Talk:Kinetic energy/Archive 1

This is the archive file "Talk:Kinetic energy/Archive 1".

Words for kinetic energy's meaning

"In words this means that the kinetic energy (Ek) is equal to the integral of the dot product of the velocity (v) of a body and the infinitesimal of the body's momentum (p)."

I think saying it "in words" ought to be something like "the amount of work done on a body to increase (or decrease) its speed".

Yes, to decrease the speed to zero. Please do so. But I just wanted to explain the formula a bit to people new to the field. That explanation of the formula must be retained somehow, I feel. Andries

Add units

Could someone add the proper SI units to the Simple Calculation section? I'm sure that many, if not most of the visitors to this page will be coming looking for a quick Newtonian definition of kinetic energy for a particular question/problem they have.

I added units to the top. Fresheneesz 03:34, 10 April 2006 (UTC)

Heat is not kinetic energy

It is incorrect to say that heat is kinetic energy. If you heat a solid, then then you add energy to the Phonons. In other words the heat becomes vibrational energy. In the simplest approximations half on the vibrational energy is kinetic whereas the bother half is potential (This requires a harmonic potential).

However if you heat a system at chemical equilibrium, then you may cause a chemical reaction to take place the system. This reaction may lower the vibrational energy. However it is not correct to say that the reaction lowers the heat of the system.

In conclusion the section about heat should be deleted.

In that integral, what's that "s"? - anon

I would disagree. The physical manifestation of heat is the kinetic movement of particles that make up the object. Vibrations are the result of shifting kinetic energy. In a spring, the kinetic energy is converted to PE then back to KE ad infinitem. However, in a heated object, I'm pretty sure that the "vibration" comes from collisions which doesn't directly involve PE, except on the very small scale. As for the chemical reactions, an endothermic reaction does indeed lower the heat of the system, why would it be otherwise? Fresheneesz 00:31, 8 April 2006 (UTC)

Heat is related to temperature. Temperature is the averaged random translational kinetic energy. An object at 0K moving with some velocity is still at 0K because that kinetic energy is not random. Therefore, it is both correct and incorrect to say either heat is or is not kinetic energy without a more in-depth explanation... I, personally, would just avoid the situation entirely, i.e., do not discuss "heat". DrF 17:17, 22 May 2006 (UTC) Note that if a the vibrations of atoms in a crystal were not random, but instead were in total unison, then there would be no "vibrations" at all, i.e., the entire crystal is translating back and forth with no internal vibrations. DrF 18:26, 22 May 2006 (UTC)

Kinetic energy and momentum

In the case of two bodies colliding, we know that momentum is preserved. We also know that energy is generally lost through heat, sound etc. Take 2 identical masses 'm' travelling toward each other in an inertial frame, with velocities +/-v in the frame.

So

• total momentum p = +mv -mv =0 (travelling in opp directions and p is a vector.)

• total ke, E = 0.5mv^2 + 0.5 mv^2 = mv^2 (energy is a scalar)

It is reasonable to assume that, if the masses are identical, that the velocity of each mass after collision will be equal but opposite in the frame. Also assume the masses do not change after collision. This means that the rebound velocities must be smaller that the impact velocities since there is loss of energy in the collision. Both laws can be satisfied.

However if one of the masses is at rest (in the frame) and the other mass travels at 2v toward it:

Before collision:

• total momentum p = 2mv
• total energy = 2mv^2

After the collision,

• All momentum is transferred to the stationary mass, which shoots off at 2v. (a la Newtons cradle)
• Its energy (and the total energyof the system) is therefore the same at 2 mv^2

This is the same as the initial energy. So we have NO loss of energy in the collision regardless of the duration or type of impact between the two bodies. --Light current 16:55, 20 January 2006 (UTC)

Loss of energy in elastic collisions

I dont think you necessarily have loss of energy in elastic collisions. See above post for example. --Light current 17:36, 20 January 2006 (UTC)

The definition of elastic collision means that no "loss" of energy can happen, the total kinetic energy before and after the collision must equal the total afterward. This is only if its a completely elastic collision, otherwise of course there will be energy transfer to heat. Fresheneesz 03:38, 10 April 2006 (UTC)

Inelastic collisions

Where does the energy go in inelastic collisions if say two objects collide in outer space? (In space- no one can hear you scream!!)--Light current 18:41, 20 January 2006 (UTC)

Heat?

Mainly heat, but perhaps some particles escape at high velocities, too. StuRat 08:56, 28 February 2006 (UTC)

Another formula

Is it correct to say that: kinetic energy = (mv^2)/(2*sqrt(1-(v/c)^2))

Re-arranged: 1/2 γm v^2

That is, using the formula for mass being: mass = γm where m is the rest mass

and substituting it into the formula kinetic energy = 1/2 m v ^2

If this is not true, why not?

86.20.211.26 17:33, 27 February 2006 (UTC)

This isn't true, but i'm not sure why. Probably because since kinetic energy is defined as "the energy required to speed the object up to its speed" and the mass was not initially as heavy as γ m is - the amount of energy is smaller to begin with and gets larger. Integrating xv^2 / 2 from x = m to x=γ m might give you the right answer, but i'm not sure. Fresheneesz 03:42, 10 April 2006 (UTC)

As the article says, kinetic energy is DEFINED by:

$E_{k}=\int {\vec {F}}\cdot d{\vec {x}}$

with the constant of integration determined by $E_{k}=0\!$ when ${\vec {v}}=0$. If we perform this integration classically, we get:

$E_{k}=\int {\vec {F}}\cdot {\vec {v}}dt=\int {\vec {v}}\cdot d{\vec {p}}=\int {\vec {v}}\cdot d(m{\vec {v}})={\frac {m\ {\vec {v}}\cdot {\vec {v}}}{2}}$

Relativistically, we must change the expression for linear momentum. Integrating by parts, we get:

$E_{k}=\int {\vec {v}}\cdot d(m\gamma {\vec {v}})=m\gamma {\vec {v}}\cdot {\vec {v}}-\int m\gamma {\vec {v}}\cdot d{\vec {v}}=m\gamma {\vec {v}}\cdot {\vec {v}}-{\frac {m}{2}}\int \gamma d({\vec {v}}\cdot {\vec {v}})$

Remembering that $\gamma =(1-v^{2}/c^{2})^{-1/2}\!$, we get:

$E_{k}=m\gamma v^{2}-{\frac {-mc^{2}}{2}}\int \gamma d(1-v^{2}/c^{2})=m\gamma v^{2}+mc^{2}(1-v^{2}/c^{2})^{1/2}+C$

And thus:

$E_{k}=m\gamma (v^{2}+c^{2}(1-v^{2}/c^{2}))+C=m\gamma (v^{2}+c^{2}-v^{2})+C=m\gamma c^{2}+C\!$

The constant of integration is found by observing that $\gamma =1\!$ when ${\vec {v}}=0$, so we get the usual formula:

$E_{k}=m\gamma c^{2}-mc^{2}=mc^{2}(\gamma -1)\!$

Is that clear enough? JRSpriggs 09:58, 6 October 2006 (UTC)

Negative work to decelerate?

"It is formally defined as work needed to accelerate a body from rest to a velocity v. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work would also be required to return the body to a state of rest from that velocity."

Wouldn't the amount of work required to return the state of rest be opposite that of the work needed to accelerate it? See the Mechanical work article.--Myncknm 00:16, 9 March 2006 (UTC)

Funnily, I have recently got to this article by chance, was bothered by the fact that it doesn't state that the work should be negative, and fixed it. Only now have I noticed you were also concerned about this fact. The next time you encounter something you have a good reason to believe is incorrect, be bold and fix it. Welcome to Wikipedia! -- Meni Rosenfeld (talk) 15:08, 10 March 2006 (UTC)

I'm not bold either, but I also suggest that eventually we should stop talking about "negative work." We could consider work to be a scalar, or absolute, quantity without regard to positive or negative, and then say that work must be subtracted from a moving object to slow it down. I think this gives a better idea of the relationship between work and kinetic energy. --LightSpeed 03:20, 1 March 2007 (UTC)

To Pjedicke: I think you are misunderstanding the subtle distinction between Energy and Work. Energy is a quantitative attribute that an entity possesses at a certain time. Work is the amount of that attribute which changes form or is transfered from one entity to another during a period of time. Power is the rate at which work is done. (A similar relationship exists between Momentum, Impulse, and Force.) Work is negative when the transfer is in the reverse of the normal direction. JRSpriggs 09:21, 1 March 2007 (UTC)

67.177.35.2's edits

I don't agree with the recent edits by 67.177.35.2. My reasons are:

1. Since this article does not deal with an advanced conept, The introduction should be intuitive, not rigorous. Later we can go into technical details.
2. Wrting E=mv^2/2 requires both non-relativity and the assumption of point object (otherwise rotational energy is added). Instead of specifying all the necessary assumptions, it is better in an introductory sentence to simply state that it holds in "simple cases".
3. Regarding the integral below, the focus of the article, and hence the formula, is on the concept of "kinetic energy" and not "work", so the formula should be explained in terms of kinetic energy and not work. An article about work with a similar formula would explain it in terms of work.

Therefore, I will now revert those edits. I'll be happy to hear everyone's thoughts on the matter. -- Meni Rosenfeld (talk) 19:23, 12 March 2006 (UTC)

Looks like RJN beat me to it. The discussion's still open though. -- Meni Rosenfeld (talk) 19:25, 12 March 2006 (UTC)

Speed vs. velocity

Velocity is a vector, describing an object's speed and direction of travel. Speed is the magnitude of the velocity. Kinetic energy depends only on the speed, and not in general on the velocity. So in most places in this article, discussing "speed" is more appropriate. Also, powers of vectors are generally undefined, so use of the term "velocity squared" is probably best avoided, in favor of "speed squared". -- Meni Rosenfeld (talk) 19:39, 14 March 2006 (UTC)

KE in any paricular direction depends on velocity (ie speed in that direction). What does the v stand for in 1/2 mv^2. speed or velocity?--Light current 22:13, 14 March 2006 (UTC)

If it is agreed that the "square" of a vector is equal to its scalar product with itself, then v in ${\frac {1}{2}}mv^{2}$ can stand for either speed or velocity, depending on the context. However, since general powers of vectors can't be sensibly defined this way, it is AFAIK a better practice to discuss expressions like ${\frac {1}{2}}mv^{2}$ only when v stands for speed. -- Meni Rosenfeld (talk) 09:43, 15 March 2006 (UTC)

yes and no: it's no accident that all the equations in all the physics books use v for velocity - Newtonian kinetic energy is subject to galelean relativity.

As for a `reasonable` definition for squaring vectors, you're putting more rigueur into it then has been left in; there are a lot of simplifications and conveniences in pedagogacle, and historical, and even practical Newtonian physics. Technically, the dot product is often used (e.g. in computer programs), which is the scalar product of its self in this case, and which does preserve the necessary information for galelean relativity. In fact, there are two very common vector multiplications in classical physics - dot product and cross product, and there are more types of multiplication and associate algebras for tensors, and in fact we have generalized the idea of addition and multiplication for linear algebra in fields.

