Kinetic energy - Wikipedia

In physics, the kinetic energy of an object is the energy that it possesses due to its motion. ... It is defined as the work needed to accelerate a body of a ... Kineticenergy FromWikipedia,thefreeencyclopedia Jumptonavigation Jumptosearch Energyofamovingphysicalbody KineticenergyThecarsofarollercoasterreachtheirmaximumkineticenergywhenatthebottomofthepath.Whentheystartrising,thekineticenergybeginstobeconvertedtogravitationalpotentialenergy.Thesumofkineticandpotentialenergyinthesystemremainsconstant,ignoringlossestofriction.CommonsymbolsKE,Ek,orTSI unitjoule(J)DerivationsfromotherquantitiesEk=1/2mv2 Ek=Et+Er PartofaseriesonClassicalmechanics F = d d t ( m v ) {\displaystyle{\textbf{F}}={\frac{d}{dt}}(m{\textbf{v}})} Secondlawofmotion History Timeline Textbooks Branches Applied Celestial Continuum Dynamics Kinematics Kinetics Statics Statistical Fundamentals Acceleration Angularmomentum Couple D'Alembert'sprinciple Energy kinetic potential Force Frameofreference Inertialframeofreference Impulse Inertia /Momentofinertia Mass Mechanicalpower Mechanicalwork Moment Momentum Space Speed Time Torque Velocity Virtualwork Formulations Newton'slawsofmotion AnalyticalmechanicsLagrangianmechanicsHamiltonianmechanicsRouthianmechanicsHamilton–JacobiequationAppell'sequationofmotionKoopman–vonNeumannmechanics Coretopics Dampingratio Displacement Equationsofmotion Euler'slawsofmotion Fictitiousforce Friction Harmonicoscillator Inertial /Non-inertialreferenceframe Mechanicsofplanarparticlemotion Motion (linear) Newton'slawofuniversalgravitation Newton'slawsofmotion Relativevelocity Rigidbody dynamics Euler'sequations Simpleharmonicmotion Vibration Rotation Circularmotion Rotatingreferenceframe Centripetalforce Centrifugalforce reactive Coriolisforce Pendulum Tangentialspeed Rotationalspeed Angularacceleration /displacement /frequency /velocity Scientists Kepler Galileo Huygens Newton Horrocks Halley Maupertuis DanielBernoulli JohannBernoulli Euler d'Alembert Clairaut Lagrange Laplace Hamilton Poisson Cauchy Routh Liouville Appell Gibbs Koopman vonNeumann  Physicsportal  Categoryvte ÉmilieduChâtelet(1706–1749)withapairofcompassesinherrighthand.Shewasthefirsttopublishtherelationforkineticenergy E kin ∝ m v 2 {\displaystyleE_{\text{kin}}\proptomv^{2}} .Thismeansthatanobjectwithtwicethespeedhitsfourtimes–2×2–harder.(PortraitbyMauriceQuentindeLaTour.) Inphysics,thekineticenergyofanobjectistheenergythatitpossessesduetoitsmotion.[1] Itisdefinedastheworkneededtoaccelerateabodyofagivenmassfromresttoitsstatedvelocity.Havinggainedthisenergyduringitsacceleration,thebodymaintainsthiskineticenergyunlessitsspeedchanges.Thesameamountofworkisdonebythebodywhendeceleratingfromitscurrentspeedtoastateofrest.Formally,akineticenergyisanyterminasystem'sLagrangianwhichincludesaderivativewithrespecttotime.[2][3] Inclassicalmechanics,thekineticenergyofanon-rotatingobjectofmassmtravelingataspeedvis 1 2 m v 2 {\textstyle{\frac{1}{2}}mv^{2}} .Inrelativisticmechanics,thisisagoodapproximationonlywhenvismuchlessthanthespeedoflight. Thestandardunitofkineticenergyisthejoule,whiletheEnglishunitofkineticenergyisthefoot-pound. Contents 1Historyandetymology 2Overview 3Newtoniankineticenergy 3.1Kineticenergyofrigidbodies 3.1.1Derivation 3.1.1.1Withoutvectorsandcalculus 3.1.1.2Withvectorsandcalculus 3.2Rotatingbodies 3.3Kineticenergyofsystems 3.4Fluiddynamics 3.5Frameofreference 3.6Rotationinsystems 4Relativistickineticenergy 4.1Generalrelativity 5Kineticenergyinquantummechanics 6Seealso 7Notes 8References 9Externallinks Historyandetymology TheadjectivekinetichasitsrootsintheGreekwordκίνησιςkinesis,meaning"motion".ThedichotomybetweenkineticenergyandpotentialenergycanbetracedbacktoAristotle'sconceptsofactualityandpotentiality.