|HPS 0410||Einstein for Everyone|
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John D. Norton
Department of History and Philosophy of Science
University of Pittsburgh
What use is spacetime ? It turns out to make visualizing and understanding the relativity of simultaneity a great deal easier. The judgments of simultaneity of different inertial observers correspond to slicing the spacetime up into different stacks of spaces with each space formed from a set of simultaneous events.
To see how this works, here are three observers in relative motion in a spacetime.
|First we have an observer whose worldline runs vertically up the page.|
|The next observer moves to the right with respect to the first.|
|The third observer moves to the left with respect to
Notice how differently they slice up the spacetime into spaces of simultaneous events. That difference simply is the relativity of simultaneity. It is expressed in the tilting of the hypersurfaces of simultaneity as we move the judgments of simultaneity of events from inertial observer to inertial observer.
In looking at the three slicings as they are drawn above, it is easy to fall into the trap of imagining that the first slicing is somehow the "right" one and the second and the third are distortions due to the observers' motion. That would be a mistake. The principle of relativity assures that all three observers are equally good. The judgments of simultaneity of any one is just as good a those of the other two and each of the figures is an equally good way of dividing the spacetime into sets of simultaneous events.
The fact that one observer's worldline is drawn as a vertical line and the others are oblique is just an accident of the way we chose to draw the diagram. Correspondingly, the fact that one observer's hypersurfaces are perfectly horizontal and the others are tilted is again an accident of the way we drew the figure. We could redraw the figures so that the third observer's worldline, say, is vertical. Then the third observer's hypersurfaces of simultaneity would be drawn as horizontal; the worldlines of the other two observers would be diagonals; and their hypersurfaces of simultaneity would be tilted.
|Two points to watch when
you are drawing this tilting of hypersurfaces.
First, setting an observer into motion to the right will tilt the observer's world line to the right; and the hypersurface of simultaneity will also tilt up on the right side to meet it.
Second, if one follows the usual convention of drawing light lines at 45o, then the angle of the observer's worldline to the vertical will be the same as the angle of the hypersurface of simultaneity to the horizontal.
The tilting of the hypersurfaces gives us a simple picture of how inertial observers in relative motion assign times to events.
|An inertial observer carries a clock that marks the
time of events along the observer's worldline as "1," "2," "3," ...
That settles the time of events only on the worldline for the observer.
What time should be assigned to events not
on the observer's worldline? The observer's hypersurfaces
of simultaneity answer.
Consider the hypersurface that passes through the event of the clock showing "1." All these events are simultaneous in the judgement of the observer. Therefore all these events are assigned time "1.
The same applies for the remaining hypersurfaces that pass through the events of the clock ticking "2" and "3." All the events on those hypersurfaces are assigned times "2" and "3," respectively.
|The same analysis
obtains for a new inertial observer who moves relative to the original
observer. The new observer's clock assigns times to events on the
observer's world line. The observer's hypersurfaces of simultaneity are
then used to propagate the times throughout the spacetimes.
Clearly the original and new observer will differ on the times each assigns to the same event in almost every case. Is there a sense in which one is assigning times correctly and the other not? There cannot be. The prinicple of relativity requires each observer's frame to be equivalent. If the procedure is good in one inertial frame, then it is equally good in all. This reminds us once again that there is no frame independent notion of simultaneity in a Minkowski spacetime.
Just why is it that hypersurfaces of simultaneity tilt when we change the state of motion of the observer or reference frame? A simple construction shows how it comes about.
|Imagine that some inertial observer wants to
determine which events in spacetime are simultaneous with some event O
on the observer's world line. The simple way to do it is with light
signals. Following Einstein's original idea of
1905, the observer sends out light signals, reflects them off
positions in space and then notes when they return.
In the figure opposite, there is a light signal leaving the observer's worldline, reflecting at event A and then arriving back at the observer's worldline. Since the event O is exactly midway in time between the departure and arrival events, the observer judges event A to be simultaneous with O.
Also it is obvious that light signals reflected at A' and A must arrive back at the observer at the same time, since they departed the observer at the same time. It is just symmetry.
This same reasoning applies to all the remaining events shown in the figure: B', C', B and C. In each case, there are arrival and departure events at the observer's worldline for light signals that are reflected at B', C', B and C. The event O is midway between the arrival and departure events in each case. Therefore, the observer judges each of B', C', B and C to be simultaneous with O. The totality of these events will form a flat plane perpendicular to the observer's worldline.
