Can someone explain 4th dimensional objects?
$begingroup$
I'm not sure if I should ask this in mathematics or in physics.
From what I can tell, there are only 3 dimensions: X, Y, and Z. However, I have seen a lot of things about fourth and even fifth dimensional objects. I have tried for a year or two now to wrap my mind around the concept but I never have been able to do so. Can someone please enlighten me?
geometry
$endgroup$
|
show 2 more comments
$begingroup$
I'm not sure if I should ask this in mathematics or in physics.
From what I can tell, there are only 3 dimensions: X, Y, and Z. However, I have seen a lot of things about fourth and even fifth dimensional objects. I have tried for a year or two now to wrap my mind around the concept but I never have been able to do so. Can someone please enlighten me?
geometry
$endgroup$
1
$begingroup$
Note that, in a very literal sense, four dimensions don't fit in the world that you're used to. A fourth dimension would require adding quite a lot of space to the universe. (Inb4 strong theorists: I'm trying to keep this simple.)
$endgroup$
– Akiva Weinberger
Apr 26 '15 at 0:51
17
$begingroup$
I think the book Flatland tries to wrap your head around that.
$endgroup$
– Akiva Weinberger
Apr 26 '15 at 0:53
1
$begingroup$
*string. Whoops.
$endgroup$
– Akiva Weinberger
Apr 26 '15 at 4:08
8
$begingroup$
That's simple. For $mathbb R^n$ and let $n=4$. ;-)
$endgroup$
– Vim
Apr 26 '15 at 7:52
5
$begingroup$
@Vim Well that answers one of OP's questions, as he thinks: "Yep, should've asked on physics.SE" ;)
$endgroup$
– OJFord
Apr 26 '15 at 12:32
|
show 2 more comments
$begingroup$
I'm not sure if I should ask this in mathematics or in physics.
From what I can tell, there are only 3 dimensions: X, Y, and Z. However, I have seen a lot of things about fourth and even fifth dimensional objects. I have tried for a year or two now to wrap my mind around the concept but I never have been able to do so. Can someone please enlighten me?
geometry
$endgroup$
I'm not sure if I should ask this in mathematics or in physics.
From what I can tell, there are only 3 dimensions: X, Y, and Z. However, I have seen a lot of things about fourth and even fifth dimensional objects. I have tried for a year or two now to wrap my mind around the concept but I never have been able to do so. Can someone please enlighten me?
geometry
geometry
asked Apr 25 '15 at 22:14
Dr. OwningDr. Owning
11327
11327
1
$begingroup$
Note that, in a very literal sense, four dimensions don't fit in the world that you're used to. A fourth dimension would require adding quite a lot of space to the universe. (Inb4 strong theorists: I'm trying to keep this simple.)
$endgroup$
– Akiva Weinberger
Apr 26 '15 at 0:51
17
$begingroup$
I think the book Flatland tries to wrap your head around that.
$endgroup$
– Akiva Weinberger
Apr 26 '15 at 0:53
1
$begingroup$
*string. Whoops.
$endgroup$
– Akiva Weinberger
Apr 26 '15 at 4:08
8
$begingroup$
That's simple. For $mathbb R^n$ and let $n=4$. ;-)
$endgroup$
– Vim
Apr 26 '15 at 7:52
5
$begingroup$
@Vim Well that answers one of OP's questions, as he thinks: "Yep, should've asked on physics.SE" ;)
$endgroup$
– OJFord
Apr 26 '15 at 12:32
|
show 2 more comments
1
$begingroup$
Note that, in a very literal sense, four dimensions don't fit in the world that you're used to. A fourth dimension would require adding quite a lot of space to the universe. (Inb4 strong theorists: I'm trying to keep this simple.)
$endgroup$
– Akiva Weinberger
Apr 26 '15 at 0:51
17
$begingroup$
I think the book Flatland tries to wrap your head around that.
$endgroup$
– Akiva Weinberger
Apr 26 '15 at 0:53
1
$begingroup$
*string. Whoops.
$endgroup$
– Akiva Weinberger
Apr 26 '15 at 4:08
8
$begingroup$
That's simple. For $mathbb R^n$ and let $n=4$. ;-)
$endgroup$
– Vim
Apr 26 '15 at 7:52
5
$begingroup$
@Vim Well that answers one of OP's questions, as he thinks: "Yep, should've asked on physics.SE" ;)
$endgroup$
– OJFord
Apr 26 '15 at 12:32
1
1
$begingroup$
Note that, in a very literal sense, four dimensions don't fit in the world that you're used to. A fourth dimension would require adding quite a lot of space to the universe. (Inb4 strong theorists: I'm trying to keep this simple.)
$endgroup$
– Akiva Weinberger
Apr 26 '15 at 0:51
$begingroup$
Note that, in a very literal sense, four dimensions don't fit in the world that you're used to. A fourth dimension would require adding quite a lot of space to the universe. (Inb4 strong theorists: I'm trying to keep this simple.)
$endgroup$
– Akiva Weinberger
Apr 26 '15 at 0:51
17
17
$begingroup$
I think the book Flatland tries to wrap your head around that.
$endgroup$
– Akiva Weinberger
Apr 26 '15 at 0:53
$begingroup$
I think the book Flatland tries to wrap your head around that.
$endgroup$
– Akiva Weinberger
Apr 26 '15 at 0:53
1
1
$begingroup$
*string. Whoops.
$endgroup$
– Akiva Weinberger
Apr 26 '15 at 4:08
$begingroup$
*string. Whoops.
$endgroup$
– Akiva Weinberger
Apr 26 '15 at 4:08
8
8
$begingroup$
That's simple. For $mathbb R^n$ and let $n=4$. ;-)
$endgroup$
– Vim
Apr 26 '15 at 7:52
$begingroup$
That's simple. For $mathbb R^n$ and let $n=4$. ;-)
$endgroup$
– Vim
Apr 26 '15 at 7:52
5
5
$begingroup$
@Vim Well that answers one of OP's questions, as he thinks: "Yep, should've asked on physics.SE" ;)
$endgroup$
– OJFord
Apr 26 '15 at 12:32
$begingroup$
@Vim Well that answers one of OP's questions, as he thinks: "Yep, should've asked on physics.SE" ;)
$endgroup$
– OJFord
Apr 26 '15 at 12:32
|
show 2 more comments
9 Answers
9
active
oldest
votes
$begingroup$
The thing is you're not supposed to "wrap your mind around" higher-dimensional shapes.
Instead, what happens is that we take a formalism that is made to describe 3-dimensional shapes (which we can understand more or less intuitively), and then we just see what happens when we replace all of the "$3$"s in that theory with "$4$" or "$5$" or more. The outcome of this is occasionally useful, but it's not supposed to be about things that can exist in the world -- the utility comes because the theory can be used to reason about certain problems that are not in themselves about shapes.
With a bit of experience one can gain a more or less reliable sense of when one's intuition about 3-dimensional situations gives valid results about higher dimensions, but the touchstone of that is always what one can prove in the formal model we're really speaking of.
$endgroup$
2
$begingroup$
I think the first assertion is wrong. This video, youtu.be/nz0ku71x22A , via the suggestion of Andrey Sokolov in comment on question, shows one way you can think about 4th dimensional shapes.
$endgroup$
– Tom
Apr 26 '15 at 13:25
add a comment |
$begingroup$
There have been people who reportedly can visualize things in four dimensions
as easily as other people can in three. It's rare, however.
Moreover, visualizing four dimensions may not help much when you
want to solve a problem in five dimensions or more.
So as Henning Makholm's answer states, to do anything really useful
in higher dimensions you need a mathematical
model in which you can formally work out the answers.
It is a nice mental exercise to try actually to visualize four-dimensional
objects, however, so I recommend not to stop trying.
One way to do this is to try to "construct" well-shaped
four-dimensional objects.
Consider the following pattern (from http://www.math.union.edu/~dpvc/talks/2000-11-22.funchal/cube-unfolded.html) that folds into a three-dimensional cube:
The pattern itself fits in two dimensions, in a flat plane, but in order
to assemble the cube you have to make parts of the pattern
"pop out" from that plane so that you can join the edges that
need to be joined.
By analogy, the following (from http://im-possible.info/english/articles/hypercube/) is a three-dimensional pattern from which a four-dimensional cube (known as a hypercube or tesseract) might be constructed:
In the assembled tesseract, cubes that are joined only at one edge in this
pattern need to be joined at the faces adjacent to that edge. To do this,
you have to make the cubes "pop out" of three-dimensional space.
The fun part is trying to imagine where the cubes can "pop out" to.
In Robert Heinlein's story, "And He Built a Crooked House," someone builds
a house in this shape and it folds up into a real tesseract with people inside.
Some of the details of the story explore how the rooms become connected in the folded tesseract.
There is at least one line of sight in which someone could see themselves
as if they were some distance away.
$endgroup$
add a comment |
$begingroup$
There are text explanations, so I will post some pictures.
We will try to build some intuition by comparing to zero-, one-, two-, three- and four-dimensional tables/arrays.
Introduction:
What's zero-dimensional array you ask? It is a single cell, as single point is a zero-dimensional space.
Then there is one-dimensional array and it can be split into $n$ one-less-dimensional strucutres. Similarly a segment can be split into infinitely many points.
