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Relativistic Momentum Not Conserved after a Collision?

  1. Oct 11, 2012 #1
    Hello, I've just registered to ask this question. This is something that's been bothering me for some time. I have a non-elastic collision of 2 masses in 1-dimension observed by two different frames of reference, and the total momentum of the system is conserved for one of them but not for the other. I'm probably making a mistake but I don't know where.

    Here's a picture that shows what's going on in both frames. I represented the frames of reference S and S' as dots for simplicity.


    The problem is this: according to the [itex]S[/itex], the two objects should stick together and stop after the collision, as predicted by both the classical and the relativistic definition of momentum. But the problem is, when you carry out the calculations for the velocities in the [itex]S'[/itex] frame, the sum of the momenta of the objects doesn't come out to be equal to the momentum of the combined object.

    Okay, so the definition of the relativistic momentum is this: [itex]\overrightarrow{P}=\frac{m \overrightarrow{v}}{\sqrt{1-\frac{v^2}{c^2}}}[/itex]

    In the first frame, both classical and relativistic definitions agree that the momentum of the object [itex]M_3[/itex] (let's call it [itex]P_3[/itex]) should be 0, because:
    [itex]m u + m (-u) = 0[/itex]
    and it doesn't matter if you use the relativistic definition because the factor [itex]\gamma = \frac{1}{\sqrt{1-\frac{u^2}{c^2}}}[/itex] is the same both objects, and it gives the same result if you multiply both sides by [itex]\gamma[/itex]. So we know that the objects will stop after colliding.

    Now let's consider the event in the reference frame of [itex]S'[/itex]. Using the Lorentz transformation, the relative velocities of the objects [itex]M_1[/itex] and [itex]M_2[/itex] with respect to [itex]S'[/itex] are found to be:
    [itex]u_1' = \frac{u - v}{1 - \frac{u v}{c^2}}[/itex] and [itex]u_2' = \frac{-u - v}{1 + \frac{u v}{c^2}}[/itex]

    So the momentum of the first object [itex]P_1'[/itex] should be equal to:
    [itex]\frac{m u_1'}{\sqrt{1-\frac{u_1'^2}{c^2}}}[/itex]

    To simplify it a little before substituting for [itex]u_1'[/itex], let's divide both the numerator and the denominator with [itex]|u_1|[/itex]:
    [itex]P_1 = \frac{m \frac{u_1'}{|u_1'|}}{\sqrt{\frac{1}{u_1'^2}-\frac{1}{c^2}}}[/itex]

    We know that [itex]\frac{u_1'}{|u_1'|}[/itex] is equal to the sign of [itex]u_1[/itex] which is equal to the sign of [itex]u - v[/itex], since the denominator is always positive. Now let's write [itex]u_1'[/itex] in terms of [itex]u[/itex] and [itex]v[/itex].

    [itex]P_1' = \frac{m sgn(u-v)}{\sqrt{\frac{1}{\big(\frac{u - v}{1 - \frac{u v}{c^2}}\big)^2}-\frac{1}{c^2}}} = \frac{m sgn(u-v)}{\sqrt{\frac{1-\frac{2 u v}{c^2}+\frac{(u v)^2}{c^4}}{u^2-2 u v+v^2}-\frac{1}{c^2}}} = \frac{m sgn(u-v)}{\sqrt{\frac{c^2-2 u v+\frac{(u v)^2}{c^2}-(u^2-2 u v+v^2)}{c^2 (u^2-2 u v+v^2)}}}[/itex]
    [itex] = \frac{m sgn(u-v)}{\sqrt{\frac{c^2-u^2-v^2+\frac{(u v)^2}{c^2}}{c^2 (u-v)^2}}} = \frac{m sgn(u-v) |u-v|}{\sqrt{1-\frac{u^2}{c^2}-\frac{v^2}{c^2}+\frac{(u v)^2}{c^4}}} = \frac{m (u-v)}{\sqrt{1-\frac{u^2}{c^2}-\frac{v^2}{c^2}(1-\frac{u^2}{c^2})}}[/itex]
    [itex]= \frac{m (u-v)}{\sqrt{(1-\frac{u^2}{c^2}) (1-\frac{v^2}{c^2})}}[/itex]


    Okay, that was a long calculation but now it's in a form that's easier to read. We know that the original equation we found for [itex]P_1'[/itex] was the same as the one for [itex]P_2'[/itex], except for the negated initial velocity [itex]-u[/itex]. So if we plug in [itex]-u[/itex] for [itex]u[/itex] in the equation we just derived for [itex]P_1'[/itex], we should get the equation for [itex]P_2'[/itex]:
    [itex]P_2' = \frac{m (-u-v)}{\sqrt{(1-\frac{u^2}{c^2}) (1-\frac{v^2}{c^2})}}[/itex]

