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Kinetic energy








Kinetic energy


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Kinetic energy

Wooden roller coaster txgi.jpg
The cars of a roller 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 gravitational potential energy. The sum of kinetic and potential energy in the system remains constant, ignoring losses to friction.

Common symbols

KE, Ek, or T
SI unit
joule (J)
Derivations from
other quantities

Ek = ½mv2


Ek = Et+Er






In physics, the kinetic energy of an object is the energy that it possesses due to its motion.[1]
It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body when decelerating from its current speed to a state of rest.


In classical mechanics, the kinetic energy of a non-rotating object of mass m traveling at a speed v is 12mv2displaystyle beginsmallmatrixfrac 12mv^2endsmallmatrixbeginsmallmatrixfrac 12mv^2endsmallmatrix. In relativistic mechanics, this is a good approximation only when v is much less than the speed of light.


The standard unit of kinetic energy is the joule.




Contents





  • 1 History and etymology


  • 2 Overview


  • 3 Newtonian kinetic energy

    • 3.1 Kinetic energy of rigid bodies

      • 3.1.1 Derivation



    • 3.2 Rotating bodies


    • 3.3 Kinetic energy of systems


    • 3.4 Frame of reference


    • 3.5 Rotation in systems



  • 4 Relativistic kinetic energy of rigid bodies

    • 4.1 General relativity



  • 5 Kinetic energy in quantum mechanics


  • 6 See also


  • 7 Notes


  • 8 References


  • 9 External links




History and etymology[edit]


The adjective kinetic has its roots in the Greek word κίνησις kinesis, meaning "motion". The dichotomy between kinetic energy and potential energy can be traced back to Aristotle's concepts of actuality and potentiality.[2]


The principle in classical mechanics that E ∝ mv2 was first developed by Gottfried Leibniz and Johann Bernoulli, who described kinetic energy as the living force, vis viva. Willem 's Gravesande of the Netherlands provided experimental evidence of this relationship. By dropping weights from different heights into a block of clay, Willem 's 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.[3]


The terms kinetic energy and work in their present scientific meanings date back to the mid-19th century. Early understandings of these ideas can be attributed to Gaspard-Gustave Coriolis, who in 1829 published the paper titled Du 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. 1849–51.[4][5]



Overview[edit]


Energy occurs in many forms, including chemical energy, thermal energy, electromagnetic radiation, gravitational energy, electric energy, elastic energy, nuclear energy, and rest 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.[6]


Kinetic energy may be best understood by examples that demonstrate how it is transformed to and from other forms of energy. For example, a cyclist uses chemical energy provided by food to accelerate a bicycle to a chosen speed. On a level surface, this speed can be maintained without further work, except to overcome air resistance and friction. The chemical energy has been converted into kinetic energy, the energy of motion, but the process is not completely efficient and produces heat 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 a dynamo 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 as heat.


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's frame of reference. Thus, the kinetic energy of an object is not invariant.


Spacecraft use chemical energy to launch and gain considerable kinetic energy to reach orbital 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 is elliptical or hyperbolic, then throughout the orbit kinetic and potential 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. Without 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 of billiards, 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 its speed as the kinetic energy is passed on to it. Collisions in billiards are effectively elastic collisions, in which kinetic energy is preserved. In inelastic collisions, kinetic energy is dissipated in various forms of energy, such as heat, sound, binding energy (breaking bound structures).


Flywheels have been developed as a method of energy 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 formula ½mv² given by Newtonian (classical) 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 or sub-atomic scale, quantum mechanical effects are significant, and a quantum mechanical model must be employed.



Newtonian kinetic energy[edit]



Kinetic energy of rigid bodies[edit]


In classical mechanics, the kinetic energy of a point object (an object so small that its mass can be assumed to exist at one point), or a non-rotating rigid body depends on the mass of the body as well as its speed. The kinetic energy is equal to 1/2 the product of the mass and the square of the speed. In formula form:


Ek=12mv2displaystyle E_textk=tfrac 12mv^2E_textk=tfrac 12mv^2

where mdisplaystyle mm is the mass and vdisplaystyle vv is the speed (or the velocity) of the body. In SI units, mass is measured in kilograms, speed in metres per second, and the resulting kinetic energy is in joules.


