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Kinetic energy

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From Wikipedia, the free encyclopedia
Energy of a moving physical body

Kinetic energy
The cars of aroller coaster reach their maximum kinetic energy when at the bottom of the path. When they start rising, the kinetic energy begins to be converted to gravitationalpotential energy. The sum of kinetic and potential energy in the system remains constant, ignoring losses due torolling resistance anddrag.
Common symbols
KE,Ek,K or T
SI unitjoule (J)
Derivations from
other quantities
Ek =⁠1/2⁠mv2
Ek =Et +Er
Part of a series on
Classical mechanics
F=dpdt{\displaystyle {\textbf {F}}={\frac {d\mathbf {p} }{dt}}}
Émilie du Châtelet (1706–1749) was the first to publish the relation for kinetic energyEkin∝mv2{\displaystyle E_{\text{kin}}\propto mv^{2}}, derived from the experimental observation of objects dropped into clay. (Portrait byMaurice Quentin de La Tour.)

Inphysics, thekinetic energy of an object is the form ofenergy that it possesses due to itsmotion.[1]

Inclassical mechanics, the kinetic energy of a non-rotating object ofmassm traveling at aspeedv is12mv2{\textstyle {\frac {1}{2}}mv^{2}}.[2]

The kinetic energy of an object is equal to thework, or force (F) in the direction of motion times its displacement (s), needed to accelerate the object fromrest to its given speed. The same amount of work is done by the object when decelerating from its currentspeed to a state of rest.[2]

TheSI unit of energy is thejoule, while theEnglish unit of energy is thefoot-pound.

Inrelativistic mechanics,12mv2{\textstyle {\frac {1}{2}}mv^{2}} is a goodapproximation of kinetic energy only whenv is much less than thespeed of light.

History and etymology

The adjectivekinetic has its roots in theGreek word κίνησιςkinesis, meaning "motion". The dichotomy between kinetic energy andpotential energy can be traced back toAristotle's concepts ofactuality and potentiality.[3]

The principle ofclassical mechanics thatE ∝mv2 is conserved was developed[dubious –discuss] byGottfried Leibniz andJohann Bernoulli, who described kinetic energy as theliving force orvis viva.[4]: 227 Willem 's Gravesande of the Netherlands provided experimental evidence of this relationship in 1722. By dropping weights from different heights into a block of clay, Gravesande determined that their penetration depth was proportional to the square of their impact speed.Émilie du Châtelet recognized the implications of the experiment and published an explanation.[5]

The termskinetic energy andwork in their present scientific meanings date back to the mid-19th century. Early understandings of these ideas can be attributed toThomas Young, who in his 1802 lecture to the Royal Society, was the first to use the termenergy to refer to kinetic energy in its modern sense, instead ofvis viva.Gaspard-Gustave Coriolis published in 1829 the paper titledDu Calcul de l'Effet des Machines outlining the mathematics of kinetic energy.William Thomson, later Lord Kelvin, is given the credit for coining the term "kinetic energy" c. 1849–1851.[6][7]William Rankine, who had introduced the term "potential energy" in 1853, and the phrase "actual energy" to complement it,[8] later citesWilliam Thomson andPeter Tait as substituting the word "kinetic" for "actual".[9]

Overview

Energy occurs in many forms, includingchemical energy,thermal energy,electromagnetic radiation,gravitational energy,electric energy,elastic energy,nuclear energy, andrest energy. These can be categorized in two main classes:potential energy and kinetic energy. Kinetic energy is the movement energy of an object. Kinetic energy can be transferred between objects and transformed into other kinds of energy.[10]

Kinetic energy may be best understood by examples that demonstrate how it is transformed to and from other forms of energy. For example, acyclist transferschemical energy provided by food to the bicycle and cyclist's store of kinetic energy as they increase their speed. On a level surface, this speed can be maintained without further work, except to overcomeair resistance andfriction. The chemical energy has been converted into kinetic energy, the energy of motion, but the process is not completely efficient and produces thermal energy within the cyclist.