[4] TheprincipleinclassicalmechanicsthatE∝mv2wasfirstdevelopedbyGottfriedLeibnizandJohannBernoulli,whodescribedkineticenergyasthelivingforce,visviva.Willem'sGravesandeoftheNetherlandsprovidedexperimentalevidenceofthisrelationship.Bydroppingweightsfromdifferentheightsintoablockofclay,Willem'sGravesandedeterminedthattheirpenetrationdepthwasproportionaltothesquareoftheirimpactspeed.ÉmilieduChâteletrecognizedtheimplicationsoftheexperimentandpublishedanexplanation.[5] Thetermskineticenergyandworkintheirpresentscientificmeaningsdatebacktothemid-19thcentury.EarlyunderstandingsoftheseideascanbeattributedtoGaspard-GustaveCoriolis,whoin1829publishedthepapertitledDuCalculdel'EffetdesMachinesoutliningthemathematicsofkineticenergy.WilliamThomson,laterLordKelvin,isgiventhecreditforcoiningtheterm"kineticenergy"c.1849–1851.[6][7]Rankine,whohadintroducedtheterm"potentialenergy"in1853,andthephrase"actualenergy"tocomplementit,[8]latercitesWilliamThomsonandPeterTaitassubstitutingtheword"kinetic"for"actual".[9] Overview Energyoccursinmanyforms,includingchemicalenergy,thermalenergy,electromagneticradiation,gravitationalenergy,electricenergy,elasticenergy,nuclearenergy,andrestenergy.Thesecanbecategorizedintwomainclasses:potentialenergyandkineticenergy.Kineticenergyisthemovementenergyofanobject.Kineticenergycanbetransferredbetweenobjectsandtransformedintootherkindsofenergy.[10] Kineticenergymaybebestunderstoodbyexamplesthatdemonstratehowitistransformedtoandfromotherformsofenergy.Forexample,acyclistuseschemicalenergyprovidedbyfoodtoaccelerateabicycletoachosenspeed.Onalevelsurface,thisspeedcanbemaintainedwithoutfurtherwork,excepttoovercomeairresistanceandfriction.Thechemicalenergyhasbeenconvertedintokineticenergy,theenergyofmotion,buttheprocessisnotcompletelyefficientandproducesheatwithinthecyclist. Thekineticenergyinthemovingcyclistandthebicyclecanbeconvertedtootherforms.Forexample,thecyclistcouldencounterahilljusthighenoughtocoastup,sothatthebicyclecomestoacompletehaltatthetop.Thekineticenergyhasnowlargelybeenconvertedtogravitationalpotentialenergythatcanbereleasedbyfreewheelingdowntheothersideofthehill.Sincethebicyclelostsomeofitsenergytofriction,itneverregainsallofitsspeedwithoutadditionalpedaling.Theenergyisnotdestroyed;ithasonlybeenconvertedtoanotherformbyfriction.Alternatively,thecyclistcouldconnectadynamotooneofthewheelsandgeneratesomeelectricalenergyonthedescent.Thebicyclewouldbetravelingsloweratthebottomofthehillthanwithoutthegeneratorbecausesomeoftheenergyhasbeendivertedintoelectricalenergy.Anotherpossibilitywouldbeforthecyclisttoapplythebrakes,inwhichcasethekineticenergywouldbedissipatedthroughfrictionasheat. Likeanyphysicalquantitythatisafunctionofvelocity,thekineticenergyofanobjectdependsontherelationshipbetweentheobjectandtheobserver'sframeofreference.Thus,thekineticenergyofanobjectisnotinvariant. Spacecraftusechemicalenergytolaunchandgainconsiderablekineticenergytoreachorbitalvelocity.Inanentirelycircularorbit,thiskineticenergyremainsconstantbecausethereisalmostnofrictioninnear-earthspace.However,itbecomesapparentatre-entrywhensomeofthekineticenergyisconvertedtoheat.Iftheorbitisellipticalorhyperbolic,thenthroughouttheorbitkineticandpotentialenergyareexchanged;kineticenergyisgreatestandpotentialenergylowestatclosestapproachtotheearthorothermassivebody,whilepotentialenergyisgreatestandkineticenergythelowestatmaximumdistance.Disregardinglossorgainhowever,thesumofthekineticandpotentialenergyremainsconstant. Kineticenergycanbepassedfromoneobjecttoanother.Inthegameofbilliards,theplayerimposeskineticenergyonthecueballbystrikingitwiththecuestick.Ifthecueballcollideswithanotherball,itslowsdowndramatically,andtheballithitacceleratesitsspeedasthekineticenergyispassedontoit.Collisionsinbilliardsareeffectivelyelasticcollisions,inwhichkineticenergyispreserved.Ininelasticcollisions,kineticenergyisdissipatedinvariousformsofenergy,suchasheat,soundandbindingenergy(breakingboundstructures). Flywheelshavebeendevelopedasamethodofenergystorage.Thisillustratesthatkineticenergyisalsostoredinrotationalmotion. Severalmathematicaldescriptionsofkineticenergyexistthatdescribeitintheappropriatephysicalsituation.Forobjectsandprocessesincommonhumanexperience,theformula½mv²givenbyNewtonian(classical)mechanicsissuitable.However,ifthespeedoftheobjectiscomparabletothespeedoflight,relativisticeffectsbecomesignificantandtherelativisticformulaisused.Iftheobjectisontheatomicorsub-atomicscale,quantummechanicaleffectsaresignificant,andaquantummechanicalmodelmustbeemployed. Newtoniankineticenergy Kineticenergyofrigidbodies Inclassicalmechanics,thekineticenergyofapointobject(anobjectsosmallthatitsmasscanbeassumedtoexistatonepoint),oranon-rotatingrigidbodydependsonthemassofthebodyaswellasitsspeed.Thekineticenergyisequalto1/2theproductofthemassandthesquareofthespeed.Informulaform: E k = 1 2 m v 2 {\displaystyleE_{\text{k}}={\frac{1}{2}}mv^{2}} where m {\displaystylem} isthemassand v {\displaystylev} isthespeed(magnitudeofthevelocity)ofthebody.InSIunits,massismeasuredinkilograms,speedinmetrespersecond,andtheresultingkineticenergyisinjoules. Forexample,onewouldcalculatethekineticenergyofan80 kgmass(about180 lbs)travelingat18 metrespersecond(about40 mph,or65 km/h)as E k = 1 2 ⋅ 80 kg ⋅ ( 18 m/s ) 2 = 12 , 960 J = 12.96 kJ {\displaystyleE_{\text{k}}={\frac{1}{2}}\cdot80\,{\text{kg}}\cdot\left(18\,{\text{m/s}}\right)^{2}=12,960\,{\text{J}}=12.96\,{\text{kJ}}} Whenapersonthrowsaball,thepersondoesworkonittogiveitspeedasitleavesthehand.Themovingballcanthenhitsomethingandpushit,doingworkonwhatithits.Thekineticenergyofamovingobjectisequaltotheworkrequiredtobringitfromresttothatspeed,ortheworktheobjectcandowhilebeingbroughttorest:netforce×displacement=kineticenergy,i.e., F s = 1 2 m v 2 {\displaystyleFs={\frac{1}{2}}mv^{2}} Sincethekineticenergyincreaseswiththesquareofthespeed,anobjectdoublingitsspeedhasfourtimesasmuchkineticenergy.Forexample,acartravelingtwiceasfastasanotherrequiresfourtimesasmuchdistancetostop,assumingaconstantbrakingforce.Asaconsequenceofthisquadrupling,ittakesfourtimestheworktodoublethespeed. Thekineticenergyofanobjectisrelatedtoitsmomentumbytheequation: E k = p 2 2 m {\displaystyleE_{\text{k}}={\frac{p^{2}}{2m}}} where: p {\displaystylep} ismomentum m {\displaystylem} ismassofthebody Forthetranslationalkineticenergy,thatisthekineticenergyassociatedwithrectilinearmotion,ofarigidbodywithconstantmass m {\displaystylem} ,whosecenterofmassismovinginastraightlinewithspeed v {\displaystylev} ,asseenaboveisequalto E t = 1 2 m v 2 {\displaystyleE_{\text{t}}={\frac{1}{2}}mv^{2}} where: m {\displaystylem} isthemassofthebody v {\displaystylev} isthespeedofthecenterofmassofthebody. Thekineticenergyofanyentitydependsonthereferenceframeinwhichitismeasured.However,thetotalenergyofanisolatedsystem,i.e.oneinwhichenergycanneitherenternorleave,doesnotchangeovertimeinthereferenceframeinwhichitismeasured.Thus,thechemicalenergyconvertedtokineticenergybyarocketengineisdivideddifferentlybetweentherocketshipanditsexhauststreamdependinguponthechosenreferenceframe.ThisiscalledtheObertheffect.Butthetotalenergyofthesystem,includingkineticenergy,fuelchemicalenergy,heat,etc.,isconservedovertime,regardlessofthechoiceofreferenceframe.Differentobserversmovingwithdifferentreferenceframeswouldhoweverdisagreeonthevalueofthisconservedenergy. Thekineticenergyofsuchsystemsdependsonthechoiceofreferenceframe:thereferenceframethatgivestheminimumvalueofthatenergyisthecenterofmomentumframe,i.e.thereferenceframeinwhichthetotalmomentumofthesystemiszero.Thisminimumkineticenergycontributestotheinvariantmassofthesystemasawhole. Derivation