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$$\sum_{i=0}^n m_i v_i$$ as well as $$\frac{1}{2}\sum_{i=1}^n m_i v^2_i$$ Have been proved to be conserved by Newton & Leibniz, respectively. (http://en.wikipedia.org/wiki/Vis_viva) I don't understand how one would prove these conservations mathematically (is experiment necessary?)

Also does anyone know why $$\sum_{i=1}^n m_i v^2_i$$ was found to not be conserved? Some energy was being lost, obviously. Where did it go and how did people come to the realization that the loss was half of the energy?

Edit: Seeing as how the $\frac{1}{2}$ can be derived using Galileo's formulation for the study of the motion of uniformly acceleration objects, my new question is: Why is that definition of acceleration used when generalizing the kinetic energy equation to all accelerations (using integration)? Is there not a definition of acceleration with respect to position that is not based on the idea of uniform acceleration? The definition I'm talking about $$a_{avg} = \frac{1}{2}\frac{\Delta {v^2}}{\Delta x}$$ Where the $\frac{1}{2}$ comes from the kinematic equations

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    $\begingroup$ Prove them starting from what? Also, the proof that these are conserved should be found in any standard resource on the topic of conservation laws. $\endgroup$ – ACuriousMind Mar 9 '16 at 20:19
  • $\begingroup$ Doesn't prove it in my textbook. I don't know how to prove them I searched it up I don't see any resources..If you want I can head over to mathematics SE and ask them. I want a mathematic proof with explanations given for used methods. $\endgroup$ – obliv Mar 9 '16 at 20:28
  • $\begingroup$ starting from anything. I proved uniform circular motion using geometry as well as differential calculus. I want any starting point to prove that in a system of masses those quantities are conserved. $\endgroup$ – obliv Mar 9 '16 at 20:29
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    $\begingroup$ if $a$ is constant, then so is $\frac{1}{2}a$... $\endgroup$ – AccidentalFourierTransform Mar 9 '16 at 20:39
  • $\begingroup$ Okay what are the requirements for some quantity to be conserved? There needs to be none of the quantity leaving the system right? So let's say H is some quantity. if dH/dt = 0 then is H conserved? Also, this is solely with respect to time right? How am I to prove that d{mv]/dt = 0 or d{mv^2}/dt = 0? $\endgroup$ – obliv Mar 9 '16 at 20:54
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I will give the more mathematical answer scince a very good description is already given.

Conservation of Momentum

Let's assume that we have a system of N particles. By definition these particles interact with each other and according to the Newton's 3rd Law, the forces of interaction are opposite.Therefore:

$$ \vec F_{ij}=- \vec F_{ji} $$

So, if we sum the internal Forces that are acted from every particle to every other particle of the system then we get:

$$ \sum_{i \neq j} \vec F_{ij}=\frac{1}{2}\sum_{i \neq j}( \vec F_{ij}+ \vec F_{ji})=\frac{1}{2}\sum_{i \neq j}( \vec F_{ij}- \vec F_{ij})=\vec 0 $$

From Newton's Second Law we know that:

$$ \sum \vec F_{total}=\sum \vec F_{external} + \sum \vec F_{internal}=\sum \vec F_{external}=\frac{d \sum \vec p}{dt}$$

Now if the system is isolated then $\sum \vec F_{external}=\vec 0$ so

$$\frac{d \sum \vec p}{dt}=\vec0 \Rightarrow \sum p =\sum_i m_iv_i = constant $$

As Benjamin pointed out the conservation of kinetic energy is not generally conserved. It's more complicated than that. A more fundamental theorem is the conservation of mechanical energy of a system, but only when all the forces that are acted upon it are conservative.

Potential Energy

When particles interact with conservative forces(interactions) meaning that:

$$ \oint \vec F \cdot d \vec r = 0 $$

then the field that describes the interaction can be dercribed by potential U. Because of this potential any particle/body with the right properties that enters this field gains potential energy.

We define potential energy as:

$$ V_{initial} - V_{final} = \int_{A}^{B} \vec F \cdot d \vec r $$

$$ dV = -dW $$

where A is the initial position and B is the final position.

