*The Fourth Dimension: Toward a Geometry of Higher Reality*is called "Time Travel and Telepathy." These are, of course, two of the best-known science fiction tropes.

He writes:

*Why does travel have to be so hard? The perfect vehicle is easy to imagine: a sort of automobile with some special buttons on the dash. Get in, punch the code numbers of where and when you want to be, turn the ignition key, and -- presto -- there you are in 1920's Paris, on the Great Plains before the pioneers, on the moon, or even in another galaxy.*

*Actually, such a vehicle as Rucker describes does exist -- it's called the DeLorean! Notice that Rucker first wrote this book in 1984, just barely before the world's most famous time travel movie,*

*Back to the Future*, was released -- otherwise he probably would have mentioned it here. (On the other hand, the movie

*E.T.*was released in time for Rucker to mention it in this chapter.)

The movie

*Back to the Future II*was mentioned in the news about five months ago, mainly because the date to which its main character, Marty McFly, travels happens to be October 21st, 2015. So there was much lighthearted discussion about whether any of the predictions made by the movie about the year 2015 had come true. (One notable prediction that failed was that the Chicago Cubs would win the 2015 World Series -- ironically, they were eliminated from the playoffs on October 21st! On the other hand, the movie correctly predicted the proliferation of flat-screen TV's and drones.)

Rucker's chapter, meanwhile, is more about time as the fourth dimension as well as the many paradoxes associated with time travel. (Yes, I know that David Kung's lectures were all about paradoxes, but he doesn't mention time travel.) The most famous time travel paradox is the Grandfather Paradox -- Rucker describes a version of this paradox where you attempt to kill your own past self.

There are six puzzles in this chapter. Throughout this blog post, I will make several comparisons to my own favorite sci-fi series,

*Futurama*. (This show has its own resolution of the Grandfather Paradox -- the main character Fry goes back in time to become

*his own grandfather*!)

*Puzzle 10.1:*

*If you were to have complete freedom in moving forward and backward in time, then you could duplicate most of the feats of a hyperbeing who can move*ana

*and*kata

*at will. How could you use time travel to enter a sealed room? How could you use it to remove someone's dinner from his stomach without disturbing him?*

*Answer 10.1:*

*To get into the sealed room, travel far, far into the future until you reach a time where the room's walls have crumbled. Step into the room's space and then travel back in time. Note that this is exactly analogous to moving ana, then moving across the location where the wall would be, and then coming back kata into the room's space.*

*Getting the food out of someone's stomach is a little trickier, since you yourself won't fit in the stomach.... Take a little scoop and set it right on the bed where his stomach was. Now send the scoop back a few hours, to when he was there, and then bring it back to the empty bed....*

*Puzzle 10.2:*

*Special relativity says that it's impossible to permanently label any given space location. "Right here last week," in other words, has no absolute meaning. How would the existence of a time machine go against this assumption?*

*Answer 10.2:*

*If we assume that the time machine goes straight backward in time, then "right here one week ago" is wherever the time machine appears when I send it one week back. One would not expect the Earth to still be "right here" a week ago, so traveling backward in time might well land one in empty space. Science fiction writers usually deal with this problem by somehow setting their time machine to track along the Earth's spacetime path.*

*Commentary: In this next question, Rucker explains how faster-than-light (FTL) travel is, in fact, equivalent to time travel.*

*Puzzle 10.3:*

*Here is a picture explaining how FTL travel can lead to time travel to one's own past. The traveler goes from*A

*to*B

*to*C

*,*B

*being an event on the world line of a distant galaxy that is moving away from Earth at half the speed of light. Explain how the paths*AB

*and*BC

*both can be regarded as pure FTL trips.*

*Commentary: Here is a link I found of an image that's similar to Rucker's. The left image is the one relevant to this puzzle (though Rucker alludes to the right image elsewhere). The particle begins in the upper left corner (*

*A*) and travels "superluminally" (i.e., FTL) to the upper right corner (

*B*), and then travels FTL again to the lower left corner (

*C*), which is in

*A*'s own past.

