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Is relativistic momentum conserved?

  1. Jul 12, 2011 #1
    Let's say that according to frame B, we have two identical bodies with the same invariant mass, say 1 kg, each travelling in opposite directions at .1 c, where v1 = .1 c and v2 = -.1 c, which then collide and stick together. Since the frame is homogeneous and the bodies are identical, they will become stationary within frame B, correct? Now let's look at what frame A will observe. Frame A measures frame B to have a relative speed of v = .6 c. By applying relativistic addition of speeds, the speeds frame A measures of the two bodies, then, are

    v1' = (v + v1) / (1 + v v1 / c^2) = .660377358 c
    v2' = (v + v2) / (1 + v v2 / c^2) = .531914893 c

    According to A, then, the total momentum of the system of two bodies before the collision is

    p1 + p2 = (m v1') / sqrt(1 - (v1'/c)^2) + (m v2') / sqrt(1 - (v2'/c)^2)

    = .879408087 kg m / sec + .628148633 kg m / sec

    = 1.507556721 kg m / sec

    The final momentum after the collision is

    p3 = ((2m) v) / sqrt(1 - (v/c)^2)

    = 1.5 kg m / sec

    Relativistic momentum doesn't appear to be conserved. Why not?
     
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  3. Jul 12, 2011 #2

    Doc Al

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    You have an inelastic collision, so the mass of the combined particle does not simply equal 2m.
     
  4. Jul 12, 2011 #3

    jtbell

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    Specfically, the (invariant or rest) mass of the "combined" particle is 2m + 2K/c^2, where K is the kinetic energy of each of the original particles in frame B.
     
  5. Jul 12, 2011 #4

    PAllen

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    The flaw is your computation of p3. This inelastic collision has converted some of the particle's KE into rest mass, so you can't use 2m for the mass in computing p3. In fact, the computation you've done (assuming it's correct) can be taken as a derivation of the change in rest mass of the combined mass. It is now

    2*(1.507556721/1.5) kg
     
  6. Jul 12, 2011 #5
    An inelastic collision does not conserve kinetic energy, but why not momentum? Why would the masses of the particles change? According to frame B, the momentum is conserved, so why not for frame A?
     
  7. Jul 12, 2011 #6

    Doc Al

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    You need to calculate the momentum correctly. See the other responses to your post.
     
  8. Jul 12, 2011 #7

    PAllen

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    It does conserve momentum. Kinetic energy is converted to mass. Once you use the correct mass, momentum is conserverd.
     
  9. Jul 12, 2011 #8
    Okay, thanks, so let's see. Using m' = m + KE / c^2 for what A measures, we gain

    m' = m + KE / c^2 = m + [(m c^2) (1 / sqrt(1 - (v/c)^2) - 1)] / c^2

    = m / sqrt(1 - (v/c)^2)

    So applying that to the momentum, we get

    p = m1' v / sqrt(1 - (v/c)^2)

    = m v / (1 - (v/c)^2)

    So for what A measures using that momentum, we now have

    p1 = m v1' / (1 - (v1'/c)^2) = 1.171085856 kg m / sec

    p2 = m v2' / (1 - (v2'/c)^2) = .741792927 kg m / sec

    p1 + p2 = 1.912878784 kg m / sec

    p3 = (2 m) v / (1 - (v/c)^2) = 3.125 kg m /sec

    It's much further off now than it was before.
     
  10. Jul 12, 2011 #9
    Thanks. What are the correct masses that frame A should measure?
     
  11. Jul 12, 2011 #10

    jtbell

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    The invariant mass that PAllen and I gave in posts 3 and 4 is the same in both frames.
     
  12. Jul 12, 2011 #11
    The invariant mass is 1 kg, is it not? Let's say the masses are measured as 1 kg each upon coming to rest in frame B. What were they before coming to rest, travelling at .1 c according to B? What were they according to frame A? I am asking about the masses that should be applied to the formula for relativistic momentum. Wouldn't that just be the invariant mass, 1 kg, multiplied by v / sqrt(1 - (v/c)^2), or the relativistic mass, m / sqrt(1 - (v/c)^2) multiplied by v, either way, giving the the relativistic momentum expressed as p = m v / sqrt(1 - (v/c)^2)?
     
    Last edited: Jul 12, 2011
  13. Jul 12, 2011 #12

    PAllen

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    In frame B, each mass was 1 Kg before collision, and the combined object (after collision) is a little over 2 Kg.

    If frame A, exactly the same statements apply.

    In each frame, us 1 kg as rest mass before collision, and use the 'over 2 kg' number for mass after collision.