In the more advanced studies (e.g. Lagrangian Mechanics), it all takes on a whole new light, involving paths of integration and more rigueur. Blablablob (talk) 04:36, 17 September 2008 (UTC)

kinetic energy of light

Would it be correct to say the energy of a photon is its kinetic energy? Thus adding it on this page would be a good thing to do. Comments? Fresheneesz 03:47, 10 April 2006 (UTC)

I thought for photons you use the photoelectric effect equation:

E= hf

where h is Plancks constant and f is the frequency of radiation. This result due to Einsteins Nobel prize winning work.--Light current 03:54, 10 April 2006 (UTC)

I suppose so, what I was getting at is that the photoelectric effect predicts the kinetic energy of a photon. I'm not sure if this is a common or uncommon way of thinking about it, which is why I posted here. Fresheneesz 19:40, 10 April 2006 (UTC)

A photon has energy. I dont think its correct to say this is kinetic energy. After all its travelling at the speed of light and the expression for energy does not include v.

However, if you substitute for f, you get:

E = hc/λ

where c is the speed of light. How you'd get anything like 1/2 mc^2 out of that is beyond me. Unless of course:

:m = 2h/λc

:which may or may not be the case. This relates mass to wavelength using Plancks constant and the speed of light. Physicist opinion required!!! --Light current 21:25, 10 April 2006 (UTC)

Well, I'm not a physicist, but I have this to say: We are definitely not seeking to get something like 1/2 mc^2. The formula 1/2 mv^2 is good only for low speeds, and the speed of photons is definitely not low. Nor do we seek an expression like $m\left({\frac {1}{\sqrt {1-({\frac {v}{c}})^{2}}}}-1\right)$ which is obviously correct only for speeds less than c. So the formula we use to express the enrergy doesn't have much to do with the nature of that energy. But regarding the original quesion - I think it is sort of correct to look at this energy as kinetic energy, but that there are some subtelties of involved. Since we're not really sure, it's best to leave it out for now. -- Meni Rosenfeld (talk) 06:39, 11 April 2006 (UTC)

My physics professor mentioned that the equation E=hf is total energy - which i'm pretty sure means kinetic + potential. But saying that it contains potential energy makes no sense to me, because potential energy cannot be transfered to other object - it must be first converted into kinetic energy. And i'm quite sure photons give all their hf to particles that absorb them. Fresheneesz 03:26, 15 April 2006 (UTC)

I think they do give up all their energy (equal to hf) to absorbing particles. So the total energy of a photon is hf. Just remembered-- they have no kinetic energy because they are massless!--Light current 03:33, 15 April 2006 (UTC)

I doubt their lack of rest mass is reason enough to have no kinetic energy. We're talking about particles moving at speed c. So, if you try to calculate their kinetic energy by the usual $m_{0}(\gamma -1)$, you'll get $0\cdot \infty$ which is indeterminate, not necessarily zero. It is known that the observed mass of a moving particle is greater than its rest mass. In the case of photons, the value of hf / c² is sometimes used to represent their mass, though this doesn't hold in every context. All in all, I do think it is fairly accurate to say that all the photons' energy is kinetic. -- Meni Rosenfeld (talk) 16:57, 15 April 2006 (UTC)

I would have to look it up to be sure. I believe photons do have momentum and this may indicate something. But whether thay have KE I am not prepared to say ATM. I dont see how you can say that all the photons' energy is kinetic in the light(!) of the above. --Light current 17:42, 15 April 2006 (UTC)

Yes, photons definitely do have momentum of hf / c - with all the implications via momentum conservation. BTW, this is probably the main reason its mass is sometimes said to be hf / c^2 - this way, the equation p = mv holds (it holds for every finite speed, why not for c? Actually, there are reasons why not). I fail to see why shouldn't I say that photons' energy is kinetic in light of the above. In any case, I agree completely that since we're not sure, we shouldn't include it without a good reference. -- Meni Rosenfeld (talk) 17:56, 15 April 2006 (UTC)

Well in any case, I've figured that hf does not include potential energy. Of course. It shouldn't, potential energy doesn't exist until it turns into something - the reason its called potential in the first place. So if light's energy isn't kinetic, then it could be in its mass, or its heat, or its light (!). The real point is that it doesn't matter what form its in, energy is energy - there is no real difference between "different" types of energy. However, since we do have lables such as "kinetic" and "thermal" etc, I think it is a basic consensus that any type of energy (other than potential, which really isn't a type) is the result of some form of kinetic motion. And so I think its not entirely unreasonable to say the KE of light is hf, but that saying its kinetic specifically is really a moot point.

As for a source look up "kinetic energy of light" with the quotes - you'll find plenty. http://lqfp.nease.net/new-11/new_page%20107.htm . In any case, if we do plan on introducing hf as KE, we should be careful in explaining why in the article. Fresheneesz 08:00, 19 April 2006 (UTC)

Since the kinetic energy of a massive particle can be given by the equation:

$E_{k}=E_{total}-E_{rest}=\gamma mc^{2}-mc^{2}\$

and the frequency of a massive particle can be given by:

$f=\gamma f_{C}-f_{C}\$

where $f_{C}$ is the Compton frequency.

We can then say that $f$ is the frequency associated with the kinetic energy and $f_{C}$ is the frequency associated with the rest energy, and that for any particle, massless or massive:

$f_{total}=f+f_{C}\$

Since a photon does not have a Compton frequency, it is obvious that the total frequency is equal to the kinetic frequency, and therfore the energy of a photon is purely kinetic energy.

GoldenBoar 03:08, 28 August 2006 (UTC)

if you want to pass your physics tests, photons do not have kinetic energy, they do have energy and they do have momentum. The reason is calculations involving momentum are perfectly reasonable - as in reflection off a light sail; but calculations involving kinetic energy imply a change in velocity - which light cannot do. Instead it is said that the photon is created or destroyed, involving a given energy.

I was originally in here to point out the absence of a good exposition on potential wells and lagrangian mechanics - which would clear up a lot of the confusion I see involving potential energy, work and kinetic energy. Blablablob (talk) 04:51, 17 September 2008 (UTC)

In special relativity, the energy of a photon in a vacuum is entirely kinetic. In a gravitational field or in a dielectric medium, part of it becomes potential energy. JRSpriggs (talk) 08:20, 17 September 2008 (UTC)

Does all energy have to be kinetic?

Well if you want to say that photon KE = hf then I suppose I would agree. But what Fresh' was implying was the possibility that photons have hf plus some KE due to ther motion. I dont think that is correct. Its not standrd terminology to talk of the KE of a photon is it?--Light current 18:02, 15 April 2006 (UTC)

Actually I wasn't implying that : ) Fresheneesz 08:01, 19 April 2006 (UTC)

I don't think that's what he meant, since this is obviously untrue - We all agree that a photon's total energy is hf, the debate was how to call it. And I can't say I'm sure how standard this terminology is. -- Meni Rosenfeld (talk) 19:04, 15 April 2006 (UTC)

Fresh': tell us what you really meant!--Light current 19:12, 15 April 2006 (UTC)

The energy of a photon is called surprisingly enough 'photon energy'.(or phonton quantum energy) [1] Not kinetic energy because its mass cannot be determined independent of its velocity. The rest mass of a photon IS zero.

If the photon travels at light speed, (which it is compelled to do - because it is in fact em radiation) it would seem that its energy can be equated to mc^2 and so its relativistic mass is: m= Eh/c^2 ie its mass depends upon its energy. So its momentum must also depend upon its energy. Does this make any sense?--Light current 19:29, 15 April 2006 (UTC)

Of course. The momentum depends on energy for ordinary particles as well. It is well known that photons have a momentum of E / c, where E = hf is the photon's energy. Assigning a value for (relativistic) mass is trickier, and I think this is usually not done. If done, the value of E / c^2 is consistent with some formulae. Of course, the rest mass is always 0. That's part of the reason why photons must travel at c - otherwise, their total energy would be 0 times a finite quantity, which is 0, and therefore not exist. -- Meni Rosenfeld (talk) 09:24, 16 April 2006 (UTC)

THe only thing I cant quite get is why the energy should be mc^2 rather than 1/2mc^2. Perhaps this is the relativistic factor coming in from the Lorenz transformaion?--Light current 13:40, 16 April 2006 (UTC)

Why would it be 1/2mc^2? Perhaps you're mixing it up with the newtonian formula $E_{k}={\frac {1}{2}}mv^{2}$. This is correct only for $v\approx 0$; it is not correct for positive speeds, definitely not for v=c. It just happens that every particle's total energy (rest + kinetic) is mc^2, where $m={\frac {m_{0}}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}$ is the relativistic mass. It should be possible to work this out from the Lorentz transformation, though I don't recall how at the moment. -- Meni Rosenfeld (talk) 14:31, 16 April 2006 (UTC)

I am mixing it up with the newtonian formula obviously which only applies at low velocities (much less than light). OK. But is there a formula for KE (or just energy) that works for all velocities? Is it still 1/2 (m*)c^2 wher m* is the relativistic mass (not rest mass?)--Light current 19:26, 16 April 2006 (UTC)

A particle's rest energy is $m_{0}c^{2}$; Its rest+kinetic energy is $mc^{2}=m_{0}\gamma c^{2}={\frac {m_{0}c^{2}}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}$; So the kinetic energy is $(m-m_{0})c^{2}=m_{0}(\gamma -1)c^{2}=m_{0}c^{2}\left({\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}-1\right)$. This obviously can't work for v = c, since then m0 = 0 and γ = $\infty$. These should be dealt with on a case-by-case basis. For photons we have E = hf. The total energy will in any case be $mc^{2}$. For photons, the rest energy is 0, so this also gives you the kinetic energy (which perhaps is not commonly called this way). Or perhaps I am missing the point of your question? -- Meni Rosenfeld (talk) 16:38, 17 April 2006 (UTC)

You have answered my question. But Im sure there must be a simple relationship between the newtonian (kinetic) energy and the real (relativistic or total) energy. After all one is exactly twice the other!--Light current 16:45, 17 April 2006 (UTC)

I doubt this has any meaning. Looks like it's just because any expression of energy should involve a mass factor and a square of speed factor. For v=c, the only relevant speed is c so any expression of energy should include $mc^{2}$ - It's just a matter of what is the coefficient. It just happens that for the "correct" expression the coefficient is 1, and for the expression you get by forcing the newtonian formula into a case which is completely opposite the one it was meant for (v=c as opposed to v roughly 0) the coefficient is 1/2. -- Meni Rosenfeld (talk) 17:08, 17 April 2006 (UTC)

Hmmmm! I shall ponder on it.--Light current 17:32, 17 April 2006 (UTC)

Light does not have mass, and does not have kinetic energy; any discussion regarding mc² or mass in any way will lead to no where because light has no mass. The energy in a photon is of an entirely different nature. BTW, light does not have a rest mass either, as it is never at rest. Blablablob (talk) 04:56, 17 September 2008 (UTC)

KE in terms of momentum

i think it would be nice to also include the following equation

$E={\frac {p^{2}}{2m}}$

but i am not sure where it would fit best.

