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0

Never mind, I got it. You simply evaluate the two integrals, and fiddle around until you get them into an idential form.


3

You only need to include $u$ and $t$ channels when the final particles are indistinguishable (classically thinking, you don't know which blip in your detector came from which initial particle). Since $e^-$ and $e^+$ are distinguishable, you will always know which one is which and only $t$ channel contributes to the scattering amplitude.


1

That particular plot (source) seems to be a plot of the number of neutron captures as a function of energy. Each narrow peak in neutron energy corresponds to the energy of an excited state in the Th-233 or U-239 nucleus --- that is, the target nucleus with an extra neutron. The size of the capture cross section, plus the properties of the decay radiation ...


0

How can a scattering process have bound states? We are familiar with bound states from our everyday experience. For example, two hydrogen atoms interact through the Coulomb force. This leads to the formation of a bound state, namely, the hydrogen molecule. The most simple model of this situation is the square-well potential. This potential has ...


2

Starlight, as emitted by a star, comes in a wide range of colours. For instance see the picture below. Now this is a picture, and pictures can often be tricky with their representation of colour, so you'll have to take my word for it that Betelgeuse does look significantly redder to the naked eye than say Vega until you get a chance to go look yourself on ...


4

The colour of stars as observed by an observer on Earth varies just like the colour of our own Sun, depending on where in the sky the source is relative to the observer. However, the light of stars is generally too faint to notice this as clearly with the naked eye, because we cannot perceive colour for weak light sources.


3

No, Rayleigh scattering models the probability (and angle) of scattering as a function of wavelength and of the particle sizes. All wavelengths travel a long way but the path followed (scatter or nonscatter) varies. Since space is mostly "empty", there's little scattering. Beyond that, your understanding of stars is quite incomplete. THey do in fact have ...


0

First, you have to realise that in the nucleus there are two main forces that you have to consider (relevant to this question, at least): the electromagnetic force and the nuclear force. The former interests particles with an electric charge (protons, electrons), the latter holds protons and neutrons together - remember that protons would repel, so you need ...


0

In asymptotic region (high energy $s\rightarrow \infty$ and small momentum transfer $t$), the eikonal approximation means we drop out any diagram that has connections between internal lines; or it is corresponding to taking infinite ladder and cross-ladder Feynman's diagrams (including tree-level diagrams) in calculation of scattering amplitude and ...


11

Absolutely. If Compton scattering occurred in first order in $e$, the only contributing diagram would be the obvious one. Say we're in a frame with the electron initially at rest and an incoming photon in the $z$ direction. Then the electron 4-momentum is $$p^\mu_{\text{in}} = (m,0,0,0)$$ while the photon 4-momentum is $$k^\mu_{\text{in}} = ...


-1

The problem can be solved by introducing the relations \begin{equation} n_2-n_1=1,\:\:\:n_3-n_2=1 \end{equation} Since the resonance occurs for the energies $E_1$,$E_2$, and $E_3$ in a sequence the argument for the sine $n \pi$ also occur in a sequence in this case. This leads to one of possible solvable equations for the well potential \begin{equation} 2 ...



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