Introduction to Hyper Objects
Now that we’re no longer bound to 3D space, let’s put our newly empowered imaginations to work. What are 4D objects like? How about five dimensions (5D) and beyond?
Time As Displacement
When considering time as a dimension, it is convenient to have some universal standard that relates time to spatial displacement. How many kilometers are there in one second of time? At first this seems like a ridiculous question, akin to asking how many apples there are in a gallon of gasoline. But think of it like this: time and displacement can be related by speed in a sensible way, as long as the speed is known and is constant.
Suppose someone tells you, “Jimsville is an hour away from Joesville.” You’ve heard people talk like this, and you understand what they mean. A certain speed is assumed. How fast must you go to get from Jimsville to Joesville in an hour? If Jimsville and Joesville are 50 kilometers from each other, then you must travel 50 kilometers per hour in order to say that they are an hour apart. If they are only 20 kilometers apart, then you need only travel 20 kilometers per hour to make the same claim.
Perhaps you remember the following formula from elementary physics:
d = st
where d is the distance in kilometers, s is the speed of an object in kilometers per hour, and t is the number of hours elapsed. Using this formula, it is possible to define time in terms of displacement and vice versa.
Universal Speed
Is there any speed that is universal, and that can be used on that basis as an absolute relating factor between time and displacement? Yes, according to Albert Einstein’s famous relativity theory. The speed of light in a vacuum, commonly denoted c , is constant, and it is independent of the point of view of the observer (as long as the observer is not accelerating at an extreme rate or in a super-intense gravitational field). This constancy of the speed of light is a fundamental principle of the theory of special relativity. The value of c is very close to 299,792 kilometers per second; let’s round it off to 300,000 kilometers per second. If d is the distance in kilometers and t is the time in seconds, the following formula is absolute in a certain cosmic sense:
d = ct = 300,000 t
According to this model, the moon, which is about 400,000 kilometers from the earth, is 1.33 second-equivalents distant. The sun is about 8.3 minute-equivalents away. The Milky Way galaxy is 100,000 year-equivalents in diameter. (Astronomers call these units light-seconds, light-minutes , and light-years .) We can also say that any two points in time that are separated by one second, but that occupy the same xyz coordinates in Cartesian three-space, are separated by 300,000 kilometer-equivalents along the t axis.
At this instant yesterday, if you were in the same location as you are now, your location in time-space was 24 (hours per day) × 60 (minutes per hour) × 60 (seconds per minute) × 300,000 (kilometers per second), or 25,920,000,000 kilometer-equivalents away. This mode of thinking takes a bit of getting used to. But after a while, it starts to make sense, even if it’s a slightly perverse sort of sense. It is, for example, just about as difficult to jump 25,920,000,000 kilometers in a single bound, as it is to change what happened in your room at this time yesterday.
The above formula can be modified for smaller distances. If d is the distance in kilometers and t is the time in milliseconds (units of 0.001 of a second), then:
d = 300 t
This formula also holds for d in meters and t in microseconds (units of 0.000001, or 10 −6 , second), and for d in millimeters (units of 0.001 meter) and t in nanoseconds (units of 0.000000001, or 10 −9 , second). Thus we might speak of meter-equivalents, millimeter-equivalents, microsecond-equivalents , or nanosecond-equivalents .
The Four-cube
Imagine some of the simple, regular polyhedra in Cartesian four-space. What are their properties? Think about a four-cube , also known as a tesseract . This is an object with several identical 3D hyperfaces , all of which are cubes. How many vertices does a tesseract have? How many edges? How many 2D faces? How many 3D hyperfaces? How can we envision such a thing to figure out the answers to these questions?
This is a situation in which time becomes useful as a fourth spatial dimension. We can’t make a 4D model of a tesseract out of toothpicks to examine its properties, and few people (if any) can envision such a thing. But we can imagine a cube that pops into existence for a certain length of time and then disappears a little later, such that it “lives” for a length of time equivalent to the length of any of its spatial edges, and does not move during its existence. Because we have defined an absolute relation between time and displacement, we can graph a tesseract in which each edge is, say, 300,000 kilometer-equivalents long. It is an ordinary 3D cube that measures 300,000 kilometers along each edge. It pops into existence at a certain time t 0 and then disappears 1 second later, at t 0 + 1. The sides of the cube are each 1 second-equivalent in length, and the cube “lives” for 300,000 kilometer-equivalents of time.
Figure 11-4A shows a tesseract in dimensionally reduced form. Each division along the x and y axes represents 100,000 kilometers (the equivalent of 1/3 second), and each division along the t axis represents 1/3 second (the equivalent of 100,000 kilometers). Figure 11-4B is another rendition of this object, illustrated as two 3D cubes (in perspective) connected by dashed lines representing the passage of time.
