Proving Isolated Photon Doesn't Split into Two Photons

AI Thread Summary
The discussion centers on the challenge of proving that an isolated photon cannot split into two photons traveling in different directions. The initial argument highlights the relationship between energy and momentum, emphasizing that if a photon were to split, it would violate momentum conservation unless the resulting photons had mass or moved in opposite directions. Participants reference equations related to energy and momentum conservation, specifically focusing on the conservation of 4-momentum. The conversation also touches on the implications of these equations in the context of photon behavior. Ultimately, the discussion aims to clarify the fundamental principles governing photon interactions and conservation laws.
Xamfy19
Messages
60
Reaction score
0
Here is the question I had trouble with it:

How to prove that an isolated photon cannot split into two photons going in direction other than the original direction.

I thought photon doesn't have mass, then

energy = momentum.
from momentum conservation law. If photon split into two part and travel in different direction, then it must have different mass (or at opposite direction).

That's all I have for the time being.
Thanks for the help.
 
Physics news on Phys.org
p^{\mu}=p_{1}^{\mu}+p_{2}^{\mu} (1)

Workin' in Heaviside-Lorentz units,

\omega=\omega_{1}+\omega_{2} (2)

\vec{p}=\vec{p}_{1}+\vec{p}_{2} (3)

Square (2) & (3) and use \omega=|\vec{p}| (which is valid for all 3 photons).

Denote the angle between the momenta of outgoing photons by \theta

Daniel.
 
Great, but what are the p^u, w and p?

I guess
w = w1 + w2 is the energy conservation
p = p1 + p2 is the momentum conservation, and what is

the first equation ?

thanks
 
Conservation of 4-momentum,what else...?That's where the other 2 eqns come from.

Daniel.
 
thanks, Daniel
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top