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{d au}=left(gamma_udotgamma_u c,gamma_u^2mathbf a+gamma_udotgamma_umathbf u ight)


: 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} ight)^{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 au.

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 a ight)

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 }{d au} = frac{dU^lambda }{d au } + Gamma^lambda {}_{mu u}U^mu U^ u

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.

ee also

* four-vector
* four-velocity
* four-momentum
* four-force



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