# Talk:Kinetic energy

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## Post Apocalyptic

Expositions of Kinetic Energy (Newtonian) are more effective in close association to work and potential energy - a simplified version of Hamiltonian Mechanics.

The absence of such material seems to have caused a great deal of confusion in the discussion page Blablablob (talk) 05:32, 17 September 2008 (UTC)

## minor correction

in the phrase "This is done by binomial approximation. Indeed, taking Taylor expansion for square root and keeping first two terms we get:"

I found that I had to take three terms of the Taylor series, not two.

can someone confirm?

-gary —Preceding unsigned comment added by 212.179.48.132 (talk) 13:42, 25 September 2008 (UTC)

Suppose one is trying to get approximations to
${\displaystyle E={\frac {mc^{2}}{\sqrt {1-(v/c)^{2}}}}\,.}$
The essential part is
${\displaystyle (1-\beta ^{2})^{-1/2}\,}$
where ${\displaystyle \beta =v/c\,.}$ If the Taylor series is expanded using ${\displaystyle x=\beta \,}$ as the independent variable in which you are expanding, then three terms are required to get
${\displaystyle {\frac {d(1-x^{2})^{-1/2}}{dx}}=-{\frac {1}{2}}(1-x^{2})^{-3/2}\cdot -2x=x(1-x^{2})^{-3/2}\,}$
${\displaystyle {\frac {d^{2}(1-x^{2})^{-1/2}}{{dx}^{2}}}=(1-x^{2})^{-3/2}+3x^{2}(1-x^{2})^{-5/2}\,}$
${\displaystyle (1-\beta ^{2})^{-1/2}\approx 1+0\cdot \beta +{\frac {1}{2}}\cdot \beta ^{2}\,.}$
However, if you use ${\displaystyle x=\beta ^{2}\,}$ as the variable in which you are expanding, then only two terms are needed to get
${\displaystyle {\frac {d(1-x)^{-1/2}}{dx}}=-{\frac {1}{2}}(1-x)^{-3/2}\cdot -1={\frac {1}{2}}(1-x)^{-3/2}\,}$
${\displaystyle (1-\beta ^{2})^{-1/2}\approx 1+{\frac {1}{2}}\cdot \beta ^{2}\,.}$
So the two different ways of expanding give different lengths. OK? JRSpriggs (talk) 17:46, 25 September 2008 (UTC)

Yes, I think that's fine. Thank you for taking the time to explain. Gary —Preceding unsigned comment added by 84.108.230.143 (talk) 06:51, 26 September 2008 (UTC)

## Relativistic KE

The idea that the amount of kinetic energy in a body "depends on the relationship between that object and the observer" is a difficult one, and the article currently doesn't come close to explaining why and how this should be so in a way that an educated layman can understand. I'd like to see a much clearer and more accessible explanation of this, and I'd encourage those who understand the subject to address this issue. —Preceding unsigned comment added by 86.161.43.41 (talk) 23:42, 15 January 2009 (UTC)

Yes, this paragraph was a mess. I cleaned it up and removed that "sea also below"-thing and the "gravitational"-remark. I.m.o. This is not the place for this. DVdm (talk) 18:41, 16 January 2009 (UTC)

As a non-mathematician, I once had a teacher walk me through this concept. I recall he pointed out the COMBINED kinetic energy of the earth and a one-kilogram weight impacting, and the concept of the relative speeds. I recall the moment of clarity when I did the math. Unfortunately, this is lost in the dim recesses of my mind! A similar example should be appreciated by many students. Mydogtrouble (talk) 15:19, 24 August 2009 (UTC)

Is it possible to be clearer that the mass referred to is rest mass? A '0' subscript maybe? —Preceding unsigned comment added by 194.94.232.72 (talk) 12:19, 28 August 2009 (UTC)

The 0-subscript is a bit old-fashioned. I made a little change, moving the reference to rest mass up a bit. DVdm (talk) 13:41, 28 August 2009 (UTC)

${\displaystyle E_{kin}=m\cdot c^{2}\cdot \left({\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}-1\right)={\frac {m\cdot v^{2}}{1-{\frac {v^{2}}{c^{2}}}+{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}\sim {\frac {m\cdot v^{2}}{2}}\quad {\text{for v}}<

If you would like to check it yourself. Put simply the expression in the round brackets on a break and extend it by ${\displaystyle 1+{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}$

--Willi windhauch (talk) 10:42, 31 August 2011 (UTC)

${\displaystyle E_{k}=m_{r}c^{2}-mc^{2}=m_{r}v^{2}\cdot {\frac {m_{r}}{m_{r}+m}}\,}$; where ${\displaystyle m_{r}}$ is relativistic mass and ${\displaystyle m}$ is rest mass; For a small velocity ${\displaystyle v< we can't see any changes in mass therefore ${\displaystyle m_{r}=m}$, then: ${\displaystyle E_{k}={\frac {mv^{2}}{2}}}$ ; The foe (talk) 23:49, 22 January 2014 (UTC)

## KE equation

Correct me if I'm wrong, but isn't it true that the 1/2 in (1/2)mv^2 is a proportionality constant that only applies to the SI system of units? If you wanted to express kinetic energy in the british engineering system you would use (1.29e-3)mv^2 = KE in BTU's. This article makes it seem as if the formula for KE is derived when in fact it is an empirical formula. The only thing we know regardless of units is that KE is proportional to mv^2. I think it's important to try to make it clear where that 1/2 comes from. I got most of this information from the external link at the bottom of the article on the conservation of energy.--70.185.112.226 (talk) 22:33, 15 February 2009 (UTC)

There are other kinds of energy than kinetic, for example thermal energy, and work energy. You can choose to define energy with any proportionality constant you like, but once you define it in any one area, you're now stuck in all other others. Thus, if you define heat energy as C*ΔT, for example (3/2)nRΔT for ideal gases, or work energy as Force*Distance (both of which quantities are equal to each other) then that forces you to define THAT energy as 1/2 mV^2 in the same units. SBHarris 22:51, 15 February 2009 (UTC)

Something is bothering me. Now Leibniz knows well that mgh=1/2mv^2. Why does he insist KE is mv^2. To me his aim is not to derive an equation that describe how object falls but rather the philosophy of KE. —Preceding unsigned comment added by 68.11.142.236 (talk) 15:08, 17 February 2009 (UTC)

• Perhaps so. But with Count Rumford figuring out that F*D = Q in mechanical work, and Joule verifying that plus also that mgh = Q (I think he did that on his honeymoon at some Falls), we can either call this "vis-viva" mv^2 and say then that it's twice the heat, 2Q, or we can bite the bullet and say it's the same as Q, and thus 1/2 mv^2. Clearly the important relationship, is that the quantity in motion that scales with heat goes as v^2, not v. After you know that, you can chose your scale factor, but if you chose it as 1/2, you can simply say it's all the same thing; the motion equivalent of heat and work. Instead of needing two names and a conversion factor of 2 between them, which seems silly. SBHarris 02:09, 18 February 2009 (UTC)

## Lp space

In Lp space the Lp norm of x is given by

${\displaystyle \ \|x\|_{p}=\left(|x_{1}|^{p}+|x_{2}|^{p}+\cdots +|x_{n}|^{p}\right)^{1/p}}$

(so the L2 norm is the familiar Euclidean norm (see Pythagorean theorem), while the distance in the L1 norm is known as the Manhattan distance).
AFAIK, in Lp space KE=kv^p and PE=F^(p-1) * D
just-emery (talk) 23:58, 22 June 2009 (UTC)

