Relativistic inelastic collisions?

[wil3]

Hello. I am trying to better understand relativistic inelastic collisions for objects with mass and the setup for them.

Let's say that I have two objects with different masses, one of which collides with the other, which is stationary.

Without even specifying a reference frame, can I solve for the final velocity of the composite using just momentum conservation? Ie, as long as I apply the Lorentz correction to momentum, won't it be conserved (although the initial and final momenta will have different velocity-dependent Lorentz factors)

In this case, I would assume that the rest mass of the composite is just the sum of the initial rest masses, since no photons escaped. Is this a valid assumption to make? This troubles me, because it does not seem to require specifying a reference frame.

Thanks very much.

 

[Dale]

Are you familiar with the four-momentum? That handles both conservation of energy and conservation of momentum in one nice neat package and it makes transforming to different reference frames and determining masses very easy. I would recommend it for this type of problem.

http://en.wikipedia.org/wiki/Four-momentum

 

[bcrowell]

Note that since it's inelastic, the rest masses will be different after than before the collision. That means that the approach wil3 originally proposed, of just using conservation of relativistic 3-momentum, won't be sufficient.

Actually, it seems to me that the relativistic version of this problem has no unique solution. Let's go into the c.m. frame, and take the x-axis to be the axis along which the motion occurs. Then we have four unknowns in the final state: v1, v2, m1, and m2. Conservation of the t and x components of four-momentum is not sufficient to determine four unknowns. You could get a unique solution if, for example, you were given that the collision was totally inelastic, or if you had some way to predict that a certain amount of heat would be produced in each body.

 

[wil3]

My apologies for the lack of clarity. By "composite," I meant to imply that the collision was completely inelastic, with one body "sticking" to the other. Even through no other particles are emitted, am I still wrong in saying that the rest mass of the composite is simply the sum of the rest mass of the two objects from which it originated?

By rest mass, I mean the mass before it is multiplied by the lorentz gamma. The quantities that I think should be equated look something like this:

<br /> <br /> \gamma_0 \, m_1 \, v_o = \gamma_f \, (m_1 + m_2) \, v_f<br /> <br />

where M1 is initially moving, and M2 is initially stationary. By solving for the final velocity, I would have vf of the composite in the frame where M2 was initially stationary. However, the assumption I make to do this is that the rest mass of the composite will be the sum of the initial rest masses.

Still working on four-vectors. I'd like ot make sure I understand how to do this without them before I move on to them.

Thanks for your replies.

 

[bcrowell]

If it's completely inelastic, then there is definitely a unique solution, which is that they stop dead in the c.m. frame. That's essentially the definition of a completely inelastic collision.

However, the assumption I make to do this is that the rest mass of the composite will be the sum of the initial rest masses.

I don't think that assumption is valid. You convert KE into heat (or other forms of internal energy), which are equivalent to mass.

 

[Dale]

However, the assumption I make to do this is that the rest mass of the composite will be the sum of the initial rest masses.

This assumption is most definitely incorrect. In addition to the initial rest masses, the kinetic energy of the masses is also converted to mass in the CoM frame. That is how very massive particles are created in a particle collider.

Still working on four-vectors. I'd like ot make sure I understand how to do this without them before I move on to them.

I wouldn't hesitate like that. It is so much simpler with the four-momentum that the effort of learning it will pay off right away. In fact, I don't even know how I would figure it out without four-vectors.

 

[K^2]

Dale, sorry, but what you just said is nonsense. Kinetic energy does not contribute to rest mass. Hence the name. Rest mass.

What will happen to the rest mass and energy depends on many factors. If there is binding energy involved, for example, you might end up with rest mass that's lower than the sum of the initial masses. If the final particle ends up in an excited state, the rest mass might be higher, at least, temporarily.

In any case, 4-momentum is conserved. Always. That's why in any two particle collision you'll have two particles going in and two or more particles going out. If the collision happens to be perfectly inelastic, then the other particle going out will be a photon carrying out excess energy and some momentum to make conservation work.

 

[bcrowell]

Dale, sorry, but what you just said is nonsense. Kinetic energy does not contribute to rest mass. Hence the name. Rest mass.

Dale is correct. For example, a significant chunk of the rest mass of a helium atom is due to the kinetic energy of the neutrons and protons.

 

[K^2]

Bad example, first of all. Helium's nucleons are all in equivalent of the S1 shell. They have zero angular momentum. If I ask you to define kinetic energy of a nucleon that's not moving, you are going to have a bit of a difficulty. Secondly, you have to use the same references. If you are talking about rest mass of a nucleus, you are talking about kinetic energy of the same nucleus. Movement of nucleons does not contribute to it. If you are talking about movement of individual nucleons, you have to be talking about rest masses of individual nucleons.

