Q What is that star photograph that forms the background of the book's cover?
A It is the famous picture of a cluster of galaxies in Centaurus, photographed from the Southern Hemisphere at Cerro Tololo, Chile, on the 4-meter Mayall telescope. The photo is owned by the taxpayers of the United States of America, and is used with the kind permission of the National Optical Astronomy Observatories.
Q If I have a light-pulse clock and a quartz crystal Timex, how will they differ in the time that they keep?
A The light-pulse clock and the quartz crystal watch keep exactly the same time.
The reason for designing a ligh-pulse clock (designing it suffices -- it is not necessary to build one) is that it is possible to tell, from relativity theory, how its time will be altered by the motion of the person observing it. The Timex would change equally under the same circumstances, but there is no theory for predicting that change.
Once the light-pulse clock's concept has been used for calculating what the change is, there is absolutely no further use for the light-pulse clock. Don't let anyone try to sell you one!
In the diagram at the left, both the light-pulse clock and the Timex keep the same time. When read by a moving observer, they both read the same.
Unless the "tick" time is corrected, both clocks read "wrong." It's the ticks that change with the speed of the observer, and clock readouts tell only how many ticks, not how long each tick took. (Say that fast!)
Q If you and I are moving rapidly relative to each other, the theory is that each of us can regard ourselves as being at rest while the other is moving. Now that would mean that your clock is slower than mine, and my clock is slower than yours, and that doesn't seem possible.
A We tell you at every chance we get that the clocks don't slow down at all. Each of us will be carrying perfectly good clocks that are not themselves affected by the motion. It is the process of observing the clocks (and the persons who carry them) that makes time progress more slowly in the observer's frame of reference than in the frame of reference of the person being observed.
Time goes more slowly on Timus Ybatu's watch than in the clocks of the judges click here to review Timus' mile run.
But notice something: it took two clocks to make a reading of the time of Timus' run in the judges' frame. It is not really possible to compare one clock in one frame of reference with one other clock in another frame of reference. It is only in Timus' frame of reference that the same clock is present both at the start and the end. To measure the time difference in the stationary frame, the judges can not do it with one clock; if they did, that clock would have to be moving along with Timus. To measure the time difference in the stationary frame, they need one clock at the place where Timus starts, and another clock at the place where he ends.
If you are asking, "Then, isn't this all just an illusion?" the answer is no. In the case of the muon falling to earth, things actually happen to the muon because of time dilation, that in classical (non-relativistic) physics we would predict should not happen. click here to review the muon story. The effects are real...
...but Read On
for a more detailed answer to this question in the answer to Dan Maas' question next:
Q  Here is the question from Dan Maas:
He wrote us, "The lack of reciprocity in perceived time dilation between intertial frames in Chapter 7 on pages 94-95 troubles me.
"If we were to set up the kind of light-pulse clock that you diagram on pages 84-85, but set it up on the platform (instead of on the train car as you show), and thus reverse everything around, would not the person on the train car (who perceives the platform swiftly passing him/her) observe a reciprocal (longer) distance that light travels for the platform clock, just as you diagram in Figure 7.7 for the reverse situation?
"It seems impossible that either a person on the platform observing a pulse clock on the train car, or a person in the train car observing a pulse clock on the platform, would fail to observe the time dilation for the other person."
A  Dan, good question. Let's give it a try:
You are right that, given the reciprocal situation you describe, you would get the reciprocal time dilation relation. This is a problem only if you hold the mistaken view that the time dilation equation describes the relation between two clocks, one on the platform and one on the train.
It is easy to get this wrong idea from having heard it said that "moving clocks move slower." They do not. Clocks are not at all affected by (uniform) motion. Moreover, the rates of clocks moving at different speeds can not be compared.
This is not what the time dilation equation does. Time dilation describes the relation between two measurements of a time interval between a pair of events, measured by observers fixed in two different frames of reference.
Time, in relativity, is always a time interval between two events.
The time dilation equation is a comparison between two time intervals, both being time intervals between the same pair of events.
When you set up the reciprocal situation, you change what the two events are. This naturally gives a different result.