When we go into second dimension, we can split in different ways: either rows of columns first. Still, whatever way we do it, a 2D array is still a collection of $n$ 1D arrays, which can be split up into $n^2$ two-less-dimensional structures.
This pattern holds in three dimensions, where we can split and split and split our structures. In other words, a 3D array is a collection of $n$ 2D arrays, or a collection of $n^2$ 1D arrays, etc.
That is, we can think of 3D array as 1D array of 2D arrays...
... or 2D array of 1D arrays, etc.
The fourth dimension:
Now, let's go up one dimension:
This is a 1D array of 3D array, or (1+3) = 4D structure. It's hard to imagine visually, but you could try scaling the cubes into the inside (and ignore that tiny little voice which says "such overlaps cannot happen in reality"). Another way to think about it is to mentally remove the inside of the cube and leave only the shell (6 faces), then such a thing can be scaled and combined with no overlaps. Trying with wireframes only it looks like this:
Or we could move it to the side:
However, in 4D we can do yet something else, namely 2D array of 2D arrays:
This is also (2+2) = 4D structure, and you can think of it as a product of two squares (with full insides). That would also be a bit hard to imagine, so to make it simple, take squares with no insides (only the edges). To further boost our intuition, observe that product of two circles gives you a torus. The wireframe for the 4D cube will look a bit alike, although remember that these are two very different objects.
I hope this helps $ddotsmile$
$endgroup$
add a comment |
$begingroup$
David Foster Wallace in everything and more, a compact history of ∞:
There is something I "know," which is that spatial dimensions beyond the Big 3 exist. I can even construct a tesseract or a hypercube out of cardboard. A weird sort of cube-within-a-cube, a tesseract is a 3D projection of a 4D object in the same way that "" is a 2D projection of a 3D object. The trick is imagining the tesseract's relevant lines and planes at 90° to each other (it's the same with "" and a real cube), because the 4th spatial dimension is on that somehow exists at perfect right angles to the length, width, and depth of our regular visual field. I "know" all this, just as you probably do . . . but now try to really picture it. Concretely. You can feel, almost immediately, a strain at the very root of yourself, the first popped threads of a mind starting to give at the seams.
$endgroup$
3
$begingroup$
This was supposed to be a comment to Henning Makholm's nice (and accurate) answer, but it was too long, and I needed the cube drawings as well. If anyone know of a 3D cube unicode symbol, I'd be very happy to hear of it.
$endgroup$
– Pål GD
Apr 26 '15 at 3:43
add a comment |
$begingroup$
Dimension usually is just the number of 'components' of some piece of information. 3 dimensions are just nice for describing a position in (Euclidean) space, but you definitely need 4 dimensions if you want to include the time also. Now you are in the room. A while later you are not. Your position has changed over time, so if we want to describe your path, we need to describe each point on the path, and each point consists of both 3 space coordinates and 1 time coordinate, so this path is 4-dimensional.
In mathematics we have a general kind of space called a vector space. The 3d Euclidean space described above is $mathbb{R}^3$, which is simply the set of all vectors with $3$ real coordinates. Spacetime could be represented as $mathbb{R}^4$. Time is special in the sense that we can only move forward in time whereas we can move in any direction in space, but this has nothing to do with dimension.
In a vector space, we would like to represent each vector as a combination of some basic vectors. Of course we want to have as few basic vectors as possible. For example $mathbb{R}^3$ has a basis with 3 vectors, ${(1,0,0),(0,1,0),(0,0,1)}$, which you could imagine stands for "1 unit in front of you", "1 unit above you" and "1 unit to your right", if you happen to stand at the centre $(0,0,0)$. Any position in space can indeed by expressed as a linear combination "go $x$ units forward, then $y$ units up, then $z$ units right". And you cannot make do with less. So we say that $mathbb{R}^3$ has dimension $3$. In general $mathbb{R}^n$ has dimension $n$.
A general vector space $V$ is defined over a field $F$ (a field permits $+,-,times,div$ except division by zero). A linear combination is defined to mean a weighted sum of finitely many vectors, for example $2x+(-3)y+5z$ is a linear combination of $x,y,z$. An $F$-linear combination is defined as a linear combination with weights in $F$. A spanning set $S$ for $V$ over $F$ is defined to be a set of vectors such that every vector in $V$ can be written as a $F$-linear combination of finitely many vectors from $S$, that is, for any vector $v in V$, we have $v = a_1 x_1 + a_2 x_2 + ... + a_n x_n$ for some $a_1,a_2,...,a_n in F$ and $x_1,x_2,...,x_n in S$. Then we can define a basis for $V$ over $F$ to mean a spanning set with minimum size and the dimension of $V$ over $F$ to be that size.
For another example, a displacement in 3d space requires 6 real values to describe it. The displacement involves both position and orientation. 3 real values are needed for the translation (change in position), and 3 real values are need for the rotation (change in orientation). The most frequently used representation is often called the 6 degrees of freedom in motion, which are up/down, left/right, forward/backward, pitch, yaw, roll. Note that you can represent any displacement using this 6 real values, and conversely any combination of these 6 values gives a unique displacement. But given two displacements, combining them cannot be computed by just adding the corresponding values. It adds for translation but not for rotation. In fact, the order of rotations in 3d space makes a difference! So this vector space structure does not fully reflect the structure of motion. What would be needed for this is transformations on $mathbb{R}^3$, which can be represented by matrices, but that is another topic.
$endgroup$
add a comment |
$begingroup$
Multidimensional spaces occur naturally all around us. Take the possible positions of your arm, for example.
- You can rotate your shoulder with two degrees of freedom;
- You can bend your elbow with one degree of freedom;
- You can rotate your wrist with two degrees of freedom
That's a total of five degrees of freedom for the possible positions of your arm.
In other words, the space of positions of your arm, a space with which you are personally familiar on an intimate level, is a 5-dimensional object.
And I'm ignoring the fact that twisting your forearm and flexing your hand bones and bending your finger bones adds at least another twenty or more dimensions.
$endgroup$
add a comment |
$begingroup$
I am a firm believer in the idea that if something in math doesn't make sense to you, you just haven't found the proper way of thinking about it yet. And in some (or maybe most?) cases, there isn't anybody on earth who has found the right way. And this way is probably different for different people.
Now, to answer your question, I will tell you how I think of higher dimensional objects, as someone who prefers to visualize everything. Some of the other answers describe how from one $(x)$ to two $(x,y)$ dimensions we just "add a perpendicular line" to everything and call something in that new direction $y$. If that works that is fine. It makes sense for the transition from 2 to 3 dimensions also, for most people. When we go from 3 to 4, we have to get creative. Instead of a "adding a new line", I usually think about adding the color spectrum to each point in 3 space. So, at each point, we have $(x,y,z)$ just like in normal 3 space, and lets call it $c$, for color. To draw points in 4 space, now you need to place it in 3 space and also assign some color to it, to tell us where in the spectrum it is.
A line is the same way, but at each point in the line, we can pick a color. If the color varies continuously through the spectrum, the line is connected. But an abrupt change from red to green, for example, is like having a break in the line. In the picture below, the green line lives in 3 space, since it only uses one color, the rainbow line travels through the whole spectrum. Also, notice that the rainbow line in three dimensions crosses itself. But it doesnt actually intersect because it has two different colors.
So, this rainbow line lives in 4 dimensions right now, and the green line is only using 3 dimensions. This analogy still has drawbacks though. The three directional unit vectors make sense, but when you try to draw the new color one, you find yourself with "line" of colors, but in the 3 dimesions we are looking at, it is only at the origin, so it looks like a point to us.
One last example, and this might be a stretch: The 3-sphere, which lives in 4 dimensions. First a quick idea of how we make this. Just like in flatland, when the sphere (which in topology we call the 2-sphere) comes to visit, it shows up a a circle (or 1-sphere), picutred below. But it has a lot of circles of different size, depending on where the plane is intersecting the sphere. A second sphere is higher than the first, so it intersects a smaller circle in the plane.
So when the 3-sphere comes to 3 dimensions, we will only see it as a normal sphere (2-sphere). If that was confusing, I apologize. Here are some pictures where we get the normal unit sphere when we are green, which I am equating to $c=0$. but when we change colors, our spheres will get smaller and when the are red and violet, they will only be a point. That is because the color is "farther" from green, which we are thinking of as the 3 space we live in, and we need to satisfy the equation $x^2+y^2+z^2+c^2=1$. Well, I hope this helps.
$endgroup$
add a comment |
$begingroup$
When somebody says a statement about the fourth dimension that you think is clearly false such as the fact that a closed string has only one possible knot, the unknot, the usual way to interpret what they're saying is a statement that is false. You're right that it's technically false to say that a string cannot be knotted. What they're really doing is thinking of a totally different statement and using that English sentence to mean something the sentence doesn't actually mean.