The kinetic energy in the moving cyclist and the bicycle can be converted to other forms. For example, the cyclist could encounter a hill just high enough to coast up, so that the bicycle comes to a complete halt at the top. The kinetic energy has now largely been converted to gravitational potential energy that can be released by freewheeling down the other side of the hill. Since the bicycle lost some of its energy to friction, it never regains all of its speed without additional pedaling. The energy is not destroyed; it has only been converted to another form by friction. Alternatively, the cyclist could connect adynamo to one of the wheels and generate some electrical energy on the descent. The bicycle would be traveling slower at the bottom of the hill than without the generator because some of the energy has been diverted into electrical energy. Another possibility would be for the cyclist to apply the brakes, in which case the kinetic energy would be dissipated through friction asheat.

Like any physical quantity that is a function of velocity, the kinetic energy of an object depends on the relationship between the object and the observer'sframe of reference. Thus, the kinetic energy of an object is notinvariant.

Spacecraft use chemical energy to launch and gain considerable kinetic energy to reachorbital velocity. In an entirely circular orbit, this kinetic energy remains constant because there is almost no friction in near-earth space. However, it becomes apparent at re-entry when some of the kinetic energy is converted to heat. If the orbit iselliptical orhyperbolic, then throughout the orbit kinetic andpotential energy are exchanged; kinetic energy is greatest and potential energy lowest at closest approach to the earth or other massive body, while potential energy is greatest and kinetic energy the lowest at maximum distance. Disregarding loss or gain however, the sum of the kinetic and potential energy remains constant.

Kinetic energy can be passed from one object to another. In the game ofbilliards, the player imposes kinetic energy on the cue ball by striking it with the cue stick. If the cue ball collides with another ball, it slows down dramatically, and the ball it hit accelerates as the kinetic energy is passed on to it.Collisions in billiards are effectivelyelastic collisions, in which kinetic energy is preserved. Ininelastic collisions, kinetic energy is dissipated in various forms of energy, such as heat, sound and binding energy (breaking bound structures).

Flywheels have been developed as a method ofenergy storage. This illustrates that kinetic energy is also stored in rotational motion.

Several mathematical descriptions of kinetic energy exist that describe it in the appropriate physical situation. For objects and processes in common human experience, the formula⁠1/2⁠mv2 given byclassical mechanics is suitable. However, if the speed of the object is comparable to the speed of light,relativistic effects become significant and the relativistic formula is used. If the object is on the atomic orsub-atomic scale,quantum mechanical effects are significant, and a quantum mechanical model must be employed.

Kinetic energy for non-relativistic velocity

Treatments of kinetic energy depend upon the relative velocity of objects compared to the fixedspeed of light. Speeds experienced directly by humans arenon-relativisitic; higher speeds require thetheory of relativity.

Kinetic energy of rigid bodies

Inclassical mechanics, the kinetic energy of apoint object (an object so small that its mass can be assumed to exist at one point), or a non-rotatingrigid body depends on themass of the body as well as itsspeed. The kinetic energy is equal to half theproduct of the mass and the square of the speed. In formula form:

Ek=12mv2{\displaystyle E_{\text{k}}={\frac {1}{2}}mv^{2}}

wherem{\displaystyle m} is the mass andv{\displaystyle v} is the speed (magnitude of the velocity) of the body. InSI units, mass is measured inkilograms, speed inmetres per second, and the resulting kinetic energy is injoules.

For example, one would calculate the kinetic energy of an 80 kg mass (about 180 lbs) traveling at 18 metres per second (about 40 mph, or 65 km/h) as

Ek=12⋅80kg⋅(18m/s)2=12,960J=12.96kJ{\displaystyle E_{\text{k}}={\frac {1}{2}}\cdot 80\,{\text{kg}}\cdot \left(18\,{\text{m/s}}\right)^{2}=12,960\,{\text{J}}=12.96\,{\text{kJ}}}

When a person throws a ball, the person doeswork on it to give it speed as it leaves the hand. The moving ball can then hit something and push it, doing work on what it hits. The kinetic energy of a moving object is equal to the work required to bring it from rest to that speed, or the work the object can do while being brought to rest:net force × displacement = kinetic energy, i.e.,

Fs=12mv2{\displaystyle Fs={\frac {1}{2}}mv^{2}}

Since the kinetic energy increases with the square of the speed, an object doubling its speed has four times as much kinetic energy. For example, a car traveling twice as fast as another requires four times as much distance to stop, assuming a constant braking force. As a consequence of this quadrupling, it takes four times the work to double the speed.

