Consider a physical process such as the oscillation of a pendulum, or the collision between two billiard balls, or the flight of a dart as it approaches the board. For any such process, we tend to think that its duration does not depend on the point of view from which it is observed. We think that, for example, the person throwing the dart and a person passing by in a train nearby both agree about the time it takes for the dart to reach the board once it is thrown. Similarly, we think that if a glass takes one second to reach the floor according to a very precise measurement made by the person next to the glass, any other observer, regardless of their state of motion, will also say that the glass took one second to reach the floor if they perform a precise measurement. In general, it is natural to think that the duration of a physical process does not depend on the point of view from which it is measured, as long as the measurement is a good one. However, as we will see in this post, this is simply not true.
In this note, I will show that in Special Relativity, the duration of a physical process depends on the state of motion between the observer and the system. I will show, in particular, that the faster the speed between the system and the observer is, the longer the physical process takes according to that observer. For example, my measurement of the time the dart takes to reach the board will produce a smaller outcome (less seconds) compared to that one made by a person passing by inside a high speed train. Crucially, this is the case even if both measurements are assumed to involve no experimental error whatsoever!
Although there are countless derivations of this result in the literature, in this post I will try to be as explicit as possible about various central assumptions that are often left implicit. Of particular importance in this derivation is the comparison of the case of light with that of a very simple mechanical system (a tennis ball). The comparison has helped me better understand some key concepts in the derivation, and better appreciate why the case of light is so peculiar. I hope it does the same for you!
Set-up
Consider a simple spaceship moving uniformly in outer space, far enough from any other bodies (this means that no external forces are acting on it). In the middle of the cabin, a little device can fire upwards, toward the roof, either a tennis ball or a laser beam. The tennis ball is always fired with speed \(v_b\) relative to the device, and the laser beam with speed \( c\) (the speed of light in vacuum). Finally, on the top of the cabin, there is a mirror situated a distance \(h\) from the device. The mirror not only reflects light perfectly, but it is also resistant enough to allow the tennis ball to bounce back without breaking in the process (the ball bounces elastically, so no energy is lost during the collision). The following animation illustrates the functioning of this device. Notice that you must click on any animation on this page to start it, and can click again to pause it.
Our goal is to study the motion of the ball and the beam as seen from two different perspectives: (1) the perspective of an observer inside the cabin, and (2) that one of an external observer relative to which the spaceship is moving uniformly to the right (as if it was moving to the right of the screen in front of you). For simplicity, we will say that the observer inside the train is in an “internal frame” and the external observer in an “external frame”. Our specific goal is to answer the following two questions from the perspective of each frame:
How much time does the ball take going from the device to the mirror, and then back to the device? How much time does the laser beam take going from the device to the mirror, and then back to the device?
We will start with the case of the tennis ball, and then we will tackle the case of the beam.
A short note about observers and frames. In physics, observers need not involve real people or agents. Rather, an observer can correspond to a measurement device like a radar, to a simple system like a little ball with no sensors at all, or even to hypothetical systems that are not part of the situation we are studying but which could allow us to define certain properties such as distances and velocities. For example, we can imagine that there is a little stick next to the device in the center of the cabin at some time, and then say things like this: “If there was a stick next to the device at the initial time, then the ball would have moved five meters relative to that stick after five seconds”. Once we choose an observer (real or not, a machine or an agent, etc.) with respect to which we can describe the motion of the physical system, we use a coordinate system that allows us to keep track of the specific motion of the system relative to this observer. One can think of choosing a frame as being the same thing as choosing a referent object (e.g., a stick, a star, a tree, a real person, a fictional ball) together with choosing a coordinate system (say a Cartesian coordinate system, a polar one, etc.). For simplicity, this referent object is typically chosen so that it sits at the origin of the coordinate system.
Tennis Ball
Observer inside the cabin
For an observer inside the spaceship, the cabin is not moving relative to them (the observer is moving together with the cabin). For example, a person inside an airplane will not see the airplane moving relative to them but rather will see the ground moving. A person on the ground, however, will see the plane moving relative to them. This illustrates the fact that motion is relative in this sense: whether or not an object is moving (or how fast is moving) depends on the point of reference. The airplane is not moving relative to the passenger, but it is moving relative to a person on the ground.
