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March 4, 2015

March 4, 2015

Posted by **Mischa** on Wednesday, May 7, 2008 at 6:08pm.

I got an answer for (a), but when i use it to find (b) the solution does not exist. I clearly did something wrong, and I can't figure it out.

- quantum physics -
**Count Iblis**, Wednesday, May 7, 2008 at 9:19pma) follows from conservation of energy:

Photon energy + m = photon energy after collision plus gamma m

(in c = 1 units)

In b) you have to take into account that both the photon after the collision and the electron after the collison are not moving in the same direction the incident photon was moving.

The simplest way to solve this problem is by writing conservation of energy and momentum as a single four-momentum equation:

pf + pe = qf + qe (1)

where

pf = four-mometum of photon before collision

pe = four-momentum of electron before collision

qf = four-mometum of photon after collision

qe = four-momentum of electron after collision

We are not interested in qe, so we want to eliminate it from the equation. What we do is we write (1) as:

qe = pf + pe - qf

We then square both sides:

qe^2 = (pf + pe - qf)^2 =

pf^2 + pe^2 + qf^2 + 2 pf dot pe +

- 2 pf dot qf -2 pe dot qf

Next substitute the energy momentum relation p^2 = m^2, so we have:

qe^2 = m^2

pe^2 = m^2

pf^2 = 0

qf^2 = 0

and the equation simplifies to:

2 pf dot pe - 2 pf dot qf -2 pe dot qf = 0

Pf dot qf contains the angle you want to know. Pe = (m,0,0) because the electron before the collision was at rest, so the Lorentz inner product of the two other terms are just the electron mass times the photon energies.

- quantum physics -
**Anonymous**, Saturday, May 7, 2011 at 10:44amHow much force will be needed to give a 10kg body resting on the ground an upward acceleration of 5m/s if acceleration due to gravity at the place is 10.02m/s?

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