Why is kinetic energy conserved?

[Yoni]

For instance in an elastic collision of two indivisible particles...

Why does m v^{2} stay constant? Why not m v^{3}?
[That is, where other type of energy are unavailable (potential, electrical, mass, etc.)]

In other words, why does the physical unit: kg \cdot m^{2} / s^{2}, or Joule, stay constant?

Please keep the discussion in classical terms...

 

[Drakkith]

Because that's simply what we've observed. If energy wasn't conserved, it would lead to things occurring for no reason, such as water suddenly freezing without transferring its heat somewhere first.

 

[DrClaude]

For instance in an elastic collision of two indivisible particles...

Why does m v^{2} stay constant? Why not m v^{3}?
[That is, where other type of energy are unavailable (potential, electrical, mass, etc.)]

In other words, why does the physical unit: kg \cdot m^{2} / s^{2}, or Joule, stay constant?

Please keep the discussion in classical terms...

Read up on Noether's theorem. Conservation of energy follows from the fact that the equations of motion are invariant under a translation in time (if you move the origin of time, $t=0$, you don't change the result). In the kind of collision you are considering, by construction all other energies are constant, therefore kinetic energy has to be conserved for the total energy to be conserved.

 

[sophiecentaur]

For instance in an elastic collision of two indivisible particles...

Why does m v^{2} stay constant? Why not m v^{3}?
[That is, where other type of energy are unavailable (potential, electrical, mass, etc.)]

In other words, why does the physical unit: kg \cdot m^{2} / s^{2}, or Joule, stay constant?

Please keep the discussion in classical terms...

That's the definition of 'elastic'.

Momentum (mv, at its simplest) is also conserved - elastic or not.

 

[Yoni]

Because that's simply what we've observed. If energy wasn't conserved, it would lead to things occurring for no reason, such as water suddenly freezing without transferring its heat somewhere first.

Interesting... But WHY? How would the world would look if a different physical scalar was conserved?? (for example m v^{3}?)

That's the definition of 'elastic'.

Can a collision between elementary particles NOT be 'elastic'?

I still don't see why should kinetic energy, as it is defined, be conserved...

Read up on Noether's theorem. ...

I looked into Noether's theorem, and I can see why you would get the dimensions of energy when taking "s" as time, i.e. you assume a temporal symmetry. However, something in his logic feels cyclic to me...
I would like to know more: How would we get the Lagrangian in the first place (such that its derivative with position gives you Momentum, and its derivative with velocity gives you Force)? What does the momentum that we get represent? What is the Force that we get? Why should such a function (L) exist at all, and what is it? Why should this entity be symmetric in time? Why would not other functions be symmetric in time, providing us with m v^{3} being constant?

I also need to explain this to a student... How would you go about explaining this? Is energy conservation a product of the conservation of linear momentum, perhaps? I can see some connection via the Pythagorean Theorem...

 

[BruceW]

Why should such a function (L) exist at all, and what is it? Why should this entity be symmetric in time? Why would not other functions be symmetric in time, providing us with m v^{3} being constant?

exactly. the Lagrangian is the most fundamental thing. (as far as I know). And we essentially just have to choose our Lagrangian such that it agrees with the physical kinds of symmetry that we experimentally observe in our universe. Another example is that we experimentally observe symmetry in each of the 3 spatial directions (x,y,z) (i.e. there is nothing special about any particular direction we choose). And so if we create a Lagrangian that does not care about which direction we call x and which we call y, then we automatically get conservation of angular momentum.

I also need to explain this to a student... How would you go about explaining this? Is energy conservation a product of the conservation of linear momentum, perhaps? I can see some connection via the Pythagorean Theorem...

no, because inelastic collisions don't conserve kinetic energy. I think it's best to just view things like this: a general collision will neither be fully elastic, nor fully inelastic, it will be somewhere in-between. But, in some cases, there will be collisions that are so close to being completely elastic, that we can say they are effectively completely elastic. These collisions are the ones where very little heat/sound/deformation of the object occurs.

Interesting... But WHY? How would the world would look if a different physical scalar was conserved?? (for example mv³?)