Going back to the case at hand, for an observer inside the cabin, the cabin is at rest. This means that when the device fires the tennis ball, this observer will see the ball moving in a straight line upwards, and then, after bouncing, in a straight line downwards (in other words, the ball has no horizontal motion whatsoever according to this observer). Importantly, the ball always keeps the same speed, as no external forces are acting on it (no gravity or something else) and the collision with the mirror is elastic. The following animation illustrates the ball’s motion as seen by an observer (frame) inside the cabin:
So how much time does the ball take going from the device to the mirror, and then back to the device? This is easy to answer given that we know the speed of the ball relative to the device, and the total distance. In particular, we know that the speed of an object moving uniformly is defined as the distance \( d\) it travels over the time interval \( \Delta t\) it takes to travel that distance: \( v=d/ \Delta t \). Thus, the time interval for an object moving with uniform velocity is given by
\begin{equation} \Delta t =\frac{d}{v} \tag{1} \end{equation}
Now, in the present case, we know that the total distance traveled by the ball is \( 2h\) (it travels \(h\) on the way to the mirror and \(h\) again going back to the device). We also know that the speed relative to the observer inside the cabin is \(v_b\). Therefore, the following equation gives us the time interval in terms of known variables:
\begin{equation} \Delta t =\frac{2h}{v_b} \tag{2} \end{equation}
For example, if the ball is fired with a speed of ten meters per second, and the height is 5 meters, then the time interval is one second.
Observer outside the cabin
Now imagine that we study the same situation, but now from the perspective of an external observer (in an external frame) that is not moving with the cabin. In particular, we will consider the point of view of an observer who sees the cabin moving with constant speed \(v_s \) toward the right. If it helps, imagine an astronaut floating in space looking at the cabin as it passes in front of them from their left to their right.
According to this external observer, how much time does the ball take going from the device to the mirror, and then back to the device? Once again, we can use the fact that for an object moving with uniform velocity such as the ball, time equals distance over speed, as equation (1) says. But before we continue, it is worth stressing that this is only true if the observer (frame) is also moving uniformly. As it turns out, the ball is moving with uniform velocity relative to any frame or observer that is also moving uniformly, and this is true even if the ball’s velocity is different relative to the different observers (as the relativity of motion requires). This might seem obvious, but the fact that all inertial frames agree about the fact that an isolated object, such as the ball, has uniform velocity is a deep fact about Newton’s laws tightly connected to the so-called “Galilean Relativity Principle” (this principle says, roughly, that the laws of mechanical systems are the same in all inertial frames). For instance, if the external observer was accelerating (this means that they would be in a non-inertial frame), then they would not see the ball as moving with uniform velocity, and so they would not be able to use equation (1) (that equation only “works” for objects in uniform motion). It is precisely because we are assuming that this external observer is moving uniformly (i.e., it is in an inertial frame) the reason why that observer can still use equation (1) to compute the ball’s time.