The kinetic energy of an object is related to itsmomentum by the equation:Ek=p22m{\displaystyle E_{\text{k}}={\frac {p^{2}}{2m}}}

where:

For thetranslational kinetic energy, that is the kinetic energy associated withrectilinear motion, of arigid body with constantmassm{\displaystyle m}, whosecenter of mass is moving in a straight line with speedv{\displaystyle v}, as seen above is equal to

Et=12mv2{\displaystyle E_{\text{t}}={\frac {1}{2}}mv^{2}}

where:

The kinetic energy of any entity depends on the reference frame in which it is measured. However, the total energy of an isolated system, i.e. one in which energy can neither enter nor leave, does not change over time in the reference frame in which it is measured. Thus, the chemical energy converted to kinetic energy by a rocket engine is divided differently between the rocket ship and its exhaust stream depending upon the chosen reference frame. This is called theOberth effect. But the total energy of the system, including kinetic energy, fuel chemical energy, heat, etc., is conserved over time, regardless of the choice of reference frame. Different observers moving with different reference frames would however disagree on the value of this conserved energy.

The kinetic energy of such systems depends on the choice of reference frame: the reference frame that gives the minimum value of that energy is thecenter of momentum frame, i.e. the reference frame in which the total momentum of the system is zero. This minimum kinetic energy contributes to theinvariant mass of the system as a whole.

Derivation

Without vector calculus

The work W done by a forceF on an object over a distances parallel toF equals

W=F⋅s.{\displaystyle W=F\cdot s.}

UsingNewton's second law

F=ma{\displaystyle F=ma}

withm the mass anda theacceleration of the object and

s=at22{\displaystyle s={\frac {at^{2}}{2}}}

the distance traveled by the accelerated object in timet, we find withv=at{\displaystyle v=at} for the velocityv of the object

W=maat22=m(at)22=mv22.{\displaystyle W=ma{\frac {at^{2}}{2}}={\frac {m(at)^{2}}{2}}={\frac {mv^{2}}{2}}.}

With vector calculus

The work done in accelerating a particle with massm during the infinitesimal time intervaldt is given by the dot product offorceF and the infinitesimaldisplacementdx

F⋅dx=F⋅vdt=dpdt⋅vdt=v⋅dp=v⋅d(mv),{\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} )\,,}

where we have assumed the relationshipp = m v and the validity ofNewton's second law. (However, also see the special relativistic derivationbelow.)

Applying theproduct rule we see that:

d(v⋅v)=(dv)⋅v+v⋅(dv)=2(v⋅dv).{\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 so thatdm = 0), we have

v⋅d(mv)=m2d(v⋅v)=m2dv2=d(mv22).{\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 atotal 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:

Ek=∫v1v2p⋅dv=∫v1v2mv⋅dv=mv22|v1v2=12m(v22−v12).{\displaystyle E_{\text{k}}=\int _{v_{1}}^{v_{2}}\mathbf {p} \cdot d\mathbf {v} =\int _{v_{1}}^{v_{2}}m\mathbf {v} \cdot d\mathbf {v} ={mv^{2} \over 2}{\bigg \vert }_{v_{1}}^{v_{2}}={1 \over 2}m(v_{2}^{2}-v_{1}^{2}).}

This equation states that the kinetic energy (Ek) is equal to theintegral of thedot product of themomentum (p) of a body and theinfinitesimal change of thevelocity (v) of the body. It is assumed that the body starts with no kinetic energy when it is at rest (motionless).

Rotating bodies

If a rigid body Q is rotating about any line through the center of mass then it hasrotational kinetic energy (Er{\displaystyle E_{\text{r}}\,}) which is simply the sum of the kinetic energies of its moving parts, and is thus given by:

Er=∫Qv2dm2=∫Q(rω)2dm2=ω22∫Qr2dm=ω22I=12Iω2{\displaystyle E_{\text{r}}=\int _{Q}{\frac {v^{2}dm}{2}}=\int _{Q}{\frac {(r\omega )^{2}dm}{2}}={\frac {\omega ^{2}}{2}}\int _{Q}{r^{2}}dm={\frac {\omega ^{2}}{2}}I={\frac {1}{2}}I\omega ^{2}}

where:

(In this equation the moment ofinertia must be taken about an axis through the center of mass and the rotation measured by ω must be around that axis; more general equations exist for systems where the object is subject to wobble due to its eccentric shape).