Now let us consider a second case in which a new inertial observer moves relative to the original inertial observer. The new observer's worldline will be drawn as a tilted line. Which events will that observer judge as simultaneous with an event O on that observer's worldline? Although the worldline is now tilted, the same procedure just described can also be used to pick out the events simultaneous with O. Indeed the principle of relativity requires this, for otherwise we would have some intrinsic difference between the first inertial frame and the second; only in the first could this procedure be used.
|The construction proceeds as before and it is drawn
for you at left. As long as we adhere to the light
postulate and draw our lightlike curves at 45 degrees to the
vertical, we will end up plotting out events A', B', C', A, B and C
that lie on a tilted hypersurface.
This happens because the departure and arrival events for the light have been displaced to the left and right respectively. We now need to locate the bends in the lightlike curves--the reflection events--in such a way that, as before, the light signals from A and A' return at the same arrival event (and so on for B and B' and for C and C'). That can only happen if we displace the reflection events into the tilted hypersurface shown.
If you are having any trouble seeing this, the simplest remedy is to draw the figure for yourself, being careful to keep all light signals propagating along lines that are at 45 degrees to the vertical.
We can use spacetime diagrams to give us a much simpler, geometric picture of how it is possible that two moving observers can each judge the other's clocks to have slowed. The set-up employs just half of the construction that is used in the twin effect to be described below. So I call it the "half twin effect."
We imagine that twins A and B set off at the same speed but in opposite directions with respect to our vantage point on earth. Each twin carries a clock. As they move away from each other, each twin judges the rate of the other's clock. How this is done is a detail we need not fuss with. They might use light signals, for example, and correct for the time of flight of the signals to figure out what was the other twin's clock reading at each instant.
The figure shows the essential result. Even before we do any detailed analysis, a quick glance at the figure shows a perfect symmetry between the twins A and B. So we already expect that whatever judgment A may make of B's clock, then B will make the same judgment of A's clock. (The clock readings are the numbers next to each twin's worldline.)
Twin A wants to determine the rate of B's clock. Twin A will note that both twin's clocks had the same zero reading when they began their outward journeys. Then twin A asks: when A's clock reads 1, 2, 3, 4, ... what does B's clock read? Answering requires that A make judgments of distant simultaneity. To do this, A must use the hypersurfaces of simultaneity appropriate to A's motion. Because of the way those hypersurfaces tilt, A will judge B's clock to read earlier times. For example, when A's clock reads "4," twin A will judge that B's clock reads only "3." As a result, twin A will judge that B's clock runs slower. For 4 units of time have elapsed on A's clock, while only 3 have elapsed on B's clock.
Twin B will give the same analysis. However, since twin B is in motion relative to A, twin B's analysis will make different judgments concerning which events are simultaneous. These different judgments are represented by B's hypersurfaces of simultaneity. When B's clock reads "4," twin B will judge A's clock to read only "3." As a result, twin B will determine that A's clock runs slower.
Thus each infers the other twin's clock is running slower. Of course, an observer on earth with our vantage point finds the twins to recede with equal speed in opposite directions and would judge that the rates of both twin's clocks to be the same.
You may wonder if things will work out the same way if the twin's approach rather than recede. Drawing the spacetime diagram shows that they do work out the same way. In the diagram, it is convenient to count down the times read by A's and B's clocks as -4, -3, -2, -1, 0, so both clocks read zero when the twins meet at the same event.
Twin A use's A's hypersurface of simultaneity to determine that, when A's clock reads "-4" ,B's clock reads "-2". When the twins meet, both clocks read "0". Hence twin A infers that 4 units of time passed on the A clock, while only 2 units passed on the B clock. Hence twin A infers that the B clock runs slower than the A clock. Twin B makes the corresponding inference about the A clock.
Essentially the same analysis applies to relativistic length contraction. Each of the observers A and B carry a rod, where the rods have the same length when compared in one frame of reference. Each observer will judge the other's rod to have contracted. That is puzzling until we recognize that each of these judgments makes essential use of a judgment of simultaneity. Once again, when we take into account how the observers' judgments of simultaneity differ, we can see how each ends up judging the other's rod to have shrunk.