Also in the relativistic section the expansion of

$E={\sqrt {p^{2}c^{2}+m^{2}c^{4}}}-mc^{2}$

to give

$E={\frac {p^{2}}{2m}}-{\frac {1}{2mc^{2}}}\left({\frac {p^{2}}{2m}}\right)^{2}+...$

is quite interesting and useful.

--80.41.16.92 15:11, 18 May 2006 (UTC)

I've added most of the above. I don't think the explicit Taylor series is that important, though. -- Meni Rosenfeld (talk) 17:51, 18 May 2006 (UTC)

thanks. shouldn't the $m_{0}$s just be $m$, as at the top of the relativity section, $m$ is defined as the rest mass. --80.41.16.92 20:11, 18 May 2006 (UTC)

You're right. I've changed the section to use m₀ consistently. -- Meni Rosenfeld (talk) 13:38, 19 May 2006 (UTC)

Making the relativistic equation look more like the classical one

Let us manipulate the relativistic formula for kinetic energy in terms of momentum to make it look more like the classical formula. Begin with:

$E_{k}={\sqrt {p^{2}c^{2}+m^{2}c^{4}}}-mc^{2}$

Move the rest energy to the left side and square to get:

$(E_{k}+mc^{2})^{2}=p^{2}c^{2}+m^{2}c^{4}\!$

Expanding and cancelling the square of the rest energy, we get:

$E_{k}^{2}+2E_{k}mc^{2}=p^{2}c^{2}\!$

Add the square of the kinetic energy to both sides:

$2E_{k}^{2}+2E_{k}mc^{2}=p^{2}c^{2}+E_{k}^{2}\!$

Factor out the kinetic energy on the left side and divide:

$E_{k}={\frac {p^{2}c^{2}+E_{k}^{2}}{2E_{k}+2mc^{2}}}\!$

Divide numerator and denominator by $c^{2}\!$ and rearrange:

$E_{k}={\frac {p^{2}+E_{k}^{2}/c^{2}}{2(m+E_{k}/c^{2})}}\approx {\frac {p^{2}}{2m}}\!$

Repeated substitution of approximations to the kinetic energy into right side this will converge to the true kinetic energy. The denominator on the right side is twice the "relativistic mass" and the numerator is the square of the linear momentum plus the square of something like the time component of the linear momentum (except for the rest energy). Neat huh? JRSpriggs 07:04, 5 October 2006 (UTC)

Quantum mechanical kinetic energy

In quantum wave-mechanics, the expectation value of the kinetic energy, $<{\hat {T}}>$, for a system of electrons described by the wavefunction $\vert \psi >$ is a sum of 1-electron operators (written here in atomic units)

$<{\hat {T}}>=<\psi \vert \sum _{i=1}^{N}-{\frac {1}{2}}\nabla _{i}^{2}\vert \psi >$

where $\nabla _{i}^{2}$ is the Laplacian operator acting upon the coordinates of the i'th electron and the summation runs over all electrons.

(Writing it out in SI units is quite silly, IMO).

The density functional formalism of quantum mechanics requires knowledge of the electron density only, i.e., it formally does not require knowledge of the wavefunction. Given an electron density $\rho (\mathbf {r} )$, the N-electron kinetic energy functional is still unknown; however, for the specific case of a 1-electron system, the kinetic energy can be written as

$T[\rho ]=\int {\frac {1}{8}}{\frac {\nabla \rho (\mathbf {r} )\cdot \nabla \rho (\mathbf {r} )}{\rho (\mathbf {r} )}}d^{3}r$

where $T[\rho ]$ is known as the Weizsacker kinetic energy functional.

Perhaps this is beyond the intended scope of this article?

DrF 19:03, 22 May 2006 (UTC)

It is perfectly normal for an article to start out with elementary concepts, and continue towards more sophisticated ones. Adding a section "Kinetic energy in quantum mechanics" with this content would probably be appropriate. -- Meni Rosenfeld (talk) 15:17, 23 May 2006 (UTC)

Vandilization...?

The xmen section seems... inappropriate I think..

It's borderline. It's okay to have articles discuss appearances of their subject in popular culture and fiction, but this particular one doesn't seem to make the cut. I've removed it. -- Meni Rosenfeld (talk) 16:45, 28 May 2006 (UTC)

Simple formula for kinetic energy in the introduction

Take a look at Enormousdude's and my recent diffs, which I explained in the comment. Any further editing of this bit should be discussed here first. -- Meni Rosenfeld (talk) 08:43, 6 August 2006 (UTC)

Mass is not a form of energy

First, I am glad you have decided to discuss the matter in a civilized fashion, instead of just rudely making the same edit every time without explanation. One other thing - please sign your comments in talk pages by typing ~~~~.

I am having trouble understanding your statement "Mass is energy, but it is not a form of energy." Do you mean that mass and energy are the same thing? This doesn't seem likely. Or do you mean that mass is a type of energy? That is exactly the same as saying that it is a form of energy. What about antimatter annihilation? This is a process in which energy is converted from one form (the mass of a particle and anti-particle) is converted to another form (photons, aka EM radiation) and can then be converted to other forms (kinetic, etc.) Does it not serve as an indication that mass is, indeed, a form of energy? -- Meni Rosenfeld (talk) 06:30, 22 September 2006 (UTC)

Mass can be converted into energy, and vice versa, according to E=mc².. For example, this happens in a process like $e^{+}e^{-}\to \gamma \gamma$ (or the rather unlikely reverse). From there, it's a matter of definition whether mass is a form of energy; but that is how physicists define it, because it allows conservation of energy to be a useful concept in modern physics. If you claimed mass was not a form of energy, then there would simply be a new conserved quantity that was something like mass plus energy, and that would be silly. -- SCZenz 07:18, 22 September 2006 (UTC)

Mass is a form of energy if rest mass or invariant mass is the only correct notion of mass. In the fission of U-235, the sum of rest masses of the fission product is 3.915 X 10^-25 kg. The rest mass sum of U-235 and a neutron is 3.918 X 10^-25 kg. As you have seen, some rest mass has been "converted" into 2.69 X 10^-11 J of energy. In the annihilation of matter and antimatter, suppose both particle remain at rest relative to each other. They have been converted into photon with equal energy which can be calculated with $E=mc^{2}$.

It will be a totally different story if relativistic mass is not an outdated notion of mass and still valid to be a true notion of mass. In my opinion, I suppose relativistic mass is the correct notion of mass although it may be highly disagreed by many modern physicists. Invariant mass is still useful. Both types of mass can be calculated through Lorentz Transformation. I then suppose the amount of mass is the amount of energy. Amount of energy is the amount of mass. An object's inertia depends on its energy content. Every types of energy posses mass at the same time. Some people could simply say photon has no effective mass since we could never measure it when it remains at rest. When we assume an object as a system of matter and energy, the rest mass of the whole system can be measured when the whole system remains at rest relative to the observer. In the system, particles need not to be remain at rest relative to the whole system. They can move around freely in the system. As I suppose energy posses mass, particle with kinetic energy and potential energy will increase its relativistic mass but not its rest mass. When the system remains close to the outside world, the rest mass of the system remains constant. The rest mass of the system is exactly the sum of relativistic mass of the particles in it. Photon in this term possesses mass too because it will surely have certain amount of energy. Absorbsion and emission of photons will affect the rest mass of the system if the system is not close. You may not measure the mass of kinetic energy or heat directly but both of them contributes to the rest mass of a system.

That's what I mean. Mass needs not to be a form of energy. Indeed, all form of energy is mass. Kinetic energy, potential energy, photon and other forms is mass. Vice versa. The mass is the energy that can do work. Annihilation of matter and antimatter, if we calculate the sum of relativistic masses, it remains constant as long as the whole system is close. It is nor created or destoyed. As I regard relativistic mass as mass, there is no change in mass.

We cannot say the energy that carried by photon is kinetic energy. Take a look to this equation: $E=(\gamma -1)m_{0}c^{2}$

$\gamma$will be ill-defined when the velocity is the velocity of the light in vacuum. This subsequently makes the whole equation invalid. What does this situation means? Opposed to Newtonian Mechanics, the matter possesses energy as their mass when it remains at rest relative to a inertia frame of reference. The matter will get extra energy namely kinetic energy if it move at a certain velocity relative to the same inertia frame of reference. We should be careful it may remain at rest relative to another inertia frame of reference. So, the kinetic energy the matter gets in this case is frame-dependent. As it gets extra energy, it will get extra mass since energy and mass is equivalent and identical. The velocity of light however is frame-independent. So, how can you define the kinetic energy that the photon may possess? To an inertia frame of reference, what is the difference of its rest mass and its moving mass since photon never remains at rest to all inertia frame of reference! Don't crack your mind! Photon simply carries its own form of energy like other waves do. Thljcl 05:28, 23 September 2006 (UTC)

It looks like there are two possible views. According to one, only rest mass is taken into account, and then mass is one of many forms of energy. According to another, relativistic mass is considered, and then, arguably, mass and energy are the same thing. But it is still correct that mass is a form of energy - it is just the only form. So, it is always true that mass is a form of energy, and whether other forms of energy exist depend on our interpretation. I therefore see no reason to exclude mass from the list of forms of energy. Perhaps, as a compromise, we can replace "mass" in the list with "rest mass" (that should satisfy you - the total relativistic energy of an object is its rest mass energy, plus its kinetic energy, plus others). -- Meni Rosenfeld (talk) 07:06, 23 September 2006 (UTC)

We should avoid overusing the term rest-mass where no ambiguity exists; it is not in line with modern physics usage, because "mass" is almost always used only to mean invariant mass. No ambiguity exists here because Thljcl's quibble is wrong; or, more importantly for Wikipedia, it is not reflected anywhere in the literature. Even if there is no such thing as kinetic energy, relativistic mass (and rest mass) are still forms of energy. SCZenz 20:51, 23 September 2006 (UTC)