The mathematics is well-known to us. But the real world (where "kinetic energy" has a meaning) has no relationship to Lp for p≠2. You cannot just make up stuff and put it in an article. JRSpriggs (talk) 09:27, 23 June 2009 (UTC)
I didnt say that the real world wasnt L2. I am simply pointing out that the reason the equations take the form they do in the real world is because the real world is L2. This may be well known to you but its certainly not common knowledge. I cant imagine why you would think I was 'making stuff up' but of course I'm sure that you acted in good faith so you must have just misunderstood me. just-emery (talk) 16:20, 23 June 2009 (UTC)
Yes, the real world is L2 (in the 3 known spatial dimensions, at least).
If you say that "the reason the equations take the form they do in the real world is because the real world is L2.", then you are also saying implicitly that the equations would have a different form in Lp for p≠2 (indeed, you specified the supposed form that they would have). However, you provided no evidence that kinetic energy would have such a form, or even that the notion of kinetic energy could be defined at all in such a hypothetical world. JRSpriggs (talk) 20:26, 7 July 2009 (UTC)

## Simpler Derivation

The derivation of the formula ${\displaystyle E_{k}={\frac {1}{2}}mv^{2}}$ that I was shown seems much simpler than the one currently in the article. I'm wondering what others think. It is as follows:

${\displaystyle W=F\times x\,}$

${\displaystyle E_{k}=\int F\,dx}$

${\displaystyle E_{k}=\int ma\,dx}$

Assuming constant mass then

${\displaystyle E_{k}=m\int a\,dx}$

${\displaystyle E_{k}=m\int {\frac {dv}{dt}}\,dx}$

${\displaystyle E_{k}=m\int {\frac {dx}{dt}}\,dv}$

${\displaystyle E_{k}=m\int v\,dv}$

${\displaystyle E_{k}={\frac {1}{2}}mv^{2}}$

A few steps could probably be omitted, I just wanted to show each step explicitly. Any thoughts or corrections would be appreciated. Ginogrz (talk) 03:05, 7 July 2009 (UTC)

Hi Ginogrz. I like your alternative derivation. It is simpler and clearer. One major objection I have about the present derivation is that it is not supported by any in-line citations to allow independent verification, as is required by WP:Verifiability. I encourage you to substitute your derivation, but before you do so please establish a suitable source. When you add your derivation be sure to include at least one in-line citation to show where your information comes from. Information on citing sources is available at WP:CITE. Dolphin51 (talk) 06:17, 7 July 2009 (UTC)
One reason for the current derivation is that the first few steps can also be used as part of the relativistic derivation, which is not possible with your version. This helps to make clear that the key change is in the definition of linear momentum.
Also notice that historically, Newton actually defined force as the time derivative of linear momentum, not as mass times acceleration. See the second law at Newton's laws. JRSpriggs (talk) 20:14, 7 July 2009 (UTC)
Ah, I can see how that is beneficial for setting up the reader for the relativistic derivation. However, it seems like it might be better to explain the ${\displaystyle E_{k}=\int \mathbf {v} \cdot d\mathbf {p} }$ statement in the relativistic section rather than the Newtonian derivation since kinetic energy is a pretty fundamental concept in beginning physics courses, which often do not require a knowledge of dot product properties or vector calculus. Also the current derivation doesn't seem to flow well with the preceding section in which the formula is introduced (a fairly uncomplicated section), being that it is in a totally different realm of knowledge. This is only my opinion though, any other thoughts would be nice.
I found a source for the derivation I suggested in a textbook, however since the section centers around Newton, it would only make sense to use his historical definition of force like you said. You could then just simply change the derviation to be
${\displaystyle E_{k}=\int F\,dx}$
${\displaystyle E_{k}=\int {\frac {d(mv)}{dt}}\,dx}$
Constant mass
${\displaystyle E_{k}=m\int {\frac {dv}{dt}}\,dx}$
${\displaystyle E_{k}=m\int {\frac {dx}{dt}}\,dv}$
${\displaystyle E_{k}=m\int v\,dv}$
${\displaystyle E_{k}={\frac {1}{2}}mv^{2}}$
Although I suppose I would have to find a new source for this one. Ginogrz (talk) 03:02, 8 July 2009 (UTC)
I prefer the approach suggested by Ginogrz, rather than the one suggested by JRSpriggs. The former approach is much more compatible with the principle of make technical articles accessible. Wikipedia articles should begin with the simplest information so that it is accessible to the greatest number of readers, and work towards the more complex information later in the article. Students and others are introduced to the concept of kinetic energy long before they need to assimilate the concept of scalar products, vector products and relativity. Dolphin51 (talk) 03:50, 8 July 2009 (UTC)
Perhaps you should add another step, otherwise it is not entirely obvious that
${\displaystyle m\int {\frac {dv}{dt}}\,dx=m\int {\frac {dv}{dt}}{\frac {dx}{dt}}\,dt=m\int {\frac {dx}{dt}}\,dv\,.}$ JRSpriggs (talk) 04:29, 9 July 2009 (UTC)
I don't know if that would help considering ${\displaystyle {\frac {dv}{dt}}\,dx={\frac {dx}{dt}}\,dv}$ is a pretty simple identity and adding those extra steps would only elongate the derivation. Ginogrz (talk) 06:38, 15 July 2009 (UTC)

## when should be easier drive bike?

Isn't according kinetic energy formula should be easier to drive bike on slow pedals rotation-fast wheel rotation mode instead fast pedals rotation-slow wheel rotation mode? How such example fits here? And why flying stone in kosmos have about square bigger kinetic energy than momentum? If object to object transmiting all momentum to eatch over, so how is transmitted kinetic energy? Supose 1 kg mass stone flying in cosmos at 100 m/s transmit his energy to don't moving relatively to this stone, to over bigger stone with mass 10 kg momentum. So smaller stone stop and bigger stone got speed 10 m/s. How supose here fit in kinetic energy? —Preceding unsigned comment added by Taupusensteinas (talkcontribs) 09:11, 15 October 2009 (UTC)

Regarding the bicycle question : see this. Bicycle stability physics is actually quite complex. With regard to your second question : kinetic energy is defined to be the integral of momentum with respect to changes in velocityUser A1 (talk) 10:09, 15 October 2009 (UTC)
To Taupusensteinas: I presume that you are not a native speaker of English since your grammar is so poor that it is hard to understand you.
If the collision of the stones is elastic, then the 1 kg stone cannot just stop when it hits the 10 kg stone (assumed to be initially at rest). If the collision is inelastic, then the lost kinetic energy becomes a mixture of: heat, sound, and damage to the structure of the stones (chemical energy). JRSpriggs (talk) 10:51, 15 October 2009 (UTC)
From article:

The work done accelerating a particle during the infinitesimal time interval dt is given by the dot product of force and displacement:

${\displaystyle \mathbf {F} \cdot d\mathbf {x} =\mathbf {F} \cdot \mathbf {v} dt={\frac {d\mathbf {p} }{dt}}\cdot \mathbf {v} dt=\mathbf {v} \cdot d\mathbf {p} =\mathbf {v} \cdot d(m\mathbf {v} ).}$

Applying the product rule we see that:

${\displaystyle d(\mathbf {v} \cdot \mathbf {v} )=(d\mathbf {v} )\cdot \mathbf {v} +\mathbf {v} \cdot (d\mathbf {v} )=2(\mathbf {v} \cdot d\mathbf {v} ).}$

Therefore (assuming constant mass), the following can be seen:

${\displaystyle \mathbf {v} \cdot d(m\mathbf {v} )={\frac {m}{2}}d(\mathbf {v} \cdot \mathbf {v} )={\frac {m}{2}}dv^{2}=d\left({\frac {mv^{2}}{2}}\right).}$

Since this is a total differential (that is, it only depends on the final state, not how the particle got there), we can integrate it and call the result kinetic energy:

${\displaystyle E_{k}=\int \mathbf {F} \cdot d\mathbf {x} =\int \mathbf {v} \cdot d\mathbf {p} ={\frac {mv^{2}}{2}}.}$

This equation states that the kinetic energy (Ek) is equal to the integral of the dot product of the velocity (v) of a body and the infinitesimal change of the body's momentum (p). It is assumed that the body starts with no kinetic energy when it is at rest (motionless).

