The Galilean transformation $x\prime =x-vt$ of classical relativity lets us calculate the position $x\prime$ of an event in frame $\text{S}\prime$ if we know its position x in frame S. Classical relativity, of course, assumes that $t\prime =t$. Is there a similar transformation in relativity that would allow us to calculate an event’s spacetime coordinates $(x\prime ,t\prime )$ in frame $\text{S}\prime$ if we know their values $(x,t)$ in frame S? Such a transformation would need to satisfy three conditions:

1. Agree with the Galilean transformations in the low-speed limit $v\ll c$.

2. Transform not only spatial coordinates but also time coordinates.

3. Ensure that the speed of light is the same in all reference frames.

We’ll continue to use reference frames in the standard orientation of FIGURE 36.29. The motion is parallel to the x- and $x\prime \text{-}\text{axes,}$ and we define $t=0$ and $t\prime =0$ as the instant when the origins of S and $\text{S}\prime$ coincide.

FIGURE 36.29 The spacetime coordinates of an event are measured in inertial reference frames S and S$\prime$.

The requirement that a new transformation agree with the Galilean transformation when $v\ll c$ suggests that we look for a transformation of the form

where $\gamma$ is a dimensionless function of velocity that satisfies $\gamma \to 1$ as $v\to 0$.

To determine $\gamma$, we consider the following two events:

• Event 1: A flash of light is emitted from the origin of both reference frames $(x=x\prime =0)$ at the instant they coincide $(t=t\prime =0)$.

• Event 2: The light strikes a light detector. The spacetime coordinates of this event are $(x,t)$ in frame S and $(x\prime ,t\prime )$ in frame $\text{S}\prime$.

Light travels at speed c in both reference frames, so the positions of event 2 are $x=ct$ in S and $x\prime =ct\prime$ in $\text{S}\prime$. Substituting these expressions for x and $x\prime$ into Equation 36.20 gives

We solve the first equation for $t\prime$, by dividing by c, then substitute this result for $t\prime$ into the second:

The t cancels, leading to

Thus the $\gamma$ that “works” in the proposed transformation of Equation 36.20 is

You can see that $\gamma \to 1$ as $v\to 0$, as expected.

The transformation between t and $t\prime$ is found by requiring that $x=x$ if you use Equation 36.20 to transform a position from S to $\text{S}\prime$ and then back to S. The details will be left for a homework problem. Another homework problem will let you demonstrate that the y and z measurements made perpendicular to the relative motion are not affected by the motion. We tacitly assumed this condition in our analysis of the light clock.

The full set of equations is called the Lorentz transformations. They are

The Lorentz transformations transform the spacetime coordinates of one event. Compare these to the Galilean transformation equations in Equations 36.1.