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 Eq. (1389)]. Since $ds\propto d\tau$ it follows that

$$U^\mu = \frac{dx^\mu}{d\tau}$$ (1426)

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

$$A^\mu = \frac{d^2x^\mu}{d\tau^2} = \frac{dU^\mu}{d\tau}$$ (1427)

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

$$U^\mu = \frac{dx^\mu}{d\tau} = \frac{dx^\mu}{dt}\frac{dt}{d\tau} = \gamma(u) ({\bf u}, c),$$ (1428)

where use has been made of Eq. (1422). 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

$$A^\mu = \frac{dU^\mu}{d\tau} = \gamma \frac{dU^\mu}{dt} =\g... ...dt} {\bf u} + \gamma {\bf a}, c \frac{d\gamma}{dt}\right),$$ (1429)

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

$$U_\mu A^\mu = 0$$ (1430)

(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.