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

[reilly]

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)):

\ddot x = f(x, \dot x, t)

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:
\dot x_1 = f(x_1, x_2, t)
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
t \rightarrow t + t_0
preserves the motion. So then we must have:
\dot x = f(x_1, x_2, t) = f(x_1, x_2, t + t_0)
The only way this can be true is if f is actually not a function of t at all. So we get:
\dot x_1 = f(x_1, x_2)

Now, if we also accept space homogeneity, then
\dot x = f(x_1, x_2) = f(x_1 + x_0, x_2 + x_0)
The only way this can be true is if
\dot x_1 = f(x_1 - x_2)
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:
x_i \rightarrow x_i + kt
Where k is some nonzero constant vector.
Applying this to the EOM:
\dot x + k = f(x_1 + kt - x_2 - kt)
\dot x + k = f(x_1 - x_2)
But by invariance, we should still have
f(x_1 - x_2) = \dot x
So then
\dot x + k = \dot x
k = 0

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]

Might help me

Order,

Over the past 2 years, I've thought about several classical mechanics mysteries--including kinetic energy KE = \frac{1}{2}mv^2--all the time.

[...]

Hope this helps!

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]

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?

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.

 

[robousy]

Hey Order,

I'm really glad you asked this question.
Its been on my mind now for...maybe 7 years!'

I am at graduate school now but I remember thinking about a slight variant of your initial paradox (if that's the right word to use here) when I was an undergrad.

When I explained it to a few people I never received a satisfactory answer and because the question seems so basic I never really pursued it anymore thinking that there was something silly that I was just missing on not thinking about. Anyhow I am very glad you brought this up!


btw - just out of interest, the way I used to think about it was in terms of being able to determine an absolute frame of reference (obviously contradicting spec. rel.)

Consider a particle colliding with a spring, both of mass 1 say in some empty space with no reference markers.

The combined speed is 10 - but is the KE 1^2+9^2 \: or \: 2^2+8^2 \:etc...

Technically one could observe the compression of the spring and determine an 'absolute' reference frame just by observing the compression of the spring.

Same idea as yours - just put in a different way.
I'll be checking back regularly in the hope of finding an intuitive way of understanding KE! ;)

 

[Order]

I was wrong

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?

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.

Oh man, what a big mistake. No, I was wrong, of course the energy change will be independent of the observer. It is only when you look at one body the energy will be different. All observers will agree about the energy change in the system as a whole (for example in a two body system). Wohu, that one was embarrassing indeed

 

[Order]

The purpose of this thread right now

Hey Order,

I'm really glad you asked this question.
Its been on my mind now for...maybe 7 years!'
...

Hey there! Good to see you have also thought about this paradox. You talk about spring energy and I talked about fuel energy. Surely we are onto the same paradox. I suppose you also know the paradox has already been solved in this thread, so you did not come to ask any specific question. Personally I think the solution gives an understanding of the depth of Newtonian mechanics.

But I think there are other things to be understood. People reading this thread might wonder why I just don’t say thanks and goodbye when the solution has been posted long time ago, even by me. Also, the question about the relation between work and energy has been touched upon. So what is left to say? Why is there still a wish to understand something more? What is this “more”? Let me try to explain.

There are many interesting thoughts in this thread, well worth thinking about. But one aspect of the problem that really struck me was FunkHaus approach (although the thought has appeared in other posts as well). In this thread I have been trying to make paradoxes to see if the laws of physics violates logic. FunkHaus went the other way around. He stated logic propositions and, from them, tried to deduce the laws of physics. One can easily see that the second approach is at a higher level of thinking. I mean, it should call for a better understanding to build something than to check if it is properly built.

So basically, what I think would be really nice to know is the logical pillars of mechanics (for example invariance under boosts (what I call Galilei transformations)) and how the math is deduced from these pillars. There should be books written about this, right? Maybe there are? I don’t know.

My thought was therefore that the true understanding of Newtonian mechanics (including the proportionality to velocity squared) would come not just by checking it, but by constructing it. So maybe that is why I wait for a better day to come? And maybe this is a too big question to ask? But then at least I know I ask too big questions, and that is fine.