Withoutvectorsandcalculus TheworkWdonebyaforceFonanobjectoveradistancesparalleltoFequals W = F ⋅ s {\displaystyleW=F\cdots} . UsingNewton'sSecondLaw F = m a {\displaystyleF=ma} withmthemassandatheaccelerationoftheobjectand s = a t 2 2 {\displaystyles={\frac{at^{2}}{2}}} thedistancetraveledbytheacceleratedobjectintimet,wefindwith v = a t {\displaystylev=at} forthevelocityvoftheobject W = m a a t 2 2 = m ( a t ) 2 2 = m v 2 2 . {\displaystyleW=ma{\frac{at^{2}}{2}}={\frac{m(at)^{2}}{2}}={\frac{mv^{2}}{2}}.} Withvectorsandcalculus TheworkdoneinacceleratingaparticlewithmassmduringtheinfinitesimaltimeintervaldtisgivenbythedotproductofforceFandtheinfinitesimaldisplacementdx F ⋅ d x = F ⋅ v d t = d p d t ⋅ v d t = v ⋅ d p = v ⋅ d ( m v ) , {\displaystyle\mathbf{F}\cdotd\mathbf{x}=\mathbf{F}\cdot\mathbf{v}dt={\frac{d\mathbf{p}}{dt}}\cdot\mathbf{v}dt=\mathbf{v}\cdotd\mathbf{p}=\mathbf{v}\cdotd(m\mathbf{v})\,,} wherewehaveassumedtherelationshipp = m vandthevalidityofNewton'sSecondLaw.(However,alsoseethespecialrelativisticderivationbelow.) Applyingtheproductruleweseethat: d ( v ⋅ v ) = ( d v ) ⋅ v + v ⋅ ( d v ) = 2 ( v ⋅ d v ) . {\displaystyled(\mathbf{v}\cdot\mathbf{v})=(d\mathbf{v})\cdot\mathbf{v}+\mathbf{v}\cdot(d\mathbf{v})=2(\mathbf{v}\cdotd\mathbf{v}).} Therefore,(assumingconstantmasssothatdm=0),wehave, v ⋅ d ( m v ) = m 2 d ( v ⋅ v ) = m 2 d v 2 = d ( m v 2 2 ) . {\displaystyle\mathbf{v}\cdotd(m\mathbf{v})={\frac{m}{2}}d(\mathbf{v}\cdot\mathbf{v})={\frac{m}{2}}dv^{2}=d\left({\frac{mv^{2}}{2}}\right).} Sincethisisatotaldifferential(thatis,itonlydependsonthefinalstate,nothowtheparticlegotthere),wecanintegrateitandcalltheresultkineticenergy.Assumingtheobjectwasatrestattime0,weintegratefromtime0totimetbecausetheworkdonebytheforcetobringtheobjectfromresttovelocityvisequaltotheworknecessarytodothereverse: E k = ∫ 0 t F ⋅ d x = ∫ 0 t v ⋅ d ( m v ) = ∫ 0 t d ( m v 2 2 ) = m v 2 2 . {\displaystyleE_{\text{k}}=\int_{0}^{t}\mathbf{F}\cdotd\mathbf{x}=\int_{0}^{t}\mathbf{v}\cdotd(m\mathbf{v})=\int_{0}^{t}d\left({\frac{mv^{2}}{2}}\right)={\frac{mv^{2}}{2}}.} Thisequationstatesthatthekineticenergy(Ek)isequaltotheintegralofthedotproductofthevelocity(v)ofabodyandtheinfinitesimalchangeofthebody'smomentum(p).Itisassumedthatthebodystartswithnokineticenergywhenitisatrest(motionless). Rotatingbodies IfarigidbodyQisrotatingaboutanylinethroughthecenterofmassthenithasrotationalkineticenergy( E r {\displaystyleE_{\text{r}}\,} )whichissimplythesumofthekineticenergiesofitsmovingparts,andisthusgivenby: E r = ∫ Q v 2 d m 2 = ∫ Q ( r ω ) 2 d m 2 = ω 2 2 ∫ Q r 2 d m = ω 2 2 I = 1 2 I ω 2 {\displaystyleE_{\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: ωisthebody'sangularvelocity risthedistanceofanymassdmfromthatline I {\displaystyleI} isthebody'smomentofinertia,equalto ∫ Q r 2 d m {\textstyle\int_{Q}{r^{2}}dm} . (Inthisequationthemomentofinertiamustbetakenaboutanaxisthroughthecenterofmassandtherotationmeasuredbyωmustbearoundthataxis;moregeneralequationsexistforsystemswheretheobjectissubjecttowobbleduetoitseccentricshape). Kineticenergyofsystems Asystemofbodiesmayhaveinternalkineticenergyduetotherelativemotionofthebodiesinthesystem.Forexample,intheSolarSystemtheplanetsandplanetoidsareorbitingtheSun.Inatankofgas,themoleculesaremovinginalldirections.Thekineticenergyofthesystemisthesumofthekineticenergiesofthebodiesitcontains. Amacroscopicbodythatisstationary(i.e.areferenceframehasbeenchosentocorrespondtothebody'scenterofmomentum)mayhavevariouskindsofinternalenergyatthemolecularoratomiclevel,whichmayberegardedaskineticenergy,duetomoleculartranslation,rotation,andvibration,electrontranslationandspin,andnuclearspin.Theseallcontributetothebody'smass,asprovidedbythespecialtheoryofrelativity.Whendiscussingmovementsofamacroscopicbody,thekineticenergyreferredtoisusuallythatofthemacroscopicmovementonly.However,allinternalenergiesofalltypescontributetoabody'smass,inertia,andtotalenergy. Fluiddynamics Influiddynamics,thekineticenergyperunitvolumeateachpointinanincompressiblefluidflowfieldiscalledthedynamicpressureatthatpoint.