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Homework Help: Neutron collision from com frame

  1. Dec 31, 2009 #1
    1. The problem statement, all variables and given/known data

    A neutron with kinetic energy T= 2mc^2 (m=rest mass),strikes another stationary neutron.What will be the combined kinetic energy of both the neutrons in the frame of their centre of inertia and the momentum p of each neutron in that frame.what is the velocity of the centre of inertia of this system of particles.

    2. Relevant equationsi think they are E=mc^2 +T
    MOMENTUM 4 VECTOR(E/C, p)



    3. The attempt at a solutioni am first trying to calculate the velocity of centre of mass by using v com= (m1 v1 +m2v2)/(m1+m2)
    and this is where i get stuck. in relativity, what values of m1 and m2 will i use? rest mass/the mass of a moving particle ie. relativistic mass), also how will i find v, i am given the relativistic kinetic energy 2mc^2= (gamma-1)mc^2, so gamma factor=3, which gives v=root(8)/3 c.
    a side question coming to my head is that does the total energy E contain all energies - ie electrostatic, magnetic, gravitational?(THIS IS NOT RELATED TO THIS PROBLEM, THOUGH)
     
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  3. Jan 1, 2010 #2

    vela

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    One suggestion I have is to avoid working with velocities in special relativity if you can, and stick with working with energy and momentum. Another suggestion is to work with four-vectors and Lorentz invariants.

    If you can answer these questions, it'll help you get started:

    1. What is the square of a neutron's four-momentum equal to?
    2. What is the total three-momentum in the center-of-mass frame?

    Start off by writing down the four-momentum of each neutron in the two frames and calculate the four-momentum of the system in both frames.
     
  4. Jan 1, 2010 #3
    Thanks for the reply. I just began relativity few days ago. see if i am in right track,please.

    ans1) i think p^2=m^2c^2 gamma^2-m^2v^2gamma^2=m^2c^2 where m is the rest mass.
    it is coming a constant.(because rest mass is the same in all frames,right?)

    ans2) 3 momentum-does this mean the normal momentum? well ,in centre of mass frame i will see both the neutrons moving,But to find their velocities in com frame, i need to know the velocity of the com itself, after that i will subtract it to find the relative velocities. but wait! that will be galilean. i will have to apply einstein velocity addition rule. It is getting complicated.......
    ok,i am writing 4 momentum in lab frame first....
    (3mc, 3 mv)-for the moving one
    (mc, 0)- for the sitting one
    so total=(4mc, 3mv)
    now in com frame..........
    here the gamma factor will be different, but i need it to know the velocity of com to know it. this is where i am getting stuck,again.:cry:
     
    Last edited: Jan 1, 2010
  5. Jan 1, 2010 #4

    vela

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    Looks good so far!

    I said "three-momentum" just to make clear I'm not referring to the four-momentum. The three-momentum is the usual (relativistic) momentum [tex](p_x,p_y,p_z)[/tex].

    Let me repeat, avoid thinking in terms of velocity. Energy and momentum are the more useful quantities. Using velocities just makes things more complicated algebraically because you have all these v's and [tex]\gamma[/tex]'s flying around. If you can figure out E and p for an object, where E is the total energy and p is the (relativistic) three-momentum, you can always get its velocity because [tex]\beta = pc/E[/tex]. (Sorry, if I seem to screw up factors of [tex]c[/tex] here and there. I'm used to setting [tex]c=1[/tex] and not having them appear in the equations, and I may put them back in incorrectly.)

    Your expression for the square of the four-momentum of the neutron is correct. In terms of energy E and three-momentum p, you can write that as [tex](E/c)^2 - p^2 = (mc)^2[/tex]. The four-momenta in the lab frame are correct as well, but again, instead of writing 3mv for the momentum of the moving neutron, just call it p. By squaring the four-momentum of the moving neutron, you can solve for its momentum p in terms of its mass.

    In the center-of-mass frame, as you noted, both neutrons will be moving. How are their (three-) momenta related in this frame? What about their energies?
     
  6. Jan 1, 2010 #5
    many thanks again:smile:
    well, atleast i got the first part correct.(i think)
    in lab frame, total momentum vector=(4mc,p) where p=root over(8)mc(since for moving neutron 9m^2- p^2=m^2, gives p=root 8 mc)
    so the dot gives 16m^2c^2-p^2=8m^2c^2(i do not know why this thing is not coming 4m^2c^2 since the total rest mass is 2m????) which should be the same in com frame.
    now in com frame, both the neutrons are moving with equal velocities in opposite directions(though they may not be v/2 as in classical mechanics), hence their momenta cancel out and so i can write the total mom vector in com frame as (E prime/c, 0) where E prime is the new total energy.
    so (E prime^2/c^2-0)= 8m^2c^2(the old thing i found in lab frame)=8m^2c^2
    so new total energy is root(8)mc^2) which surprises me because this was infact the old momentum in lab frame.(apart from a factor of c)
    well, so new total energy= rest mass+ new total k.e
    =2mc^2+ new total ke, solving k.e in com frame=(root(8)-2)mc^2 =779MEV which is pretty close to the answer given as 777MEV.(guess the neutron mass is behind this, i took it as 1.672 E-27 kg)


    i can find the velocity of the centre of inertia as well, because Eprime^2-p prime^2=(2m)^2 c^2=4m^2 c^2(since toatal rest mass is 2m)
    putting in E prime, i get p prime=2mc
    so the total momentum vector in com frame=(root8 mc, 2mc)
    now the ratio of the spatial and energy part will give me v/c.(=beta), this gives velocity of centre of mass which is matching the answers.
    but a second thought tells me i am in deep trouble........
    you see, i assumed total spatial p =0 in the first part and now, the second part ,i used total new spatial p=2mc
    i never knew relativity is so confusing.:confused:
     
  7. Jan 2, 2010 #6
    maybe the problem is wrong. however it is an IRODOV problem.-1.384. i still think my concepts are going wrong somewhere......
     