Okay, so it is clear that equation (1) remains valid for the external frame. Can’t we just use equation (2), then? No! The distance \(d\) in equation (1) will not be \(2h\) according to this external frame. The reason is that the cabin is moving to the right relative to this observer, and so all the contents, including the ball, are also moving with it. Hence, as seen from the perspective of this external frame, the ball’s motion is a bit more complex than a simple vertical line. As the ball moves upwards with uniform speed, it also moves to the right with uniform speed (the same speed the cabin has), so the ball really moves diagonally. Similarly, after the ball bounces on the mirror, it starts to move down with uniform speed, but it also keeps moving to the right with the speed of the cabin. In other words, the total motion of the ball can be thought of as consisting of a horizontal motion with the speed of the cabin, \(v_s\), and a vertical motion with speed \(v_b\) (the external observer is not moving upwards or downwards relative to the ship, and so that observer agrees with the internal one that the vertical speed is \(v_b\) ). The idea that the total motion of an object can be composed of two or more motions (here a vertical motion and a horizontal one) is often known as the “Composition of Motion”, another important concept studied by Galileo (we will discuss it on another occasion). Hence, as seen from the perspective of the external observer, the ball travels a longer distance than just \(2h\) because it moves up and down but also from left to right. The following animation shows the total motion of the ball as seen by this external observer:
From the triangle at the end of the animation, it becomes clear that the total distance traveled by the ball is \(d= 2l\). It is also clear that we can compute \( l\) in terms of \(w \) and \(h\) if we use the Pythagorean Theorem, where \( w\) is the total horizontal distance traveled by the cabin (and the ball) during the time under consideration. In particular, from the triangle at the end of the animation, it is clear that
\begin{equation} l^2=h^2+(w/2)^2 \tag{3} \end{equation}
Hence, the total distance traveled by the ball as seen by the external observer is
\begin{equation} d=2l=2 \sqrt{h^2+\frac{w^2}{4} } \tag{4}, \end{equation}
where the factor of 2 indicates that the ball travels \(l \) two times (a distance \(l\) going to the mirror and a distance \(l\) going back). Can we then divide this distance by the speed \( v_b\) to get an expression for the total time interval? No! When we say that time equals distance over speed (as in equation (1)), we are referring to the total speed of the system. In this case, the total speed of the ball as seen by the internal observer is \(v_b\) because the ball only moves up and down according to them. However, as seen in the external frame, the ball is moving both upwards with speed \(v_b\) and towards the right with speed \(v_s\) (the same horizontal speed as the speed of the cabin), so the total speed will be a function of these two speeds. In particular, just as we can use the Pythagorean Theorem to calculate the total distance, we can also use it to calculate the total speed in terms of the horizontal and vertical speeds, as illustrated here (this is not an animation!):
The diagram illustrates that the total velocity of the ball can be seen as the vector sum of the horizontal and vertical velocity of the ball (this is a consequence of the composition of motion mentioned earlier). What matters to us is that the total speed (the magnitude of the total velocity) can be calculated using the Pythagorean Theorem. In particular, as seen from the perspective of the external frame, the ball’s total speed is given by
\begin{equation} v’_b=\sqrt{v_b^2+{v_s}^2} \tag{5}, \end{equation}
where we have used \(v’_b\) for the total speed of the ball as seen from the external perspective (not to confuse with \(v_b\), the ball’s speed as seen from the inside).
Now that we have an expression for both the total distance and the total speed as seen from the perspective of the external frame, we can finally use equation (1) to calculate the total time interval as seen in the external frame:
\begin{equation} \Delta t’=\frac{2l}{v_b’}=\frac{2\sqrt{h^2+\frac{w^2}{4} }}{\sqrt{v_b^2+{v_s}^2} } \tag{6} \end{equation}
Now, notice that equations (2) and (6) look very different from one another. Does this mean that the time interval for the ball is different when measured from the perspective of the observer inside the cabin and when measured from the perspective of an external observer not moving with the cabin? Hopefully not, as in Newtonian mechanics (and in ordinary life) the duration of physical processes is not supposed to depend on the frame! So what is going on? How come we got two different answers for the ball’s time interval? Well, perhaps these equations are equivalent even if they do not seem so. It turns out that we can simplify equation (6) much further, and if we do so, we will find that \( \Delta t’=2h/v_b\).