*Answer 10.3:*

*The point is that "instantaneous" is a relative concept. Relative to Earth, B happens at the same time as A, so one can travel instantaneously from A to B. Relative to the distant galaxy that moves away from us, C is simultaneous with B, so one can travel instantaneously from B to C. Combining the two trips take one from A to C, and thus into one's own past.*

*Puzzle 10.4:*

*Not only does FTL travel lead to time travel, the converse is true as well. Time travel leads to FTL travel. Given a rocket and a time machine, how could you send a probe all the way around the galaxy, yet have it get back the same day?*

*Answer 10.4:*

*Equip the rocket with a good robot brain so that after its hundreds-of-thousands-of-years-long journey, it can use the time machine to jump hundreds of thousands of years back in time and find the Earth. Once it finds the Earth again, it makes a small time jump to the right day (the day of launch) and lands.*

*Commentary: The TV show*

*Futurama*avoids the problems with FTL travel to distant galaxies by declaring that in a few centuries, the speed of light will have increased!

*Puzzle 10.5:*

*If time itself were bent into a vast circle, then one might hope to reach the past by traveling "around" time. But thinking about a universe in which time is a vast circle leads to some strange problems. Say, for instance, that you build a very durable radio beacon and set it afloat in space near the Earth. Is it possible that this beacon can last all the way around time. If it does, then once you set one afloat, how many more should you be able to detect? What if you decide to set your beacon afloat if and only if you detect no beacons out there before the launch?*

*Answer 10.5:*

*The idea of a beacon that lasts all the way around time would lead to difficulties. If it drifted away from Earth never to return, then there would be, it seems, endlessly many of them out there, as a consequence of the one launch! This seems nonsensical. The situation is particularly vicious if we suppose that B_1 sends out a signal that can inhabit the launch of B_0. A yes-and-no paradox!*

*...It is, in other words, impossible to build a truly indestructible object if time is circular! For anything you build must in time disintegrate into pieces so that you can "again" build it.*

*Commentary: As I mentioned in my last post, one particular*

*Futurama*episode, "The Late Phillip J. Fry," explores the idea that time is circular. In this episode, Fry, the robot Bender, and the Professor travel in a time machine that can only travel to the future, not the past. So in order to return home, the crew must travel all the way around time. So the time machine itself is

*B_0*, an object that travels the entire wheel of time.

Here's how the episode avoids the paradox -- upon arriving, the time machine

*B_0*lands right on top of time machine

*B_1*, thereby destroying it. (Actually, it's

*B_2*, not

*B_1*-- the crew accidentally overshoots its home year and must make a second lap around time.) With

*B_2*destroyed, we no longer have to worry about there being infinitely many time machines around.

*Puzzle 10.6:*

*We discussed a number of paradoxes that arise from an ability to travel to the past. But just being able to*communicate

*with the past leads to paradoxes as well. Suppose, for instance, that I have a magic telephone with the following properties: Whenever I pick up the receiver and dial "1," the magic telephone rings an hour earlier. I am thus able to telephone my past self. If I happen to hear the phone ringing, and pick up the receiver, I can expect to hear the voice of my future self. By now, what if at 9:00 I decide I will dial "1" at 11:00 unless I have received a call at 10:00?*

*Answer 10.6:*

*We have a yes-and-no paradox here. I dial "1" at 11:00 if and only if I don't get a call at 10:00, but I get a call at 10:00 if and only if I dial "1" at 11:00. In other words, I get a call at 10:00 if and only if I don't get a call at 10:00. This particular paradox was first raised in a paper by G. Benford, D. Book, and W. Newcomb, "The Tachyonic Antitelephone" (1970). Gregory Benford, is, by the way, a science fiction writer, as well as being a physicist. The "tachyons" his paper refers to are hypothetical particles that, unlike ordinary mass particles, always go faster than light. Benford's paper argues that since tachyons could be used to send messages into one's own past, it must be, even in principle, impossible to detect them. If, then, tachyons are real, they fill a sort of undetectable ghost universe whose time direction is in a sense perpendicular to our own time direction.*

Commentary:

*Futurama*also has a method of avoiding time paradoxes. In the

*Futurama*movie

*Bender's Big Score*, some of the characters go back in time to tell their past selves not to travel back in time. This is a paradox as now there are now two copies of the character. The Professor explains that the universe corrects the paradox by destroying one of the time copies. This follows Rucker when he writes, "Or maybe the cosmos would, in the interest of self-preservation, strike dead anyone about to perform a paradoxical time travel experiment!"

