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Tough relativity problem

  1. Mar 13, 2008 #1
    Two particles of masses m1 and m2 are travelling at v1 and v2, and, collide. after the collision, their speeds become v3 and v4 (every speed w.r.t. ground frame). now, if i watch this collision in a moving reference frame with velocity 'v' (in the opposite direction to the particles)

    Laws used:
    Conservation of momentum (classical and relativistic)
    conservation of total energy(relativistic only)

    'c' = speed of light in vacuum.
    bi = sqrt(1-vi^2/c^2) => b1 = sqrt(1-v1^2/c^2)

    vi' = speed vi as seen in the moving frame of reference.
    classical: vi' = vi + v
    relativistic: vi' = (vi+v)/(1+(vi*v/c^2))
    classical collision:
    momentum conservation:
    gnd frame:
    m1*v1+m2*v2 = m1*v3+m2*v4.
    knowing v3, i can uniquely find v4.
    moving frame:
    m1*v1'+m2*v2' = m1*v3'+m2*v4'

    knowing v3', i can uniquely find v4'. then, transform it to gnd frame (v4), and these two things match up. no problem here.

    now, coming to relativistic collision:
    gnd frame:
    momentum conservation:
    m1*v1/b1 + m2*v2/b2 = m1'*v3/b3 + m2'*v4/b4. yes, the rest masses change due to collision. so, even if we know v3 (we choose it arbitrarily), we cannot solve this equation of 3 unknowns.
    so, we go for TOTAL energy conservation.

    m1*c^2/b1 + m2*c^2/b2 = m1'*c^2/b3 + m2'*c^2/b4.

    i.e., m1/b1 + m2/b2 = m1'/b3 + m2'/b4.

    now, we have 2 equations and 4 unknowns. (remember, v3 is also unknown)
    for perfectly elastic collision, m1'=m1, m2'=m2. so, 2 equations, 2 unknowns. these are trancedental equations. solved in both the frames of reference, and then, converting moving frame speeds to ground frame speeds, answers match up.

    for perfectly inelastic collision, m1' and m2' are not there, it is only M (both the bodies are attached together and travel as one). also, M>m1+m2, and v3=v4. so, 2 equations, 2 unknowns, simple equations. solving them in both frames, transferring moving frame speeds to ground frame, the results match up.

    now, for partially inelastic collisions:
    for classical:
    we have coefficient of restitution. this is a measure of loss of kinetic energy in the collision. (actually, this is the total energy. no potential energy, and, of course, no mass energy)

    e = (v3-v4) / (v1-v2)

    in moving frame:
    e = (v3'-v4')/(v1'-v2') = (v3+v -(v4+v))/(v1+v -(v2+v)) = e in gnd frame.

    so, it is a unique number.

    in relativistic collision:
    we are already conserving the total energy. so, wot other property can we have for such collisions
  2. jcsd
  3. Mar 13, 2008 #2


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    Maybe [tex]\frac{m_1'+m_2'-m_1-m_2}{M-m_1-m_2}[/tex] is ok. M=rest mass of the system.
  4. Mar 13, 2008 #3
    In relativity you no longer have separate energy and momentum conservation equations. You only have one equation, for the conservation of the 4-vector energy-momentum. See top of page 115 in Taylor-Wheeler "SpacetimePhysics".
  5. Mar 22, 2008 #4
  6. Mar 22, 2008 #5
    and, can u tell me what is M ?
  7. Mar 22, 2008 #6


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    can u plz rite normally, and wud u plz reed "M=rest mass of the system".
  8. Mar 22, 2008 #7
    I am sorry, but, I am new over here, infact, just registered yesterday. and, it would be very nice of you to elaborate on what exactly is considered 'NORMAL' here. I thought that English was normal, with some SMS abbrevations thrown in, but then, now it seems that I am wrong!

    Yes, M = Rest Mass of the System, but is it before or after collision?

    The rest mass itself changes. The kinetic Energy lost in the collision is converted to rest mass of the individual particles. so, (m1'+m2'-m1-m2)/(M-m1-m2) got me confused.
  9. Mar 23, 2008 #8


    Staff: Mentor

    Most of us who answer questions on a regular basis are crochety old farts, and the SMS abbreviations really irritate us (remind us of our advancing age, decaying health, lost vitality, irritable bowel, etc. :smile:).

    Have you ever heard of the four-momentum? If not, you should look into it. The four-momentum relates energy and momentum in the same way that time and space are related in relativity. There are three nice things about the four-momentum.

    First, that it transforms by the Lorentz transform, so if you know the four-momentum in one reference frame you can easily determine the four-momentum in any other frame. Since the components of the four-momentum are related to energy and classical momentum you can therefore determine those in any reference frame.