Kinetic energy of systems

A system of bodies may have internal kinetic energy due to the relative motion of the bodies in the system. For example, in theSolar System the planets and planetoids are orbiting the Sun. In a tank of gas, the molecules are moving in all directions. The kinetic energy of the system is the sum of the kinetic energies of the bodies it contains.

A macroscopic body that is stationary (i.e. a reference frame has been chosen to correspond to the body'scenter of momentum) may have various kinds ofinternal energy at the molecular or atomic level, which may be regarded as kinetic energy, due to molecular translation, rotation, and vibration, electron translation and spin, and nuclear spin. These all contribute to the body's mass, as provided by the special theory of relativity. When discussing movements of a macroscopic body, the kinetic energy referred to is usually that of the macroscopic movement only. However, all internal energies of all types contribute to a body's mass, inertia, and total energy.

Fluid dynamics

Influid dynamics, the kinetic energy per unit volume at each point in an incompressible fluid flow field is called thedynamic pressure at that point.[11]

Ek=12mv2{\displaystyle E_{\text{k}}={\frac {1}{2}}mv^{2}}

Dividing by V, the unit of volume:

EkV=12mVv2q=12ρv2{\displaystyle {\begin{aligned}{\frac {E_{\text{k}}}{V}}&={\frac {1}{2}}{\frac {m}{V}}v^{2}\\q&={\frac {1}{2}}\rho v^{2}\end{aligned}}}

whereq{\displaystyle q} is the dynamic pressure, and ρ is the density of the incompressible fluid.

Frame of reference

The speed, and thus the kinetic energy of a single object is frame-dependent (relative): it can take any non-negative value, by choosing a suitableinertial frame of reference. For example, a bullet passing an observer has kinetic energy in the reference frame of this observer. The same bullet is stationary to an observer moving with the same velocity as the bullet, and so has zero kinetic energy.[12] By contrast, the total kinetic energy of a system of objects cannot be reduced to zero by a suitable choice of the inertial reference frame, unless all the objects have the same velocity. In any other case, the total kinetic energy has a non-zero minimum, as no inertial reference frame can be chosen in which all the objects are stationary. This minimum kinetic energy contributes to the system'sinvariant mass, which is independent of the reference frame.

The total kinetic energy of a system depends on theinertial frame of reference: it is the sum of the total kinetic energy in acenter of momentum frame and the kinetic energy the total mass would have if it were concentrated in thecenter of mass.

This may be simply shown: letV{\displaystyle \textstyle \mathbf {V} } be the relative velocity of the center of mass framei in the framek. Since

v2=(vi+V)2=(vi+V)⋅(vi+V)=vi⋅vi+2vi⋅V+V⋅V=vi2+2vi⋅V+V2,{\displaystyle v^{2}=\left(v_{i}+V\right)^{2}=\left(\mathbf {v} _{i}+\mathbf {V} \right)\cdot \left(\mathbf {v} _{i}+\mathbf {V} \right)=\mathbf {v} _{i}\cdot \mathbf {v} _{i}+2\mathbf {v} _{i}\cdot \mathbf {V} +\mathbf {V} \cdot \mathbf {V} =v_{i}^{2}+2\mathbf {v} _{i}\cdot \mathbf {V} +V^{2},}

Then,

Ek=∫v22dm=∫vi22dm+V⋅∫vidm+V22∫dm.{\displaystyle E_{\text{k}}=\int {\frac {v^{2}}{2}}dm=\int {\frac {v_{i}^{2}}{2}}dm+\mathbf {V} \cdot \int \mathbf {v} _{i}dm+{\frac {V^{2}}{2}}\int dm.}

However, let∫vi22dm=Ei{\textstyle \int {\frac {v_{i}^{2}}{2}}dm=E_{i}} the kinetic energy in the center of mass frame,∫vidm{\textstyle \int \mathbf {v} _{i}dm} would be simply the total momentum that is by definition zero in the center of mass frame, and let the total mass:∫dm=M{\textstyle \int dm=M}. Substituting, we get:[13]