To simplify matters, we should use some information that we have not used yet, namely, the fact that during the time it takes the ball to move from the device to the mirror and back to the device (this is the time \( \Delta t’\) that we are trying to find), the ship and the ball have moved a distance of \( w= v_s \Delta t’\) to the right, as seen in the triangle from the last animation. For example, if \( \Delta t’ \) is one second, and the cabin’s speed is one meter per second, then the cabin (and the ball) would have traveled one meter to the right. So let’s use \( w= v_s \Delta t’\) in equation (6) to get:
\begin{equation} \Delta t’=\frac{2\sqrt{h^2+\frac{v_s^2 \Delta t’^2}{4} }}{\sqrt{v_b^2+{v_s}^2} } \tag{7} \end{equation}
This seems even more complex than before, but now we can start the simplification process because we have fewer unknown variables. Let’s square both sides:
\begin{equation} {\Delta t’}^2=\frac{4({h^2+\frac{v_s^2 \Delta t’^2}{4} })}{{v_b^2+{v_s}^2} } \tag{8} \end{equation}
Now multiply by \( ({v_b}^2+{v_s}^2) \) to get
\begin{equation} (v_b^2+{v_s}^2 ){\Delta t’}^2=4(h^2+\frac{v_s^2 \Delta t’^2}{4}) \tag{9} \end{equation}
Now distribute the various terms to get
\begin{equation} v_b^2{\Delta t’}^2+{v_s}^2 {\Delta t’}^2=4h^2+{v_s^2 \Delta t’^2} \tag{10} \end{equation}
After canceling the term \( {v_s}^2 {\Delta t’}^2\) (repeated in both sides), we get
\begin{equation} v_b^2{\Delta t’}^2=4h^2 \tag{11} \end{equation}
Thus,
\begin{equation} {\Delta t’}^2=\frac{4h^2}{v_b^2} \tag{12} \end{equation}
Finally, if we take the square root, we get
\begin{equation} {\Delta t’}=\frac{2h}{v_b}=\Delta t \tag{13} \end{equation}
So we have recovered the answer obtained in equation (2)! This means that the two frames agree about the time the ball takes to go from the device to the mirror and back to the device even though they disagree both about the total distance traveled by the ball and its total speed. Notice, in particular, that equation (13) does not refer to the speed \(v_s\) between the cabin and the external observer. This is expected and perhaps even obvious; the duration of a physical process should not depend on the relative speed between the system in the process (here the ball) and the observer (the frame). Obvious or not, we will see that things are not so simple with light.
Light Beam
Now we want to find how much time the beam takes to go from the device to the mirror, and then back to the device as seen by both an internal observer and an external one. As before, we will start with the case of the observer inside the cabin.
Observer inside the cabin
As it happened with the case of the ball, the time as measured by the internal observer is rather easy to calculate. As seen inside, the beam is moving completely vertically with a uniform velocity of magnitude \(c\), and so the total time it takes between the moment it is fired and the moment it returns to the device is given by the total distance (here \(2h\)) over the speed (here \(c\)). So we can use equation (2), but now adapted for the speed of light:
\begin{equation} \Delta t =\frac{2h}{c} \tag{14} \end{equation}
Nothing else changes!
Observer outside the cabin
As seen by the external observer, the cabin and all the objects inside of it are moving to the right with speed \(v_s\). So for the external observer, we might be tempted to reuse equation (7) except for the fact that we should use \( c\) instead of \(v_b\). This is tempting, but it would be wrong to do so. To see why, let’s start by discussing what things are analogous to the case of the ball and what things are not.
As seen from the perspective of an external observer, the light will trace a triangle just like the ball did because it will be moving both upwards at the same time it is moving to the right with the cabin. Notice that the light beam has to have some motion to the right, otherwise it would not hit the mirror in the middle (the mirror is moving to the right). Also, it is obvious that the light has to have some vertical motion, otherwise, it would not reach the mirror. But why do we know that in the external frame, light reaches the mirror and hits it on the middle? It turns out that if an event occurs in one frame, it must occur in other frames as well.
(the underlying assumption is that if an event occurs in one frame, it must also occur in other frames, as we will discuss later). So just as was the case with the ball, the motion of light can be thought of as being composed of some horizontal motion (determined by the cabin’s motion) and some vertical motion, as this animation shows:
Hence, the total distance traveled by the light beam can be calculated using equation (4):
\begin{equation} d=2l=2 \sqrt{h^2+\frac{w^2}{4} } \tag{15} \end{equation}
So far, this is just as with the case of the ball. What about the speed of the beam as seen from the external perspective? Recall that for the ball, the speed is also understood as being composed of two things, the horizontal speed \(v_s\) that moves the ball to the right with the cabin, and the vertical speed \(v_b\) that moves the ball towards the mirror. So the total speed of the ball was given using the Pythagorean Theorem, as we did with equation (5): \( v’_b=\sqrt{v_b^2+{v_s}^2}\). The idea, then, is that we use this equation for the total speed of light as seen from the external frame: \( c’=\sqrt{c^2+{v_s}^2}\).
