Can Two Colliding Photons Create a New Particle?

[budafeet57]

Homework Statement

Imagine that we discover a new particle, called P particle by a head-on collision of two photons of energies 500 Mev and 200 MeV. The photons are annihilated in the process.
a) what is the mass of the newly discovered particle P?
b) what is the kinetic energy of the P particles produced in this reaction?
c) what is the momentum of the P particles?
d) what is the velocity of the P particles(magnitude and direction)

Homework Equations

E = mc^2
K = p^2/2m
momentum: p1 + p2 = pf, where p1 and p2 have opposite signs
so E1/c + E2/c = 500/c + -200/c = pf = 300/c

The Attempt at a Solution

a) m = 1.24 * 10^-27 kg = 700 MeV/c^2

one possibility because all energy goes into creating particle:
b) KE = 0
c) p = 0
d) v = 0

second possibility
b) KE = 700 MeV
c) p = 300 MeV /c
d)

 

[Simon Bridge]

Imagine that we discover a new particle, called P particle by a head-on collision of two photons of energies 500 Mev and 200 MeV

Since $E=h\nu = pc$ then the photons don't have equal momenta.

Since no other particles are involved, can you end up with zero net momentum for the particle?

In photon-photon collisions - how many conservation laws are involved?

 

[budafeet57]

Since $E=h\nu = pc$

Since the photons have different momenta, and no other particles are involved, can you cannot end up with zero net momentum for the particle?

In photon-photon collisions - how many conservation laws are involved?

right... it cannot be zero momentum.

conservation of energy:
E1 + E2 = Ef
200 + 500 = 700 MeV
E = K + mc^2 where mc^2 = 700 MeV and K = 0?

and conservation of momentum:
p1 + p2 = pf

are these all?

oh, there's E^2 = (pc)^2 + (mc^2)^2

 

[frogjg2003]

Where did you get mc^2=700Mev?
You know the initial energies and momenta, so you can calculate the final energy and momentum. You haven't done anything with mass yet.

 

[Simon Bridge]

right... it cannot be zero momentum.

conservation of energy:
E1 + E2 = Ef
200 + 500 = 700 MeV
E = K + mc^2 where mc^2 = 700 MeV and K = 0?

If K=0, then p=0 (since $K=p^2/2m$), and you just said that momentum cannot be zero.

This last relation is just one equation with two variables ... rest-mass and momentum You need another equation with momentum in it to get a unique solution.

and conservation of momentum:
p1 + p2 = pf
are these all?

Conservation laws? Charge - lepton number - angular momentum... etc etc... how many can you think of? All of them have to be obeyed.

The point is that you have to consider more than just conservation of energy

oh, there's E^2 = (pc)^2 + (mc^2)^2

Total energy ... yep. Also $E=\gamma mc^2$ ... the equation you did before was for the non-relativistic case.

 

[budafeet57]

Where did you get mc^2=700Mev?
You know the initial energies and momenta, so you can calculate the final energy and momentum. You haven't done anything with mass yet.

I see my problem after a nice sleep lol.

I originally thought all energy from photon goes into creating the particle P, but because there's momentum left so that assumption was wrong and I cannnot use E = mc^2 where E is 700 MeV and thus m = 700MeV/c^2.

 

[budafeet57]

If K=0, then p=0 (since $K=p^2/2m$), and you just said that momentum cannot be zero.

This last relation is just one equation with two variables ... rest-mass and momentum You need another equation with momentum in it to get a unique solution. Conservation laws? Charge - lepton number - angular momentum... etc etc... how many can you think of? All of them have to be obeyed.

The point is that you have to consider more than just conservation of energy
Total energy ... yep. Also $E=\gamma mc^2$ ... the equation you did before was for the non-relativistic case.

So instead I should use E^2 = (pc)^2 + (mc^2)^2 to get mc^2, and get K = E - mc^2, then v.
now I get:
mc^2 = 632 MeV
K = 68 MeV
v = 0.43c

 

[Simon Bridge]

Cool.
Remains only to double-check your units and figure the direction :)

Aside:- I'll show you something:
once you also know
$E_{tot}=\gamma E_0$
it is easy to get gamma a lot becomes quite easy.

eg.
$(v/c)=\sqrt{1-(1/\gamma)^2}$

$K=(1-\gamma)E_0$

 

[budafeet57]

Cool.
Remains only to double-check your units and figure the direction :)

Aside:- I'll show you something:
once you also know
$E_{tot}=\gamma E_0$
it is easy to get gamma a lot becomes quite easy.

eg.
$(v/c)=\sqrt{1-(1/\gamma)^2}$

$K=(1-\gamma)E_0$

Thank you.

direction of particle should be positive x direction because the head-on collision is 1D and velocity is positive.

 

[Simon Bridge]

well ... as written you have two directions of motion in that 1D and you did not define a +ve direction. Which direction is each photon headed in? (If the problem in front of you does not explicitly define a direction to be positive, you should do so at the start. It's usually worth an extra mark in an exam.)

Asking for direction in (d) is a little redundant since the momentum, required in (c), is also a vector. Notice how the answer to part (c) is, along with the total energy, the first of these things you calculate? This is an important lesson for real life too:

You don't have to solve problems in the order presented.

;)

 

[budafeet57]

well ... as written you have two directions of motion in that 1D and you did not define a +ve direction. Which direction is each photon headed in? (If the problem in front of you does not explicitly define a direction to be positive, you should do so at the start. It's usually worth an extra mark in an exam.)

Asking for direction in (d) is a little redundant since the momentum, required in (c), is also a vector. Notice how the answer to part (c) is, along with the total energy, the first of these things you calculate? This is an important lesson for real life too:

You don't have to solve problems in the order presented.

;)

I can define +x and +v to the right side and the 500MeV photon is heading to the right while the 200MeV photon is heading to the left. The final momentum of the particle P will have positive value, which indicates that it's moving toward the right. This is definitely what I should think of in the beginning.

Thank you for very much. I have learned a lot!

 

[Simon Bridge]

No worries - you've been receptive.