  8. Jan 2, 2010 #7

    vela

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    Your instinct is correct. The fact you got the correct answer is a fluke.

    The rest mass of the two-particle system isn't the sum of the masses; it's the total energy of the system in its rest frame divided by [tex]c^2[/tex], i.e. [tex]E'/c^2[/tex].

    You have three pairs of four-vectors, and the four-vectors in each pair are related to each other by a Lorentz transformation. You can use any pair to calculate the velocity that produces the appropriate boost.

    You could also consider the neutron at rest in the lab frame. In the center-of-mass frame, it's moving, and its velocity is easily found. How is its velocity in CM frame related to the velocity of the center of mass in the lab frame?
     
  9. Jan 2, 2010 #8

    vela

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    Yup.
     
  10. Jan 2, 2010 #9
    calculating.......
    velocity of com wrt lab frame is coming c/root(2)
    vel of the static neutron wrt com frame= -c/root(2)
     
    Last edited: Jan 2, 2010
  11. Jan 2, 2010 #10
    is that wrong? will i show my cals?:frown:
     
  12. Jan 2, 2010 #11

    vela

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    Oh, I don't know. I haven't worked that part out. How did you calculate it?
     
  13. Jan 2, 2010 #12
    i used the total momentum vector in lab frame=(4mc,root8 mc) and the same thing in com frame(root8mc, 0) and related them by lorentz transformation assuming an arbitrary beta and solved it out. beta came 1/root2.
    and guess what- all the answers are matching now. I HAVE GOT HOLD OF THE PROBLEM.i do not know how to express my gratitude to you.:smile:
     
  14. Jan 2, 2010 #13
    will you please tell me one thing? in com frame , i used spatial p of the total momentum 4 vector =0applying instincts from the classical version of the problem. but how can u guarantee that the two p s will cancel out relativistically??
    and also i asked another question in my original post about relativistic energy. can you helpme there a bit. (if time permits ,of course). u have already helped me a lot.
    thanks.
     
  15. Jan 2, 2010 #14
    ok, another question is that say ,the problem asked me to find the rest mass of the system.now acc. to u , it will be total energy in rest frame/c^2) which i also agree. but what will be the rest frame in this case... ie. one mass is moving at v (say) and other is at rest. ??
     
  16. Jan 2, 2010 #15

    vela

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    Glad you got the problem figured out. I too found the same velocity you did.

    It's essentially the definition of the COM frame: the reference frame where the total momentum is zero. If the momenta didn't cancel, the system would have a net momentum, and its center of mass would be moving, which can't happen in the COM frame.

    In a sense, E does include various forms of energy because the mass of, for instance, a hydrogen atom is the sum of the mass of the proton, electron, and the binding energy, but I'm guessing that's not what you meant by your question. I'm pretty sure E is just the rest energy plus the kinetic energy.
     
  17. Jan 2, 2010 #16
    thanks.
    but yes, i meant exactly that. ie. in my book it says total relativistic energy of a particle- now when i say TOTAL , then i generally mean everytype of energy .but in the formulasion, i have only the rest energy and the kinetic energy. where are the others taken care of?

    say, when i drop a ball from roof, its potential energy is decreasing, / or when a charge accelerates, it is radiating energy, how is that accounted by the formula? now one explanasion can be that these effects change the velocity, and thus are incorporated in the formulasion of E=gamma m c^2.
    thanks.:smile:
     
  18. Jan 2, 2010 #17

    vela

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    It would be the COM frame. The velocity of a system is the velocity of its center of mass, so a rest frame of a system is one where this velocity is zero, which occurs in a COM frame.

    Think about a brick, for example. We can consider a brick to be at rest even though all of the atoms that make up the brick are moving. In its rest frame, the momentum of the brick vanishes, and the momentum of the brick is just the sum of the momenta of its atoms. In other words, the rest frame of the brick system is the frame where the momenta of the atoms sum to zero.
     
  19. Jan 2, 2010 #18

    vela

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    I was trying to make the distinction between the internal energy of an object and energy the object may have due to an external field. The mass of the object includes the internal energy: an object has more mass when it's hot than when it's cold, for instance. So this internal energy, due to whatever interaction, is accounted for in the rest mass of an object. But say you have a charged particle in an electrostatic field. It has a potential energy associated with its position in the field. This energy is not included in E. Consider the fact that since the electric force is conservative, the sum of the kinetic and potential energies is constant. If the potential energy were included in E, E wouldn't change even as the particle accelerates. If E and m are constant, then p must be as well, which implies that the particle isn't accelerating, which would clearly not be the case. Total energy just means kinetic plus rest energy, as opposed to just kinetic energy.

    With the dropped ball, the decrease in potential energy appears as an increase in kinetic energy and therefore in the total energy as well. For a charged particle radiating, the photon will carry off energy and momentum, and this will be matched exactly by a decrease in the energy and momentum of the particle.
     
  20. Jan 2, 2010 #19
    confusion cleared. thanks!!
     
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