Take $mathbb{R}^4$, the set of all ordered quadruplets of real numbers. Now, we can invent a definition of the distance from any member of $mathbb{R}^4$ to any other member of $mathbb{R}^4$ as follows. For any points in $mathbb{R}^4$, $(w_1, x_1, y_1, z_1)$ and $(w_2, x_2, y_2, z_2)$, the distance from $(w_1, x_1, y_1, z_1)$ to $(w_2, x_2, y_2, z_2)$ is $sqrt{(w_2 - w_1)^2 + (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$. We can invent a binary operation * on that space that functions essentially the same as quaternion multiplication. Let's define a function $f$ from $mathbb{R}^4$ to itself to be an origin rotation if and only if $exists y in mathbb{R}^4exists z in mathbb{R}^4$ such that distance from the origin to $y$ and the distance from the origin to $z$ are 1 and $forall x in mathbb{R}^4, f(x) = y * x * z$. Let's define any rotation in $mathbb{R}$ to be an operation such that there exists a translation $T$ and an origin rotation $f$ such that it can be expressed as $T circ f circ T^{-1}$. Although this definition of a rotation works only for $mathbb{R}^4$, we can show that it satisfies all the basic intuitive properties that the set of all rotations in the third dimension satisfies.
$endgroup$
add a comment |
$begingroup$
This man can ! https://www.youtube.com/watch?v=M9sbdrPVfOQ :)
But I never understand I think we have to have a brain in 4D to understand.
You can also see that :
A conference really interesting Here
And a short video Here
$endgroup$
add a comment |
protected by Zev Chonoles May 15 '15 at 15:52
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
9 Answers
9
active
oldest
votes
9 Answers
9
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The thing is you're not supposed to "wrap your mind around" higher-dimensional shapes.
Instead, what happens is that we take a formalism that is made to describe 3-dimensional shapes (which we can understand more or less intuitively), and then we just see what happens when we replace all of the "$3$"s in that theory with "$4$" or "$5$" or more. The outcome of this is occasionally useful, but it's not supposed to be about things that can exist in the world -- the utility comes because the theory can be used to reason about certain problems that are not in themselves about shapes.
With a bit of experience one can gain a more or less reliable sense of when one's intuition about 3-dimensional situations gives valid results about higher dimensions, but the touchstone of that is always what one can prove in the formal model we're really speaking of.
$endgroup$
2
$begingroup$
I think the first assertion is wrong. This video, youtu.be/nz0ku71x22A , via the suggestion of Andrey Sokolov in comment on question, shows one way you can think about 4th dimensional shapes.
$endgroup$
– Tom
Apr 26 '15 at 13:25
add a comment |
$begingroup$
The thing is you're not supposed to "wrap your mind around" higher-dimensional shapes.
Instead, what happens is that we take a formalism that is made to describe 3-dimensional shapes (which we can understand more or less intuitively), and then we just see what happens when we replace all of the "$3$"s in that theory with "$4$" or "$5$" or more. The outcome of this is occasionally useful, but it's not supposed to be about things that can exist in the world -- the utility comes because the theory can be used to reason about certain problems that are not in themselves about shapes.
With a bit of experience one can gain a more or less reliable sense of when one's intuition about 3-dimensional situations gives valid results about higher dimensions, but the touchstone of that is always what one can prove in the formal model we're really speaking of.
$endgroup$
2
$begingroup$
I think the first assertion is wrong. This video, youtu.be/nz0ku71x22A , via the suggestion of Andrey Sokolov in comment on question, shows one way you can think about 4th dimensional shapes.
$endgroup$
– Tom
Apr 26 '15 at 13:25
add a comment |
$begingroup$
The thing is you're not supposed to "wrap your mind around" higher-dimensional shapes.
Instead, what happens is that we take a formalism that is made to describe 3-dimensional shapes (which we can understand more or less intuitively), and then we just see what happens when we replace all of the "$3$"s in that theory with "$4$" or "$5$" or more. The outcome of this is occasionally useful, but it's not supposed to be about things that can exist in the world -- the utility comes because the theory can be used to reason about certain problems that are not in themselves about shapes.
With a bit of experience one can gain a more or less reliable sense of when one's intuition about 3-dimensional situations gives valid results about higher dimensions, but the touchstone of that is always what one can prove in the formal model we're really speaking of.
$endgroup$
The thing is you're not supposed to "wrap your mind around" higher-dimensional shapes.
Instead, what happens is that we take a formalism that is made to describe 3-dimensional shapes (which we can understand more or less intuitively), and then we just see what happens when we replace all of the "$3$"s in that theory with "$4$" or "$5$" or more. The outcome of this is occasionally useful, but it's not supposed to be about things that can exist in the world -- the utility comes because the theory can be used to reason about certain problems that are not in themselves about shapes.
With a bit of experience one can gain a more or less reliable sense of when one's intuition about 3-dimensional situations gives valid results about higher dimensions, but the touchstone of that is always what one can prove in the formal model we're really speaking of.
answered Apr 25 '15 at 22:25
Henning MakholmHenning Makholm
239k17304541
239k17304541
2
$begingroup$
I think the first assertion is wrong. This video, youtu.be/nz0ku71x22A , via the suggestion of Andrey Sokolov in comment on question, shows one way you can think about 4th dimensional shapes.
$endgroup$
– Tom
Apr 26 '15 at 13:25
add a comment |
2
$begingroup$
I think the first assertion is wrong. This video, youtu.be/nz0ku71x22A , via the suggestion of Andrey Sokolov in comment on question, shows one way you can think about 4th dimensional shapes.
$endgroup$
– Tom
Apr 26 '15 at 13:25
2
2
$begingroup$
I think the first assertion is wrong. This video, youtu.be/nz0ku71x22A , via the suggestion of Andrey Sokolov in comment on question, shows one way you can think about 4th dimensional shapes.
$endgroup$
– Tom
Apr 26 '15 at 13:25
$begingroup$
I think the first assertion is wrong. This video, youtu.be/nz0ku71x22A , via the suggestion of Andrey Sokolov in comment on question, shows one way you can think about 4th dimensional shapes.
$endgroup$
– Tom
Apr 26 '15 at 13:25
add a comment |
$begingroup$
There have been people who reportedly can visualize things in four dimensions
as easily as other people can in three. It's rare, however.
Moreover, visualizing four dimensions may not help much when you
want to solve a problem in five dimensions or more.
So as Henning Makholm's answer states, to do anything really useful
in higher dimensions you need a mathematical
model in which you can formally work out the answers.
It is a nice mental exercise to try actually to visualize four-dimensional
objects, however, so I recommend not to stop trying.
One way to do this is to try to "construct" well-shaped
four-dimensional objects.
Consider the following pattern (from http://www.math.union.edu/~dpvc/talks/2000-11-22.funchal/cube-unfolded.html) that folds into a three-dimensional cube:
The pattern itself fits in two dimensions, in a flat plane, but in order
to assemble the cube you have to make parts of the pattern
"pop out" from that plane so that you can join the edges that
need to be joined.
By analogy, the following (from http://im-possible.info/english/articles/hypercube/) is a three-dimensional pattern from which a four-dimensional cube (known as a hypercube or tesseract) might be constructed:
In the assembled tesseract, cubes that are joined only at one edge in this
pattern need to be joined at the faces adjacent to that edge. To do this,
you have to make the cubes "pop out" of three-dimensional space.
The fun part is trying to imagine where the cubes can "pop out" to.
In Robert Heinlein's story, "And He Built a Crooked House," someone builds
a house in this shape and it folds up into a real tesseract with people inside.
Some of the details of the story explore how the rooms become connected in the folded tesseract.
There is at least one line of sight in which someone could see themselves
as if they were some distance away.
$endgroup$
add a comment |
$begingroup$
There have been people who reportedly can visualize things in four dimensions
as easily as other people can in three. It's rare, however.
Moreover, visualizing four dimensions may not help much when you
want to solve a problem in five dimensions or more.
So as Henning Makholm's answer states, to do anything really useful
in higher dimensions you need a mathematical
model in which you can formally work out the answers.
It is a nice mental exercise to try actually to visualize four-dimensional
objects, however, so I recommend not to stop trying.
One way to do this is to try to "construct" well-shaped
four-dimensional objects.
Consider the following pattern (from http://www.math.union.edu/~dpvc/talks/2000-11-22.funchal/cube-unfolded.html) that folds into a three-dimensional cube:
The pattern itself fits in two dimensions, in a flat plane, but in order
to assemble the cube you have to make parts of the pattern
"pop out" from that plane so that you can join the edges that
need to be joined.
By analogy, the following (from http://im-possible.info/english/articles/hypercube/) is a three-dimensional pattern from which a four-dimensional cube (known as a hypercube or tesseract) might be constructed:
In the assembled tesseract, cubes that are joined only at one edge in this
pattern need to be joined at the faces adjacent to that edge. To do this,
you have to make the cubes "pop out" of three-dimensional space.
The fun part is trying to imagine where the cubes can "pop out" to.
In Robert Heinlein's story, "And He Built a Crooked House," someone builds
a house in this shape and it folds up into a real tesseract with people inside.
Some of the details of the story explore how the rooms become connected in the folded tesseract.
There is at least one line of sight in which someone could see themselves
as if they were some distance away.
$endgroup$
add a comment |
$begingroup$
There have been people who reportedly can visualize things in four dimensions
as easily as other people can in three. It's rare, however.
Moreover, visualizing four dimensions may not help much when you
want to solve a problem in five dimensions or more.
So as Henning Makholm's answer states, to do anything really useful
in higher dimensions you need a mathematical
model in which you can formally work out the answers.