 

[FunkHaus]

In this thread I have been trying to make paradoxes to see if the laws of physics violates logic. FunkHaus went the other way around. He stated logic propositions and, from them, tried to deduce the laws of physics.

Actually Order, I think we were kind of doing the same thing!
You were asking things like, is the definition
KE = \frac{1}{2}mv^2
consistent with the "intuitive" idea that the change in kinetic energy is invariant with respect to different reference frames (ie invariant with respect to boosts), etc. What I was doing was saying that basically, the definition follows precisely because of reasons like that.

But here's the key thing to keep in mind--in order to understand where math ends and physics begins, you must in a way part with the idea of "intuition". In fact, invariance with respect to inertial reference frames is not intuitive; well, I guess it depends on your definition of "intuitive". But the point is that complex and interesting worlds that do not exhibit this phenomenon could and do easily exist (I know, I've been to some of them)--we simply find experimentally that our world has this property.

In classical mechanics all we have is a set of world lines of particles, or if you will, a single curve in a set \Re^{3N} (N is the number of particles, 3 is the dimension of space) which contains an amount of information equivalent to that which we believe our universe to contain. This an "almost" purely general mathematical statement about the containment of every piece of information in our world. Physics is, in the most basic sense, the application of mathematical constraints to this curve, or this set. Well, at least that's how I think about it (hopefully none of my profs. are reading this thinking I'm crazy).

When I talked of symmetries, I was talking of constraints. Basically, they're the same thing. None of these symmetries are "logical". Logic deals with the validity of mathematical theorems. Physics is different. It is, to a mathematician, totally arbitrary. That is to say, the constraints are arbitrary. When I say that (classical) boost invariance, and spatial homogeneity are symmetries, I mean that we determine empirically that they are constraints as to what may happen in our specific world.

The truth, at least as I've found, is that there are not many good classical mechnics textbooks that actually start from the ground up with a basic set and use nothing but pure empirical symmetries to reveal "why" the theorems of physics are as they are. Most books take stuff for granted. In fact, most books (Goldstein is a good example) just start off by saying
\frac{d \vec p}{dt} = \vec F
is the equation of motion. OK, except what is momentum? Mass times velocity? OK, what is mass? Seems like some sort of extra empirically determined parameter. "Well if that's the case, then tell me about it!", I say. But most books don't. They just kind of take it for granted. Which is in my opinion, kind of a bummer.

But, I found this one great book with a whole different (and rather mathematically formal) perspective--V.I. Arnold's "Mathematical Methods of Classical Mechanics". Check it out! It goes through basically some of the same symmetry arguments I've laid down above.

Anyway, I'd love to talk about this more with you. I must say, you've started up quite a good thread. (Also, speaking of CM books, I've been working on a rather brief book of my own. It definitely won't cover all the theorems about rigid body motion and perturbation theory and all that other stuff, but it will (hopefully) take a fresh perspective).


--funkhaus

 

[Order]

Thoughts on doing physics



Hey Funkhaus,

I rarely use the word intuition; the reason being it is not clearly defined. Yet, there is a need for clarification here, and I thought it was rewarding reading and thinking about your post for a while.

Yes, I believe it could pose a risk to, somewhere in the back of the mind, thinking that principles magically can derive themselves. This fault could be due to laziness, not wanting to invest hard work in the analysis, always taking refuge to excess rationalization and generalization, pretending to master what one does not.

Anyway, the wonderful thing about physics is that the constraints or symmetries do not follow directly from observation. It takes logic to see them. (Nothing new about this, but it is good to remind oneself now and then.) After having the constraints it is possible to look at the implications – the laws at other levels. But I think these two parts of the analysis – finding the constraints and applying them – shares the same essence. “Getting to know this essence” is a possible definition of (empirically based) intuition then. From this viewpoint mathematics is nothing but a "sixth sense", just another way to feel physics. I have no idea if this analysis is excess rationalization.

I checked my local library and they had the Arnold but not the FunkHaus book. Unfortunately some jerk is not returning it in time. (I am never on time myself, but hey.) It could take me a while reading it, by the way.

