3.4: Measuring Lengths and Coincidences

Simultaneity

Let's continue our discussion of how to measure coordinate time by synchronizing clocks at all the lattice points in a reference frame. Suppose Ann and Bob are moving past each other along the \(x\)-axis, and at the moment that their origins coincide, they start their clocks at the origin. Then each of them synchronizes all the clocks on their lattice with the clock at the origin. Doesn't this mean that all of Ann's clocks are synchronized with all of Bob's clocks? And if so, doesn't this mean that they should measure the same coordinate times between events, in contradiction to everything we have said fo far? Such a conundrum calls for a thought experiment!

Let's suppose Ann decides to synchronize two clocks using a flash from her clock at the origin:

Figure 2.3.6 – Simultaneous Events for Ann

Does Bob agree that the two clocks (located at the two detectors) are synchronized? Let's look at what Bob sees:

Figure 2.3.7 – Ann's Synchronized Events Seen by Bob (a)

The detectors Ann is using are fixed in the lattice points in her frame, so they move along with her, according to Bob. When the light flashes, it takes time for the wave to get to the detectors, and while this time passes, the detectors move, according to Bob:

Figure 2.3.8 – Ann's Synchronized Events Seen by Bob (b)

As you can see, the detector trailing Ann receives the signal before the other detector, according to Bob.

Figure 2.3.9 – Ann's Synchronized Events Seen by Bob (c)

Far from seeing the two events simultaneously, Bob measures a time difference between them. This means that when he looks at all the clocks in Ann's lattice, he sees them all out of sync, with the times getting later the farther the clock is on the positive side of the origin. We therefore find that the concept of simultaneous events is relative (frame-dependent).

Alert

It is important to keep in mind that when we are talking simultaneous events in one frame, we are not talking about about simply seeing two things occur at a different time. For example, if Ann happened to be standing close to detector #1, then the light from the flag that pops up there would reach her sooner than the light coming from the flag at detector #2, and she would witness the two flags popping at different times, but the two events would still be simultaneous in her frame.

Example \(\PageIndex{1}\)

We found in the light clock thought experiment that the relationship between Ann's and Bob's time measurements is given by Equation 1.2.2. If Ann's two clocks are synchronized, then the time between the two events that occur when the flash reaches both detectors is zero. So why don't we find that for those same two events viewed by Bob, the time interval is also zero?

\[\Delta t' = \gamma_v \Delta t = 0\nonumber\]

Solution

The equation quoted assumed that the time measured by Ann was the proper time, since the two events occurred at the same position in her inertial frame. The synchronized events in this case do not occur at the same position, so it is the coordinate time that she measures to be zero. One way to avoid this confusion is to write the time dilation formula of Equation 1.2.2 explicitly in terms of the proper time:

\[\Delta t' = \gamma_v \Delta \tau\nonumber\]

In the case above, the comparison is not between a coordinate time and a proper time, but two different coordinate times.

Length Contraction

Instead of comparing time spans between two frames in relative motion, let’s compare distance spans. To do this, we need to first figure out what it means to measure the length of an object (say a meter stick). As we know, whatever we do in relativity must be in terms of spacetime events. We can’t simply say that the length of an object is the distance between events that occur at the object’s endpoints, because the object might move after one event occurs and before the second one occurs. So clearly to define the length of an object, we need to stipulate that the two events that occur at the endpoints of the object being measured occur at the same time. But since observers in two frames in relative motion will not agree to what events are simultaneous, it stands to reason that they might not agree to length measurements.

We consider The following scenario: Ann lays a rod down in her stationary frame along her \(x\)-axis. Bob passes close by moving in the +x direction at a speed v relative to Ann. At either end of the rod is a device that will emit a bright spark when illuminated with a certain type of laser. As Bob passes the rod, he shines a laser at it, so that as he passes each end of the rod creates a spark (constituting a spacetime event) when it coincides with a device stationary in Bob's frame that we will call a "sparker."

Figure 2.3.10 – Ann Measures the Length of the Rod

Ann measures the length of her rod to be the speed of Bob's sparker multiplied by the coordinate time she measures between the two events.

