Editing Four-acceleration (section)

== Four-acceleration in inertial coordinates ==

In inertial coordinates in [[special relativity]], four-acceleration <math>\mathbf{A}</math> is defined as the rate of change in [[four-velocity]] <math>\mathbf{U}</math> with respect to the particle's [[proper time]] along its [[worldline]].  We can say:
<math display="block">\begin{align}
  \mathbf{A} = \frac{d\mathbf{U}}{d\tau}
    &= \left(\gamma_u\dot\gamma_u c,\, \gamma_u^2\mathbf a + \gamma_u\dot\gamma_u\mathbf u\right) \\
    &= \left(
         \gamma_u^4\frac{\mathbf{a}\cdot\mathbf{u}}{c},\,
         \gamma_u^2\mathbf{a} + \gamma_u^4\frac{\mathbf{a}\cdot\mathbf{u}}{c^2}\mathbf{u}
       \right) \\
    &= \left(
         \gamma_u^4\frac{\mathbf{a}\cdot\mathbf{u}}{c},\,
         \gamma_u^4\left(\mathbf{a} + \frac{\mathbf{u}\times \left(\mathbf{u}\times\mathbf{a}\right)}{c^2}\right)
       \right),
\end{align}</math>
where
* <math>\mathbf a = \frac{d\mathbf u}{dt}</math> , with <math>\mathbf a </math> the three-acceleration and <math>\mathbf u </math> the three-velocity, and 
* <math>\dot\gamma_u
  = \frac{\mathbf a \cdot \mathbf u}{c^2} \gamma_u^3
  = \frac{\mathbf a \cdot \mathbf u}{c^2} \frac{1}{\left(1 - \frac{u^2}{c^2}\right)^{3/2}},
</math> and
* <math>\gamma_u</math> is the [[Lorentz factor]] for the speed <math>u</math> (with <math>|\mathbf{u}| = u</math>).  A dot above a variable indicates a derivative with respect to the coordinate time in a given reference frame, not the proper time <math>\tau</math> (in other terms, <math display="inline">\dot\gamma_u = \frac{d\gamma_u}{dt}</math>).

In an instantaneously co-moving inertial reference frame <math>\mathbf u = 0</math>, <math>\gamma_u = 1 </math> and <math>\dot\gamma_u = 0</math>, i.e. in such a reference frame 
<math display="block">\mathbf{A} = \left(0, \mathbf a\right) .</math>

Geometrically, four-acceleration is a [[curvature vector]] of a worldline.<ref>{{cite book| author=Pauli W.|title=Theory of Relativity |edition=1981 Dover|publisher=B.G. Teubner, Leipzig|year=1921|pages=74|isbn=978-0-486-64152-2}}</ref><ref>{{cite book| author1=Synge J.L.|author2=Schild A.|title=Tensor Calculus|edition=1978 Dover|publisher=University of Toronto Press| year=1949| isbn=0-486-63612-7|pages=[https://archive.org/details/tensorcalculus00syng/page/149 149, 153 and 170]|url-access=registration| url=https://archive.org/details/tensorcalculus00syng/page/149}}</ref>

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 worldline. A worldline having constant four-acceleration is a Minkowski-circle i.e. hyperbola (see [[hyperbolic motion (relativity)|''hyperbolic motion'']])

The [[scalar product]] of a particle's [[four-velocity]] and its four-acceleration is always 0.

Even at relativistic speeds four-acceleration is related to the [[four-force]]: <math display="block"> F^\mu = m A^\mu ,</math> where {{mvar|m}} is the [[invariant mass]] of a particle.

When the [[four-force]] is zero, only gravitation affects the trajectory of a particle, and the four-vector equivalent of Newton's second law above reduces to the [[geodesic equation]]. The four-acceleration of a particle executing geodesic motion is zero.  This corresponds to gravity not being a force.  Four-acceleration is different from what we understand by acceleration as defined in Newtonian physics, where gravity is treated as a force.