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We have seen that the quantity
transforms as a 4-vector under
a general Lorentz transformation [see Equation (1686)]. Because
it follows that
![$\displaystyle U^{\,\mu} = \frac{dx^{\,\mu}}{d\tau}$](img3701.png) |
(1721) |
also transforms as a 4-vector. This quantity is known as the 4-velocity.
Likewise, the quantity
![$\displaystyle A^{\,\mu} = \frac{d^{\,2}x^{\,\mu}}{d\tau^{\,2}} = \frac{dU^{\,\mu}}{d\tau}$](img3702.png) |
(1722) |
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
![$\displaystyle U^{\,\mu} = \frac{dx^{\,\mu}}{d\tau} = \frac{dx^{\,\mu}}{dt}\frac{dt}{d\tau} = \gamma(u) \,({\bf u},\, c),$](img3703.png) |
(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
![$\displaystyle A^{\,\mu} = \frac{dU^{\,\mu}}{d\tau} = \gamma\,\frac{dU^{\,\mu}}{...
...\frac{d\gamma}{dt} \,{\bf u} + \gamma\,{\bf a},\, c\,\frac{d\gamma}{dt}\right),$](img3704.png) |
(1724) |
where
is the 3-acceleration.
In the rest frame of the particle,
and
. It follows that
![$\displaystyle U_\mu \,A^{\,\mu} = 0$](img3708.png) |
(1725) |
(note that
is an invariant quantity).
In other words, the 4-acceleration of a particle is always orthogonal
to its 4-velocity.
Next: Current Density 4-Vector
Up: Relativity and Electromagnetism
Previous: Proper Time
Richard Fitzpatrick
2014-06-27