Conservation of energy and momentum in collisions

[krd]

The old Hydrogen Atom stuff we start with is just not enough to deal with anything more complicated - but, deep down, we want it to.

Just a question. I'm just getting back into re-learning my physics, and learning physics I never had a chance to study.

But...Do you know?... If two atoms - with velocities - collide and recoil, how much energy is lost from their velocities/momentum?

Is there a formula?...Does it have ugly partial derivatives? (I've completely forgotten how to do them - they look like ugly squiggles to me)

 

[sophiecentaur]

I thought that you could mostly treat those collisions as elastic. (As long as they don't form a molecule)
Somebody may well put me right on that, though!

 

[krd]

I thought that you could mostly treat those collisions as elastic. (As long as they don't form a molecule)
Somebody may well put me right on that, though!

Some momentum would have to be lost in the collision. By entropy alone.

Chemical reactions aside...van der Waals aside (if they're going at just the right velocity even gas molecules will bind under Van der Waals.)

Energy is always conserved...but a fundamental rule of the universe is, that momentum is not (in a collision)

 

[sophiecentaur]

What universe do you live in where momentum is not conserved?

 

[Simon Bridge]

Entropy does not have to increase - it's just not allowed to decrease.
Momentum is strictly conserved in fundamental interactions. Even when momentum isn't conserved (as in some classical physics examples) you can still get a good agreement with observation in many cases by treating an interaction as elastic.

In a solid - an atom absorbs a photon and the atom recoils (conservation of momentum) against the lattice (think of the atoms as being joined by little springs). The springs communicate the motion partly to the adjacent atoms - like if you tapped a lump of jelly - and makes the whole thing jiggle about: heat.

Thus not all the energy of the collision is available to the re-emmitted photon. It may be that all the energy (and momentum) gets distributed to the solid as heat.
At this scale both energy and momentum is conserved.

 

[krd]

What universe do you live in where momentum is not conserved?

Momentum is conserved...As long as there isn't a collision...some momentum is conserved, some is lost.

Otherwise the ball bearings in Newton's Cradle, would keep going forever, and eva.

 

[krd]

Entropy does not have to increase - it's just not allowed to decrease.
Momentum is strictly conserved in fundamental interactions. Even when momentum isn't conserved (as in some classical physics examples) you can still get a good agreement with observation in many cases by treating an interaction as elastic.

An elastic interaction doesn't mean all the momentum is conserved. It's not.

In a solid - an atom absorbs a photon and the atom recoils (conservation of momentum) against the lattice (think of the atoms as being joined by little springs). The springs communicate the motion partly to the adjacent atoms - like if you tapped a lump of jelly - and makes the whole thing jiggle about: heat.

You're probably going to hate me for this. But. That jelly, is the electrostatic force - and when it jiggles - the vibration which you can call "heat"....Is a photon - and not a virtual photon either.

Thus not all the energy of the collision is available to the re-emmitted photon. It may be that all the energy (and momentum) gets distributed to the solid as heat.
At this scale both energy and momentum is conserved.

Are there formulas that give the breakdown of the distribution. I'm just getting back into my physics - I've a lot of maths to re-learn. And lots of physics I never studied - and didn't really have the brain for understanding when I was.

 

[xAxis]

An elastic interaction doesn't mean all the momentum is conserved. It's not.

That's exactly what it means :)

 

[tiny-tim]

… a collision...some momentum is conserved, some is lost.

sorry, krd, but that's impossible …

in a collision (whether elastic or not), momentum is always conserved, in any direction (if there are no external forces in that direction)

Otherwise the ball bearings in Newton's Cradle, would keep going forever, and eva.

Newton's cradle slows down only for the same reason a pendulum slows down …

the external forces of air resistance, and friction at the pivot ​

(and, in Newton's cradle, that's between collisions … each collision itself does conserve momentum)

 

[sophiecentaur]

the external forces of air resistance, and friction at the pivot

Those forces serve to transfer momentum to the air and to the rest of the Earth. Momentum is conserved in the wider situation.

