromagnetic wave including the motion of photons, which moves at the speed of light.
고전적 결과는 상대론적 영역에서도 이미 잘 작동한다. 이는 맥스웰 방정식이 상대론적 속도에서도 잘 적용된다는 사실은 놀라울 것이 없다. 무엇보다도 맥스웰 방정식은 빛의 속도로 움직이는 광자의 전자기 파동을 잘 기술 한다.
- We also do not need to take into account nuclear interactions between projectile and target, since the alpha particle will never penetrate into the target nucleus, even for head-on collision. Only the electromagnetic force acts outside the nuclear volume.
입사 입자와 목표 입자 사이의 핵반응을 고려할 필요가 없다. 입사하는 알파 입자가 정면 충돌이라 해도 핵을 뚫고 지나가지 못하기 때문이다. 전자기력이 핵의 외부 주변에서도 강력히 작동하기 때문이다.
But until now, we have discussed the scattering process in a language adapted to classical physics. Now, we will change toolkit to discuss the quantum approach.
While the classical results stays valid in the quantum regime, the interpretation is totally different. Particles may well be point-like but they move like a probability wave.
(입자의 산란은 강체의 탄성 충돌로 보고 뉴튼 역학 운동에너지로 푼)고전적 결과가 양자론과도 일치하지만 그 해석은 완전히 다르다. 양자론의 영역에서는 점입자(point-like particle)인 된시에 확률 파동(probability wave)의 성격을 더 띄고 있다고 본다.
- Because of Heisenberg's principle, the impact perimeter b has no more role in our consideration. We must use a scenario of Huygen to understand how scattering works. This means that the target particle will be the origin of a new scattered wave, which will add to the incoming wave of the projectile.
- The classical trajectory is replaced by the wave function ψ(x), as we've mentioned earlier. It contains the particle aspect, energy and momentum, (E, p) and the wave expect, frequency and wave number, (ω,k) of the kinematics at the same time.
- Before the scattering, the projectile is described by a plane wave in the x direction in our little example. Here we have also normalized the amplitude to one for simplicity.
- Huygens' principle says that a small fraction f of the incident amplitude will be spherically re-emitted by the target. The factor 1/r in the scattered amplitude is necessary. It ensures that the number of scattered particles, ψ*ψ, remains independent of distance.
- The remaining amplitude (1-f) continues as an unperturbed plane wave.
As a consequence of the wave approach we must also reinterpret the cross section in terms of intensities of incoming and scattered wave.
- This means that we now look for a relation between the cross section and the scattered amplitude f(θ,φ).
- The square of the second term is proportional to the number of particles scattered into a volume subtended by a solid angle dΩ, and the radial thickness dr. It represents the scattered flux.
- For a normalized incident wave the particle density is ρ=1. Thus the incident flux is simply ρ times velocity or the velocity itself. I = ρv = v
- The ratio between the scattered and the incoming flux is precisely the cross section.
- We find a simple result, f is the probability amplitude for our scattering process. And σ = f^2 is the scattering probability itself.
- The amplitude f is calculable if we know the potential generated by the target, just like in the classical sense.
To visualize reactions between particles and calculate their probability, we use a very useful tool which is called Feynman diagrams.
- They represent lines of propagation of particles in coordinates E and p. Not in space-time but in momentum space.
- They also represent the vertices of an interaction, that is the points where a force particle is emitted or absorbed to transmit energy and momentum.
- Finally they represent the virtual particles living between two vertices. Those have the properties of their real counter part, but not the mass of a free particle.
- Feynman diagrams are a useful visualization of a reaction but also a prescription for
calculating their probability amplitude. One does that by applying what is called Feynman rules.
- At each vertex, energy-momentum and quantum numbers are rigorously conserved.
- The virtual particle transfers energy momentum from one of these particles to the other one, thus a force act between the projectile and the target.
- Details on how to construct a Feynman diagram will be discussed in module 4 when we talk about electromagnetic interactions in a quantum way.
- For each type of elementary interaction there is a spin 1 boson transmitting the force as we have explained before. They are collectively called gauge bosons:
> The photon transmits electromagnetic interactions.
> The W and Z bosons transmit the two forms of weak interactions.
> The eight different gluons are responsible for transmitting strong interactions.
- To absorb or emit one of these bosons, a particle must carry the required type of charge. In our jargon we say that the boson couples to a given charge. The charge is thus a coupling constant.
> It needs electrical charge Q to couple to the photon. Q has 1 component.
> It needs weak isospin T and T_3 to couple to W and Z. This charge has 2 components.
> It needs color charge R, G or B to couple to gluons. This charge has 3 components.
- The probability amplitude to emit or absorb a gauge boson is proportional to the charge of the particle. The probability is thus proportional to the square of the charge as we found for Rutherford scattering already.
- In the Feynman diagram at each vertex energy and momentum are conserved. But do not forget that virtual particles which relate two vertices do not necessarily have the mass of their real counterpart nor even a real number as their mass.
- At each vertex charges as well as baryon and lepton number are conserved. Flavor is conserved by electromagnetic and strong interactions as we stated before but not by the weak one. Charge weak interactions transmitted by a W boson in fact change a particle flavor.
- Here are a few examples of allowed and forbidden vertices.
> In this top left diagram, neither the photon nor the Z boson can change the flavor of the lepton. The electron must thus stay an electron.
> In the second top diagram from the left, conservation of charge requires that the emitted W boson has a negative charge. We can only emit a negative boson. This transforms the electron into a neutrino.
> On the right-hand diagram, we show a diagram that does simply not exist. Since the electron does not carry color charge, it cannot emit or absorb a gluon so this diagram is simply non-existant.
> On the bottom left you find a u quark that emits a photon or a Z. Neither of these two particles can change the flavor of the quark such that it must stay a u quark and cannot become any other quark.
> In the middle diagram, the flavor changes. The u quark becomes a d quark by emitting a W. And if you make the balance of the charges you start out with a plus two-third charge and you come out with a minus one-third charge quark. So you must emit a W^+ to conserve charge.
> The right-most diagram is a diagram of strong interactions. The incoming red d quark emits a gluon and becomes a green d quark. And in order to conserve color the quantum numbers carried by the gluon must be red, anti-green. On the other hand, the gluon is not able to change the flavor of the particles such that the d quark must indeed stay at d quark.
In the next video we will visit the laboratory of the nuclear physics course at University of Geneva to see how our students go about to measure the Rutherford cross-section.
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