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

[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 won't 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 won't 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.

 

[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.

 

[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

 

[Order]

What about the conceptual interpretation

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

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

 

[Order]

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

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.

 

[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".

 

[Hurkyl]

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

 

[HallsofIvy]

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 $$/alpha= 2$$.

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

 

[Order]

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

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?

 

[Order]

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

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?

 

[Slinkie]

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".

 

[krab]

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.

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

 

[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.

 

[mustafa]

The difficulty now lies in analyzing why it was more energy consuming to accelerate from v to 2v than from 0 to v.

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 traveled 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 traveling 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.

Is it impossible to understand physics intuitively?

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

 

[learningphysics]

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. /Order

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.

 

[CarlB]

I am one of those who cannot really go into advanced calculations and leave the simple questions behind.

This is good.

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?

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

 

[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.

 

[Hans de Vries]

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.

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 $p_x$ 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:

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

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.

 

[Order]

Multiple answers

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

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.

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

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

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

The difficulty now lies in analyzing why it was more energy consuming to accelerate from v to 2v than from 0 to v.

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.

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

You agree that work is Fd, right?

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.

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

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]

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.

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

 

[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 ass tonight, but I saw a lot 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...

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

 

[HallsofIvy]

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...

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

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]

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

That's a little sloppy.

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

 

[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 high school 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]

The elegance of conserved quantities

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. 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)):

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

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:

$$\ddot x = F(x)$$ <-- F is called the "Force"

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

$$E(x(t)): \Re^3 \rightarrow \Re$$

which are time independent (ie conserved). That is,
$$\frac{dE}{dt} = 0$$.
(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
$$U = - \int F(x) dx$$.
Then
$$\frac{dU}{dT} = \frac{dU}{dx}\frac{dx}{dt} = -F\frac{dx}{dt} = -Fv$$

Where we call $$\frac{dx}{dt}$$ 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:
$$\frac{dE}{dt} = \frac{dU}{dt} + \frac{dT}{dt} = 0$$

From this condition we find:
$$-Fv + \frac{dT}{dt} = 0$$
$$\frac{dT}{dt} = Fv = av$$

(where $$a = \frac{d^2 x}{dt^2}$$)

So $$T = \int avdt = \int vdv = \frac{1}{2} v^2 + C$$

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 $$T = \frac{1}{2} m v^2$$.

Hope this helps!

 

[matthewkeating]

Wk= Work done in accelerating object

Work done = Force x distance

If body a is traveling v m/s after 1 second, it is intuitive that it would be traveling 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 traveling 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 traveled 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 argument for a body traveling 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]

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! ;)