My quibble is wrong? "Mass" is always used only to mean invariant mass? What about the inertial mass which can be deduced from the equation $F=ma$? In the low-speed approximation where we do not need to consider about relativistic effect, inertial mass that is defined is the measurement of the resistance to change its velocity relative to an inertial frame of reference. The definition of gravitational mass can be deduced from the equation $|F|={GM_{A}M_{B} \over |r_{AB}|^{2}}$ where object A and object B have gravitational masses M_A and M_B respectively. Inertial mass is equivalent to gravitational mass according to Equivalence Principle. Before the introducing of relativity, inertial mass and gravitational mass may be the only correct notions of mass. Mass has been assumed to be invariant to all inertial frame of reference. This invariant mass is basically true when the velocity of an object relative to an inertial frame of reference is far lower than the speed of light in vacuum. From $F=ma$, F that is given to an object is proportional to the acceleration of the object, provided the mass of the object invariant. However, we could also interpret this equation as mass of an object is inversely proportional to the acceleration of the object, provided a constant force is provided. Since the acceleration of an object decrease steadily when a constant force continually to be provided relative to an inertial frame of reference, this prohibits an object to exceed the speed of light in vacuum. Just as the definition of inertial mass above, inertial mass apparently increase because the decrease of acceleration with the same force provided. This implies that inertial mass is actually relativistic mass and not invariant mass. We should know that relativistic effect could not be noticed in day-to-day life. If invariant mass which is frame-independent being used as the only true definition of mass, mass of an object has nothing to do with the resistance to change the velocity of the object relative to an inertial frame of reference. Also, as I have mentioned, equivalence of mass and energy is only valid if mass that is defined is relativistic mass. You can refer E=mc² and Mass-energy equivalence to know more about the equivalence of mass and energy. If you accept relativistic mass as the definiton of mass, invariant mass or rest mass of an object is exactly the mass of an object divided by $\gamma$ where $\gamma$ stands for Lorentz Factor. Rest mass will be exactly same with relativistic mass when an object remains at rest relative to an inertial frame of reference. Thljcl 05:08, 24 September 2006 (UTC)

KE by integration of v.dp

I think that this equation:

$E_{k}=\int \mathbf {F} \cdot \mathrm {d} \mathbf {s} =\int \mathbf {v} \cdot \mathrm {d} \mathbf {p}$

would be easier to understand if we also included Newton's Second Law:

${\vec {F}}={\mathrm {d} {\vec {p}} \over \mathrm {d} t}$.

Yes?—anskas 23:22, 9 December 2006 (UTC)

This might be a stupid question

Hello,

In the section marked Definition, where does the 1/2 come from when you have v.d(mv) = m/2 d(v.v), Shouldn't the 1/2 appear up when evaluating the integral of m.v? User A1 05:32, 9 January 2007 (UTC)

Never Mind. d(v.v) = dv^2 = 2v , so d(1/2 (v.v)) = v

Applying the product rule: $d(v\cdot v)=(dv)\cdot v+v\cdot (dv)=2v\cdot dv$ JRSpriggs 10:26, 9 January 2007 (UTC)

An inadequate definition of kinetic energy, or the frustrated student

• Physics student: I have a pistol out in space and it fires a bullet. Or even better, since I want this to be elastic, an atomic nucleus which fires off an alpha particle. Now I want to know about kinetic energy! All the potential energy has been converted to kinetic, right?
• Prof Yes.
• Physics student Well, don't keep me in suspense. How do I calculate it, if I don't know the potential, but only the velocities? The nuclear potential lost is the kinetic energy gained. I read this here Wiki article, and it gave me a lot of equations with velocities in them, but none I can use. Okay, look---If I stand on the decayed nucleus and watch the alpha go away, can I use any of these equations to calculate it's "kinetic energy"?
• Prof: Relative to what? The nucleus? Sure.
• Student: Will that represent all the energy transformed?
• Prof: No.
• Student: How about if I do it the other way, and stand on the alpha, and calculate the kinetic energy of the nucleus?
• Prof: No, that won't work, either. You see, both particles have some of the kinetic energy, sort of. But the answer is different depending on which you pick and where you are.
• Student: Well, then how about if I take those numbers for each system and just sum them up: say kinetic of bullet relative to gun, and gun relative to bullet? Or Alpha relative to nucleus plus vice versa?
• Prof: No. THAT won't work. Won't even work with Newton, let alone Einstein.
• Student: Well, you've got me, then. Where should I go, what reference do I use, to get the right answer from the velocities, about these kinetic energies? The article defines kinetic energy so it never talks about that! I kid you not! It just says kinetic energy exists as work to accelerate an object to a velocity. Not where you go to figure it out from velocities so you get the right conservation answer.
• Prof: Well, the answer is for this problem, you need to be in some reference frame that isn't at the alpha or the nucleus. Or at the gun or the bullet. Depending on their masses and velocities. You see, the kinetic energy isn't really IN the bullet or the pistol. The proper amount of it, is only located for the observer who isn't standing still with regard to either one of them.
• Student: Where??
• Prof: We don't go into that. It's complicated.
• Student: But, but, I see all these special relativity integral calculus thingies here in this Wiki! More complicated than THAT?
• Prof: No.
• Student: Well, then what gives?
• Prof: Badly written article. Committee, probably.
• Student: Why don't you fix it, then?
• Prof: Seems if you do that, people who want to "simplify" it, just keep reverting it to the way it was, which as you note, is REALLY complicated.
• Student: Come again?
• Prof: Forget it. Sit down, and I'll show you the way you can get the answer. And let Wikipedia go ahead and continue to run in incomprehensible circles. SBHarris 13:38, 2 February 2007 (UTC)

I see a lot of revert warring going on between you and JRSpriggs, but this is the first attempt of communication I see. Perhaps, instead of writing Wikipedia off, it is better to use it properly (discussing changes, requesting for comment if that doesn't work, etc.)? -- Meni Rosenfeld (talk) 13:33, 3 February 2007 (UTC)

I'm willing. Springgs has called my edits confusing and irresponsible, but I see no math and no references from him to back his case. I have books on basic and advanced dry old kinematics, the Feynman lectures, Taylor and Wheeler, MTW, and so on. We can start anywhere. The kinetic energy which results from potential energy converted from the potential in a system is not (in general) "co-located" in any given object, but in fact, isn't located *anywhere.* That does not mean it's now somehow arbitrary, since energy is conserved so long as you don't switch frames. The total energy (including the kinetic energy) you get can be calculated from an inertial frame (the COM frame) but that frame usually doesn't correspond with the motion of any particular object (exceptions are where one object vastly out-masses the other). But it is wrong to say such kinetic energy is entirely frame dependent, as though it was now some kind of relative thing with lessreal existence than the potential energy it came from. It continues to have a quantitative existence. You merely need to go to the proper frame to find it. This article does not say anything like this, but it's an important concept. Now, Spriggs, where would you like to begin the math. You're going to be teaching me physics formally, now, remember. Otherwise I'm going to be embarrassing you by teaching YOU physics. A lot. You don't just get to sidestep the whole question by making revisions with rude edit summaries. Continue to do that with no attempt to discuss the physics here, and we'll have a 3R war, and that means ArbCom. SBHarris 14:05, 3 February 2007 (UTC)

Doesn't 3R usually mean just a block from an admin, with ArbCom being for more serious issues like rogue admins and such? In any case, you would probably both be better off by bringing this up on Wikipedia talk:WikiProject Physics. -- Meni Rosenfeld (talk) 22:00, 3 February 2007 (UTC)

According to WP:3RR, "An editor must not perform more than three reversions, in whole or in part, on a single page within a 24 hour period.". I have no intention of reverting Sbharris more than once per day on this article. In fact, I cannot remember every reverting any article more than twice in an editing session (excepting clear vandalism which is excluded from the count). We have already talked a little on my talk page, see User talk:JRSpriggs#Kinetic energy. However, I feel that trying to talk to him is futile and I do not have enough time to waste it on educating fools. JRSpriggs 11:12, 4 February 2007 (UTC)

• Well, that was uncivil. Even worse, uncalled for and unreferenceable. As it is, this article ends with a statement that is at best in need of long qualification, and at worst is simply wrong (the kinetic energy is colocated with the object and contributes to its gravitational field). And then you've accused ME of confusing the student! You might as well say that "The velocity of an object is colocated with the object". How helpful is that? Does that make velocity a property of objects? If I catch up to an object which I think has some kinetic energy, where does that energy go? In what sense is it real? In systems of moving objects, some system kinetic energy can be got rid of by choosing a different reference frame. But very often, some cannot. Some kinetic energies of objects in a system contribute to the invariant mass of the system, and some do not. But the student will not discover this from reading this article. SBHarris 15:56, 4 February 2007 (UTC)

I think that your statement "Otherwise I'm going to be embarrassing you by teaching YOU physics. A lot. You don't just get to sidestep the whole question by making revisions with rude edit summaries. Continue to do that with no attempt to discuss the physics here, and we'll have a 3R war, and that means ArbCom." is both uncivil and threatening. JRSpriggs 07:04, 5 February 2007 (UTC)

This has nothing to do with what you were talking about, but was it necessary to swear? Weirdy 06:32, 20 February 2007 (UTC)

I changed the title of this section. I hope that satisfies you, Weirdy. JRSpriggs 07:50, 20 February 2007 (UTC)

Four times the distance to stop

Hello,

The article suggests that a car traveling at 2v would require four times the distance to stop as a car traveling at v. However I don't think that this can be stated so simply as the work done by the brakes when traveling at v is not the same as at 2v, surely? Would it not be more correct to say that four times the work needs to be done? User A1 00:40, 11 February 2007 (UTC)

I think it is assumed that the brakes are applied in such a way as to exert a certain maximum tolerable braking force.JRSpriggs 06:12, 12 February 2007 (UTC)

Well yes..WolfKeeper 06:35, 12 February 2007 (UTC)

If the force was more than that, the driver would experience excessive discomfort or the brakes would evolve heat so rapidly that they would fail. Wind and rolling resistance are usually greater at higher speeds, but that is a small effect by comparison. So for low speed at least, this is a good approximation. (By the way the work is done ON the brakes, not BY the brakes.) JRSpriggs 06:12, 12 February 2007 (UTC)

It's simpler than that. Brakes can only exert a certain braking force, before the tyre skids, and that force is independent of speed (due to the nature of friction between the tyre and the road). Therefore the decelleration is independent of speed. It takes twice as long to stop at twice the initial speed (since you have twice the speed to lose), and the average speed is twice as high. So that's 4x the distance. Alternatively from the energy point of view, energy is force times distance, and you have 4x the energy to lose, so your stopping distance must be four times higher (since brakes supply a constant deceleration force). Actually, it's not incorrect to talk about the work done by the brakes in stopping a car.WolfKeeper 06:35, 12 February 2007 (UTC)

Good point about the static friction of the tyre,once you have the situation where the brakes exceed the wheel-road force limit then you have an entirely different problem. This would make it a fair comment. Thanks User A1 12:46, 12 February 2007 (UTC)

All these arguments sound plausible, but does anyone have a reference to hard data obtained using real cars? Maybe 4x the distance is a good approximation, but the real world tends to be more complicated than expected. :-) --Itub 09:38, 19 June 2007 (UTC)

I found an interesting article that uses some real data to come up with a third degree polynomial model. The linear term is due to the reaction time of the driver (it could be dropped if you are just measuring the physical breaking distance, rather that the actual stopping distance). The square term is due to the justifications given above. And the cubic term is attributed to a variation of the breaking force as a function of velocity. See D. A. Lawson and J. H. Tabor. Stopping distances: an excellent example of empirical modelling. Teaching Mathematics and its Applications 2001, 20, 66-74. doi:10.1093/teamat/20.2.66. --Itub 09:58, 19 June 2007 (UTC)

missing step?

