36.8.4 The Lorentz Velocity Transformations

FIGURE 36.33 shows an object that is moving in both reference frame S and reference frame $S'$ . Experimenters in frame S determine that the object’s velocity is u, while experimenters in frame $S'$ find it to be $u'$ . For simplicity, we’ll assume that the object moves parallel to the x- and $x' - axes$ .

The figure consists of two parts that illustrate the velocity of a moving object with respect to two reference frames.

FIGURE 36.33 Full Alternative Text

The Galilean velocity transformation $u' = u - v$ was found by taking the time derivative of the position transformation. We can do the same with the Lorentz transformation if we take the derivative with respect to the time in each frame. Velocity $u'$ in frame $S'$ is

u' = \frac{d x'}{d t'} = \frac{d [γ (x - v t)]}{d [γ (t - v x / c^{2})]}

(36.27)

where we’ve used the Lorentz transformations for position $x'$ and time $t'$ .

Carrying out the differentiation gives

u' = \frac{γ (d x - v d t)}{γ (d t - v d x / c^{2})} = \frac{d x / d t - v}{1 - v (d x / d t) / c^{2}}

(36.28)

But $d x / d t$ is u, the object’s velocity in frame S, leading to

u' = \frac{u - v}{1 - u v / c^{2}}

(36.29)

You can see that Equation 36.29 reduces to the Galilean transformation $u' = u - v$ when $v ≪ c$ , as expected.

The transformation from $S'$ to S is found by reversing the sign of v. Altogether,

u' = \frac{u - v}{1 - u v / c^{2}} and u = \frac{u' + v}{1 + u' v / c^{2}}

(36.30)

Equations 36.30 are the Lorentz velocity transformation equations.

NOTE It is important to distinguish carefully between v, which is the relative velocity between two reference frames, and u and $u'$ , which are the velocities of an object as measured in the two different reference frames.

EXAMPLE 36.9 A really fast bullet

A rocket flies past the earth at 0.90c. As it goes by, the rocket fires a bullet in the forward direction at 0.95c with respect to the rocket. What is the bullet’s speed with respect to the earth?

MODEL The rocket and the earth are inertial reference frames. Let the earth be frame S and the rocket be frame $S'$ . The velocity of frame $S'$ relative to frame S is $v = 0.90 c$ . The bullet’s velocity in frame $S'$ is $u' = 0.95 c$ .

SOLVE We can use the Lorentz velocity transformation to find

u = \frac{u' + v}{1 + u' v / c^{2}} = \frac{0.95 c + 0.90 c}{1 + (0.95 c) (0.90 c) / c^{2}} = 0.997 c

The bullet’s speed with respect to the earth is 99.7% of the speed of light.

NOTE Many relativistic calculations are much easier when velocities are specified as a fraction of c.

REVIEW In Newtonian mechanics, the Galilean transformation of velocity would give $u = 1.85 c$ . Now, despite the very high speed of the rocket and of the bullet with respect to the rocket, the bullet’s speed with respect to the earth remains less than c. This is yet another indication that objects cannot travel faster than the speed of light.

Suppose the rocket in Example 36.9 fired a laser beam in the forward direction as it traveled past the earth at velocity v. The laser beam would travel away from the rocket at speed $u' = c$ in the rocket’s reference frame $S'$ . What is the laser beam’s speed in the earth’s frame S? According to the Lorentz velocity transformation, it must be

u = \frac{u' + v}{1 + u' v / c^{2}} = \frac{c + v}{1 + c v / c^{2}} = \frac{c + v}{1 + v / c} = \frac{c + v}{(c + v) / c} = c

(36.31)

Light travels at speed c in both frame S and frame $S'$ . This important consequence of the principle of relativity is “built into” the Lorentz transformations.