Information About

Four-acceleration

:

A =rac{dU}{d	au}=\left(\gamma_u\dot\gamma_u c,\gamma_u^2\mathbf a+\gamma_u\dot\gamma_u\mathbf u
ight)

where

:

\mathbf a = {d\mathbf u \over dt}

and

\dot\gamma_u = rac{\mathbf{a \cdot u}}{c^2} \gamma^3 = rac{\mathbf{a \cdot u}}{c^2} rac{1}{\left(1-rac{u^2}{c^2}
ight)^{3/2}}= {u\dot 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 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

\dot\gamma_u = 0

, i.e. in such a reference frame
:

A =\left(0, \mathbf a
ight)

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^\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 := rac{DU^\lambda }{d	au} = rac{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 along, and the four-vector version of Newton's second law above reduces to the Geodesic Equation .

SEE ALSO

Four-vector

Four-velocity

Four-momentum

Four-force

REFERENCES

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