Question about Particle disintegration (Landau mechanics page 43)

[Scott75]

On page 41 for the spontaneous disintegration of a particle into two, Landau states the total momentum in the C system is zero.

On page 43 for the disintegration of many particles into two, Landau states: In the C system... every resulting particle (of a given kind) has the same energy...

I thought the kinetic energies of the resulting particles are different when viewed from the C system. So by C system does he still mean a system where a primary particle has zero momentum? From each of these systems, the resulting particles of a given kind will have the same energy.

 

[Doc Al]

So by C system does he still mean a system where a primary particle has zero momentum?

Yes. By "C system" he means the center of mass system (frame of reference), where the momentum of the primary particle is zero and thus where the total momentum of the resulting particles is also zero. (As opposed to the "L system", the laboratory frame.)

 

[Scott75]

Thanks for the reply.

But this means V in eqn 16.8 isn't constant.

I'm beginning to think the wording and equations suggest that a C system is where the total momentum of the whole system is zero and, due to the primary particles being isotropically distributed, the K.E. of resulting particles of a given kind are equal in the C system. I'm assuming isotropically distributed means the particles are randomly traveling in all directions. I've also assumed the speeds of the primary particles are equal.

The resulting paricles of a given kind would all have the same K.E. in the C system if they made all made equal angles with the velocity vector of the primary particle. But this means all primary particles disintegrate in the same way but is there a reason why they should?

 

[Doc Al]

Perhaps I'm misinterpreting what Landau means, but I assumed that when he speaks of many particles that they all have the same velocity V. (Think of a beam of particles.) Those primary particles are randomly oriented, thus the resulting particles shoot off in random directions.

 

[Scott75]

Perhaps I'm misinterpreting what Landau means, but I assumed that when he speaks of many particles that they all have the same velocity V. (Think of a beam of particles.) Those primary particles are randomly oriented, thus the resulting particles shoot off in random directions.

Can it be a beam of primary particles if they are randomly oriented? I assumed randomly oriented meant the primary particles are traveling in random directions.

 

[Doc Al]

I assumed randomly oriented meant the primary particles are traveling in random directions.

I assumed that randomly oriented meant that the particles have random orientations (their axes point in random directions) and that the orientation of the primary particle determines the direction that the daughter particles will shoot off in.

 

[etotheipi]

I assumed that randomly oriented meant that the particles have random orientations (their axes point in random directions) and that the orientation of the primary particle determines the direction that the daughter particles will shoot off in.

@Doc Al this makes sense, but how do we define the orientation of a particle in this case? Crudely I imagine two spheres glued together, and the orientation refers to the orientation of the vector joining the centres of the two spheres. Then, when the particle disintegrates, in the $C$ system the two spheres move off parallel to the line of centres.

But I imagine a real particle will be more uniform, and it will not be evident how to define an 'orientation'?

 

[Scott75]

I assumed that randomly oriented meant that the particles have random orientations (their axes point in random directions) and that the orientation of the primary particle determines the direction that the daughter particles will shoot off in.

I don't think it's possible to define the orientation that way. When considering only one primary particle there is not enough information to determine the direction of the resulting particles. All that can be said from the C system is the resulting particles travel in opposite directions with momentum of equal magnitude. All directions are equally likely.

 

[Scott75]

@Doc Al this makes sense, but how do we define the orientation of a particle in this case? Crudely I imagine two spheres glued together, and the orientation refers to the orientation of the vector joining the centres of the two spheres. Then, when the particle disintegrates, in the $C$ system the two spheres move off parallel to the line of centres.

But I imagine a real particle will be more uniform, and it will not be evident how to define an 'orientation'?

Agree with your last point. All that can be said is a primary particle spontaneously decays into two.

 

[Doc Al]

@Doc Al this makes sense, but how do we define the orientation of a particle in this case? Crudely I imagine two spheres glued together, and the orientation refers to the orientation of the vector joining the centres of the two spheres. Then, when the particle disintegrates, in the $C$ system the two spheres move off parallel to the line of centres.