    Second, the four-momentum is conserved. Not only is it conserved in collisions, but it is even conserved in nuclear reactions or other particle decay or creation events.

    Finally, the norm of the four-momentum is the "rest mass" or "rest energy" (I prefer the term "rest energy"). Since it is the norm of a four-vector it is a frame-invariant quantity. That means that all observers will agree on its value regardless of their state of relative motion.
  10. Mar 23, 2008 #9
    Oh Yes, thank you. Indeed, I have heard of Four Momentum, and the way as I see it, it is just a vector, 4D, much like space and time. Like time co-ordinate is imaginary, so is the Energy Coefficient. And, essentially, conservation of 4-momentum IS conservation of energy and linear momentum separately, as these are just 'components'. I mean, since Energy and Momentum transform in the Lorentz way, the 4 momentum too has to transform. It is just clubbing, so that the mathematical formulation is easier.

    like, if a=c, and b=d, then, a^2-b^2 has to be equal to c^2-d^2.

    I hope I am right in this regard.
  11. Mar 23, 2008 #10


    Staff: Mentor

    Yes, you are right.

    So if a0 and b0 are the four-momenta of A and B before the collision, and if a1 and b1 are the four-momenta after the collision then the conservation of total energy and the conservation of momentum can be summarized as:

    a0 + b0 = a1 + b1
  12. Mar 23, 2008 #11


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    :smile: Perfect! :smile:
  13. Mar 23, 2008 #12
    You have already been told this sometme ago at post #3.
  14. Mar 23, 2008 #13


    Staff: Mentor

    Yeah, but you gave a reference to an actual book, I gave a Wikipedia reference! :biggrin:
    Last edited: Mar 23, 2008
  15. Mar 23, 2008 #14
    Perfectly Inelastic Collision

    ok people, I guess I will post all the maths I have worked out. Probably then, this might be helpful in solving my problem.

    Note: I have read the 4-momentum, and, thank you very much for once again bringing it to my notice. All I would like to say is that, I am more comfortable conserving 3-momentum and Mass equivalent + Energy separately (Cannot get rid off of the classical mould. Basically, I am from Electrical Engineering background, and so, I am not as much comfortable with the higher maths. But, because the physical meaning of conserving 4 momentum and 3-momentum + Energy separately is same, I am sticking to the latter. I apologize for this.)

    (ALL symbols consistent with the original post)

    suppose m1 = 1Kg, m2 = 2Kg. taking c = 1, v1 = 0.85, v2 = 0.8 given that the collision is completely inelastic, find out the final velocity of the block.

    for this, let M be the final mass. (we shall see if it is m1+m2 or not)

    b1 = sqrt(1-v1^2/c^2) = 0.5267827
    b2 = sqrt(1-v2^2/c^2) = 0.6

    1*0.85/0.5267827 + 2*0.8/0.6 = M*v/b, -----(1)

    where b = sqrt(1-v^2/c^2)
    now, conserving TOTAL ENERGY (rest mass +kinetic, i.e., mass equivalent + energy)
    sigma(mc^2/b) is const.

    => 1*1^2/0.5267827 + 2*1^2/0.6 = M*1^2/b.
    => M/b = 5.2316493----(2)

    putting (2) in (1):
    4.280235 = 5.2316493 * v
    => v = 0.8181426

    => b = sqrt(1-v^2/c^2) = 0.5750154
    => M = 5.2316493 * 0.5750154 = 3.008279

    what this means is that the rest mass did not simply add up. the rest masses change due to collision. so, rest mass is also unknown.
    now, if i watch this in another frame of reference, moving with say V = 0.98 in the opposite direction:

    v1' = (v1+V)/(1+v1*V/c^2) = 0.9983633
    v2' = 0.9977578

    now, in my reference frame, i have only these velocities. mind u, the rest masses do NOT change.

    so, b1' = sqrt(1-v1'^2/c^2) = 0.05719022
    b2' = 0.066927406

    again, conserving linear momentum:

    (1*0.9983633/0.05719022) + (2*0.9710856/0.066927406) = M*v/b

    and, conserving energy, we have:
    (1/0.05719022)+(2/0.066927406) = M/b
    => M/b = 47.36883868

    => 47.27321797 = 47.36883868*v => v = 0.99798136

    b = 0.06350755

    and, M = 3.008279, within a reasonable degree of approximation. (M > m1 + m2. always!)

