4-Velocity and 4-Acceleration

We have seen that the quantity $dx^{\,\mu}/ds$ transforms as a 4-vector under a general Lorentz transformation [see Equation (1686)]. Because $ds\propto d\tau$ it follows that

$$U^{\,\mu} = \frac{dx^{\,\mu}}{d\tau}$$ (1721)

also transforms as a 4-vector. This quantity is known as the 4-velocity. Likewise, the quantity

$$A^{\,\mu} = \frac{d^{\,2}x^{\,\mu}}{d\tau^{\,2}} = \frac{dU^{\,\mu}}{d\tau}$$ (1722)

is a 4-vector, and is called the 4-acceleration.

For events along the world-line of a particle traveling with 3-velocity ${\bf u}$ , we have

$$U^{\,\mu} = \frac{dx^{\,\mu}}{d\tau} = \frac{dx^{\,\mu}}{dt}\frac{dt}{d\tau} = \gamma(u) \,({\bf u},\, c),$$ (1723)

where use has been made of Equation (1719). This gives the relationship between a particle's 3-velocity and its 4-velocity. The relationship between the 3-acceleration and the 4-acceleration is less straightforward. We have

$$A^{\,\mu} = \frac{dU^{\,\mu}}{d\tau} = \gamma\,\frac{dU^{\,\mu}}{... ...\frac{d\gamma}{dt} \,{\bf u} + \gamma\,{\bf a},\, c\,\frac{d\gamma}{dt}\right),$$ (1724)

where ${\bf a} = d{\bf u}/{dt}$ is the 3-acceleration. In the rest frame of the particle, $U^{\,\mu}=({\bf0},\, c)$ and $A^{\,\mu} = ({\bf a},\,0)$ . It follows that

$$U_\mu \,A^{\,\mu} = 0$$ (1725)

(note that $U_\mu \,A^{\,\mu}$ is an invariant quantity). In other words, the 4-acceleration of a particle is always orthogonal to its 4-velocity.