[11] E k = 1 2 m v 2 {\displaystyleE_{\text{k}}={\frac{1}{2}}mv^{2}} DividingbyV,theunitofvolume: E k V = 1 2 m V v 2 q = 1 2 ρ v 2 {\displaystyle{\begin{aligned}{\frac{E_{\text{k}}}{V}}&={\frac{1}{2}}{\frac{m}{V}}v^{2}\\q&={\frac{1}{2}}\rhov^{2}\end{aligned}}} where q {\displaystyleq} isthedynamicpressure,andρisthedensityoftheincompressiblefluid. Frameofreference Thespeed,andthusthekineticenergyofasingleobjectisframe-dependent(relative):itcantakeanynon-negativevalue,bychoosingasuitableinertialframeofreference.Forexample,abulletpassinganobserverhaskineticenergyinthereferenceframeofthisobserver.Thesamebulletisstationarytoanobservermovingwiththesamevelocityasthebullet,andsohaszerokineticenergy.[12]Bycontrast,thetotalkineticenergyofasystemofobjectscannotbereducedtozerobyasuitablechoiceoftheinertialreferenceframe,unlessalltheobjectshavethesamevelocity.Inanyothercase,thetotalkineticenergyhasanon-zerominimum,asnoinertialreferenceframecanbechoseninwhichalltheobjectsarestationary.Thisminimumkineticenergycontributestothesystem'sinvariantmass,whichisindependentofthereferenceframe. Thetotalkineticenergyofasystemdependsontheinertialframeofreference:itisthesumofthetotalkineticenergyinacenterofmomentumframeandthekineticenergythetotalmasswouldhaveifitwereconcentratedinthecenterofmass. Thismaybesimplyshown:let V {\displaystyle\textstyle\mathbf{V}} betherelativevelocityofthecenterofmassframeiintheframek.Since v 2 = ( v i + V ) 2 = ( v i + V ) ⋅ ( v i + V ) = v i ⋅ v i + 2 v i ⋅ V + V ⋅ V = v i 2 + 2 v i ⋅ V + V 2 , {\displaystylev^{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, E k = ∫ v 2 2 d m = ∫ v i 2 2 d m + V ⋅ ∫ v i d m + V 2 2 ∫ d m . {\displaystyleE_{\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}}\intdm.} However,let ∫ v i 2 2 d m = E i {\textstyle\int{\frac{v_{i}^{2}}{2}}dm=E_{i}} thekineticenergyinthecenterofmassframe, ∫ v i d m {\textstyle\int\mathbf{v}_{i}dm} wouldbesimplythetotalmomentumthatisbydefinitionzerointhecenterofmassframe,andletthetotalmass: ∫ d m = M {\textstyle\intdm=M} .Substituting,weget:[13] E k = E i + M V 2 2 . {\displaystyleE_{\text{k}}=E_{i}+{\frac{MV^{2}}{2}}.} Thusthekineticenergyofasystemislowesttocenterofmomentumreferenceframes,i.e.,framesofreferenceinwhichthecenterofmassisstationary(eitherthecenterofmassframeoranyothercenterofmomentumframe).Inanydifferentframeofreference,thereisadditionalkineticenergycorrespondingtothetotalmassmovingatthespeedofthecenterofmass.Thekineticenergyofthesysteminthecenterofmomentumframeisaquantitythatisinvariant(allobserversseeittobethesame). Rotationinsystems Itsometimesisconvenienttosplitthetotalkineticenergyofabodyintothesumofthebody'scenter-of-masstranslationalkineticenergyandtheenergyofrotationaroundthecenterofmass(rotationalenergy): E k = E t + E r {\displaystyleE_{\text{k}}=E_{\text{t}}+E_{\text{r}}} where: Ekisthetotalkineticenergy Etisthetranslationalkineticenergy Eristherotationalenergyorangularkineticenergyintherestframe Thusthekineticenergyofatennisballinflightisthekineticenergyduetoitsrotation,plusthekineticenergyduetoitstranslation. Relativistickineticenergy Seealso:MassinspecialrelativityandTestsofrelativisticenergyandmomentum Ifabody'sspeedisasignificantfractionofthespeedoflight,itisnecessarytouserelativisticmechanicstocalculateitskineticenergy.Inspecialrelativitytheory,theexpressionforlinearmomentumismodified. Withmbeinganobject'srestmass,vandvitsvelocityandspeed,andcthespeedoflightinvacuum,weusetheexpressionforlinearmomentum