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=12960J=12.96kJdisplaystyle E_textk=frac 12cdot 80,textkgcdot left(18,textm/sright)^2=12960,textJ=12.96,textkJE_textk=frac 12cdot 80,textkgcdot left(18,textm/sright)^2=12960,textJ=12.96,textkJ

When you throw a ball, you do work on it to give it speed as it leaves your 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=12mv2displaystyle Fs=tfrac 12mv^2Fs=tfrac 12mv^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 its momentum by the equation:


Ek=p22mdisplaystyle E_textk=frac p^22mE_textk=frac p^22m

where:



pdisplaystyle p;p; is momentum


mdisplaystyle m;m; is mass of the body

For the translational kinetic energy, that is the kinetic energy associated with rectilinear motion, of a rigid body with constant mass mdisplaystyle m;m;, whose center of mass is moving in a straight line with speed vdisplaystyle v;v;, as seen above is equal to


Et=12mv2displaystyle E_textt=tfrac 12mv^2E_textt=tfrac 12mv^2

where:



mdisplaystyle m;m; is the mass of the body


vdisplaystyle v;v; is the speed of the center of mass of the body.

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 the Oberth 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 the center of momentum frame, i.e. the reference frame in which the total momentum of the system is zero. This minimum kinetic energy contributes to the invariant mass of the system as a whole.



Derivation[edit]


The work done in accelerating a particle with mass m during the infinitesimal time interval dt is given by the dot product of force F and the infinitesimal displacement dx


F⋅dx=F⋅vdt=dpdt⋅vdt=v⋅dp=v⋅d(mv),displaystyle mathbf F cdot dmathbf x =mathbf F cdot mathbf v dt=frac dmathbf p dtcdot mathbf v dt=mathbf v cdot dmathbf p =mathbf v cdot d(mmathbf v ),,mathbf F cdot dmathbf x =mathbf F cdot mathbf v dt=frac dmathbf p dtcdot mathbf v dt=mathbf v cdot dmathbf p =mathbf v cdot d(mmathbf v ),,

where we have assumed the relationship p = m v and the validity of Newton's Second Law. (However, also see the special relativistic derivation below.)


Applying the product rule we see that:


d(v⋅v)=(dv)⋅v+v⋅(dv)=2(v⋅dv).displaystyle d(mathbf v cdot mathbf v )=(dmathbf v )cdot mathbf v +mathbf v cdot (dmathbf v )=2(mathbf v cdot dmathbf v ).d(mathbf v cdot mathbf v )=(dmathbf v )cdot mathbf v +mathbf v cdot (dmathbf v )=2(mathbf v cdot dmathbf v ).

Therefore, (assuming constant mass so that dm=0), we have,


v⋅d(mv)=m2d(v⋅v)=m2dv2=d(mv22).displaystyle mathbf v cdot d(mmathbf v )=frac m2d(mathbf v cdot mathbf v )=frac m2dv^2=dleft(frac mv^22right).mathbf v cdot d(mmathbf v )=frac m2d(mathbf v cdot mathbf v )=frac m2dv^2=dleft(frac mv^22right).

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. Assuming the object was at rest at time 0, we integrate from time 0 to time t because the work done by the force to bring the object from rest to velocity v is equal to the work necessary to do the reverse:


Ek=∫0tF⋅dx=∫0tv⋅d(mv)=∫0vd(mv22)=mv22.displaystyle E_textk=int _0^tmathbf F cdot dmathbf x =int _0^tmathbf v cdot d(mmathbf v )=int _0^vdleft(frac mv^22right)=frac mv^22.E_textk=int _0^tmathbf F cdot dmathbf x =int _0^tmathbf v cdot d(mmathbf v )=int _0^vdleft(frac mv^22right)=frac mv^22.

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).