    The over 2kg number can be derived more easily in B frame as jtbell suggested. Using .1 c, you get:

    2.01007563... Kg

    which happens to be exactly what I suggested by exploiting your computations in frame A.
     
    Last edited: Jul 12, 2011
  14. Jul 12, 2011 #13
    Okay, so both frames measure m = 1 kg for the mass of each body before the collision and 2 m' = 2 m / sqrt(1 - (v1/c)^2) after, using the speed that B measures of the objects, so giving 2 m' = 2.010075631 kg. Applied to the relativistic momentum, then, that gives

    p1 = m v1' / sqrt(1 - (v1'/c)^2) = .879408087 kg m / sec

    p2 = m v2' / sqrt(1 - (v2'/c)^2) = .628148633 kg m / sec

    p3 = (2 m') v / sqrt(1 - (v/c)^2) = 1.50755672 kg m / sec

    p1 + p2 = 1.50755672 kg m /sec

    so it works out mathematically in this manner, yes, but the reasoning does not make sense to me. At any rate, since it works for these two frames, let's check the result with frame C, measuring B to be moving at v = .5 c, so measuring the speeds of the bodies before the collision to be

    v1' = .571428571 c
    v2' = .421052631 c

    with masses before the collision of m = 1 kg and after the collision of 2 m' = 2.010075631 kg, so

    p1 = m v1' / sqrt(1 - (v1'/c)^2) = .696310623 kg m / sec

    p2 = m v2' / sqrt(1 - (v2'/c)^2) = .464207082 kg m / sec

    p3 = (2 m') v / sqrt(1 - (v/c)^2) = 1.160517707 kg m / sec

    p1 + p2 = 1.160517707 kg m / sec

    Again, it works out mathematically, so that's good, but I still don't get the reasoning. 2m' is the relativistic mass 2 m / sqrt(1 - .1^2). Why does the invariant mass apply in frame B when the objects are moving at .1 c but the relativistic mass applies when they are at rest? Isn't that the opposite of what one normally considers to be relativistic mass, whereas the invariant mass applies at rest and the relativistic mass applies while in motion, or is this some different concept altogether?
     
  15. Jul 12, 2011 #14

    PAllen

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    Actually, there is dispute about how useful relativistic mass is as a concept. I never use it (in the case of force, you end up defining transverse versus parallel relativistic mass, which I find borders on the absurd). I prefer to think of only invariant mass. The original bodies are 1 kg in all frames. The merged body is 2.010075631 kg in all frames, and this mass comes from conversion of kinetic energy to mass (E=mc^2). Momentum is *not* relativistic mass time velocity; instead, it is invariant mass time v*gamma. Kinetic energy is simply total energy less rest (invariant) energy. Total energy is just mass (invariant) * c^2 * gamma.

    The elegant way to organize this is the energy-momentum 4-vector. This encompasses energy and momentum conservation; its norm (norm of a vector is invariant) is the invariant mass. It covariant derivative by proper time is 4-force.
     
  16. Jul 13, 2011 #15
    Apparently, the total energy is also conserved in relativity. So to conserve total energy, the particles while traveling at .1 c according to frame B with 1 kg mass each have a total energy of

    E = m c^2 + KE, where KE = [1 / sqrt(1 - (v/c)^2) - 1] m c^2, so the total energy becomes

    E = m c^2 / sqrt(1 - (v/c)^2)

    Since this is conserved when the particles collide, the masses of each particle after the collision must be

    E / c^2 = m / sqrt(1 - (v/c)^2), or m' = 1.005037815 kg each

    Thing is, I've heard about relativistic mass becoming greater with greater speed, but it seems to be just the opposite. Frame B measures the masses to be m' when at rest, but m = m' sqrt(1 - (v1/c)^2) when travelling at v1. In a way that makes sense, though, since apparently it would mean that an object travelling at light speed to B has zero mass, as all objects that travel at c do. But in another way, it still doesn't make sense to me, because it seems to be saying that the frame that the masses were in originally, say frame C moving at .1 c to frame B and travelling with the masses before the collision, would measure m = 1 kg directly while the masses are at rest in that frame, while after coming to rest in frame B that frame C measures at .1 c, then, the masses would be m / sqrt(1- (v/c)^2) as directly measured by B. So one frame measures a lesser mass when then object is in motion and the other frame measures a greater mass. This also seems to go against the first postulate, that the physics is the same in every inertial frame. For instance, the mass of an electron is different as measured in frame C than in frame B, so there is still something I am missing.
     
  17. Jul 14, 2011 #16

    PAllen

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    You are confusing so many things here, you would clearly benefit reading a basic introduction to SR.

