Inertial Frames

Transformation Function

Relative Velocity

Light Speed

Length Contraction

The Inverse Transform

Time and Distance

Questions

We have the following apparatus for measuring the speed of light. It contains a clock that measures time, *t*, and a ruler that measures position, *x*. A laser at position *x* = 0 m turns on at time *t* = 0 s. Its light propagates towards a target at position *x* = *L*. The light arrives at the detector at time *t* = *a*. The speed of light is *c* = *L/a*.

We set up the apparatus in our laboratory on the surface of the Earth. We point the laser in the direction of the Earth's 30 km/s orbital velocity. We measure *c* = 299,792 km/s. We rotate the apparatus and point the laser in the opposite direction. Some of us are expecting to get an answer 60 km/s higher this time, because the Earth is carrying the target towards the advancing light, while it carried the target away from the advancing light in the first measurement. But we measure *c* = 299,792 km/s again. We repeat our measurement in many orientations and we always get the same answer. We repeat our measurement in the same orientation a hundred times throughout the day, and we always get the same answer.

What we have described is a simplified version of the series of experiments performed by Michelson and Morley in 1887, and repeated in a variety of ways in the decades that followed. The speed of light is the same for all observers moving at constant velocity, regardless of how fast they move, the direction of the light, or which observer holds the source of the light.

In the following sections, we will determine how it is possible for different observers moving at different velocities to measure the same beam of light to be moving at the same speed. For brevity and clarity, but without loss of generality, we will conduct our investigation in one spatial dimension. We will arrive at the one-dimensional Lorentz Transformation, which is easily extended to three dimensions.

For our purposes, an *inertial frame of reference* is a ruler that is not accelerating, accompanied by a set of synchronized clocks. The ruler measures position, *x*, along a line. In one direction, *x* increases and in the other it decreases. Along the ruler we have the clocks. Observers who are stationary with respect to the ruler agree that the clocks read exactly the same time. We say the clocks are *synchronized* within the frame of reference. We can check the synchronisation of the clocks at any time with a machine that travels along our ruler at a constant velocity *u*. We release this machine from *x* = 0 m at time *t* = *b* s. When it passes a clock at position *x* = *L*, the clock should read *b + L/u*. If the clock is wrong, we correct it.

The following diagram shows two frames of reference, *F* and *F ^{ |}*. We could place them right on top of one another, so that their rulers were along the same line, but this would be hard to draw. Let us imagine that their rulers are parallel and close together. The frames are moving with respect to one another. So far as frame

Suppose something happens in the space between the two rulers, so that observers on both rulers have a clear view of the event. This event could be a beam of light being emitted by a laser or a beam of light striking a target. When this event occurs, it does so at a position *x* in *F* and *x ^{ |}* in

The values of *t* and *t ^{ |}* could be different because one clock is far behind the other. But let us suppose, without loss of generality, that at

We are inclined to assume that the clocks facing one another across the small space between the rulers will always read the same time. And we are inclined to assume that the 1-mm divisions on the two rulers will have the same length. But we will make neither assumption. We will assume only that one frame is moving at *v* with respect to the other and the speed of light will be *c* in both frames. These two observations are so constraining that they already dictate what we will see on either side of the small gap between the rulers, so we cannot make any further assumptions. All we can do is figure out the consequences of the assumptions we have already made.

Consider an event that occurs at position *x* and time *t* in frame *F*. In frame *F ^{ |}* the same event has position

Suppose we keep *t* constant. The slope of a graph of *x ^{ |}* versus

Suppose the transformation function were not linear. Suppose the slope of the graph of *x ^{ |}* versus

We can use the same argument for any curvature in the graph of *x ^{ |}* versus

The transformation function is linear. Our job is to deduce the constants *p*, *q*, *r*, and *s* using the following constraints: the transform will obey the principle of relativity, the transform will predict that light will travel at speed *c* in both frames, and the transform will agree that one frame is moving at velocity *v* with respect to the other.

The point *x ^{ |}* = 0 m moves with velocity

We are sitting on a train looking outside. We are *F ^{ |}* and the world outside is

Observers in both *F* and *F ^{ |}* will measure the speed of a beam of light to be

At time *t* = 0 s, *t ^{ |}* = −

Consider the segment of the ruler in *F* that extends from *x* = 0 m to *x* = *L*. At time *t* = 0 s our transformation function tells us that one end of this segment is at *x ^{ |}* = 0 m and the other is at

When we measure the length of a moving object, we mark the position of its front end and its back end on our ruler. We must make these marks simultaneously, or else the object will move between the time we make the first mark and the time we make the second mark. But simultaneity is not conserved between frames. If simultaneity is not conserved, nor will length be conserved.

If the ruler in *F ^{ |}* appears contracted by a factor of

The constant *p* is called *gamma* by physicists, denoted γ. When *v* << *c*, γ ≈ 1. When *v* is larger, γ > 1. For example, when *v* = *c*/2, γ = 1.15. Because γ is a function of *v ^{2}*, it does not matter if

The transformation function from *F* to *F ^{ |}* is below. Its constants are a function of

The transformation function from *F ^{ |}* to

The only difference between the original transform and its inverse is that the terms in *v* have changed sign. The sign change is a consequence of the fact that *F* moves at −*v* in *F ^{ |}*, while

Suppose *v* = *c*/2. We have γ = 1.15. In *F*, at *t* = 0 s we note the value of *x ^{ |}* that coincides with

The observer in *F ^{ |}*, however, sees us marking the position of

The *Lorentz Contraction* is what we call the shrinking of the length of moving objects in their direction of motion. We used large distances in our example above, but all lengths are contracted by γ. A 1-m ruler passing by at *c*/2 will appear to be only 0.67 m long. We don't usually observe rulers going by at the half the speed of light, but we do observe electrons going at 99% of the speed of light. The electron itself has no length, but it produces an electric field, and this field is compressed in the direction of motion by the Lorentz contraction. The voltage such an electron induces in a nearby wire is indeed exactly predicted by the contraction.

Let us explore the transformation of time in more detail. Consider an observer in *F ^{ |}* at

The *twin paradox* refers to the apparent violation of relativity that occurs when one observer experiences a different passage of time from another, even though they end up in the same place. But we see that this is not a paradox at all, because there is no symmetry between the two observers. One undergoes dramatic acceleration several times, and the other does not. If a spaceship accelerates to *c*/2 and travels to the stars and back over twenty three Earth years, when it returns its crew will have aged by only twenty years. At 99% of the speed of light, γ = 7.1. If the ship is gone for 71 years, its crew will have aged only 10 years. We call this difference in age between two coincident observers *time dilation*.

Time does not slow down for any observer, no matter how fast they move. As one observer moves with respect to a frame of reference, she moves into the future of that frame of reference at a faster rate than she moves into the future of her own frame of reference. By changing her velocity, she can return to where she started and be younger than a twin she left behind.

We have not sent space ships on journeys approaching the speed of light. But we have sent a cesium clock to fly around in circles for a few days in a jet plane, and then found that less time has passed on the cesium clock than its partner down on Earth, with the difference consistent with the Lorentz contraction. We have many other observations of time dilation, such as the increase in the half-life of unstable particles traveling at close to the speed of light.

Here are some exercises.

- A
*muon*is a heavy cousin of the electron. It is unstable. When it is at rest, its half-life is 2.2 μs. Suppose a muon is traveling at 99.99% the speed of light. In its own frame of reference (the frame in which it is stationary), what is its half-life? In our frame of reference (the one in which it is moving at 99.99% the speed of light), what is its half-life? - More questions coming soon...