Rotating bodies[edit]


If a rigid body Q is rotating about any line through the center of mass then it has rotational kinetic energy (Erdisplaystyle E_textr,E_textr,) 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=12Iω2displaystyle E_textr=int _Qfrac v^2dm2=int _Qfrac (romega )^2dm2=frac omega ^22int _Qr^2dm=frac omega ^22I=beginmatrixfrac 12endmatrixIomega ^2E_textr=int _Qfrac v^2dm2=int _Qfrac (romega )^2dm2=frac omega ^22int _Qr^2dm=frac omega ^22I=beginmatrixfrac 12endmatrixIomega ^2

where:


  • ω is the body's angular velocity


  • r is the distance of any mass dm from that line


  • Idisplaystyle I,I, is the body's moment of inertia, equal to ∫Qr2dmdisplaystyle int _Qr^2dmint _Qr^2dm.

(In this equation the moment of inertia 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[edit]


A system of bodies may have internal kinetic energy due to the relative motion of the bodies in the system. For example, in the Solar 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's center of momentum) may have various kinds of internal 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 body's mass, inertia, and total energy.



Frame of reference[edit]


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 suitable inertial 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.[7] 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's invariant mass, which is independent of the reference frame.


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


This may be simply shown: let Vdisplaystyle textstyle mathbf V textstyle mathbf V be the relative velocity of the center of mass frame i in the frame k.
Since v2=(vi+V)2=(vi+V)⋅(vi+V)=vi⋅vi+2vi⋅V+V⋅V=vi2+2vi⋅V+V2displaystyle textstyle v^2=(v_i+V)^2=(mathbf v _i+mathbf V )cdot (mathbf v _i+mathbf V )=mathbf v _icdot mathbf v _i+2mathbf v _icdot mathbf V +mathbf V cdot mathbf V =v_i^2+2mathbf v _icdot mathbf V +V^2textstyle v^2=(v_i+V)^2=(mathbf v _i+mathbf V )cdot (mathbf v _i+mathbf V )=mathbf v _icdot mathbf v _i+2mathbf v _icdot mathbf V +mathbf V cdot mathbf V =v_i^2+2mathbf v _icdot mathbf V +V^2,


Ek=∫v22dm=∫vi22dm+V⋅∫vidm+V22∫dm.displaystyle E_textk=int frac v^22dm=int frac v_i^22dm+mathbf V cdot int mathbf v _idm+frac V^22int dm.E_textk=int frac v^22dm=int frac v_i^22dm+mathbf V cdot int mathbf v _idm+frac V^22int dm.

However, let ∫vi22dm=Eidisplaystyle int frac v_i^22dm=E_iint frac v_i^22dm=E_i the kinetic energy in the center of mass frame, ∫vidmdisplaystyle int mathbf v _idmint mathbf v _idm would be simply the total momentum that is by definition zero in the center of mass frame, and let the total mass: ∫dm=Mdisplaystyle int dm=Mint dm=M. Substituting, we get:[8]


Ek=Ei+MV22.displaystyle E_textk=E_i+frac MV^22.E_textk=E_i+frac MV^22.

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 the center of mass frame or any other center 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 the center of momentum frame is a quantity that is invariant (all observers see it to be the same).



Rotation in systems[edit]


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+Erdisplaystyle E_textk=E_t+E_textr,E_textk=E_t+E_textr,

where:



Ek is the total kinetic energy


Et is the translational kinetic energy


Er is the rotational energy or angular 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 of rigid bodies[edit]



If a body's speed is a significant fraction of the speed of light, it is necessary to use relativistic mechanics to calculate its kinetic energy. In special relativity theory, the expression for linear momentum is modified.


With m being an object's rest mass, v and v its velocity and speed, and c the speed of light in vacuum, we use the expression for linear momentum p=mγvdisplaystyle mathbf p =mgamma mathbf v mathbf p =mgamma mathbf v , where γ=1/1−v2/c2displaystyle gamma =1/sqrt 1-v^2/c^2gamma =1/sqrt 1-v^2/c^2.