p = m γ v {\displaystyle\mathbf{p}=m\gamma\mathbf{v}} ,where γ = 1 / 1 − v 2 / c 2 {\textstyle\gamma=1/{\sqrt{1-v^{2}/c^{2}}}} . Integratingbypartsyields E k = ∫ v ⋅ d p = ∫ v ⋅ d ( m γ v ) = m γ v ⋅ v − ∫ m γ v ⋅ d v = m γ v 2 − m 2 ∫ γ d ( v 2 ) {\displaystyleE_{\text{k}}=\int\mathbf{v}\cdotd\mathbf{p}=\int\mathbf{v}\cdotd(m\gamma\mathbf{v})=m\gamma\mathbf{v}\cdot\mathbf{v}-\intm\gamma\mathbf{v}\cdotd\mathbf{v}=m\gammav^{2}-{\frac{m}{2}}\int\gammad\left(v^{2}\right)} Since γ = ( 1 − v 2 / c 2 ) − 1 2 {\displaystyle\gamma=\left(1-v^{2}/c^{2}\right)^{-{\frac{1}{2}}}} , E k = m γ v 2 − − m c 2 2 ∫ γ d ( 1 − v 2 c 2 ) = m γ v 2 + m c 2 ( 1 − v 2 c 2 ) 1 2 − E 0 {\displaystyle{\begin{aligned}E_{\text{k}}&=m\gammav^{2}-{\frac{-mc^{2}}{2}}\int\gammad\left(1-{\frac{v^{2}}{c^{2}}}\right)\\&=m\gammav^{2}+mc^{2}\left(1-{\frac{v^{2}}{c^{2}}}\right)^{\frac{1}{2}}-E_{0}\end{aligned}}} E 0 {\displaystyleE_{0}} isaconstantofintegrationfortheindefiniteintegral. Simplifyingtheexpressionweobtain E k = m γ ( v 2 + c 2 ( 1 − v 2 c 2 ) ) − E 0 = m γ ( v 2 + c 2 − v 2 ) − E 0 = m γ c 2 − E 0 {\displaystyle{\begin{aligned}E_{\text{k}}&=m\gamma\left(v^{2}+c^{2}\left(1-{\frac{v^{2}}{c^{2}}}\right)\right)-E_{0}\\&=m\gamma\left(v^{2}+c^{2}-v^{2}\right)-E_{0}\\&=m\gammac^{2}-E_{0}\end{aligned}}} E 0 {\displaystyleE_{0}} isfoundbyobservingthatwhen v = 0 ,   γ = 1 {\displaystyle\mathbf{v}=0,\\gamma=1} and E k = 0 {\displaystyleE_{\text{k}}=0} ,giving E 0 = m c 2 {\displaystyleE_{0}=mc^{2}} resultingintheformula E k = m γ c 2 − m c 2 = m c 2 1 − v 2 c 2 − m c 2 = ( γ − 1 ) m c 2 {\displaystyleE_{\text{k}}=m\gammac^{2}-mc^{2}={\frac{mc^{2}}{\sqrt{1-{\frac{v^{2}}{c^{2}}}}}}-mc^{2}=(\gamma-1)mc^{2}} Thisformulashowsthattheworkexpendedacceleratinganobjectfromrestapproachesinfinityasthevelocityapproachesthespeedoflight.Thusitisimpossibletoaccelerateanobjectacrossthisboundary. Themathematicalby-productofthiscalculationisthemass-energyequivalenceformula—thebodyatrestmusthaveenergycontent E rest = E 0 = m c 2 {\displaystyleE_{\text{rest}}=E_{0}=mc^{2}} Atalowspeed(v≪c),therelativistickineticenergyisapproximatedwellbytheclassicalkineticenergy.ThisisdonebybinomialapproximationorbytakingthefirsttwotermsoftheTaylorexpansionforthereciprocalsquareroot: E k ≈ m c 2 ( 1 + 1 2 v 2 c 2 ) − m c 2 = 1 2 m v 2 {\displaystyleE_{\text{k}}\approxmc^{2}\left(1+{\frac{1}{2}}{\frac{v^{2}}{c^{2}}}\right)-mc^{2}={\frac{1}{2}}mv^{2}} So,thetotalenergy E k {\displaystyleE_{k}} canbepartitionedintotherestmassenergyplustheNewtoniankineticenergyatlowspeeds. Whenobjectsmoveataspeedmuchslowerthanlight(e.g.ineverydayphenomenaonEarth),thefirsttwotermsoftheseriespredominate.ThenexttermintheTaylorseriesapproximation E k ≈ m c 2 ( 1 + 1 2 v 2 c 2 + 3 8 v 4 c 4 ) − m c 2 = 1 2 m v 2 + 3 8 m v 4 c 2 {\displaystyleE_{\text{k}}\approxmc^{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}}}} issmallforlowspeeds.Forexample,foraspeedof10 km/s(22,000 mph)thecorrectiontotheNewtoniankineticenergyis0.0417 J/kg(onaNewtoniankineticenergyof50 MJ/kg)andforaspeedof100 km/sitis417 J/kg(onaNewtoniankineticenergyof5 GJ/kg). Therelativisticrelationbetweenkineticenergyandmomentumisgivenby E k = p 2 c 2 + m 2 c 4 − m c 2 {\displaystyleE_{\text{k}}={\sqrt{p^{2}c^{2}+m^{2}c^{4}}}-mc^{2}} ThiscanalsobeexpandedasaTaylorseries,thefirsttermofwhichisthesimpleexpressionfromNewtonianmechanics:[14] E k ≈ p 2 2 m − p 4 8 m 3 c 2 . {\displaystyleE_{\text{k}}\approx{\frac{p^{2}}{2m}}-{\frac{p^{4}}{8m^{3}c^{2}}}.