It seems that a factor 1/2 is suddenly introduced to these calculations with no explanation:

$\mathbf {F} \cdot d\mathbf {x} =\mathbf {F} \cdot \mathbf {v} dt={\frac {d\mathbf {p} }{dt}}\cdot \mathbf {v} dt=\mathbf {v} \cdot d\mathbf {p} =\mathbf {v} \cdot dm\mathbf {v} ={\frac {m}{2}}d(\mathbf {v} \cdot \mathbf {v} )={\frac {m}{2}}dv^{2}=d({\frac {mv^{2}}{2}})$

I'm no expert with physics but i fail to see how

$\mathbf {v} \cdot dm\mathbf {v} ={\frac {m}{2}}d(\mathbf {v} \cdot \mathbf {v} )$

Even if the answer can be considered trivial to some, i'm sure i'm not the only one confused about it. And since the whole point is to show how :${\frac {mv^{2}}{2}}$ is deduced, i think some sort of explanation is needed. Otherwise people (like me) won't find it very useful.

86.52.162.3 17:44, 18 February 2007 (UTC)

Hi, i had the same problem here... See above help by JRSpriggs, 9 Jan. User A1 06:21, 19 February 2007 (UTC)

Yes. As I said at Talk:Kinetic energy#This might be a stupid question, applying the product rule: $d(v\cdot v)=(dv)\cdot v+v\cdot (dv)=2v\cdot dv.$ If you then divide by two and turn it around, you get $v\cdot dv={\frac {1}{2}}d(v\cdot v).$ OK? JRSpriggs 09:23, 19 February 2007 (UTC)

How about dv^2/dv = 2v, therefore dv^2 = 2vdv 81.152.87.111 (talk) 01:14, 13 September 2008 (UTC)

Heat is not kinetic energy

I have deleted a paragraph that implied heat is kinetic energy. It also implied that potential energy is related to position. A brief consideration of chemical potential energy in a battery shows this to be untrue. JMcC 10:31, 10 May 2007 (UTC)

The paragraph in question said

Alternately kinetic energy can be understood as all energy that is shown through motion. This includes both heat energy (the motion of molecules), and the motion of macroscopic and microscopic objects. By contrast potential energy can be understood as all energy that results from relative position. For example gravitational energy is created by relative position "above" a massive body, such as the earth.

Yes, I think the division specified by this paragraph is improperly drawn. Not necessarily for the reasons which you state, but because the energy of the electromagnetic field

$U={\frac {\varepsilon _{0}}{2}}\mathbf {E} ^{2}+{\frac {1}{2\mu _{0}}}\mathbf {B} ^{2}$

cannot be divided so easily between kinetic and potential. JRSpriggs 07:06, 11 May 2007 (UTC)

Part of heat is kinetic energy. In an ideal gas, ALL of the heat is kinetic energy. In most other objects (and certainly in solids and liquids), about half the heat is kinetic energy, and the other half is potential energy. Why not say it like that? SBHarris 02:52, 19 June 2007 (UTC)

Location of kinetic energy??

Somebody seems to think that kinetic energy is located in space, since they keep insisting that kinetic energy of objects is "co-located" with them (which means in the same physical place as they are).

If you consider the kinetic energy (heat) present in a container of hot gas, however, you'll see this isn't so. That kinetic energy contributes to the mass of the gas and certainly to its gravitational field, but it has no location. An observer moving with any given gas molecule will see the total kinetic energy of the system located everywhere BUT with his particular molecule. Moreover, that energy will generally be quite different than the minimal kinetic energy of the system, which is that in the center-of-mass frame, which is the ONLY kinetic energy which contributes to the invariant mass of the system (this contribution of kinetic energy, and ONLY this contribution, is frame-independent). From any given gas molecule's perspective, the whole system will be moving quite fast in the opposite direction, and its kinetic energy will be much greater than the system COM kinetic energy.

In summary, kinetic energy is a system property, and has no location. If you have object A and B moving away from each other, where is the kinetic energy located? From A, it's all in B. Seen from B, it's all in A. What is the "truth"? SBHarris 03:01, 19 June 2007 (UTC)

Effectively, you are saying that because kinetic energy is not invariant that it is unphysical and has no location. That does not follow. To A, the location of the energy is at B. To B, it is at A. But they are talking about different quantities. Each is correct with respect to what he is talking about. It is all relative. Read that section again carefully. JRSpriggs 08:30, 19 June 2007 (UTC)

I think that in Newtonian mechanics, kinetic energy is not colocated with the object. I'm not an expert in General Relativity, but I know that in General Relativity an accelerated object has an increased effective mass. So it seems to me that in General Relativity the energy is colocated with the object (as in E=mc²), but is still frame dependent :-).WolfKeeper 03:50, 20 June 2007 (UTC)

To Sbharris: Let me add a little clarification to what I said earlier. Quantities which are not invariant can be made into invariants by internalizing their context. For example, one can take the inner product of the time component of a vector (which depends on one's choice of reference frame) with another (unit) vector which defines the time direction of the reference frame. The result is a scalar invariant can be understood as the physical meaning of that time component, that is, it is the part of the vector quantity which points in the time direction specified.

Regarding the gravitational field, the field (or equivalent curvature) is itself a directional entity and the kinetic energy may be regarded as affecting it by changing its direction rather than its magnitude.

To Wolfkeeper: In general relativity, energy warps space-time so it must have a location. In special relativity, that is not a factor. However, energy does contribute to what I personally call rapidity momentum (similar to angular momentum, but associated with boosts rather than rotations) which is not usually referred to as such but can be considered to be related to the motion of the center of mass of a composite system. So even in SR, energy has a location. In Newtonian physics, a similar quantity can be defined. Conclusion, all energy has location including kinetic energy. JRSpriggs 06:00, 20 June 2007 (UTC)

No, in Newtonian mechanics it makes no sense to do that; and I don't agree that a similar quantity can be usefully defined there. These 'boosts' seem to be OR, and if you are basing your edits on that, then they will be removed.WolfKeeper 10:12, 20 June 2007 (UTC)

I was basing it on general relativity since that is the most comprehensive theory available. The use of the word "boost" for Lorentz transformations is common, not OR. See Lorentz transformation#General boosts. JRSpriggs 11:06, 20 June 2007 (UTC)

• REBUTT: Okay, this is my penalty for talking about the two different kinds of kinetic energy, which really relate to the two different definitions of mass we've been arguing so long about (relativistic mass and invariant mass). One kind of kinetic energy is sort of a "relativistic" kinetic energy due to its being relative to the observer. It's the only kinetic energy you find as a property of single massive particles, and it's totally frame-dependent (not invariant). You can make it go entirely away by going to the rest frame of the (massive) particle. The OTHER kind of kinetic energy, however, is a property of particle systems (like 2 photons moving in different directions, or two massive particles also which move with relation to each other). This kinetic energy has a part which is frame-dependent, but ALSO a part which is frame-independent, or invariant. The invariant part is the system kinetic energy contribution to the invariant mass of a composite system which still contains kinetic energy in its CM inertial frame. Like the heat enegy of hot ideal gas in a can, say, or the kinetic energy of quarks moving around inside a resting proton, helping to define its "rest" mass. The CM (or COM) frame is the frame which MINIMIZES this sort of system kinetic energy, but often cannot make it go away completely. This invariant kinetic energy (what which you see as a KE contribution to system "rest mass" in the CM frame) is (of course) invariant. All inertial observers measure its contribution to the system invariant mass, to be the same. So our generalities about kinetic energy (as in this article) DO NOT APPLY TO IT.

  Now, to be sure, in addition, any system also will have a component of non-invariant KE, because you can always increase the total (relativistic) energy of a system, by flying by it faster. However, (again) while that extra kinetic energy you get by moving to a faster moving frame is not invariant, that fact does NOT mean that ALL system KE IS frame/observer dependent. Often, some is not. In heated systems, some fraction of system KE is always invariant. Whatever heat you are able (in theory) to weigh in a system on a scale (meaning that you're in the system's CM frame), is in large fraction this invariant KE (for an ideal gas, it is ALL of heat, but for solids, it is only about half of it, with the rest potential E).

  Okay, now that we have definitions out of the way, I'm going to argue that both of these types of KE (the invariant and the noninvariant one) are nonlocal (undefinable locally or at a point). Clearly, the invarient component of system KE (i.e., the total additive system KE that appears in the CM frame) is non-local, because-- well-- it has no location! But since it IS also clearly invariant, it is therefore physical and real and quantitable. It generates gravity and momentum and weight. It's the same quantity when seen from any particle in the system, and by any observer, but yet it's always distributed in a different fashion, and always located SOMEWHERE ELSE (i.e., it resides entirely in the motions of all the OTHER system particles, when seen from the rest frame of any given particle which by definition is now at rest-- though admitedly this invariant KE is no longer the simple sum of the other particle KE's. Moveover, the argument (given above by JRSpriggs above), that we're really measuring something different when we go to various particle frames and measure the system (invariant) KE (which is the simple sum of KEs in the CM frame, though often not in the other frames), is wrong. Since the quantity of invariant mass due to KE, and the fraction of the TOTAL invariant mass of the system due to its invariant KE (however you measure THAT) never varies no matter what frame you go to, OBVIOUSLY you're measuring the same thing in all these frames! (Otherwise it gets pretty hard to say why you keep coming up with same invariant number). This interesting quantity keeps moving around on you, always appearing after calculating the motions of particles, but never distributed the same, when seen from different particles. If you insist it resides in the motions of particles, they are different motions, depending on which particles you pick, but interestingly, after you get done, you always get the same number.

  Physicists, BTW, have the same frustrating experience trying to point to the location of the energy in a GR gravitational wave. Clearly the energy exists (we see it tapped out of pulsar binaries as they orbit each other), but it has no LOCATION in space (no quantitative number for its energy density at any POINT) because if it did, you could always use the equivalence principle to pick an accelerated frame to make it disappear THERE at that that point (see MTW for a nice summary on this point, and why trying to find gravity wave E density is a wrong-headed enterprise). For system KE in special relativity, the process and reasoning are the same, but even simpler: you can always use the principle of galilean relativity to make the LOCAL KE of any part of the system go away, by picking the proper inertial frame at rest WRT this part of the system, and yet there is an invariant KE there anyway, which is as real as any property of any system gets. And one which can only be minimized by proper choice of inertial frame (the CM frame) but never made to disappear entirely-- thus demonstrating its physicality (for more, see below).

  Now, what about the relativistic KE-- the one you can make into any arbitrary number for any given particle, just by choosing your inertial frame? Well, if invariant KE has no location in systems, I can't see that the considerably less "real" or physical relativistic KE has any location. Feel free to argue, but if you accept the point for invariant KE, you certainly have to accept it for the variant (observer dependent) types.

