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

1. Jun 9, 2005

### Order

I am one of those who cannot really go into advanced calculations and leave the simple questions behind. I have to understand physics at a basic level. For that reason I am glad some members at this forum don’t hesitate to ask very fundamental questions. For years I have just taken everything for granted, being unaware of the rewards of asking the almost childish and simple questions.

Why is, for example, kinetic energy proportional to velocity squared? One might have grown accustomed to this over the years, but are there any simple explanations of this fact to a beginning student of physics?

I have thought about work (W=Fd) and the fact that it is not invariant under Galilei Transformation (d is not invariant). This is contra intuitive for the reason that we would believe that the energy consumed by an accelerating spacecraft would be independent of the observer. However, if you look at the kinetic energy of the rocket fuel (perhaps dropping in speed when leaving the rocket engine) there is no paradox. In fact one can show (I have shown it in a simple case below) that in a system of particles, under the valid assumption that momentum is conserved, the energy difference from one moment to the next will be independent of the observer. (This would also apply to the spacecraft of course.) The conclusion is that the energy change (to or from kinetic energy) is independent of the observer. If energy change is Galilei invariant this would strongly suggest that the idea of velocity squared is correct.

However, I don’t know if it is possible to explain this to a school student. And it gets worse if you look at a car accelerating. Here the energy is transformed into rotational energy in the wheels, which in its turn is transformed into kinetic energy of the car and rotational energy of the planet. But I guess it is possible (doing the calculations was too much for me) to reduce this to the particle interacting case, the earth being a big particle and the car being a small. I wont evaluate my thoughts about this right here, but can say that crucial to analyzing this situation is that you can never fix the inertial system to the ground, since the whole planet will be affected. In school books they never discuss this fact, of course.

And yet, although I know that the acceleration of a car does not violate Galilei invariance, and although I would agree that the high speed of a car is “energy in itself” and this is why energy is proportional to velocity squared, it is still not obvious to me why it takes more fuel to accelerate from 50 km/h 100 km/h than from 0 km/h to 50 km/h. I think my problem is that in physics you always lock everything into an inertial frame, but in reality there are no obvious inertial frames. And this is why I would falter if I where to explain why this is so to someone else. Are there any good physicists out there who can explain why I never seem to reach a full understanding of this energy formula?

/Order

PS. I know the question of energy has been in focus before at this forum. If anyone has a really good link where all my questions have an answer, it is ok to report it here. DS.

PPS. I wont give a full mathematical explanation of why energy change is galilei invariant, but one can look at a simple example: suppose you have a ball of mass 2m moving with a velocity v (this fixes the inertial frame in one dimension). Then suddenly this ball is split into two equally massive parts. So $$m_1 = m$$ and $$m_2 = m$$. Particle 1 is boosted with a velocity h. Therefore, in order to concerve momentum, particle 2 is boosted with a velocity -h. The splitting energy then is
$$\Delta E = \frac{m_1 (v + h)^2}{2} + \frac{m_2 (v - h)^2}{2} - \frac{m_1 v^2}{2} - \frac{m_2 v^2}{2}= \frac{m}{2} (-2 v^2 + 2 v^2 + 2 h^2) = m h^2$$
Since the splitting energy is independent of the initial velocity v it is Galilei invariant. Therefore logic is not violated.

One can extend this calculation to involve different masses and different initial velocities of the two particles. But the interaction energy will still be Galilei invariant. DDS.

2. Jun 9, 2005

### Crosson

There are very elegant ways to derive the classical expression for kinetic energy. This is the simplese one I know of:

In the case of a constant force, one can show the following equations:

$$x(t)=x_0 +v_0 t + \frac{1}{2} a t^2$$

$$v(t)=v_0 + a t$$

Combining these equations to elimanate t:

$$\delta v^2 = 2 a \delta x$$

If we multiply through by m this is the work-kinetic energy theorem. Its only proven for the case of a constant force, but because the above equation holds for each infinitesimal interval dx in the integral definition of work, we can see why the theorem is true in general.