My great wish is that textbooks are clear about what ideas are being presented through the mathematics. If they clearly explain what constraints there are behind the derivations, then this should work for me.

Maybe this is the end of this thread :’)

 

[kthouz]

Now as you found out how kinetic energy is proportional to the square of the velocity, can you try to get the kinetic energy of a system of 3 particles in terms of their relative velocities each other. For example v(1,2) is the velocity of particle 1 relative to particle 2.?

 

[ZapperZ]

Now as you found out how kinetic energy is proportional to the square of the velocity, can you try to get the kinetic energy of a system of 3 particles in terms of their relative velocities each other. For example v(1,2) is the velocity of particle 1 relative to particle 2.?

Are you aware that you are responding to a thread that had a last post in 2005?

Zz.

 

[Doc Al]

And that kinetic energy is frame dependent?

 

[RonL]

Are you aware that you are responding to a thread that had a last post in 2005?

Zz.

Not all is lost!,
This is a thread that I needed to read, and might not have found it on my own. There are things mentioned that will serve as guide points to some of the basics I need to learn.

Thanks to all

Ron

 

[Electrica]

Can I just say this... (and I'm not offering any explanations but offering a possible path for thought). I believe you might by underestimating the reaction of inertia and it's direct connection to kinetic energy. If you think about (before you add the math) why kinetic energy increases at the SQUARE of velocity and it's not getting it from the energy input (i.e. gas, actually when you increase from 10-20 mph your mpg goes down ergo you require less power to attain the 2nd 10 mph) why is the kinetic energy growing exponentially? Why the sudden drop off after 65 or so? How much does the car's manufacturing and technology, as well as aerodynamics, road friction, car lift, etc have to play into it? Take an "out of the box" fresh look at inertia. If you explained it to a school student would it really add up? The last sentence suggested to inspire a more simplistic review. Like Sherlock Holmes said, after you remove everything else, whatever remains no matter how unlikely, must be the truth.

 

[Lsos]

The way I came to understand it is like it has been mentioned a few times in this thread already...

Say you're using the same force to accelerate from 0-10 as from 10-20. Except, when you're accelerating from 10-20...you're going twice as fast, therefore covering twice the distance. Applying the force for twice the distance means you'll need twice the energy.

Kinetic energy being proportional to speed also means that a car's brakes overheat much faster at higher speeds. Braking from 100-0 might require the same force as braking from 200-100. Indeed, the brake calipers will squeeze just as hard on the disks...but think about how much faster the disk is spinning between those pads. How much more distance the pads scrape against.

When finish braking from 100-0, the disk comes to a complete stop between the pads. As you complete braking from 200-100...that thing is STILL spinning at 100! Same change in speed, but a HUGE difference in the heat produced.

Sure enough, when some proper racing is going on, disks can get glowing red hot.

 

[Prologue]

...It is not easy to explain why it takes more energy to increase the relative velocity between two objects (in for example free space) by a certain amount, if the two objects already have a high relative velocity...

If by 'explain' you mean 'appeal to experiences that people are accustomed to' then it is easy to find an example of this playing out in real life. Have you ever tried to push a skateboard as fast as you could? The difference in velocity between you and the ground is a huge factor in how much velocity you can actually gain. You can gain a few mphs easily when starting from rest, but trying to go really fast is extremely difficult (and air resistance isn't the issue, the speeds are low for a skate board, surface friction can be the issue but find a nice polished concrete floor and that argument goes out the window). While going fast you can push just as hard over the same distance (about half of your normal stride) and you will gain very little velocity. Therefore KE is not linear, but increases with a power greater than 1.

What is the "exact" expression? (not counting special relativity).

I think the method of playing with differentials is the key. Although, it isn't as much of a game as it seems. Define these quantities:

v=\frac{\Delta x}{\Delta t}

\Delta t= The \:time\:that\:you\:are\:pushing

\Delta x=The \:distance\:that\:you\:are\:pushing

m\frac{dv}{dt}=F=m\frac{\Delta v}{\Delta t}=The \:constant \:force\:that\:you\:can\:apply

\Delta v=The \:change\:in\:your\:velocity

dW\equiv Fdx

or for our case

\Delta W=F\Delta x

So, we would like to find the change in energy due to one push. And then sum all of the pushes up to find the total energy at the end.

