Why is kinetic energy conserved in collisions?

[Gerenuk]

Why is it that kinetic energy is conserved - for example during collisions?

Or can one prove in general that for
<br /> m_1\frac{\mathrm{d}v_1}{\mathrm{d}t}=\frac{\alpha(\vec{s}_2-\vec{s}_1)}{|\vec{s}_1-\vec{s}_2|^3}<br />
<br /> m_2\frac{\mathrm{d}v_2}{\mathrm{d}t}=\frac{\alpha(\vec{s}_1-\vec{s}_2)}{|\vec{s}_1-\vec{s}_2|^3}<br />
the term
<br /> m_1\frac{v_1^2}{2}+m_2\frac{v_2^2}{2}-\frac{\alpha}{|\vec{s}_1-\vec{s}_2|}<br />
is conserved?

 

[Doc Al]

Why is it that kinetic energy is conserved - for example during collisions?

In general, kinetic energy is not conserved during a collision.

Or can one prove in general that for
<br /> m_1\frac{\mathrm{d}v_1}{\mathrm{d}t}=\frac{\alpha(\vec{s}_2-\vec{s}_1)}{|\vec{s}_1-\vec{s}_2|^3}<br />

Can you explain what you mean by this.

 

[D H]

Kinetic energy is not always conserved during collisions. It is conserved during purely elastic collisions, but that is tautological. The definition of a purely elastic collision is one in which kinetic energy is conserved. Kinetic energy is not conserved, for example, if the colliding objects stick together. (This is a purely inelastic collision).

 

[Gerenuk]

In general, kinetic energy is not conserved during a collision.
Can you explain what you mean by this.

Ah, I see the questions.
I meant the purely microscopical collisions where charge collide subject to the Coulomb force. (Collision of macroscopic objects are a combination of purely elastic microscopic collisions)

The above equations are the equations of motion for two charges. I believe it should be possible to derive the law of conservation of energy from these two equations.

 

[D H]

Conservation of energy pertains to more than just the Coulomb force. Conservation of energy can be derived, but not from the Coulomb force. (What about other forces?) Conservation of energy results from the homogeneity of time. (Google Noether's theorem.)

That the Coulomb force is a conservative force is a direct consequence of the fact that the Coulomb force can be represented as the gradient of a potential.

 

[Gerenuk]

Conservation of energy pertains to more than just the Coulomb force.

OK, then I'm just talking about interacting charges. Can the above expression shown to be conserved?

Conservation of energy results from the homogeneity of time. (Google Noether's theorem.)

As far as I remember that doesn't apply in this case, as there are many particles and nothing is really homogeneous.

 

[denisv]

Equations of motion describe motion, not collisions.

 

[cepheid]

OK, then I'm just talking about interacting charges. Can the above expression shown to be conserved?As far as I remember that doesn't apply in this case, as there are many particles and nothing is really homogeneous.

What do you mean? He is not talking about the homogeneity of your system (or lack thereof). Homogeneity of time refers to the idea that if, when constructing your mathematical foundations of classical mechanics, you impose the condition that the laws of physics themselves do not change with time, the conservation of energy follows as a result. I am no expert and I am speaking loosely. I would take his advice and Google for more details.

 

[Bob S]

Momenta are vector quantities, and only their components (e.g., p_x, p_y, and p_z) can be added, not their magnitudes. The magnitudes of kinetic energies, which are not vectors, can be added. Momentum is a conserved quantity, but kinetic energy is not. If you can hear a click as two billiard balls collide, the collision is not conserving kinetic energy.

 

[D H]

Suppose you are shown a pair of simulated movies, one in which time is going forward and the other, time going backwards. In one movie you see the two separate balls coated with some sticky substance put on a collision course. Some globs of sticky stuff flies off as a result of the collision, but the two balls stick and start rotating about their center of mass. The total kinetic energy after the collision is less than the total kinetic energy prior to the collision. In the other movie you see a pair of stuck-together balls rotating about their center of mass. Several globs of sticky stuff comes from off screen and collide with the balls simultaneously. At the moment of impact, the balls spring apart with an increase in total kinetic energy. Which is the time-reversed movie?

Now suppose you are shown another pair of simulated movies, again differing only in the arrow of time. In this case multiple charged particles interact solely through the Coulombic interaction between pairs of particles. In this case, you cannot tell which movie has time going forward and which has time going backwards. The Coulomb force and Newton's equations of motion are time symmetric, and it is this time symmetry that makes the Coulomb force coupled with Newton's equations of motion form a system that conserves energy.

 

[Gerenuk]

Equations of motion describe motion, not collisions.

They do. Bodies collide through Coulomb interactions.