 

[Simon Bridge]

There are often text-book examples where students do a calculation for a collision and find that momentum or energy is not conserved ... but this just means that something got left out. As has been pointed out in the Newton's Cradle example - there is an energy and momentum transfer to the air (sound, drag, heat etc) and the structure of the cradle (friction/heat) and so on.

Note that this example is also at odds with krd's earlier assertion that energy is always conserved: the same argument that momentum is not conserved here also means that energy isn't either ... any explanation of where the energy goes also works for momentum.

One I've seen in textbooks as the case of an object striking a spring+ratchet mechanism. The spring compresses, bringing the object to rest and the ratchet prevents the spring from re-extending. Momentum before = mv, momentum after = 0.

However, this calculation is done in the frame of reference of the spring+Earth. This is not an inertial reference frame since the Earth has accelerated (recoil from the collision). We don't expect momentum to be conserved in a non-inertial reference frame.

Put another way: the original description has left out the reaction of the Earth to the collision.

An elastic interaction doesn't mean all the momentum is conserved. It's not.

No - the term "elastic collision" is formally defined in physics to mean that momentum and energy are conserved. What we are debating is if elastic collisions ever occur.

You're probably going to hate me for this. But. That jelly, is the electrostatic force - and when it jiggles - the vibration which you can call "heat"....Is a photon - and not a virtual photon either.

Not exactly but OK, and that very real photon has momentum p_\gamma = h\nu/c which it got from conservation of momentum in the collision. Can you show otherwise?

Are there formulas that give the breakdown of the distribution.

...yes, and those formulas rely on energy and momentum being conserved in collisions.

I'm just getting back into my physics - I've a lot of maths to re-learn. And lots of physics I never studied - and didn't really have the brain for understanding when I was.

Well one of the things you'll have to learn is that energy and momentum are always conserved.

Also see what happens when a collision is not elastic :)

 

[krd]

sorry, krd, but that's impossible …

in a collision (whether elastic or not), momentum is always conserved, in any direction (if there are no external forces in that direction)

If you consider energy and momentum to be the same thing, both are conserved. If you limit your momentum to two colliding particles - or the balls in the cradle, then momentum is not conserved. The total momentum of all the particles is conserved.

Newton's cradle slows down only for the same reason a pendulum slows down …

the external forces of air resistance, and friction at the pivot ​

The momentum of the ball bearings is not being conserved. The click clack of the balls costs energy, that energy comes from the momentum of the balls.

(and, in Newton's cradle, that's between collisions … each collision itself does conserve momentum)

Momentum is only conserved if you consider the momentum of all the atoms, and all the other energies created in the system.

If you collide two atoms, you will create a photon. That photon will carry away some of the momentum. Only if you rule out these photons, is linear motion ever conserved. If in a solid, these photons will move as phonons through the electron cloud - cause other atoms to jiggle and release photons. The momentum will be distributed to all the atoms - and this is how thermal equilibrium occurs.

If you define an elastic collision as one where all momentum is conserved, fine - but they don't happen in this universe. Nothing has a Coefficient of restitution of 1.

Now, what I don't know, and I'm curious to know, is how much energy is in the photon when two atoms collide. I have an idea that its' wavelength is directly related to the recoil -(I have a funny feeling there's something really fundamental about the nature of light wrapped up in the recoil). I did a google for the Coefficient of restitution for atoms and got nada.

 

[tiny-tim]

The momentum of the ball bearings is not being conserved. The click clack of the balls costs energy, that energy comes from the momentum of the balls.

sorry, krd, but that's completely wrong …

the final velocities {v_i} can have the same ∑m_iv_i, but different ∑m_iv_i² …

the difference in energy does not come from the momentum, it comes from adjusting the relative velocities while keeping the total momentum the same​

 

[krd]

There are often text-book examples where students do a calculation for a collision and find that momentum or energy is not conserved ... but this just means that something got left out. As has been pointed out in the Newton's Cradle example - there is an energy and momentum transfer to the air (sound, drag, heat etc) and the structure of the cradle (friction/heat) and so on.

