# Law of impulse

## Main Question or Discussion Point

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:
mytextbook said:
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??

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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.

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?

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 enery and c the speed of light (in vacuum).

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

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.

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

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

is it same with conservation of energY?

jtbell
Mentor
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.

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

malawi_glenn
Homework Helper
Physicsissuef said:
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

Physicsissuef said:
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.

Physicsissuef said:
Why $$e$$ is written like $$^{0}_{-1}e$$
instead of $$e^-^1$$
or something??
They follow the nuclear notion, as pam pointed out.

E is energy, so what is the difference?

malawi_glenn
Homework Helper
i) momentum is a vector, energy is a scalar.

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

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

malawi_glenn
Homework Helper
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

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

malawi_glenn
Homework Helper
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.

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
Homework Helper
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, lets 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.

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
Homework Helper
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.

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
Homework Helper
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.

and what is p?