- Four-acceleration
In
special relativity , four-acceleration is afour-vector and is defined as the change infour-velocity over the particle'sproper time :: mathbf{A} =frac{dmathbf{U{d au}=left(gamma_udotgamma_u c,gamma_u^2mathbf a+gamma_udotgamma_umathbf u ight)
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} 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 aworld line .The world lines having constant magnitude of four-acceleration are Minkowski-circles i.e. hyperbolas (see "hyperbolic motion")The
scalar product of afour-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 thefour-velocity through acovariant 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 thegeodesic equation .ee also
*
four-vector
*four-velocity
*four-momentum
*four-force References
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