Well, you can do the experiments and get the right result if you use some kind of damping co-efficient.

One I've seen in textbooks as the case of an object striking a spring+ratchet mechanism. The spring compresses, bringing the object to rest and the ratchet prevents the spring from re-extending. Momentum before = mv, momentum after = 0.

Even before you get into frames of reference - you will always lose momentum to the atoms in the spring. Momentum is not really being lost - it's just being distributed to the atoms in the spring. But it's no longer linear. Eventually it's lost to heat and sound. Even if you had some kind of frictionless swing pendulum in a vacuum, eventually you'd lose your linear momentum.

No - the term "elastic collision" is formally defined in physics to mean that momentum and energy are conserved.

You're right - though I wasn't thinking in terms of the formal definition.

What we are debating is if elastic collisions ever occur.Not exactly but OK, and that very real photon has momentum p_\gamma = h\nu/c which it got from conservation of momentum in the collision.

And the missing momentum from the two atoms after a collision should be the momentum of the photon. And I believe the frequency of the photon is directly proportional to the net velocity of the atoms at collision (I don't know if I've put that very well - imagine one part is stationary and the other particle has all the velocity).

Can you show otherwise?...yes, and those formulas rely on energy and momentum being conserved in collisions.Well one of the things you'll have to learn is that energy and momentum are always conserved.

I'm not arguing otherwise - a better way to put it, is linear momentum is never conserved - though the net momentum is conserved by all other particles in the system.

 

[krd]

sorry, krd, but that's completely wrong …

No. I'm terribly sorry...But you are completely wrong. Not even wrong. :(

The sound of the clacking, is at the cost of the momentum of the swinging balls - they lose momentum and slow down.

the final velocities {v_i} can have the same ∑m_iv_i, but different ∑m_iv_i² …

I really don't understand your formula. You're giving what you say is the final velocity the sub script i (I'd assume you'd use i for initial velocity). Then you're giving velocity a superscript 2, I'm assuming you're squaring it. Forgive my ignorance, but I don't know what you're on about.
For conserved momentum ƩMassVelocity_initial = ƩMassVelocity_final

And

ƩMassVelocity_initial² = ƩMassVelocity_final²

But really it looks something like this for ball bearing

ƩMassVelocity_{initialballbearing} = ƩMassVelocity_{finalballbearing} + ƩMassVelocity_{nonballbearingpeturbedparticles}

for atoms colliding

ƩMassVelocity_initial = ƩMassVelocity_final + MomentumOfNewPhoton.

the difference in energy does not come from the momentum, it comes from adjusting the relative velocities while keeping the total momentum the same​

Total momentum of what?...The linear momentum of the balls, or the total momentum of all particles in the system.

 

[Simon Bridge]

a better way to put it, is linear momentum is never conserved - though the net momentum is conserved by all other particles in the system

... this is quite a change from earlier statements :) sounds more like saying that kinetic energy is not always conserved.

However - the momentum that is conserved is linear momentum. The photon momentum is also linear.

Even before you get into frames of reference - you will always lose momentum to the atoms in the spring. Momentum is not really being lost - it's just being distributed to the atoms in the spring. But it's no longer linear.

Unless you have a different definition of "linear" to the rest of us?

The common text problem deals with the situation before and after the spring is compressed. During compression you get some momentum in bits of the spring. Or maybe you are thinking of heating the spring - the momentum in atoms that make up the spring as they vibrate a bit more than before? How is this not "linear"?

 

[krd]

... this is quite a change from earlier statements :) sounds more like saying that kinetic energy is not always conserved.

No, you're just trying to play Gotcha! with me.

However - the momentum that is conserved is linear momentum.

The momentum of the pendulum is linear.

