Law of Impulse Preservation in Positron-Electron Annihilation

[malawi_glenn]

Yes, in my textbook says that they are moving in opposite directions... But still can't understand. I found this site which states for mass http://www.glenbrook.k12.il.us/gbssci/Phys/Class/momentum/u4l2b.html
And it says that the momentum will be zero, if we have two same objects (same mass and same velocity, but moving in opposite directions)...

exaclty WHAT are you not understanding?

 

[Physicsissuef]

that the momentum should be 0 before and after collision...

 

[malawi_glenn]

that the momentum should be 0 before and after collision...

ok

i) conservation of linear momentum is one how the most fundamental things in physics. It is derived from translation symmetry of physical systems.

ii) Momentum is a vector, for non-relativistic particles: $$\vec{p} = m\vec{v}$$

iii) Momentum can be added, to form Total momentum of a system of particles

iv) For a system of two particle, with same mass, moving at opposite directions with equal magnitude of velocity, the Total momentum of the system is:
$$\vec{p}_{\text{tot}} = m\vec{v} + m(-\vec{v}) = \vec{0}$$

Now due to (i): Tota momentum after, must also be equal to the 0-vector.

Does this makes things clearer? If you NEVER have heard of momentum before, or have no clue what a vector is and how they work, I suggest that you consult a textbook, or google a bit more. Read, study, and practice.

 

[Physicsissuef]

Isn't momentum for mass object kg*m/s ?

 

[malawi_glenn]

Isn't momentum for mass object kg*m/s ?

What is the matter with you?

p = m*v

m is measured in kg

v is measured in m/s

It seems like you ask questions just in desperation...

 

[Physicsissuef]

So the momentum before and after the collision is zero? Why you said that it isn't?

 

[malawi_glenn]

So the momentum before and after the collision is zero? Why you said that it isn't?

I never said that, I said that the momentum of ONE gamma ray can't be equal to zero.

Post #33
"Yes. IT is gamma rays. As we said conservation of energy and conservation of momentum are two different things... So will the momentum of gamma ray be zero?"

etc.

Then look at my post #38, you see that Iam arguing that the momentum before and after the collision is indeed equal to 0.


You must use better english I think.
1 gamma ray
2 gamma rays
3 gamma rays

etc.

I mean, I answer the things that I read, so please be specific and clear when you write.

 

[Physicsissuef]

Sorry, you are too much specific :) but no problem, next time I will try to be better...
Yes, the momentum of the gamma rays will be 0, that means that they move in opposite directions, but can't somehow imagine the gamma rays, since momentum of massless objects, looks difficult to me. Are the massless objects must "target" to collision (like the mass objects do), so we can determine the momentum?

 

[malawi_glenn]

again, you must look at the GENERAL formula for momentum, post #24 and #12

So the result is, for massless particles: p = E/c

The thing that it may look difficult to you is that it comes from special relativity, which you may not have been exposed to yet in school?

So how one does, in practice, meausure the momentum of a massless particle (a photon) is related to measure its energy. And now we have moved from introductory Newtonian dynamics to detector physics.

Here you have some easy read references:
http://heasarc.nasa.gov/docs/xte/learning_center/universe/photon.html
http://en.wikipedia.org/wiki/Scintillator

Particle detectors are cool and useful tools, that is one of my favorite areas in physics (radiation & detectors). They are used in both industry, medecine and fundamental research. So have fun with those links ;-)

And at last:
I must be specific, otherwise there can be missunderstandings.

 

[Physicsissuef]

Do the rays must interact to have reversible reaction?

 

[malawi_glenn]

Do the rays must interact to have reversible reaction?

you mean if you can have:

$$\gamma + \gamma \rightarrow e^- + e^+$$ (eq 1)

??

I know at least that this reaction is possible, if it take place near a nucleus or an atom:
$$\gamma \rightarrow e^- + e^+$$ (eq 2)

For (eq 1), I really don't know if two gammas can combine to form a pair of electron and positron. Must check that one up.

Perhaps ask it yourself in a new thread?

 

[Physicsissuef]

Ok, I did it. Thanks for advising...