Conservation of mechanical energy

[Mr. Peterman]

Hi. I've been questioning the conservation of 1/2 mv^2 because I've heard some decent arguments about it. Anyway I'm not sure if it is true, but I do realize that it is equivalent to the conservation of work. So I would like to have a rigorous proof of the conservation of work. I've been looking at the two body problem in one dimension for help, but in my textbook a perfectly elastic collision is described by the conservation of momentum equation and the conservation of kinetic energy equation. So I'm trying to prove the conservation of kinetic energy without invoking the conservation of kinetic energy so this approach won't work. I know that a near perfectly elastic bouncing ball and earth system will satisfy the conservation of work idea but I consider this a special case with a small mass and an infinitely larger mass, so I would like to prove the conservation of work or lack thereof in a case where the two masses are comparable and unequal without first invoking the very thing I am trying to prove or disprove. Any ideas?

[Mr. Peterman]

I think I got this one figured out.

[UncleAl]

1) Time is homogeneous.

2) Noether's theorems.

3) Local mass-energy is conserved. Inescapable.

 

(mv^2)/2 alone is obviously NOT conserved. Drop an initially stationary baseball. Kinetic energy = zero at the start. Circumstances matter. The relativistic four-vector of momentum requires local conservation of energy. General Relativity has no global conservation laws, but there are no proposals for local violation stratetgies or tactics.

 

If you collide two objects you must do bookkeeping for both linear and angular terms. Conservation of angular momentum arises from vacuum isotropy and Noether's theorem. Vacuum isotropy in the massed sector could be testably broken, the naughty link in the signature below. In the massless sector things are observationally tight,

 

[0706.2031] Relativity tests by complementary rotating Michelson-Morley experiments

[0801.0287] Data Tables for Lorentz and CPT Violation

 

--

Uncle Al

UNDER SATAN'S LEFT FOOT

[CraigD]

Sorry to have missed this thread for so long. It brings up a pretty cool question.

Hi. I've been questioning the conservation of 1/2 mv^2 because I've heard some decent arguments about it. Anyway I'm not sure if it is true, but I do realize that it is equivalent to the conservation of work.

The law of conservation of energy (recall that, in physics, a law is simply an description of something that appears true, without the systematic explanation of why that would allow it to be part of a theory) simply states that the total energy of a system is constant.

 

In a system with no forces other than instantaneous elastic collisions (which could also be called “instantaneous changes in velocity in which energy and momentum are conserved”), conservation of energy would be equivalent to conservation of kinetic energy (your “conservation of [math]\frac12 mv^2[/math]”). For an n-body system, it would be fully described by

[math]2k_E = \sum_{i=1}^n m_i v_i^2[/math], where [math]k_E[/math] is a scalar constant.

 

In real-world situations, even involving only classical mechanics, there’re non-instantaneous forces, many at great distances (gravity and magnetism), which introduce potential energy terms to the equation, complicating the description of the systems constant energy [math]k_E[/math]. In these cases, conservation of kinetic energy is no longer equivalent to the law of conservation of energy, and isn’t even approximately true.

I've been looking at the two body problem in one dimension for help, but in my textbook a perfectly elastic collision is described by the conservation of momentum equation and the conservation of kinetic energy equation. So I'm trying to prove the conservation of kinetic energy without invoking the conservation of kinetic energy so this approach won't work.

I understand your concern. Convervation of energy is a physical law, so attempting to prove it with a mechanical formalism that includes it as a law is circular reasoning, which as you say, doesn’t work.

 

It’s worth noting that you need both conservation of energy and conservation of momentum –

[math]k_P = \sum_{i=1}^n m_i v_i[/math], where [math]k_P[/math] is a vector constant

– to describe motion is a system with instantaneous elastic collisions

 

To the best of my knowledge, the key to describing a mechanical system in which conservation of energy can be derived from simpler principles is to eliminate instantaneous elastic collisions. If you describe attractive and/or repulsive forces between bodies, such as by giving each body positive electric charge proportional to its mass and/or gravity, both of which follow a [math]F=\frac{Km_1m_2}{r^2}[/math] “inverse distance squared” force law, changes in velocity are gradule, and you don’t need to assume conservation of energy to derive conservation of energy or momentum. You need only assume its force laws, and the 3 laws of motion.

So I would like to have a rigorous proof of the conservation of work.

Unfortunately, even for the simplest non-trivial system, a 2-body one, a rigorous calculus derivation isn’t easy, and is beyond my ability to quickly do. However, if you have a computer and some skill at writing or finding gravity/magnetic simulators, you can observer conservation of energy and momentum to arbitrary precision empirically arising from its force formulae.

 

In short if you implement a simulation of a system in which bodies attract and/or repel one another according to inverse distance square force laws (or any distance-determined force laws, I’m pretty sure), conservation of energy and momentum arise from it “like magic”. :)

[modest]

Hi. I've been questioning the conservation of 1/2 mv^2 because I've heard some decent arguments about it. Anyway I'm not sure if it is true, but I do realize that it is equivalent to the conservation of work.

 

Yes. I think you are saying that the change in kinetic energy (1/2mv²) is equal to work. When dealing with an elastic collision or a conservative force that is true. Work is the force acting on an object times the distance over which the object moves which is equal to the change in kinetic energy. Mathematically:

[math]F \cdot D = W = \Delta(1/2mv^2)[/math]

The symbol [math]\Delta[/math] means "change in" and 1/2mv² is kinetic energy. Another way of writing change in kinetic energy would be:

[math]\Delta KE = 1/2m{v_2}^2 - 1/2m{v_1}^2[/math]

where v₂ is the final velocity and v₁ is the starting velocity.

I've been looking at the two body problem in one dimension for help, but in my textbook a perfectly elastic collision is described by the conservation of momentum equation and the conservation of kinetic energy equation. So I'm trying to prove the conservation of kinetic energy without invoking the conservation of kinetic energy so this approach won't work. I know that a near perfectly elastic bouncing ball and earth system will satisfy the conservation of work idea but I consider this a special case with a small mass and an infinitely larger mass

 

I think the bouncing ball example from your textbook is probably saying that the work done by the ball in getting to height D is equal to the kinetic energy it had at the surface. The reason they use this example is that the force of gravity acting on a ball is constant over such a small distance. You can then use the equation above.

 

So I would like to have a rigorous proof of the conservation of work... Any ideas?

 

You can start with the equation of "work" and derive the expression for "change in kinetic energy". I'm not sure if that's exactly what you're looking for, but I will demonstrate.

 

Starting with the equation for work (work equals force times distance):

[math]W = F \cdot D[/math]

F is "force". According to Newton's second law of motion we know that force is equal to mass times acceleration (F = m • a). This means we can put "m • a" in the work equation instead of F.

[math]W = (m \cdot a) \cdot D[/math]

We can make another substitution for acceleration. Acceleration can be expressed in terms of distance and velocity by the kinematic equation:

[math]a = \frac{{v_2}^2 - {v_1}^2}{2D}[/math]

Substituting this in the the last work equation for 'a' you get:

[math]W = m \left( \frac{{v_2}^2 - {v_1}^2}{2D} \right) \cdot D[/math]

The two D's algebraically cancel while the M and the 2 can be algebraically distributed giving:

[math]W = 1/2m{v_2}^2 - 1/2m{v_1}^2[/math]

And that is the mathematical expression meaning work equals change in kinetic energy. Assuming Newton's second law is true and applicable to the situation, force times distance is the same as the difference in kinetic energy.

 

~modest