Fig. 11-4 . At A, a dimensionally reduced plot of a time-space tesseract. At B, another rendition of a tesseract, portraying time as lateral motion.
The Rectangular Four-prism
A tesseract is a special form of the more general figure, known as a rectangular four-prism or rectangular hyperprism . Such an object is a 3D rectangular prism that abruptly comes into existence, lasts a certain length of time, disappears all at once, and does not move during its “lifetime.” Figure 11-5 shows two examples of rectangular four-prisms in dimensionally reduced time-space.
Fig. 11-5 . Dimensionally reduced plots of two rectangular hyperprisms in time-space.
Suppose the height, width, depth, and lifetime of a rectangular hyperprism, all measured in kilometer-equivalents, are h , w , d , and t , respectively. Then the 4D hypervolume of this object (call it V 4D ), in quartic kilometer-equivalents , is given by the product of them all:
V 4D = hwdt
The mathematics is the same if we express the height, width, depth, and lifetime of the object in second-equivalents; the 4D hypervolume is then equal to the product hwdt in quartic second-equivalents .
Impossible Paths
Certain paths are impossible in Cartesian 4D time-space as we’ve defined it here. According to Einstein’s special theory of relativity, nothing can travel faster than the speed of light. This restricts the directions in which line segments, lines, and rays can run when denoting objects in motion.
Consider what happens in 4D Cartesian time-space when a light bulb is switched on. Suppose the bulb is located at the origin, and is surrounded by millions of kilometers of empty space. When the switch is closed and the bulb is first illuminated, photons (particles of light) emerge. These initial, or leading, photons travel outward from the bulb in expanding spherical paths. If we dimensionally reduce this situation and graph it, we get an expanding circle centered on the time axis, which, as time passes, generates a cone as shown in Fig. 11-6. In true 4D space this is a hypercone or four-cone . The surface of the four-cone is 3D: two spatial dimensions and one time dimension.
Imagine an object that starts out at the location of the light bulb, and then moves away from the bulb as soon as the bulb is switched on. This object must follow a path entirely within the light cone defined by the initial photons from the bulb. Figure 11-6 shows one plausible path and one implausible path.
Fig. 11-6 . Dimensionally reduced plot of the leading photons from a light bulb. Paths outside the light cone represent speeds greater than c (the speed of light) and are therefore implausible.
General Time-space Hypervolume
Suppose there is an object—any object—in 3D space. Let its spatial volume in cubic kilometer-equivalents be equal to V 3D . Suppose that such an object pops into existence, lasts a certain length of time t in kilometer-equivalents, and then disappears. Suppose that this object does not move with respect to you, the observer, during its “lifetime.” Then its 4D time-space hypervolume V 4D is given by this formula:
V 4D = V 3D t
That is to say, the 4D time-space hypervolume of any object is equal to its spatial volume multiplied by its lifetime, provided the time and displacement are expressed in equivalent units, and as long as there is no motion involved.
If an object moves, then a correction factor must be included in the above formula. This correction factor does not affect things very much as long as the speed of the object, call it s , is small compared with the speed of light c . But if s is considerable, the above formula becomes:
V 4D = V 3D t (1 − s 2 / c 2 ) 1/2
The correction factor, (1 − s 2 / c 2 ) 1/2 , is close to 1 when s is a small fraction of c , and approaches 0 as s approaches c . This correction factor derives from the special theory of relativity. (It can be proven with the help of the Pythagorean theorem. The proof is not complicated, but getting into it here would take us off the subject. Let’s just say that objects are “spatially squashed” at extreme speeds, and leave it at that.)
In this context, speed s is always relative. It depends on the point of view from which it is observed, witnessed, or measured. For speed to have meaning, we must always add the qualifying phrase “relative to a certain observer.” In these examples, we envision motion as taking place relative to the origin of a 3D Cartesian system, which translates into lines, line segments, or rays pitched at various angles with respect to the time axis in a 4D time-space Cartesian system.
If you’re still confused about kilometer-equivalents and second-equivalents, you can refer to Table 11-1 for reference. Keep in mind that time and displacement are related according to the speed of light:
d = ct
where d is the displacement (in linear units), t is the time (in time units), and c is the speed of light in linear units per unit time. Using this conversion formula, you can convert any displacement unit to an equivalent time interval, and any time unit to an equivalent displacement.
Table 11-1 . Some displacement and time equivalents. Displacement equivalents are accurate to three significant figures.
PROBLEM 1
How many second-equivalents are there in 1 kilometer?
SOLUTION 1
We know that the speed of light is 300,000 kilometers per second, so it takes 1/300,000 of a second for light to travel 1 kilometer. That is approximately 0.00000333 seconds or 3.33 microseconds. One kilometer is 0.00000333 second-equivalents, or 3.33 microsecond-equivalents.
Practice problems for these concepts can be found at: Hyperspace And Warped Space Practice Test.
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