Conservation of kinetic energy between neutral fundamental particles in a vacuum?

one_raven

one_raven

God is a Chinese Whisper
Valued Senior Member
Hey… Feel up for a set of quantum mechanics thought experiments?

Disregarding, at least for the moment, the question of whether time, space and energy are quantized, let’s imagine two quantum particles in a perfect vacuum.

Particle B is at rest. Particle A is launched at particle B at speed s. Assuming a direct, head-on collision between two inelastic particles and conservation of energy, we get a classic Newton’s cradle…
Particle B will be ejected at speed s. Particle A will be at rest in the position where it struck particle B.

Now, let’s further assume the speed of transfer of kinetic energy is c.

What happens when s > c? (Greater rate of speed than it takes to transfer its kinetic energy.)
My intuition wants to say that particle A will not come to rest because it did not transfer all its kinetic energy so B will be ejected at c, and A will continue its trajectory at s-c, but something feels missing…

Next, we have 3 particles. Particles B and C are at rest, and in contact. Particle A is launched at speed s for a direct hit. Again, we have Newton’s cradle. (Heisenberg’s cradle?)
Particle C gets ejected at speed s, particle B remains at rest, particle a comes to rest alongside particle B.

Again, what happens when s > c?

Finally, we have the inverse of above… C is at rest. B and A are traveling together. Perfect collision… This time, A comes to rest. B and C fly off into the sunset together.

What happens if B and A were traveling together at a speed greater than c when they struck particle C?
 
one_raven said:
Hey… Feel up for a set of quantum mechanics thought experiments?

Disregarding, at least for the moment, the question of whether time, space and energy are quantized, let’s imagine two quantum particles in a perfect vacuum.

Particle B is at rest. Particle A is launched at particle B at speed s. Assuming a direct, head-on collision between two inelastic particles and conservation of energy, we get a classic Newton’s cradle…
Particle B will be ejected at speed s. Particle A will be at rest in the position where it struck particle B.

Now, let’s further assume the speed of transfer of kinetic energy is c.

What happens when s > c? (Greater rate of speed than it takes to transfer its kinetic energy.)
My intuition wants to say that particle A will not come to rest because it did not transfer all its kinetic energy so B will be ejected at c, and A will continue its trajectory at s-c, but something feels missing…

Next, we have 3 particles. Particles B and C are at rest, and in contact. Particle A is launched at speed s for a direct hit. Again, we have Newton’s cradle. (Heisenberg’s cradle?)
Particle C gets ejected at speed s, particle B remains at rest, particle a comes to rest alongside particle B.

Again, what happens when s > c?

Finally, we have the inverse of above… C is at rest. B and A are traveling together. Perfect collision… This time, A comes to rest. B and C fly off into the sunset together.

What happens if B and A were traveling together at a speed greater than c when they struck particle C?
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But s cannot be >c, surely?
 
one_raven said:
It's a thought experiment. Run with it...
Click to expand...
You can't just run with it. The outcome hinges on whether or not exchemist's supposition is true.
If it's true, the thought experiment is trivial.

(BTW, I assume your variable c is not c: the speed of light in a vacuum. Not a good idea. Choosing a variable that, by convention, already has a value is bound to lead to tears and heartbreak.)
 
Fine. Call the speed of lossless transfer of kinetic energy x. What happens when s > x?
 
I am not convinced that 'inelastic' is a meaningful property as this scale.
I think that generic particles are insufficient to derive a meaningful answer - the devil is in the details.
For example, you would get different results depending on if the particle were an (charged) electron versus, say, a neutron.
 
Okay...

Let's call them billiard balls in a hypothetical frictionless environment with lossless transfer of kinetic energy. I know it's not possible. That's what makes it a thought experiment (use your imagination).

In the transfer of kinetic energy, we have to assume there is a speed of propagation of that energy. If the speed of the moving object exceeds the speed of energy transfer in a Newton's cradle, what happens?
 
DaveC426913 said:
You can't just run with it. The outcome hinges on whether or not exchemist's supposition is true.
If it's true, the thought experiment is trivial.

(BTW, I assume your variable c is not c: the speed of light in a vacuum. Not a good idea. Choosing a variable that, by convention, already has a value is bound to lead to tears and heartbreak.)
Click to expand...
I assumed the poster meant c was the speed of light because it was said to be the speed with which kinetic energy was transferred in the collision. I assumed c had been chosen because of the idea that information cannot travel faster than light. But if c is just some arbitrary speed then my objection falls away.
 
one_raven said:
Okay...

Let's call them billiard balls in a hypothetical frictionless environment with lossless transfer of kinetic energy. I know it's not possible. That's what makes it a thought experiment (use your imagination).