The way Bob measures the length of the rod is similar:

Figure 2.3.11 – Bob Measures the Length of the Rod

We can now use the time dilation formula that relates these two times to determine a relationship between the two measured lengths, but we have to be very careful here. Namely, we must ask ourselves, whose measures the dilated time here? Put another way, which of these two observers measures the proper time between the two events? The answer is clearly Bob, since both events occur at the end of his sparker, which is at rest in his frame, meaning that both events occur at the same place – the exact criterion for proper time. Therefore we find that it is Ann's time between events that is longer than the time measured by Bob, giving:

\[\Delta t = \gamma_v \Delta t' \;\;\;\Rightarrow\;\;\; L' = v\Delta t' = v\dfrac{\Delta t}{\gamma_v} = \dfrac{L}{\gamma_v}\]

So Bob measures the rod to be shorter than Ann measures it to be (recall \(\gamma_v>1\)). This phenomenon is known as length contraction.

A few comments about this result:

• The longest possible measurement of length occurs in the rest frame of the object whose length is being measured. This length is often referred to as the proper length.
• We are accustomed to attributing different visual observations of the same object to optical illusions. This is not one of those cases. The same rod is shorter for Bob than it is for Ann. Length is not an intrinsic property – it is observer-dependent.
• If Bob zooms by Ann with an identical rod, then Bob will measure Ann’s rod to be shorter than his own, and Ann will measure Bob’s rod to be shorter than hers. We will explore this seeming paradox soon, but the short answer is that length is not a quality that is inherent to an object, so the fact that the rods are identical does not mean that their lengths are. Most people are not bothered by the fact that two identical rods may have different colors (due to red/blue shift), because it isn’t too difficult to accept that color is not a property inherent to objects, but length is significantly tougher to swallow. If it helps, it’s okay to say that identical objects have equal proper lengths.
• If we consider examining the same rod in both frames when it is aligned along the \(y\)-axis (while relative motion is still along the \(x\)-axis), we will fairly easily find that both observers agree on the length. Put another way, lengths contract only along the dimension parallel to the direction of relative motion.
• In terms of the lattice points for the two frames, we would say that Bob sees Ann's lattice points ate closer together along the \(x\)-direction that his own, and Ann would conclude the same about Bob's lattice points. This is because each can measure the rod in terms of these points, so if the length of the rod is relative, so is the entire space of the moving item.

Angles

Consider our usual setup with Ann and Bob in relative motion along their common \(x\)-axis. We know that there is a contraction of length along the \(x\)-axis when an object is moving relative to the frame it is viewed in. Furthermore, we know that lengths along the \(y\) and \(z\) axes do not contract. Suppose Ann has a right-triangular wooden block at rest in her frame as in the figure below, and measures the angle it makes with the \(x\)-axis to be \(\theta\). What angle does Bob measure?

Figure 2.3.12 – Ann and Bob Measure an Angle

With only the side of the triangle along the \(x\)-axis contracting, the angle must change. If we call the length of the base of the triangle in Ann's frame \(x\) and the height \(y\), then we get:

\[\theta ' = \tan^{-1}\left(\dfrac{y'}{x'}\right) = \tan^{-1}\left(\dfrac{y}{\frac{x}{\gamma_v}}\right) = \tan^{-1}\left(\gamma_v\dfrac{y}{x}\right) = \tan^{-1}\left(\gamma_v \tan\theta\right)\]

Example \(\PageIndex{1}\)

Compute Bob's unanswered question in the figure above – what does he measure for the hypotenuse of the triangle, in terms of the hypotenuse and angle measured by Ann?

Solution

Start with the Pythagorean theorem:

\[H' = \sqrt{x'^2+y'^2} = \sqrt{\dfrac{x^2}{\gamma_v^2}+y^2} = \sqrt{\left(1-\frac{v^2}{c^2}\right)x^2+y^2} = \sqrt{x^2+y^2-\frac{v^2}{c^2}x^2}\nonumber\]

Now write Ann's values of \(x\) and \(y\) in terms of \(H\) and \(\theta\):

\[\left. \begin{array}{l} H^2=x^2+y^2 \\ x = H\cos\theta \end{array} \right\}\;\;\; H' = \sqrt{H^2-\frac{v^2}{c^2}H^2\cos^2\theta} = H\sqrt{1-\frac{v^2}{c^2}\cos^2\theta}\nonumber\]