Integrating by parts yields


Ek=∫v⋅dp=∫v⋅d(mγv)=mγv⋅v−∫mγv⋅dv=mγv2−m2∫γd(v2)displaystyle E_textk=int mathbf v cdot dmathbf p =int mathbf v cdot d(mgamma mathbf v )=mgamma mathbf v cdot mathbf v -int mgamma mathbf v cdot dmathbf v =mgamma v^2-frac m2int gamma d(v^2)E_textk=int mathbf v cdot dmathbf p =int mathbf v cdot d(mgamma mathbf v )=mgamma mathbf v cdot mathbf v -int mgamma mathbf v cdot dmathbf v =mgamma v^2-frac m2int gamma d(v^2)

Since γ=(1−v2/c2)−1/2displaystyle gamma =(1-v^2/c^2)^-1/2!gamma =(1-v^2/c^2)^-1/2!,


Ek=mγv2−−mc22∫γd(1−v2/c2)=mγv2+mc2(1−v2/c2)1/2−E0displaystyle beginalignedE_textk&=mgamma v^2-frac -mc^22int gamma d(1-v^2/c^2)\&=mgamma v^2+mc^2(1-v^2/c^2)^1/2-E_0endalignedbeginalignedE_textk&=mgamma v^2-frac -mc^22int gamma d(1-v^2/c^2)\&=mgamma v^2+mc^2(1-v^2/c^2)^1/2-E_0endaligned

E0displaystyle E_0E_0 is a constant of integration for the indefinite integral.
Simplifying the expression we obtain


Ek=mγ(v2+c2(1−v2/c2))−E0=mγ(v2+c2−v2)−E0=mγc2−E0displaystyle beginalignedE_textk&=mgamma (v^2+c^2(1-v^2/c^2))-E_0\&=mgamma (v^2+c^2-v^2)-E_0\&=mgamma c^2-E_0endalignedbeginalignedE_textk&=mgamma (v^2+c^2(1-v^2/c^2))-E_0\&=mgamma (v^2+c^2-v^2)-E_0\&=mgamma c^2-E_0endaligned

E0displaystyle E_0E_0 is found by observing that when v=0, γ=1displaystyle mathbf v =0, gamma =1!mathbf v =0, gamma =1! and Ek=0displaystyle E_textk=0!E_textk=0!, giving


E0=mc2displaystyle E_0=mc^2,E_0=mc^2,

resulting in the formula


Ek=mγc2−mc2=mc21−v2/c2−mc2displaystyle E_textk=mgamma c^2-mc^2=frac mc^2sqrt 1-v^2/c^2-mc^2E_textk=mgamma c^2-mc^2=frac mc^2sqrt 1-v^2/c^2-mc^2

This formula shows that the work expended accelerating an object from rest approaches infinity as the velocity approaches the speed of light. Thus it is impossible to accelerate an object across this boundary.


The mathematical by-product of this calculation is the mass-energy equivalence formula—the body at rest must have energy content


Erest=E0=mc2displaystyle E_textrest=E_0=mc^2!E_textrest=E_0=mc^2!

At a low speed (vdisplaystyle vv<<cdisplaystyle cc), the relativistic kinetic energy is approximated well by the classical kinetic energy. This is done by binomial approximation or by taking the first two terms of the Taylor expansion for the reciprocal square root:


Ek≈mc2(1+12v2/c2)−mc2=12mv2displaystyle E_textkapprox mc^2left(1+frac 12v^2/c^2right)-mc^2=frac 12mv^2E_textkapprox mc^2left(1+frac 12v^2/c^2right)-mc^2=frac 12mv^2

So, the total energy Ekdisplaystyle E_kE_k can be partitioned into the rest mass energy plus the Newtonian 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+12v2/c2+38v4/c4)−mc2=12mv2+38mv4/c2displaystyle E_textkapprox mc^2left(1+frac 12v^2/c^2+frac 38v^4/c^4right)-mc^2=frac 12mv^2+frac 38mv^4/c^2E_textkapprox mc^2left(1+frac 12v^2/c^2+frac 38v^4/c^4right)-mc^2=frac 12mv^2+frac 38mv^4/c^2

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


The relativistic relation between kinetic energy and momentum is given by


Ek=p2c2+m2c4−mc2displaystyle E_textk=sqrt p^2c^2+m^2c^4-mc^2E_textk=sqrt p^2c^2+m^2c^4-mc^2

This can also be expanded as a Taylor series, the first term of which is the simple expression from Newtonian mechanics:[9]


Ek≈p22m−p48m3c2.displaystyle E_textkapprox frac p^22m-frac p^48m^3c^2.displaystyle E_textkapprox frac p^22m-frac p^48m^3c^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[edit]



Using the convention that


gαβuαuβ=−c2displaystyle g_alpha beta ,u^alpha ,u^beta ,=,-c^2g_alpha beta ,u^alpha ,u^beta ,=,-c^2

where the four-velocity of a particle is


uα=dxαdτdisplaystyle u^alpha ,=,frac dx^alpha dtau u^alpha ,=,frac dx^alpha dtau

and τdisplaystyle tau ,tau , is the proper time of the particle, there is also an expression for the kinetic energy of the particle in general relativity.