  Now, do all these KE energies interact with gravity? Well, it's very interesting! Much the same way as the two kinds of masses. Relativistic KE affects the G field, but only by distorting it (boosting your observer compresses space and causes a G field lobe to move out transversely in the observer's direction, thus increasing the field strength there). What happens is that the field increases in some directions, but decreases in others, as it is squashed by relativistic foreshortening. It's quite the same process as choosing different inertial frames to look at the field lines from a charge-- frames in motion WRT the charge give an "extra" E field (called a magnetic field), but the integrated charge as source though the whole volume, remains the same. For "mass charge," (ie G field) rather the same thing happens in moving frames-- an object passing by you generates an extra gravitomagnetic field, due to the spacial compression of the longitudinal component of its G field and extension fo the transverse one (toward the observer), but the total field flux in all directions is invariant, just as happens for charge. It is in this sense that the increase in G field due to observer motion is not objectively "real"-- and it's the underlying reason why you can't pass an object fast enough to make its G field so strong that it appears to be a black hole. An electron can go by so fast that its "relativistic mass" can can be 10 times that of our sun, and still the size of an electron, but it will never get to blackholeness that way.

  Now, what about the invariant component of a system's KE-- such as the heat in a can of hot ideal gas? Different story! *That* increases the system's invariant mass, and thus its minimal G field, for ALL observers. And it WOULD contribute to the system becoming a black hole. The invariant KE of the quarks in baryons is a large part of their (rest) mass, and no doubt a large part of the mass of neutron stars, which are made of squashed baryons. And I have little doubt it plays a significant part in the generation of the G field necessary for the compression of stars into black holes. Again, all this is consistant with this KE being a much more "real" and objective physical quantity, due to being an invariant 4-vector, than is the fly-by generated, single-particle, "total-relativistic-energy minus rest-energy" type of KE.

  Interestingly, the question of which part of the KE in a particle or system is "real," and which is "illusionary," (or observer relative) comes up in the question of which part of the energy you can extract to turn into other forms of energy. For a single particle, it turns out you can't get ANY of its kinetic energy for work. So you play around all you like with your observer frame and none of it does you any good. To get any of the KE of a particle for work or to produce other particles (as in pair production), you must introduce a second particle, which means now you have a 2-particle system. And guess which part of this system KE is available to do things like generate particle-antiparticle pairs, and so on? Yep, you guessed it. Only the system KE that is seen in the CM frame-- the invariant part of the system KE I've been calling objectively "real." For example, if you bang one proton against another which is at rest, it isn't the KE of the incoming proton in the lab frame that is important to particle production. It's only the system KE appearing in the CM frame (i.e., that summed KE of both particles appearing in the frame moving at half the velocity of the incoming proton) which is available to make new stuff. And that's NOT the KE of the proton in the lab frame, but a lot less. That is the sense that only this KE is "real". But of course, this measure of KE does not reside in the incoming proton, or in the proton it hits, because it can't be calculated from either of their velocities in either of their frames (in Newtonian or SR physics). What's your thought on where this manifestly real and available energy DOES reside?? SBHarris 22:00, 20 June 2007 (UTC)

To Wolfkeeper: After thinking about it, I realize that you are correct about Newtonian physics. I usually think in terms of relativity, and I jumped too quickly to the conclusion that the same argument would work classically. So classical physics lacks the ability to locate energy. But I still maintain that all energy has location in both SR and GR.

To Sbharris: It would help if you could try to be brief in your statements and bullet-ize them. I think that you are trying to take the distinction between different kinds of energy (mass versus potential versus kinetic; variant kinetic versus invariant kinetic (heat)) much too seriously. Energy cannot be subdivided so easily. The formulas for calculating it are not always simple linear combinations of terms.

However, one must distinguish gravity from other force-fields. The physically significant quantities in other force-fields are tensors (or spinors). But in gravity, they do not have to be. Gravitational force (Christoffel symbol) and gravitational energy have more complicated transformation laws (not the simple local homogeneous linear transformations of tensors). As a result, it is even more difficult to talk in a sensible way about gravitational energy than about other forms of energy.

If you want to understand the physical world, you need to be more consistent about only comparing quantities which are defined with respect to the same coordinate system. JRSpriggs 02:31, 21 June 2007 (UTC)

• Disagree. If you do that, you're going to be regarding time as a universal, which ticks at the same rate, forever. If you really want to understand the physical world, you need to pay attention to quantities which are universal-- things that are the same no matter what observer happens to view them. The 3-D length of a tree, for example, looks very different to different observers-- what is the deep underlying reality? It would be silly to give up on this question too early, and simply say there is no such thing as the length of a tree, because every observer sees something different. The 3-D "reality" of the length of the tree is not what various people see, but what you get with a tape measure. But even this is in the limit of low velocity. Extend this to 4 dimentions and the corresponding realities (at any velocities) are 4-D objects which are invariant, and even length itself is not one of these, but merely a shadow of it, just as tree-shadow lengths are not the 3-D length of a tree but only a component of a deeper invariant.

  I don't know why you think I need bullet points. I write clearly enough: use my paragraph divisions.

  Executive summary of my previous point on energy is that the total kinetic energy of moving quarks in a proton, or molecules in a can of hot gas, contribute to the object's "rest mass," but that kinetic energy has no location (either in classical or relativistic physics). It is a real thing, and quantifiable by weight or any other measure of mass, but it is distributed, with no particular location. Some kinds of energy simply do NOT have a location in SR, and kinetic energy contained in systems is one of these. So we cannot blather on about how it is "colocated" anywhere. I can quote from textbooks on SR if you like: The physics FAQ still stays that Taylor and Wheeler are the best introduction to SR out there [2]. T&W are very explicit that heat and the internal kinetic energy that makes for heat, are system properties with no location. Here you are, from page 225 of Spacetime Physics:

  No one with any detective instincts will rest content with the vague thought that heat has mass. Where within our stuck together wads of chewing gum or Rumford's barrel of water or Braginski's quartz pellet [previously used examples] is the mass located? In random motions of the atoms? Nonsense. Each atom has mass, yes. But does an atom aquire additional mass by virtue of any motion? Does motion have mass? No. Absolutely not. Then where, and in what form, does the extra mass reside? Answer: not in any part, but in the system.

  Heat resides not in the particles individually, but in the system of particles. Heat arrises not from the motion of one particle, but from the relative motions of two or more particles. Heat is a system property [emphasis in original].

  So, there you are. T&W go on about this quantitatively for a whole chapter, but the message is the same. If you want to keep at this, then quote your own textbook on the subject. Where is the extra mass of a two particle system in motion, located? You can "dis" Taylor and Wheeler, but how many texts have YOU written on this subject? Why are you convinced that you know more about the professors of physics who are explaining special relativity to the student, as quantitatively as they can? When these profs say "no," "absolutely not" and "nonsense" to the idea that heat kinetic energy has location in space in the theory of SR--- how much more specific can they BE? SBHarris 23:12, 21 June 2007 (UTC)

As an example of what I meant by "Energy cannot be subdivided so easily.", see Kepler problem in general relativity where it gives the energy of a massive particle near a star as

$E=mc^{2}\left(1-{\frac {r_{s}}{r}}\right){\frac {dt}{d\tau }}$.

How would you divide this into your different types of energy?

Regarding your professors and authors, there are tens of thousands (if not more) of people in the world who claim to be authorities on this subject. Many of them are either incompetent or cranks. I have never seen a book which did not contain some errors. So frankly, I am not impressed by your appeal to authority. And your argument from incredulity (saying that it cannot have a location simply because you do not believe that it can have a location) is also logically invalid. You have not refuted my arguments that energy does have a location. JRSpriggs 10:24, 22 June 2007 (UTC)

If I may interrupt this debate for a moment, just to discuss the procedure rather than the physics... the very fact that this level of discussion is going on means that there is something wrong from the point of view of Wikipedia policies. Unfortunately, Wikipedia does require an appeal to authority, due to its policies regarding original research and verifiability. It is of course possible that you are correct regarding the physics; I frankly don't know because I don't know much about relativity, but that doesn't matter. What matters is what can you cite about it. If X reputable physicist says kinetic energy has a location, we can cite him. If Y reputable physicist says it doesn't, we can cite him too. It is quite possible that respected physicists disagree on this topic. But there's no point with trying to solve the scientific problem here in the talk page by recurring to complicated arguments and equations. Unless y'all are having the argument just for fun and not to decide what goes into the article (but in that case this may not be the best place to have the discussion). --Itub 11:26, 22 June 2007 (UTC)

The way that the wikipedia works is WP:NPOV and WP:RS. You need to come up with a notable example of a reliable reference that proves your case. Rightly or wrongly SBHarris's position seems to be the one that is notable in the literature.WolfKeeper 12:14, 22 June 2007 (UTC)

This is one of those things which is taken for granted and thus not talked about by the real experts usually. Can you find a quotation from the head of the National Weather Service saying that "water is wet"? Can you find a quotation from Einstein saying that space-time is curved (I had to look for such a quote once due to being challenged by an idiot)? It is very difficult to find such quotations. These facts are implicit in everything they do, but not discussed as such because they are taken for granted.

None the less, if you look on page 467 of gravitation (book) in the second paragraph, they say:

To ask for the amount of electromagnetic energy and momentum in an element of 3-volume makes sense. First, there is one and only one formula for this quantity. Second, and more important, this energy-momentum in principle "has weight." It curves space. It serves as a source term on the righthand side of Einstein's field equations. It produces a relative geodesic deviation of two nearby world lines that pass through the region of space in question. It is observable. ....

Generalizing from this, any form of energy which contributes to the stress-energy tensor, which is the source of the gravitational field, has a definite location. This covers all forms of energy other than gravitational energy.

Now I must confess that this paragraph only appears in that book at all because they are contrasting this to the case of the energy of the gravitational field itself which is not a tensor and does not enter into the stress-energy tensor. They are arguing that gravitational energy is an exception which lacks a definite location. However, there is a pseudo-tensor expression for gravitational energy density which is just a function of the Christoffel symbols and the metric. So it is defined locally. They are just saying that it can be shifted from one position to another by transforming the coordinates in a non-linear way. (But remember that the Lorentz transformation is a linear transformation.) With respect to any set of coordinate systems which are related by linear transformations, even gravitational energy has a definite location. JRSpriggs 07:59, 23 June 2007 (UTC)

Critical comparison of this article to the Potential Energy article

The information here is commendable, as is the discussion on this page.

However (!!), it seems reasonable that since there are relationships between kinetic and potential energies, there should be a relationship between their two respective articles here at Wikipedia.

The potential energy article lists, explains, and hypertext references all the types of potential energy; this article does not, which reduces its audience to only people who already fully understand the principles.

Um. Isn't there only one type of kinetic energy- stuff that moves?WolfKeeper 22:50, 1 November 2007 (UTC)

I am creating articles on Wikiveristy as springboard for potential scientists at the middle school level; this potential knowledge energy, or work, is stored, of course, as elastic energy.