He is not talking about the homogeneity of your system (or lack thereof). ... I am no expert and I am speaking loosely. I would take his advice and Google for more details.

I'm not an expert either, but the last expert who used this approach failed to explain why kinetic energy is conserved. I know the simplified version of this theorem and it doesn't seem to have a connection here. But I'm be glad to read one. In that case I would appreciate a reference to a proof rather than I comment that's it's surely somehow possible.

Momentum is a conserved quantity, but kinetic energy is not.

Where do you think the kinetic energy goes? It goes to the kinetic and potential energy of the individual atoms, which is basically the statement I made above (for the case of two particles).

Suppose you are shown a pair of simulated movies, one in which time is going forward and the other, time going backwards. In one movie you see the two separate balls coated with some sticky substance put on a collision course.

All the physics here is only Coulomb interactions, no matter how complex or sticky the objects are.

 

[D H]

All the physics here is only Coulomb interactions, no matter how complex or sticky the objects are.

I guess that means that all that quantum mechanics nonsense I learned (and mostly forgot; my specialty is pretty much applied Newtonian mechanics) back in college is just that -- nonsense. Or is it perhaps that the quoted statement is nonsense?

Even from a classical perspective, energy is conserved in an inelastic collision. The reason it appears not to be conserved is because looking at only at the macroscopic-scale kinetic energy does not provide a full accounting of the total energy of the system. In an inelastic collision, some of the energy goes to changing the potential and thermal energy of the system. Colliding objects bend, break, and heat up during an inelastic collision. If you did a careful accounting of all of the energy at a quantum level. Nobody in their right mind would think of modeling the collision between, for example a car and a bus, at the quantum level. Instead we use short cuts such as coefficients of restitution and often ignore the energy that went into bending, breaking, and heating. Auto collision analysts of course do look at the bending and breaking as a sign of how fast the colliding car and bus were traveling.

 

[denisv]

They do. Bodies collide through Coulomb interactions.

No. Formally, the configuration space of the two-body problem is R^3 x R^3 - {(x1,x2), x1==x2}. Behavior of collision (i.e. x1==x2) is undefined and extra assumptions must be made to handle it. You can't use the equations of motion to analize collisions.

Pay attention next time.

 

[D H]

Behavior of collision (i.e. x1==x2)

In the parlance of quantum physics, the term "collision" has a broader meaning than point masses that are instantaneously collocated. Suppose one particle is stationary at the origin of a reference frame and another particle has coordinates (x,y,0) and velocity (-v_x,0,0), with x large, y small but non-zero. Sans any interaction between the particles, the second particle will merely zip past the stationary particle with a closest approach distance of y. Suppose the particles interact via an inverse square force law and suppose v_x is sufficiently large. The time interval over which the interaction between the particles has any significant impact on the trajectories will be rather small. If y is sufficiently small, the peak interaction will be quite large. Even though the particles never touch, the overall interaction between these particles is called a collision.

Pay attention next time.

Yes, please do.

 

[tiny-tim]

the only show in town …

Why is it that kinetic energy is conserved …

I meant the purely microscopical collisions where charge collide subject to the Coulomb force. (Collision of macroscopic objects are a combination of purely elastic microscopic collisions)

Hi Gerenuk!

I think you're asking a non-question.

Total energy is always conserved …

if we ever found a situation where it wasn't, we'd look for where the missing energy had gone, and we'd call that energy also, to balance the books!

If two elementary charges interact, and there's no molecules or nuclei for vibrational energy to go into, and no radiation emitted, then there's no way of "losing" enery, and kinetic energy will be conserved, simply because kinetic is the only show in town.

 

[cepheid]

If two elementary charges interact, and there's no molecules or nuclei for vibrational energy to go into, and no radiation emitted, then there's no way of "losing" enery, and kinetic energy will be conserved, simply because kinetic is the only show in town.

What do you mean kinetic energy is the only show in town? There is a Coulomb potential present. Even if we go along with the OP's scenario and consider two particles interacting by the Coulomb force only, it doesn't change what everybody has been saying, which is that in all instances, total energy is what is conserved, and kinetic energy is not (in general).

Take the simplest case, a 1D problem where two opposite charges q1 and q2 are at rest and begin to accelerate towards each other, and consider the energy of the system before (at separation R1), where all of the energy was potential, and after i.e. some time later at separation R2 < R1, where some of the energy has been converted into kinetic energy of the charges. Clearly kinetic energy was not conserved, because it started at zero (no motion), but at the later time, is non-zero (due to motion!).

However, if you do the math you will find that:

Initial PE = total energy

Final PE + increase in KE of charge 1 + increase in KE of charge 2 = total energy

Two charges interacted. The total energy of the system is what was conserved. Kinetic was not. So what are you (and OP) saying?