    Now let's find the momentum of [itex]M_3[/itex]. We know that it is at rest for [itex]S[/itex], so for [itex]S'[/itex] it must be moving at [itex]v[/itex] along with [itex]S[/itex]:

    [itex]P_3' = \gamma (2m) (-v) = \frac{-2 m v}{\sqrt{1-\frac{v^2}{c^2}}}[/itex]

    If the momentum is conserved, [itex]P_1' + P_2'[/itex] should be equal to [itex]P_3'[/itex].
    [itex]P_1' + P_2' = \frac{m (u-v)}{\sqrt{(1-\frac{u^2}{c^2}) (1-\frac{v^2}{c^2})}} + \frac{m (-u-v)}{\sqrt{(1-\frac{u^2}{c^2}) (1-\frac{v^2}{c^2})}} = \frac{-2m v}{\sqrt{(1-\frac{u^2}{c^2}) (1-\frac{v^2}{c^2})}}[/itex]
    and
    [itex]P_3' = \gamma (2m) (-v) = \frac{-2 m v}{\sqrt{1-\frac{v^2}{c^2}}}[/itex]

    Here [itex]P_3'[/itex] is missing the factor [itex]\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}[/itex] that [itex]P_1' + P_2'[/itex] has so they can't be equal.

    I appreciate it if you bothered to read this far. I don't understand what I may have done wrong. I've already asked a few people but couldn't get an answer. If you have an idea, please drop a comment below. Thank you.
     
    Last edited: Oct 11, 2012
  2. jcsd
  3. Oct 11, 2012 #2
    Four-momentum combines ordinary momentum and energy. Inelastic collisions don't conserve energy, so how can the four-momentum be conserved?

    Edit: similarly, if the total four-momentum is not conserved, then boosting to some other frame will have some effect on the ordinary momentum.
     
  4. Oct 11, 2012 #3
    I've never heard of four momentum before, all I know is that in classical mechanics, momentum is conserved in non-elastic collisions even though energy isn't. If momentum isn't conserved in this case, how would [itex]S'[/itex] explain the motion of the combined mass after the collision?
     
  5. Oct 11, 2012 #4
    When you work in 3d space, you work with vectors. A vector's components depend on the basis you choose, but regardless of the choice of basis, we view the vector as being the same object. This is the theory of passive transformations.

    When you work in relativity, true vectors have four components instead of three, so ordinary momentum cannot be a vector in relativity. Physicists rather uncreatively call the relativistic analogue the four-momentum, and as it happens, this combines ordinary momentum and energy into a single vector. When you switch between reference frames, you're choosing a different basis to evaluate the components of vectors on. Hence, any switching between frames will inevitably mix up momentum and energy components, just as choosing a rotated set of axes in 3d will mix up x, y, and z components of a vector.

    Anyway, after doing some reading on relativistic inelastic collisions, it seems that total relativistic four-momentum is conserved in such collisions, but as in Newtonian mechanics, kinetic energy is not conserved. However, the total energy is, and what this means is that the combined object may have a different mass than the sum of the two masses.

    Example: particle A has four-momentum [itex]p_A = m c e_0[/itex] (that is, it goes in the time direction, so it is "at rest"), and particle B has four-momentum [itex]p_B = m \gamma c (e_0 + \beta e_1)[/itex]--that is, it moves with some velocity [itex]\beta c[/itex] in the x direction.

    When these two particles collide, the four-momentum is conserved. The total four-momentum is

    [tex]p = p_A + p_B = mc ( [\gamma + 1] e_0 + \gamma \beta e_1)[/tex]

    The magnitude of the four-momentum gives us the mass of the composite particle.

    [tex]p^2 = [-(\gamma + 1)^2 + \gamma^2 \beta^2] m^2 c^4 = [-\gamma^2 + \gamma^2 \beta^2 - 1 - 2 \gamma] m^2 c^4 = - 2 (1 + \gamma) m^2 c^4[/tex]

    This tells us the total mass [itex]M = m\sqrt{2(1+\gamma)} \geq 2 m[/itex], strictly greater as long as the incoming particle has nonzero velocity.
     
  6. Oct 11, 2012 #5

    Dale

    Staff: Mentor

    Yes, they do. They don't conserve KE because some of the KE is converted to other forms of energy, but they do conserve E. The E in the four momentum includes everything, not just KE, so it is conserved in an inelastic collision.
     