It is a nice mental exercise to try actually to visualize four-dimensional
objects, however, so I recommend not to stop trying.
One way to do this is to try to "construct" well-shaped
four-dimensional objects.
Consider the following pattern (from http://www.math.union.edu/~dpvc/talks/2000-11-22.funchal/cube-unfolded.html) that folds into a three-dimensional cube:
The pattern itself fits in two dimensions, in a flat plane, but in order
to assemble the cube you have to make parts of the pattern
"pop out" from that plane so that you can join the edges that
need to be joined.
By analogy, the following (from http://im-possible.info/english/articles/hypercube/) is a three-dimensional pattern from which a four-dimensional cube (known as a hypercube or tesseract) might be constructed:
In the assembled tesseract, cubes that are joined only at one edge in this
pattern need to be joined at the faces adjacent to that edge. To do this,
you have to make the cubes "pop out" of three-dimensional space.
The fun part is trying to imagine where the cubes can "pop out" to.
In Robert Heinlein's story, "And He Built a Crooked House," someone builds
a house in this shape and it folds up into a real tesseract with people inside.
Some of the details of the story explore how the rooms become connected in the folded tesseract.
There is at least one line of sight in which someone could see themselves
as if they were some distance away.
$endgroup$
There have been people who reportedly can visualize things in four dimensions
as easily as other people can in three. It's rare, however.
Moreover, visualizing four dimensions may not help much when you
want to solve a problem in five dimensions or more.
So as Henning Makholm's answer states, to do anything really useful
in higher dimensions you need a mathematical
model in which you can formally work out the answers.
It is a nice mental exercise to try actually to visualize four-dimensional
objects, however, so I recommend not to stop trying.
One way to do this is to try to "construct" well-shaped
four-dimensional objects.
Consider the following pattern (from http://www.math.union.edu/~dpvc/talks/2000-11-22.funchal/cube-unfolded.html) that folds into a three-dimensional cube:
The pattern itself fits in two dimensions, in a flat plane, but in order
to assemble the cube you have to make parts of the pattern
"pop out" from that plane so that you can join the edges that
need to be joined.
By analogy, the following (from http://im-possible.info/english/articles/hypercube/) is a three-dimensional pattern from which a four-dimensional cube (known as a hypercube or tesseract) might be constructed:
In the assembled tesseract, cubes that are joined only at one edge in this
pattern need to be joined at the faces adjacent to that edge. To do this,
you have to make the cubes "pop out" of three-dimensional space.
The fun part is trying to imagine where the cubes can "pop out" to.
In Robert Heinlein's story, "And He Built a Crooked House," someone builds
a house in this shape and it folds up into a real tesseract with people inside.
Some of the details of the story explore how the rooms become connected in the folded tesseract.
There is at least one line of sight in which someone could see themselves
as if they were some distance away.
answered Apr 26 '15 at 12:09
David KDavid K
53.1k341115
53.1k341115
add a comment |
add a comment |
$begingroup$
There are text explanations, so I will post some pictures.
We will try to build some intuition by comparing to zero-, one-, two-, three- and four-dimensional tables/arrays.
Introduction:
What's zero-dimensional array you ask? It is a single cell, as single point is a zero-dimensional space.
Then there is one-dimensional array and it can be split into $n$ one-less-dimensional strucutres. Similarly a segment can be split into infinitely many points.
When we go into second dimension, we can split in different ways: either rows of columns first. Still, whatever way we do it, a 2D array is still a collection of $n$ 1D arrays, which can be split up into $n^2$ two-less-dimensional structures.
This pattern holds in three dimensions, where we can split and split and split our structures. In other words, a 3D array is a collection of $n$ 2D arrays, or a collection of $n^2$ 1D arrays, etc.
That is, we can think of 3D array as 1D array of 2D arrays...
... or 2D array of 1D arrays, etc.
The fourth dimension:
Now, let's go up one dimension:
This is a 1D array of 3D array, or (1+3) = 4D structure. It's hard to imagine visually, but you could try scaling the cubes into the inside (and ignore that tiny little voice which says "such overlaps cannot happen in reality"). Another way to think about it is to mentally remove the inside of the cube and leave only the shell (6 faces), then such a thing can be scaled and combined with no overlaps. Trying with wireframes only it looks like this:
Or we could move it to the side:
However, in 4D we can do yet something else, namely 2D array of 2D arrays:
This is also (2+2) = 4D structure, and you can think of it as a product of two squares (with full insides). That would also be a bit hard to imagine, so to make it simple, take squares with no insides (only the edges). To further boost our intuition, observe that product of two circles gives you a torus. The wireframe for the 4D cube will look a bit alike, although remember that these are two very different objects.
I hope this helps $ddotsmile$
$endgroup$
add a comment |
$begingroup$
There are text explanations, so I will post some pictures.
We will try to build some intuition by comparing to zero-, one-, two-, three- and four-dimensional tables/arrays.
Introduction:
What's zero-dimensional array you ask? It is a single cell, as single point is a zero-dimensional space.
Then there is one-dimensional array and it can be split into $n$ one-less-dimensional strucutres. Similarly a segment can be split into infinitely many points.
When we go into second dimension, we can split in different ways: either rows of columns first. Still, whatever way we do it, a 2D array is still a collection of $n$ 1D arrays, which can be split up into $n^2$ two-less-dimensional structures.
This pattern holds in three dimensions, where we can split and split and split our structures. In other words, a 3D array is a collection of $n$ 2D arrays, or a collection of $n^2$ 1D arrays, etc.
That is, we can think of 3D array as 1D array of 2D arrays...
... or 2D array of 1D arrays, etc.
The fourth dimension:
Now, let's go up one dimension:
This is a 1D array of 3D array, or (1+3) = 4D structure. It's hard to imagine visually, but you could try scaling the cubes into the inside (and ignore that tiny little voice which says "such overlaps cannot happen in reality"). Another way to think about it is to mentally remove the inside of the cube and leave only the shell (6 faces), then such a thing can be scaled and combined with no overlaps. Trying with wireframes only it looks like this:
Or we could move it to the side:
However, in 4D we can do yet something else, namely 2D array of 2D arrays:
This is also (2+2) = 4D structure, and you can think of it as a product of two squares (with full insides). That would also be a bit hard to imagine, so to make it simple, take squares with no insides (only the edges). To further boost our intuition, observe that product of two circles gives you a torus. The wireframe for the 4D cube will look a bit alike, although remember that these are two very different objects.
I hope this helps $ddotsmile$
$endgroup$
add a comment |
$begingroup$
There are text explanations, so I will post some pictures.
We will try to build some intuition by comparing to zero-, one-, two-, three- and four-dimensional tables/arrays.
Introduction:
What's zero-dimensional array you ask? It is a single cell, as single point is a zero-dimensional space.
Then there is one-dimensional array and it can be split into $n$ one-less-dimensional strucutres. Similarly a segment can be split into infinitely many points.
When we go into second dimension, we can split in different ways: either rows of columns first. Still, whatever way we do it, a 2D array is still a collection of $n$ 1D arrays, which can be split up into $n^2$ two-less-dimensional structures.
This pattern holds in three dimensions, where we can split and split and split our structures. In other words, a 3D array is a collection of $n$ 2D arrays, or a collection of $n^2$ 1D arrays, etc.
That is, we can think of 3D array as 1D array of 2D arrays...
... or 2D array of 1D arrays, etc.
The fourth dimension:
Now, let's go up one dimension:
This is a 1D array of 3D array, or (1+3) = 4D structure. It's hard to imagine visually, but you could try scaling the cubes into the inside (and ignore that tiny little voice which says "such overlaps cannot happen in reality"). Another way to think about it is to mentally remove the inside of the cube and leave only the shell (6 faces), then such a thing can be scaled and combined with no overlaps. Trying with wireframes only it looks like this:
Or we could move it to the side:
However, in 4D we can do yet something else, namely 2D array of 2D arrays:
This is also (2+2) = 4D structure, and you can think of it as a product of two squares (with full insides). That would also be a bit hard to imagine, so to make it simple, take squares with no insides (only the edges). To further boost our intuition, observe that product of two circles gives you a torus. The wireframe for the 4D cube will look a bit alike, although remember that these are two very different objects.
I hope this helps $ddotsmile$
$endgroup$
There are text explanations, so I will post some pictures.
We will try to build some intuition by comparing to zero-, one-, two-, three- and four-dimensional tables/arrays.
Introduction:
What's zero-dimensional array you ask? It is a single cell, as single point is a zero-dimensional space.
Then there is one-dimensional array and it can be split into $n$ one-less-dimensional strucutres. Similarly a segment can be split into infinitely many points.
When we go into second dimension, we can split in different ways: either rows of columns first. Still, whatever way we do it, a 2D array is still a collection of $n$ 1D arrays, which can be split up into $n^2$ two-less-dimensional structures.
This pattern holds in three dimensions, where we can split and split and split our structures. In other words, a 3D array is a collection of $n$ 2D arrays, or a collection of $n^2$ 1D arrays, etc.
That is, we can think of 3D array as 1D array of 2D arrays...
... or 2D array of 1D arrays, etc.