But light is rather special in this respect: unlike any other (massive) objects in the universe, the speed of light is the same in all inertial frames. That is, one and the very same light beam will be seen to travel with speed \(c\) in every inertial frame, even in frames moving at very fast velocities relative to one another. This, of course, is very different from the case of the tennis ball. Choose a frame that is moving very fast to the left of the cabin, and the ball will be seen to move very fast to the right of that frame, and the faster the frame moves to the left, the faster the ball is seen to move to the right. But with light, this is not the case. The very same light beam will be seen to move with speed \( c \) in all frames, even in frames moving close to the speed of light relative to one another. In short, then, the (total) speed of light cannot be computed using \( c’=\sqrt{c^2+{v_s}^2}\) because that speed does not depend on the velocity of the frame and the system (here the cabin). Or, to put it differently, given that all frames agree about the speed of a light beam, it follows that \( c’ = c\) in all frames (the speed measured in one frame is the same as the one measured in any other one), and so equation \( c’=\sqrt{c^2+{v_s}^2}\) yields the wrong result. Yes, this is very puzzling because it means that the cabin’s speed to the right adds nothing to the total speed of light, and it is even more puzzling given that the cabin’s motion to the right does make light acquire some horizontal motion relative to the external observer (recall that for the internal observer, light only moves up and down). In simple words, the cabin moving to the right of the observer does make light move partially to the right, but the cabin’s speed to the right does not affect in any way the total speed of light relative to this observer!
So, to sum up, the total distance traveled by light can be computed using the Pythagorean Theorem (equation (15)), as it was with the case of the ball, but the total speed of light (\( c\)) is the same for the internal observer and the external one, unlike it was with the case of the ball. Hence, to find the time interval for the light beam as seen by the external frame, we can reuse equation (6) except for the fact that we need to use the speed \(c\) in the denominator:
\begin{equation} \Delta t’=\frac{2\sqrt{h^2+\frac{w^2}{4}}}{c } \tag{16} \end{equation}
This looks very different when compared with equation (14), but as we showed in the case of the ball, appearances can be misleading. Maybe, after simplification, this equation yields the same result as the one given by equation (14). As was the case with the tennis ball, we can use that the horizontal distance traveled by the cabin (and light) is \( w= v_s \Delta t’\) because the cabin is moving with constant speed \( v_t\) relative to the external observer (the fact that we are talking about light in nothing changes the fact that the cabin is moving to the right with constant speed \(v_s\) relative to the external observer). So if we use this result in equation (16), we get:
\begin{equation} \Delta t’=\frac{2\sqrt{h^2+\frac{v_s^2 \Delta t’^2}{4} }}{c} \tag{17} \end{equation}
Let’s now square both sides:
\begin{equation} {\Delta t’}^2=\frac{4(h^2+\frac{v_s^2 \Delta t’^2}{4}) }{c^2} \tag{18} \end{equation}
Now distribute the terms on the right side:
\begin{equation} {\Delta t’}^2=4\frac{h^2}{c^2}+\frac{v_s^2 \Delta t’^2}{c^2} \tag{19} \end{equation}
Move the terms with \( {\Delta t’}\) to the left:
\begin{equation} {\Delta t’}^2-\frac{v_s^2 \Delta t’^2}{c^2}=4\frac{h^2}{c^2} \tag{20} \end{equation}
Factorize \( {\Delta t’}\):
\begin{equation} {\Delta t’}^2(1-\frac{v_s^2 }{c^2})=4\frac{h^2}{c^2} \tag{21} \end{equation}
Now divide by \((1-\frac{v_s^2 }{c^2})\):
\begin{equation} {\Delta t’}^2=\frac{1}{(1-\frac{v_s^2 }{c^2})}4\frac{h^2}{c^2} \tag{22} \end{equation}
Finally, take square root on both sides:
\begin{equation} {\Delta t’}=\frac{1}{\sqrt{(1-\frac{v_s^2 }{c^2})}}2\frac{h}{c} \tag{23} \end{equation}
Now compare this with equation (14). The two equations are identical except for a factor that depends on \(v_s\) and \(c\). Specifically, we can use equation (14) in equation (23) to obtain:
\begin{equation} {\Delta t’}=\frac{1}{\sqrt{(1-\frac{v_s^2 }{c^2})}}\Delta t \tag{24} \end{equation}
So the two observers disagree about the time light takes to go from the device to the mirror and back to the device! And in particular, the external observer will see light take longer to complete the trip, as the factor in front of \( \Delta t\) in equation (24) is always greater than 1 (one minus a number less than one is always less than one, and one over the square root of a number less than one is always more than one). For example, if the cabin is moving at around 86% the speed of light to the right of the external observer (i.e., if \( v_s=0.86 c\)), then it follows from equation (24) that \(\Delta t’ \approx 2 \Delta t \). In other words, in that case, light will take twice as much time to complete the whole trip according to the external observer when compared to the time it takes according to the internal observer!