    Frame B
    --------

    In terms of invariant mass, each object is 1 kg, period, no matter its motion. After collision and merger, a new, different object is produced converting kinetic energy to mass. Mass of this merged object is 2.01... kg, independent of its state of motion.

    If you insist using relativistic mass, each object weighs 1 kg at rest, 1.005... at .1c. Merged object weighs 2.01... kg at rest, more if it were moving. Merged object is a new body, not the same as each original body at rest.

    Frame C
    --------
    Using invariant, mass, same as for frame B: each is 1 kg before collision, 2.01.. after collision, independent of state of motion.

    Using relativistic mass, each is 1 kg if brought to rest. The one moving at a little less .2 c (relativistic velocity addition) weighs a little less than 1/sqrt(.96) kg. The merged body would weight 2.01...kg at rest. Moving at .1c after collision, it weighs 2.01.../sqrt(.99) kg.
     
  18. Jul 14, 2011 #17
    Right, I am agreeing with you. I am not saying it is relativistic mass, I'm saying that ironically it is the opposite of that, at least according to frame B. Anyway, you seem to be saying that all frames measure 1 kg each for the separate masses and 2.01... kg for the combined mass, as if it were a different kind of object, the merged object. I'm saying that if we were to now separate the objects again in the rest frame of B where still they remain stationary but at a distance to each other, the new mass of each individual object is now 1.005... kg even when no longer merged together, is that right?

    In other words, if we had identical ships in frames C and D that were travelling with a relative speed to frame B at 3 * 10^(-7) c, about as fast as a race car, and each ship's mass is 100,000 kg as measured in frames C and D, then when they collide and merge in frame B, by locking bumpers sorta speak at that speed or with intertwined metal, the mass is now about 9 * 10^(-9) kg greater, right? Now, if the material of one of the ships is red and the material of the other ship is blue, then if we separated all of the pieces from each ship into separate piles after the collision the best we can and measure the mass in frame B for each pile separately, each pile would have a mass of approximately 10000 + 4.5 * 10(-9) kg, meaning the components of each individual ship now have greater mass, correct?
     
    Last edited: Jul 14, 2011
  19. Jul 14, 2011 #18

    PAllen

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    It is a different object. Kinetic energy has been converted to rest mass. If you split it apart after the inelastic collision, each piece would be 1.005... kg. The specific form of this extra mass would be mostly heat. How much? Get ready for this: if you inelastically collide two 1 kg masses at .1 c, the merged object will be hotter by the energy equivalent of 10 Nagasiki atom bombs (about the energy of 200 kilotons of TNT). In reality, any real objects would be vaporized. If you could somehow avoid this, and assuming specific heat for typical matter, you would be talking about 500 billion degrees C (many orders of magnitude hotter than the center of the sun). Wouldn't that feel like a different object if you touched it:wink:
    Yes, the super hot bodies would be heavier. After they radiate away all this energy (vaporizing any nearby city), they would be back to 1 kg.
    No need to comment on this. It's all about conversion of kinetic energy to rest mass in the form of heat (mostly).
     
  20. Jul 14, 2011 #19
    Okay great, thanks. :) that's the way I was thinking about it. Which is why I also decided to lower that speed quite a bit to about that of a typical race car speed.

    Okay, now this is what I was really wondering about and was leading toward. I figured there were two possibilities here.

    One, the extra energy becomes heat as you said and is radiated away until the masses become 1 kg again. But in this case, nothing really happens to the matter of the objects themselves during the collision, then, I would think, since it generally remains constant, but the extra mass comes solely from the kinetic energy being converted to heat energy, which apparently can be measured as having mass. But then again, heat is basically the matter itself moving around or vibrating much faster within the object, isn't it, so it's almost like the kinetic energy is still there in the stationary merged object as heat energy, which can be still be measured as the object having extra mass until the heat dissipates, is that about right?

    Two, the extra mass is binding energy, as with the binding energy of sub-atomic particles, the "missing" mass. This wouldn't be considered so much as heat, it would seem, so merged particles continue to have that extra mass and it doesn't dissipate, is that right? Are both of these possibilities more or less true?
     
  21. Jul 14, 2011 #20

    PAllen

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    Your discussion of heat is generally ok.

    Binding energy isn't appropriate in this context[EDIT: binding energy make an object weigh less, necessitating input of energy to separate its parts]. What could an alternative conversion of collision energy to new particles, as happens in accelerators. However, since there has to be a balance of particles and anti-particles, if held together, you eventually get back to radiation. Another variant (assuming less extreme collision), of course, is breaking bonds and formation of new compounds. Still, I think most of the collision energy has to end up as heat.
     
    Last edited: Jul 14, 2011
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