The final explanation to why kinetic energy is proportional to velocity squared

Hans de Vries

Proportionality of KE to velocity squared.....

Understanding follows from the greater picture. Fundamentally, the Energy is
dependent on the first order of the speed, like the momentum, not the
second order. The kinetic Energy in any direction x,y,z is related to the
momentum as:

[tex]KE_x = p_xc,\ \ KE_y = p_yc,\ \ KE_z = p_zc[/itex].

Where [itex]p_x[/itex] is the momentum [itex]mv_x[/itex]. To add the speeds in the x,y and z
direction you need to use Phytagoras. The total Energy is:

[tex]E \ = \ \sqrt{m_0^2c^4 + p_x^2c^2 + p_y^2c^2 + p_z^2c^2 } [/tex]

Were the first term comes from the rest mass. Now the apparent second
order effect is the result of an approximation of the addition via Phytagoras.
if [itex]\epsilon[/itex] is sufficiently small then:


[tex]\sqrt{1 + \epsilon^2 }\ \approx \ 1 + \frac{1}{2}\epsilon^2 [/tex]

It's exactly this approximation which turns [itex]KE = pc[/itex] into [itex]KE = \frac{1}{2}mv^2[/itex].
As you see, this explains both the factor 1/2 as well as the apparent
second order behavior.


Regards, Hans.

Order

Thanks for all the answers! Now I am not disappointed anymore

I had so many comments to everything that has been written, so I chose to place them all in one post.


To Slinkie
Nice to hear I have a questionbuddy and thanks for the link. Your paradox should be solved using what we have found in this thread, though (in case it was not before). You just have to make sure to measure the total energy change involved and stick to one frame. learningphysics just did it above. However, you might not be convinced by this fact. So let me say something without referring to math. The thing is that the local energy change is not the same in different frames, but the total energy change is. This is why it seems so unintuitive! It is more intuitive from a higher perspective, I guess. Using momentum as energy seems like a good idea, but it leads to absurdities, as you may already know.

What we are discussing right now in this thread is different perspectives of why KE is proportional to velocity squared. By looking at it from different angles we will understand better why this is so.


To Krab
Ok, it is true you always learn something good. But let me tell you why I think Slinkie has got a good point. When I was a student I learned the math and how to apply it to different problems. But a big piece of understanding was missing. Much of my work was therefore in vain. Time was spent and I made no real progress. I never realized the power of physics. It was more of a mathgame. Of course I have myself to blame to some extent. But I hope that some day we will have better physics teachers who will be able to resolve the basic paradoxes.


To learningphysics

Yes you are right in the sense that energy change is frame independent, but this was not the actual situation I was really proposing in the quote. In your example you simply changed the initial speed of the rocket (v). The fuel also had the same speed (v). In my example I was rather proposing a case where, comparing with the rocket situation, the rocket had a speed v and the fuel had an initial speed 0. I don’t know how to technically achieve this, but according to the equations it will be much more energy consuming for the rocket to accelerate using a fuel that already has a velocity relative to the rocket (-v).

In the car example I said it was more difficult to push the ground backwards (in order to accelerate) if the ground already has a negative velocity (i.e. the car has a positive velocity relative to the ground). This was also discussed in Slinkie’s thread “KE puzzle”.

So this thing was not about changing the observer (as you basically did). If you change the observer you have the same energy change. This is also a very interesting fact discussed in this thread. But when I said “accelerate from v to 2v instead of 0 to v” then I really meant “accelerate from v to 2v instead of 0 to v whilst ‘the ground velocity’/’initial velocity of exhaust’ is 0. (The reason I did not say this more exact phrase was me being confused at the time of writing.)

PS. By the way, I understand KE much better by saying that it is basically nonlocal. One can not say that KE goes into an object, because another observer might say that KE decreased and therefore went out of the object. So all one can say is that KE increased in the total physical system and affected a few particles in that system. DS.


To mustafa
To some degree. In the mornings I do, in the evenings I don’t. I need a better understanding of work, although the next step, after you have admitted this definition is correct, is also interesting.

That was interesting. I believe Faraday is my/our hero then.


To Carl
Although I sometimes say I don’t like math, I really do. And this seems like good math. I have three questions:
1. Why is continuity not sufficient?
2. Why does n have to be an integer?
3. Why would you have to define it “something like this”?


To Claude
I almost forgot that work is equated using a vector formula. The world of physics is much richer when force can be perpendicular to displacement.

Yes, this derivation seems “fairly intuitive” to me too. It would be worthwhile thinking about it.


To Hans
Ok, this was a very impressing derivation. More pedagogical than the one you posted before. I am confused and impressed. It does not say much about the limit when m goes to zero. In this case the energy also goes to zero.