This is an important result, one that shows that two different observers in relative motion to one another will disagree about the duration of a physical process. In particular, this result shows that the duration of a physical process depends on the speed between the system and the frame (here, the speed \(v_s\) between the cabin and the external observer), something that is rather puzzling (recall that we did not get this result for the ball). This dependence of time intervals on the relative speed between the system and the observer is known as time dilatation.
The factor in front of \( \Delta t \) in equation (24) appears in so many relativistic calculations that it has a name, the “Gamma factor” (also the “Lorentz factor” because Lorentz derived it), and it is defined in this manner:
\begin{equation} {\gamma}=\frac{1}{\sqrt{(1-\frac{v_s^2 }{c^2})}} \tag{25} \end{equation}
Thus, we can rewrite equation (24) in this way:
\begin{equation} {\Delta t’}=\gamma\Delta t \tag{26} \end{equation}
How general is this result?
Let’s end this post by addressing a possible source of confusion. One might think that what we proved in this discussion is that the duration of physical processes associated with light depends on the motion of the system and the frame, but not so for physical processes not associated with light, such as the tennis ball. After all, in the first part of this discussion, we showed that different observers in different frames do agree about the time the ball takes to complete the trip. Perhaps, then, all we have established upon this point is this: the duration of a physical process not involving light is the same in all inertial frames, but the duration of a physical process involving light is not the same in all inertial frames. This might be a natural thing to think, but I will end this discussion by showing that it is wrong. In particular, I will give an argument that aims to show that the relationship between the time intervals as measured by two different observers in relative motion \( v_s\) is indeed given by equation (14) regardless of what the physical process is (this includes things such as the time the tennis ball takes to complete a trip).
First, let’s start with what I think is a bad argument, one that is unfortunately widespread in many presentations of Special Relativity. The argument starts by saying that we can use a set-up like the one here discussed, with a light beam being reflected from a mirror, as a clock. If we do that, then this clock will tick at different rates depending on the motion between the cabin and the frame, just as equation (14) says. Since we are free to use this set-up as a clock to measure anything we want, and since the clock itself ticks at different rates depending on its motion relative to the frame, it will follow that the duration of any process measured by this clock will also depend on the velocity relative to the frame. In simpler words, if a clock slows down, it slows down also when used for measuring any other physical process including those not involving light. Thus, physical processes also slow down, or so the argument goes.
But this argument is problematic for at least two reasons. First, it seems arbitrary to use this set-up with light as a clock: why not use the set-up with the tennis ball, instead, as a clock? After all, we showed earlier that all observers agree about how long the ball takes to complete a trip (we will see soon that this is not quite correct). Second, and relatedly, if one already thinks that the duration of physical processes not involving light does not depend on the motion between the system and the frame, and one also thinks that the duration of physical processes involving light does depend on such motion, then it is very natural to think that it is a mistake to use a physical process involving light to measure the duration of a physical process not involving light (one slows down, but the other one does not!). What we want is an argument telling us why we are justified in using a light clock (such as the one here) to measure the time interval of all physical processes, and so it is not helpful to simply stipulate that one can do just that. In what follows, I will introduce one argument, but I warn the reader that this is just a first sketch of an argument (a more complete argument will require concepts beyond the contents of this note).