Ek=Ei+MV22.{\displaystyle E_{\text{k}}=E_{i}+{\frac {MV^{2}}{2}}.}

Thus the kinetic energy of a system is lowest to center of momentum reference frames, i.e., frames of reference in which the center of mass is stationary (either thecenter of mass frame or any othercenter of momentum frame). In any different frame of reference, there is additional kinetic energy corresponding to the total mass moving at the speed of the center of mass. The kinetic energy of the system in thecenter of momentum frame is a quantity that is invariant (all observers see it to be the same).

Rotation in systems

It sometimes is convenient to split the total kinetic energy of a body into the sum of the body's center-of-mass translational kinetic energy and the energy of rotation around the center of mass (rotational energy):

Ek=Et+Er{\displaystyle E_{\text{k}}=E_{\text{t}}+E_{\text{r}}}

where:

  • Ek is the total kinetic energy
  • Et is the translational kinetic energy
  • Er is therotational energy orangular kinetic energy in the rest frame

Thus the kinetic energy of a tennis ball in flight is the kinetic energy due to its rotation, plus the kinetic energy due to its translation.

Relativistic kinetic energy

See also:Mass in special relativity andTests of relativistic energy and momentum

If a body's speed relative to an inertial frame is a significant fraction of thespeed of light, it is necessary to use relativistic mechanics to calculate its kinetic energy relative to that frame.In relativistic mechanics, energy combines with momentum in a way analogous to the combination of time and space intospacetime. The energy component in the inertial frame can be expressed as:[14]: 195 E=γm,{\displaystyle E=\gamma m,}whereγ=1/1−v2/c2{\textstyle \gamma =1/{\sqrt {1-v^{2}/c^{2}}}} andm{\displaystyle m} is the object's(rest) mass,v{\displaystyle v} speed, andc the speed of light in vacuum. If the body is at rest in the inertial frame,v=0{\displaystyle v=0} andγ=1{\displaystyle \gamma =1}, soErest=m{\displaystyle E_{\textrm {rest}}=m}.[14]: 201  The rest energy of an object is its rest mass in units wherec=1{\displaystyle c=1}. InSI units,Erest=mc2{\displaystyle E_{\textrm {rest}}=mc^{2}}. The kinetic energy can then be defined as the total energy minus the rest energy:[14]: 201 (kinetic energy)=(energy)−(rest energy).{\displaystyle ({\textrm {kinetic}}\ {\textrm {energy}})=({\textrm {energy}})-({\textrm {rest}}\ {\textrm {energy}}).}

Another way to describe the kinetic energy is to start with the total energy given by theenergy-momentum relation:

E2=(pc)2+(mc2)2{\displaystyle E^{2}=(p{\textrm {c}})^{2}+\left(m{\textrm {c}}^{2}\right)^{2}\,}

Herep=mγv{\displaystyle p=m\gamma v} is the relativistic expression for linear momentum, whereγ=1/1−v2/c2{\textstyle \gamma =1/{\sqrt {1-v^{2}/c^{2}}}} andm{\displaystyle m} is the object's(rest) mass,v{\displaystyle v} speed, andc the speed of light in vacuum.Then kinetic energy is the total relativistic energy minus the rest energy:EK=E−m0c2=(pc)2+(mc2)2−m0c2{\displaystyle E_{K}=E-m_{0}c^{2}={\sqrt {(p{\textrm {c}})^{2}+\left(m{\textrm {c}}^{2}\right)^{2}}}-m_{0}c^{2}}

At low speeds, the square root can be expanded and the rest energy drops out, giving the Newtonian kinetic energy.

Log of relativistic kinetic energy versus log relativistic momentum, for many objects of vastly different scales. The intersections of the object lines with the bottom axis approaches the rest energy. At low kinetic energy the slope of the object lines reflect Newtonian mechanics. As the lines approachc{\displaystyle c} the slope bends at the lightspeed barrier.