The fourth dimension:
Now, let's go up one dimension:
This is a 1D array of 3D array, or (1+3) = 4D structure. It's hard to imagine visually, but you could try scaling the cubes into the inside (and ignore that tiny little voice which says "such overlaps cannot happen in reality"). Another way to think about it is to mentally remove the inside of the cube and leave only the shell (6 faces), then such a thing can be scaled and combined with no overlaps. Trying with wireframes only it looks like this:
Or we could move it to the side:
However, in 4D we can do yet something else, namely 2D array of 2D arrays:
This is also (2+2) = 4D structure, and you can think of it as a product of two squares (with full insides). That would also be a bit hard to imagine, so to make it simple, take squares with no insides (only the edges). To further boost our intuition, observe that product of two circles gives you a torus. The wireframe for the 4D cube will look a bit alike, although remember that these are two very different objects.
I hope this helps $ddotsmile$
answered Apr 27 '15 at 14:40
dtldarekdtldarek
32.2k743100
32.2k743100
add a comment |
add a comment |
$begingroup$
David Foster Wallace in everything and more, a compact history of ∞:
There is something I "know," which is that spatial dimensions beyond the Big 3 exist. I can even construct a tesseract or a hypercube out of cardboard. A weird sort of cube-within-a-cube, a tesseract is a 3D projection of a 4D object in the same way that "" is a 2D projection of a 3D object. The trick is imagining the tesseract's relevant lines and planes at 90° to each other (it's the same with "" and a real cube), because the 4th spatial dimension is on that somehow exists at perfect right angles to the length, width, and depth of our regular visual field. I "know" all this, just as you probably do . . . but now try to really picture it. Concretely. You can feel, almost immediately, a strain at the very root of yourself, the first popped threads of a mind starting to give at the seams.
$endgroup$
3
$begingroup$
This was supposed to be a comment to Henning Makholm's nice (and accurate) answer, but it was too long, and I needed the cube drawings as well. If anyone know of a 3D cube unicode symbol, I'd be very happy to hear of it.
$endgroup$
– Pål GD
Apr 26 '15 at 3:43
add a comment |
$begingroup$
David Foster Wallace in everything and more, a compact history of ∞:
There is something I "know," which is that spatial dimensions beyond the Big 3 exist. I can even construct a tesseract or a hypercube out of cardboard. A weird sort of cube-within-a-cube, a tesseract is a 3D projection of a 4D object in the same way that "" is a 2D projection of a 3D object. The trick is imagining the tesseract's relevant lines and planes at 90° to each other (it's the same with "" and a real cube), because the 4th spatial dimension is on that somehow exists at perfect right angles to the length, width, and depth of our regular visual field. I "know" all this, just as you probably do . . . but now try to really picture it. Concretely. You can feel, almost immediately, a strain at the very root of yourself, the first popped threads of a mind starting to give at the seams.
$endgroup$
3
$begingroup$
This was supposed to be a comment to Henning Makholm's nice (and accurate) answer, but it was too long, and I needed the cube drawings as well. If anyone know of a 3D cube unicode symbol, I'd be very happy to hear of it.
$endgroup$
– Pål GD
Apr 26 '15 at 3:43
add a comment |
$begingroup$
David Foster Wallace in everything and more, a compact history of ∞:
There is something I "know," which is that spatial dimensions beyond the Big 3 exist. I can even construct a tesseract or a hypercube out of cardboard. A weird sort of cube-within-a-cube, a tesseract is a 3D projection of a 4D object in the same way that "" is a 2D projection of a 3D object. The trick is imagining the tesseract's relevant lines and planes at 90° to each other (it's the same with "" and a real cube), because the 4th spatial dimension is on that somehow exists at perfect right angles to the length, width, and depth of our regular visual field. I "know" all this, just as you probably do . . . but now try to really picture it. Concretely. You can feel, almost immediately, a strain at the very root of yourself, the first popped threads of a mind starting to give at the seams.
$endgroup$
David Foster Wallace in everything and more, a compact history of ∞:
There is something I "know," which is that spatial dimensions beyond the Big 3 exist. I can even construct a tesseract or a hypercube out of cardboard. A weird sort of cube-within-a-cube, a tesseract is a 3D projection of a 4D object in the same way that "" is a 2D projection of a 3D object. The trick is imagining the tesseract's relevant lines and planes at 90° to each other (it's the same with "" and a real cube), because the 4th spatial dimension is on that somehow exists at perfect right angles to the length, width, and depth of our regular visual field. I "know" all this, just as you probably do . . . but now try to really picture it. Concretely. You can feel, almost immediately, a strain at the very root of yourself, the first popped threads of a mind starting to give at the seams.
answered Apr 26 '15 at 3:41
Pål GDPål GD
652413
652413
3
$begingroup$
This was supposed to be a comment to Henning Makholm's nice (and accurate) answer, but it was too long, and I needed the cube drawings as well. If anyone know of a 3D cube unicode symbol, I'd be very happy to hear of it.
$endgroup$
– Pål GD
Apr 26 '15 at 3:43
add a comment |
3
$begingroup$
This was supposed to be a comment to Henning Makholm's nice (and accurate) answer, but it was too long, and I needed the cube drawings as well. If anyone know of a 3D cube unicode symbol, I'd be very happy to hear of it.
$endgroup$
– Pål GD
Apr 26 '15 at 3:43
3
3
$begingroup$
This was supposed to be a comment to Henning Makholm's nice (and accurate) answer, but it was too long, and I needed the cube drawings as well. If anyone know of a 3D cube unicode symbol, I'd be very happy to hear of it.
$endgroup$
– Pål GD
Apr 26 '15 at 3:43
$begingroup$
This was supposed to be a comment to Henning Makholm's nice (and accurate) answer, but it was too long, and I needed the cube drawings as well. If anyone know of a 3D cube unicode symbol, I'd be very happy to hear of it.
$endgroup$
– Pål GD
Apr 26 '15 at 3:43
add a comment |
$begingroup$
Dimension usually is just the number of 'components' of some piece of information. 3 dimensions are just nice for describing a position in (Euclidean) space, but you definitely need 4 dimensions if you want to include the time also. Now you are in the room. A while later you are not. Your position has changed over time, so if we want to describe your path, we need to describe each point on the path, and each point consists of both 3 space coordinates and 1 time coordinate, so this path is 4-dimensional.
In mathematics we have a general kind of space called a vector space. The 3d Euclidean space described above is $mathbb{R}^3$, which is simply the set of all vectors with $3$ real coordinates. Spacetime could be represented as $mathbb{R}^4$. Time is special in the sense that we can only move forward in time whereas we can move in any direction in space, but this has nothing to do with dimension.
In a vector space, we would like to represent each vector as a combination of some basic vectors. Of course we want to have as few basic vectors as possible. For example $mathbb{R}^3$ has a basis with 3 vectors, ${(1,0,0),(0,1,0),(0,0,1)}$, which you could imagine stands for "1 unit in front of you", "1 unit above you" and "1 unit to your right", if you happen to stand at the centre $(0,0,0)$. Any position in space can indeed by expressed as a linear combination "go $x$ units forward, then $y$ units up, then $z$ units right". And you cannot make do with less. So we say that $mathbb{R}^3$ has dimension $3$. In general $mathbb{R}^n$ has dimension $n$.
A general vector space $V$ is defined over a field $F$ (a field permits $+,-,times,div$ except division by zero). A linear combination is defined to mean a weighted sum of finitely many vectors, for example $2x+(-3)y+5z$ is a linear combination of $x,y,z$. An $F$-linear combination is defined as a linear combination with weights in $F$. A spanning set $S$ for $V$ over $F$ is defined to be a set of vectors such that every vector in $V$ can be written as a $F$-linear combination of finitely many vectors from $S$, that is, for any vector $v in V$, we have $v = a_1 x_1 + a_2 x_2 + ... + a_n x_n$ for some $a_1,a_2,...,a_n in F$ and $x_1,x_2,...,x_n in S$. Then we can define a basis for $V$ over $F$ to mean a spanning set with minimum size and the dimension of $V$ over $F$ to be that size.
For another example, a displacement in 3d space requires 6 real values to describe it. The displacement involves both position and orientation. 3 real values are needed for the translation (change in position), and 3 real values are need for the rotation (change in orientation). The most frequently used representation is often called the 6 degrees of freedom in motion, which are up/down, left/right, forward/backward, pitch, yaw, roll. Note that you can represent any displacement using this 6 real values, and conversely any combination of these 6 values gives a unique displacement. But given two displacements, combining them cannot be computed by just adding the corresponding values. It adds for translation but not for rotation. In fact, the order of rotations in 3d space makes a difference! So this vector space structure does not fully reflect the structure of motion. What would be needed for this is transformations on $mathbb{R}^3$, which can be represented by matrices, but that is another topic.
$endgroup$
add a comment |
$begingroup$
Dimension usually is just the number of 'components' of some piece of information. 3 dimensions are just nice for describing a position in (Euclidean) space, but you definitely need 4 dimensions if you want to include the time also. Now you are in the room. A while later you are not. Your position has changed over time, so if we want to describe your path, we need to describe each point on the path, and each point consists of both 3 space coordinates and 1 time coordinate, so this path is 4-dimensional.