This is wrote last year about Lesson 9-5 of the U of Chicago text, a special spring break lesson:

We already discussed translations and rotations for Common Core Geometry, but those transformations applied to the

*two*-dimensional plane. The physical world has

*three*dimensions, and so it's the 3-D transformations that actually apply to physics.

Lesson 9-5 of the U of Chicago text is on reflections in space. I find this to be an interesting topic, but since it doesn't appear on the PARCC, it would be a waste of time to cover it in class. There will be

*no*worksheet for this lesson, since it's not intended to be taught in class. This is why I waited until spring break to blog about this topic.

The text begins by defining what it means for a

*plane*in 3-D, rather than a line in 2-D, to be a perpendicular bisector:

"In general, a plane

*M*is the perpendicular bisector of a segment

*M*is perpendicular to

*M*contains the midpoint of

Now we can definition 3-D reflections almost exactly the same way we define 2-D reflections:

"For a point

*A*which is not on a plane

*M*, the reflection image of

*A*over

*M*is the point

*B*if and only if

*M*is the perpendicular bisector of

*A*on a point

*M*, the reflection image of

*A*over

*M*is

*A*itself."

If you think about it, when you (a 3-D figure!) look at yourself in a mirror, the mirror itself isn't a line, but rather a plane. Mirrors in 2-D are lines, while mirrors in 3-D are planes. So now we can define what it means for two 3-D figures to be congruent:

"Two figures F and G in space are congruent figures if and only if G is the image of F under a reflection or composite of reflections."

Most texts don't actually define what it means for two 3-D figures to be

*congruent*. We know that the traditional textbook definition, that congruent figures have corresponding sides and angles congruent, only applies to polygons. It doesn't even apply to circles, much less 3-D figures. But we can simply use the Common Core definition -- two figures are congruent if and only if there exists an isometry (i.e., a composite of reflections) mapping one to the other -- and it instantly applies to all figures, polygons, circles, and 3-D figures.

In 2-D there are only four isometries -- reflections, translations, rotations, and glide reflections. An interesting question is, how many isometries are there in 3-D?

Well, for starters, translations and rotations exist in 3-D. We can define both of these exactly the same way that we do in 2-D -- a translation is the composite of two reflections in

*parallel*planes, while a rotation is the composite of two reflections in

*intersecting*planes.

Notice that every 2-D rotation has a center -- the point of intersection of the reflecting lines. The same thing happens in 3-D, except that the intersection of two planes isn't a point, but a

*line*. Thus, a 3-D rotation has an entire

*line*as its center -- every point on this line is a fixed point of the rotation. But usually, instead of calling the line the

*center*of the rotation, we call it the

*axis*of the rotation. One 3-D object that famously rotates is the

*earth*, and this rotation has an axis -- the line that passes through the North and South Poles. Confusingly, the mirror of a 2-D reflection is often called an

*axis*-- but in some ways, these two definitions are related. One can perform a 2-D reflection by taking a 2-D figure and rotating it 180 degrees about the axis in 3-D (and recall that A Cube has reflected A Square in exactly this manner).

Glide reflections also exist in 3-D -- although these are often called glide

*planes*in 3-D. A glide reflection is the composite of a reflection and a nontrivial translation parallel to the mirror. Notice that there are

*infinitely many*directions to choose from for our translation in 3-D, but if this were a 2-D glide reflection there are only

*two*possible directions for a translation that's parallel to the mirror.

But are there any other isometries in 3-D? Well, we notice that a glide reflection is the composite of two other known isometries, a reflection and a translation. So the next natural possibility to consider is, what if we find the composite of the other two combinations? What is the composite of a reflection and a rotation, or a translation and a rotation?