Let's use the definition for F
F \Delta x=m\frac{\Delta v}{\Delta t}

To get

F \Delta x=m\frac{\Delta v}{\Delta t}\Delta xCancel the delta x's, and move delta t to the other side, to get

F \Delta t=m\Delta v

This shows an interesting relationship. As your ground speed increases, the amount of time that you can be pushing is lessened, and since the force of your leg is assumed to be constant the total quantity must be lessened. So if you are going faster, you have less time and so the quantity m \Delta v goes up less. Since your mass is constant, it means that the amount of velocity added is less and less as you speed goes up and up.

Rearrange the velocity relationship to solve for delta t

v=\frac{\Delta x}{\Delta t}

\Delta t=\frac{\Delta x}{v}

Substitute back into

F \Delta t=m\Delta v

to get

F \frac{\Delta x}{v}=m\Delta v

Solve for F delta x

F \Delta x=mv\Delta v

Replace F delta x with dW

\Delta W=mv\Delta vSo the amount of work you do in one push is given by the above expression. To get the exact instantaneous expression (no constant force) it is merely

dW=mvdv

Integrate from rest to final velocity to get

W=\frac{1}{2}mv^{2}This is an exact expression with the assumptions we have made (classical galilean stuff), not an approximation. When you account for special relativity then it is just an approximation.

 

[ftfaaa]

F=m(dv/dt)
F=m(dv/dx)*(dx/dt)
F=m(dv/dx) *v
F*dx=mdv *v w =∑F*dx=∑mdv *v=1/2 mv^2-1/2mv0^2
then we define 1/2mv^2 as the kinetic energy

 

[jtbell]

Somebody already did this in post #4 of this thread, four and a half years ago.

 

[Dale]

Hehe, I give this thread the zombie "night of the living thread" award. It first died in 2005, came back to life for a day in 2008 and promptly died again, and then came back to life again in 2010 where it has been terrorizing the villagers for a couple of weeks now!

 

[Prologue]

Somebody already did this in post #4 of this thread, four and a half years ago.

True they did the chain rule/short version. But sometimes that doesn't show why it is relevant, that is why I spelled it all out.

 

[dnquark]

This thread keeps coming back because nobody gave an entirely satisfactory explanation that motivates v^2 proportionality. (I can't give one either, I came here in search of one). What puzzles me is all the explanations that involve work-energy theorem. They make sense mathematically, but they require you to define an auxiliary quantity called "work" with dW = F dx. How is that definition more intuitive than defining energy as ~ m v^2?..

I think the original question should be rephrased as "what is the minimum set of assumptions one needs to make in order to identify 1/2 m v^2 as a conserved quantity?"

One candidate answer is assuming the principle of least action, from which energy conservation (and KE expression) can be derived using Noether's theorem. However, that's probably not the only possible answer. For instance, Leibnitz and du Chatelet probably had other motivations when they posited v^2 dependency of energy.

 

[Hurkyl]

This thread keeps coming back because nobody gave an entirely satisfactory explanation that motivates v^2 proportionality.

We saw it when we measured things.

I think the original question should be rephrased as "what is the minimum set of assumptions one needs to make in order to identify 1/2 m v^2 as a conserved quantity?"

This one's easy. You just need one assumption: "1/2 m v^2 is a conserved quantity".

 

[Driftwood1]

Interesting...


Why isn't Kinetic Energy proportional to velocity cubed, or to the power of 1.35?

Why is velocity proportional to the Kinetic Energy raised to the power of 0.5?

 

[Dale]

Because the units wouldn't work out otherwise. When I was taking freshman physics that is what my professor pounded into our heads the first week or two: "always check the units". Kinetic energy couldn't possibly be anything other than kmv² where k is some unitless constant.

 

[Dadface]

I haven't the patience to read through all responses to this lengthy and ancient thread but surely the question has already been answered.If not here goes:
From the conservation of energy the KE of an object will be equal to the work done in bringing that object from rest to a velocity v(or to bring it from a velocity v to rest) and is independent of the method by which that work is done.
Work done=force*distance
=mass* acceleration*distance(Mad)...(1)
For uniform acceleration from rest to a velocity v in time t, a=v/t and d=(vt)/2
Substitute a and d into equation (1) and we get the well known non relativistic KE equation.