But I imagine a real particle will be more uniform, and it will not be evident how to define an 'orientation'?

Yeah, my attempt at making sense of Landau's remarks was something of a stretch.

 

[etotheipi]

Yeah, my attempt at making sense of Landau's remarks was something of a stretch.

I still think your interpretation is correct, though, as it is surely the most reasonable explanation. In reality, I think it suffices to say that the directions of motion are isotropically distributed due to symmetry considerations in the frame C, and that when L&L mention 'orientation' of the primary particles it is just a more pedagogical explanation

 

[Doc Al]

When considering only one primary particle there is not enough information to determine the direction of the resulting particles. All that can be said from the C system is the resulting particles travel in opposite directions with momentum of equal magnitude. All directions are equally likely.

I think is Landau's only point. That the particles shoot off in uniformly random directions (in the C frame).

 

[Scott75]

I think is Landau's only point. That the particles shoot off in uniformly random directions (in the C frame).

He also says they all have the same energy when viewed from the C system which is what I'm struggling with.

 

[Doc Al]

I still think your interpretation is correct, though, as it is surely the most reasonable explanation. In reality, I think it suffices to say that the directions of motion are isotropically distributed due to symmetry considerations in the frame C, and that when L&L mention 'orientation' of the primary particles it is just a more pedagogical explanation

Agreed!

 

[Doc Al]

He also says they all have the same energy when viewed from the C system which is what I'm struggling with.

Note that he says all resulting particles of a given kind have the same energy. Even that requires some justification. I imagine each primary particle disintegrates via the same process into its daughter particles. If so, then all daughter particles of the same kind should have the same energy.

 

[Scott75]

I still think your interpretation is correct, though, as it is surely the most reasonable explanation. In reality, I think it suffices to say that the directions of motion are isotropically distributed due to symmetry considerations in the frame C, and that when L&L mention 'orientation' of the primary particles it is just a more pedagogical explanation

What are those symmetry conditions? I can think of symmetry conditions where the resulting particles (of a given kind) have different energies but satisfy conservation of energy and momemtum

 

[etotheipi]

What are those symmetry conditions? I can think of symmetry conditions where the resulting particles (of a given kind) have different energies but satisfy conservation of energy and momemtum

It is from the invariance of Physical laws under rotations. The particle is at rest in the frame C, and every direction is equally likely.

On your point about the conservation laws... conservation of momentum in frame C dictates that the two fragments must have velocities $\vec{v}$ and $-\vec{v}$. It follows that their kinetic energies in the frame C will also be equal, specifically to $T = \frac{1}{2}m\vec{v} \cdot \vec{v}$. Their energies in this frame will only differ if they carry away different amounts of the remaining internal energy.

In the frame L, the velocities of the two fragments and thus kinetic energies are not necessarily equal any more (and generally, they will not be). You can check this by explicitly transforming the velocities back into the frame L by adding back $\vec{V}$.

 

[Scott75]

Note that he says all resulting particles of a given kind have the same energy. Even that requires some justification. I imagine each primary particle disintegrates via the same process into its daughter particles. If so, then all daughter particles of the same kind should have the same energy.

Yes of a given kind. I would say given kind implies the same mass and internal energy. Agreed the same process happens but the direction of the the resulting particles is random as long as conservation of momentum and energy are satisfied. Hence the circle with radius equal to the velocity of one of the resulting particles

 

[Scott75]

It is from the invariance of Physical laws under rotations. The particle is at rest in the frame C, and every direction is equally likely.

On your point about the conservation laws... conservation of momentum in frame C dictates that the two fragments must have velocities $\vec{v}$ and $-\vec{v}$. It follows that their kinetic energies in the frame C will also be equal, specifically to $T = \frac{1}{2}m\vec{v} \cdot \vec{v}$. Their energies in this frame will only differ if they carry away different amounts of the remaining internal energy.

In the frame L, the velocities of the two fragments and thus kinetic energies are not necessarily equal any more (and generally, they will not be). You can check this by explicitly transforming the velocities back into the frame L by adding back $\vec{V}$.