In 2-D, the composite of a translation and a nontrivial rotation is another rotation. This is possible to prove, as follows: let T be a translation and R be a nontrivial rotation, and G be the composite of T following R. Both translations and rotations preserve orientation, and so their composite G must preserve orientation as well. In 2-D, three mirrors suffice -- that is, every isometry is the composite of at most three reflections. Since G preserves orientation, it must be the composite of an even number of rotations. Therefore G is the composite of two reflections (or the identity transformation -- the transformation that maps every point to itself), and so G is either a translation or a rotation.

Let's try an indirect proof -- assume that G is a translation. That is:

T o R = G

Since T is a translation, it has a translation vector

**t**, and as we're assuming that G is a translation, it must also have a translation vector

**g**. Now let U be the inverse translation of T -- that is, the translation whose vector is -

**t**, the additive inverse of

**t**. We now compose U on both sides:

U o T o R = U o G

Now since U and T are inverses, U o T must be I, the identity transformation:

I o R = U o G

Since I is the identity transformation, I o R must be R:

R = U o G

Notice that G and U are both translations, whose vectors are

**g**and -

**t**, respectively. Then their composite must be another translation V whose vector is

**g**-

**t**. So now we have:

R = V

that is, a rotation equals a translation. Now no rotation can equal a translation (as the former has a fixed point, while the latter has no fixed point) unless both are the identity -- which contradicts the assumption that R is a nontrivial rotation (i.e., not the identity). Therefore, the composite of a translation and a nontrivial rotation isn't a translation, so it must be a rotation. QED

But this proof is invalid in 3-D. This is because the proof uses a step that only works in 2-D -- namely that three mirrors suffice. We must show how many mirrors suffice in 3-D.

Let's recall why mirrors suffice in 2-D. Let G be any 2-D isometry, and let

*A*,

*B*, and

*C*be three noncollinear points whose images under G are

*A'*,

*B*', and

*C'*. The first mirror maps

*A*to

*A'*, the second mirror fixes

*A'*and maps

*B*to

*B'*. It could be that the image of

*C*under both mirrors is already

*C'*, otherwise a third mirror maps it to

*C'*. Notice that the proof of the existence of these mirrors is nontrivial and depends on theorems such as the Perpendicular Bisector Theorem, since reflections are defined using perpendicular bisectors.

As it turns out,

*four*mirrors suffice in 3-D. To prove this, we let G be any 3-D isometry, and let

*A*,

*B*,

*C*, and

*D*be four noncoplanar points. The first mirror maps

*A*to

*A'*, the second mirror fixes

*A'*and maps

*B*to

*B'*, the third mirror fixes

*A'*and

*B'*and maps

*C*to

*C'*, and the fourth, if necessary, fixes

*A'*,

*B'*, and

*C'*and maps

*D*to

*D'*.

And so this opens the door for there to be a new transformation in 3-D, one that is the composite of a translation and a rotation as well as the composite of four reflections. We can imagine twisting an object like a screw. A screwdriver rotates the screw about its axis, but then it's being translated into the wall in the same direction as that axis. And because of this, this new transformation is often called a

*screw motion*.

We still have one last combination, the composite of a reflection and a rotation. It is subtle why the composite of a reflection and a rotation in 2-D is usually a

*glide reflection*-- why should the composite of a reflection and a rotation equal the composite of a (different) reflection and a

*translation*? And it's even subtler why the composite of a reflection and a rotation may be a new transformation in 3-D.

But the simplest example of this

*roto-reflection*is the inversion map. In 3-D coordinates, we map the point (

*x*,

*y*,

*z*) to its opposite point (-

*x*, -

*y*, -

*z*). This map is the composite of three reflections -- the mirrors are the three coordinate planes (

*xy*,

*xz*, and

*yz*). As the composite of an odd number of reflections, it must reverse orientation. Yet it can't be a reflection, since it has only a single fixed point (0, 0, 0) and not an entire plane. Similarly, it can't be a glide plane because glide planes don't have any fixed points at all.

