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Next: The current density 4-vector Up: Relativity and electromagnetism Previous: Proper time

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. (2.47)). Since $ds\propto d\tau$ it follows that
\begin{displaymath}
U^\mu = \frac{dx^\mu}{d\tau}
\end{displaymath} (114)

also transforms as a 4-vector. This quantity is known as the 4-velocity. Likewise, the quantity
\begin{displaymath}
A^\mu = \frac{d^2x^\mu}{d\tau^2} = \frac{dU^\mu}{d\tau}
\end{displaymath} (115)

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

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

\begin{displaymath}
U^\mu = \frac{dx^\mu}{d\tau} = \frac{dx^\mu}{dt}\frac{dt}{d\tau}
= \gamma(u) ({\bfm u}, c),
\end{displaymath} (116)

where use has been made of Eq. (2.78). 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
\begin{displaymath}
A^\mu = \frac{dU^\mu}{d\tau} = \gamma\,\frac{dU^\mu}{dt}
=\g...
...dt} \,{\bfm u} + \gamma{\bfm a}, c\,\frac{d\gamma}{dt}\right),
\end{displaymath} (117)

where ${\bfm a} = d{\bfm u}/{dt}$ is the 3-acceleration. In the rest frame of the particle $U^\mu=({\bf0}, c)$ and $A^\mu = ({\bfm a}, 0)$. It follows that
\begin{displaymath}
U_\mu A^\mu = 0
\end{displaymath} (118)

(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.


next up previous
Next: The current density 4-vector Up: Relativity and electromagnetism Previous: Proper time
Richard Fitzpatrick 2002-05-18