The photon momentum is also linear.

Not necessarily.

Unless you have a different definition of "linear" to the rest of us?

Linear in a line. And the basic equations for the motion of objects are one dimensional linear equations. You can make them two dimensional, you can make them three dimensional.

The common text problem deals with the situation before and after the spring is compressed. During compression you get some momentum in bits of the spring. Or maybe you are thinking of heating the spring - the momentum in atoms that make up the spring as they vibrate a bit more than before? How is this not "linear"?

You can describe the motion of all the particles in a system as linear but if you consider the atoms in the spring, very quickly your lines just turn into a chaotic fuzz. And if you start with a one dimensional linear equation, it breaks down the moment you hit the spring. You could account for the loss of momentum by a damping co-efficient.

 

[Simon Bridge]

No, [this is not quite a change from earlier statements from me] you're just trying to play Gotcha! with me.

It was an observation ... when someone changes position in an argument it is well to acknowledge this. In this case:

The earlier statement (post #17) was:
Energy is always conserved...but a fundamental rule of the universe is, that momentum is not (in a collision)

That was the statement that caused all this outcry... since it is so at odds with the fact that momentum is always conserved in fundamental interactions.
That would make momentum conservation a fundamental rule.

The recent statement was:
linear momentum is never conserved - though the net momentum is conserved by all other particles in the system

Perhaps we can take it that this is a clarification of a position you have held all along?

The photon momentum is also linear.

Not necessarily.

... can you give an example of non-linear photon momentum? Or do you mean that a large system of photons in a solid or gas is better treated statistically?

You can describe the motion of all the particles in a system as linear but if you consider the atoms in the spring, very quickly your lines just turn into a chaotic fuzz. And if you start with a one dimensional linear equation, it breaks down the moment you hit the spring. You could account for the loss of momentum by a damping co-efficient.

Well yes - any very large linear system can be described statistically. It is often more convenient to do that. Does not change the fact that all the momentum transfers are linear momentum.

Perhaps you mean that it is not always easy to track all the changes in momentum in every situation so it may appear that momentum is not conserved?

I feel that you are trying very hard to hang on to a half-baked idea about momentum. However - we are also very far off topic. I think anyone googling to this thread with questions about color or about conservation laws will have enough information to go on with. Cheers.

 

[vela]

krd, it sounds like you have energy and momentum backwards. Momentum is always conserved in collisions, but kinetic energy isn't except in elastic collisions.

 

[krd]

krd, it sounds like you have energy and momentum backwards.

No, I don't.

Momentum is always conserved in collisions

Linear momentum is only conserved in an abstract sense - for any collision in the real world it is not.

, but kinetic energy isn't except in elastic collisions.

But a perfect elastic collision is something that only exists in the abstract sense. In reality they do not exist.


I know these explanations are good for rote learning and giving kids something to do in exams. But they are incomplete descriptions of reality. Newton's cradle is not a perpetual motion device - and you can witness the loss in momentum with your own eyes - you don't need to build a big tunnel under Switzerland to see it.

 

[Dale]

But a perfect elastic collision is something that only exists in the abstract sense. In reality they do not exist.

At the atomic level, as in the OP, they most certainly do exist.

A macroscopic object has a lot of internal degrees of freedom where kinetic energy can be turned into internal energy of one form or another. An atom has relatively few internal degrees of freedom, and those degrees of freedom are generally quantized. If the collision does not change one of those quantized degrees of freedom (as many do not) then it is perfectly elastic.

Similarly with momentum.

Your reasoning based on classical macroscopic objects simply doesn't apply to quantum mechanical situations.

 

[krd]

At the atomic level, as in the OP, they most certainly do exist.

A macroscopic object has a lot of internal degrees of freedom where kinetic energy can be turned into internal energy of one form or another. An atom has relatively few internal degrees of freedom, and those degrees of freedom are generally quantized. If the collision does not change one of those quantized degrees of freedom (as many do not) then it is perfectly elastic.