In the original setup, with the light pulse clock on the moving train, the pair of events is e1, the emission of the light pulse, and e2, the return of the light pulse to the clock. These are the events, regardless of whether you observe them on the train or on the platform.
The two events are at the same location when observed in the train (in frame Ko).
What makes the time dilation relation a special case is that in the frame of the train both events occur at the same location. This means that the distance between the events in the train is zero.
If you jump to chapter 9, you can express the invariant space-time interval for the time dilation derivation of chapter 7 in this way:
With just a single space dimension (x), the value of xo , the distance between the events e1 and e2 in the train frame, Ko , is 0. This makes Ko the "proper frame," and it makes the time interval in that frame the "proper time," to .
Then the invariant space-time interval (squared) is
(tau)²  =   to²  - (0)²   =  to²
(tau)² is the same for all frames of reference from which these two events are observed. This is the meaning of the word "invariant" here.
In all frames, then, the value of [(t)² - (x)²] is the same and is equal to (tau)²
This makes a couple things easy to confirm. First, to is the smallest value of t possible. And second, in any frame in which x² is greater than 0, t will be larger than to. This is the essence of the message of time dilation.
Now, what happens if you set up the reciprocal situation? Again there will be two events, e1 , the emission of the light pulse, and e2 , the return of the light pulse to the clock. These two events will be at the same location in the frame of the light pulse clock, this time on the platform. This will make the platform the proper frame, and the one in which the value of the time interval is the least.
And that is perfectly all right.
No one is saying that there are two clocks, each of which goes slower than the other, which would indeed be impossible.
This is not a comparison between two clocks. In the first place it takes at least three clocks to do the experiment.
In the frame of reference in which the events are not at the same location, it is not possible to make the time interval measurement with one clock. The way that one might attempt to do it with one clock would involve moving that single clock from the location of event e1 to the event e2 in the available time. This is, of course, possible to do. However, in that case the clock which is being thus moved is not a fixed observer in the frame where it is physically located, the frame in which the light pulse clock is not (which is now the train); it is in fact a fixed observer in the other frame (the platform frame), because it is moving to keep up with the moving platform and its light pulse clock to be at event e2 when and where it occurs.
In practice, in the train frame, in which the events are (now) not in the same place, one must have two clocks. These two clocks are synchronized with each other, but not with the clock in the platform frame.
It is not possible to synchronize two clocks in different frames of reference. It is possible to set them to the same "zero" time setting when they are next to each other, but that does not make them synchronized.
Q How did Einstein discover his two postulates of relativity?
A He didn't discover them at all. Everybody knew them already. Scientists found a contradiction between the two, and had decided there must be something wrong with one of them. They couldn't find out what it was that was wrong with them, nor which one was "at fault."
This was the big mystery. Einstein's theory of relativity maintained that both could be true, and that the contradiction was based on the idea that time and space are separate. Einstein said, let us suppose that time and space are related, no longer separate. That solved the mystery, but created an even bigger problem: to figure out what the world would be like if time and space were related in just such a way as to make Maxwell's law of the propagation of light be true.
Q If two events, A and B, are simultaneous in a particular frame of reference, then in other frames of reference the order in which they occur may be "A then B"; but the reverse, "B then A" may also occur. If event A is the cause of event B, then must not event A precede event B?
A The answer to your two questions is "That's right," and, "Yes." You are right that if two events are simultaneous in a particular frame, their order may be either way in other frames.
The spacetime interval between a pair of events that can be simultaneous in any frame is "spacelike," which means that the space separation between them in all frames is too great to be bridged by any signal, no matter how fast it travels, even light.
Thus, two such events can not be causally related; that means neither can be the cause of the other.
Events whose spacetime separation is "timelike" can not be simultaneous in any frame; in such a pair one can be the cause of the other, but it is also true that in such a pair the order in which they occur is the same in all frames of reference. See?
Q What's a transformation?
A We're sending you to the page for chapter 12 CLICK HERE for What's a Transformation?.
Q This space is reserved for YOUR so-far unanswered question.
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