Ok, so the other day I’m reading this book, Chaos. No ,it’s not about anarchy, it’s about, surprise surprise, physics. Anyway, something in it got me thinking, and I realized I understood why the fourth dimension is time. Let me explain, first. See, I’d always thought of dimensions as ways of drawing shapes. You know, 1 dimension= a line, 2=a rectangle, 3= a cube, 4= ??? Then it occured to me- they’re not ways of creating euclidean shapes, they’re for locating points! See, with the first dimension you know where any point is along the x axis, but not along the y or z. With the second dimension, you’re only missing the z axis, with the 3rd, you should be all set, right? Nuh-uh! Look 10 feet in front of you. Record the time. Now, throw a baseball. When it’s ten feet in front of you, look at it- exact same point in 3 dimensions, but the 2 points are different, unless there’s a baseball hanging from the ceiling. So, without including time as a coordinate, it becomes very hard to locate things. Hmm, what do you guys think?

Yep sounds right. Might I add that the super string theory suggests at least 10 dimensions.

cool…

Ya think that’s cool?

It also suggests that as many as 6 of the 10 dimensions have been “rolled up” to the size of a single point (or possibly smaller.)

Kids, don’t try this at home.

-David

Argeable, in relativity a point in space-time is often called an ‘event’. I.e. there is the event at 12:00 PM with no baseball at point XYZ, and there is an event at 12:01 PM with a baseball at point XYZ. It might make more sense to think of the trajectory of a batted ball - there is a ‘hit’ event at time t and location x,y,z, and a ‘catch’ event at time t’, location x’, y’, z’.

This doesn’t really change anything, it just makes it easier for me to think about it.

Oh, and what are we debating?

No debate as I see it but . . .

Ever read Edwin Abbot’s * Flatland* ? Sounds much like your epiphany, without the allegorical representation of Victorian society (the book was written in the late 1800’s). You may enjoy it. If you’ve never read it (or heard of it) basically it’s about a two dimenional world (Flatland) in which the narrator, A. Square, is visited by beings from another dimension and allowed to see them.

Try it, you might like it!

Bah humbug. *Flatland* can’t hold a candle to A. K. Dewdney’s 1984 masterpiece *The Planiverse*. Unlike Abbot’s work, which features pure geometic shapes (with only one eye apiece) floating around in a 2-dimensional space with houses in it, Dewdney’s work takes the idea of a 2-dimensional universe to its logical conclusion. He has circular stars, orbited by circular planets, on the 2-dimensional “surface” of which are whole biospheres full of the kinds of organisms that can evolve in a 2-dimensional universe. He gets around the problem of not being able to have “tube”-shaped structures in 2 dimensions by inventing a “zipper joint” that can open and close in small segments to allow materials to pass through.

With regards to the OP… I dunno, something doesn’t seem to fit…

In my mind (this is only my opinion here… nothing more), time doesn’t fit the bill as a “dimension”. Here’s why: With the three “basic” dimensions (height, breadth, width), each dimension can, on paper, at least, exist without any of the others. But no dimension can exist without Time, since things can’t exist for NO amount of time. Likewise, Time cannot exist on its own since it needs another dimension to act as a “starting point” for its own measurement.

So instead of acting like a “dimension”, Time acts like a “dimensional constant”, a prerequisite for any sort of existence at all.

Time is a perfectly good dimension … as long as you are willing to extend the concept of “dimension” to something more abstract than a simple spatial measure. Time enters the equations of relativity just like the spatial dimensions *with some notable differences*, such as a minus sign instead of a plus sign. The classic Pythagorean theorem for the shortest distance “d” between two points in two dimensions is (in terms of the dstance “x” measured parallel to the X axis and the distance “y” measured parallel to the Y axis):

x[sup]2[/sup] + y[sup]2[/sup] = d[sup]2[/sup]

The extension to three dimensions is obvious:

x[sup]2[/sup] + y[sup]2[/sup] + z[sup]2[/sup]= d[sup]2[/sup]

but the extension to four-dimensional space-time is not intuitive:

(tc)[sup]2[/sup] - (x[sup]2[/sup] + y[sup]2[/sup] + z[sup]2[/sup]) = (sc)[sup]2[/sup]

Because of this difference, we don’t use “d” on the right side of the equals sign, we use “s”. S is the “separation” (IIRC, it’s also often called the “interval”).

So time is a dimension because, in the mathematics, it’s obviously a dimension. But when you start trying do dig deep into the analogy you do get problems, because it’s not exactly analogous to spatial dimensions.

Of course, if you define time in units of the square root of -1, then the minus signs go away …

Then there’s fractal geometry, in which the number of dimensions is seldom and integer …

But it’s all useful, and it works.

Wait…does that make sense? I don’t think so- you have two legs and a hypotneuse for “x[sup]2[/sup] + [sup]y2[/sup] = [sup]d2[sup/]” (c/d=hypotneuse). But what do you have in that other, 3 dimensional equation? The proof used to solve find that equation can’t really be extended to three dimensions. (Besides, it logically doesn’t make sense- if x[sup]2[/sup] + [sup]y2[/sup] = [sup]d2[sup/], then why would x[sup]2[/sup] + [sup]y2[/sup] plus another term still equal the third thing? The actual equaction would be, in a 3D triangle (its name escapes me) with all one 90 degree angle, two thrity degree angles, and…well, it’s not possible to describe without a diagram, but I’ll try anyway: Take a 30-60-90 triangle, and stand it up so that it’s longer leg is on the ground, then do the same with another, congruent, triangle. Lean the two towards eachother so that their hypotneuses are touching, and the following can be said of them:

Ok, after spending quite a while fiddling around with a calculator, I’ve come to the conclusion that there is no algebraically expressable relationship between the 2 hypotneuses and 4 legs (or at least not between those of a 5-12-13 triangle, the one which I used…if anyone wants to solve this, you’re welcome to, but it ain’t easy.) However, I will spend the rest of tonight trying to figure this out, just for you dopers!