As I see the force is itself particle with velocity v, so this derivations shows not kinetic energy, but what velocity will reach particle if he will fly with acceleration, so velocity will be quadratic per linear time and liner distance x. Let me try to explain. F=dp/dt, so F already is particle of mass m flying at velocity v in small time interval delta-t=dt. And now again we see v*dt, which means distance traveled per very small time. So here is multiplication of distance traveled at speed v with mass and with rising speed. Time also rising. If we take constant time say dt=10 seconds. And constant speed v=100 m/s. m=1 kg. And dv/dt is average speed. So this equation describes what will be if object of mass m=1 kg and flying at acceleration dv/dt=100/10=10 m/s/s will be multiplicated with distance, what he fly through 10 seconds at speed 100 m/s. This distance is 1000 meters. So this distance is muliplied by mass and acceleration. Acceleration times distance and mass and only 10000 kg remaining from all this trick. From formula should be taken dv. Acceleration may be imposible to write in math phorm and aliens etc don't using it in math like symbol (dv/dt). And it have tricky phorm m/s^2, which frozen in time.

${\displaystyle \mathbf {F} \cdot d\mathbf {x} =\mathbf {F} \cdot \mathbf {v} dt={\frac {m}{dt}}\cdot \mathbf {v} dt=\mathbf {v} \cdot m.}$

So m do not moving yet, but why not it can't be devided by dt. Mass m egzist in cosmos in whatever time frame. And it multiplicated with v gives us old good mv momentum. How object can have already some speed dv if it do not have? Or if object have already acceleration then he have already energy before he was multiplied by v and before goting kinetic energy. There somthing dirty with this derivation of kinetic energy and acceleration dimension.

But here good question: Who will move faster after collision, bed (of mass m=40 kg) in which metal ball of mass 1 kg stroke at 10 m/s in soft side or same mass as bed metal box say of mass m=40 kg? Ball almost do not reflecting from bed soft side, so isn't ball transmit bigger momentum and kinetic energy to soft bed than to metal box? Or this ball stoping springs force is converted to heat and like that? Who will got bigger velocity: bed or box after collision with ball? —Preceding unsigned comment added by Taupusensteinas (talkcontribs) 10:25, 17 October 2009 (UTC)

The center of mass moves with a speed of 10/41 m/s before and after collision, in both cases, so if the ball stays with the bed each of the two have this speed. If the ball bounces back the box will have a higher speed than 10/41 m/s. The kinetic energy is more in the second case, in the first more is converted to heat etc.--Patrick (talk) 07:49, 7 February 2010 (UTC)

## Relativistic kinetic energy example

I calculated this as an example of an object at relativistic speeds: 1 kg object at 0.8 c would have kinetic energy of about 60 PJ (accurately it would be 59917011915787842 and two thirds J). That would make the kinetic energy in such an object little less than a third of the yield of Tsar Bomba. Is this anywhere near the actual value? How would I calculate the "yield" of this kind of projectile? --Khokkanen (talk) 23:35, 7 June 2010 (UTC)

This Discussion page is primarily about the article Kinetic energy. If you don't receive a prompt answer to your question I suggest you delete it and re-post it at WP:Reference desk/Science. Dolphin (t) 02:20, 8 June 2010 (UTC)
${\displaystyle E_{k}={\frac {mc^{2}}{\sqrt {1-(v/c)^{2}}}}-mc^{2}}$. The factor gamma-1 at 0.8 c is 2/3. Thus an object will have a kinetic energy 2/3 that of mc^2. A kg is 21 megatons, so this energy is 2/3 of that, or 14 megatons. This is 14/50 = 28% of Tsar bomba, so your calculation is indeed correct (quite a glowbomb, as Vinge would say). Another way of looking at this is that the 50 megaton yeild of Tsar Bomba is 50/21 = 2.38 kilograms of heat, light, and other forms of active energy. SBHarris 02:52, 8 June 2010 (UTC)

## Edit Requested

{{editsemiprotected}}

Please consider changing "The kinetic energy of an object is the extra energy..." to "The kinetic energy of an object is the energy..."

Adding "extra" prior to energy is confusing and is not consistent with other accepted definitions of KE available. TSRibeye (talk) 07:22, 11 June 2010 (UTC)

No. I think that the word "extra" is necessary for correctness. Objects have energy even when they are not moving. I want to make it clear that the kinetic energy is distinct from that rest energy. JRSpriggs (talk) 07:42, 11 June 2010 (UTC)
While it is true and correct that objects do have energy when not moving, generally accepted and used definitions of KE do not include the word "extra". The definition of KE includes the phrase "..the energy of motion.." and potential energy. "Motion" removes any question as to KE referring to energy "at rest". At a minimum it may be helpful to remove "extra" from the initial definition and then discuss how this "extra" energy is different.
Spigotmap- I was following Wikipedia's directions for suggesting edits to protected content by using the editsemiprotected template. If I'm doing this the wrong way, please let me know how else to go about suggesting edits and correct the Wiki directions. (TSRibeye (talk) 21:31, 14 June 2010 (UTC))

Not done: please establish a consensus for this alteration before using the {{edit semi-protected}} template. SpigotMap 12:19, 11 June 2010 (UTC)

To TSRibeye: I do not see what your problem is with that word. "Extra" works with "due to its motion" to express an idea. Some people define "kinetic energy" in a way which includes rest energy. That definition conflicts with the one used in this article. "Extra" helps to draw attention to the distinction.
Do not feel that we have to slavishly follow some particular source. There are many thousands of sources out there which do things in a variety of different ways. It is more important to be internally consistent and clearly present the concepts. JRSpriggs (talk) 06:57, 15 June 2010 (UTC)
My problem with the word "extra" is that it does not add anything to the definition, and may be confusing to someone looking to understand the theory. As for "slavishly" following a particular source, I have yet to see "extra" used in any other definition. If you would like I can cite the resources I used, but it may be easier if you can cite your source for the definition. I feel it is more important to be consistent across the internet while clearly explaining the concept-neither of which apply to the use of "extra". All I'm asking is it be removed from the definition, but the inclusion of energy at rest be included in the next sentence. This will allow for a definition that is in line with the majority of main stream sources, and ensure the understanding of what the extra motion means. (TSRibeye (talk) 19:37, 15 June 2010 (UTC))
Sources for defining kinetic energy taken from the first two search pages of a google search for "kinetic energy". Note none of them include the word "extra" in the definition. Please cite your reference for "extra energy". TSRibeye (talk) 21:02, 24 June 2010 (UTC)
• www.physicsclassroom.com/class/energy/u5l1c.cfm
• hyperphysics.phy-astr.gsu.edu/hbase/ke.html
• www.schools.utah.gov/curr/science/sciber00/8th/forces/.../potkin.htm
• physics.about.com/od/energyworkpower/f/KineticEnergy.htm
• scienceworld.wolfram.com/physics/KineticEnergy.html
• wiki.answers.com/Q/What_is_kinetic_energy
• http://www.wisegeek.com/what-is-kinetic-energy.htm
• www.thefreedictionary.com/kinetic+energy
I agree with TSRibeye's latest theme. If authoritative sources can be found, and cited, for use of the word extra then use of that extra word in this article is reasonable. Conversely, if no authoritative source can be found then the word extra should not be used in the article. The argument put forward by JRSpriggs is a sound one and I don't disagree with it, but in the absence of a citable source it is merely original research. At WP:V there is the instruction that The threshhold for inclusion in Wikipedia is verifiability, not truth. Dolphin (t) 02:16, 25 June 2010 (UTC)
You must be joking. Now we have to have a source for the exact choice of wording, for grammar? Do not be ridiculous. JRSpriggs (talk) 07:09, 25 June 2010 (UTC)
You can try distracting our attention by use of expressions like You must be joking and Do not be ridiculous but I won't be distracted by such things. On Wikipedia we try to discuss things objectively. TSRibeye and I have made a very simple and totally reasonable request. So far you haven't acknowledged the legitimacy of that request, nor have you identified a source for your belief that kinetic energy is extra energy.
I have checked my Physics textbook and it doesn't suggest that kinetic energy is extra energy. It is at pains to point out that we are always interested in changes of energy rather than absolute values. For example, when I am sitting at my computer I can choose a reference frame that gives me a velocity of zero and a kinetic energy of zero. Alternatively I can use some other reference frame that gives me a non-zero velocity and a non-zero kinetic energy. If I stand up and walk away at 1 metre per second my velocity and kinetic energies are different in the two reference frames but the change of kinetic energy is equal to the work done in both cases. Similarly, whether my gravitational potential energy is measured relative to the floor, or ground level, or sea level, or the center of the Earth is unimportant because it is only the change in potential energy that conforms to the laws of physics. So if I choose a reference frame in which an object's velocity is zero I am entitled to say that object's kinetic energy is zero. If I choose to say the object's temperature is equal to my arbitrary datum I am entitled to say its internal energy is zero; and I can also say its energy is zero. Consequently I think it is reasonable to challenge your view that kinetic energy should be defined to be extra energy, and ask you to identify your source. Dolphin (t) 07:53, 25 June 2010 (UTC)
"Extra energy" DOES talk about only about a changes in energy. Kinetic energy is the "extra energy" due to relative motion of an object or center of mass of a system of objects, and this is frame-dependent. There is a sort of "minimum energy" in SR which occurs when the system has zero-momentum (center of mass frame). All this neglects gravity and curved space (mass in general relativity). But what's wrong with defining kinetic energy as the change in energy due to motion. Change = delta = "extra." They are basically all the same concept, no? SBHarris 19:14, 25 June 2010 (UTC)

