Talk:Kinetic energy

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[edit] 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

[edit] 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)

[edit] 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)

[edit] 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.

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)

[edit] 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)

[edit] 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)

[edit] 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:

$E_k = \int \vec F \cdot \vec v d t = \int \vec v \cdot d \vec p = \int \vec v \cdot d (m \vec v) = \frac {m \ \vec v \cdot \vec v}{2}$

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:

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

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:

$E_k = m \gamma c^2 - m c^2 = m c^2 (\gamma - 1)\!$

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

[edit] 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)

[edit] 67.177.35.2's edits

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

Since this article does not deal with an advanced conept, The introduction should be intuitive, not rigorous. Later we can go into technical details.
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".
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)

[edit] 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)

[edit] 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 (γ - 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 m c^2 - m c^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)

[edit] 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 $m c^2 = m_0 \gamma c^2 = \frac{m_0 c^2}{\sqrt{1-\frac{v^2}{c^2}}}$ ; 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

m c 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

m c 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)

[edit] 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

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

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)

[edit] 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:

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

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

$(E_k+mc^2)^2 = p^2c^2+m^2c^4 \!$

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

$E_k^2+2E_kmc^2 = p^2c^2 \!$

Add the square of the kinetic energy to both sides:

$2E_k^2+2E_kmc^2 = p^2c^2+E_k^2 \!$

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

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

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)

[edit] 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)

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

where $\nabla^2_i$ 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

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

where $T [ρ]$ 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)

[edit] 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)

[edit] 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)

[edit] 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 $E = m c 2$ .

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: $E = (γ - 1) m 0 c 2$

$γ$ 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 $F = m a$ ? 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 $|F| = {G M_A M_B \over |r_{AB}|^2}$ 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 $F = m a$ , 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 $γ$ where $γ$ 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)

[edit] 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)

[edit] 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: $d(v \cdot v) = (d v) \cdot v + v \cdot (d v) = 2 v \cdot dv$ JRSpriggs 10:26, 9 January 2007 (UTC)

[edit] 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. S B H arris 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. S B H arris 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. S B H arris 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)

[edit] 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)

[edit] missing step?

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

$\mathbf{F} \cdot d \mathbf{x} = \mathbf{F} \cdot \mathbf{v} d t = \frac{d \mathbf{p}}{d t} \cdot \mathbf{v} d t = \mathbf{v} \cdot d \mathbf{p} = \mathbf{v} \cdot d m \mathbf{v} = \frac{m}{2} d (\mathbf{v} \cdot \mathbf{v}) = \frac{m}{2} d v^2 = d (\frac{m v^2}{2})$

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

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

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 : $\frac{m v^2}{2}$ 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: $d(v \cdot v) = (d v) \cdot v + v \cdot (d v) = 2 v \cdot dv.$ 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)

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