Law of Impulse Preservation in Positron-Electron Annihilation

[Physicsissuef]

Annihilation (connecting positron and electrons) gives us 2 gamma rays. I.e
^{0}_{-1}e+ ^{0}_{+1}e \rightarrow 2 \gamma
In my textbook says:

The pair positron-electron is processing according to the laws of preservation of the energy and the impulse...
Impulse of the gamma quant is equal to the both particles, same as the nucleus (which is in interaction with), but, the summary impulse is same before and after the process... That is confirmed in The Wilson's chamber which is inside of magnetic and electric field. In that field the electron and the positron like opposite charged particles are going into opposite directions.

Now my question, what is law of preservation of impulse??
Why e is written like ^{0}_{-1}e instead of e^-^1 or something??

 

[pam]

1. Your text must be translated from one language to another.
I think when it says "preservation of impulse", it just means conservation of momentum.
2. That clumsy notation is what is used in nuclear physics.
It means nucleon number=0. The subscripts are the charge.
That notation is almost never used for simple electrons.

 

[Physicsissuef]

1. Your text must be translated from one language to another.
I think when it says "preservation of impulse", it just means conservation of momentum.
2. That clumsy notation is what is used in nuclear physics.
It means nucleon number=0. The subscripts are the charge.
That notation is almost never used for simple electrons.

Yes, I mean conservation of momentum... What is that?

 

[eys_physics]

Conservation of momentum means that the total momentum is the same after the reaction as before. For a particle with mass the momentum is

\textbf{p}=m\textbf{v}

where m is the mass and v the velocity of the particle.
For photons:

|p|=\frac{E}{c}

where E is the energy and c the speed of light (in vacuum).

 

[Physicsissuef]

in this case, it imply for the gamma ray? So
<br /> |p|=\frac{E}{c}<br /> is the one that we look for, right?

 

[eys_physics]

Yes in the final state you have two phonons, but in the initial state you have one positron and one electron and they have mass.

 

[Physicsissuef]

so mv=\frac{E}{c}, right?

 

[Physicsissuef]

and why p is in long brackets i.e |p| ?

 

[Physicsissuef]

is it same with conservation of energY?

 

[jtbell]

and why p is in long brackets i.e |p| ?

Momentum is a vector. It has both magnitude and direction. The vertical bars indicate that we are talking about the magnitude only.

 

[Physicsissuef]

conservation of energy and conservation of impulse are same in this casE?

 

[malawi_glenn]

so mv=\frac{E}{c}, right?

No, since the general expression for momentum is related via:

E^2 = p^2c^2 + m^2c^4

So for massless particles (as the photon):

P = E/c

There is no way that you can get mv = E/c

conservation of energy and conservation of impulse are same in this casE?

Nope.

Conversvation of total energy and conservation of momentum/impulse is not the same thing.

Why e is written like ^{0}_{-1}e
instead of e^-^1
or something??

They follow the nuclear notion, as pam pointed out.

 

[Physicsissuef]

E is energy, so what is the difference?

 

[malawi_glenn]

i) momentum is a vector, energy is a scalar.

ii) you have massive particles on the Left Hand Side (LHS) of the reaction.

 

[Physicsissuef]

So conservation of energy is for the gamma ray, and conservation of momentum for the particles, right?

 

[malawi_glenn]

NO

conservation of energy for the whole reaction &
consercation of momentum for the whole reaction!

momentum(LHS) = momentum(RHS)
&
Energy(LHS) = Energy (RHS)

Both equations must be fullfilled.

RHS = right hand side
LHS = left hand side

 

[Physicsissuef]

And what is conservation of impulse, said with simpler words? Maybe some analogy?

 

[malawi_glenn]

impulse is just change of momentum. That is the real definition of it. So it can't be conserervation of impulse.

Pam wrote:
"1. Your text must be translated from one language to another.
I think when it says "preservation of impulse", it just means conservation of momentum." in pots #2

And sometimes, authors uses impulse sloppy instead of momentum. So I confirm what Pam wrote, what is meant is conservation of momentum.

 

[Physicsissuef]

but what is the impulse of the mass objects? Massless objects like gamma ray, the impulse is that line (which is not straight)

 

[malawi_glenn]

but what is the impulse of the mass objects? Massless objects like gamma ray, the impulse is that line (which is not straight)

whar are you talking about? Impuse is that line (which is not straight)?!

Instead of talking about impulse, let's just use the word momentum in the future, to avoid misunderstanding ;)

The momentum of massive obejcts is (if non relativistic) p = mv
if relativistic particles, use relativistic kinematics.

 

[Physicsissuef]

And how it says that the momentum is same before and after the reaction (when at the start we have massive objects and at the end massless objects)? First we have p=mv and second p=E/c^2, right? So those two should be equal?

 

[malawi_glenn]

well yes, in this specific case, you would have something like: 2*mv = 2E/c

I wanted to stress that (i) momentum is vectors and (ii) how momentum relations works.

You wasn't specific enough when you made your statement "mv=E/c", I think.

 

[Physicsissuef]

and why p=E/c. I don't figure out what is the logic. In m*v there is mass*change in velocity... But In p=E/c I don't see much the logic...

 

[malawi_glenn]

and why p=E/c. I don't figure out what is the logic. In m*v there is mass*change in velocity... But In p=E/c I don't see much the logic...

a photon is massless.

THIS is the formula for momentum that ALWAYS holds:

E^2 = p^2c^2 + m^2c^4

As i said in post #12 ...

E is total energy, i.e kinetic energy + rest energy (mass) and potential energy.

¤ so for massive particles, non relativistic ( v < 0.1c) we get (after some work): p = mv
¤ and for mass-less particles we get: p = E/c


How you got: "m*v there is mass*change in velocity" is for me a riddle, v is not change i velocity.. v is the velocity.

 

[Physicsissuef]

and what is p?

 

[malawi_glenn]

and what is p?

What has p resembled this entire thread?

And if I say that:

"THIS is the formula for momentum that ALWAYS holds:"
And says what E is, can't you draw the conclusion what p is then?
I even 'derived' this for you from that perticular formula:
p = mv & p = E/c

So, again, what do you think p is?

EDIT: If you want to know how E^2 = p^2c^2 + m^2c^4 is derived, borrow books in special relativity from your library and/or google.

 

[Physicsissuef]

p=change in momentum ?

 

[malawi_glenn]

post #4

 

[Physicsissuef]

Ok, my mistake, but... If the speed of the electrons and their mass is same, than the momentum will be equal to 0 kg*m/s before the collision... Also after the collision will be
0 kg*m/s if there is conservation of momentum. Will that momentum correspond to the gamma ray?

 

[malawi_glenn]

Ok you mean if their speed is equal in magnitude and in direction, and their masses is the same?

Again, one must use vector notation, momentum is really a vector (see again post #4)

\vec{p} = \sum _{i=1}^3p_i\hat{x}_i = m \vec{v}

where the hat-x is unit vector for the i:th coordinate.

So from E^2 = p^2c^2 + m^2c^4 one gets the total magnitude of momentum:
p = |\vec{p}| = \sum _{i=1}^3p_i

Then one has to perform all the vector algebras and so on.

 

[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...