I also think that we should drop the extra. Its presence would be warranted (or even required) when kinetic energy is explained in the context of rest energy or of energy in general. I think the word is highly misplaced in the first sentence of the lead of an article about kinetic energy. Surely there must be a place in the Energy article where we can refer to KE as extra energy? DVdm (talk) 08:55, 25 June 2010 (UTC)

At Energy#Forms of energy there is a list of the different forms of energy. It is clear that kinetic energy is not the only form of energy so an object can simultaneously possess energy in two or more forms. However, I very much agree with your point that the opening sentence of the article on kinetic energy is not the place to write everything that can be written about kinetic energy. The opening sentence needs to be the simplest thing that can be accurately stated about kinetic energy. The opening sentence may be the only thing read by young readers and newcomers to science. Whether they read further will depend on how meaningful the opening sentence is. Make technical articles accessible! Dolphin (t) 12:18, 25 June 2010 (UTC)
I agree with the last statement, although I don't think that this one extra word would make the opening sentence less meaningful. As far as I'm concerned the opening sentence could go like:
• The kinetic energy of an object is the extra energy which it possesses due to its motion, in addition to other possible (non-motion related) forms of energy.
I think that "extra" begs for more, and that this specification renders moot (and effectively kills) the source requirement you brought up. For me this would be a very good opening sentence as well. DVdm (talk) 12:47, 25 June 2010 (UTC)
By the way, I think that what you just did to the lead of Potential Energy will immediately scare every young reader and newcomer to science away. DVdm (talk) 12:58, 25 June 2010 (UTC)
Note - TSRibeye, I notice that you removed the word again. Please don't do this. See my message/warning on your talk page. DVdm (talk) 17:35, 25 June 2010 (UTC)

My intent behind removing "extra" from the first sentence/initial definition is that I don't want people to be confused when they are looking for a quick definition. I was having a discussion with a friend re: kinetic energy and needed a quick reference to verify I correctly remembered my college physics. The "extra energy" jumped out at me and I thought I wasn't remembering the concept correctly. A quick web search determined "extra" was not part of the generally used definition. I didn't understand the use of "extra" until JRSpriggs explained on the talk page. Solution? I propose one of my first suggestions. Remove "extra" from the first sentence, and add a sentence or two immediately following explaining the distinction between KE vs. other (extra) energy. DVdm, my apologies. How about "The kinetic energy of an object is the energy which it possesses due to its motion. While an object may simultaneously possess multiple forms of energy Energy#Forms of energy, Kinetic energy is specific to motion." Slightly redundant, but it returns the definition to its base form while distinguishing kinetic energy from rest energy (and other forms of energy). TSRibeye (talk) 20:00, 25 June 2010 (UTC)

Sure, and the current wording is ok as well for me. But if JRSpriggs really feels better with the extra word, I propose we use the above compromize (now set in bold) and move on. DVdm (talk) 20:27, 25 June 2010 (UTC)
There is an excellent, simple definition of kinetic energy in Jain's Textbook of Engineering Physics. It says The energy possessed by a body due to its motion is called its kinetic energy. It is on page 9 and immediately precedes coverage of potential energy on page 10.
Considering we are citing Jain in the opening sentence in Potential energy I suggest we do the same in the opening sentence in Kinetic energy. Dolphin (t) 12:18, 5 July 2010 (UTC)

## Ownership of article

Apparently user User:JRSpriggs is claiming ownership of this article against WP policies by reverting obvious improvements to the lead without explanation. For examples, the article stated that the body maintains this kinetic energy unless its speed changes. This is such a naive formulation, that one must ask, why would it change its speed? This is not the only condition, and equally naive would be the statement, .. unless it changed its mass. The correct statement is that it maintains it kinetic energy until another force acts on the object (to either change mass or speed). The rest of the lede had other either confusing or specialist language in poor style. Kbrose (talk) 21:43, 16 September 2010 (UTC)

Small point: This is actually a stronger/more strictly correct formulation than the force statement. You can apply forces to accelerate an object without changing its speed. I make no comment on the targeting of the lede. User A1 (talk) 22:22, 16 September 2010 (UTC)+
Perhaps you need to look at the definition of acceleration: a = v / t. If you accelerate something, you have to change its velocity (v), or shrink time. Kbrose (talk) 22:56, 16 September 2010 (UTC)
But you dont need to change speed. Circular motion User A1 (talk) 22:58, 16 September 2010 (UTC)
I can see why JRSpriggs reverted the edits by Kbrose. Some of the changes made by Kbrose were incorrect. For example, Kbrose wrote that a body maintains its kinetic energy as long as no other force acts upon it. When a body is moving along a curved path a centripetal force is acting upon it. A centripetal force does no work on a body and so does not alter the body's kinetic energy.
The statement by Kbrose would have been correct if it said as long as no other force does work upon it. Dolphin (t) 22:53, 16 September 2010 (UTC)
Kbrose has written that acceleration a = v / t. This is true in rectilinear motion but it is not true generally. For three-dimensional motion, acceleration, velocity and displacement are all vector quantities. A particle or body moving along a curved path is undergoing an acceleration, even though its speed may be unchanged. Dolphin (t) 23:08, 16 September 2010 (UTC)
Due to the incorrect nature of these changes, I have reverted them. I am in a rush right now, so later i will come back and restore the non-wrong bits. User A1 (talk) 08:34, 17 September 2010 (UTC)

I am sorry that I did not post here earlier. I meant to explain my second reversion, but I was called away from my computer. I am not claiming ownership of this article — I have allowed many changes that I did not entirely like over the years that it has been on my watch list. But I feel that Kbrose should show more respect for the fact that many others have already polished the article.
The part of Kbrose's edit affecting the lead contained these changes which I felt were objectionable:

• He changed "unless its speed changes" to "as long as no other forces act upon it". This obscures the fact that kinetic energy is a function of speed. And implicitly raising issues about mass-change in the lead is inappropriate for the intended audience of this article who are probably at high school level or below.
• He introduced the words "the same bullet has zero kinetic energy with respect to an observer riding the bullet". This is ludicrous and will confuse or put off the reader.
• He said "at the same velocity and in the same direction". This is redundant because velocity includes direction as well as speed.