} Thissuggeststhattheformulaeforenergyandmomentumarenotspecialandaxiomatic,butconceptsemergingfromtheequivalenceofmassandenergyandtheprinciplesofrelativity. Generalrelativity Seealso:Schwarzschildgeodesics Usingtheconventionthat g α β u α u β = − c 2 {\displaystyleg_{\alpha\beta}\,u^{\alpha}\,u^{\beta}\,=\,-c^{2}} wherethefour-velocityofaparticleis u α = d x α d τ {\displaystyleu^{\alpha}\,=\,{\frac{dx^{\alpha}}{d\tau}}} and τ {\displaystyle\tau} isthepropertimeoftheparticle,thereisalsoanexpressionforthekineticenergyoftheparticleingeneralrelativity. Iftheparticlehasmomentum p β = m g β α u α {\displaystylep_{\beta}\,=\,m\,g_{\beta\alpha}\,u^{\alpha}} asitpassesbyanobserverwithfour-velocityuobs,thentheexpressionfortotalenergyoftheparticleasobserved(measuredinalocalinertialframe)is E = − p β u obs β {\displaystyleE\,=\,-\,p_{\beta}\,u_{\text{obs}}^{\beta}} andthekineticenergycanbeexpressedasthetotalenergyminustherestenergy: E k = − p β u obs β − m c 2 . {\displaystyleE_{k}\,=\,-\,p_{\beta}\,u_{\text{obs}}^{\beta}\,-\,m\,c^{2}\,.} Considerthecaseofametricthatisdiagonalandspatiallyisotropic(gtt,gss,gss,gss).Since u α = d x α d t d t d τ = v α u t {\displaystyleu^{\alpha}={\frac{dx^{\alpha}}{dt}}{\frac{dt}{d\tau}}=v^{\alpha}u^{t}} wherevαistheordinaryvelocitymeasuredw.r.t.thecoordinatesystem,weget − c 2 = g α β u α u β = g t t ( u t ) 2 + g s s v 2 ( u t ) 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}\,.} Solvingforutgives u t = c − 1 g t t + g s s v 2 . {\displaystyleu^{t}=c{\sqrt{\frac{-1}{g_{tt}+g_{ss}v^{2}}}}\,.} Thusforastationaryobserver(v=0) u obs t = c − 1 g t t {\displaystyleu_{\text{obs}}^{t}=c{\sqrt{\frac{-1}{g_{tt}}}}} andthusthekineticenergytakestheform E k = − m g t t u t u obs t − m c 2 = m c 2 g t t g t t + g s s v 2 − m c 2 . {\displaystyleE_{\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}\,.} Factoringouttherestenergygives: E k = m c 2 ( g t t g t t + g s s v 2 − 1 ) . {\displaystyleE_{\text{k}}=mc^{2}\left({\sqrt{\frac{g_{tt}}{g_{tt}+g_{ss}v^{2}}}}-1\right)\,.} Thisexpressionreducestothespecialrelativisticcasefortheflat-spacemetricwhere g t t = − c 2 g s s = 1 . {\displaystyle{\begin{aligned}g_{tt}&=-c^{2}\\g_{ss}&=1\,.\end{aligned}}} IntheNewtonianapproximationtogeneralrelativity g t t = − ( c 2 + 2 Φ ) g s s = 1 − 2 Φ c 2 {\displaystyle{\begin{aligned}g_{tt}&=-\left(c^{2}+2\Phi\right)\\g_{ss}&=1-{\frac{2\Phi}{c^{2}}}\end{aligned}}} whereΦistheNewtoniangravitationalpotential.Thismeansclocksrunslowerandmeasuringrodsareshorternearmassivebodies. Kineticenergyinquantummechanics Furtherinformation:Hamiltonian(quantummechanics) Inquantummechanics,observableslikekineticenergyarerepresentedasoperators.Foroneparticleofmassm,thekineticenergyoperatorappearsasatermintheHamiltonianandisdefinedintermsofthemorefundamentalmomentumoperator p ^ {\displaystyle{\hat{p}}} .Thekineticenergyoperatorinthenon-relativisticcasecanbewrittenas T ^ = p ^ 2 2 m . {\displaystyle{\hat{T}}={\frac{{\hat{p}}^{2}}{2m}}.} Noticethatthiscanbeobtainedbyreplacing p {\displaystylep} by p ^ {\displaystyle{\hat{p}}} intheclassicalexpressionforkineticenergyintermsofmomentum, E k = p 2 2 m . {\displaystyleE_{\text{k}}={\frac{p^{2}}{2m}}.} IntheSchrödingerpicture, p ^ {\displaystyle{\hat{p}}} takestheform − i ℏ ∇ {\displaystyle-i\hbar\nabla} wherethederivativeistakenwithrespecttopositioncoordinatesandhence T ^ = − ℏ 2 2 m ∇ 2 . {\displaystyle{\hat{T}}=-{\frac{\hbar^{2}}{2m}}\nabla^{2}.