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Next: 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 Equation (1686)]. Because $ ds\propto d\tau$ it follows that

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

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

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

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

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

$\displaystyle U_\mu \,A^{\,\mu} = 0$ (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.


next up previous
Next: Current Density 4-Vector Up: Relativity and Electromagnetism Previous: Proper Time
Richard Fitzpatrick 2014-06-27