3. Jun 9, 2005

### Hans de Vries

The expression for the kinetic ernergy:

$$K = \frac{1}{2} m_0 v^2$$

is only the first order approximation of the precise formula:

$$K = \frac{1}{2} m_0 v^2\ +\ \frac{1\cdot 3}{2\cdot 4} m_0 \frac{v^4}{c^2}\ +\ \frac{1\cdot 3 \cdot 5}{2\cdot 4\cdot 6} m_0 \frac{v^6}{c^4}\ +\ \frac{1\cdot 3 \cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8} m_0 \frac{v^8}{c^6}\ +\ .......$$

Kinetic energy is the total energy minus the rest mass energy:

$$K\ =\ mc^2 -m_0c^2$$

where the total energy E depends on the speed as:

$$E\ =\ mc^2 \ =\ \gamma m_0c^2$$

where:

$$\gamma\ = \ \frac{1}{\sqrt{1-v^2/c^2}} \ = \frac{1}{2} \frac{v^2}{c^2}\ +\ \frac{1\cdot 3}{2\cdot 4} \frac{v^4}{c^4}\ +\ \frac{1\cdot 3 \cdot 5}{2\cdot 4\cdot 6} \frac{v^6}{c^6}\ +\ \frac{1\cdot 3 \cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8} \frac{v^8}{c^8}\ +\ .......$$

So the classical formula of the kinetic energy is a first order approximation only.

Regards, Hans

Last edited: Jun 9, 2005
4. Jun 11, 2005

### Order

Yes, that theorem seems worthwhile to think about. In fact I thought about it (and I will think a little more about it later as well ) and found that if you are allowed to play with differentials, you can prove the work-energy theorem in three steps:
$$m\frac{dv}{dt}dx = m\frac{dx}{dt}dv \Rightarrow F dx = m v dv \Rightarrow \int Fdx = \frac{m v^2}{2} \arrowvert^{end}_{beginning}$$​
That’s a nice one. However, what I think is hard to understand is not the relation between work and kinetic energy. It is more difficult to understand why work is equal to energy in the first place. (Some people even claim that momentum is energy.) As I pointed out above, work is not Galilei invariant. Then you have to think about what happens when you switch frames. This is very contra intuitive.

This is a paradox: John wants to save fuel when driving. So when he accelerates he switches inertial frames. (As I pointed out before, energy change is Galilei invariant, so this is ok.) To reach the speed 2v he first accelerates to v, then switches to his rest frame and accelerates to v in this frame. According to those who are in the first frame his energy is $$2mv^2$$, but according to John he only used up $$mv^2$$. By swithching frames a little more often he can even save more fuel, he claims.

The solution to this paradox is of course that he did not take the energy of earth into account. So let us see how much energy really is converted when he accelerates in the second frame. The earth has a mass $$m_{earth} = M$$ and the car has a mass $$m_{car} = m$$, where M>>m. To simplify I don’t take any rotation into account. All I say is that when the car is boosted v, the earth is boosted –mv/M. In the second frame the earth has a speed -v, and the the car starts out with zero speed. Then you can see that the energy change from fuel energy to kinetic energy during acceleration is:
$$\frac{M(-v-\frac{m}{M}v)^2}{2}+\frac{m v^2}{2}-\frac{M v^2}{2} \approx \frac{M 2 v^2 \frac{m}{M}}{2}+\frac{m v^2}{2} = m v^2 +\frac{m v^2}{2} = \frac{3 m v^2}{2}$$​

One can see that the energy is compensated even if you switch frames during acceleration. Thus, there is no paradox. Now this was just math and math is the easy part. The difficulty now lies in analyzing why it was more energy consuming to accelerate from v to 2v than from 0 to v. I think that one way to understand this conceptually is that it is more difficult to push earth backwards when it is already moving with a certain speed under your feet. This is like if you are riding a snowboard and try to speed up by pulling a lift pole. This is more difficult or energy consuming if you already have a high speed, even if you are a skilful pole grabber. This is my explanation, but I don’t know if it is right. Does anyone else have any thoughts on this one? What is difficult is that you have different interpretations at different inertial frames.

