Introduction to The Fourth Dimension and Beyond
There is no limit to the number of dimensions that can be defined using the Cartesian scheme. There can be any positive whole number of dimensions. Time can be (but does not have to be) one of them.
Cartesian N -space
A system of rectangular coordinates in five dimensions defines Cartesian five-space . There are five number lines, all of which intersect at a point corresponding to the zero point of each line, and such that each of the lines is perpendicular to the other four. The resulting axes can be called v , w , x , y , and z . Alternatively they can be called x _{1} , x _{2} , x _{3} , x _{4} , and x _{5} . Points are identified by ordered quintuples such as ( v , w , x , y , z ) or ( x _{1} , x _{2} , x _{3} , x _{4} , x _{5} ). The origin is defined by (0,0,0,0,0).
A system of rectangular coordinates in Cartesian n -space (where n is any positive integer) consists of n number lines, all of which intersect at their zero points, such that each of the lines is perpendicular to all the others. The axes can be named x _{1} , x _{2} , x _{3} , . . ., and so on up to x _{n} . Points in Cartesian n -space can be uniquely defined by ordered n -tuples of the form ( x _{1} , x _{2} , x _{3} ,. . ., x _{n} ).
Imagine a tesseract or a rectangular four-prism that pops into existence at a certain time, does not move, and then disappears some time later. This object is a rectangular five-prism . If x _{1} , x _{2} , x _{3} , and x _{4} represent four spatial dimensions (in kilometer-equivalents or second-equivalents) of a rectangular four-prism in Cartesian four-space, and if t represents its “lifetime” in the same units, then the 5D hypervolume (call it V _{5D} ) is equal to the product of them all:
V _{5D} = x _{1} x _{2} x _{3} x _{4} t
This holds only as long as there is no motion. If there is motion, then the relativistic correction factor must be included.
Dimensional Chaos
There is nothing to stop us from dreaming up a Cartesian 25-space in which the coordinates of the points are ordered 25-tuples ( x _{1} , x _{2} , x _{3} ,. . ., x _{25} ), none of which are time. Alternatively, such a hyperspace might have 24 spatial dimensions and one time dimension. Then the coordinates of a point would be defined by the ordered 25-tuple ( x _{1} , x _{2} , x _{3} ,. . ., x _{24} , t).
Some cosmologists—scientists who explore the origin, structure, and evolution of the cosmos—have suggested that our universe was “born” with 11 dimensions. According to this hypothesis, not all of these dimensions can be represented by Cartesian coordinates. Some of the axes are “curled up” or compactified as if wrapped around tiny bubbles. Some mathematicians have played with objects that seem to be 2D in some ways and 3D in other ways. How many dimensions are there in the complicated surface of a theoretical foam, assuming each individual bubble is a sphere of arbitrarily tiny size and with an infinitely thin 2D surface? Two dimensions? In a way. Three? In a way. How about two and a half?
The examples we have looked at here are among the simplest. Imagine the possible ways in which a 4D parallelepiped might exist, or a 4D sphere . How about a 5D sphere , or a 7D ellipsoid ? Let your mind roam free. But don’t think about this stuff while driving, operating heavy equipment, cycling, or walking across a street in traffic.
Distance Formulas Beyond the Fourth Dimension
In n -dimensional Cartesian space, the shortest distance between any two points can be found by means of a formula similar to the distance formulas for 2D and 3D space. The distance thus calculated represents the length of a straight line segment connecting the two points.
Suppose there are two points in Cartesian n -space, defined as follows:
P = ( x _{1} , x _{2} , x _{3} ;. . . x _{n} )
Q = ( y _{1} , y _{2} , y _{3} ;. . . y _{n} )
The length of the shortest possible path between points P and Q , written | PQ |, is equal to either of the following:
| PQ | = [( y _{1} − x _{1} ) ^{2} + ( y _{2} − x _{2} ) ^{2} + ( y _{3} − x _{3} ) ^{2} + . . . + ( y _{n} − x _{n} ) ^{2} ] ^{1/2}
| PQ | = [( x _{1} − y _{1} ) ^{2} + ( x _{2} − y _{2} ) ^{2} + ( x _{3} − y _{3} ) ^{2} + . . . + ( x _{n} − y _{n} ) ^{2} ] ^{1/2}
Beyond Four Dimensions Problems and Solutions
PROBLEM 1
Find the distance | PQ | between the points P = (4,−6,−3,0) and Q = (−3,5,0,8) in Cartesian four-space. Assume the coordinate values to be exact; express the answer to two decimal places.
