5.6 Relativistic Velocity Transformation

Velocity Transformations

Imagine a car traveling at night along a straight road, as in Figure 5.19. The driver sees the light leaving the headlights at speed c within the car’s frame of reference. If the Galilean transformation applied to light, then the light from the car’s headlights would approach the pedestrian at a speed $u=v+c,$ contrary to Einstein’s postulates.

Both the distance traveled and the time of travel are different in the two frames of reference, and they must differ in a way that makes the speed of light the same in all inertial frames. The correct rules for transforming velocities from one frame to another can be obtained from the Lorentz transformation equations.

Relativistic Transformation of Velocity

Suppose an object P is moving at constant velocity $u=\left(u_x^{\prime },u_y^{\prime },u_z^{\prime }\right)$ as measured in the $S\prime$ frame. The $S\prime$ frame is moving along its $x\prime \text{-axis}$ at velocity v. In an increment of time $dt\prime$, the particle is displaced by $dx\prime$ along the $x\prime \text{-axis.}$ Applying the Lorentz transformation equations gives the corresponding increments of time and displacement in the unprimed axes:

$$\begin{array}{ccc}\hfill dt& =\hfill & \gamma \left(d{t}^{\prime }+vd{x}^{\prime }\text{/}{c}^2\right)\hfill \\ \hfill dx& =\hfill & \gamma \left(d{x}^{\prime }+vd{t}^{\prime }\right)\hfill \\ \hfill dy& =\hfill & d{y}^{\prime }\hfill \\ \hfill dz& =\hfill & d{z}^{\prime }.\hfill \end{array}$$

The velocity components of the particle seen in the unprimed coordinate system are then

$$\begin{array}{ccc}\hfill \frac{dx}{dt}& =\hfill & \frac{\gamma \left(dx\prime +vdt\prime \right)}{\gamma \left(dt\prime +vdx\prime \text{/}{c}^2\right)}=\frac{\frac{dx\prime }{dt\prime }+v}{1+\frac{v}{c^2}\frac{dx\prime }{dt\prime }}\hfill \\ \hfill \frac{dy}{dt}& =\hfill & \frac{dy\prime }{\gamma \left(dt\prime +vdx\prime \text{/}{c}^2\right)}=\frac{\frac{dy\prime }{dt\prime }}{\gamma \left(1+\frac{v}{c^2}\frac{dx\prime }{dt\prime }\right)}\hfill \\ \hfill \frac{dz}{dt}& =\hfill & \frac{dz\prime }{\gamma \left(dt\prime +vdx\prime \text{/}{c}^2\right)}=\frac{\frac{dz\prime }{dt\prime }}{\gamma \left(1+\frac{v}{c^2}\frac{dx\prime }{dt\prime }\right)}.\hfill \end{array}$$

We thus obtain the equations for the velocity components of the object as seen in frame S:

$$u_x=\left(\frac{u_x^{\prime }+v}{1+v{u}_x^{\prime }\text{/}{c}^2}\right),u_y=\left(\frac{u_y^{\prime }\text{/}\gamma }{1+v{u}_x^{\prime }\text{/}{c}^2}\right),u_z=\left(\frac{u_z^{\prime }\text{/}\gamma }{1+v{u}_x^{\prime }\text{/}{c}^2}\right).$$

Compare this with how the Galilean transformation of classical mechanics says the velocities transform, by adding simply as vectors:

$$u_x=u_x^{\prime }+u,u_y=u_y^{\prime },u_z=u_z^{\prime }.$$

When the relative velocity of the frames is much smaller than the speed of light, that is, when $v\ll c,$ the special relativity velocity addition law reduces to the Galilean velocity law. When the speed v of $S\prime$ relative to S is comparable to the speed of light, the relativistic velocity addition law gives a much smaller result than the classical (Galilean) velocity addition does.

Velocities cannot add to greater than the speed of light, provided that v is less than c and $u\prime$ does not exceed c. The following example illustrates that relativistic velocity addition is not as symmetric as classical velocity addition.

Summary

• With classical velocity addition, velocities add like regular numbers in one-dimensional motion: $u=v+u\prime ,$ where v is the velocity between two observers, u is the velocity of an object relative to one observer, and $u\prime$ is the velocity relative to the other observer.
• Velocities cannot add to be greater than the speed of light.
• Relativistic velocity addition describes the velocities of an object moving at a relativistic velocity.

Problems