In the transfer of kinetic energy, we have to assume there is a speed of propagation of that energy. If the speed of the moving object exceeds the speed of energy transfer in a Newton's cradle, what happens?
Click to expand...
If you are assuming a classical billiard ball analogy (i.e. a perfectly elastic collision), then the energy is transferred at a rate determined by the deformation and restitution of the two balls, as they come into contact and spring apart. It is fairly meaningless to talk of a "speed" of transfer, though you could speak of a time interval over which it occurs.

However at the start you said the objects were inelastic. That means they absorb some of the energy of the collision:https://en.wikipedia.org/wiki/Inelastic_collision

Did you mean that?
 
exchemist said:
If you are assuming a classical billiard ball analogy (i.e. a perfectly elastic collision), then the energy is transferred at a rate determined by the deformation and restitution of the two balls, as they come into contact and spring apart.
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Agree. If we are divorcing ourselves from any basis in reality, then presumably they collide and rebound depending on their material. I would have thought one would assume elastic collision, but you assert inelastic, so the energy transfer is not complete. Both particles continue to move.

Still, you cannot ignore nuclear forces at this scale. Two neutrons colliding will stick together because they're close enough for the strong nuclear force to come into play.

IOW, I don't think a generic particle-particle thought experiment is meaningful.
 
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one_raven said:
Particle B will be ejected at speed s. Particle A will be at rest in the position where it struck particle B.
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I read about this when I was about age 9 70 odd years ago in a comic book

In the comic book version did not have particles

The immovable object was a moon looking object, the irresistible force was long object, like ʻOumuamua but surface bit rougher sharper at direction of travel wider at backend

As you posted when irresistible force hit immovable object they swapped characteristics

The 9 year old thought the solution brilliant and still do even though reality does not work like that

Irresistible force cannot stop in an instant

Immovable object cannot obtain instant top speed

Also there is a loss of kinetic energy from Irresistible force during its collision with Immovable object hence does not contribute to the speed of the Immovable object

The lost kinetic energy has been / is converted into heat energy warming up the rocks at collision site

Still a good example of a science puzzle to to put out there

:)
 
Apologies for my confusion over elastic vs inelastic. It was my understanding that only inelastic particles could have a perfectly elastic collision, because there is no energy loss due to absorption, deformity, heat transfer, etc.

So, yes... In a perfectly elastic, lossless transfer of kinetic energy between a stationary object and a moving object, if the speed of the moving object exceeds the kinetic energy transfer duration from that object to the stationary one, what happens?
 
DaveC426913 said:
IOW, I don't think a generic particle-particle thought experiment is meaningful.
Click to expand...

It is to me, if it helps my understand what I'm puzzling over. I thought I was simplifying it with a hypothetical that would remove all energy sinks, but apparently that made it far LESS simple, unfortunately.
 
DaveC426913 said:
Agree. If we are divorcing ourselves from any basis in reality, then presumably they collide and rebound depending on their material. I would have thought one would assume elastic collision, but you assert inelastic, so the energy transfer is not complete. Both particles continue to move.

Still, you cannot ignore nuclear forces at this scale. Two neutrons colliding will stick together because they're close enough for the strong nuclear force to come into play.

IOW, I don't think a generic particle-particle thought experiment is meaningful.
Click to expand...
There are all sorts of issues if we deal with quantum objects. I think it best to get a classical situation straight first, so we can at least understand what issues that scenario presents. Thinking a bit further about it, it seems to me that the deformation of the billiard balls will in effect travel as a sound wave, at the speed of sound in the solid of which they are composed. So maybe after all we can speak of the relevant speed of sound as the speed of transmission of energy.

If that is right, then I think what would happen in the case of the impact velocity exceeding that speed, is that the impact would be supersonic and a shock wave would be created in the material of the billiard ball. However this shock wave would presumably still only travel at the speed of sound. I am honestly not sure, in such a situation, how the shock wave would be able to impart an impulse sufficient to give the second ball a speed exceeding the speed of sound in the material.
 
So, the stationary ball would be deflected at the speed of sound? If so, what happens to the ball that struck it? Would it bounce back in the direction of its source at some fraction its original speed, minus the speed of sound? Would it continue on its current course at its original speed minus the speed of sound?
 
one_raven said:
So, the stationary ball would be deflected at the speed of sound? If so, what happens to the ball that struck it? Would it bounce back in the direction of its source at some fraction its original speed, minus the speed of sound? Would it continue on its current course at its original speed minus the speed of sound?
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Well no, using your Newton's cradle analogy the first ball would be brought to a halt, if the collision was perfectly elastic.
 
Then kinetic energy would not be conserved. Isn't that right? The kinetic energy of the ball exiting at the speed of sound would be less than the kinetic energy of the ball that struck it, since it has supersonic speed at the point of collision.
 
one_raven said:
The kinetic energy of the ball exiting at the speed of sound
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Not sure if this was missed but "speed of sound" refers to speed of sound in the material of the ball.

That is the max rate at which kinetic energy can be transferred between atoms within the ball.

So, if the ball were made of diamond, for example, that is about 12 times the speed of sound through air.
 
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