If the force between two point particles is a conservative force, then the collision will be elastic. i.e. if the two point particles start off very far from each other, the potential energy between them is initially zero. Then as they get close to each other, some of the energy gets stored as potential energy. And when they get far away from each other, all the potential energy is converted back into kinetic energy. So the total kinetic energy is conserved. This is essentially due to the work-energy theorem. For example, two positively charged particles that are moving toward each other (but not head-on) will have the same final total KE as the initial total KE.

edit: so, in other words, a collision is not elastic if there is a non-conservative force, or if some of the kinetic energy has been stored as potential energy, or given away as some other form of energy, like sound.

 

[Dale]

Hi Yoni,

In the end, all of the answers that you can get for this question will always boil down to Drakkith's response: that is simply what we have observed as an experimental fact about the world.

From Noether's theorem we know that conserved quantities are related to symmetries of the Lagrangian. So how do we find the Lagrangian? By doing experiments and determining what matches observation. When we do that we find that the Lagrangian's that match observation are time-symmetric meaning that there is a corresponding conserved quantity which we call energy.

But again, the fact that the world can be described using Lagrangians with time symmetry is just what we observe. We don't know why it is that way.

 

[Dadface]

If the kinetic energy of collision is less than a certain minimum given by quantum theory there will be no energy conversions and KE will be conserved. The best example I can think of is when gas atoms collide with a total kinetic less than the minimum excitation energy of either one of the atoms. But then the question becomes "Why is energy conserved? That's been answered above

 

[Andrew Mason]

For instance in an elastic collision of two indivisible particles...

Why does m v^{2} stay constant? Why not m v^{3}?
[That is, where other type of energy are unavailable (potential, electrical, mass, etc.)]

In other words, why does the physical unit: kg \cdot m^{2} / s^{2}, or Joule, stay constant?

Please keep the discussion in classical terms...

The reason KE is constant has to do with the relationship between Force, acceleration and mass.

KE is defined as the ability of a body, by virtue of its motion, to do work. Work is defined as the application of a force through a distance. So when we say that a body has a KE of x we MEAN that it has the ability to apply a force F over the distance s of x = Fs

So let's see how what x is for a body of mass m moving at velocity v.

The first thing to note is that Newton's third law says that a body applying a force to another will experience an equal and opposite force on itself.

So if a body moving at velocity v applies a force F to another body, it will experience a force -F itself. Suppose it experiences a force -F over a distance of Δs. How much momentum will it lose? It will lose Δp = mΔv = -FΔt where Δt is the time it takes for the body to cover the distance Δs.

Now we know that it can't apply a force after it stops, so this force can only last while it is moving. That means that the maximum time is Δp = m(0 - v) = -FΔt so:

\Delta t = mv/F

The average speed over that time is v_avg = Δs/Δt so Δs = v_avgΔt = v_avg(mv)/F

Since in this example we are assuming that it is slowing down at a constant deceleration (F is constant) v_avg is just (v-0)/2 = v/2 and we end up with:

\Delta s = v\Delta t = \frac{mv^2}{2F} so: F\Delta s = \frac{mv^2}{2}

So a body with mass m and speed v is capable of applying a force through a distance equal to mv²/2 before it stops.

AM

 

[Yoni]

The reason KE is constant has to do with the relationship between Force, acceleration and mass.

...

AM

Wonderful!
This is exactly what I will to explain to the student. Thanks, Andrew!

I understand that you only relied on the laws of motion (definition of velocity, and such), Newton's third law, and also Newton's second law which defines *Force*. Then you asked what would the work be for a moving object at velocity *v*, which gave you the known \frac{mv^2}{2}.

You also assumed a constant Force, or constant deceleration, but then again one can take this to a small enough change and calculate the work for small intervals when the deceleration can be assumed to be constant...
So the only remaining question is: Why can we assume that motion under small enough time intervals is defined by a *constant acceleration*??

Why not constant velocity, or constant "derivative by time of the acceleration"?

 

[BruceW]

more generally, the force will not be constant. But this is a nice specific example. Also, a constant force is a particular example of a conservative force. This is the most important thing I think (that the force is conservative). If the force is not conservative, then the kinetic energy which is lost, will not be stored as potential energy.

 

[voko]

As was mentioned above, our laws of motion are based on experimental facts. That applies to the Lagrangian form, and that applies to the Newtonian form. In fact, these two forms are equivalent and one can be derived from another. So no matter how you slice it, you will always end up with conservation of energy, because that is also an experimental fact.