 

[PAllen]

Bad example, first of all. Helium's nucleons are all in equivalent of the S1 shell. They have zero angular momentum. If I ask you to define kinetic energy of a nucleon that's not moving, you are going to have a bit of a difficulty. Secondly, you have to use the same references. If you are talking about rest mass of a nucleus, you are talking about kinetic energy of the same nucleus. Movement of nucleons does not contribute to it. If you are talking about movement of individual nucleons, you have to be talking about rest masses of individual nucleons.

Suppose I have two identical perfect clay balls that hit each other dead on, each at the same speed (in this c.m. frame), resulting in a single blob that radiates nothing. Then all of the kinetic energy each had before has been converted to rest mass of the composite blob.

 

[K^2]

Except, that never actually happens with particles. There are only certain energy levels that are available, meaning only certain possible rest masses for combined particle, and even these are unstable and usually extremely short lived. A perfectly inelastic collision is a pink elephant.

The way that heavy particles are actually created in accelerator is by smashing two particles hard enough to create showers of particle-anti-particle pairs.

 

[PAllen]

Except, that never actually happens with particles. There are only certain energy levels that are available, meaning only certain possible rest masses for combined particle, and even these are unstable and usually extremely short lived. A perfectly inelastic collision is a pink elephant.

The way that heavy particles are actually created in accelerator is by smashing two particles hard enough to create showers of particle-anti-particle pairs.

Original post never said anything about particles (they said objects), or that they wanted to worry about quantum corrections to classical theory (which, of course, prevents many things allowed by classical theories).

A perfectly inelastic collision is a pink elephant in the real, quantum world; it is trivially consistent in the idealized classical world.

 

[yuiop]

Suppose I have two identical perfect clay balls that hit each other dead on, each at the same speed (in this c.m. frame), resulting in a single blob that radiates nothing. Then all of the kinetic energy each had before has been converted to rest mass of the composite blob.

Would I right in thinking that the temperature of the final blob is greater than the temperature of the original clay balls and this heat energy is the origin of the additional rest mass?

 

[PAllen]

Would I right in thinking that the temperature of the final blob is greater than the temperature of the original clay balls and this heat energy is the origin of the additional rest mass?

Yes, to the partial extent you could get this to happen in the real world, the composite object would be hotter.

 

[K^2]

Original post never said anything about particles (they said objects), or that they wanted to worry about quantum corrections to classical theory (which, of course, prevents many things allowed by classical theories).

A perfectly inelastic collision is a pink elephant in the real, quantum world; it is trivially consistent in the idealized classical world.

And that's accounted for in my general description. Now read DaleSpam's post and see that it's still nonsense. That's the context in which all of the quotes you replied to were. Not the OP.

 

[PAllen]

And that's accounted for in my general description. Now read DaleSpam's post and see that it's still nonsense. That's the context in which all of the quotes you replied to were. Not the OP.

Why isn't it reasonable to view the mass of produced particle/antiparticle pairs as having come from the kinetic energy of the original colliding particles? Kinetic energy has been converted to rest mass (of different particles, because a single composite particle is prohibited).

 

[K^2]

It is. It simply has nothing to do with two particles colliding and sticking together.

 

[PAllen]

I cannot comment on the helium nucleus case, but consider the case off box of hot gas versus a box of cold gas. Unless you inist that rest mass can only be used in the sense of an intrinsic property of a fundamental particle, one must admit that the box of hot gas has higher rest mass than the box of cold gas due to the kinetic energy of the gas molecules. Both boxes are at rest in my lab; one weighs more; somehow I know they have the same number of molecules.

 

[bcrowell]

Bad example, first of all. Helium's nucleons are all in equivalent of the S1 shell. They have zero angular momentum. If I ask you to define kinetic energy of a nucleon that's not moving, you are going to have a bit of a difficulty.

Having zero angular momentum does not mean the same thing as "not moving" or having zero kinetic energy. By the Heisenberg uncertainty principle, it is not possible for the neutrons and protons to be "not moving." For example, a commonly used model of the nucleus uses a harmonic oscillator potential, so that each neutron and proton has energy (3/2)\hbar\omega in the ground state. Half of that energy is potential and half is kinetic.

Secondly, you have to use the same references. If you are talking about rest mass of a nucleus, you are talking about kinetic energy of the same nucleus. Movement of nucleons does not contribute to it. If you are talking about movement of individual nucleons, you have to be talking about rest masses of individual nucleons.