JRSpriggs (talk) 10:10, 17 September 2010 (UTC)

## "Relativistic kinetic energy of rigid bodies" section

In the "Relativistic kinetic energy of rigid bodies" section, it is assumed that ${\displaystyle p=\gamma mv}$, and then the kinetic energy ${\displaystyle E_{k}}$ is derived from that, somewhat laboriously. Why not instead just start with the assumption that ${\displaystyle E=\gamma mc^{2}}$ and skip all that derivation? IMHO, the expression for ${\displaystyle p}$ is no more or less intuitive than that for ${\displaystyle E}$, so I see no advantage to starting with ${\displaystyle p}$ rather than ${\displaystyle E}$ in that sense. Also, IMHO, the derivation itself isn't that enlightening given that the more transparent, non-relativistic derivation is well discussed earlier in the article. Quantling (talk) 20:09, 24 September 2010 (UTC)

In Newtonian physics, conservation of momentum (defined as mv) is a given. Conservation of energy including kinetic energy is derived.
I think that we need to follow the same pattern in special relativity, if only to make it clear where Newtonian physics goes wrong. JRSpriggs (talk) 07:41, 27 September 2010 (UTC)

Okay, I am giving it another try with the article edits. Feel free to modify my changes. Especially if you choose to revert, please leave detailed comments here. Thanks — Quantling (talk) 18:39, 27 September 2010 (UTC)

I very much prefer the original derivation for the reasons given by JRSpriggs. This new derivation uses the expression ${\displaystyle \mathbf {v} ={\frac {\mathbf {p} c^{2}}{\sqrt {m^{2}c^{4}+p^{2}c^{2}}}}}$ which seems to be falling from the sky to only serve our purpose in the imminent derivation. Do we have a proper source for either derivation? If not, I think we should find a good source and stick with it. This is a source with a good derivation, but it uses yet another slightly different approach. Thoughts? DVdm (talk) 19:19, 27 September 2010 (UTC)

Interesting. The expressions, ${\displaystyle \mathbf {v} ={\frac {\mathbf {p} c^{2}}{\sqrt {m^{2}c^{4}+p^{2}c^{2}}}}}$ and ${\displaystyle \mathbf {p} ={\frac {m\mathbf {v} }{\sqrt {1-v^{2}/c^{2}}}}}$, are mathematically equivalent. Yet I take it that you consider the latter to be more accessible to the lay reader. I guess I could agree with that. On the flip side, the ${\displaystyle \int \mathbf {v} \cdot d\mathbf {p} }$ formulation calls out for doing the math as a function of p rather than of v, yes? (For instance, if there were yet another way to relate momentum to velocity, kinetic energy might most easily be derived by first writing v in terms of p and then integrating.) Maybe there is some way to have it both ways; perhaps by first proving the formula "that seems to be falling from the sky" and then proceeding with p. Thoughts? Quantling (talk) 20:39, 27 September 2010 (UTC)

Sure, I know they are equivalent, but I have never seen the expression in the literature. Here in Wikipedidia we need verifiable sources - see wp:V and wp:RS. We also are not allowed to create and present a quick proof here, as that would be orginal research (see wp:NOR), and note that this is not just a trivial calculation in the spirit of wp:CALC. So, unless we have a source for the expression, I'm afraid it will have to go. On the other hand, I have seen int{ v dp } been done as int{ v d(m γ v) } etc... many times in the literature, so I guess a source can easily be found for it.

Meanwhile, since throwing a {{cn}}-tag at the expression looks a bit silly, I have restored the "original" version. I suggest we keep it that way, and first discuss what we do with the section on the talk page. DVdm (talk) 21:24, 27 September 2010 (UTC)

What if we introduced the integration by parts

${\displaystyle \int \mathbf {v} \cdot d\mathbf {p} =\mathbf {p} \cdot \mathbf {v} -\int \mathbf {p} \cdot d\mathbf {v} }$

in the non-relativistic section before we plugged in ${\displaystyle \mathbf {p} =m\mathbf {v} }$? Then we could use

${\displaystyle E_{k}=\mathbf {p} \cdot \mathbf {v} -\int \mathbf {p} \cdot d\mathbf {v} }$

in the relativistic section by directly plugging in ${\displaystyle \mathbf {p} ={\frac {m\mathbf {v} }{\sqrt {1-v^{2}/c^{2}}}}\,.}$ Quantling (talk) 13:12, 28 September 2010 (UTC)

I would not want to complicate the non-relativistic section in order to simplify the relativistic section because many more people will look at the former than at the latter. JRSpriggs (talk) 14:09, 28 September 2010 (UTC)
The IPB is very nice, but whatever we do, I think it should be solidly sourced. Otherwise the section will never stabilise. DVdm (talk) 16:22, 28 September 2010 (UTC)

FWIW, ${\displaystyle H=\int \mathbf {v} \cdot d\mathbf {p} }$ is the kinetic part of a Hamiltonian, though this relationship is usually expressed as ${\displaystyle dH/dp_{i}=v_{i}}$. Similarly, ${\displaystyle L=\int \mathbf {p} \cdot d\mathbf {v} }$ is the kinetic part of a Lagrangian, though this relationship is usually expressed as ${\displaystyle dL/dv_{i}=p_{i}}$. The integration by parts to give the Hamiltonian in terms of the Lagrangian is over a century old. — Quantling (talk) 20:22, 28 September 2010 (UTC)

Afaiac, put it in the article, perhaps in a separate little section, but please make sure to include a proper source citation. Thx - DVdm (talk) 20:55, 28 September 2010 (UTC)

## Kinetic and total energy in General Relativity

I think that in full general relativity we cannot say that total energy is conserved, even locally; and I am removing the part of the article that indicates that it is. My reasoning is as follows. In general relativity, the covariant conservation of total energy, including the contribution from kinetic energy is expressed as ${\displaystyle \nabla _{\alpha }T^{\alpha \beta }=0}$ for the symmetric stress-energy tensor ${\displaystyle T^{\alpha \beta }\,}$. At any given point ${\displaystyle p\,}$ in space-time, it is possible to find a coordinate system in which the metric tensor ${\displaystyle g_{\alpha \beta }\,}$ equals the Minkowski metric ${\displaystyle \eta _{\alpha \beta }\,}$ and for which the gradient of metric tensor ${\displaystyle g_{\alpha \beta ,\gamma }\,}$ is zero. In this coordinate system, the covariant conservation of energy appears as ${\displaystyle \partial _{\alpha }T^{\alpha \beta }=0}$. If we had this last expression true not just at ${\displaystyle p\,}$ but throughout a neighborhood of ${\displaystyle p\,}$, then we could use the divergence theorem to conclude that the total four-flux of the energy (or any component of momentum) through the boundary of the neighborhood (or any smaller neighborhood of ${\displaystyle p\,}$) is zero. That is, we can conclude that any energy that enters such a neighborhood eventually leaves the neighborhood; and this is what we mean by conservation of energy. However, if space-time is curved and continuous at the point ${\displaystyle p\,}$ then there will not be a coordinate frame for which ${\displaystyle \partial _{\alpha }T^{\alpha \beta }=0}$ throughout some neighborhood of ${\displaystyle p\,}$. Thus the total four-flux through the boundary of any neighborhood of ${\displaystyle p\,}$ is not forced to zero, and we do not have conservation of energy in the ordinary sense. This applies even for very small "local" neighborhoods of ${\displaystyle p\,}$. —Quantling (talk | contribs) 16:22, 11 November 2010 (UTC)

