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Four-acceleration

In special relativity, four-acceleration is a four-vector and is defined as the change in four-velocity over the particle's proper time:

mathbf{A} =frac{dmathbf{U}}{dtau}=left(gamma_udotgamma_u c,gamma_u^2mathbf a+gamma_udotgamma_umathbf uright)

where

mathbf a = {dmathbf u over dt}

and

dotgamma_u = frac{mathbf{a cdot u}}{c^2} gamma_u^3 = frac{mathbf{a cdot u}}{c^2} frac{1}{left(1-frac{u^2}{c^2}right)^{3/2}}= {udot u/c^2 over (1 - u^2/c^2)^{3/2}}

and $gamma_u$ is the Lorentz factor for the speed $u$ . It should be noted that a dot above a variable indicates a derivative with respect to the coordinate time in a given reference frame, not the proper time $tau$ .

In an instantaneously co-moving inertial reference frame $mathbf u = 0$ , $gamma_u = 1$ and $dotgamma_u = 0$ , i.e. in such a reference frame

mathbf{A} =left(0, mathbf aright)

Therefore, the magnitude of the four-acceleration (which is an invariant scalar) is equal to the proper acceleration that a moving particle "feels" moving along a world line. The world lines having constant magnitude of four-acceleration are Minkowski-circles i.e. hyperbolas (see hyperbolic motion)

The scalar product of a four-velocity and the corresponding four-acceleration is always 0.

Even at relativistic speeds four-acceleration is related to the four-force such that

F^mu = mA^mu

where m is the invariant mass of a particle.

In general relativity the elements of the acceleration four-vector are related to the elements of the four-velocity through a covariant derivative with respect to proper time.

A^lambda := frac{DU^lambda }{dtau} = frac{dU^lambda }{dtau } + Gamma^lambda {}_{mu nu}U^mu U^nu

This relation holds in special relativity too when one uses curved coordinates, i.e. when the frame of reference isn't inertial.

When the four-force is zero one has gravitation acting alone, and the four-vector version of Newton's second law above reduces to the geodesic equation.