  7. Oct 11, 2012 #6

    PeterDonis

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    This is true, as is the following information you give about 4-momentum, but it doesn't invalidate what the OP was trying to do. Within a given inertial frame, both energy and ordinary momentum (or 3-momentum, to differentiate it from 4-momentum) are conserved separately. The OP was asking why that appears to hold in frame S but not in frame S'. You only really need to look at 4-momentum directly when you are trying to write your results in a frame-independent form (i.e., without specifying a frame).

    This is the real answer to the OP's question: his formula for [itex]P_{3} '[/itex] doesn't take this into account. The velocity of the final object in frame S' is [itex]- v[/itex], but the rest mass of that object is *not* [itex]2m[/itex]; it is *larger*, because some of the original kinetic energy of the two objects that collided got converted to heat, as DaleSpam pointed out, and that heat has to be counted as part of the final object's rest mass, since the final object has zero kinetic energy in frame S, so its rest mass *is* the total energy of the system in frame S after the collision.

    This effect occurs in frame S too, but in frame S it doesn't make any difference for momentum conservation because the velocity of the final object in frame S is zero. However, if the OP were to compute energy conservation in frame S, he would find that the rest mass of the final object, which is the total energy in frame S after the collision, would have to be greater than [itex]2m[/itex] because, as just mentioned, that total energy has to include the kinetic energy of the two original objects that collided, as well as their rest energy.
     
    Last edited: Oct 11, 2012
  8. Oct 11, 2012 #7
    I don't really understand why the heat produced after the collision has to be counted as part of the total mass, they seem like totally unrelated things to me. That's probably because I'm missing some of the fundamental concepts like four-momentum. I've done a quick search about it but the results that came up all go over my head. I guess it's too "advanced" for me for now.

    So I guess what I did in my post above can only be applied to perfectly elastic collisions, where no kinetic energy is lost to heat or radiation, is that right?
     
  9. Oct 11, 2012 #8

    PAllen

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    Ever heard of E=mc^2 ? In relativity (and in our world) energy counts as mass and vice versa. A hot brick weighs more than a cold brick.

    With the corrections pointed out in this thread, you can apply it to all collisions - you just have to account for all energy and momentum.
     
  10. Oct 11, 2012 #9
    I know about the mass-energy equivalence, it just didn't feel natural to apply it here. If some of the kinetic energy was lost as radiation, why would the energy of that radiation contribute to the object's mass?
     
  11. Oct 11, 2012 #10

    PAllen

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    Work through the steps:

    1) Inelastic collision leading to zero momentum (in this frame). All kinetic energy, at this moment is converted to rest mass of combined object.

    2) Some amount is radiated away. The rest mass of the combined object now decreases by <radiated energy>/ c^2. Note: the radiated energy carries momentum as well as energy. The momentum of the radiation will be isotropic in this frame, but not in other frames. Thus, if you want to demonstrate momentum conservation in a different frame after radiation has occurred, you must account for energy and momentum of the radiation.

    It is much easier to compute the situation before radiation (1), where all the former KE is in mass of the composite object.
     
  12. Oct 11, 2012 #11
    Well I guess this explains it. I really should learn more about these because there is so many things I don't know. Thank you for your answers, they were helpful.
     
  13. Oct 12, 2012 #12

    PAllen

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    One thing to put in perspective is how large certain effects are in the regime where relativistic mechanics differs substantially from Newtonian mechanics. Suppose two one gram objects are moving towards each other at 86% light speed ins some frame.

    1) In that frame, how much energy would it take to get each one gram object (about 1/2 a soda bottle cap) moving at .86c, starting from rest? Answer: the total energy of the Nagasaki nuclear bomb would have to delivered to each one gram chunk, converted completely to motion, to accomplish this.

    2) So, when we talk about heat released in the inelastic collision, we are talking about 2 city destroying nuclear bombs worth of energy. In terms of mass, at the moment of merger, the two 1 gram chunks moving at .86c produce a blob with rest mass of 4 grams, not 2 grams. This 4 gram invariant mass, as the name suggests, is true in any other reference frame as well.

    3) If you realize how much radiation there is (enough to destroy two cities) it becomes less paradoxical to see that perhaps there is some momentum to worry about in the radiation.
     
  14. Oct 12, 2012 #13

    robphy

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    What might be helpful here is a diagram of the collision in energy-momentum space.

    Conservation of 4-momenta
    (componentwise: Conservation of relativistic-energy and Conservation of relativistic momentum)
    says:
    (Sum of 4-momenta before)=(Sum of 4-momenta after)

    As a diagram, this is a polygon, with timelike sides for the massive particles and lightlike sides for photons. In a totally-inelastic collision of two massive particles, the polygon degenerates into a triangle.

    Under a Lorentz boost, the polygon transforms into a related polygon. Thus, the total 4-momenta is conserved in all frames.
     
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