The atom has lots of degrees of freedom. If it didn't radio waves couldn't pass through walls, and light couldn't pass through solids or liquids.

Also you wouldn't be able to create radio broadcasts at the frequency of your choosing.

And I think you're confusing some principles. Light is not simply emitted and absorbed in atomic spectral lines - it's also created through inter-atomic collisions. For line spectra it's limited degrees of freedom. For inter-atomic collisions, it's unlimited degrees of freedom - well the limit is - I'm not actually sure - the top is obviously particle collisons at relativistic speeds - the bottom would be glancing slow collisions creating photons with wavelengths miles long or more even.

Here let's look at a graph of spectra just for the heck of it. If photon emission could only occur on the spectral lines, the world would be a very dark place. And there wouldn't be much choice of frequencies to broadcast radio on.

Your reasoning based on classical macroscopic objects simply doesn't apply to quantum mechanical situations.

Quantum, Smwantum. Dragging something into the realm of the quantum is often the trick used when someone's understanding is incoherent. Here have a wave equation, a delta, and a squiggle, ψ Δ ζ and a snake ∫...two snakes about to fight ∫ ζ

At a certain level atoms can be treated as semi-classical. And the electrostatic force between atoms as classical liquid or better a jelly that wobbles. For certain purposes the classical rules breakdown - but for others they don't.

Transmitting a radio single from an antenna can be explained as simply wobbling the atoms in the antenna at the desired frequency.

When it comes to the conservation of linear momentum, your reasoning is mistaking abstract objects for real classical macroscopic ones.

So, you are wrong. Or maybe you're in a superposition of correctness, both simultaneously right and wrong. Or maybe in another one of the multi-verses you're right and I'm wrong.

 

[mfb]

Ok, first thing: In an isolated system (no external forces, nothing leaves/enters the system), the vector sum of the momenta of all objects cannot change. At the same time, the sum of all energy components in this system cannot change.
These are the fundamental laws of energy and momentum conservation.

The momentum and energy of a single particle in this system can change, the sum of absolute momenta can change - that has nothing to do with momentum or energy conservation!

In inelastic collisions, some kinetic energy is converted to other forms of energy, like sound and heat. As you usually do not want to keep track of them, it is possible to treat this as an energy loss - in this case, "energy is not conserved" with the meaning that the kinetic energy is not constant. Those things do not carry significant momentum, therefore you cannot do the same trick with momentum - it is conserved even for inelastic collisions.Collisions between uncharged atoms are usually elastic, as they have no good way to distribute energy to something else apart from kinetic energy. With molecules, things are different.
At the sun, you are looking at a plasma, with free charges flying around everywhere. Usually, electrons which are bound to specific atoms have a small contribution to emission and absorption of light.

Transmitting a radio single from an antenna can be explained as simply wobbling the atoms in the antenna at the desired frequency.

You mean the free electrons in the conductor.

 

[Dale]

The atom has lots of degrees of freedom.

Incorrect.

If it didn't radio waves couldn't pass through walls, and light couldn't pass through solids or liquids.

Actually, this contradicts your point. The more degrees of freedom an atom or molecule has then the greater likelyhood that a photon would be absorbed. So the fact that a radio wave does pass through a wall implies that there is not a degree of freedom available to absorb the energy. This example contradicts your position.

And I think you're confusing some principles. Light is not simply emitted and absorbed in atomic spectral lines - it's also created through inter-atomic collisions.

When you are talking about atoms rather than molecules, then you should think of a monoatomic gas. In a monoatomic gas light is, in fact, emitted and absorbed in characteristic atomic spectral lines. That is what gives such gasses their optical transparency or characteristic color. The collisions broaden the spectral lines through a phenomenon called Doppler broadening, but they do not change the fact that energy is absorbed only in discrete amounts governed by the available internal quantized degrees of freedom of the atom.