If the particle has momentum


pβ=mgβαuαdisplaystyle p_beta ,=,m,g_beta alpha ,u^alpha p_beta ,=,m,g_beta alpha ,u^alpha

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


E=−pβuobsβdisplaystyle E,=,-,p_beta ,u_textobs^beta E,=,-,p_beta ,u_textobs^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_textobs^beta ,-,m,c^2,.E_k,=,-,p_beta ,u_textobs^beta ,-,m,c^2,.

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


uα=dxαdtdtdτ=vαutdisplaystyle u^alpha =frac dx^alpha dtfrac dtdtau =v^alpha u^t,u^alpha =frac dx^alpha dtfrac dtdtau =v^alpha u^t,

where vα 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(u^t)^2+g_ssv^2(u^t)^2,.-c^2=g_alpha beta u^alpha u^beta =g_tt(u^t)^2+g_ssv^2(u^t)^2,.

Solving for ut gives


ut=c−1gtt+gssv2.displaystyle u^t=csqrt frac -1g_tt+g_ssv^2,.u^t=csqrt frac -1g_tt+g_ssv^2,.

Thus for a stationary observer (v= 0)


uobst=c−1gttdisplaystyle u_textobs^t=csqrt frac -1g_tt,u_textobs^t=csqrt frac -1g_tt,

and thus the kinetic energy takes the form


Ek=−mgttutuobst−mc2=mc2gttgtt+gssv2−mc2.displaystyle E_textk=-mg_ttu^tu_textobs^t-mc^2=mc^2sqrt frac g_ttg_tt+g_ssv^2-mc^2,.E_textk=-mg_ttu^tu_textobs^t-mc^2=mc^2sqrt frac g_ttg_tt+g_ssv^2-mc^2,.

Factoring out the rest energy gives:


Ek=mc2(gttgtt+gssv2−1).displaystyle E_textk=mc^2left(sqrt frac g_ttg_tt+g_ssv^2-1right),.E_textk=mc^2left(sqrt frac g_ttg_tt+g_ssv^2-1right),.

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


gtt=−c2displaystyle g_tt=-c^2,g_tt=-c^2,

gss=1.displaystyle g_ss=1,.g_ss=1,.

In the Newtonian approximation to general relativity


gtt=−(c2+2Φ)displaystyle g_tt=-left(c^2+2Phi right),g_tt=-left(c^2+2Phi right),

gss=1−2Φc2displaystyle g_ss=1-frac 2Phi c^2,g_ss=1-frac 2Phi c^2,

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



Kinetic energy in quantum mechanics[edit]



In quantum mechanics, observables like kinetic energy are represented as operators. For one particle of mass m, the kinetic energy operator appears as a term in the Hamiltonian and is defined in terms of the more fundamental momentum operator p^displaystyle hat phat p. The kinetic energy operator in the non-relativistic case can be written as


T^=p^22m.displaystyle hat T=frac hat p^22m.hat T=frac hat p^22m.

Notice that this can be obtained by replacing pdisplaystyle pp by p^displaystyle hat phat p in the classical expression for kinetic energy in terms of momentum,


Ek=p22m.displaystyle E_textk=frac p^22m.E_textk=frac p^22m.

In the Schrödinger picture, p^displaystyle hat phat p takes the form −iℏ∇displaystyle -ihbar nabla -ihbar nabla where the derivative is taken with respect to position coordinates and hence


T^=−ℏ22m∇2.displaystyle hat T=-frac hbar ^22mnabla ^2.hat T=-frac hbar ^22mnabla ^2.