If you find the Newtonian picture easiest to comprehend and explain to students, then here is how it attains conservation of energy. Newton's 2 law: $$ m {\mathrm d^2 \vec r \over \mathrm d^2 t} = \vec F .$$ The force is not assumed to be constant; the equation is vectorial. Using the scalar (dot) product, multiply that with velocity: $$ {\mathrm d \vec r \over \mathrm d t} \cdot m {\mathrm d^2 \vec r \over \mathrm d^2 t} = {\mathrm d \vec r \over \mathrm d t} \cdot \vec F.$$ This is the same as $$ m {\mathrm d \over \mathrm d t} \left( \frac 1 2 {\mathrm d \vec r \over \mathrm d t} \cdot {\mathrm d \vec r \over \mathrm d t} \right) = {\mathrm d \vec r \over \mathrm d t} \cdot \vec F.$$ This in turn is $$ {\mathrm d \over \mathrm d t} \left( {mv^2 \over 2 } \right) = \vec v \cdot \vec F .$$ If we integrate that along the trajectory of motion, we end up with $$ {mv_1^2 \over 2 } - {mv_0^2 \over 2 } = \int \mathrm d \vec r \cdot \vec F .$$ The left hand side is the change in kinetic energy; the right hand side the work of force.

A force is called potential if its work over any path depends only on the endpoints of the path. Then its work over any path is $$ \int \limits _{\vec r_0}^{\vec r_1} \mathrm d \vec r \cdot \vec F = V(\vec r_1) - V(\vec r_0) .$$ Denoting further $U = -V$, the final result is $$ {mv_1^2 \over 2 } + U(\vec r_1) = {mv_0^2 \over 2 } + U(\vec r_0) .$$ $U$ is known as potential energy, and all together this is conservation of total mechanical energy.

 

[BruceW]

right. nice derivation. the key step (in my opinion) is the assertion that the force is conservative. Without this, we just have the statement "change in kinetic energy is equal to work done on particle". But at this step, "work done" has no special meaning. It's only when we state that the force is conservative, that we can equate "work done" to "loss of potential energy". And therefore, we get KE+PE = constant.

If the force is not conservative, then at best, you can get "d/dt(KE+PE) = power flow into the system". Which I suppose is still useful, for example, you could calculate the rate at which energy is lost as heat, or such.

 

[voko]

That is true. However, in the setting of the original question, which deals with the unit "Joule" and the constancy of something it is applied to, something, namely KE - Work, is conserved anyway.

 

[Dadface]

I think some posts here have strayed from the original question which referred to elastic collisions, ones where kinetic energy is conserved. Elastic collisions do not happen with macroscopic objects but only with microscopic objects and only under certain conditions. The OPs request to keep the discussion at a classical level makes the question difficult to answer without the use of simplifying assumptions,some of which are difficult to justify. For example during the impact time of two macroscopic objects there will be a distortion of the objects, this resulting in a transfer of K.E.

 

[BruceW]

yeah, elastic collision in this context means to me simply that the distortion of the objects and heat and sound generated is negligible.

 

[Yoni]

I thank you all for your awesome answers! I understand it all much better now, and I'm sure my student will be very happy.

 

[Philip Wood]

I think one can show that (apart from conservation of m and mv) only conservation of mv² is possible as general law, without straying beyond collision dynamics and the principle of galilean relativity

Now we know that if momentum is conserved in one frame it is conserved in another frame moving at velocity w with respect to the first, because
if \sum{mu} = \sum{mv}
Then \sum{m(u-w)} = \sum{m(v-w)} is required in the new frame
That is \sum{mu} - w\sum{m} = \sum{mv} - w\sum{m}
and the sides DO equal each other if both momentum and mass are conserved in the FIRST frame. that is \sum{mu} = \sum{mv} and \sum{m} = \sum{m}.

Now suppose that we believe that mu² is conserved in the first frame. It's easy to show by the same method (expansion of (u-w)² and (v-w)²) that provided momentum and mass are also conserved in the first frame, then conservation of mu² also holds in the second frame, because if \sum{mu^2} = \sum{mv^2}
then in a frame moving at velocity w relative to the first then \sum{m(u-w)^2} = \sum{m(v-w)^2} is required.
So \sum{mu^2} + 2w \sum{mu} + w^2\sum m = \sum{mv^2} + 2w \sum {mv} + w^2 \sum m
which will be the case if m, mu and mu² are both conserved in the first frame, and is therefore a possible respectable law of Physics.