Hans de Vries

For low speeds [itex] v<< c [/itex] and thus [itex] m = m_0 [/itex] we may write

[tex]E \ = \ \sqrt{m_0^2c^4 + p_x^2c^2 + p_y^2c^2 + p_z^2c^2 } [/tex]

as:

[tex]E \ = \ m_0c \sqrt{c^2 + v_x^2 + v_y^2 + v_z^2 } [/tex]

We see that the approximation which results in [itex]KE\ =\ \frac{1}{2}mv^2[/itex] is
independent of the mass at low speeds.


For very high speeds where [itex] m >> m_0 [/itex] the formula becomes:

[tex]E \ = \ mc \sqrt{v_x^2 + v_y^2 + v_z^2 }\ = \ \frac{m_0c}{\sqrt{1-v^2/c^2}} \sqrt{v_x^2 + v_y^2 + v_z^2 }[/tex]

This can be simplified to:

[tex]E \ = \ mvc\ =\ pc[/tex]

Which is the right formula for (massless) photons relating Energy and momentum.


Regards, Hans

da_willem

So I guess now the question becomes, why is kinetic energy proportional (affine) to the velocity...?

(thanks for the insight though, I never heard this before)

Felix83

I admit I didn't read the whole thread cause I'm feeling like a lazy *** tonight, but I saw alot of math, and in physics I like to think about why things occur without using equations. Of course you need them to get exact answers in problems, but its nice to just think about the concept to see if it makes sense. Heres how I see it (I apologize if its been posted already, again - I'm feeling lazy):

Earlier it was posted that it's confusing as to why it takes more energy to accelerate a car from say 60mph to 120mph than 0 to 60mph (at the same accel rate). It takes more work/energy to apply a constant force while moving fast than it does while moving slow (and 0 when not moving, hence why a floor does no work :) ). This is similar to the torque of a cars engine. Imagine your are accelerating in first gear and the engine is at 5000rpm. Your engine produces say 300hp (for fun :) ) anytime it is at 5000rpm at WOT (disregard air density etc). Now as you shift through your gears at full throttle, each time you hit 5000rpm, the car is providing 300hp, yet as you go up in the gears, your level of acceleration decreases. It takes more energy to produce 200 ft/lbs of torque at 5000rpm than it does at 2000rpm. This is why an engine with the same amount of torque at a higher engine speed has more hp. It's pretty simple, it basically comes down to the fact that it takes more energy to apply a force/torque the faster you are moving/spinning.

da_willem

Yes, and the question of why kinetic energy is proportional to the velocity squared is equivalent to why there is a larger froce needed to accelerate something at a higher velocity...in equations...

[tex] W=\vec{F} \cdot d \vec{s} \rightarrow P= \vec{F} \cdot d \vec{v}=m\frac{d \vec{v}}{dt} \vec{v}=\frac{d(\frac{1}{2}mv^2)}{dt}[/tex]

HallsofIvy

Am I misunderstanding you? Unless you are talking about relativity, which doesn't seem to be the case, a larger force is not needed to accelerate something at a higher velocity. The same force applied to the same mass will result in the same acceleration no matter what the velocity is. The equation you gives shows that the change in kinetic energy is equal to the work done on the mass but doesn't seem to be connected with your first claim.

krab

That's a little sloppy.

[tex] W=\int\vec{F} \cdot d \vec{s},\ \ dW=\vec{F} \cdot d \vec{s}[/tex]
therefore,
[tex]P= dW/dt=\vec{F} \cdot \vec{v}=m\frac{d \vec{v}}{dt} \cdot\vec{v}=\frac{d(\frac{1}{2}mv^2)}{dt}[/tex]

reilly

The reason that KE is proportional to v squared in non-relativistic physics is purely practical, as is the idea of force and potential, mass ... That is, it is a concept that proves to be very useful to the point where it is a dominant idea -- it works. Think of it as: if we define KE as (m/2)V*V, then we get an enormous bag of useful tools, rules,and, yes, insight into mechanics. KE is just another product of human creativity and genius, much like "1" and zero. Enjoy.

Said as gently as possible, the basic theory and description of KE is not exactly what one would term advanced math -- this theory is found in both highschool and freshman physics classes.

Part of the purpose of such courses is to help introduce students to the art and science of using mathematics as a descriptive language, often done through storybook algebra problems. And, as any successful student of physics and any physics professor will tell you, based on experience: you develop intuition and understanding by doing the homework problems, more than assigned if possible. If you think V*V is tough, wait until you get to Heisenberg's matrix mechanics, or the theory of diffraction.