 

[Gerenuk]

I guess that means that all that quantum mechanics nonsense I learned (and mostly forgot; my specialty is pretty much applied Newtonian mechanics) back in college is just that -- nonsense. Or is it perhaps that the quoted statement is nonsense?

Even from a classical perspective, energy is conserved in an inelastic collision. The reason it appears not to be conserved is because looking at only at the macroscopic-scale kinetic energy does not provide a full accounting of the total energy of the system...

Not sure why there is a comment about quantum mechanics.
And the second paragraph is completely correct and exactly what I am saying here! So basically I only need to prove conservation of energy for two charges. That brings me back to the equations in the original question.

Behavior of collision (i.e. x1==x2) is undefined and extra assumptions must be made to handle it. You can't use the equations of motion to analize collisions. Pay attention next time.

They equations are basic school physics and yet it doesn't seem many people here understand them. Obviously two charges will never occupy the same position. They will always scatter - that what means collisions in that sense.

Hi Gerenuk!
I think you're asking a non-question.
Total energy is always conserved …
if we ever found a situation where it wasn't, we'd look for where the missing energy had gone, and we'd call that energy also, to balance the books!

That's an OK reasoning, however I believe one shouldn't accept too many unneccessary postulates. It should be possible to derive the conservation of energy (in this case at least) from the equations of motion. And we cannot call every missing bit of energy being somewhere else. Then at some point we would have virtual latent energy all of the place and the concept of energy would be effectively useless.

 

[tiny-tim]

What do you mean kinetic energy is the only show in town? There is a Coulomb potential present.
…
Two charges interacted. The total energy of the system is what was conserved. Kinetic was not. So what are you (and OP) saying?

oh, I meant kinetic energy is conserved if you compare "before" the collision with "after" the collision …

ie, when they're so far apart that you can ignore the interaction force.

 

[cepheid]

oh, I meant kinetic energy is conserved if you compare "before" the collision with "after" the collision …

ie, when they're so far apart that you can ignore the interaction force.

Oh. Okay. Makes sense!

 

[denisv]

They equations are basic school physics and yet it doesn't seem many people here understand them.

Yes. It's bizarre, quite frankly.

 

[diazona]

I thought it wouldn't be hard to present an actual proof but then I got bogged down in icky math and hidden unstated conditions on formulas I thought I liked But anyway, for your amusement... (this is sort of a hack):

The Lagrangian for two charged particles is

L = \frac{1}{2}m_1 \dot{r}_1^2 + \frac{1}{2}m_2 \dot{r}_2^2 + \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{\vert \vec{r}_1 - \vec{r}_2\vert}

A consequence of Noether's theorem is that for a Lagrangian which is not directly dependent on time, a conserved quantity h can be obtained from

h = \sum_k \frac{\partial L}{\partial\dot{r_k}} \frac{\mathrm{d} r_k}{\mathrm{d} t} - L

Thus a conserved quantity is given by

\begin{align*}h &amp;= \frac{\partial L}{\partial\dot{r}_1} \frac{\mathrm{d} r_1}{\mathrm{d} t} + \frac{\partial L}{\partial\dot{r}_2} \frac{\mathrm{d} r_2}{\mathrm{d} t} - L\\<br /> &amp;= m_1 \dot{r_1} \dot{r}_1 + m_2 \dot{r}_2 \dot{r}_2 - \frac{1}{2}m_1 \dot{r}_1^2 - \frac{1}{2}m_2 \dot{r}_2^2 - \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{\vert \vec{r}_1 - \vec{r}_2\vert}\\<br /> &amp;= \frac{1}{2}m_1 \dot{r}_1^2 + \frac{1}{2}m_2 \dot{r}_2^2 - \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{\vert \vec{r}_1 - \vec{r}_2\vert}\\<br /> \end{align*}
which is the energy. Ipso facto energy is conserved.

I pulled this mostly out of Goldstein's Classical Mechanics chapters 2 and 13, if anyone wants to follow up...

 

[Gerenuk]

The Lagrangian for two charged particles is ...

It seems just by using the Lagrangian one already assumes a concept of energy which is mv^2/2. But that the quantity that I want to derive first.

So I wonder if the kinetic energy can be derived from the equations of motion. Actually that's the equations that I posted in the first post. Does anyone see a way to derive the statement with these assumptions and the given conclusion?

 

[D H]

It seems just by using the Lagrangian one already assumes a concept of energy which is mv^2/2.

