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Why is kinetic energy conserved?

  1. Jan 18, 2014 #1
    For instance in an elastic collision of two indivisible particles...

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

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

    Please keep the discussion in classical terms...
  2. jcsd
  3. Jan 18, 2014 #2


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    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.
  4. Jan 18, 2014 #3


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    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.
  5. Jan 19, 2014 #4


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    That's the definition of 'elastic'.

    Momentum (mv, at its simplest) is also conserved - elastic or not.
  6. Jan 19, 2014 #5
    Interesting... But WHY? How would the world would look if a different physical scalar was conserved?? (for example [itex]m v^{3}[/itex]?)

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

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

    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 [itex]m v^{3}[/itex] 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...
  7. Jan 19, 2014 #6


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    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.

    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.

    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.
  8. Jan 19, 2014 #7


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    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.
  9. Jan 19, 2014 #8
    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
  10. Jan 19, 2014 #9

    Andrew Mason

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    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:

    [itex]\Delta t = mv/F[/itex]

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

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

    [itex]\Delta s = v\Delta t = \frac{mv^2}{2F}[/itex] so: [itex]F\Delta s = \frac{mv^2}{2}[/itex]

    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.

  11. Jan 20, 2014 #10
    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 [itex]\frac{mv^2}{2}[/itex].

    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"?
  12. Jan 20, 2014 #11


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    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.
  13. Jan 20, 2014 #12
    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.
  14. Jan 20, 2014 #13


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    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.
  15. Jan 20, 2014 #14
    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.
  16. Jan 20, 2014 #15
    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.
  17. Jan 20, 2014 #16


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    yeah, elastic collision in this context means to me simply that the distortion of the objects and heat and sound generated is negligible.
  18. Jan 20, 2014 #17
    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.
  19. Jan 20, 2014 #18

    Philip Wood

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    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 [itex]\sum{mu} = \sum{mv}[/itex]
    Then [itex]\sum{m(u-w)} = \sum{m(v-w)}[/itex] is required in the new frame
    That is [itex]\sum{mu} - w\sum{m} = \sum{mv} - w\sum{m}[/itex]
    and the sides DO equal each other if both momentum and mass are conserved in the FIRST frame. that is [itex]\sum{mu} = \sum{mv}[/itex] and [itex]\sum{m} = \sum{m}[/itex].

    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 [itex]\sum{mu^2} = \sum{mv^2}[/itex]
    then in a frame moving at velocity w relative to the first then [itex]\sum{m(u-w)^2} = \sum{m(v-w)^2}[/itex] is required.
    So [itex]\sum{mu^2} + 2w \sum{mu} + w^2\sum m = \sum{mv^2} + 2w \sum {mv} + w^2 \sum m[/itex]
    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!
    Last edited: Jan 20, 2014
  20. Jan 20, 2014 #19


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    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.
  21. Jan 21, 2014 #20

    Philip Wood

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    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.
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