/Order

5. Jun 11, 2005

### Order

That’s a good reminder. Some people say that relativity is more logical than Newtonian physics, but I don’t know if they’re right.

/Order

PS. I don't know what my nick means, but it could be that I like David Bohm, or it could mean that I like to bring conceptual order in the theory. Sorry for mentioning it DS.

6. Jun 19, 2005

### Order

New question

Big dissapointment
I am a little dissapointed no physicist was able to answer all or at least some of my main questions in this thread. But that is ok. I have taken time instead to think it through myself.

Proportionality to velocity squared follows from logic
One of my most interesting findings, is that the law that kinetic energy is proportional to velocity squared, follows almost immediately from logic. Assuming energy change/transformation in a system of particles is Galilei invariant (which is logical; all observers should agree the same amount of energy is put in or that the same amount of fuel is used up by a car) and assuming energy is proportional to $$mv^\alpha$$ one can show that $$\alpha=2$$. So one doesn’t really need to refer to experiments to derive this proportionality.

But this is just a mathematical proof
Still things are bothering me. This proof was done using mathematics and it seems almost impossible to get an intuitive feel for the logic in it. 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. (Although a relativist should be more accostumed to this fact, as Hans pointed out above.) Mathematically one can sea this is logical, thinking about different aspects of the problem, but it is difficult to have an intuitive feel for it.

Is it impossible to understand physics intuitively?
If no one here on this forum can answer this question, at least someone could answer this question: is it impossible to really understand physics? Are physicists just believers in math, never trying to understand what the formulas mean in a deeper sense? Of course one can learn to picture what the equations mean in a certain physical situation, but it is more difficult to have this deeper understanding. Put another way: one can understand how things are, but not why things are so. It is difficult to reach the point where one can say "this is not just the way it is, this is also the way it has to be".

7. Jun 19, 2005

### Hurkyl

Staff Emeritus
Intuition is something one can develop. Something that seems unintuitive to you now can become intuitive if you study it enough.

8. Jun 19, 2005

### HallsofIvy

Staff Emeritus
Okay, I'll bite: how does one show that?

9. Jun 20, 2005

### Order

Ok, I van show it. It is quite straightforward from the physical point of view. Thanks for asking by the way.

Suppose an object explodes into two parts. The first part has a mass m and the second has a mass M. If the first object has a boost v, then the other has a boost
$$-\frac{m}{M}v$$​
according to conservation of momentum. According to the suggested energy formula the explosion energy is
$$\Delta E_1 = \frac{mv^\alpha}{2} +\frac{M \biggl[ - \frac{m}{M}\biggr]^\alpha}{2}$$​
But according to an observer that moves with a relative velocity $$-\omega$$, the explosion energy is
$$\Delta E_2 = \frac{m(v+ \omega)^\alpha}{2}+ \frac{M \biggl[ - \frac{m}{M}+\omega \biggr]^\alpha}{2} -\frac{m \omega^\alpha}{2} -\frac{M \omega^\alpha}{2}$$​
Assuming they both measure the same explosion energy (Galilei invariance):
$$mv^\alpha + M \biggl[ - \frac{m}{M}\biggr]^\alpha = m(v+ \omega)^\alpha + M \biggl[ - \frac{m}{M}+\omega \biggr]^\alpha - m \omega^\alpha - M \omega^\alpha$$​
Now this is a messy relation, but using the binomial theorem one can simplify this to
$$\sum^{\alpha-i}_{i=1} {\alpha \choose i}\omega^i v^{\alpha-i} \biggl[ m+M \biggl[ –- \frac{m}{M} \biggr]^{\alpha-i} \biggr]=0$$​
Analyzing the bracketed right expression "one can see that" $$\alpha=2$$ is the only solution that holds generally.
Now this was fun, wasn’t it ? I must agree this is not the most general case, but for those who enjoy mathematics it should be possible to include the case when the two objects have a relative velocity to start with, or the (maybe unphysical) case when $$\alpha$$ is not an integer.

So what do you say HallsofIvy? Is this an interesting proof or is it rubbish?