As we can see, potential can only calculate differences between the initial and the final state. Therefore we can give any value to the initial state or the final. We chose to give the value 0 to the initial state and what we now get is:

$$ V_{final}= - \int_{A}^{B} \vec F \cdot d \vec r $$

Conservation of Mechanical Energy

Now lets assume that we have a system of particles and they interact with forces that we dont know if they are conservative or not. Then the work that all the internal forces produce is:

$$ W_{total} = W_{interior} + W_{exterior}$$ $$ \int [ \sum \vec F_{in-conserv} + \sum \vec F_{in-non-conserv} + \sum \vec F_{ext-conserv} + \sum \vec F_{ext-non-conserv}]\cdot d \vec r $$

$$ \int \sum \vec F_{in-conserv} \cdot d \vec r + \int \sum \vec F_{in-non-conserv} \cdot d \vec r +\int \sum \vec F_{ext-conserv} \cdot d \vec r + \int \sum \vec F_{ext-non-cons} \cdot d \vec r $$

$$ W_{total} = W_{in-conserv} + W_{ext-conserv} + W_{in-non-conserv} + W_{ext-non-conserv} $$

$$ K_{int-f} - K_{int-in} + K_{ext-f} - K_{ext-in} = V_{int-in} - V_{int-f} + V_{ext-in} + V_{ext-f} +W_{int-non-cons}+W_{ext-non-cons} $$

If all the forces are conservative then:

$$ K_{int-f} + V_{int-f} + K_{ext-f} + V_{ext-f} = K_{int-in} + V_{int-in} + K_{ext-in} + V_{ext-in} $$

We define Mechanical energy as $E_M = K+V$ so:

$$ E_{M-ext-in} + E_{M-int-in} = E_{M-ext-f} + E_{M-int-f} $$

$$ E_{M-total-in} = E_{M-total-f} $$

So if all the forces that are acting on a system are conservative the total mechanical energy is conserved. Otherwise it's not.

Conservation of Kinetic Energy

Now, when the potential energy during a phenomenon does not change then we can talk about conservation of kinetic energy.

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  • $\begingroup$ If you want me to show you the last part, tell me to edit. $\endgroup$ – George Smyridis Mar 9 '16 at 23:47
  • $\begingroup$ Thanks George for being more mathematically involved. I tried to avoid it. I usually try helping individuals by statements. But, some people appreciate the formalism more. Good Job, $\endgroup$ – Benjamin Mar 10 '16 at 6:02
  • $\begingroup$ If you could explain how one would go about conserving kinetic energy i'll accept your answer. Thanks for the mathematical form of benjamin's answer though, it helps. $\endgroup$ – obliv Mar 11 '16 at 15:20
  • $\begingroup$ Made an edit. I hope it helps. $\endgroup$ – George Smyridis Mar 18 '16 at 17:00
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For the first summation (which corresponds to the Conservation of Total Linear Momentum) the condition to be met is that the "system" is expected to be isolated meaning that there should be no external forces acting from the rest of "Universe" onto the "system". This can be achieved from starting with the Newton's Second Law expressed in the form of:

The total net force acting on a system is equal to the rate of change of the total linear momentum of the system.

Where the total net force is expressed as a summation of internal and external forces. Since we assumed the system is isolated, the external forces would add up to zero. And, by Newton's Third Law you can show that the internal forces would also add up to zero. (because internal forces would add up to zero mutually according to Newton's Third Law). Hence, as you can see, the left-hand side which is the total net force would be identically zero. This means that: The rate of change of Total Linear Momentum is zero which translates into Conservation of Total Linear Momentum.

The same method of approach also holds for the second summation (which corresponds to the Conservation of Total Kinetic Energy). However, this is more restrictive and tricky to prove because it requires more assumptions to be made about the "system;" it is usually not conserved for a "system" just because there is loss of kinetic energy into other forms of energies such as heat, sound and...This proof will be different than Conservation of Total Energy of the System which is more fundamental in Nature where ALL sorts of energies involved in the "system," are taken care of in the same manner as the first proof.

I hope this will give you a better picture of approaching the proofs. Thanks,

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    $\begingroup$ That makes sense intuitively for the first one, thanks. I just realized how Leibniz came to think of the second conservation. He basically used free-fall as an example. nature.berkeley.edu/departments/espm/env-hist/articles/2.pdf (page 5 of pdf) although it was not a real proof it explained his intuition and how he came to the observation. $\endgroup$ – obliv Mar 9 '16 at 21:35

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