 

[stevenb]

Without taking away from some of the good answers above, I'll add that Landau and Lifgarbagez answer this question right in the beginning of their famous Classical Mechanics books.

This thread needed only one response.

"Go look there and come back with any questions if it's not clear."

Their explanation is clear and elegant. "Classics" are classics for the reason that everyone should read them.

 

[Lsos]

It's because if you want to make something go twice as fast, you have to either push it 4x as hard, 4x as far...or some combination of these. Whichever path you take, you'll get 4x as tired. There's no other way around it.

Since energy is force x distance, it stands to reason that as speed increases then so will the distance. Simple math show exactly how much the distance increases.

 

[Dadface]

This thread keeps coming back because nobody gave an entirely satisfactory explanation that motivates v^2 proportionality. (I can't give one either, I came here in search of one). What puzzles me is all the explanations that involve work-energy theorem. They make sense mathematically, but they require you to define an auxiliary quantity called "work" with dW = F dx. How is that definition more intuitive than defining energy as ~ m v^2?..

I think the original question should be rephrased as "what is the minimum set of assumptions one needs to make in order to identify 1/2 m v^2 as a conserved quantity?"

One candidate answer is assuming the principle of least action, from which energy conservation (and KE expression) can be derived using Noether's theorem. However, that's probably not the only possible answer. For instance, Leibnitz and du Chatelet probably had other motivations when they posited v^2 dependency of energy.

I have just scanned through other posts on this thread and your post,in particular,has caused me to re-evaluate my post 55.I agree that if energy is defined with respect to work then it is no more intuitive than defining it with respect to (mv^2)/2 or with respect to anything else that is relevant such as gravitational potential energy.It's the chicken and egg,circular argument syndrome.Some people may have an intuitive feeling for energy but I think that it is partly(perhaps fully)because those people have gained some familiarity with the subject.We observe energy but I don't think we really understand it.Why then is KE proportional to v squared?I now think that a possible best answer is quite simply that "it is because observations and experiments show it to be so".

 

[D H]

This thread keeps coming back because nobody gave an entirely satisfactory explanation that motivates v^2 proportionality.

Read the thread. It follows directly from the definition of work. The derivation has been given many times; [post=691978]post #25[/post] gives a very nice and succinct derivation.

This begs the question, why define work that way? As is the case with many of the widely used concepts in physics (why is force defined to be mass times acceleration?), the answer is that we use this definition because it is so very useful and so very simple. We physicists are very much enamored with utility and simplicity, particularly when expressed mathematically, and particularly so when it describes some aspect of reality.

The concept of energy was an important one even in basic Newtonian mechanics. The concept took on an even bigger role with the developments of thermodynamics and in the development of the Lagrangian/Hamiltonian formulations of Newtonian mechanics. The concept of energy plays a predominant role in quantum mechanics. Noether's theorems pretty much iced the cake.

The concept does require a bit of modification with regard to special relativity. With these mods it remains a very important, if not the central, concept of physics.

Bottom line: We physicists use the concept of energy because it works so very nicely for us (pun intended).

 

[Dadface]

Read the thread. It follows directly from the definition of work. The derivation has been given many times; [post=691978]post #25[/post] gives a very nice and succinct derivation.

This begs the question, why define work that way? As is the case with many of the widely used concepts in physics (why is force defined to be mass times acceleration?), the answer is that we use this definition because it is so very useful and so very simple. We physicists are very much enamored with utility and simplicity, particularly when expressed mathematically, and particularly so when it describes some aspect of reality.



Bottom line: We physicists use the concept of energy because it works so very nicely for us (pun intended).

I have to agree.Experiments reveal that there is a quantity which we call energy which can appear in several different forms and which is conserved and which has other observeable properties.It may be possible to define energy in different ways but given the choice we should go for the definition which is the most elegant,most useful and most simple to use and I think the definition of "energy"in terms of "work" meets these criteria.