# Quadratic forms and kinetic energy

1. Aug 9, 2015

### exponent137

I heard that proportionality of kinetic energy with square of velocity, $E_k\propto v^2$, can be derived with help of quadratic forms.

It goes like: we guess that $E_k\propto v^2$ and we assume that momentum $p\propto v$, then equation is valid in another inertial system. And so on. The we try with $E_k\propto v^3$ and it does not work on this way.(I will not write all equations, for now.)

I think that I understand this principle, but I wish some references, where this is explained. And, I wish a more general explanation, maybe my explanation is too specific.

2. Aug 14, 2015

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Aug 15, 2015

### exponent137

I will write more, that it will be clear what I thing with quadratic forms: We have two balls which collide. Let us respect conservation of momentum:
$m_1(v_1'-v_1)=-m_2(v_2'-v_2)$
$v_{1,2}$ are velocities before collision and $v'_{1,2}$ are velocities after collision. $m_{1,2}$ are masses of both balls. I guess conservation of energy at collision, where part of energy may change in internal energy $\Delta W$.
$1/2m_1(v_1^2-v_1'^2)+1/2m_2(v_2^2-v_2'^2)=\Delta W$
The first equation is put into the second one, what gives:
$1/2m_1(v_1-v_1')(v_1+v_1'-v_2-v_2')=\Delta W$
If $\Delta W=0$ The values 0 in the first (...) means that collision had not happened, and the values 0 in the second (...) is essential for description of our collision.

Internal energy should be the same, if we are in another inertial system, which moves with velocity $u$, thus to every velocity it is added velocity $u$:
$1/2m_1(v_1+u-v_1'-u)(v_1+u+v_1'+u-v_2-u-v_2'-u)=\Delta W$
It can be evident, that all velocities $u$ are anihilate, thus value for $\Delta W$ is the same as before.

Let us try still with energy with the power of 3:
$1/3m_1(v_1^3-v_1'^3)+1/3m_2(v_2^3-v_2'^3)=\Delta W$
It is obtained:
$1/3m_1(v_1-v_1')(v_1^2+v_1 v_1'+v_1'^2-v_2^2-v_2 v_2'-v_2'^2)=\Delta W$
If now $u$ has been added to every velocity, the value $\Delta W$ is not the same. The value is also not the same, if we use $v^4$ instead of $v^2$, and so on.

This is one form of proof of $W_{kin}\propto v^2$ with quadratic form. I am interested for more generalized proof of this and for references and links for quadratic forms calculations?

4. Aug 15, 2015

### Staff: Mentor

It looks like you are using energy conservation to prove energy is conserved. That result should not be surprising.

5. Aug 15, 2015

### brainpushups

The proportionality constant of 1/2 comes from the choice of units.

6. Aug 15, 2015

### exponent137

??
Maybe you want to say, that $E_{kin}\propto v^2$ comes from $E=F dx=v dv$?
But $E=Fdx$ come empirically from observations, it is not derived from the first principles.

7. Aug 25, 2015

### robphy

8. Aug 27, 2015

### exponent137

9. Aug 28, 2015

### robphy

The link to "Tensor Analysis on Manifolds, p. 265" (the 4th link) is relevant.
While the metric is "a thing of SR", a metric is found in many places.
Given a vector space, there is no natural notion of dot-product. One has to be introduced as extra structure, usually via a metric [although one might not refer to it explicitly].

A variation of the above search is

The search is inspired by this statement:
"Kinetic energy is a Riemann Metric on Configuration Space"
- Geometrical Mechanics, Lectures by Saunders MacLane, Dept. of Math, U. Chicago, 1968.
http://www.worldcat.org/title/geome...-of-chicago-winter-quarter-1968/oclc/63616841

10. Aug 29, 2015

### exponent137

Yes, it is a nice link, but I thought something more simple, a similar derivation, which I gave in post #3 in this topic.

11. Aug 29, 2015

### RMalayappan

A simple derivation of $W \propto v^2$? I always thought energy was just a convenient quantity that we defined from the following:

From kinematics we have $$v_f^2- v_i^2 =2a \Delta x$$
Multiplying through by $\frac{1}{2}m$ gives $$\frac{1}{2}m(v_f^2-v_i^2) = \Delta (\frac{1}{2}mv^2) = ma \Delta x = F \Delta x$$
For an object being pushed by a variable force along a path in 1D, we can say that $$\Sigma \Delta (\frac{1}{2}mv^2) = \Sigma F_i \Delta x$$
Taking the limits as $\Delta x \rightarrow 0$, $$\int_{x_i}^{x_f} d(\frac{1}{2}mv^2) = \frac{1}{2}m(v_f^2-v_i^2) = \int_{x_i}^{x_f} Fdx$$
so we define the left quantity to be the change in the kinetic energy and the right integral to be work. This becomes $\int_C \vec{F} \cdot \hat{T}ds$ in higher dimensions because the only part of the force that contributes to a change in velocity is the force that is projected along the path.

12. Aug 30, 2015

### exponent137

Kinematics only gives the relation between force and energy. At this we should assume that energy is proportional with Fdx, what is an empirial fact, because of this conservation of energy followed. But, kinetic energy conservation should be derived without this empirical fact, only with a similar conservation as momentum conservation.

Last edited: Aug 30, 2015
13. Aug 30, 2015

### RMalayappan

Is the statement that energy is proportional to Fdx an assumption? I just showed that the left hand quantity is proportional to Fdx, and that is what I defined to be the kinetic energy.

14. Aug 30, 2015

### Staff: Mentor

You have to define "energy" somehow. You can use the integral over F dx or something else as definition, and then show everything else can be related to this quantity.

15. Aug 31, 2015

### exponent137

Your left and right sides are only relations, they do not give, that energy is conserved, neither kinetic (left) not potential energy (right).

16. Aug 31, 2015

### exponent137

More precisely, conservation of energy is important, not only energy per se.
Derivation #3 gives, that we need only law about conservation of momentum and independence of internal energy of inertial system (u). We do not need Fdx etc.