I believe that "The Road to Reality" by Roger Penrose makes this argument, though I don't have the book in front of me and can't give you the page number. —Quantling (talk | contribs) 16:34, 11 November 2010 (UTC)

I have removed the (now moot) source and cn-tagged the statement. Surely someone can provide a good source. Otherwise we'll have to remove it. Cheers - DVdm (talk) 16:38, 11 November 2010 (UTC)
See Stress-energy-momentum pseudotensor. If we are contained in a region of space which (for practical purposes) is asymptotically Minkowskian as the distance from the Sun increases to infinity, then one can define gravitational potential energy in such a way that the total energy is globally conserved. JRSpriggs (talk) 20:23, 11 November 2010 (UTC)
It looks like that definition of gravitational energy achieves conservation by subtracting off all the other forms of energy, so that when the gravitational energy is added to these other energies, the total energy is a function of the metric tensor only. To me it looks like one gets conservation of total energy by effectively ignoring the kinetic energy, electromagnetic energy etc. that one usually computes. —Quantling (talk | contribs) 20:54, 11 November 2010 (UTC)
As point 4 of the verification says, "the explicit second derivative terms in the pseudotensor cancel with the implicit second derivative terms contained within the Einstein tensor". So it only depends on the metric and its first partial derivatives. An explicit expression is given lower down in the article. The Einstein field equations are used to convert material stress-energy into an expression which does not refer to matter. This is no different than what we do whenever we try to determine a formula for a new kind of energy — we have to make sure that it increases when known forms decrease and vice-versa by the same amount. JRSpriggs (talk) 22:02, 11 November 2010 (UTC)
Because the "non-conservation in GR" material is controversial and not all that relevant to the article, I removed it on 15 November, 2010. —Quantling (talk | contribs) 18:54, 18 November 2010 (UTC)

### The new section on kinetic energy in GR

Recently the General relativity section has been changed to be less clear. The use of ${\displaystyle p_{\beta }=mg_{\alpha \beta }u^{\alpha }}$ is unnecessary since ${\displaystyle g_{\alpha \beta }u^{\alpha }}$ is simply a change of base and can do without. On top of that, the sum over the beta's is ambiguous, since the metric is not specified, and should be left as I had it. I also think it is important to include a section on stationary observes in a metric since it can be intuitively understood and provides insight. For these reasons I will be editing the General relativity section so as to contain those items again. —Preceding unsigned comment added by 24.121.104.225 (talk) 21:30, 9 April 2011 (UTC)

There are a number of problems with the subsection as you wrote it, including: (1) it fails to distinguish between covariant and contravariant indices consistently; (2) it ignores the fact that linear momentum is a covariant vector rather than a contravariant vector; and (3) 1/(gtt) may not be the same as gtt, if g is not a diagonal matrix.
Furthermore, the sum over the betas is NOT ambiguous if one is covariant and one is contravariant, as I had it. No metric is needed in that case.
It appears to me that your understanding of general relativity is limited, and thus you should be cautious in editing things on that subject. JRSpriggs (talk) 00:44, 11 April 2011 (UTC)
Thank you for pointing out the diagonal aspect for my expression. I'm a little confused though, I did not have that gtt=gtt, I had that they were inverses of each other. Is gtt not the inverse of gtt? Second, are you saying pb and pa are not both covariant? In my case momentum was a covariant vector, following the well know chant "co is low". Are you saying gabua does not equal ub? Finally, I am also unfamiliar with writing gamma and way you have in the formula for the stationary observer, can you explain why you wrote it the way you did? Thank you. — Preceding unsigned comment added by Jfy4 (talkcontribs) 05:16, 11 April 2011 (UTC)
Notice that gαβ is the inverse matrix of gαβ, that is,
${\displaystyle g^{\alpha \beta }g_{\beta \gamma }=\delta _{\gamma }^{\alpha }\,}$
where δαβ is the Kronecker delta. So the corresponding element on the diagonal of the inverse will not be the inverse of the diagonal generally, but usually only when the off-diagonal elements are zero.
Unlike some people, I prefer to show all raising and lowering of indices explicitly. To answer your questions:
• No, it is not.
• No, I am not saying that.
• I would not use uβ at all.
• See the following calculation:
${\displaystyle u^{\alpha }={\frac {dx^{\alpha }}{dt}}{\frac {dt}{d\tau }}=v^{\alpha }u^{t}\,.}$
Thus for a diagonal and spatially isotropic metric
${\displaystyle -c^{2}=g_{\alpha \beta }u^{\alpha }u^{\beta }=g_{tt}(u^{t})^{2}+g_{ss}v^{2}(u^{t})^{2}\,.}$
Solving for ut gives
${\displaystyle u^{t}=c{\sqrt {\frac {-1}{g_{tt}+g_{ss}v^{2}}}}\,.}$
Is that clear? JRSpriggs (talk) 10:44, 11 April 2011 (UTC)
Yes, that is clear. Thank you for explaining it to me. I'll be careful in the future. — Preceding unsigned comment added by Jfy4 (talkcontribs) 02:30, 12 April 2011 (UTC)

──────────────────────────────────────────────────────────────────────────────────────────────────── I am torn about whether we should just delete the subsection or try to finish fixing it. It could be argued that the subsection is unnecessary because according to the equivalence principle the locally measured kinetic energy should be unaffected by where it is in a gravitational field. On the other hand, if one wants to express the kinetic energy in terms of a given coordinate system (in which there is a non-zero gravitational potential), then it might be useful to have an expression. See Parameterized post-Newtonian formalism#Alpha-zeta notation and Talk:Schwarzschild geodesics#Is the energy formula correct?. Of course, if we were playing by the official rules around here, we should come up with a citation to a reliable source (journal or book) which gives the formulas instead of trying to derive them ourselves (see WP:OR). JRSpriggs (talk) 05:15, 12 April 2011 (UTC)

It seems to me that the subsection on GR, as now included, addresses a rather trivial problem that only applies to spacial points (single particles), anyway. Kinetic energy of systems (even in SR) could use some work here, as in SR the spacially-extended-system KE = Ek is (Total energy system/c^2) - (E contribution from massless particles/c^2) - (sum of potential energies/c^2) - (sum of rest masses of particles in system). For one massive particle, that is a lot simpler and you lose the middle terms: Ek = sqrt(E^2 + p^2)- m = (E/c^2)-m = is what is in this article.

In GR, as you know, the problem of KE suffers from the problem of total E and only arises in spacially-extended systems, not locally. Since indeed, the equivalence principle always garantees that locally you can choose a frame that gives you no gravitational energy and locally flat space, so the problem for a point, in the inertial frame, always reduces to the SR answer for point particles (as seen in this article). Extend beyond the point and now you have potentials you can't get rid of (any given point in a gravitational wave can have no energy for some inertial observer at that point, and yet the wave still has energy as an extended-system property because you can't go around choosing inertial observers everywhere). So you can't use the familiar [total energy] minus [other energies except kinetic] formula, because there's no well-defined way to figure total energy. In this article the "GR approach" messes around at a single point, merely reproducing the SR single-point answer (which in all fairness certainly includes terms EM potentials and Newtonian G potentials, too! This is basically use of the Hamiltonian, with an extra term for rest-mass energy). If you go to any point-solutions that need a semi-GR approach due to the high G involved (like near a neutron star) you'll simply find that the very high non-Newtonian G potential is given by GR gravitational-potential equations, and the rest of the thing again looks just like SR, but with a GR term for the G-potential (basically this is the 3+1 split, made possible by no spacial-extension). However, with spacially-extended systems, it seems to me that you run into the same problems that GR does for total mass and total energy, so you might as well give up, unless you want to reproduce the entire mass in general relativity article, with some [sum of rest masses] and [sum of EM potentials] terms subtracted from all the kinds of GR mass defined there. No? SBHarris 18:32, 12 April 2011 (UTC)

## kinetic energy wrong for rocket in space?

Possibly kinetic energy was created, when wasn't rockets, which can fly to Mars. Also kinetic energy for gravity possibly was addapted later and whats why kinetic energy formula mgh is wrong.