My work at the wikiversity is my attempt to objectify and reuse my studying for teacher exams, and, frankly, I am overwhelmed by the effort. If anybody here wants to collaborate on that project, please do not hesitate. I am filling the content quickly, but fact checking (including copyright checking), and citing are going to be additional chores. Quoting a scientist I know from the L4 OS project, "by the time they get to high school, it is already too late." So please help, --John van v 19:12, 10 September 2007 (UTC)

Section on Kinetic Energy of Systems Needs to be Revised

The section is very badly written , both from a mathematical and from a text standpoint. DS1000 (talk) 17:51, 30 July 2008 (UTC)

Kinetic energy increase caused by a push depending on frame of reference?

Can someone help or explain the following to me.

KE = ½ mv2

Say I have a 1kg mass on a ‘stationary’ table on earth and give it exactly the KE required in a push so that it is now moving at 10m/s (relative to the table on Earth).

That is the mass has gained 1/2x1x10x10 J of KE , = 50J

Now we all know the earth is not stationary but moving. Say it is moving at 100m/s (relative to ‘space’).

Then the mass, before I gave it a push, had 1/2x1x100x100J of KE, = 5000J

After I gave it a push, its velocity (relative to ‘space’) is 110m/s.

Its new KE is 1/2x1x110x110J = 6050J

The increase in KE is therefore 6050J-5000J = 1050J

So in this example, there is a discrepancy of 1000J in KE simply by calculating from 2 different reference frames (which are both true).

Exactly how much energy did I give to the mass?

Could someone please explain? 81.152.87.111 (talk) 01:39, 13 September 2008 (UTC)

As the article says,

The kinetic energy of a system thus depends on the inertial frame of reference and it is lowest with respect to the center of mass reference frame, i.e., in a frame of reference in which the center of mass is stationary. In any other frame of reference there is an additional kinetic energy corresponding to the total mass moving at the speed of the center of mass.

The change of the total kinetic energy (object and rest of the Earth) is (1-1/M)50J, where M is the mass of the Earth including the object, independent of the choice of the inertial frame of reference. The faster the Earth moves, the more the rest of the Earth loses kinetic energy.--Patrick (talk) 07:07, 13 September 2008 (UTC)

Thanks. well I guess (1-1/M)50J is very nearly 50J. In the last statement of your answer, what would happen to the 'lost' KE? 81.159.84.158 (talk) 00:48, 14 September 2008 (UTC)

The kinetic energy the rest of the Earth loses accounts for the discrepancy of 1000J you found.--Patrick (talk) 06:50, 14 September 2008 (UTC)

I don't think so, because the energy can theoretically come from a source (such as the solar wind) which does not change the speed of the moving earth (ie due to recoil). 81.159.84.158 (talk) 23:03, 14 September 2008 (UTC)

Kinetic energy is not a scalar, it is part of a vector; or more precisely, it is the change in the time component of a vector. So your question is rather like asking

I have a rod one meter long pointing East. If I turn it 30 degrees counterclockwise, then its East-ward extension is reduced by (2-√3)/2 = 0.134 . But if it was already turned 30 degrees to the North and I turn it another 30 degrees, then its East-ward extension is reduced by 1/2 = 0.500 . However, 0.134 + 0.134 = 0.268 ≠ 0.500 , so where did the other 0.500 - 0.268 = 0.232 go?

Notice that the signs in this example are different because it is in Euclidean space rather than Minkowski space. JRSpriggs (talk) 07:46, 13 September 2008 (UTC)

My question is nothing of the sort. My question is exactly how much energy (a scalar) was given to the 1kg mass. For simplicity, let's assume everything is moving in a straight line, ie the mass is moving in a straight line relative to the earth, and the earth and the mass are moving away from me in a straight line relative to me sitting stationary wrt 'space', along this imaginery line (thus avoiding the requirement of vector analyses). I want to know how to calculate the Law of Conservation of Energy, ie I have given an amount of energy to the mass to increase its velocity by 10m/s in both frames of reference, but how much? 81.159.84.158 (talk) 00:48, 14 September 2008 (UTC)

No. Energy IS NOT A SCALAR EVER! Your example IS like mine. Motion in a straight line is analogous to a rotated line. The only difference is that one involves a space dimension and a time dimension while the other involves two space dimensions.

The fact that you are getting a contradiction in your calculation proves that you have made a mistake. The mistake is in your assumption that energy is a scalar. JRSpriggs (talk) 02:41, 14 September 2008 (UTC)

Energy is a scalar. You are thinking of momentum which is a vector. However you can specify energy in a given direction which is a vector. In my case, I have simplified by restricting all travel to one direction. 81.159.84.158 (talk) 23:31, 14 September 2008 (UTC)

My answer was in terms of classical mechanics. Your answer seems to be related to

In special relativity energy is also a scalar (although not a Lorentz scalar but a time component of the energy-momentum 4-vector). In other words, energy is invariant with respect to rotations of space, but not invariant with respect to rotations of space-time (= boosts).

in the article Energy, but it is not very clear to me. Could you make the answer more specific for the example of the question?--Patrick (talk) 08:00, 14 September 2008 (UTC)

(unindent)

"My question is exactly how much energy (a scalar) was given to the 1kg mass"".

First, a scalar is defined as a numeric quantity that has the same value in all references frames, so kinetic energy does not apply. But that is about names.

Second, there is no law of conservation of kinetic energy. There is a law of conservation of total energy.

But what really matters here is the way you apply the law to the situation.

In the frame of the earth where the mass was pushed from speed 0 to speed v, you say the mass gained DE = 1/2 m v^2.

In the frame where the earth is moving at speed V, you say the mass gained DE' = 1/2 m (V+v)^2 - 1/2 m V^2 = 1/2 m v^2 + m v V, and you wonder why enery is not conserved.

The thing is, this has nothing to do with energy conservation to begin with. You could just as well say that the law is violated because in the first frame it had kinetic energy 0 and in the second it had energy 1/2 m V^2.

Conservation of energy pertains to the total energy a system has now, compared to the value it has later, both measured in one and the same reference frame. If you want to check the law, you must compare the total energy of the system mass+pusher+earth before you set the mass in motion, versus after you have set it in motion. You can do this is the two frames separately. And of course, kinetic energy is understood to be just a small part of the total energy. DVdm (talk) 09:25, 14 September 2008 (UTC)

OK let's consider the total energy of the system. The system consists of a 1kg mass sitting on a body moving at 100m/s wrt to 'stationary space'. Let's say on the body is a battery with stored electrical energy. The battery gives 50J to the 1kg mass which is converted to KE. It so happens this amount of KE changes the velocity of the 1kg mass by 10m/s relative to the body. The battery suffers no significant mass change, and stays stationary wrt the body. That is the total energy of the system remains the same: The body and the battery show no change in velocity. The battery loses 50J of electrical energy, the 1kg mass gained 50J of KE. Total energy conserved.

Now looking at this from the stationary outside observer's point. He sees the 1kg mass increase its velocity from 100m/s to 110m/s. By his calculation this corresponds to an increase in 1050J of KE. This energy must be compensated for somewhere. The body is still moving at 100m/s, so no change in KE there. He sees a battery being used, so from his observations, did the battery supply 1050J of energy?

Please is there anyone there, a professor of physics or the like who can explain this? 81.159.84.158 (talk) 03:39, 15 September 2008 (UTC)

The tricky thing is that one may forget the extremely small change in speed of the rest of the Earth, or think that it is negligible, while this is not the case in a frame of reference where the Earth had already some speed.--Patrick (talk) 10:03, 14 September 2008 (UTC)

But this does not explain the fact that the energy can come from the battery which does not involve in slowing down the rest of the earth. It is easy to say a very large mass moved by a very small amount is the source of the compensation, but if the distance involved is much smaller than the diameter of a proton or the hydrogen nucleus from the calculations, can you actually say anything's actually moved? We are dealing with sub-quantum dimesions. 81.159.84.158 (talk) 12:24, 16 September 2008 (UTC)

The battery provides the 50J, the slowing down is independent of using a battery, and provides the 1000J.

I am talking in terms of classical mechanics. I am not very familiar with quantum mechanics. Are you saying that a very small change in speed is not possible?--Patrick (talk) 15:44, 16 September 2008 (UTC)

I am saying that there comes a point when it is uncertain. Energy is quantised, and therefore according to 1/2xmxv^2, v has at some point to fall lower than a quantum of energy, so cannot take place. Can you think of an experiment which could measure a change of 50J or 1000J in the KE of the mass of the entire earth? Your explanation that the 1000J comes from the slowing down of the earth is unconvincing. The only explanation I can think of is that KE is not a state function, but depends on the observer's frame of reference. I was hoping that someone here would tell us how to convert the measurement from one frame to the other, and what it actually means. 81.159.84.158 (talk) 18:34, 16 September 2008 (UTC)

Sure, but it is also not the case in the frame of reference where the Earth had no speed. The other much less tricky thing is to ignore other forms of energy. And of course the least tricky thing is to ignore what conservation actually means :-) DVdm (talk) 10:14, 14 September 2008 (UTC)

In the example the change of speed of the Earth is 1/M. In the frame of reference where the Earth initially had no speed this gives the Earth a kinetic energy of 1/(2M). This is negligible compared to the 50J of the object.--Patrick (talk) 10:38, 14 September 2008 (UTC)

But who said anything about thrusting the 1kg mass against the earth to get the 10m/s increase in speed? The energy could have come from an extra-terrestrial source such as say solar wind. What I want to know is how much energy I need to increase the speed of the 1kg mass in the direction of travel. And because KE is dependent on v^2 and not v, the answer seems to depend on what v the mass is assumed to be moving at initially. Although the value for energy cannot be measured, the change in energy can. I want to measure the value of the change in energy (the kinetic kind), which instinctively one would assume to be the same for both scenarios, but they are not because KE depends on v^2 and not v. But why? 81.159.84.158 (talk) 23:22, 14 September 2008 (UTC)

Arbitrary break to assist editing

“Although the total energy of a system does not change with time, its value may depend on the frame of reference. For example, a seated passenger in a moving airplane has zero kinetic energy relative to the aeroplane, but non-zero kinetic energy relative to the earth.”

I have copied the above from the ‘Energy’ article in Wiki. Perhaps I could rephrase the above question using the aeroplane instead of the earth.

Let’s use the aeroplane example. Let’s assume the aeroplane is flying close to the surface of the earth in a straight line at uniform velocity.

Let this velocity be 100m/s. Inside the aeroplane is a 1kg mass. I decide to make the 1kg mass move at 10m/s in the plane, in the direction of travel of the plane (although I would not know the plane is moving). I can make the 1kg mass move in such a manner because I have a mechanism of giving the 1kg mass, energy (specifically kinetic energy). This energy can be transferred from a battery in the form of electrical energy converting to mechanical energy, or it could come from chemical energy in the form of an explosion, or gravitational energy by lifting a weight and letting it fall whilst connected to a mechanism to drive the 1kg mass, or any number of ways. In other word, I can measure the energy change from one form to another (in a thermodynamic way).

