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We have seen that the quantity
transforms as a 4-vector under
a general Lorentz transformation [see Eq. (1389)]. Since
it follows that
![\begin{displaymath}
U^\mu = \frac{dx^\mu}{d\tau}
\end{displaymath}](img2943.png) |
(1426) |
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}](img2944.png) |
(1427) |
is a 4-vector, and is called the 4-acceleration.
For events along the world-line of a particle traveling with 3-velocity
, we have
![\begin{displaymath}
U^\mu = \frac{dx^\mu}{d\tau} = \frac{dx^\mu}{dt}\frac{dt}{d\tau}
= \gamma(u) ({\bf u}, c),
\end{displaymath}](img2945.png) |
(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
![\begin{displaymath}
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),
\end{displaymath}](img2946.png) |
(1429) |
where
is the 3-acceleration.
In the rest frame of the particle,
and
. It follows that
![\begin{displaymath}
U_\mu A^\mu = 0
\end{displaymath}](img2950.png) |
(1430) |
(note that
is an invariant quantity).
In other words, the 4-acceleration of a particle is always orthogonal
to its 4-velocity.
Next: The current density 4-vector
Up: Relativity and electromagnetism
Previous: Proper time
Richard Fitzpatrick
2006-02-02