Kinetic energy formula may be correct for calculating how much car engine energy, which is Force ma was wasted to accelerate car to speed v. Also kinetic energy may be correct for caluculating, what distance car will brake if it goes 2 times faster, so 4 times bigger distance. Why 4 times? Gravity force 9,8*m, where m is mass, on car braking acting some amount of time, and this time is shorter for faster car. For example to deaccelerate from 40 km/h, to 30 km/h (and say, then car moves distance 10 meters), need less, time, than to deaccelerate from 20 km/h, to 10 km/h (and say, then car moves distance 10 meters), so gravity force F=9,8*(mass) accting less time. And thats why possible it is answer why 2 times faster car need 4 times bigger distance to deaccelerate, but in this case wheels absolutely do not rotating during braking. If there not wheels tyres braking, but wheels still rotating; but braking disks with friction into wheels, then amount of time, is at bigger speed, shorter of car deaccelerating from 40 km/h to 30 km/h (distance, say, car moves 10 meters), than of car, which deaccelerating from 20 km/h to 10 km/h (say deacceleration distance is 10 meters). Shorter time means less (smaller) braking discs friction (those discs are made for braking car), but in this case is harder to understand, than with shorter time gravity force [down] on do not rotating whells (braking car into tyres). So there can be even two possible solution of how much bigger distance 2 times faster car will deaccelerate. If car wheels slowly rotating during deacceleration then 2 times car will longer distance brake (if car is 2 times faster). And if car wheels do not rotating during deaccelration, then 4 times longer distance 2 times faster car will brake. — Preceding unsigned comment added by Versatranitsonlywaytofly (talkcontribs) 19:19, 2 March 2011 (UTC)

No, in general it doesn't matter what kind of braking system is used, if we only assume that it is able to apply a maximum FORCE (F) on the wheels (and thus on the car), when you brake as hard as you can. Energy = Force x distance = F*d = 1/2 * m v^2. So if your velocity increases by 2, your kinetic energy increases by a factor of 4, and with the same braking force, so does the distance through which the force needs to be applied to stop you. Non-slip braking systems may be able to bring to bear more stopping force than sliding friction in a full skid, but whatever that force is, it's still a maximum and doesn't go up when you go faster. So you might be able to stop in a smaller distance for the SAME speed with non-skid anti-lock brakes, but THAT (smaller) distance, whatever it is, will still increase by about a factor of 4, if you double your speed. SBHarris 22:11, 2 March 2011 (UTC)

## Original Research - again

User:Peter.sujak has repeatedly been adding [[WP:OR}} and has been warned. Starting discussion here of content added. Asked multiple times for user to discuss (to no effect). Best, 22:46, 20 August 2011 (UTC)

## Relativistic kinetic energy

I'm not a physicist, I assume some of you must be, if we take the second term of the integration by parts - the integral of m gamma v by dv - the formula given has a maximum value of mc squared (0 to c)- but if you draw the graph, it tends to infinity as v goes to c. I've seen this derivation in other places on the web, is this taught at uni? Is it a recognized mathematical trick? If so maybe an explanation is needed. AbsoluteZero01 (talk) 10:11, 24 November 2011 (UTC)

The integral, when written down as a definite integral would have (for instance) V as the integration variable, and [ 0, v ] as the integration interval, not [ 0, c ], so there is no infinity involved. If we do the gamma integral with (for instance) Γ as the integration variable, the integration interval is [ 1, γ ]. Of course no infinity problem there either.

I agree that this derivation should be properly sourced, if not seriously trimmed down. I have tagged the section. - DVdm (talk) 10:50, 24 November 2011 (UTC)

Thanks for the reply, it would be a shame to lose it as I enjoyed reading it and following the steps and the insights, but I've just looked it up on Wiki and I think the integral is an improper integral with a vertical asymptote, I may be missing something really obvious, like it is OK to truncate it or something.AbsoluteZero01 (talk) 12:53, 24 November 2011 (UTC)
I see no asymptote. For instance take ∫ γ d(v2) = 2 ∫ γ v dv = 2 ∫ v/√(1-v2/c2) dv. If you plot the integrand v/√(1-v2/c2) with v between 0 and some value < c, there is no asymptote in that interval and the kinetic energy is finite. There's only an asymptote at v=c, meaning that kinetic energy at speed c is infinite. - DVdm (talk) 15:41, 24 November 2011 (UTC)
The area under the graph of f(v) = mγv between 0 and c is infinite, however if you plug in the range 0 to v = c for the anti derivative [-mc2/γ] you get - 0 - - mc2 = mc2.AbsoluteZero01 (talk) 19:36, 24 November 2011 (UTC)
Ha, yes. Actually, the area under the graph of f(v) = mγv between 0 and c is finite, contrary to what I said before—sorry for that. Its value is mc2. The infinity of the kinetic energy at v=c stems from the (partial integration) first term mγv2. I have struck my mistake in my previous message. - DVdm (talk) 20:10, 24 November 2011 (UTC)

────────────────────────────────────────────────────────────────────────────────────────────────────I don't understand this whole argument. For example, the integral of dx/x is ln(x). Using a definite integral with 0 as one of your endpoints is definitely therefore going to get you into trouble, but that's because 1/x is not defined for x = 0. That only means that particular integral is improper, not that the integration itself is improper. You can do the integration from x1 to x2 and get perfectly valid answers. f(x) = ln(x1)-ln(x2). You must stay away from x=0 because it isn't defined in the original function, either. SBHarris 20:16, 24 November 2011 (UTC)

Indeed, there is no problem with the integration. And... once again we see why all content should be properly sourced. - DVdm (talk) 20:22, 24 November 2011 (UTC)
I think what I'm trying to say is that the area under the graph say from 0 to 0.99999c is larger than mc2, meaning that [-mc2/γ] is an approximation that works at lower speeds but it is not the true area under the graph when you get to speeds approaching c, so I was wondering whether this was a recognized mathematical trick. As I said I could be missing something.AbsoluteZero01 (talk) 07:41, 25 November 2011 (UTC)
To AbsoluteZero01: You seem to be assuming that the integral from 0 to c of a function which goes to infinity at c must be infinite. That is not true. It may be infinite for some functions, but not for all. And in this particular case it is not true. JRSpriggs (talk) 08:37, 25 November 2011 (UTC)
I must admit that I can't see how it is finite in this case, especially as you can actually plot it and look at it (I used a free piece of software called Graph 4.3, set m=4, c=2), as I said maybe an explanation is needed, rather than just the statement that it is finite. Thanks.AbsoluteZero01 (talk) 09:41, 25 November 2011 (UTC)
Note that the integral of 1/x between 0 and 1 is infinite, whereas the integral of 1/√x over the same interval is finite (value=2). If this really makes no sense to you, and you still want to talk about it on Wiki, you'll have to go to WP:Reference desk/Mathematics, since this article talk page is really for discussing the article, not the subject or the mathematics behind some step in some derivation. Cheers - and good luck. - DVdm (talk) 09:55, 25 November 2011 (UTC)
OK.AbsoluteZero01 (talk) 10:24, 25 November 2011 (UTC)