Low speed limit

The mathematical by-product of this calculation is themass–energy equivalence formula, that mass and energy are essentially the same thing:[15]: 51 [16]: 121 

Erest=mc2{\displaystyle E_{\text{rest}}=mc^{2}}

At a low speed (v ≪c), the relativistic kinetic energy is approximated well by the classical kinetic energy. To see this, apply thebinomial approximation or take the first two terms of theTaylor expansion in powers ofv2{\displaystyle v^{2}} for the reciprocal square root:[15]: 51 

Ek≈mc2(1+12v2c2)−mc2=12mv2{\displaystyle E_{\text{k}}\approx mc^{2}\left(1+{\frac {1}{2}}{\frac {v^{2}}{c^{2}}}\right)-mc^{2}={\frac {1}{2}}mv^{2}}

So, the total energyEk{\displaystyle E_{k}} can be partitioned into the rest mass energy plus the non-relativistic kinetic energy at low speeds.

When objects move at a speed much slower than light (e.g. in everyday phenomena on Earth), the first two terms of the series predominate. The next term in the Taylor series approximation

Ek≈mc2(1+12v2c2+38v4c4)−mc2=12mv2+38mv4c2{\displaystyle E_{\text{k}}\approx mc^{2}\left(1+{\frac {1}{2}}{\frac {v^{2}}{c^{2}}}+{\frac {3}{8}}{\frac {v^{4}}{c^{4}}}\right)-mc^{2}={\frac {1}{2}}mv^{2}+{\frac {3}{8}}m{\frac {v^{4}}{c^{2}}}}

is small for low speeds. For example, for a speed of 10 km/s (22,000 mph) the correction to the non-relativistic kinetic energy is 0.0417 J/kg (on a non-relativistic kinetic energy of 50 MJ/kg) and for a speed of 100 km/s it is 417 J/kg (on a non-relativistic kinetic energy of 5 GJ/kg).

The relativistic relation between kinetic energy and momentum is given by

Ek=p2c2+m2c4−mc2{\displaystyle E_{\text{k}}={\sqrt {p^{2}c^{2}+m^{2}c^{4}}}-mc^{2}}

This can also be expanded as aTaylor series, the first term of which is the simple expression from Newtonian mechanics:[17]

Ek≈p22m−p48m3c2.{\displaystyle E_{\text{k}}\approx {\frac {p^{2}}{2m}}-{\frac {p^{4}}{8m^{3}c^{2}}}.}

This suggests that the formulae for energy and momentum are not special and axiomatic, but concepts emerging from the equivalence of mass and energy and the principles of relativity.

General relativity

See also:Schwarzschild geodesics

Using the convention that

gαβuαuβ=−c2{\displaystyle g_{\alpha \beta }\,u^{\alpha }\,u^{\beta }\,=\,-c^{2}}

where thefour-velocity of a particle is

uα=dxαdτ{\displaystyle u^{\alpha }\,=\,{\frac {dx^{\alpha }}{d\tau }}}

andτ{\displaystyle \tau } is theproper time of the particle, there is also an expression for the kinetic energy of the particle ingeneral relativity.

If the particle has momentum

pβ=mgβαuα{\displaystyle p_{\beta }\,=\,m\,g_{\beta \alpha }\,u^{\alpha }}

as it passes by an observer with four-velocityuobs, then the expression for total energy of the particle as observed (measured in a local inertial frame) is

E=−pβuobsβ{\displaystyle E\,=\,-\,p_{\beta }\,u_{\text{obs}}^{\beta }}

and the kinetic energy can be expressed as the total energy minus the rest energy:

Ek=−pβuobsβ−mc2.{\displaystyle E_{k}\,=\,-\,p_{\beta }\,u_{\text{obs}}^{\beta }\,-\,m\,c^{2}\,.}

Consider the case of a metric that is diagonal and spatially isotropic (gtt,gss,gss,gss). Since

uα=dxαdtdtdτ=vαut{\displaystyle u^{\alpha }={\frac {dx^{\alpha }}{dt}}{\frac {dt}{d\tau }}=v^{\alpha }u^{t}}

wherevα is the ordinary velocity measured w.r.t. the coordinate system, we get

−c2=gαβuαuβ=gtt(ut)2+gssv2(ut)2.{\displaystyle -c^{2}=g_{\alpha \beta }u^{\alpha }u^{\beta }=g_{tt}\left(u^{t}\right)^{2}+g_{ss}v^{2}\left(u^{t}\right)^{2}\,.}