For inter-atomic collisions, it's unlimited degrees of freedom - well the limit is - I'm not actually sure - the top is obviously particle collisons at relativistic speeds - the bottom would be glancing slow collisions creating photons with wavelengths miles long or more even.

Please provide a peer-reviewed reference supporting this point. Remember, we are talking about internal degrees of freedom for an isolated atom. Not molecules or solids, and not external degrees of freedom like momentum.

Quantum, Smwantum... Here have a wave equation, a delta, and a squiggle, ψ Δ ζ and a snake ∫...two snakes about to fight ∫ ζ

this is a rather immature comment, and it casts the remainder of your comments in a very poor light.

Transmitting a radio single from an antenna can be explained as simply wobbling the atoms in the antenna at the desired frequency.

You mean wobbling the free electrons, the atoms in an antenna don't wobble in relation to the transmission. But sure, which is why I limited my comments to atoms, which have fewer degrees of freedom than molecules or solids.

When it comes to the conservation of linear momentum, your reasoning is mistaking abstract objects for real classical macroscopic ones.

Your OP was about atoms, not classical macroscopic objects. So this comment is irrelevant to the topic at hand, as are all of your continued references to other macroscopic phenomena such as antennas.

 

[krd]

When you are talking about atoms rather than molecules, then you should think of a monoatomic gas. In a monoatomic gas light is, in fact, emitted and absorbed in characteristic atomic spectral lines. That is what gives such gasses their optical transparency or characteristic color.

The collisions broaden the spectral lines through a phenomenon called Doppler broadening, but they do not change the fact that energy is absorbed only in discrete amounts governed by the available internal quantized degrees of freedom of the atom.

Mono-atomic gaslight is not limited to it's spectral lines. And if I think I've got your thinking right, you think that the continuous spectra of black body radiation, which all bodies and gases emit is the result of Doppler broadening. Well, if that's what you're thinking, then you're wrong. The continuous spectra would look completely different, its intensity would be bumpy. Even at high temperatures the lines do not become that broad.

For the mechanism that governs the emission and absorption of line spectra there are limited degrees of freedom. By comparison the mechanism that creates continuous spectra is nearly unlimited.

 

[Dale]

Mono-atomic gaslight is not limited to it's spectral lines. And if I think I've got your thinking right, you think that the continuous spectra of black body radiation, which all bodies and gases emit is the result of Doppler broadening.

Not all bodies and gases emit blackbody radiation, only blackbodies do. That is why it is called blackbody radiation. Many non-blackbodies are reasonable approximations to blackbodies, but monoatomic gasses are not even close.

Well, if that's what you're thinking, then you're wrong. The continuous spectra would look completely different, its intensity would be bumpy.

It is "bumpy", in fact your picture in post 22 showed quite clearly the difference between the continuous blackbody spectrum and the "bumpy" spectrum of a typical monoatomic gas, Neon. Even materials which are reasonable approximations to blackbodies will have "bumpy" spectra in regions where they deviate from a blackbody.

For the mechanism that governs the emission and absorption of line spectra there are limited degrees of freedom. By comparison the mechanism that creates continuous spectra is nearly unlimited.

Agreed, which is why monoatomic gasses are not blackbodies.

krd, one piece of advice. You have some extremely severe misconceptions about physics. You have made some physics mistake with almost every single post you have written in this thread. You would be much better served by asking questions and learning than making statements and arguing.

 

[Simon Bridge]

Stars are black bodies. And everything is actually a black body. The name is misleading. As everything emits radiation. Cooler objects you can't see it, because the intensity of the light is very weak in the visible spectrum. But as they get hotter, the intensity of the light increases in the visible spectrum and you can see the light. First they glow red and then white hot (white hot as it's called - and if it is glowing white then it is hot).

There is always visible light everywhere - there's just not always enough to see it. Even deep space is full of starlight.

from another thread.

Should shed some light on krd's thinking here.
I suspect the idea is that gasses have a line spectra and a black body spectra - but the bb one is too weak to see.