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Closed system Energy transformation

  1. Aug 29, 2015 #1
    Lets suppose we have a close system.In this system we have a particles.The total mass of particles is M.
    The total energy of system will be ##E_x+E_M=E_t## (I made the system like this) Here ##E_x## is just energy,its not important.##E_m## is the energy of masses.Now I want to move this system and I will gonna make ##E_x## zero.But for that as i said above I need to move the system.Then ##E_x=1/2Mv^2##.

    Now the problem exist when we use ##E_x## so large.Then v must be so high.Then the problem becomes relativity problem.I used Newtonian equation but then I noticed that v is so high (of course less then c like 0.96c).Now what should I do ? Should I change my equation an write ##E_x=(γ-1)Mc^2## ? But I didnt know that v will be so high.Is that a mistake ? Or its not important cause v is small than c.

    The other problem is the If I put this kinetic energy to the place of ##E_x## and then we will gonna have extra mass.This mass is this energy (movement energy) isnt it ?

    I means the mass of object will be γM and the energy will be ##γMc^2## ##Mc^2+1/2Mv^2## (Which I asked before in detail)

    Thank you
     
  2. jcsd
  3. Aug 29, 2015 #2

    Dale

    Staff: Mentor

    If your system is isolated then you need to account for momentum as well as energy. The easiest way to do this is to use the four-momentum. In units where c=1, you have the initial four momentum is ##(E,\vec{p})=(E_t,0)## and the final four momenta are ##(E_t /2,\vec p)## and ##(E_t /2,-\vec p)##
     
  4. Aug 30, 2015 #3
    Whats the ##E_t## here.Ok the logical thing is use ##(E_x+Mc^2)## 'c' is one so the initial momentum is ##(E_x+M,0)##.Lets suppose objecr moves only x direction.Then then four momentum is ##[(E_x+M)/2,γMv,0,0]## and the other one is ##[(E_x+M)/2,-γMv,0,0]##

    Am I right ?.If I am right what should I do next ?
     
  5. Aug 30, 2015 #4

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    How did you "make the system like this"? What property of the system, other than mass, gives it the energy ##E_x##? Also, in what frame are these energies being measured? Are all the masses at rest in that frame?

    This implies that it wasn't moving before. If so, once again, where does the energy ##E_x## come from?

    How will moving the system make ##E_x## zero?
     
  6. Aug 30, 2015 #5
    My main language is not English andI wrote this question when I was drunk.So let me clear things.Total energy must be conserved.Thus is the key of the question.

    Now the question is think a system In this system we have particles The total Energy of particles are ##E_M=Mc^2##.And we are observing aome extra energy but we dont know its type or thing.Just we now that.Lets call it ##E_x##.Now this time as you can see particles or system is not moving.(Cause the referance frame is inside the system or thats the choosen frame)

    Some physicist makes experiments and says look the ##E_x## can be the kinetic energy of system(physicist change the referance frame and he thought that system can move which will cause extra energy and thats the ##E_x##) the physicist use ##1/2Mv^2=E_x## but he saw that v is too high (near c).

    Hear my first problem Is Higher v a problem ? Why cause we used Newtonian equation but we get Higher v which I thought I used the equation wrong ?
     
  7. Aug 30, 2015 #6

    PeterDonis

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    2016 Award

    Staff: Mentor

    In a given frame, yes. But if you change frames, you change the total energy, because total energy is frame-dependent. To find an actual conserved quantity that is frame-invariant, you need to look at the 4-momentum, as DaleSpam said; the norm of the 4-momentum vector is an invariant (like the norm of any 4-vector).

    You mean, the total rest energy of particles.

    The obvious source of this extra energy is kinetic energy of the particles; that is, they are not at rest in the center of mass frame of the system as a whole. This is most commonly due to thermal vibrations.

    The system is not moving in this frame by definition. But that does not mean the individual particles are not moving. They could be; in fact, if the system has a temperature above absolute zero they will have to be.

    He doesn't have to change frames; the individual particles can be moving in the frame in which the center of mass of the system is at rest. See above. In any case, you defined ##E_x## as the extra energy, over and above the rest energy of its particles, that the system has in its own rest frame (i.e., the frame in which its center of mass is at rest). So ##E_x## can't be energy due to the motion of the system as a whole; that's inconsistent with how you defined it.

    If you change frames, then the total energy of the system will be larger because the system's center of mass is moving in the new frame. The new total energy will be ##\gamma \left( E_M + E_x \right)##, where ##\gamma = 1 / \sqrt{1 - v^2 / c^2}##. (The momentum of the system will also be nonzero in this frame, where it was zero in its own rest frame; so the norm of the 4-momentum will be the same.)

    Yes, when you should have used the relativistic equation. See above.
     
  8. Aug 30, 2015 #7

    Dale

    Staff: Mentor

    ##E_t## is your notation from the first post.

    Yes.

    What do you want to do next? I am not sure what your goal is. The next step depends on your goal.
     
    Last edited: Aug 30, 2015
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