Solving forut gives

ut=c−1gtt+gssv2.{\displaystyle u^{t}=c{\sqrt {\frac {-1}{g_{tt}+g_{ss}v^{2}}}}\,.}

Thus for a stationary observer (v = 0)

uobst=c−1gtt{\displaystyle u_{\text{obs}}^{t}=c{\sqrt {\frac {-1}{g_{tt}}}}}

and thus the kinetic energy takes the form

Ek=−mgttutuobst−mc2=mc2gttgtt+gssv2−mc2.{\displaystyle E_{\text{k}}=-mg_{tt}u^{t}u_{\text{obs}}^{t}-mc^{2}=mc^{2}{\sqrt {\frac {g_{tt}}{g_{tt}+g_{ss}v^{2}}}}-mc^{2}\,.}

Factoring out the rest energy gives:

Ek=mc2(gttgtt+gssv2−1).{\displaystyle E_{\text{k}}=mc^{2}\left({\sqrt {\frac {g_{tt}}{g_{tt}+g_{ss}v^{2}}}}-1\right)\,.}

This expression reduces to the special relativistic case for the flat-space metric where

gtt=−c2gss=1.{\displaystyle {\begin{aligned}g_{tt}&=-c^{2}\\g_{ss}&=1\,.\end{aligned}}}

In the Newtonian approximation to general relativity

gtt=−(c2+2Φ)gss=1−2Φc2{\displaystyle {\begin{aligned}g_{tt}&=-\left(c^{2}+2\Phi \right)\\g_{ss}&=1-{\frac {2\Phi }{c^{2}}}\end{aligned}}}

where Φ is the Newtoniangravitational potential. This means clocks run slower and measuring rods are shorter near massive bodies.

Kinetic energy in quantum mechanics

Further information:Hamiltonian (quantum mechanics)

Inquantum mechanics, observables like kinetic energy are represented asoperators. For one particle of massm, the kinetic energy operator appears as a term in theHamiltonian and is defined in terms of the more fundamental momentum operatorp^{\displaystyle {\hat {p}}}. The kinetic energy operator in thenon-relativistic case can be written as

T^=p^22m.{\displaystyle {\hat {T}}={\frac {{\hat {p}}^{2}}{2m}}.}

Notice that this can be obtained by replacingp{\displaystyle p} byp^{\displaystyle {\hat {p}}} in the classical expression for kinetic energy in terms ofmomentum,

Ek=p22m.{\displaystyle E_{\text{k}}={\frac {p^{2}}{2m}}.}

In theSchrödinger picture,p^{\displaystyle {\hat {p}}} takes the form−iℏ∇{\displaystyle -i\hbar \nabla } where the derivative is taken with respect to position coordinates and hence

T^=−ℏ22m∇2.{\displaystyle {\hat {T}}=-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}.}

The expectation value of the electron kinetic energy,⟨T^⟩{\displaystyle \left\langle {\hat {T}}\right\rangle }, for a system ofN electrons described by thewavefunction|ψ⟩{\displaystyle \vert \psi \rangle } is a sum of 1-electron operator expectation values:

⟨T^⟩=⟨ψ|∑i=1N−ℏ22me∇i2|ψ⟩=−ℏ22me∑i=1N⟨ψ|∇i2|ψ⟩{\displaystyle \left\langle {\hat {T}}\right\rangle =\left\langle \psi \left\vert \sum _{i=1}^{N}{\frac {-\hbar ^{2}}{2m_{\text{e}}}}\nabla _{i}^{2}\right\vert \psi \right\rangle =-{\frac {\hbar ^{2}}{2m_{\text{e}}}}\sum _{i=1}^{N}\left\langle \psi \left\vert \nabla _{i}^{2}\right\vert \psi \right\rangle }

whereme{\displaystyle m_{\text{e}}} is the mass of the electron and∇i2{\displaystyle \nabla _{i}^{2}} is theLaplacian operator acting upon the coordinates of theith electron and the summation runs over all electrons.