The expectation value of the electron kinetic energy, ⟨T^⟩displaystyle langle hat Trangle langle hat Trangle , for a system of N electrons described by the wavefunction |ψ⟩displaystyle vert psi rangle 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 langle hat Trangle =bigg langle psi bigg vert sum _i=1^Nfrac -hbar ^22m_textenabla _i^2bigg vert psi bigg rangle =-frac hbar ^22m_textesum _i=1^Nbigg langle psi bigg vert nabla _i^2bigg vert psi bigg rangle langle hat Trangle =bigg langle psi bigg vert sum _i=1^Nfrac -hbar ^22m_textenabla _i^2bigg vert psi bigg rangle =-frac hbar ^22m_textesum _i=1^Nbigg langle psi bigg vert nabla _i^2bigg vert psi bigg rangle

where medisplaystyle m_textem_texte is the mass of the electron and ∇i2displaystyle nabla _i^2nabla _i^2 is the Laplacian operator acting upon the coordinates of the ith electron and the summation runs over all electrons.


The density functional formalism of quantum mechanics requires knowledge of the electron density only, i.e., it formally does not require knowledge of the wavefunction. Given an electron density ρ(r)displaystyle rho (mathbf r )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)d3rdisplaystyle T[rho ]=frac 18int frac nabla rho (mathbf r )cdot nabla rho (mathbf r )rho (mathbf r )d^3rT[rho ]=frac 18int frac nabla rho (mathbf r )cdot nabla rho (mathbf r )rho (mathbf r )d^3r

where T[ρ]displaystyle T[rho ]T[rho ] is known as the von Weizsäcker kinetic energy functional.



See also[edit]



  • Escape velocity

  • Joule

  • KE-Munitions

  • Kinetic energy per unit mass of projectiles

  • Kinetic projectile

  • Parallel axis theorem

  • Potential energy

  • Recoil


Notes[edit]




  1. ^ Jain, Mahesh C. (2009). Textbook of Engineering Physics (Part I). p. 9. ISBN 978-81-203-3862-3. , Chapter 1, p. 9


  2. ^ Brenner, Joseph (2008). Logic in Reality (illustrated ed.). Springer Science & Business Media. p. 93. ISBN 978-1-4020-8375-4.  Extract of page 93


  3. ^ Judith P. Zinsser (2007). Emilie du Chatelet: Daring Genius of the Enlightenment. Penguin. ISBN 0-14-311268-6. 


  4. ^ Crosbie Smith, M. Norton Wise. Energy and Empire: A Biographical Study of Lord Kelvin. Cambridge University Press. p. 866. ISBN 0-521-26173-2. 


  5. ^ John Theodore Merz (1912). A History of European Thought in the Nineteenth Century. Blackwood. p. 139. ISBN 0-8446-2579-5. 


  6. ^ "Khan Academy". Khan Academy. Retrieved 2016-10-09. 


  7. ^ Sears, Francis Weston; Brehme, Robert W. (1968). Introduction to the theory of relativity. Addison-Wesley. p. 127. , Snippet view of page 127


  8. ^ Physics notes - Kinetic energy in the CM frame Archived 2007-06-11 at the Wayback Machine.. Duke.edu. Accessed 2007-11-24.


  9. ^ Fitzpatrick, Richard (20 July 2010). "Fine Structure of Hydrogen". Quantum Mechanics. Retrieved 20 August 2016. 




References[edit]



  • Physics Classroom (2000). "Kinetic Energy". Retrieved 2015-07-19. 


  • Oxford Dictionary 1998


  • School of Mathematics and Statistics, University of St Andrews (2000). "Biography of Gaspard-Gustave de Coriolis (1792-1843)". Retrieved 2006-03-03. 


  • Serway, Raymond A.; Jewett, John W. (2004). Physics for Scientists and Engineers (6th ed.). Brooks/Cole. ISBN 0-534-40842-7. 


  • Tipler, Paul (2004). Physics for Scientists and Engineers: Mechanics, Oscillations and Waves, Thermodynamics (5th ed.). W. H. Freeman. ISBN 0-7167-0809-4. 


  • Tipler, Paul; Llewellyn, Ralph (2002). Modern Physics (4th ed.). W. H. Freeman. ISBN 0-7167-4345-0. 


External links[edit]



  • Media related to Kinetic energy at Wikimedia Commons


  • kinetic energy - what it is and how it works








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