Four-acceleration

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

$A =\frac{dU}{d\tau}=\left(\gamma_u\dot\gamma_u c,\gamma_u^2\mathbf a+\gamma_u\dot\gamma_u\mathbf u\right)$

where

$\mathbf a = {d\mathbf u \over dt}$ and $\dot\gamma_u = \frac{\mathbf{a \cdot u}}{c^2} \gamma^3 = \frac{\mathbf{a \cdot u}}{c^2} \frac{1}{\left(1-\frac{u^2}{c^2}\right)^{3/2}}= {u\dot u/c^2 \over (1 - u^2/c^2)^{3/2}}$

and $γ u$ 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 time in a given reference frame, not the proper time $τ$ .

In an instantly co-moving inertial reference frame $\mathbf u = 0$ , $γ u = 1$ and $\dot\gamma_u = 0$ , i.e. in such a reference frame

$A =\left(0, \mathbf a\right)$

Therefore, the four-acceleration 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. hyperboles (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 μ = m A μ

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.

$A^\lambda := \frac{DU^\lambda }{d\tau} = \frac{dU^\lambda }{d\tau } + \Gamma^\lambda {}_{\mu \nu}U^\mu U^\nu$

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 along, and the four-vector version of Newton's second law above reduces to the geodesic equation.