# Independence of Energy and Momentum Conservation

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1. Mar 21, 2015

### nayanm

Suppose we take the three Newton’s Laws as axioms.
1. Existence of inertial reference frames
2. F = ma
3. F(A on B) = -F(B on A)
Also suppose also we are considering purely classical mechanical processes on point particles (no heat transfer, etc.).

It is clear to me that the conservation of momentum directly follows from Newton's Laws. (If a force is the time rate of momentum transfer and forces come in pairs, then the momentum increase caused by one force is balanced by the momentum decrease caused by its reaction force.)

But here are my questions:
• If I take Newton's Laws as axioms, is the conservation of energy a derivable theorem or an independent axiom? In other words, does the conservation of energy directly follow from Newton's Laws just like the conservation of momentum does?
• Are equations written for the conservation of energy and conservation of momentum independent? (I.E. are they merely two representations of the same thing, or does one give additional information over the other?)
From what little I know about more advanced physics, I have a feeling that the two are independent (since energy and linear momentum conservations are consequences of spatial and time invariance). I was hoping someone could give me more insight on the matter.

Thanks!

2. Mar 21, 2015

### Abtinnn

My knowledge is limited, but from what I think I know, the concept of work can be derived from newton's laws. It can then be said that when a work is done on a system, the energy of the system changes.

Wapp= ΔE

The conservation of energy is very specific. In the case of an object falling on the surface of Earth, the work done on the object changes its kinetic energy:

W= -ΔU = ΔK

Where U is the potential energy (gravity is a conservative force) and K is the kinetic energy. Therefore

ΔEsystem = ΔU + ΔK = 0

(mgh2 - mgh1) + (½mv22 - ½mv12) = 0

∴ mgh1 + ½mv12 = mgh2 + ½mv22

The conservation of energy for the falling object can then be derived from there.

The thing with momentum is that it is a much more "fundamental" concept than conservation of energy. Momentum can be applied in many broader situations. Conservation of energy can be applied to very specific ones. It's just that for a certain problem, you can solve it in an easier way by, say, using the concept of momentum, and another problem can be solved faster when you use conservation of energy.

3. Mar 21, 2015

### A.T.

Yes.

4. Mar 21, 2015

### nayanm

Thanks for the replies!

Well, if both are a direct consequence of Newton's Laws, then both should be applicable at least whenever Newton's Laws are. I don't see how, in the consideration of only classical "slow-moving" point particles undergoing solely mechanical interactions, one can be more fundamental than the other.

Then, assuming the invariance of both space and time, why can we derive one from the other? As in here: http://www.cs.utep.edu/vladik/2013/tr13-53.pdf

5. Mar 21, 2015

### Abtinnn

Let's take the case of the inelastic collision of two balls.
Ball one is moving to the right and ball two is moving to the left. After the collision, ball one is moving to the left with a different velocity and ball two is moving to the right with yet another velocity.

Now it is wrong to say that the conservation of energy does not apply here, because the system is closed and no external work is being done on it (also energy cannot be created or destroyed). HOWEVER, in the collision, some of the energy transforms into other forms of energy (such as heat, sound and sometimes light). The difficulty with these types of energy is that they are hard to calculate. Sure it is possible, but it is most likely a long and inefficient thing to do. So energy is conserved and conservation of energy CAN be applied, but it just is not worth it when there is a better way. And that better way is the concept of conservation of momentum, which is kinda the broader and bigger picture.

6. Mar 21, 2015

### nayanm

Everything you said is right, of course. Sure it may not be "worth it" to consider energy conservation in an elastic collision, but that is not a purely mechanical process (energy is dissipated as heat).

I'm really asking this as an exercise in mathematical physics than in application. You are pre-supposing the conservation of energy by saying energy cannot be created nor destroyed. True. Also, momentum cannot be created nor destroyed. Also true. But my question is: are these independent axioms or consequences of each other.

I have a feeling that the answer involves more advanced physics concepts such as Noether's Theorem and the symmetry of space and time, but I do not know enough to address this.

7. Mar 22, 2015

### anorlunda

Your feelings are correct. If classical physics interests you, then you may find study of Noether's theorum very rewarding, and perhaps not as diffcult as you imagine. An excellent way to do that is through the series of video lectures by Professor Leonard Susskind, available on youtube and itunesU.

links to lecture 1 of 10.

8. Mar 22, 2015

### A.T.

Yes you can derive both conservation laws from more general assumptions. But that doesn't mean that one conservation law implies the other.

9. Mar 22, 2015

### stevendaryl

Staff Emeritus
Newton's equations of motion don't imply conservation of energy by themselves. You can see that by considering dissipative forces (such as friction or viscous drag). Of course, in those cases, the loss of energy due to dissipative forces all goes into heat energy, but you can't prove that from Newton's laws alone.

10. Mar 22, 2015

### andresB

Stevendaryl is right. You can have a mechanical system where the mechanical energy is not conserved. The same happen for the momentum (and angular momentum) and the electromagnetic interaction , the momentum will not be conserved if you only take into consideration the newton equations for the charges.