But surely we can go on and do the same for mv³, i.e use the same technique (expansion of (u-w)³ and (v-w)³) to show that if m, mv, mv² and mv³ are conserved in one frame, they'll be conserved in another? We can, but it's a blind alley, because conservation of mv³ can be disproved by a simple counter-example…
Suppose m₁=1, m₂=2, u₁=2, u₂=1, v₁=-2v, v₂=v. The momentum is zero before and after in this frame of reference. If we seek a value of v for which mv³ is conserved, we find v = -1 is the only real root. This gives the bodies the same velocities after the 'collision' as before; in other words they haven't collided at all. They certainly haven't exerted forces on each other.

So conservation of mv³ can't be a general law. So we can't appeal to it if we try and show by the usual technique that it or conservation of mvⁿ for n>2 is consistent with the relativity principle; indeed we know that it can't be in the general case!

 

[BruceW]

I think you can go on and do the same for mv³ and all the higher powers. In the counter-example, you also find that the magnitude of v must be 1, just due to conservation of kinetic energy.

 

[Philip Wood]

Yes. (I think it's 'yes'!) Because conservation of mv³ can't hold (you can show this for any mass ratio), then conservation of mvⁿ, although it could hold in the centre-of-mass frame for even powers of n, can't also hold in other frames, for n ≥ 3, as shown by the argument based on expansion of (v - w)ⁿ.

 

[BruceW]

no, I mean we can go on and do the same for mv³, i.e use the same technique (expansion of (u-w)³ and (v-w)³) to show that if m, mv, mv² and mv³ are conserved in one frame, they'll be conserved in another. Your counter-example is not a counter-example.

 

[Philip Wood]

Your counter-example is not a counter-example.

That's a very blunt statement! Haven't I shown that mvⁿ isn't conserved in the centre of mass (zero total momentum) frame?

 

[BruceW]

no, because of what I said in post #19. conservation of momentum and conservation of kinetic energy means that the speed is determined. this is not something special for mvⁿ with n>2.

 

[Philip Wood]

The speed is indeed determined. But there are no sensible final velocities if one demands that mv³ is conserved (as well as momentum and mass). In other words, it can't be conserved (if momentum and mass are conserved).

 

[voko]

I think one can show that (apart from conservation of m and mv) only conservation of mv² is possible as general law, without straying beyond collision dynamics and the principle of galilean relativity

Just for record, Huygens derived conservation of energy in elastic collisions using solely Galilean relativity and some basic hypotheses such as that two identical rigid bodies moving at equal speeds move at the same speed after the collision.

But I have trouble following your argument. Some of that stems from non knowing what your symbols $v, u, w$ denote and how the are related.

Now we know that if momentum is conserved in one frame it is conserved in another frame moving at velocity w with respect to the first, because
if \sum{mu} = \sum{mv}

Assuming $u$ and $v$ are velocities in two different frames, this is not correct. For example, two masses at rest in one frame: total momentum in that frame is zero, in any other frame it is not. But perhaps I totally misinterpret what you are saying. Please clarify.

 

[BruceW]

The speed is indeed determined. But there are no sensible final velocities if one demands that mv³ is conserved (as well as momentum and mass). In other words, it can't be conserved (if momentum and mass are conserved).

why not? in your counter-example, it seemed like you did find a sensible final velocity where mv³ was also conserved.

edit: I'm just playing devil's advocate, I'm not really sure about the answer. But since you seem sure, I'm trying to find out.

 

[Philip Wood]

voko. Profuse apologies. The notation is a bit naff. Sigma mu is the sum of initial momenta, Sigma mv is the final momenta. w is the velocity of the new frame relative to the old.

Interesting about Huygens. Underrated. His collision dynamics was clearly a lot better than that of Descartes!

 

[Philip Wood]

why not? in your counter-example, it seemed like you did find a sensible final velocity where mv³ was also conserved.

edit: I'm just playing devil's advocate, I'm not really sure about the answer. But since you seem sure, I'm trying to find out.

Hello Bruce. The answer wasn't sensible: the only final velocities that fitted were the bodies' initial velocities: in other words the bodies hadn't collided at all!

