Kinetic energy

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:

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

where ${\displaystyle m}$ is the mass and ${\displaystyle v}$ is the speed (magnitude of 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

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

When a person throws a ball, the person does work 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.,

${\displaystyle Fs={\frac {1}{2}}mv^{2}}$

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 its momentum by the equation:

${\displaystyle E_{\text{k}}={\frac {p^{2}}{2m}}}$

where:

• ${\displaystyle p}$ is momentum
• ${\displaystyle m}$ 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 ${\displaystyle m}$, whose center of mass is moving in a straight line with speed ${\displaystyle v}$, as seen above is equal to

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

where:

• ${\displaystyle m}$ is the mass of the body
• ${\displaystyle v}$ 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.

The work W done by a force F on an object over a distance s parallel to F equals

${\displaystyle W=F\cdot s}$.

Using Newton's Second Law

${\displaystyle F=ma}$

with m the mass and a the acceleration of the object and

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

the distance traveled by the accelerated object in time t, we find with ${\displaystyle v=at}$ for the velocity v of the object

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

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

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

where we have assumed the 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:

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

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

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

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

${\displaystyle E_{\text{k}}=\int _{0}^{t}\mathbf {F} \cdot d\mathbf {x} =\int _{0}^{t}\mathbf {v} \cdot d(m\mathbf {v} )=\int _{0}^{t}d\left({\frac {mv^{2}}{2}}\right)={\frac {mv^{2}}{2}}.}$

This equation states that the kinetic energy (E_k) 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).

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

${\displaystyle E_{\text{r}}=\int _{Q}{\frac {v^{2}dm}{2}}=\int _{Q}{\frac {(r\omega )^{2}dm}{2}}={\frac {\omega ^{2}}{2}}\int _{Q}{r^{2}}dm={\frac {\omega ^{2}}{2}}I={\frac {1}{2}}I\omega ^{2}}$

where:

• ω is the body's angular velocity
• r is the distance of any mass dm from that line
• ${\displaystyle I}$ is the body's moment of inertia, equal to ${\textstyle \int _{Q}{r^{2}}dm}$.

(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).

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 a body's mass, inertia, and total energy.

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

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

Dividing by V, the unit of volume:

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

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

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.[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'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 ${\displaystyle \textstyle \mathbf {V} }$ be the relative velocity of the center of mass frame i in the frame k. Since

${\displaystyle v^{2}=\left(v_{i}+V\right)^{2}=\left(\mathbf {v} _{i}+\mathbf {V} \right)\cdot \left(\mathbf {v} _{i}+\mathbf {V} \right)=\mathbf {v} _{i}\cdot \mathbf {v} _{i}+2\mathbf {v} _{i}\cdot \mathbf {V} +\mathbf {V} \cdot \mathbf {V} =v_{i}^{2}+2\mathbf {v} _{i}\cdot \mathbf {V} +V^{2},}$

Then,

${\displaystyle E_{\text{k}}=\int {\frac {v^{2}}{2}}dm=\int {\frac {v_{i}^{2}}{2}}dm+\mathbf {V} \cdot \int \mathbf {v} _{i}dm+{\frac {V^{2}}{2}}\int dm.}$

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

${\displaystyle E_{\text{k}}=E_{i}+{\frac {MV^{2}}{2}}.}$

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 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).

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):

${\displaystyle E_{\text{k}}=E_{\text{t}}+E_{\text{r}}}$

where:

• E_k is the total kinetic energy
• E_t is the translational kinetic energy
• E_r 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.

Using the convention that

${\displaystyle g_{\alpha \beta }\,u^{\alpha }\,u^{\beta }\,=\,-c^{2}}$

where the four-velocity of a particle is

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

and ${\displaystyle \tau }$ 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

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

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

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

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

${\displaystyle E_{k}\,=\,-\,p_{\beta }\,u_{\text{obs}}^{\beta }\,-\,m\,c^{2}\,.}$

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

${\displaystyle u^{\alpha }={\frac {dx^{\alpha }}{dt}}{\frac {dt}{d\tau }}=v^{\alpha }u^{t}}$

where v^α is the ordinary velocity measured w.r.t. the coordinate system, we get

${\displaystyle -c^{2}=g_{\alpha \beta }u^{\alpha }u^{\beta }=g_{tt}\left(u^{t}\right)^{2}+g_{ss}v^{2}\left(u^{t}\right)^{2}\,.}$

Solving for u^t gives

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

Thus for a stationary observer (v = 0)

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

and thus the kinetic energy takes the form

${\displaystyle E_{\text{k}}=-mg_{tt}u^{t}u_{\text{obs}}^{t}-mc^{2}=mc^{2}{\sqrt {\frac {g_{tt}}{g_{tt}+g_{ss}v^{2}}}}-mc^{2}\,.}$

Factoring out the rest energy gives:

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

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

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

In the Newtonian approximation to general relativity

${\displaystyle {\begin{aligned}g_{tt}&=-\left(c^{2}+2\Phi \right)\\g_{ss}&=1-{\frac {2\Phi }{c^{2}}}\end{aligned}}}$

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