10. Jun 20, 2005

### Order

This sounds hopeful. My problem is that I have studied the math, but does not know how to go beyond the math. But I might be able to do that.

May I ask if your own "intuition" ever evolved, by the way? Did you ever understand anything that seemed completely impossible to understand?

11. Jul 1, 2005

I've asked the same question regarding KE on this forum and got basically the same response. Do a search for "KE Puzzle" for fun.

It seems as though no one can get "outside" of the dogmatic definitions and explain the results of the equations intuitively. I'm leaning towards regarding this a semantic problem where KE is really some type of expression of "motion" and momentum is "energy" (as suggested above - flame away). Yes, I'm aware of some of the implications of this re-labeling (other areas of physics would have to be semantically synchronized) and I'm not suggesting this path should be taken. However this is what makes sense "intuitively".

Regarding developing intuition, what a joke. Sounds more like learning a pattern and by repetition convincing oneself that it's real. Accept it, repeat it, stop questioning it -- now it seems intuitive.

Sorry, I'm still waiting for a clear interpretation for KE as "energy" and why this form of "energy" transfer is proportional to V^2. The derivation and application of the formulas are straight forward, but I too am still trying to reconcile the math with "intuition".

12. Jul 1, 2005

### krab

So your position is that you are born with intuition and die with the same. Now that's a joke!

13. Jul 1, 2005

### learningphysics

I'm probably repeating what you've already posted in post #4 Order. Anyway, after I read your first post I just went through this little analysis. Essentially the trick (as you already know) is remembering that energy is going into the car (or ship.. or whatever) and the object the vehicle is exerting the force on.

I'm ignoring relativity here.

Example: Rocketship... the exhaust gas (thrust) pushes the rocket forward... the rocket pushes the gas backward. The rocket is in space and fires its thrust till it reaches a velocity of $$v_r$$. $$v_e$$ is the final velocity of the exhaust.

Using conservation of momentum:
$$0 = m_r v_r + m_e v_e$$

$$v_e = -m_r v_r / m_e$$

So using conservation of energy
Initial energy=final energy
energy of the fuel = $$(1/2)m_r (v_r)^2 + (1/2)m_e (v_e) ^2$$

Substitue for v_e and we get
energy of the fuel =
$$(1/2)m_r(v_r)^2 + (1/2)m_e m_r^2 v_r^2/m_e^2$$
$$=(1/2)(m_r + m_r^2/m_e)v_r^2$$

Now let's suppose that the rocket is initially at velocity v_r and goes up to 2v_r

$$(m_r + m_e)v_r = m_r (2v_r) + m_e v_e$$

$$v_e = (m_e - m_r)v_r/m_e$$

Now use conservation of energy:
$$chemical energy + (1/2)(m_r + m_e)v_r ^ 2 = (1/2)m_r(2v_r)^2 + (1/2)m_e(v_e)^2$$

plug in the formula for v_e and solve for chemical energy

$$chemical energy = (-1/2)(m_r+m_e)v_r^2 + 2 m_r v_r^2 + (1/2)(m_e - m_r)^2 v_r^2/m_e$$

So chemical energy simplies to $$(1/2)(m_r + m_r^2/m_e)v_r^2$$

So it takes the same amount of chemical energy to go from 0 to v_r that it does to go from v_r to 2v_r. Switch to different frames of reference (ignoring special relativity) for either case and the chemical energy will come out the same.

Last edited: Jul 1, 2005
14. Jul 2, 2005

### mustafa

You agree that work is Fd, right?

Now you know that the same force is required for accelerating from 0 to v and from v to 2v in the same time. But if you consider the distance travelled in a given time, it will be definitely more in the velocity range v to 2v than in 0 to v. So more work is to be done to accelerate from v to 2v than from 0 to v.

Also, one more confusion was there regarding switching frames. You have a body of mass m travelling at velocity v relative to some inertial reference frame. Its kinetc energy in that frame is 1/2 mv^2. For a person moving in an inertial reference frame having velocity -u relative to the first frame, the body has a kinetic energy 1/2 m(v+u)^2. The explanation to this can be as follows: If you extract the energy from the body by bringing it to a rest relative to the first frame you will be able to extract 1/2 mv^2 energy, whereas if you are in the second reference frame you will be able to extract more energy from the body. This is analogous to relativity in gravitational potential energy: If you are on a plateau having 300m height from sea level you may consider an object placed at your level to have zero potential energy even though it has high potential energy relative to sea level.