 

[BruceW]

you will get the same result as you got, if you use just conservation of kinetic energy. It is conservation of kinetic energy which means that the final speeds must be the same as the initial speeds. It has nothing to do with mv³.

 

[Philip Wood]

My stuff about mv³ was in response to the original question in this thread. Have you read it?

 

[voko]

Interesting about Huygens. Underrated. His collision dynamics was clearly a lot better than that of Descartes!

Yes, he is clearly underrated. It was a very clever examination of a collision from two frames moving with respect to each other; very modern. I think his proof of conservation of energy was the first time ever proof of that. His other works are also first rate. It is a bit of tragedy that Newton wrote such an excellent work on dynamics that it overshadowed everything that existed just before it. It should, however, be noted that Huygens was among the very few names mentioned in Newton's work, so Newton clearly realized the Huygens's contribution to science.

Regrading your proof, one problem with your approach is that you treat velocities as scalars. That is OK when you form first and seconds order invariants because that can be trivially re-written in a vectorial form. But as you go to third and higher degrees, there is no standard interpretation of such invariants. This actually might be a fundamental reason why such invariants are not known - they simply cannot be formulated in a physically meaningful way.

 

[Philip Wood]

An unusually generous acknowledgment by Newton?

I'm treating a head-on, straight line collision, and my velocities are actually velocity components (dot products with the unit vector in one direction along the straight line). As such they can be raised to any power, but I do see the point of your remark. It's all very well to cube a velocity component, but cubing the velocity itself isn't a defined operation. I think this may well go to the heart of the matter.

 

[ChrisVer]

One more problem the OP should solve for him/her self is that in classical mechanics, the Lagrangian and its formalism is equivalent to the Newton's law... It's just another way of seeing it (since by Hamilton's principle you'll get the Newton's laws), which sometimes can be more simple or more complicated-depending on the problem.

So "why or what is the Lagrangian?" and feeling the explanations coming from it as "weird", while on the other hand accepts the Newtonian approach, shows exactly that.

So in this case, I'd refer you into studying Lagrangian from classical mechanics books, and see how they deal with it, and how the Newton's laws come naturally out of it.

 

[rkmurtyp]

voko. Profuse apologies. The notation is a bit naff. Sigma mu is the sum of initial momenta, Sigma mv is the final momenta. w is the velocity of the new frame relative to the old.

Interesting about Huygens. Underrated. His collision dynamics was clearly a lot better than that of Descartes!

I read about Huygens analysis of elastic collisions. I understand that Descartes analysed the collision process using conservation of 'quantity of motion' (product of mass and speed - a scalar quantity), and Huygens corrected it by changing conservation of quantity of motion to conservation of moementum. Some say that Huygens solution applies to one dimensional collisions only.

In the light of the above would you please let me know:

1. Where we go wrong by using Descartes analysis instead of Huygens analysis, with a numerical example?
2. Does Huygens solution not apply to multidimentional elastic collisions?

 

[voko]

Huygens used "quantity of motion" as well. He, however, proved that the quantity of motion can change in a collision, while Descartes posited it would be constant. He also stated, although I am not sure whether a proof was ever published, that momentum is conserved, even though he did not use the term "momentum"; specifically he stated that the common centre of gravity of two or more bodies moves uniformly in a straight lines before and after collision, and he stated that he could prove that for spherical bodies, for elastic and inelastic collisions, for head-on and oblique collisions.

For a particular example, say there is a stationary 2 kg ball, and a 1k ball moving toward the first one at 2 m/s. The total quantity of motion is 2 kg.m/s before collision. According to Descartes, it should stay that way after the collision.

Huygens described a method of determining the speeds after the collision: $$ {2 \ \text{kg} + 1 \ \text{kg} \over 2 \cdot 1 \ \text{kg} } = {2 \ \text{m/s} \over v} $$ so v = 4/3 m/s, where v is the speed the 2 kg ball after the collision; the speed of the 1 kg ball is such that the relative speeds before and after the collision be equal, so it is 2/3 m/s. The resultant speeds are consistent with conservation of energy and momentum.

The total quantity of motion after collision is 10/3 kg.m/s, which is greater than the initial quantity 2 kg.m/s.

A remark on terminology. Some languages now use "quantity of motion" in sense of "momentum". This message uses the historic definition, where "speed" is strictly the magnitude of velocity and is always positive.