    from this v, if i find the ground frame v from (v-V)/(1-v*V), i expect to get the previous answer, which i do get. so, this is all valid.
  16. Mar 23, 2008 #15
    Perfectly Elastic Collision

    same case:
    m1 = 1kd, m2 = 2kg, v1=0.85, v2=0.8, v=0.98
    now, for perfectly elastic collision, the rest masses do not change, as there is no conversion of kinetic energy into rest mass.

    so, (m1*v1/b1) + (m2*v2/b2) = (m1*v3/b3) + (m2*v4/b4)
    => 1*0.85/0.5267827 + 2*0.8/0.6 = (1*v3/b3) + (2*v4/b4)


    (m1/b1) + (m2/b2) = (m1/b3) + (m2/b4).
    1/0.5267827 + 2/0.6 = 1/b3 + 2/b4.

    b3 and b4 are functions of v3 and v4 respectively.
    hence, solving the above two equations iteratively, with a step of 1e-7:
    i.e., start v3 = 0, find v4 from momentum equation, then, find the difference between RHS and LHS of the energy equation. if it is less than, say, 10^-7, i am saying that convergence has been met.
    update v3 by 10^-9.

    by doing this, i get v3 and v4 as: 0.7803223 and 0.83479063.

    now, solving the same problem in a moving frame of reference:

    v1' = (v1+V)/(1+v1*V/c^2) = 0.9983633
    v2' = 0.9977578

    (1*0.9983633/0.05719022) + (2*0.9710856/0.066927406) = (1*v3'/b3')+(2*v4'/b4')

    (1/0.05719022)+(2/0.066927406) = (1/b3') + (2/b4')

    again, setting the same criteria:
    i get:

    v3' = 0.99751029 and v4' = 0.99818262

    converting this v3' and v4' back to v3 and v4, and comparing them with the values i got from the first iteration(ground frame), the difference between the values is of the order 10^-8.

    hence, this is a valid result.
    2 equations, 2 unknowns, and, solving them, getting a unique solution.
  17. Mar 23, 2008 #16
    THE Question

    Like demonstrated in the above example, we can solve the two extreme cases with just these two equations (or conserving 4 momentum only). But, when we come to cases which are neither perfectly elastic, nor perfectly inelastic (for the time being, please leave out the disintegration/explosion case), the masses change, but they remain separate.

    Now, the main question is: to define HOW MUCH elastic a collision was, the only data that can be given is: how much KINETIC ENERGY has been converted to REST MASS energy. This data is to be defined, and, at low speeds, should converge to classical mechanics' 'coefficient of restitution'.

    Note: I am trying to avoid the concept of Relativistic Mass. So, all the masses in my solution are Rest Mass only.
  18. Mar 24, 2008 #17


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    … "inelastic" means kinetic energy is not conserved …

    Hi prasad! :smile:

    "elastic" means kinetic energy is conserved.

    "inelastic" means kinetic energy is not conserved.

    If a problem constrains two things to end up with the same velocity (for example, a bullet embedding itself in a piece of wood), then that constraint itself is a third constraint, which cannot be consistent with the other two constraints of conservation of energy and of momentum.

    This is the case both in ordinary and in relativistic mechanics.

    Any constraint on the result will destroy conservation of energy. :smile:
  19. Mar 24, 2008 #18
    Yes, but, for any collision, the MASS+ENERGY of the entire system has to be conserved.
    elastic => kinetic energy is conserved => No energy being put in the mass => no change in mass.

    inelastic=> kinetic energy is not conserved => kinetic energy is lost! Where did it go? obviously into mass, because, in relativistic mechanics, not just kinetic energy but, rest mass energy and kinetic energy is conserved. Hence, that third constraint is NOT in conflict with the first two constraints.

    Agreed that in classical mechanics, kinetic energy is not conserved. The energy is lost as 'heat' during the collision.
  20. Mar 24, 2008 #19


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    … ah … !

    ah, now I see what you're getting at …

    Your point is in your post #16 "THE question", and is:

    When two particles collide, they may either combine masses or exchange parts of their mass, and if they do so, then some rest mass will be lost.

    The loss of rest mass is a measure of the classical elasticity, or "coefficient of restitution", of the collision.

    The loss of rest mass is balanced by some additional form of energy.​

    Have I got that right? :smile:
  21. Mar 24, 2008 #20
    Oh yes! Almost! only a few details:
    The mass is not lost, but added.
    The mass is not EXCHANGED. If the mass was exchanged, then, mass itself would be conserved separately. What is being implied is that THAT part of kinetic energy which went unaccounted for in classical collision, is now being accounted for in the form of rest mass increase.
    It is the other way round. The loss of some energy is balanced by some additional mass (read rest mass. Like I said, I am totally avoiding relativistic mass. so, any mass==rest mass for me).​

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