Law of Impulse Preservation in Positron-Electron Annihilation

[Physicsissuef]

How will I know all those this for the gamma ray? I mean, will the momentum of the gamma ray be 0?

 

[malawi_glenn]

How will I know all those this for the gamma ray? I mean, will the momentum of the gamma ray be 0?

You have one more equation; conservation of energy...

We have as many unknowns as equations -> the system can be described completely.

Now this thread has gone to discussing basic fundamental Newtonian mechanics. I really don't have the time to explain this for just one person when there are thousands of textbooks and internet tutorials about linear Newtonian dynamics.

Ps. it is GAMMA RAYS, you have two of them, not just one.

 

[Physicsissuef]

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?

 

[malawi_glenn]

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?

yes they are completley different things, but since both must be fullfilled, we can get information of the quantities need in the equation for momentum conservation (and vice versa, it depends on which quantities that are given in the orginal problem).

have you not done physics courses yet?..

No, the momentum of a gamma ray will not be zero, such things don't even exists.

Think about it, if momentum is zero, then E is zero. If E = 0, then it does not exists. E of a gamma ray is also related according to:

E = h\nu, where \nu is the frequency of that gamma ray. h is a constant. So \nu = 0 , and the gamma ray does not exist.

Even if it WAS possible to have zero energy photons, energy will not be conserved in this reaction.

HOWEVER the total momentum of the RHS of the reaction must be equal to the 0-vector:
\vec{0} = \vec{p}_{\gamma 1} + \vec{p}_{\gamma 2}
(if now the electron and positron on the LHS has the same mass and moves with equal magnitude of velocity and in exactly opposite directions)

As an example, this might be the case, that from the conservation of energy equation that:
E_{\gamma 1} = E_{\gamma 2}

Then the gamma rays must be moving at exactly opposite directions in order to maintain momentum conservation, i.e:

\hat{x}_{\gamma 1} = - \hat{x}_{\gamma 2}

Where \hat{x} is an arbritrary direction-vector.

I don't think I can't make myself clearer without writing a short book about this topic.

 

[Physicsissuef]

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)...

 

[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...