SOLUTION 1
Assign the numbers in these ordered quadruples the following values:
Then plug these values into either of the above two distance formulas. Let’s use the first formula:
PROBLEM 2
How many vertices are there in a tesseract?
SOLUTION 2
Imagine a tesseract as a 3D cube that lasts for a length of time equivalent to the linear span of each edge. When we think of a tesseract this way, and if we think of time as flowing upward from the past toward the future, the tesseract has a “bottom” that represents the instant it is “born,” and a “top” that represents the instant it “dies.” Both the “bottom” and the “top” of the tesseract, thus defined, are cubes. We know that a cube has eight vertices. In the tesseract, there are twice this many vertices. The eight vertices of the “bottom” cube and the eight vertices of the “top” cube are connected with line segments that run through time.
Another way to envision this is to portray a tesseract as a cube-within-a-cube (Fig. 11-7). This is one of the most popular ways that illustrators try to draw this strange 4D figure. It isn’t a true picture, of course, because the “inner” and the “outer” cubes in a real tesseract are the same size. But this rendition demonstrates that there are 16 vertices in the tesseract. Look at Fig. 11-7 and count them!
Fig. 11-7 . The cube-within-a-cube portrayal of a tesseract shows that the figure has 16 vertices.
PROBLEM 3
What is the 4D hypervolume, V _{4D} , of a rectangular four-prism consisting of a 3D cube measuring exactly 1 meter on each edge, that “lives” for exactly 1 second, and that does not move? Express the answer in quartic kilometer-equivalents and in quartic microsecond-equivalents.
SOLUTION 3
We must find the 4D hypervolume of a 3D cube measuring l × 1 × 1 meter (whose 3D volume is therefore 1 cubic meter) that “lives” for 1 second of time.
To solve the first half of this problem, note that light travels 300,000 kilometers in one second, so the four-prism “lives” for 300,000, or 10 ^{5} , kilometer-equivalents. That can be considered its length. Its cross section is a cube measuring 1 meter, or 0.001 kilometer, on each edge, so the 3D volume of this cube is 0.001 × 0.001 × 0.001 = 0.000000001 = 10 ^{−9} cubic kilometers. Therefore, the 4D hypervolume ( V _{4D} ) of the rectangular four-prism in quartic kilometer-equivalents is:
V _{4D} = 300,000 × 0.000000001
= 3 × 10 ^{5} × 10 ^{−9}
= 3 × 10 ^{−4}
= 0.0003 quartic kilometer-equivalents
To solve the second half of the problem, note that in 1 microsecond, light travels 300 meters, so it takes light 1/300 of a microsecond to travel 1 meter. The 3D volume of the cube in cubic microsecond-equivalents is therefore (1/300) ^{3} = 1/27,000,000 = 0.00000003704 = 3.704 × 10 ^{−8} . The cube lives for 1 second, which is 1,000,000, or 10 ^{6} , microseconds. Therefore, the 4D hypervolume V _{4D} of the rectangular four-prism in quartic microsecond-equivalents is:
V _{4D} = 0.00000003704 × 1,000,000
= 3.704 × 10 ^{−8} × 10 ^{6}
= 3.704 × 10 ^{−2}
= 0.03704 quartic microsecond-equivalents
PROBLEM 4
Suppose the four-prism described in the previous problem moves, during its brief lifetime, at a speed of 270,000 kilometers per second relative to an observer. What is its 4D hypervolume ( V _{4D} ) as seen by that observer? Express the answer in quartic kilometer-equivalents and in quartic microsecond-equivalents.
SOLUTION 4
The object moves at 270,000/300,000, or 9/10, of the speed of light relative to the observer. If we let s represent its speed, then s/c = 0.9, and s ^{2} / c ^{2} = 0.81. We must multiply the answers to the previous problem by the following factor:
(1 − s ^{2} / c ^{2} ) ^{1/2}
= (1 − 0.81) ^{1/2}
= 0.19) ^{1/2}
= 0.436)
This gives us:
V _{4D} = 0.0003 × 0.436 = 0.0001308 quartic kilometer-equivalents
V _{4D} = 0.03704 × 0.436 = 0.01615 quartic microsecond-equivalents
Practice problems for these concepts can be found at: Hyperspace And Warped Space Practice Test.