In mathematics we have a general kind of space called a vector space. The 3d Euclidean space described above is $mathbb{R}^3$, which is simply the set of all vectors with $3$ real coordinates. Spacetime could be represented as $mathbb{R}^4$. Time is special in the sense that we can only move forward in time whereas we can move in any direction in space, but this has nothing to do with dimension.
In a vector space, we would like to represent each vector as a combination of some basic vectors. Of course we want to have as few basic vectors as possible. For example $mathbb{R}^3$ has a basis with 3 vectors, ${(1,0,0),(0,1,0),(0,0,1)}$, which you could imagine stands for "1 unit in front of you", "1 unit above you" and "1 unit to your right", if you happen to stand at the centre $(0,0,0)$. Any position in space can indeed by expressed as a linear combination "go $x$ units forward, then $y$ units up, then $z$ units right". And you cannot make do with less. So we say that $mathbb{R}^3$ has dimension $3$. In general $mathbb{R}^n$ has dimension $n$.
A general vector space $V$ is defined over a field $F$ (a field permits $+,-,times,div$ except division by zero). A linear combination is defined to mean a weighted sum of finitely many vectors, for example $2x+(-3)y+5z$ is a linear combination of $x,y,z$. An $F$-linear combination is defined as a linear combination with weights in $F$. A spanning set $S$ for $V$ over $F$ is defined to be a set of vectors such that every vector in $V$ can be written as a $F$-linear combination of finitely many vectors from $S$, that is, for any vector $v in V$, we have $v = a_1 x_1 + a_2 x_2 + ... + a_n x_n$ for some $a_1,a_2,...,a_n in F$ and $x_1,x_2,...,x_n in S$. Then we can define a basis for $V$ over $F$ to mean a spanning set with minimum size and the dimension of $V$ over $F$ to be that size.
For another example, a displacement in 3d space requires 6 real values to describe it. The displacement involves both position and orientation. 3 real values are needed for the translation (change in position), and 3 real values are need for the rotation (change in orientation). The most frequently used representation is often called the 6 degrees of freedom in motion, which are up/down, left/right, forward/backward, pitch, yaw, roll. Note that you can represent any displacement using this 6 real values, and conversely any combination of these 6 values gives a unique displacement. But given two displacements, combining them cannot be computed by just adding the corresponding values. It adds for translation but not for rotation. In fact, the order of rotations in 3d space makes a difference! So this vector space structure does not fully reflect the structure of motion. What would be needed for this is transformations on $mathbb{R}^3$, which can be represented by matrices, but that is another topic.
$endgroup$
add a comment |
$begingroup$
Dimension usually is just the number of 'components' of some piece of information. 3 dimensions are just nice for describing a position in (Euclidean) space, but you definitely need 4 dimensions if you want to include the time also. Now you are in the room. A while later you are not. Your position has changed over time, so if we want to describe your path, we need to describe each point on the path, and each point consists of both 3 space coordinates and 1 time coordinate, so this path is 4-dimensional.
In mathematics we have a general kind of space called a vector space. The 3d Euclidean space described above is $mathbb{R}^3$, which is simply the set of all vectors with $3$ real coordinates. Spacetime could be represented as $mathbb{R}^4$. Time is special in the sense that we can only move forward in time whereas we can move in any direction in space, but this has nothing to do with dimension.
In a vector space, we would like to represent each vector as a combination of some basic vectors. Of course we want to have as few basic vectors as possible. For example $mathbb{R}^3$ has a basis with 3 vectors, ${(1,0,0),(0,1,0),(0,0,1)}$, which you could imagine stands for "1 unit in front of you", "1 unit above you" and "1 unit to your right", if you happen to stand at the centre $(0,0,0)$. Any position in space can indeed by expressed as a linear combination "go $x$ units forward, then $y$ units up, then $z$ units right". And you cannot make do with less. So we say that $mathbb{R}^3$ has dimension $3$. In general $mathbb{R}^n$ has dimension $n$.
A general vector space $V$ is defined over a field $F$ (a field permits $+,-,times,div$ except division by zero). A linear combination is defined to mean a weighted sum of finitely many vectors, for example $2x+(-3)y+5z$ is a linear combination of $x,y,z$. An $F$-linear combination is defined as a linear combination with weights in $F$. A spanning set $S$ for $V$ over $F$ is defined to be a set of vectors such that every vector in $V$ can be written as a $F$-linear combination of finitely many vectors from $S$, that is, for any vector $v in V$, we have $v = a_1 x_1 + a_2 x_2 + ... + a_n x_n$ for some $a_1,a_2,...,a_n in F$ and $x_1,x_2,...,x_n in S$. Then we can define a basis for $V$ over $F$ to mean a spanning set with minimum size and the dimension of $V$ over $F$ to be that size.
For another example, a displacement in 3d space requires 6 real values to describe it. The displacement involves both position and orientation. 3 real values are needed for the translation (change in position), and 3 real values are need for the rotation (change in orientation). The most frequently used representation is often called the 6 degrees of freedom in motion, which are up/down, left/right, forward/backward, pitch, yaw, roll. Note that you can represent any displacement using this 6 real values, and conversely any combination of these 6 values gives a unique displacement. But given two displacements, combining them cannot be computed by just adding the corresponding values. It adds for translation but not for rotation. In fact, the order of rotations in 3d space makes a difference! So this vector space structure does not fully reflect the structure of motion. What would be needed for this is transformations on $mathbb{R}^3$, which can be represented by matrices, but that is another topic.
$endgroup$
Dimension usually is just the number of 'components' of some piece of information. 3 dimensions are just nice for describing a position in (Euclidean) space, but you definitely need 4 dimensions if you want to include the time also. Now you are in the room. A while later you are not. Your position has changed over time, so if we want to describe your path, we need to describe each point on the path, and each point consists of both 3 space coordinates and 1 time coordinate, so this path is 4-dimensional.
In mathematics we have a general kind of space called a vector space. The 3d Euclidean space described above is $mathbb{R}^3$, which is simply the set of all vectors with $3$ real coordinates. Spacetime could be represented as $mathbb{R}^4$. Time is special in the sense that we can only move forward in time whereas we can move in any direction in space, but this has nothing to do with dimension.
In a vector space, we would like to represent each vector as a combination of some basic vectors. Of course we want to have as few basic vectors as possible. For example $mathbb{R}^3$ has a basis with 3 vectors, ${(1,0,0),(0,1,0),(0,0,1)}$, which you could imagine stands for "1 unit in front of you", "1 unit above you" and "1 unit to your right", if you happen to stand at the centre $(0,0,0)$. Any position in space can indeed by expressed as a linear combination "go $x$ units forward, then $y$ units up, then $z$ units right". And you cannot make do with less. So we say that $mathbb{R}^3$ has dimension $3$. In general $mathbb{R}^n$ has dimension $n$.
A general vector space $V$ is defined over a field $F$ (a field permits $+,-,times,div$ except division by zero). A linear combination is defined to mean a weighted sum of finitely many vectors, for example $2x+(-3)y+5z$ is a linear combination of $x,y,z$. An $F$-linear combination is defined as a linear combination with weights in $F$. A spanning set $S$ for $V$ over $F$ is defined to be a set of vectors such that every vector in $V$ can be written as a $F$-linear combination of finitely many vectors from $S$, that is, for any vector $v in V$, we have $v = a_1 x_1 + a_2 x_2 + ... + a_n x_n$ for some $a_1,a_2,...,a_n in F$ and $x_1,x_2,...,x_n in S$. Then we can define a basis for $V$ over $F$ to mean a spanning set with minimum size and the dimension of $V$ over $F$ to be that size.
For another example, a displacement in 3d space requires 6 real values to describe it. The displacement involves both position and orientation. 3 real values are needed for the translation (change in position), and 3 real values are need for the rotation (change in orientation). The most frequently used representation is often called the 6 degrees of freedom in motion, which are up/down, left/right, forward/backward, pitch, yaw, roll. Note that you can represent any displacement using this 6 real values, and conversely any combination of these 6 values gives a unique displacement. But given two displacements, combining them cannot be computed by just adding the corresponding values. It adds for translation but not for rotation. In fact, the order of rotations in 3d space makes a difference! So this vector space structure does not fully reflect the structure of motion. What would be needed for this is transformations on $mathbb{R}^3$, which can be represented by matrices, but that is another topic.
answered Apr 26 '15 at 11:42
user21820user21820
38.8k543153
38.8k543153
add a comment |
add a comment |
$begingroup$
Multidimensional spaces occur naturally all around us. Take the possible positions of your arm, for example.
- You can rotate your shoulder with two degrees of freedom;
- You can bend your elbow with one degree of freedom;
- You can rotate your wrist with two degrees of freedom
That's a total of five degrees of freedom for the possible positions of your arm.
In other words, the space of positions of your arm, a space with which you are personally familiar on an intimate level, is a 5-dimensional object.
And I'm ignoring the fact that twisting your forearm and flexing your hand bones and bending your finger bones adds at least another twenty or more dimensions.
$endgroup$
add a comment |
$begingroup$
Multidimensional spaces occur naturally all around us. Take the possible positions of your arm, for example.
- You can rotate your shoulder with two degrees of freedom;
- You can bend your elbow with one degree of freedom;
- You can rotate your wrist with two degrees of freedom
That's a total of five degrees of freedom for the possible positions of your arm.