The rest mass of any object is defined as m=\sqrt{E^2-p^2} (in units with c=1). The kinetic energies of the neutrons and protons contributes to the E of the nucleus, and therefore increase m.

 

[bcrowell]

Except, that never actually happens with particles. There are only certain energy levels that are available, meaning only certain possible rest masses for combined particle, and even these are unstable and usually extremely short lived. A perfectly inelastic collision is a pink elephant.

The way that heavy particles are actually created in accelerator is by smashing two particles hard enough to create showers of particle-anti-particle pairs.

This objection is irrelevant, because the OP referred to objects, not particles, and in any case "particle" does not necessarily mean a fundamental particle. For example, in a cosmological model a "particle" can be a galaxy. For that matter, neutrons and protons are not fundamental particles.

The objection is also incorrect. As a counterexample, in low-energy nuclear physics experiments such as the ones I've worked on, a beam nucleus collides with a target nucleus, and the collision is perfectly elastic (meaning that the fused system recoils with a velocity equal to the center-of-mass velocity). Quantization of energy does not prevent the process from occurring, as you seem to believe, because the density of states is very high, and the natural width of each state is big enough to make it overlap with its neighbors. The fused system reaches a thermal equilibrium and can be characterized by a temperature (although thermal fluctuations are much bigger in a system with ~100 particles than in one with Avogadro's number worth of particles).

 

[K^2]

A perfectly elastic collision is very likely at low energies. That's very different from perfectly inelastic collision which has to absorb an exactly the right amount of energy, and thus, never happens. Ever. There is always a photon emitted when a nucleus absorbs a massive particle. Always.

And again, I was replying to Dale. OP did not specify to particles. Dale did.

Having zero angular momentum does not mean the same thing as "not moving" or having zero kinetic energy.

It is not moving in any conventional sense. Expectation of velocity/momentum is precisely zero in CM frame.

The rest mass of any object is defined as (in units with c=1). The kinetic energies of the neutrons and protons contributes to the E of the nucleus, and therefore increase m.

Except these aren't really kinetic energies of nucleons either. These are combinations of various energies in the gluon and quark fields, and you wouldn't even be able to pin this onto either one, because none of them have a particular momentum. And then we have to pin all mass to chiral symmetry breaking, and then none of the particles have a rest mass.

Except they do. A helium atom is still a bound state with a singularity in a propagator at the mass shell. And the kinetic energy of the helium atom does not contribute to its mass.

 

[bcrowell]

There is always a photon emitted when a nucleus absorbs a massive particle. Always.

This is incorrect, as you can convince yourself by looking at any nuclear physics textbook, e.g., DeShalit and Feshbach, vol. 1, p. 87, which shows data from the reaction 27Al(p,alpha)24Mg.

It is not moving in any conventional sense. Expectation of velocity/momentum is precisely zero in CM frame.

Yes, <v>=0 and <p>=0, but the kinetic energy is not zero. This is discussed in any elementary quantum mechanics textbook.

And the kinetic energy of the helium atom does not contribute to its mass.

If you're still claiming that the kinetic energies of the neutrons and protons does not contribute to the rest mass of a helium atom, then you should spend some time with any textbook on special relativity. There is a good discussion of this on pp. 246-249 of Spacetime Physics by Taylor and Wheeler.

 

[Dale]

Dale, sorry, but what you just said is nonsense. Kinetic energy does not contribute to rest mass. Hence the name. Rest mass.

Sorry about any confusion that I caused due to poor wording. My statements are essentially correct, but I should have been clear about which masses I was referring to. I should have said something more like:

In addition to the masses* of the initial objects, the kinetic energy of the objects also contributes to the mass of the of the system. After the collision the mass of the composite object is equal to the mass of the system, not the sum of the masses of the initial objects. That is how very massive particles are created in a particle collider.

*here the word mass always refers to the invariant mass. The mass of a system is equal to the Minkowski norm of the sum of the four-momenta of the individual objects.

In any case, 4-momentum is conserved. Always. That's why in any two particle collision you'll have two particles going in and two or more particles going out. If the collision happens to be perfectly inelastic, then the other particle going out will be a photon carrying out excess energy and some momentum to make conservation work.

I agree that the 4-momentum is conserved. But that is not a counter example to my statement above. There are many reactions where the product particles are more massive than the sum of the masses of the reactant particles. That extra mass came from the KE of the reactant particles. That is, in fact, the whole point of building larger and more energetic colliders. If the KE of the accelerated particles didn't contribute to the mass of the products then building a bigger collider would not help discover more massive particles.