While it is not impossible to understand physics intuitively, it is indeed difficult. Just think about relativity theory - Do you think it is possible for you to imagine intuitively that length and mass of a body may change with its velocity? However there are scientists like Faraday who used their intuition to visualise and discover physical phenomenae without much use of mathematics.

Intuition can lead you to visualize new things or existing things from a different perspective but you need mathematics and scientific reasoning to check their validity

15. Jul 2, 2005

### learningphysics

From a simplistic viewpoint using just conservation of linear momentum and conservation of translational kinetic energy and taking into account both bodies.... it is not more energy consuming. It takes the same amount of energy.

16. Jul 3, 2005

### CarlB

This is good.

Sure. First, let's agree that KE is some smooth (all derivatives exist) function of velocity, defined for all velocity, increasing, and that velocity is a vector, and that the KE must not depend on the direction in which the velocity is pointed.

The nth power of the magnitude of a vector is given by:

$$|v|^n = (v_x^2+v_y^2+v_z^2)^{n/2}$$

and this is smooth (over all $$(v_x,v_y,v_z)$$ only for the even powers of n. The lowest power that is nontrivial is n=2, and this gives the usual formula for KE (with mass as the constant of proportionality).

In short, the formula for KE is derivable from the assumption of smoothness and lowest nontrivial order of approximation. Note that this derivation is compatible with special relativity and also note that in special relativity, only the even powers of v appear in the formula.

If you wanted to have a smooth function of velocity that used the 1st power of the magnitude of velocity, you would have to define it by something like this:

$$U^2 = \kappa (v_x^2+v_y^2+v_z^2)$$

This would make perfect sense, but note that the definition is double valued in that $$\pm U$$ are solutions. So part of the reason for KE not having linear dependence is the requirement that we have a formula for it that is single valued.

Carl

17. Jul 3, 2005

### Claude Bile

The formula for KE is dependant upon our definition of Work. If you can justify the definition of work, then the formula for KE follows. As we all (should) know, the formula for work is simply;

$$W=\vec{F}.\vec{s}$$

or, more generally;

$$W=\int \vec{F} d\vec{s}$$

Work must necessarily be dependant on force and displacement, since when either of these quantities is zero, the work done is zero. It makes sense too, to use a dot product since this is the only form of vector multiplication that yields a scalar as a result. A linear depedance on either quantity is fairly intuitive, and there is no need for a constant of proportionality because of how the units for all these quantities are defined.

Regards,
Claude.

18. Jul 4, 2005

### 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:

$$KE_x = p_xc,\ \ KE_y = p_yc,\ \ KE_z = p_zc[/itex]. Where $p_x$ is the momentum $mv_x$. To add the speeds in the x,y and z direction you need to use Phytagoras. The total Energy is: [tex]E \ = \ \sqrt{m_0^2c^4 + p_x^2c^2 + p_y^2c^2 + p_z^2c^2 }$$

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 $\epsilon$ is sufficiently small then:

$$\sqrt{1 + \epsilon^2 }\ \approx \ 1 + \frac{1}{2}\epsilon^2$$

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

Regards, Hans.

19. Jul 5, 2005

### 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.

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.

20. Jul 6, 2005

### Hans de Vries

For low speeds $v<< c$ and thus $m = m_0$ we may write

$$E \ = \ \sqrt{m_0^2c^4 + p_x^2c^2 + p_y^2c^2 + p_z^2c^2 }$$

as:

$$E \ = \ m_0c \sqrt{c^2 + v_x^2 + v_y^2 + v_z^2 }$$

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

For very high speeds where $m >> m_0$ the formula becomes:

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

This can be simplified to:

$$E \ = \ mvc\ =\ pc$$

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

Regards, Hans

Last edited: Jul 6, 2005