Solve Conservation of Energy & Momentum: Photon & Electron Collision

  • Thread starter Thread starter Minus1
  • Start date Start date
  • Tags Tags
    Relativity
Minus1
Messages
16
Reaction score
0
A photon of energy E traveling in the +ve x direction collides elastically with an electron of mass m, moving in the opposite direction. After the collision the photon travels back along the -ve x direction with the same energy E.

Use the conservation of energy and momentum to demonstrate that the initial and final momenta of the electron are equal and opposite and of magnitude E/c.

This question is worth 10 marks and I am kinda lost on what to do soany help at all would be greatly appreciated.
 
Physics news on Phys.org
Welcome to PF!

Hi Minus1! Welcome to PF! :wink:

The photon has energy E, so what is its momentum?

Now suppose the electron has momentum p before, and momentum q after …

what is the energy of the electron before and after? :smile:
 
Minus1 said:
A photon of energy E traveling in the +ve x direction collides elastically with an electron of mass m, moving in the opposite direction. After the collision the photon travels back along the -ve x direction with the same energy E.

Use the conservation of energy and momentum to demonstrate that the initial and final momenta of the electron are equal and opposite and of magnitude E/c.

This question is worth 10 marks and I am kinda lost on what to do soany help at all would be greatly appreciated.

In an elastic collision, the total kinetic energy of the colliding bodies won't change. So the conservation law of momentum gives
P_1+P_2 = P'_1+ P'_2,
that is, for the collision of an electron [with all specifications indexed by 1] and a photon [with all quantities indexed by 2]. Since the direction of motion of photon has been reversed, so
P'_2 = -P_2
which its insertion into the preceding equation gives
P_1-P'_1 = -2P_2.
Now we switch to the conservation law of energy that, assuming electron is moving at low speed, says
P^2_1/2m + |P_2|c =P'^2_1/2m + |P'_2|c.
By assumption, |P_2|c = |P'_2|c. Thus from the above equation we can get P'_1 = +- P_1. From this point I'll leave the remainder for you ro gain the required result.

AB
 
Thanks guys, I wish I could take you in the exam but I don't think they'd allow it, lol
 


tiny-tim said:
Hi Minus1! Welcome to PF! :wink:

The photon has energy E, so what is its momentum?

Now suppose the electron has momentum p before, and momentum q after …

what is the energy of the electron before and after? :smile:


Thanks for the warm welcome, I feel at home already
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top