When measuring from the inside the aeroplane, the KE I give the mass would be 1/2x1x10x10J = 50J. I am asking the contributors here whether this is indeed correct. That is I need to get 50J from somewhere to give to the 1kg mass and it will move at 10m/s inside the plane in the stated fashion.

Now lets see from the observer stationary wrt the earth. The aeroplane and the 1kg mass were stationary to start off with, then acquired energy (the kinetic kind) so that they both move at 100m/s.

For the stationary observer, the 1kg mass has KE of 1/2x1x100x100J = 5000J whilst moving at 100m/s.

The next question is when the 1kg mass gained an extra 10m/s whilst in the plane. Assume the plane is transparent, the stationary observer observes that the 1kg mass is now moving at 110m/s, so its KE must be 1/2x1x110x110J =6050J

So the stationary observer notes that the 1kg mass had gained a further (6050-5000)J = 1050J.

However on the plane, if my interpretation of the KE law is correct, I simply supplied 50J to the 1kg mass to get it to move 10m/s faster.

The question is the stationary observer thinks the 1kg mass has gained 1050J of KE in going from 100m/s to 110m/s. On the plane I think I have given the 1kg mass 50J to make it go from 0m/s to 10m/s.

Can someone tell me exactly how much energy or kinetic energy changed hands in the process, and please note we are not even talking of speeds close to that of light. And on an application of this, is this the reason why rockets have several stages of boosters, because it takes ‘less’ energy to increase velocity of the vehicle, if the vehicle already has velocity? 81.159.84.158 (talk) 00:10, 15 September 2008 (UTC)

Due to the conservation of momentum there is always recoil, i.e. mass (Earth, solar wind, plane, etc.) is slowed down in the same direction or accelerated in the opposite direction. You can increase the energy of the 1kg mass by 1050J while only producing 50J because the other mass loses 1000J. Producing the 50J kinetic energy from a battery etc, is a real physical process, but the 1000J "transfer of kinetic energy" is not, since it depends on the frame of reference; it is just a matter of bookkeeping by the observer.--Patrick (talk) 06:23, 15 September 2008 (UTC)

It is advantageous to boost a spaceship when it is moving fast, see gravity drag and Oberth effect.--Patrick (talk) 06:10, 15 September 2008 (UTC)

Can you prove what you say? The transfer of electrical energy from a battery into kinetic energy is a real process. So is lifting a weight and letting it fall and transfer the kinetic energy to another object. 81.159.84.158 (talk) 23:09, 15 September 2008 (UTC)

I am inclined to say that if a quantity depends on the frame of reference, it is not a property of the process itself. Transfer of momentum can be considered a property of the process itself, but the amount of kinetic energy transferred would not be a property of the process itself.--Patrick (talk) 04:21, 16 September 2008 (UTC)

"The kinetic energy of an object is the extra energy which it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its current velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. Negative work of the same magnitude would be required to return the body to a state of rest from that velocity."

The above is taken at the beginning of the article. The work required to accelerate the the 1kg mass from 0m/s to 10m/s is very different to that of from 100m/s to 110m/s as exemplified by my question. One is 50J and the other is 1050J. But if this depends on where the observer is located, what does work actually mean, do you do less work if you are observed to be moving slower? I think, if you do the experiment, you will find that to make the 1kg mass move 10m/s relative to the plane (travelling at 100m/s) in the direction of travel will require 1050J and not 50J, and that this is what tells the observer that the plane is moving and not standing still relative to the earth. Otherwise you will get free energy, for example going from 0m/s to 10m/s will take 50J, then from 10m/s to 20m/s will take another 50J (provided you calculate it in a 10m/s frame), etc and finally going from 90m/s to 100m/s will take another 50J. Total in going from 0m/s to 100m/s in 10 successive step will require 500J. Then let the 1kg mass moving at 100m/s smash into something to bring it to 0m/s will yield 5000J (1/2 x 1 x 100 x100 J). Ie put in 500J, and get out 5000J. Has anyone done this experiment? Or can anyone confirm the physics and maths of this and the reason for a special frame of reference? 81.159.84.158 (talk) 18:55, 16 September 2008 (UTC)

Two remarks:

1. I will repeat something here with some emphasis added: Conservation of energy pertains to the total energy a system has now, compared to the value it has later, both measured in one and the same reference frame. If you want to check the law, you must compare the total energy of the system mass+pusher+earth before you set the mass in motion, versus after you have set it in motion. You can do this is the two frames separately. And of course, kinetic energy is understood to be just a small part of the total energy.
2. An article's talk page (see guidelines) is not a place to discuss one's misconceptions and misunderstandings of the articles' subject. It is a place where the content and the form of the article is discussed. Feel free to continue this in your user talk pages, or better, please take this some some Usenet newsgoup like sci.physics, where this very problem is treated on a daily basis.

Good luck. DVdm (talk) 20:35, 16 September 2008 (UTC)

I was trying to get clarification so the article could be improved. Unfortunately you don't seem to have any understanding of the subject and cannot make a convincing explanation. I have pointed out movements without impacting on the larger body such as the earth. Usual explanations of energy conservation are heat generation and sound. Go on explain why apparently if moving a 1kg mass at 10m/s on the plane moving at 100m/s takes 50J of energy, have you measured it or not. All I want to know is the 50J calculated according to KE law is correct or not. If it is then fine, if not what should the value be? 81.159.84.158 (talk) 00:27, 17 September 2008 (UTC) Continued on User talk:81.159.84.158.

The short answer is it takes 50 J of energy (force applied times distance) for the person on the plane. But the person in the control tower sees the object gain far more kinetic energy than that, and the airplane lose far more than that, and the (total airplane plus everything in it), still gains only 50 J. So no energy is created just because the plane is seen from a "ground" frame moving with respect to the airplane, if you don't forget to include the airplane. Example: If it takes 10 seconds to move the mass to that speed, the acceleration is 1 m/sec^2. Force required is 10 N. In 10 seconds the plane moves 1000 m and the object moves ahead 1/2at^2 = 50 m. The 10 N force applied through that much longer distance (10050 m) is 10,050 J, which is what the object gets in the control tower frame. But, the plane experiences a back reaction force of 10 N through 1000 m of flight, and thus loses 10,000 J of energy (though its velocity hardly changes due to it being so massive). So the control tower sees the plane gain the same net 50 J as if it had been sitting on the runway and the object inside had been rolled. Yes, the control tower sees 10,000 J transfered from plane to object that the person in the plane doesn't. But that's okay. Both of them see the same net energy change, which is 50 J for the system if you don't count the energy loss in the muscles of the mover, and zero if you do, as you should in a closed system (as the 50 J comes out of chemicals in somebody's muscles). SBHarris 04:56, 17 September 2008 (UTC)

Your explanation could be correct, but I think you are barking up the wrong tree. I had already considered it. The change in velocity of the aeroplane or the earth (the larger object) is very small, and thus its change in KE, compared to the 1kg mass (the smaller object) on board is also very very small.

The reason is total momentum is conserved in all cases.

In this example:

In the frame of the aeroplane: Total momentum before = 0 therefore Total momentum after also = 0.

Assuming, as everyone here say the 1kg mass thrusting forward thrusts the aeroplane (or the earth) backwards:

Momentum of 1kg mass after = 1kg x 10m/s = 10kgm/s Momentum of plane = - 10kgm/s.

Let say mass of plane is 10^6 kg

Then 10^6kg x v = -10kgm/s

Therefore v = -10^ -5 m/s for the plane

Therefore KE of plane = 1/2mv^2 = ½ x 10^6 x (-10^ -5m/s) x (-10^ -5m/s) = ½ x 10^ -4 J, which is small compare to the 50J of KE of the 1kg mass.

If the earth is the larger body, then the change in v is even smaller, and therefore the KE imparted on it would be even smaller.

You can do the momentum law in the frame of the stationary observer, and come to the same conclusion.

In the vast majority of cases, conservation of momentum leads to loss of kinetic energy, and this is usually attributed to the generation of heat, noise, a flash of light etc. —Preceding unsigned comment added by 81.159.84.158 (talk) 07:34, 18 September 2008 (UTC)

No, do the calculation in the frame of the stationary observer, and you will see.--Patrick (talk) 09:08, 18 September 2008 (UTC)

Why don't you show us? You said the other body loses 1000J, because it has been slowed down, and I've just proved it doesn't. 81.159.84.158 (talk) 23:42, 18 September 2008 (UTC)

You only considered the case in the frame of the aeroplane.--Patrick (talk) 20:08, 19 September 2008 (UTC)

Patrick this is what you wrote earlier: "You can increase the energy of the 1kg mass by 1050J while only producing 50J because the other mass loses 1000J." Could you please prove it with some simple arithmetic calclations? (Assuming you know what you are talking about). 81.159.84.158 (talk) 21:43, 19 September 2008 (UTC)

• outdent: Patrick is right. In the frame of the plane the plane does not get much energy, but then that doesn't happen on the ground when this happens, either. Most energy is always translated to the light object, in the COM frame. But nothing wrong with that, as energy is still conserved, due to the fact we didn't count the small energy lost from the muscles.

Now, you just looked at the small velocity change for the plane (10^-5 m/sec) and ASSUMED that this would amount to a miniscule amount of KE change when seen from the control tower, but you're quite wrong. Had you actually done the math, which you challenged Patrick to do, you see this isn't so. KE change for the plane of mass M as seen in the ground frame is:

KE(plane) = 1/2 M * ([v^2]-[v-δ]^2) = 1/2M * (v^2-v^2+2δv+δ^2) [should be -δ^2; also, to clarify, this 1/2M * (2δv-δ^2) is the decrease of KE(plane).--Patrick (talk) 07:33, 20 September 2008 (UTC)]

Assuming δ^2 is small enough to neglect WRT 2vδ, which is surely the case:

KE(plane) = Mvδ = 10^6 kg * 100 m/sec * 10^-5 m/sec = 1000 J.

Same number as the extra that the ground sees the object gain. That's the energy the plane loses, as seen by the ground, even from that small velocity change.

The reason the ground sees the same momentum changes as the plane, but very different energy changes (though all are conserved in each frame) is because the same force acts for the same TIME in both frames, but through very different DISTANCES in both frames. But in no case is there anything but conservation. I wish you'd go through the Freshman Physics calculations before coming here with your attitude. SBHarris 22:28, 19 September 2008 (UTC)

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Repeating (slightly modified):

An article's talk page (see guidelines) is not a place to discuss one's (mis)conceptions and (mis)understandings of the articles' subject. It is a place where the content and the form of the article is discussed. Feel free to continue this in your talk page), or better, please take this to some Usenet newsgoup like sci.physics, where this very problem is treated on a daily basis. People will we able to able to help you there.

Good luck and thank you - DVdm (talk) 10:32, 18 September 2008 (UTC)