 

[wil3]

...I think I should probably go learn 4-vectors now.

 

[Dale]

Let us know if you have any questions. I think you will find it very worthwhile.

 

[wil3]

Update: I did derive it by assuming rest mass was not conserved and then introducing a new conservation equation. But the method does seem clumsy, so I'll try out 4-vectors.

 

[K^2]

Sorry about any confusion that I caused due to poor wording. My statements are essentially correct, but I should have been clear about which masses I was referring to. I should have said something more like:

In addition to the masses* of the initial objects, the kinetic energy of the objects also contributes to the mass of the of the system. After the collision the mass of the composite object is equal to the mass of the system, not the sum of the masses of the initial objects. That is how very massive particles are created in a particle collider.

Maybe it is organized better in your head, but when you write it, you are not making a clear enough distinction between inertial mass and rest mass. All energy contributes to inertial mass, but kinetic energy specifically does not contribute to rest mass by very definition of rest mass. It can be later converted into rest mass. Simplest example is an object escaping a gravitational well. It converts kinetic energy into potential energy, and that's rest mass already.

And yes, part of internal energy of the body could be in kinetic energy of components, but again, that's not kinetic energy of the body, which is due to CoM motion alone.

I agree that the 4-momentum is conserved. But that is not a counter example to my statement above. There are many reactions where the product particles are more massive than the sum of the masses of the reactant particles. That extra mass came from the KE of the reactant particles. That is, in fact, the whole point of building larger and more energetic colliders. If the KE of the accelerated particles didn't contribute to the mass of the products then building a bigger collider would not help discover more massive particles.

No. The high-energy colliders are not designed to make things stick together. They are designed to blast things apart. It's a very different type of collision. KE is converted to mass here, but via particle-anti-particle creation. The only type of reactions where what you say is actually true are neutron and alpha activation reactions, and these are not done by accelerators. They are done with simple neutron/alpha sources. And again, these collisions are never perfectly inelastic. There is always a photon ejected in the process, and the resulting nucleus is almost always unstable.

...I think I should probably go learn 4-vectors now.

Very sensible idea.

 

[PAllen]

And yes, part of internal energy of the body could be in kinetic energy of components, but again, that's not kinetic energy of the body, which is due to CoM motion alone.

I don't see anyone implying that KE of a body as whole contributes to its rest mass. The claim is the KE of components may be said to contribute to rest mass of composite system, defined as the inertial mass of the composite system at rest. You may refuse to use the term rest mass for a composite system, but this seems an unconventional position. Taken to extreme, it would suggest you should not refer to the rest mass of proton because it is complex system whose constituents contribute in such complex ways to its inertial mass at rest that it took multiple supercomputers huge effort to calculate it to 1% accuracy.

 

[yuiop]

The rest mass of a composite system is a function of all the inertial mass and the momentum of all its components. By this definition, even two photons going in opposite directions have a rest mass as a system. To try and make that clearer, the total rest mass of a system is defined as:

m_{tot} = \sqrt{E_{tot}^2/c^4 - p_{tot}^2/c^2}

where E_{tot} is the sum of all the individual energies of the particles (m_o \gamma c^2) for massive particles and (hf) for photons and p_{tot} is the sum of all the individual momenta of the particles (m_o\gamma vc) and (hf/c) for massive and massless particles respectively.

For a pair of photons of equal energy (hf) and equal in magnitude but opposite in direction momentum (hf/c) the rest mass of the two photons is:

m_{tot} = \sqrt{E_{tot}^2/c^4 - p_{tot}^2/c^2} = \sqrt{(2hf)^2/c^4 - (0)^2/c^2} = 2hf/c^2

(if I have done it right).

This rest mass of the system as a whole is a conserved quantity. If a massive particle and its massive antiparticle counterpart collide and annihilate to produce two photons, the rest mass of the system before and after the annihilation is unchanged, because the two photons have a rest mass as a composite whole, even though they have zero rest mass as individual particles.

 

[Dale]

Maybe it is organized better in your head, but when you write it, you are not making a clear enough distinction between inertial mass and rest mass. All energy contributes to inertial mass, but kinetic energy specifically does not contribute to rest mass by very definition of rest mass.

I didn't use either term. Why should I make a distinction between terms I didn't use? I was unambiguous about the terms I did use.

No. The high-energy colliders are not designed to make things stick together. They are designed to blast things apart.

Are you claiming that high-energy colliders do not wind up with any particles which are more massive (invariant mass) than the sum of the invariant masses of the original particles? That all of the products have smaller masses, and producing smaller-mass products is indeed their designed purpose?