How Do You Calculate the Kinetic Energy of a Meson That Decays into Two Photons?

[Chip90]

Homework Statement

A meson decays into two photos, each at 150 MeV in its rest frame. Find the KE of the meson if the photons decay at an angle of 60 degrees between them.

I have attached a picture for reference.

Homework Equations

Pi=Pf and Ei=Ef

The Attempt at a Solution

So for Pi=Pf I used

Pi= gamma*m*v
Pf= 2*h*(w/c)*cos30

where gamma is 1/(1- [v^2/c^2])^(1/2); m is mass of meson; v is velocity of meson
h is Planck's constant; w is angular frequency; c is speed of light; cos30 is for angle between photon and meson's path

i think mass of meson is 300Mev/c^2= 3.33e-9 eV/c^2 but this could be wrong since its not going at c
not sure what velocity is or how i find it

So for Ei=Ef

Since E^2=(mc^2)^2 + (pc)^2
Ei=[(mc^2)^2 + (pc)^2]^(1/2)

Since E=h*w
Ef=2*h*w

[(mc^2)^2)+(pc)^2]^(1/2)=2*h*w

So I then substitute h*w from momentum for gamma*m*v*c/sqrt(3)

then square both sides

(mc^2)^2 + (pc)^2=4*m^2*v^2*c^2/3


I am not sure if I am going in the right direction with this. I still don't know what is v for original meson. Thanks for the help

 

[vela]

Looks okay. What is p equal to in your last equation? Write it in terms of m, v, and gamma, and you should be able to solve for v.

 

[Chip90]

(mc^2)^2 + (pc)^2=4*m^2*v^2*c^2/3

this p in this? that would be the momentum of the meson, which is gamma*m*v so

m^2*c^4 + gamma^2*m^2*v^2*c^2=(4/3)*m^2*v^2*c^2

not sure what do to now.. is the mass correct that i calculated? if so, why, it isn't traveling at the speed of light, yet that equation would hold? (i mean that dividing Mev by c^2 gives mass)..

 

[vela]

Yes, the mass is correct. In the rest frame of the particle, all of its mass is converted into energy of the two photons. Just because c appears in the units doesn't mean the object is moving at the speed of light. In fact, because it has mass, it can't move at the speed of light.

In your most recent equation, you can write gamma in terms of v, so the only unknown left is v. Solve for it. Once you have that, you can calculate the kinetic energy of the meson.

 

[Chip90]

ok i entered in gamma and after a bunch of equations, i come up with this

v^4 - v^2 *c^2=-3/4 * c^4

does that seem right? the masses ended up canceling...

ok just tried that in my ti-89, it says false..

ok, i just entered my that equation in post 3 and it says false..

 

[vela]

The mass should cancel. You seem to have accidentally dropped the γ on the righthand side when you squared the equation. If you fix that, I think you'll get it to work out.

 

[Chip90]

ok, can't believe i missed that.

now i tried it again and get 0.72c

so how do i get the ke now?

do i just use

KE= (gamma -1)*m*c^2 ?

 

[vela]

Calculate the meson's total energy and subtract off the rest energy. What's left over is the kinetic energy.

 

[Chip90]

ok so total E is

E^2=(mc^2)^2+(pc)^2 and rest energy is (mc^2)^2 so

KE= E^2- (mc^2)^2=(pc)^2

so KE is just (pc)^2= (gamma*m*v*c)^2

that gives me 9.67e16 joules?

 

[vela]

No, K=E-mc². Also, it's better to keep the answer in units of MeV. For particle physics, it's a much more convenient unit than joules.

 

[Chip90]

ok so then I get 132MeV. Does that sound right?

and thank you so much for your help so far.. I've been a little stuck on it

 

[vela]

No. You should find the kinetic energy is equal to the rest energy.

 

[Chip90]

? i thought

KE= E- mc^2

where KE in kinetic energy and E is total energy right?

 

[vela]

Yes. What did you find for E? What did you get for γ?

 

[Chip90]

I got E is 431.86 MeV and gamma is 1.44

 

[vela]

Recheck your calculation of γ.

 

[Chip90]

hmm v is 0.72c and gamma is 1/sqrt(1-(v/c)^2) correct?

i get 1.44 still.. did i somehow mess up v?

.72^2=0.5814

1-0.5814=0.4816

sqrt 0.4816= 0.693974063

and 1/0.693974063 is 1.44

 

[vela]

Looks like it. I got v/c=sqrt(3)/2. Is your v actually v²?

 

[Chip90]

well i just plugged that eq into my ti-89 and it solved v for 216281140.43179 and I just divided by 3e8 and got 0.72c.

with the v you calculated i get

E=480.41 MeV and KE=180.7 MeV...

 

[vela]

well i just plugged that eq into my ti-89 and it solved v for 216281140.43179 and I just divided by 3e8 and got 0.72c.

Apparently, you plugged the wrong equation into your calculator.

with the v you calculated i get

E=480.41 MeV and KE=180.7 MeV...

I have no idea how you managed to get those numbers.

 

[Chip90]

ok so if v=sqrt(3)/2=0.87c

E^2=(mc^2)^2+(pc)^2

for p=gamma*m*v=1.44*3.33e-9*0.87*3.8=1.246

E^2=(3.33e-9*3e8*3e8)^2 + (1.246*3e8)^2
E^2=8.982e16 + 1.397e17
E^2=2.295e17
E=479,082,550 eV = 479.1 MeV

Then KE= E = mc^2
KE=479,082,550 - (3.33e-9*3e8*3e8)
KE=179,382,550 eV =179.4 Mev

I still get the same answer..

 

[vela]

You need to recalculate γ. You're still using the value for v/c=0.72.

 

[Chip90]

oh no, you're right gamma is actually 2

so

E=601.5 MeV

KE=301.76 MeV

That seems correct..

Just a question, what mattered if they separated at a higher angle, isn't the point of the problem to show that Energy is conserved?

 

[vela]

For a different angle, you'd find the energy of the photons have changed. It changes in exactly the right way so that both total energy and momentum are conserved.

Another way you can solve this problem is to look at it initially in the center-of-mass frame. You can immediately write down the energy and momentum of the photon in this frame. You can then calculate the Lorentz boost required to get the desired angle for the photons' momenta in the lab frame. That same boost applied to the meson would allow you to calculate its energy.