Completey agree but isn't that for the C system applied to one primary particle that is at rest? For more than one particle I thought the C system is where the total momentum is zero. So why can't the resulting particles of a given kind have different energies as their speeds don't have to be equal when viewed from the C system?
I'm assuming the directions of motion of the resulting particles can't be predicted.

 

[etotheipi]

I think you are mis-understanding the setup. As @Doc Al mentioned, with scattering experiments like this you are sending a collimated beam of primary particles into the chamber. You can consider each one individually, as all are at rest initially in the frame C, and the direction of scattering is random for each pair. This means that in the frame C, the distribution of scattering angles will tend toward being uniform. The total momentum of each pair will be zero, and the total momentum of the whole ensemble will also be zero.

So why can't the resulting particles of a given kind have different energies as their speeds don't have to be equal when viewed from the C system?

The speeds of any two particles in the same pair will be equal. Without any further knowledge on the mechanism of the disintegration, I don't think it is fruitful to speculate as to how the internal energy is divided up in any given disintegration (i.e. we are not told if the ratio of kinetic energy to internal energy is the same for all disintegrations).

If by "the same kind" you mean a subset of disintegrations which share the same ratio of kinetic energy to internal energy, then I disagree with your statement. Their speeds and kinetic energies will of course be equal. Think about what you are saying...

 

[Scott75]

I understood all of that before posting the original question. So perhaps I have misunderstood the setup. I could see how it worked nicely with a beam but ruled it out as took 'randomly oriented' primary particles to mean they travel in random directions. On page 1 he describes a particles as a body whose dimensions may be neglected and you mentioned previously that a primary particle will be more uniform so not evident how to define its 'orientation'

 

[Scott75]

If only they had written the word 'beam'!

 

[etotheipi]

Completey agree but isn't that for the C system applied to one primary particle that is at rest? For more than one particle I thought the C system is where the total momentum is zero. So why can't the resulting particles of a given kind have different energies as their speeds don't have to be equal when viewed from the C system?
I'm assuming the directions of motion of the resulting particles can't be predicted.

Actually, I think I understand what you were trying to say. Would I be right in thinking you were assuming the primary particles were incoming from random directions (with a not necessarily uniform distribution) so that the primary particles are not necessarily initially at rest in the centre of mass frame? Yes in that case, the two fragments will not necessarily have the same magnitude of momentum or kinetic energy in the centre of mass frame of the whole system.

But that is a much more complicated scenario, and is not what I think L&L are describing. Their argument rests on all of the primary particles initially being at rest in the centre of mass frame, and then them disintegrating in random directions, so the fragments off any given primary particle have equal magnitudes of momentum and kinetic energies in the centre of mass frame. Then we can try and predict an angular probability distribution for the scattering

 

[Scott75]

Actually, I think I understand what you were trying to say. Would I be right in thinking you were assuming the primary particles were incoming from random directions (with a not necessarily uniform distribution) so that the primary particles are not necessarily initially at rest in the centre of mass frame? Yes in that case, the two fragments will not necessarily have the same magnitude of momentum or kinetic energy in the centre of mass frame of the whole system.

But that is a much more complicated scenario, and is not what I think L&L are describing. Their argument rests on all of the primary particles initially being at rest in the centre of mass frame, and then them disintegrating in random directions, so the fragments off any given primary particle have equal magnitudes of momentum and kinetic energies in the centre of mass frame. Then we can try and predict an angular probability distribution for the scattering

Yes that is what I meant but couldn't get off that train of thought as misunderstood the meaning of 'randomly oriented'. I could see it working if the primary particles had the same speed, all secondary particles had equal masses and their velocities all made an equal angle to their primary particle's velocity. That is a very special case though.

Thanks for clearing up my misunderstanding. Thanks too to @Doc Al who also pointed out it was a beam.

 

[Doc Al]

If only they had written the word 'beam'!

I feel your pain. I recall (many years ago!) having to chew through L&L's Quantum Mechanics. Rough going.

Many thanks to @etotheipi for jumping in and clarifying things nicely.