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

T[ρ]=18∫∇ρ(r)⋅∇ρ(r)ρ(r)d3r{\displaystyle T[\rho ]={\frac {1}{8}}\int {\frac {\nabla \rho (\mathbf {r} )\cdot \nabla \rho (\mathbf {r} )}{\rho (\mathbf {r} )}}d^{3}r}

whereT[ρ]{\displaystyle T[\rho ]} is known as thevon Weizsäcker kinetic energy functional.

See also

Notes

  1. ^Jain, Mahesh C. (2009).Textbook of Engineering Physics (Part I). PHI Learning Pvt. p. 9.ISBN 978-81-203-3862-3.Archived from the original on 2020-08-04. Retrieved2018-06-21.,Chapter 1, p. 9Archived 2020-08-04 at theWayback Machine
  2. ^abResnick, Robert and Halliday, David (1960)Physics, Section 7-5, Wiley International Edition
  3. ^Brenner, Joseph (2008).Logic in Reality (illustrated ed.). Springer Science & Business Media. p. 93.ISBN 978-1-4020-8375-4.Archived from the original on 2020-01-25. Retrieved2016-02-01.Extract of page 93Archived 2020-08-04 at theWayback Machine
  4. ^Feather, Norman (1959).An Introduction to the Physics of Mass Length and Time (Hardcover ed.). Edinburgh University Press. RetrievedJuly 9, 2025.
  5. ^Judith P. Zinsser (2007).Emilie du Chatelet: Daring Genius of the Enlightenment. Penguin.ISBN 978-0-14-311268-6.
  6. ^Crosbie Smith, M. Norton Wise (1989-10-26).Energy and Empire: A Biographical Study of Lord Kelvin. Cambridge University Press. p. 866.ISBN 0-521-26173-2.
  7. ^John Theodore Merz (1912).A History of European Thought in the Nineteenth Century. Blackwood. p. 139.ISBN 0-8446-2579-5.{{cite book}}:ISBN / Date incompatibility (help)
  8. ^William John Macquorn Rankine (1853)."On the general law of the transformation of energy".Proceedings of the Philosophical Society of Glasgow.3 (5).
  9. ^"... what remained to be done, was to qualify the noun 'energy' by appropriate adjectives, so as to distinguish between energy of activity and energy of configuration. The well-known pair of antithetical adjectives, 'actual' and 'potential,' seemed exactly suited for that purpose. ... Sir William Thomson and Professor Tait have lately substituted the word 'kinetic' for 'actual.'"William John Macquorn Rankine (1867)."On the Phrase "Potential Energy," and on the Definitions of Physical Quantities".Proceedings of the Philosophical Society of Glasgow.VI (III).
  10. ^Goel, V. K. (2007).Fundamentals Of Physics Xi (illustrated ed.). Tata McGraw-Hill Education. p. 12.30.ISBN 978-0-07-062060-5.Archived from the original on 2020-08-03. Retrieved2020-07-07.Extract of page 12.30Archived 2020-07-07 at theWayback Machine
  11. ^A. M. Kuethe and J. D. Schetzer (1959).Foundations of Aerodynamics, 2nd edition, p.53. John Wiley & SonsISBN 0-471-50952-3
  12. ^Sears, Francis Weston; Brehme, Robert W. (1968).Introduction to the theory of relativity. Addison-Wesley. p. 127.,Snippet view of page 127Archived 2020-08-04 at theWayback Machine
  13. ^Physics notes – Kinetic energy in the CM frameArchived 2007-06-11 at theWayback Machine.Duke.edu. Accessed 2007-11-24.
  14. ^abcTaylor, Edwin F.; Wheeler, John Archibald (1992).Spacetime physics: introduction to special relativity (2nd ed.). New York: W.H. Freeman.ISBN 978-0-7167-2327-1.
  15. ^abEinstein, Albert (1922).The Meaning of Relativity: Four Lectures Delivered at Princeton University, May, 1921. Methuen & Company Limited. pp. 51–52.
  16. ^Resnick, Robert (1968).Introduction to special relativity. New York: Wiley.ISBN 978-0-471-71725-6. Retrieved2024-11-17.
  17. ^Fitzpatrick, Richard (20 July 2010)."Fine Structure of Hydrogen".Quantum Mechanics.Archived from the original on 25 August 2016. Retrieved20 August 2016.

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