Setting things up is a little more complicated, because measuring length is more complicated than comparing the time read by clocks. We start with two rods, A and B, moving in opposite directions. We then ask how observers moving with each would judge the length of the other rod. The most direct way is for observer A to have a very long ruler that is at rest with respect to A and long enough to reach the other rod B. The B rod will be moving along A's ruler. Observer A arranges for a measurement of the B rod's length using the procedure described in an earlier chapter. Similarly, observer B will have a very long ruler that extends all the way to the A rod and will use it to measure the length of the A rod.
This arrangment is shown in the figure. The A rod and observer A's ruler are moving together to the left. The length of the A rod is unity, as measured on A's ruler. Similarly, the B rod and observer B's ruler are moving to the right. The length of the B rod is unity as measured on B's ruler.
What length will observer A judge for rod B? It is hard to tell from the ordinary space figure above. The B rod is somewhere around the 6 and 7 marks on A's ruler. But since the B rod is moving rapidly over observer A's ruler, we need to proceed carefully. To know exactly what length observer A would find, we would need to knowhow the two ends of B's rod align with the marks on A's ruler at the same moment. That crucial condition--"at the same moment"--is where A's judgments of simultaneity enter. Precisely the same considerations apply for observer B's assessment of the length of the A rod.
A spacetime diagram lets us see at leisure how these crucial judgments of simultaneity enter. The diagram shows the worldlines of the two ends of observer A's rod. The rod itself and the long ruler carried by A are also shown. They are shown at one instant, corresponding to a hypersurface of simultaneity of observer A's frame of reference. Observer B's rod and ruler are shown in corresponding fashion.
As before, the symmetry of the figure already shows that whatever observer A finds for the B rod, observer B will find for the A rod. So if observer A judges the B rod to have shrunk, observer B must judge the A rod to have shrunk. But will observer A judge the B rod to have shrunk?
Observer A judges the length of a rod by checking where the worldlines of the ends of the rod intersect with A's ruler. These intersections are events in spacetime. The essential condition is that these intersection events happen at the same time; that is, they are simultaneous. The ends of the A rod intersect with observer A's ruler in the two simultaneous events shown in the figure below as the two small red circles on the left. (Remember: they are simultaneous events since they lie on one of observer A's hypersurfaces of simultaneity.) These two events are at the 1 and 2 mark on the ruler, affirming that observer A will judge the A rod to be of unit length.The ends of the B rod intersect with observer A's ruler at the two events marked by red circles on the right. These two events lie between the 6 and 7 marks on observer A's ruler. Hence observer A will judge the B rod to be shorter than unity in length. That is, observer A will judge it to have shrunk.
Observer B will carry out an analogous analysis and judge that the B-rod is of unit length, but that the A-rod is shorter than unit length. Below is the figure showing the corresponding intersection events for observer B's measurements of the lengths of the A and the B rod.
Just what is so special about a Minkowski spacetime? One might think that is it the idea of representing space and time together in a four dimensional geometry, where the four dimensionality of the geometry outstrips our immediate powers of visualization. It is certainly the case that the four dimensionality if both interesting and hard to visualize. But there is nothing inherently relativistic about it. One can take all the physics of Newton and re-express it in four dimensional terms.
The big difference between Newtonian and relativistic spacetimes lies in how they are sliced up into three dimensional spaces. That slicing is done by picking out sets of simultaneous events to form three dimensional spaces.
In Newtonian spacetimes, there is only one way to do this, so a Newtonian spacetime unstacks into a unique set of spaces. In this sense, space and time remain distinct even if we represent the physics in a spacetime.
In a relativistic (i.e. Minkowski) spacetime, the relativity of simultaneity tells us that there are many ways to do this; there is no unique, preferred unstacking. In this sense, space and time get fused together and this fusion is the real novelty of the spacetime approach in relativity theory.
This novelty is surely what Hermann Minkowski had in mind when he wrote in the introduction to his famous lecture "Space and Time" of 1908:
|"The views of space and time which I wish to lay before you have sprung from the soil of experimental physics and therein lies their strength. They are radical. Henceforth space by itself and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality."|
Copyright John D. Norton. January 2001, September 2002; July 2006; February 3, 2007; January 23, September 24, 2008; February 2, 2010; February 7, 2012.