Roto-reflections are formed when the axis of the rotation intersects the reflecting plane in a single point -- and of course, this single point is the only fixed point of the roto-reflection.

So now we have six isometries in 3-D -- reflections, translations, rotations, glide planes, screw motions, and roto-reflections. Are there any others? As it turns out, these six are all of them -- and the proof depends on the fact that four mirrors suffice in 3-D.

Returning to the U of Chicago text, we have the definition of a reflection-symmetric figure:

"A space figure is F is a reflection-symmetric figure if and only if there is a plane

*M*such that the reflection of F in

*M*is F."

Similarly, a figure can be rotation-symmetric, as well as roto-reflection-symmetric. In the text, figures such as the right cylinder have reflection, rotation, and roto-reflection symmetry.

But a figure can't have translation symmetry unless it's infinite, as translations lack fixed points. So likewise, figures that have glide reflection or screw symmetry must also be infinite, as these transformations are based on translations.

Of course, there exist dilations in 3-D space as well. There is little discussion of similarity in 3-D, except to compare the surface areas and volumes of 3-D figures.

Now that we are reading about the fourth dimension, the next thing we wonder is, what would a reflection in 4-D look like? We follow the same pattern that we followed for 2-D and 3-D -- we begin with the perpendicular bisector of a segment. In 2-D the perpendicular bisector is a 1-D line, in 3-D it's a 2-D plane, and so in 4-D it's a 3-D

*hyperplane*. This hyperplane

*M*is the mirror of a 4-D reflection, mapping each point

*A*to its image

*B*if and only if

*M*is the perpendicular bisector of

~~AB~~

(and of course mapping each point on the hyperplane to itself).

We know that the 3-D isometries include all of the 2-D isometries (translations, rotations, reflections, and glide reflections) plus two new isometries (screws and roto-reflections). So we expect the 4-D isometries to include all the 3-D isometries plus two new ones.

One of these new isometries is called a

*double rotation*, as it is the composite of two rotations. Notice that the center of a 2-D rotation is a 0-D point, and the center (axis) of a 3-D rotation is a 1-D line, and so the axis of a 4-D rotation is a 2-D plane. The double rotation is the composite of two rotations whose respective axes (planes) intersect in a single point. A point inversion is a double rotation.

The other new 4-D isometry is a roto-glide reflection. It is the composite of a roto-reflection (which requires three dimensions) and a translation in the fourth dimension. Just as three mirrors suffice in 2-D and four mirrors suffice in 3-D, we see that five mirrors suffice in 4-D.

But since today we read Rucker's chapter on time travel, we may ask, how do 4-D reflections work if we take

*time*to be the fourth dimension? That is, suppose the mirror hyperplane is a single instant in time (that is, it contains only the three spatial dimensions), then what would a reflection look like?

In today's Rucker chapter, the author writes that a small particle called a

*positron*may in fact be the time-reflection image of an electron:

*If we look at it in terms of the "moving Now" viewpoint, it seems a little surprising that the electron and positron manage so neatly to appear and disappear together. But according to physicist Richard Feynman, one can take a spacetime viewpoint and regard the positron as an electron that is traveling backward in time. From this standpoint we simply have a nice little closed causal loop.*

*Rucker then follows this with a short story he wrote, called "A New Experiment With Time," in which an entire*

*person*undergoes a time-reflection! The main character, Maisie Gleaves, is a young woman who misses high school so much that she wants to go back in time to her school days. So she undergoes a time-reflection. Like all reflections, a time-reflection reverses orientation -- except that it's the time dimension that is reversed. Maisie sees everyone seeming to walk and talk backwards -- and to everyone else, Maisie is the one walking and talking backwards.

Before I conclude this post, let me announce that tonight on PBS, the TV show

*NOVA*is airing a repeat of last year's episode, "The Great Math Mystery." I described this episode in detail on the blog almost exactly a year ago (back in April). I highly recommend watching it if you missed it last year!

Thus concludes my spring break post. I'll post once more during the vacation period.