Note, in special relativity the energy of explosion is not an invariant, but varies from observer to observer. This is important, for example in dealing with radioactive decays of particles like neutrons, pions, and so forth.

Regards,
Reilly Atkinson

FunkHaus

Order,

Over the past 2 years, I've thought about several classical mechanics mysteries--including kinetic energy [tex]KE = \frac{1}{2}mv^2[/tex]--all the time. In class when I'm supposed to be learning other things, on the crapper, on dates...you name it. And finally, after TONS of brain straining, I feel like I may have figured it out. So here's the deal:

There are certain symmetries associated with the 3-dimensional set of real triples R^3 we call "space". All of these symmetries share the common quality that they preserve the "interval" (distance between two points assuming a Euclidean metric). When we apply these symmetries to the equations of motion, we obtain constraints on the form of this equation. Invariance with respect to spatial and uniform velocity translations beg us to use a second order differential equation as our equation of motion (a first order equation could not be invariant with respect to uniform velocity translations (boosts)):

[tex]\ddot x = f(x, \dot x, t)[/tex]

Any second order diff eq can be written in this form.
Now, applying more symmetries (time homogeneity, spatial homogeniety, etc.) we find that f must be a function of x only (actually, differences in x, like x1 - x2 for 2 different particles, etc.). So we have:

[tex]\ddot x = F(x)[/tex] <-- F is called the "Force"

Now, like many differential equations, this equation implies that there are functions of position

[tex] E(x(t)): \Re^3 \rightarrow \Re[/tex]

which are time independent (ie conserved). That is,
[tex]\frac{dE}{dt} = 0[/tex].
(Imagine we're in actually only in one dimension this whole time so I don't have to use vectors and dot products).

So let's see if we can find these conserved quantities, shall we?
Call
[tex] U = - \int F(x) dx [/tex].
Then
[tex]\frac{dU}{dT} = \frac{dU}{dx}\frac{dx}{dt} = -F\frac{dx}{dt} = -Fv[/tex]

Where we call [tex]\frac{dx}{dt}[/tex] the velocity, "v".

Now, is there a quantity, call it T, that we can add to U such that T + U = E is CONSTANT??
Yes--In fact, let's find it. We require that:
[tex]\frac{dE}{dt} = \frac{dU}{dt} + \frac{dT}{dt} = 0 [/tex]

From this condition we find:
[tex] -Fv + \frac{dT}{dt} = 0 [/tex]
[tex]\frac{dT}{dt} = Fv = av [/tex]

(where [tex]a = \frac{d^2 x}{dt^2}[/tex])

So [tex] T = \int avdt = \int vdv = \frac{1}{2} v^2 + C [/tex]

And that's kinetic energy. So now total energy E is conserved! And notice how E = T + U + C is conserved for any constant C.

Now of course, there's also another parameter besides spatial position in mechanics. It's called "mass". It's a little more tricky, but can also be formulated in terms of symmetries, and will again yield [tex] T = \frac{1}{2} m v^2 [/tex].

Hope this helps!

matthewkeating

Wk= Work done in accelerating object

Work done = Force x distance

If body a is travelling v m/s after 1 second, it is intuitive that it would be travelling with an average speed of v/2 m/s, so in one second it will travel a distance of v/2 m.

Now if body B was accelerated under the same force, it should also be intuitive that it will be travelling at twice the velocity after 2 seconds (velocity 2v). Over the 2 seconds it will have an average velocity of v m/s, and it follows that it will travel a distance of 2v m in that time. So body B has travelled 4 times the distance of body A in order to achieve twice the velocity, with the same applied force, thus the work done on body B is 4x the work done on body A. Now the work done on each body equates to the kinetic energy gained by each. Assuming no other external forces.

So.... doubleing the velocity causes the Ke to increase by a factor of four. You could show the same sort of arguement for a body travelling at 3v, 4v etc and you will arrive at the result that KE is proprtional to the square of the velocity.

Sorry if this is too non-mathematical

matthewkeating

With regard to your question about ke rockets and fuel... there is no "absolute" value of kinetic energy... we can only observe changes. W = fd is not really correct. Should be delta W = delta (Fd). Is it true that all observers, irrespective of their frame of reference would be able to measure and agree on changes in KE?

Claude Bile

I said in my previous post that the question can be reduced to why work is given by the dot product of force and displacement.

The formula for work is a definition and as such is not derived from a more fundamental law, however it can be justified using simple arguments.

The answer to this question is not complicated.

Claude.

reilly

[QUOTE=FunkHaus]Order,


There are certain symmetries associated with the 3-dimensional set of real triples R^3 we call "space". All of these symmetries share the common quality that they preserve the "interval" (distance between two points assuming a Euclidean metric). When we apply these symmetries to the equations of motion, we obtain constraints on the form of this equation.