11. Mar 23, 2015

### nayanm

Thanks for the suggestion! I will definitely take a look at the videos when I get a little bit of time. Is there perhaps a "short answer" to my momentum and energy independence question, or does it maybe depend upon the system under consideration?

Yes, but in the traditional derivation of the work-energy theorem, I don't see any "more general assumptions" other than mass is constant.

∫(F)dr = ∫(ma)dr = m∫(dv/dt)dr = m∫(dv/dr)(dr/dt)dr = m∫v dv = 1/2*m*Δ(v^2)

It is merely plugging in Newton's second law into the integral an rearranging using the Calculus Chain Rule and definitions from kinematics. How then can the result be independent?

That's definitely a good point. But considering the derivation above, how can it simultaneously be true that Newton's Laws hold, yet the conservation of mechanical energy (which, considering the derivation above, is nothing but Newton's Second Law combined with kinematics manipulations) does not?

12. Mar 23, 2015

### A.T.

That is not a derivation of general Energy Conservation.

13. Mar 23, 2015

### nayanm

So then can it be said that mechanical energy conservation (when it applies) is a direct consequence of Newton's Laws, whereas total energy conservation is a more fundamental and independent postulate?

14. Mar 23, 2015

### stevendaryl

Staff Emeritus
I want you to think about this situation: You throw a ball straight up into the air. It reaches a maximum height, then drops again. Obviously, the kinetic energy at the top of the motion is zero. So kinetic energy is not conserved.

Of course, we say that the total energy is conserved, because we add the kinetic energy to the potential energy, and the potential energy is greatest at the top of the motion. But in order to have a "total energy" that is conserved, there must be a "potential energy". But only certain forces (conservative forces) have an associated potential energy. In particular, there is no potential energy associated with the friction force.

15. Mar 23, 2015

### stevendaryl

Staff Emeritus
That's an argument that the change in kinetic energy is due to the work done by forces. That is pretty general. But it doesn't imply conservation of energy, except in the special case where F is a conservative force.

16. Mar 23, 2015

### mac_alleb

Excuse me, could you explicitly derive momentum conservation equation (don't forget, momentum is vector value). Many thanks before.

17. Mar 23, 2015

### stevendaryl

Staff Emeritus
Well, if $\vec{P}$ is the total momentum, then, in Newtonian mechanics,

$\vec{P} = \sum_j m_j \vec{v}_j$ (where $j$ is the index over all particles)

So

$\frac{d\vec{P}}{dt} = \sum_j m_j \frac{d\vec{v}_j}{dt}$

By Newton's laws,
$m_j \frac{d\vec{v}_j}{dt} = \vec{F}_j = \sum_k \vec{F}_{jk}$

where $\vec{F}_{jk}$ is the force on particle $j$ due to particle $k$. So we have:

$\frac{d\vec{P}}{dt} = \sum_j \sum_k \vec{F}_{jk}$

At this point, we use that $\vec{F}_{jk} = - \vec{F}_{kj}$. The force on particle $j$ due to particle $k$ is equal and opposite to the force on particle $k$ due to particle $j$. So in the sum over all $j$ and $k$, we can add them in pairs $\vec{F}_{jk} + \vec{F}_{kj}$. Those pairs add up to zero. So the right-hand side is 0:

$\frac{d\vec{P}}{dt} = 0$

This proof only works for point-particles exerting instantaneous forces on each other, with no external forces (that is, the only forces are between particles). But the proof generalizes (Noether's theorem) to more complicated interactions involving fields and extended bodies.

18. Mar 23, 2015

### mac_alleb

1) Strange tricking:
R = r1 + r2
dR / dt = r1/dt + r2/dt
V = v1 + v2
dV / dt = dv1/dt + dv2/dt
A = a1 + a2
Proceed?
2) How the j sum become jk sum?

19. Mar 23, 2015

### stevendaryl

Staff Emeritus
Let's consider just two particles, to make things simple. Then let $\vec{F_{12}}$ be the force on particle #1 caused by particle #2. Let $\vec{F_{21}}$ be the force on particle #2 caused by particle #1. By Newton's 3rd law,

$\vec{F_{21}} = - \vec{F_{12}}$

So we have two equations:

$\vec{F_{12}} = \frac{d \vec{P_1}}{dt}$
$\vec{F_{21}} = \frac{d \vec{P_2}}{dt}$

where $\vec{P_1}$ is the momentum of the first particle, and $\vec{P_2}$ is the momentum of the second particle.

So if the total momentum is given by:

$\vec{P} = \vec{P_1} + \vec{P_2}$, then

$\frac{d \vec{P}}{dt} = \frac{d \vec{P_1}}{dt} + \frac{d \vec{P_2}}{dt}$
$= \vec{F_{12}} + \vec{F_{21}}$
$= 0$

That generalizes to any number of particles.

20. Mar 23, 2015

### mac_alleb

Now you simply took from the very beginning that
P1 = F12 P2= F21
I.e. P1 = -P2
and then easily concluded
P1 + P2 = 0
Seems like it's NOT quite correct.