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Addition of Velocities

Figure 15.2: $P$ has ${\bf u}' = (u',\theta',\phi')$ in $K'$ frame. $K'$ is moving in the 1 direction at $v = c\beta $. $\gamma (v)$ changes frames. We want ${\bf u}(u,\theta,\phi)$.

If we form the infinitesimal version of the Lorentz transformation of coordinates:

$\displaystyle dx_0$ $\textstyle =$ $\displaystyle \gamma (dx_0' + \beta dx_1')$ (15.66)
$\displaystyle dx_1$ $\textstyle =$ $\displaystyle \gamma (dx_1' + \beta dx_0')$ (15.67)
$\displaystyle dx_2$ $\textstyle =$ $\displaystyle dx_2'$ (15.68)
$\displaystyle dx_3$ $\textstyle =$ $\displaystyle dx_3'$ (15.69)

Point $P$ is moving at velocity ${\bf u}'$ in frame $K'$, which is in turn moving at velocity ${\bf v} = v \hat{1}$ with respect to the ``rest'' frame $K$. We need to determine $u$ (the velocity of $P$ in $K$). We will express the problem, as usual, in coordinates $\parallel$ and $\perp$ to the direction of motion, exploiting the obvious azimuthal symmetry of the transformation about the $\hat{1}$ direction.

Note that

\begin{displaymath}
u_i = c \frac{dx_i}{dx_0}
\end{displaymath} (15.70)

for $i = 0 \ldots 3$. Then
$\displaystyle u_\parallel$ $\textstyle =$ $\displaystyle c \frac{\gamma(dx_1' + \beta dx_0')}{\gamma(dx_0' +
\beta dx_1')}$  
  $\textstyle =$ $\displaystyle c \frac{\left\{ \frac{dx_1'}{dx_0'} + \beta
\right\}}{\left\{ 1 + \beta \frac{dx_1'}{dx_0'} \right\}}$  
  $\textstyle =$ $\displaystyle \frac{u_\parallel + v}{1 + \frac{{\bf u}' \cdot {\bf
v}}{c^2}} .$ (15.71)

Similarly, $u_\perp$ (e.g. -- $u_2$) is given by
$\displaystyle u_2$ $\textstyle =$ $\displaystyle \frac{c dx_2'}{\gamma(dx_0' + \beta dx_1')}$  
  $\textstyle =$ $\displaystyle \frac{u_2'}{\gamma(1 + \frac{{\bf u}' \cdot {\bf v}}{c^2}}$ (15.72)

or
\begin{displaymath}
{\bf u}_\perp = \frac{{\bf u}_\perp}{\gamma \left\{ 1 + \frac{{\bf u}'
\cdot {\bf v}}{c^2} \right\}} .
\end{displaymath} (15.73)

We see, then, that the velocity changes in both the $\parallel$ and the $\perp$ directions.

Note also that if $\left\vert \bf u' \right\vert$ and $\left\vert \bf v \right\vert << c$, then

\begin{displaymath}
\frac{{\bf u}' \cdot {\bf v}}{c^2} << 1
\end{displaymath} (15.74)

and
\begin{displaymath}
\gamma \approx 1
\end{displaymath} (15.75)

so that we recover the Gallilean result,
$\displaystyle {\bf u}_\parallel$ $\textstyle =$ $\displaystyle {\bf u}_\parallel' + {\bf v}$ (15.76)
$\displaystyle {\bf u}_\perp$ $\textstyle =$ $\displaystyle {\bf u}_\perp'.$ (15.77)

What about the other limit? If $\left\vert \bf u' \right\vert = c$, then

\begin{displaymath}
\left\vert \bf u \right\vert = c
\end{displaymath} (15.78)

as you should verify on your own. This is Einstein's second postulate! We have thus proven explicitly that the speed of light (and the speed of anything else travelling at the speed of light) is invariant under Lorentz coordinate transformations. This is their entire motivation.

We observe that the three spatial components of ``velocity'' do not seem to transform like a four vector. Both the $\parallel$ and the $\perp$ components are mixed by a boost. We can, however, make the velocity into a four vector that does. We define

$\displaystyle U_0$ $\textstyle =$ $\displaystyle \frac{dx_0}{d\tau} = \frac{dx_0}{dt} \frac{dt}{d \tau}$  
  $\textstyle =$ $\displaystyle c \gamma(u)$ (15.79)
$\displaystyle {\bf U}$ $\textstyle =$ $\displaystyle \frac{d{\bf x}}{d\tau} = \frac{d{\bf x}}{dt}
\frac{dt}{d\tau}$  
  $\textstyle =$ $\displaystyle {\bf u} \gamma(u)$ (15.80)

where $\gamma(u)$ is evaluated using the magnitude of u. It is an exercise to show that this transforms like the coordinate 4-vector $x$.

Now we can ``guess'' that the 4-momentum of a particle will be $ \sim m U$. To prepare us for this, observe that

\begin{displaymath}
U = (U_0,{\bf U}) = (\gamma_u c, \gamma_u {\bf u})
\end{displaymath} (15.81)

are just the $\gamma_u$-scaled ``velocities'' of the particle:

Figure 15.3: Note that $\gamma _u \ge 1$ so that each component of the 4-velocity is always ``larger'' than associated Cartesian components, even though (as usual) the length of the four velocity is invariant. What is its invariant length?


next up previous contents
Next: Relativistic Energy and Momentum Up: Special Relativity Previous: Proper Time and Time   Contents
Robert G. Brown 2014-08-19