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Law of impulse

  1. Apr 6, 2008 #1
    Annihilation (connecting positron and electrons) gives us 2 gamma rays. I.e
    [tex]^{0}_{-1}e+ ^{0}_{+1}e \rightarrow 2 \gamma[/tex]
    In my textbook says:
    Now my question, what is law of preservation of impulse??
    Why [tex]e[/tex] is written like [tex]^{0}_{-1}e[/tex] instead of [tex]e^-^1[/tex] or something??
     
  2. jcsd
  3. Apr 6, 2008 #2

    pam

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    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.
     
  4. Apr 7, 2008 #3
    Yes, I mean conservation of momentum... What is that?
     
  5. Apr 7, 2008 #4
    Conservation of momentum means that the total momentum is the same after the reaction as before. For a particle with mass the momentum is

    [tex]\textbf{p}=m\textbf{v}[/tex]

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

    [tex]|p|=\frac{E}{c}[/tex]

    where E is the enery and c the speed of light (in vacuum).
     
  6. Apr 7, 2008 #5
    in this case, it imply for the gamma ray? So
    [tex]
    |p|=\frac{E}{c}
    [/tex] is the one that we look for, right?
     
  7. Apr 9, 2008 #6
    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.
     
  8. Apr 9, 2008 #7
    so mv=[itex]\frac{E}{c}[/itex], right?
     
  9. Apr 13, 2008 #8
    and why p is in long brackets i.e |p| ?
     
  10. Apr 15, 2008 #9
    is it same with conservation of energY?
     
  11. Apr 15, 2008 #10

    jtbell

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    Staff: Mentor

    Momentum is a vector. It has both magnitude and direction. The vertical bars indicate that we are talking about the magnitude only.
     
  12. Apr 15, 2008 #11
    conservation of energy and conservation of impulse are same in this casE?
     
  13. Apr 15, 2008 #12

    malawi_glenn

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

    [tex] E^2 = p^2c^2 + m^2c^4 [/tex]

    So for massless particles (as the photon):

    [tex] P = E/c [/tex]

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

    Nope.

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


    They follow the nuclear notion, as pam pointed out.
     
  14. Apr 15, 2008 #13
    E is energy, so what is the difference?
     
  15. Apr 15, 2008 #14

    malawi_glenn

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

    ii) you have massive particles on the Left Hand Side (LHS) of the reaction.
     
  16. Apr 15, 2008 #15
    So conservation of energy is for the gamma ray, and conservation of momentum for the particles, right?
     
  17. Apr 15, 2008 #16

    malawi_glenn

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    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
     
  18. Apr 15, 2008 #17
    And what is conservation of impulse, said with simpler words? Maybe some analogy?
     
  19. Apr 15, 2008 #18

    malawi_glenn

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    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.
     
  20. Apr 15, 2008 #19
    but what is the impulse of the mass objects? Massless objects like gamma ray, the impulse is that line (which is not straight)
     
  21. Apr 15, 2008 #20

    malawi_glenn

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

    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.
     
  22. Apr 16, 2008 #21
    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?
     
  23. Apr 16, 2008 #22

    malawi_glenn

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    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.
     
  24. Apr 16, 2008 #23
    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...
     
  25. Apr 16, 2008 #24

    malawi_glenn

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    a photon is massless.

    THIS is the formula for momentum that ALWAYS holds:

    [tex] E^2 = p^2c^2 + m^2c^4 [/tex]

    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.
     
  26. Apr 16, 2008 #25
    and what is p?
     
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