Imagine the following modification of the set-up: the person inside the cabin changes the device so that when it fires a light beam, it also fires a tennis ball to the right with speed \(v_r\) (the light beam and the ball are fired at the same time). Furthermore, the person inside calibrates the speed of the ball and the distance between the device and the right wall so that the total time the ball takes to go to the right wall, bounce, and return to the device is the same as the time the light takes to go from the device to the mirror and back to the device (yes, light is fast, but one can suppose that \(h\) is very big and the ball is very close to the right wall and \(v_r\) is also big). Furthermore, the person calibrates things so that the device explodes if and only if it receives the light beam and the tennis ball at exactly the same time (if it receives the light beam first, or the tennis ball first, it will not explode). The person inside wants the device to explode, so they make sure that the ball takes exactly a time of \( \Delta t = 2\frac{h}{c}\) to complete the whole trip, as this is the time light takes to complete a trip as seen from inside. The person starts the experiment and, as predicted, the device explodes some instants of time afterward when receiving the light beam and the ball simultaneously (we will discuss simultaneity on another occasion, but for now it suffices to note that all frames agree about what events are simultaneous if they are sufficiently localized).
How do things look from the external perspective? Well, if we assume that the duration of physical processes such as those involving tennis balls do not depend on the velocity relative to the frame, it must be that the external observer agrees with the internal one about the time the ball takes to complete the trip. So the person outside the cabin must agree that the ball takes a time \( \Delta t = 2\frac{h}{c}\) to complete the trip, as this is the time it takes as seen by the person inside. However, we are also assuming that for this external observer, processes involving light do slow down according to \({\Delta t’}=\gamma\Delta t \). And so, in particular, this person will see the ball reaching the device after time \( \Delta t = 2\frac{h}{c}\) and will see the beam reaching the device after a time \({\Delta t’}=\gamma\Delta t \). Since \({\Delta t’}>\Delta t \), this person will see the beam reaching the device sometime after the ball does. But if the two objects do not reach the device at the same time, the device should not explode as it is programmed to explode only if it receives the beam and the ball at the same time. Hence, the external observer will not see an explosion. But of course, either there is an explosion or there is not! Changes in frames are not the kinds of things that can change what things actually happen in the world, not at least when we are talking about localized events (we will save this notion of localization for another post!). If we bite the bullet and say that the external observer does see an explosion, that also would not help much because it would mean that they disagree with the observer in the cabin about the program running the device (either the program has a code that triggers an explosion when some conditions are met, or it does not).
It seems, then, that if we want to preserve agreement about what events do happen according to different frames, we must accept that time dilatation affects all physical processes, not just those involving light. In this case, we must say that according to the external observer, the ball took a time \({\Delta t’}=\gamma\Delta t \) to complete the trip, just as the light beam did. If we treat time intervals for light differently from time intervals for other physical systems, we fall prey to inconsistencies in our descriptions of what happens in the world in a way that threatens even our descriptions of the most basic physical processes and measurements. The price to pay to save ourselves from such inconsistencies is to accept that the duration of all physical processes (including tennis balls) does depend on the motion between the physical systems and the observer as prescribed by
\begin{equation} {\Delta t’}=\gamma\Delta t \end{equation}
Remember that we proved earlier that the time the ball takes to complete a trip (equation 13) was the same in the two frames? Well, that proof is correct in a Newtonian world, but it is not quite correct in a relativistic one. The observer outside will say that if the ball took time \( \frac{2h}{v_b}\) to complete the trip as seen by someone inside, then the time it takes as seen by the external observer is
\begin{equation} {\Delta t’}=\gamma \frac{2h}{v_b} \end{equation}
The difference, however, is only noticeable when the speed between the two frames is considerable, way more than the speeds we are used to deal with on Earth!