No! The Lagrangian certainly exists for non-conservative systems. It will just be time-varying, and explicitly so. Read what diazona said:

A consequence of Noether's theorem is that for a Lagrangian which is not directly dependent on time, a conserved quantity h can be obtained from

h = \sum_k \frac{\partial L}{\partial\dot{r_k}} \frac{\mathrm{d} r_k}{\mathrm{d} t} - L

Conservation of energy in this problem is a consequence of Noether's theorem. Note well: It is Noether's theorem, not Noether's theory.

 

[Gerenuk]

I have not yet seen a convincing argument why the Noether theorem should be applied to two particles. But through some simple calculation it is possible to derive from the conservation of momentum that for two particles the total energy defined by
<br /> E_\text{kin}(p_1)+E_\text{kin}(p_2)+E_\text{pot}=\text{const}<br />
<br /> E_\text{kin}\equiv\int \frac{p\mathrm{d}p}{m(p)}<br />
<br /> E_\text{pot}\equiv-\int \vec{F}\mathrm{d}\vec{r}<br />
is conserved.

So it seems, conservation of energy follows from conservation of momentum. The mass may depend on momentum as in special relativity.

 

[D H]

I have not yet seen a convincing argument why the Noether theorem should be applied to two particles.

Just because you do not understand something does not mean it is not true.

But through some simple calculation it is possible to derive from the conservation of momentum ...

You did not derive conservation of energy from conservation of momentum. You instead derived conservation of energy by implicitly assuming that energy is conserved. Here is where you did that:

<br /> E_\text{kin}\equiv\int \frac{p\mathrm{d}p}{m(p)}<br />

That is not the definition of kinetic energy. It is instead an alternate expression for work. You implicitly assumed the Work-Energy Principle to equate work to change in kinetic energy. The work-energy principle is a consequence of conservation of energy. In short, you derived conservation of energy by assuming conservation of energy. That of course is not a valid derivation.

 

[Gerenuk]

Just because you do not understand something does not mean it is not true.

Sure. But as long as no-one can explain it, I don't need to accept it. Smart-*** comments such as "read it up the Noethers theorem" don't help, when the person doesn't give an explanation himself. It often turns out that the reference is some random science which is not related.

You did not derive conservation of energy from conservation of momentum. You instead derived conservation of energy by implicitly assuming that energy is conserved.

How can you know that if you haven't even seen my derivation yet? The definitionof kinetic energy I use can be derived by a simple integration. And this definition is valid for classical mechanics and special relativity as you can check.

So please think before you complain! And as I myself don't want to state unsupported claims here is the derivation for a conservative force F=-\nabla V and conservation of momentum p=m\dot{r}. No more assumptions:
<br /> \dot{p}_1=-\nabla V(r_2-r_1)<br />
<br /> \dot{p}_2=\nabla V(r_2-r_1)<br />
<br /> \frac{p_2}{m_2}\dot{p}_2+\frac{p_1}{m_1}\dot{p}_1=\nabla V(r_2-r_1)\left(\frac{p_2}{m_2}-\frac{p_1}{m_1}\right)<br />
<br /> \frac{p_2}{m_2}\dot{p}_2+\frac{p_1}{m_1}\dot{p}_1-\nabla V(r_2-r_1)(\dot{r}_2-\dot{r}_1)=0<br />
<br /> \frac{p_2}{m_2}\mathrm{d}p_2+\frac{p_1}{m_1}\mathrm{d}p_1-\nabla V(r_2-r_1)\mathrm{d}(r_2-r_1)=0<br />
<br /> \int\frac{p_2}{m_2}\mathrm{d}p_2+\int\frac{p_1}{m_1}\mathrm{d}p_1-V(r_2-r_1)=\text{const}<br />
The first terms are identified as kinetic energy.

 

[D H]

You are assuming a conservative force. A conservative force conserves mechanical energy. You are assuming conservation of energy to prove conservation of energy.

You are also acting like a crackpot, which is not a good road to go down if you want to proceed in science.

 

[Gerenuk]

Do you actually understand any of the equations above? I do not assume conservation of energy at any point.

The fact that a conservative force conserves energy is what I'm actually proving whereas you assume it to be a law of physics.

And the thread asked to prove conservation of energy starting from only the equations in the first post, which apparently is possible.

Maybe you get confused by the term "conservative force". I only mean the the force is the mathmatical expression -\nabla V - no more. That energy is conserved is proved afterwards.

 

[D H]

Conservative forces can always be expressed as the gradient of a potential function. Nonconservative forces cannot. You have assumed conservation of energy by assuming the force is the gradient of some scalar potential.

 

[Gerenuk]

You have assumed conservation of energy by assuming the force is the gradient of some scalar potential.

OK, then tell me why a force expressed as a gradient conserves energy! You will say because you learned it at school. Well, I do not make use of this knowledge. Instead I prove it.