In other words, the space of positions of your arm, a space with which you are personally familiar on an intimate level, is a 5-dimensional object.
And I'm ignoring the fact that twisting your forearm and flexing your hand bones and bending your finger bones adds at least another twenty or more dimensions.
$endgroup$
add a comment |
$begingroup$
Multidimensional spaces occur naturally all around us. Take the possible positions of your arm, for example.
- You can rotate your shoulder with two degrees of freedom;
- You can bend your elbow with one degree of freedom;
- You can rotate your wrist with two degrees of freedom
That's a total of five degrees of freedom for the possible positions of your arm.
In other words, the space of positions of your arm, a space with which you are personally familiar on an intimate level, is a 5-dimensional object.
And I'm ignoring the fact that twisting your forearm and flexing your hand bones and bending your finger bones adds at least another twenty or more dimensions.
$endgroup$
Multidimensional spaces occur naturally all around us. Take the possible positions of your arm, for example.
- You can rotate your shoulder with two degrees of freedom;
- You can bend your elbow with one degree of freedom;
- You can rotate your wrist with two degrees of freedom
That's a total of five degrees of freedom for the possible positions of your arm.
In other words, the space of positions of your arm, a space with which you are personally familiar on an intimate level, is a 5-dimensional object.
And I'm ignoring the fact that twisting your forearm and flexing your hand bones and bending your finger bones adds at least another twenty or more dimensions.
edited Apr 26 '15 at 21:49
TRiG
199112
199112
answered Apr 26 '15 at 13:27
Lee MosherLee Mosher
48.4k33681
48.4k33681
add a comment |
add a comment |
$begingroup$
I am a firm believer in the idea that if something in math doesn't make sense to you, you just haven't found the proper way of thinking about it yet. And in some (or maybe most?) cases, there isn't anybody on earth who has found the right way. And this way is probably different for different people.
Now, to answer your question, I will tell you how I think of higher dimensional objects, as someone who prefers to visualize everything. Some of the other answers describe how from one $(x)$ to two $(x,y)$ dimensions we just "add a perpendicular line" to everything and call something in that new direction $y$. If that works that is fine. It makes sense for the transition from 2 to 3 dimensions also, for most people. When we go from 3 to 4, we have to get creative. Instead of a "adding a new line", I usually think about adding the color spectrum to each point in 3 space. So, at each point, we have $(x,y,z)$ just like in normal 3 space, and lets call it $c$, for color. To draw points in 4 space, now you need to place it in 3 space and also assign some color to it, to tell us where in the spectrum it is.
A line is the same way, but at each point in the line, we can pick a color. If the color varies continuously through the spectrum, the line is connected. But an abrupt change from red to green, for example, is like having a break in the line. In the picture below, the green line lives in 3 space, since it only uses one color, the rainbow line travels through the whole spectrum. Also, notice that the rainbow line in three dimensions crosses itself. But it doesnt actually intersect because it has two different colors.
So, this rainbow line lives in 4 dimensions right now, and the green line is only using 3 dimensions. This analogy still has drawbacks though. The three directional unit vectors make sense, but when you try to draw the new color one, you find yourself with "line" of colors, but in the 3 dimesions we are looking at, it is only at the origin, so it looks like a point to us.
One last example, and this might be a stretch: The 3-sphere, which lives in 4 dimensions. First a quick idea of how we make this. Just like in flatland, when the sphere (which in topology we call the 2-sphere) comes to visit, it shows up a a circle (or 1-sphere), picutred below. But it has a lot of circles of different size, depending on where the plane is intersecting the sphere. A second sphere is higher than the first, so it intersects a smaller circle in the plane.
So when the 3-sphere comes to 3 dimensions, we will only see it as a normal sphere (2-sphere). If that was confusing, I apologize. Here are some pictures where we get the normal unit sphere when we are green, which I am equating to $c=0$. but when we change colors, our spheres will get smaller and when the are red and violet, they will only be a point. That is because the color is "farther" from green, which we are thinking of as the 3 space we live in, and we need to satisfy the equation $x^2+y^2+z^2+c^2=1$. Well, I hope this helps.
$endgroup$
add a comment |
$begingroup$
I am a firm believer in the idea that if something in math doesn't make sense to you, you just haven't found the proper way of thinking about it yet. And in some (or maybe most?) cases, there isn't anybody on earth who has found the right way. And this way is probably different for different people.
Now, to answer your question, I will tell you how I think of higher dimensional objects, as someone who prefers to visualize everything. Some of the other answers describe how from one $(x)$ to two $(x,y)$ dimensions we just "add a perpendicular line" to everything and call something in that new direction $y$. If that works that is fine. It makes sense for the transition from 2 to 3 dimensions also, for most people. When we go from 3 to 4, we have to get creative. Instead of a "adding a new line", I usually think about adding the color spectrum to each point in 3 space. So, at each point, we have $(x,y,z)$ just like in normal 3 space, and lets call it $c$, for color. To draw points in 4 space, now you need to place it in 3 space and also assign some color to it, to tell us where in the spectrum it is.
A line is the same way, but at each point in the line, we can pick a color. If the color varies continuously through the spectrum, the line is connected. But an abrupt change from red to green, for example, is like having a break in the line. In the picture below, the green line lives in 3 space, since it only uses one color, the rainbow line travels through the whole spectrum. Also, notice that the rainbow line in three dimensions crosses itself. But it doesnt actually intersect because it has two different colors.
So, this rainbow line lives in 4 dimensions right now, and the green line is only using 3 dimensions. This analogy still has drawbacks though. The three directional unit vectors make sense, but when you try to draw the new color one, you find yourself with "line" of colors, but in the 3 dimesions we are looking at, it is only at the origin, so it looks like a point to us.
One last example, and this might be a stretch: The 3-sphere, which lives in 4 dimensions. First a quick idea of how we make this. Just like in flatland, when the sphere (which in topology we call the 2-sphere) comes to visit, it shows up a a circle (or 1-sphere), picutred below. But it has a lot of circles of different size, depending on where the plane is intersecting the sphere. A second sphere is higher than the first, so it intersects a smaller circle in the plane.
So when the 3-sphere comes to 3 dimensions, we will only see it as a normal sphere (2-sphere). If that was confusing, I apologize. Here are some pictures where we get the normal unit sphere when we are green, which I am equating to $c=0$. but when we change colors, our spheres will get smaller and when the are red and violet, they will only be a point. That is because the color is "farther" from green, which we are thinking of as the 3 space we live in, and we need to satisfy the equation $x^2+y^2+z^2+c^2=1$. Well, I hope this helps.
$endgroup$
add a comment |
$begingroup$
I am a firm believer in the idea that if something in math doesn't make sense to you, you just haven't found the proper way of thinking about it yet. And in some (or maybe most?) cases, there isn't anybody on earth who has found the right way. And this way is probably different for different people.
Now, to answer your question, I will tell you how I think of higher dimensional objects, as someone who prefers to visualize everything. Some of the other answers describe how from one $(x)$ to two $(x,y)$ dimensions we just "add a perpendicular line" to everything and call something in that new direction $y$. If that works that is fine. It makes sense for the transition from 2 to 3 dimensions also, for most people. When we go from 3 to 4, we have to get creative. Instead of a "adding a new line", I usually think about adding the color spectrum to each point in 3 space. So, at each point, we have $(x,y,z)$ just like in normal 3 space, and lets call it $c$, for color. To draw points in 4 space, now you need to place it in 3 space and also assign some color to it, to tell us where in the spectrum it is.
A line is the same way, but at each point in the line, we can pick a color. If the color varies continuously through the spectrum, the line is connected. But an abrupt change from red to green, for example, is like having a break in the line. In the picture below, the green line lives in 3 space, since it only uses one color, the rainbow line travels through the whole spectrum. Also, notice that the rainbow line in three dimensions crosses itself. But it doesnt actually intersect because it has two different colors.
So, this rainbow line lives in 4 dimensions right now, and the green line is only using 3 dimensions. This analogy still has drawbacks though. The three directional unit vectors make sense, but when you try to draw the new color one, you find yourself with "line" of colors, but in the 3 dimesions we are looking at, it is only at the origin, so it looks like a point to us.
One last example, and this might be a stretch: The 3-sphere, which lives in 4 dimensions. First a quick idea of how we make this. Just like in flatland, when the sphere (which in topology we call the 2-sphere) comes to visit, it shows up a a circle (or 1-sphere), picutred below. But it has a lot of circles of different size, depending on where the plane is intersecting the sphere. A second sphere is higher than the first, so it intersects a smaller circle in the plane.
So when the 3-sphere comes to 3 dimensions, we will only see it as a normal sphere (2-sphere). If that was confusing, I apologize. Here are some pictures where we get the normal unit sphere when we are green, which I am equating to $c=0$. but when we change colors, our spheres will get smaller and when the are red and violet, they will only be a point. That is because the color is "farther" from green, which we are thinking of as the 3 space we live in, and we need to satisfy the equation $x^2+y^2+z^2+c^2=1$. Well, I hope this helps.
$endgroup$
I am a firm believer in the idea that if something in math doesn't make sense to you, you just haven't found the proper way of thinking about it yet. And in some (or maybe most?) cases, there isn't anybody on earth who has found the right way. And this way is probably different for different people.