} Theexpectationvalueoftheelectronkineticenergy, ⟨ T ^ ⟩ {\displaystyle\left\langle{\hat{T}}\right\rangle} ,forasystemofNelectronsdescribedbythewavefunction | ψ ⟩ {\displaystyle\vert\psi\rangle} isasumof1-electronoperatorexpectationvalues: ⟨ T ^ ⟩ = ⟨ ψ | ∑ i = 1 N − ℏ 2 2 m e ∇ i 2 | ψ ⟩ = − ℏ 2 2 m e ∑ i = 1 N ⟨ ψ | ∇ i 2 | 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} where m e {\displaystylem_{\text{e}}} isthemassoftheelectronand ∇ i 2 {\displaystyle\nabla_{i}^{2}} istheLaplacianoperatoractinguponthecoordinatesoftheithelectronandthesummationrunsoverallelectrons. Thedensityfunctionalformalismofquantummechanicsrequiresknowledgeoftheelectrondensityonly,i.e.,itformallydoesnotrequireknowledgeofthewavefunction.Givenanelectrondensity ρ ( r ) {\displaystyle\rho(\mathbf{r})} ,theexactN-electronkineticenergyfunctionalisunknown;however,forthespecificcaseofa1-electronsystem,thekineticenergycanbewrittenas T [ ρ ] = 1 8 ∫ ∇ ρ ( r ) ⋅ ∇ ρ ( r ) ρ ( r ) d 3 r {\displaystyleT[\rho]={\frac{1}{8}}\int{\frac{\nabla\rho(\mathbf{r})\cdot\nabla\rho(\mathbf{r})}{\rho(\mathbf{r})}}d^{3}r} where T [ ρ ] {\displaystyleT[\rho]} isknownasthevonWeizsäckerkineticenergyfunctional. Seealso Energyportal Escapevelocity Foot-pound Joule Kineticenergypenetrator Kineticenergyperunitmassofprojectiles Kineticprojectile Parallelaxistheorem Potentialenergy Recoil Notes ^Jain,MaheshC.(2009).TextbookofEngineeringPhysics(PartI).p. 9.ISBN 978-81-203-3862-3.Archivedfromtheoriginalon2020-08-04.Retrieved2018-06-21.,Chapter1,p.9Archived2020-08-04attheWaybackMachine ^Landau,Lev;Lifshitz,Evgeny(15January1976).Mechanics(Third ed.).p. 15.ISBN 0-7506-2896-0. ^Goldstein,Herbert.ClassicalMechanics(Third ed.).p. 62-33.ISBN 978-0201657029. ^Brenner,Joseph(2008).LogicinReality(illustrated ed.).SpringerScience&BusinessMedia.p. 93.ISBN 978-1-4020-8375-4.Archivedfromtheoriginalon2020-01-25.Retrieved2016-02-01.Extractofpage93Archived2020-08-04attheWaybackMachine ^JudithP.Zinsser(2007).EmilieduChatelet:DaringGeniusoftheEnlightenment.Penguin.ISBN 978-0-14-311268-6. ^CrosbieSmith,M.NortonWise(1989-10-26).EnergyandEmpire:ABiographicalStudyofLordKelvin.CambridgeUniversityPress.p. 866.ISBN 0-521-26173-2. ^JohnTheodoreMerz(1912).AHistoryofEuropeanThoughtintheNineteenthCentury.Blackwood.p. 139.ISBN 0-8446-2579-5. ^WilliamJohnMacquornRankine(1853)."Onthegenerallawofthetransformationofenergy".ProceedingsofthePhilosophicalSocietyofGlasgow.3(5). ^"...whatremainedtobedone,wastoqualifythenoun'energy'byappropriateadjectives,soastodistinguishbetweenenergyofactivityandenergyofconfiguration.Thewell-knownpairofantitheticaladjectives,'actual'and'potential,'seemedexactlysuitedforthatpurpose....SirWilliamThomsonandProfessorTaithavelatelysubstitutedtheword'kinetic'for'actual.'"WilliamJohnMacquornRankine(1867)."OnthePhrase"PotentialEnergy,"andontheDefinitionsofPhysicalQuantities".ProceedingsofthePhilosophicalSocietyofGlasgow.VI(III). ^Goel,V.K.(2007).FundamentalsOfPhysicsXi(illustrated ed.).TataMcGraw-HillEducation.p. 12.30.ISBN 978-0-07-062060-5.Archivedfromtheoriginalon2020-08-03.Retrieved2020-07-07.Extractofpage12.30Archived2020-07-07attheWaybackMachine ^A.M.KuetheandJ.D.Schetzer(1959)FoundationsofAerodynamics,2ndedition,p.53.JohnWiley&SonsISBN 0-471-50952-3 ^Sears,FrancisWeston;Brehme,RobertW.(1968).Introductiontothetheoryofrelativity.Addison-Wesley.p. 127.,Snippetviewofpage127Archived2020-08-04attheWaybackMachine ^Physicsnotes-KineticenergyintheCMframeArchived2007-06-11attheWaybackMachine.Duke.edu.Accessed2007-11-24. ^Fitzpatrick,Richard(20July2010)."FineStructureofHydrogen".QuantumMechanics.Archivedfromtheoriginalon25August2016.Retrieved20August2016. References PhysicsClassroom(2000)."KineticEnergy".Retrieved2015-07-19. SchoolofMathematicsandStatistics,UniversityofStAndrews(2000)."BiographyofGaspard-GustavedeCoriolis(1792-1843)".Retrieved2006-03-03. Serway,RaymondA.;Jewett,JohnW.(2004).PhysicsforScientistsandEngineers(6th ed.).Brooks/Cole.ISBN 0-534-40842-7. Tipler,Paul(2004).PhysicsforScientistsandEngineers:Mechanics,OscillationsandWaves,Thermodynamics(5th ed.).W.H.Freeman.ISBN 0-7167-0809-4. Tipler,Paul;Llewellyn,Ralph(2002).ModernPhysics(4th ed.).W.H.Freeman.ISBN 0-7167-4345-0. 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