## Simple kinetic energy equation

The current simple equation at the top of the article states: mv²/2. It seems simpler to me to me if it was written: 0.5mv². I believe it would be at least more correct to write (mv²)/2. Any reason why the notation is confusing? Veoulux (talk) 10:21, 28 November 2011 (UTC)

Yes, the way it is written now, is a bit awkward. I think it should be written as ½mv2 or as ½mv². I picked the former, but feel free to hone. - DVdm (talk) 10:35, 28 November 2011 (UTC)
Thank you for the response and the fix, but as it is, that could be taken to mean 1/2 * m * v² which is what it should be. However it could also be read as 1/(2 * m * v²) which is incorrect. With the edit you made, it is lees likely to be confused, but I think it would benefit most with some parenthesis, or a decimal rather than a fraction to eliminate any confusion. Veoulux (talk) 11:21, 28 November 2011 (UTC)
I think the current way is standard, if you check the literature. Adding parentheses would make it more awkward. Would the addition of a space ( ½ mv² ) make you feel more comfortable? - DVdm (talk) 11:30, 28 November 2011 (UTC)
It's a formatting thing from this point on, I don't believe anyone will think that mv² is part of the denominator now that there is a space. Thanks for your patience with me. Veoulux (talk) 13:12, 28 November 2011 (UTC)
No problem.  Done. Cheers - DVdm (talk) 13:15, 28 November 2011 (UTC)

## Quantum mechanics

Why does the section "Quantum mechanical kinetic energy of rigid bodies" only discuss the kinetic energy of a system of electrons? On the whole, I think this section is too specific/deep for this page. At the very least the formula

${\displaystyle {\hat {T}}={\frac {{\hat {p}}^{2}}{2m}}}$

which is the quantum-mechanical kinetic energy operator (see Hamiltonian (quantum mechanics)) should be presented first instead of that unilluminating summation. Xuanji (talk) 11:13, 8 January 2012 (UTC)

I believe that this is the most commonly used approximate method for computing the energy of a molecule. The nuclei are so much more massive (generally tens of thousands of times) than the electrons that they may be treated using classical physics. If the positions of the nuclei are taken as given, then one can find the minimum energy solution for the Schrödinger equation of the electrons taken collectively. This then gives one a classical potential energy which can be regarded as a function of the positions of the nuclei and solved classically. JRSpriggs (talk) 19:20, 8 January 2012 (UTC)
Point taken, but I still feel that this section should be a general introduction to the concept of kinetic energy in quantum mechanics rather than say a guide on how one might actually go about calculating it for the special case of a molecule. — Preceding unsigned comment added by Xuanji (talkcontribs) 07:28, 10 January 2012 (UTC)
Then have a go at changing the section. If I have a problem with it, I will let you know. JRSpriggs (talk) 16:25, 10 January 2012 (UTC)
Ok, done Xuanji (talk) 15:36, 12 January 2012 (UTC)

## case of a rolling wheel

A couple of examples in some text books tend to pivot on the point of contact with the ground to calculate MOI (parallel axis theorem) and total energy and not translation + rotational motion of the center of the wheel. The idea is that the rotational cum translation can be attributed to the continuous tipping at the point of contact. I have found this to be a very ad-hoc method though have seen some books point to it and attribute it to pure rotation about the base point. However, on an counter example, I have never seen rotational motion attributed to work done by friction. A ball can roll and slide at the same time.. any examples appreciated -Alok 17:24, 8 July 2013 (UTC)

## Is the formula written correctly?

I don't think the following formula is correctly written:
Ek ½mv'= '2
In math/algebra, when you multiply something by 1, writing the 1 is redundant.
The proposed, more correct version is: Ek = mv2 ÷ 2 (or with a line under it)

18:44, 8 February 2014 (UTC)

The mere presence of an unnecessary factor of 1 does not make the formula incorrect. Many sources choose to put the fractional coefficient out front. This is especially encouraged in this (Wikipedia) environment since the one-half "½" is just one character while doing it as you suggest would require at least two characters (or many depending on the situation). JRSpriggs (talk) 03:27, 9 February 2014 (UTC)
I said the formula was simply redundant. I still think that adding 2 characters is not really a big stuff. (2 characters in Unicode take 4 bytes. Assuming (from a quick google search) that 1 average hard drive has 500 GB, that is 7.45058*10-12% of the total storage capacity). 76.10.160.64 (talk) 00:58, 13 February 2014 (UTC)
There is no factor 1. There is a factor ${\displaystyle {\tfrac {1}{2}}}$, which in ASCII renders like ½. The formula is written correctly. - DVdm (talk) 08:26, 13 February 2014 (UTC)
Said factor is equivalent to dividing an equation by 2 76.10.160.64 (talk) 01:40, 14 February 2014 (UTC)
Equations are not divided. Factors are divided. If said factor is equivalent to dividing by two, then said factor, both as ${\displaystyle {\tfrac {1}{2}}}$ or ½, is perfectly okay here, as elsewhere in this encyclopedia. - DVdm (talk) 07:30, 14 February 2014 (UTC)

## limits of integration

Integrating the force upon displacement dx (in the proof of K.E formula) is from 0→x and this is equivalent to ingratiating {v.d(mv)} from 0→v. — Preceding unsigned comment added by Izees (talkcontribs) 17:44, 6 June 2014 (UTC)

## Charged particle KE

What is the kinetic energy of moving charged particles?--5.2.200.163 (talk) 15:24, 29 March 2016 (UTC)

Perhaps you can ask at wp:Reference desk/science. Here we can only the discuss the article, not the subject—see wp:Talk page guidelines. Good luck at the ref desk! - DVdm (talk) 15:31, 29 March 2016 (UTC)
I think this aspect would be a good (proposed) addition to article. I'll ask also there.--5.2.200.163 (talk) 15:36, 29 March 2016 (UTC)
The fact that the electric and magnetic fields surrounding a charged object contain energy and momentum does not affect the formulas in this article. They contribute proportionally to the mass and the kinetic energy of the object so that the relationship between mass and kinetic energy remains the same. JRSpriggs (talk) 01:50, 30 March 2016 (UTC)
How about the involvement of electromagnetic mass?--5.2.200.163 (talk) 15:50, 30 March 2016 (UTC)
That is what I meant by "They contribute ... to the mass ...". The mass of a clothed particle maybe considered to be its bare mass plus the equivalent mass of the energy in its electric field. However, only the clothed mass is strictly definable. Bare mass would be negative infinity, and the self-energy of the electric field would be positive infinity, if one does not cut-off the integration at some minimal distance. JRSpriggs (talk) 14:04, 31 March 2016 (UTC)

How about the motion of some magnetic balls or MHD generators? What is the involvement of electromagnetic momentum associated to electromagnetic mass in these cases?--5.2.200.163 (talk) 14:00, 4 May 2016 (UTC)

If you have some reason to think that the kinetic energy does not follow the formulas given in this article in some case, then state what it is. Otherwise, just asking "what about this? what about that?" is pointless.
Even if there were some deviation from these formulas for elementary particles, the charge-to-mass ratio is so much smaller (in magnitude) for bulk matter than it is for elementary particles that the deviation for charged or magnetized bulk matter would be negligible. JRSpriggs (talk) 18:36, 4 May 2016 (UTC)

# Comments about the non-relativistic derivation

More referencing is needed for this important section. Furthermore the following sentence may need to be clarified: "Since this is a total differential (that is, it only depends on the final state, not how the particle got there), we can integrate it and call the result kinetic energy." Here, the terms "this" and "it" could be made clearer to the reader. Additionally the phrase "that is, it only depends on the final state, not how the particle got there" could be written a bit more precisely. The reader may also have confusions with regard to the differences and similarities between the terms "total differential" and "exact differential". My suggestion is that the term "exact differential" may be introduced instead. Zyvov 06:43, 27 October 2016 (UTC) — Preceding unsigned comment added by Zyvov (talkcontribs)