Consider the scattering of a photon off a particle at rest,...

Consider the scattering of a photon off a particle at rest, or more generally the scattering of massless particles off fixed targets. Label the particle momenta $q = p_A$ (photon), $p = p_B$, $q' = p_C$ (photon), $p' = p_D$. Denote the energy of the incident photon by $\omega$, the mass of the target particle by $m$, the energy of the scattered photon by $\omega'$, and that of the scattered target particle by $E'$. Denote the scattering angle (in the lab frame) by $\theta$. Show that $\frac{1}{\omega'} = \frac{1}{\omega} + \frac{1-\cos\theta}{m}$

Transcript

00:01 And this question, you were looking at basically a campus gathering between a photon, which has the energy.

00:06 Let me see e photon, right? ei, the instant photon has an energy which is 0 .8 mini electron roots, right? and it's gathered by electron initial rest.

00:17 So let me try to draw the diagram for this scattering.

00:21 So suppose the electron is somewhere here, the electron e, and the photon comes in with energy ei, right? of course, you know, it has also weblands lambda i, which is given by, you know, i think it's given by hc divided by, divided by, er, right? right.

00:52 I think that is the, that is the wavelength of the instant light.

00:59 And then it's going to be scattered into a different direction, which, let's say, this.

01:05 Scattered wave a photon is going this way.

01:07 You have a final energy, which is ef.

01:10 And of course, you will also have a wavelength which is lamta, which is h to c over ef, right? so h is the plant constant that c is the speed of light in a vacuum, right? suppose the scattering angle, which is this is cta.

01:25 And the question says the electron is also scattered to a direction, which basically the scatter angle is the same for the scatter photo.

01:34 So this is also ceta, right? so the electron has been sketched this direction.

01:38 And especially there also has some energy that called the e of the electron.

01:43 First, we know that the momentum has to be conserved.

01:48 That means if the momentum has to be conserved, that means, you know, in particular, initially there was no momentum in this perpendicular direction, right? so these two momentum must exactly cancel it...

Question

Please give Ace some feedback