**Argeable**, the code for end of superscript is /sup, not sup/. {grin}. The effect of multiple superscripting is certainly amusing …

And you used an … er … *interesting* method of extending Pythogoras to 3D.

Here’s your entire reply, fixed:

I was actually speaking of an abstraction of the Pythagorean theorem, in which the question is “what is the shortest distance between two points in N-dimensional space?”. This can easily be proven with trangles in 3D “spatial dimensions” space.

In 2D, draw horizontal line AB and vertical line BC, with common endpoints at point B and AB perpendicular to BC. The shortest distance between points A and C is line AC, forming a triangle, and you can use the classic Pythagorean theorem to calculate the length of AC:

(AC)[sup]2[/sup] = (AB)[sup]2[/sup] + (BC)[sup]2[/sup]

Now move to 3D and add line CD, also horizontal but perpendicular to both AB and BC and sharing endpoint C with line BC. AB is parallel to the X axis, BC is parallel to the Y axis, and CD is parallel to the Z axis. What is the shortest distance from point A to point D? It’s the length of line AD.

Consider triangle ACD. It is a right triangle. (This is difficult to prove in a text presentation, but we know that BC is perpendicular to CD because we constructed CD that way. Consider rotating line BC around the axis defined by line CD; this does not change the angle between line BC and line CD, and at some point line BC will lie on top of line AC, Therefore the angle between line AC and line CD is a right angle in the plane containing those two lines.)

Therefore, by the classical Pythagorean theorem, the square of the length of line AD is:

(AD)[sup]2[/sup] = (AC)[sup]2[/sup] + (CD}[sup]2[/sup]

Substituting for (AC) from the first expression:

(AD)[sup]2[/sup] = (AB)[sup]2[/sup] + (BC)[sup]2[/sup] + (CD}[sup]2[/sup]

QED.

As you continue to pile on more * spatial* dimensions, you just keep on piling on more terms, one for each diomension. Four dimensions and up is a little more difficult to visualize, though.

However, the interesting stuff starts to happen in four dimensional space-time, because of the minus sign and because of the factor of the speed of light in vacuum ©:

(tc)[sup]2[/sup] - (x[sup]2[/sup] + y[sup]2[/sup] + z[sup]2[/sup]) = (sc)[sup]2[/sup]

(Please don’t ask me to prove that in this medium).

The shortest “distance” between two events in space time is *not* always a straight line in three-dimensional “spatial dimensions” space. There are some pairs of events in four dimensional space-time *for which there is no real number describing the shortest distance*; traveling between such events involves moving faster than the speed of light.

My feeble geometry skills cannot compare. I acceed to your greater (and much less confusing) wisdom in such matters. It all makes sense now…I must do physical labor, so I shall leave you in peace.

Agreable: your extension to 3D doesn’t really address the issue of shortest distance. Here’s another explanation that will perhaps be more understandable than JonF’s.

Imagine a 3d box. You want to know what the distance between to opposite corners is. So imagine a plane that contains both points and is at right angles to the bottom. In other words, it’s a vertical line that makes a diagonal with both the top and the bottom. Now, what’s the length of that diagonal (d1)? Well, it must be that d1^2=x^2+y^2. Now look at the plane. There is a diagonal going from the two opposite corners. This diagonal is the hypotenus of a triangle with a base of d1 and a height of z. So the length of this diagonal (d2) follows the equation d2^2=d1^2+z^2. But remember that d1^2=x^2+y^2. So d2^2=d1^2+z^2=(x^2+y^2)+z^2.

Well, so obviously, Time isn’t a spatial dimension (duh)… which is what I was trying to get at. My reference was towards H.G. Wells’ “The Time Machine”, where the main character said that one was able to move about in time just like one was able to move about in any of the other three dimensions. As much as I loved that book, the flaw in that theory was that Time didn’t act like width, breadth, or height.

However, in more abstract terms, Time definitely counts as a dimension (perhaps we can call them “Existence Dimensions”?). Of course, string theory brings the dimension count up to 11 (not 10 as previously stated WA-A-A-AY up there), so Time being a dimension isn’t exactly that big a deal.

If you want to know what the experts say, then see How many dimensions are there and what are they?.

Nope, I think it’s the crack talking. ::dodging assaults::

I heard it explained this way once, if anyone is still in search of more epiphanies: time is the fourth dimension because in order for something real to exist, physically, it has to have height, width, depth, and have them for a period of time. That is, a dimension is a property required to have physical existence.

As far as I know, this explanation is not used in any scientific or mathematical disciplines, but it’s interesting to think about.

This reminds me of one of my favorite euphemisms. Nasa doesn’t talk about two spaceships “crashing.” Instead, they become “coincident in space/time.”