Invariance with respect to spatial and uniform velocity translations beg us to use a second order differential equation as our equation of motion (a first order equation could not be invariant with respect to uniform velocity translations (boosts)):

[tex]\ddot x = f(x, \dot x, t)[/tex]

Any second order diff eq can be written in this form.

********************
Other than your unnecessary restriction to a second order DE, you have developed a very solid version of the standard approach to specifying Lagrangians in QFT. There, second order is preserved simply by fiat, not by proof. Thus you have demonstrated the enormous power of symmetry arguments.

Regards, Reilly Atkinson

FunkHaus

Reilly,

Thanks--your response is both encouraging and enlightening. I've been working over the summer and the past year or so to develop a very consistant and non circular approach to classical mechanics. I've been let down somewhat by books like Goldstein's "Classical Mechanics" which seem to take too much for granted. I hope to show through practically nothing but symmetry arguments such as those above that Newton's laws and other principle theorems of classical mechanics follow. I hope in doing so I will gain better intuition for problems in special and general relativity, as well.

So let me ask you this, if you don't mind--the more I can learn about this the better. You said that the fact that the equation of motion is second order is an unnecessary restriction. What do you mean by this? The only reason I said that the equation of motion (EOM) must be second and not first order is the following (I'm curious to know if you agree with the reasoning):

Say the EOM is first order. Say we have a system of 2 particles and we are interested in the EOM for the first one. Then we can write:
[tex]\dot x_1 = f(x_1, x_2, t)[/tex]
This would really be a vector equation but imagine we're in one dimension for simplicity.

Then, let's apply some symmetries (constraints) to the EOM. If we accept time homogeneity, then
[tex]t \rightarrow t + t_0[/tex]
preserves the motion. So then we must have:
[tex]\dot x = f(x_1, x_2, t) = f(x_1, x_2, t + t_0)[/tex]
The only way this can be true is if f is actually not a function of t at all. So we get:
[tex]\dot x_1 = f(x_1, x_2)[/tex]

Now, if we also accept space homogeneity, then
[tex]\dot x = f(x_1, x_2) = f(x_1 + x_0, x_2 + x_0)[/tex]
The only way this can be true is if
[tex]\dot x_1 = f(x_1 - x_2)[/tex]
That is, the function f is a function of the differences of the particle positions.
(By the way, these first two derivations are taken almost verbatim from V.I Arnold's "Mathematical Methods of Classical Mechanics"--a great classical mechanics text, I feel)

Now, what happens when we demand invariance under boosts? Well, the transformation in this case is:
[tex]x_i \rightarrow x_i + kt[/tex]
Where k is some nonzero constant vector.
Applying this to the EOM:
[tex]\dot x + k = f(x_1 + kt - x_2 - kt)[/tex]
[tex]\dot x + k = f(x_1 - x_2)[/tex]
But by invariance, we should still have
[tex]f(x_1 - x_2) = \dot x[/tex]
So then
[tex]\dot x + k = \dot x[/tex]
[tex]k = 0[/tex]

Since we assumed k is some nonzero constant, we have reached a contradiction! But if we assume a second order DE at the beginning we do not reach a contradiction (I feel that this may be closely related to the fact that a second order DE gives us one more "degree of freedom", in having one more constant of integration). What do you think of this? Do you agree? If not, why not? I would be very happy to learn the validity/invalidity of this. It would explain a lot. I've always hoped that there was a way to prove the second order nature of Newton's second law from symmetries, even if these symmetries are true by "fiat". Thanks again!

(Side note: Doesn't it seem intuitive that invariance with respect to spatial translations and constant velocity translations is related in some way to a second order DE? I mean, don't you think that if constant acceleration transformations preserved the motion, that the EOM would have to be third order?)

Order

Hello Funky and thanks for your posts on this subject!

I think the main idea you have seems very interesting. But I am not very accustomed at all to symmetry reasoning. I do not even know very much about properties of differential equations. But the playing of equations seems like good fun, and that is why we all do physics, isn’t it?

Anyway, when you talk about symmetries you seem to be accustomed to symmetry reasoning from other fields of physics, but I am not. So maybe I should read a book about it? Or if you like, you can always send me an email and I can ask you some questions about what from my (lower) point of view seems like logical holes in your reasoning, about why there should be certain symmetries and other stuff. Oh, and again, thanks for an original point of view.

Order

Different observers will not agree about the energy change. It is not even difficult to see why this is so. Choose a frame where the body is accelerating, then the kinetic energy increases. Or choose a frame where the body is decelerating, then the kinetic energy decreases. Obviously the two observers do not agree.