Now, to answer your question, I will tell you how I think of higher dimensional objects, as someone who prefers to visualize everything. Some of the other answers describe how from one $(x)$ to two $(x,y)$ dimensions we just "add a perpendicular line" to everything and call something in that new direction $y$. If that works that is fine. It makes sense for the transition from 2 to 3 dimensions also, for most people. When we go from 3 to 4, we have to get creative. Instead of a "adding a new line", I usually think about adding the color spectrum to each point in 3 space. So, at each point, we have $(x,y,z)$ just like in normal 3 space, and lets call it $c$, for color. To draw points in 4 space, now you need to place it in 3 space and also assign some color to it, to tell us where in the spectrum it is.
A line is the same way, but at each point in the line, we can pick a color. If the color varies continuously through the spectrum, the line is connected. But an abrupt change from red to green, for example, is like having a break in the line. In the picture below, the green line lives in 3 space, since it only uses one color, the rainbow line travels through the whole spectrum. Also, notice that the rainbow line in three dimensions crosses itself. But it doesnt actually intersect because it has two different colors.
So, this rainbow line lives in 4 dimensions right now, and the green line is only using 3 dimensions. This analogy still has drawbacks though. The three directional unit vectors make sense, but when you try to draw the new color one, you find yourself with "line" of colors, but in the 3 dimesions we are looking at, it is only at the origin, so it looks like a point to us.
One last example, and this might be a stretch: The 3-sphere, which lives in 4 dimensions. First a quick idea of how we make this. Just like in flatland, when the sphere (which in topology we call the 2-sphere) comes to visit, it shows up a a circle (or 1-sphere), picutred below. But it has a lot of circles of different size, depending on where the plane is intersecting the sphere. A second sphere is higher than the first, so it intersects a smaller circle in the plane.
So when the 3-sphere comes to 3 dimensions, we will only see it as a normal sphere (2-sphere). If that was confusing, I apologize. Here are some pictures where we get the normal unit sphere when we are green, which I am equating to $c=0$. but when we change colors, our spheres will get smaller and when the are red and violet, they will only be a point. That is because the color is "farther" from green, which we are thinking of as the 3 space we live in, and we need to satisfy the equation $x^2+y^2+z^2+c^2=1$. Well, I hope this helps.
answered May 15 '15 at 18:42
N. OwadN. Owad
4,53521232
4,53521232
add a comment |
add a comment |
$begingroup$
When somebody says a statement about the fourth dimension that you think is clearly false such as the fact that a closed string has only one possible knot, the unknot, the usual way to interpret what they're saying is a statement that is false. You're right that it's technically false to say that a string cannot be knotted. What they're really doing is thinking of a totally different statement and using that English sentence to mean something the sentence doesn't actually mean.
Take $mathbb{R}^4$, the set of all ordered quadruplets of real numbers. Now, we can invent a definition of the distance from any member of $mathbb{R}^4$ to any other member of $mathbb{R}^4$ as follows. For any points in $mathbb{R}^4$, $(w_1, x_1, y_1, z_1)$ and $(w_2, x_2, y_2, z_2)$, the distance from $(w_1, x_1, y_1, z_1)$ to $(w_2, x_2, y_2, z_2)$ is $sqrt{(w_2 - w_1)^2 + (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$. We can invent a binary operation * on that space that functions essentially the same as quaternion multiplication. Let's define a function $f$ from $mathbb{R}^4$ to itself to be an origin rotation if and only if $exists y in mathbb{R}^4exists z in mathbb{R}^4$ such that distance from the origin to $y$ and the distance from the origin to $z$ are 1 and $forall x in mathbb{R}^4, f(x) = y * x * z$. Let's define any rotation in $mathbb{R}$ to be an operation such that there exists a translation $T$ and an origin rotation $f$ such that it can be expressed as $T circ f circ T^{-1}$. Although this definition of a rotation works only for $mathbb{R}^4$, we can show that it satisfies all the basic intuitive properties that the set of all rotations in the third dimension satisfies.
$endgroup$
add a comment |
$begingroup$
When somebody says a statement about the fourth dimension that you think is clearly false such as the fact that a closed string has only one possible knot, the unknot, the usual way to interpret what they're saying is a statement that is false. You're right that it's technically false to say that a string cannot be knotted. What they're really doing is thinking of a totally different statement and using that English sentence to mean something the sentence doesn't actually mean.
Take $mathbb{R}^4$, the set of all ordered quadruplets of real numbers. Now, we can invent a definition of the distance from any member of $mathbb{R}^4$ to any other member of $mathbb{R}^4$ as follows. For any points in $mathbb{R}^4$, $(w_1, x_1, y_1, z_1)$ and $(w_2, x_2, y_2, z_2)$, the distance from $(w_1, x_1, y_1, z_1)$ to $(w_2, x_2, y_2, z_2)$ is $sqrt{(w_2 - w_1)^2 + (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$. We can invent a binary operation * on that space that functions essentially the same as quaternion multiplication. Let's define a function $f$ from $mathbb{R}^4$ to itself to be an origin rotation if and only if $exists y in mathbb{R}^4exists z in mathbb{R}^4$ such that distance from the origin to $y$ and the distance from the origin to $z$ are 1 and $forall x in mathbb{R}^4, f(x) = y * x * z$. Let's define any rotation in $mathbb{R}$ to be an operation such that there exists a translation $T$ and an origin rotation $f$ such that it can be expressed as $T circ f circ T^{-1}$. Although this definition of a rotation works only for $mathbb{R}^4$, we can show that it satisfies all the basic intuitive properties that the set of all rotations in the third dimension satisfies.
$endgroup$
add a comment |
$begingroup$
When somebody says a statement about the fourth dimension that you think is clearly false such as the fact that a closed string has only one possible knot, the unknot, the usual way to interpret what they're saying is a statement that is false. You're right that it's technically false to say that a string cannot be knotted. What they're really doing is thinking of a totally different statement and using that English sentence to mean something the sentence doesn't actually mean.
Take $mathbb{R}^4$, the set of all ordered quadruplets of real numbers. Now, we can invent a definition of the distance from any member of $mathbb{R}^4$ to any other member of $mathbb{R}^4$ as follows. For any points in $mathbb{R}^4$, $(w_1, x_1, y_1, z_1)$ and $(w_2, x_2, y_2, z_2)$, the distance from $(w_1, x_1, y_1, z_1)$ to $(w_2, x_2, y_2, z_2)$ is $sqrt{(w_2 - w_1)^2 + (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$. We can invent a binary operation * on that space that functions essentially the same as quaternion multiplication. Let's define a function $f$ from $mathbb{R}^4$ to itself to be an origin rotation if and only if $exists y in mathbb{R}^4exists z in mathbb{R}^4$ such that distance from the origin to $y$ and the distance from the origin to $z$ are 1 and $forall x in mathbb{R}^4, f(x) = y * x * z$. Let's define any rotation in $mathbb{R}$ to be an operation such that there exists a translation $T$ and an origin rotation $f$ such that it can be expressed as $T circ f circ T^{-1}$. Although this definition of a rotation works only for $mathbb{R}^4$, we can show that it satisfies all the basic intuitive properties that the set of all rotations in the third dimension satisfies.
$endgroup$
When somebody says a statement about the fourth dimension that you think is clearly false such as the fact that a closed string has only one possible knot, the unknot, the usual way to interpret what they're saying is a statement that is false. You're right that it's technically false to say that a string cannot be knotted. What they're really doing is thinking of a totally different statement and using that English sentence to mean something the sentence doesn't actually mean.
Take $mathbb{R}^4$, the set of all ordered quadruplets of real numbers. Now, we can invent a definition of the distance from any member of $mathbb{R}^4$ to any other member of $mathbb{R}^4$ as follows. For any points in $mathbb{R}^4$, $(w_1, x_1, y_1, z_1)$ and $(w_2, x_2, y_2, z_2)$, the distance from $(w_1, x_1, y_1, z_1)$ to $(w_2, x_2, y_2, z_2)$ is $sqrt{(w_2 - w_1)^2 + (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$. We can invent a binary operation * on that space that functions essentially the same as quaternion multiplication. Let's define a function $f$ from $mathbb{R}^4$ to itself to be an origin rotation if and only if $exists y in mathbb{R}^4exists z in mathbb{R}^4$ such that distance from the origin to $y$ and the distance from the origin to $z$ are 1 and $forall x in mathbb{R}^4, f(x) = y * x * z$. Let's define any rotation in $mathbb{R}$ to be an operation such that there exists a translation $T$ and an origin rotation $f$ such that it can be expressed as $T circ f circ T^{-1}$. Although this definition of a rotation works only for $mathbb{R}^4$, we can show that it satisfies all the basic intuitive properties that the set of all rotations in the third dimension satisfies.
edited Dec 15 '18 at 21:25
answered Dec 15 '18 at 21:19
TimothyTimothy
304212
304212
add a comment |
add a comment |
$begingroup$
This man can ! https://www.youtube.com/watch?v=M9sbdrPVfOQ :)
But I never understand I think we have to have a brain in 4D to understand.
You can also see that :
A conference really interesting Here
And a short video Here
$endgroup$
add a comment |
$begingroup$
This man can ! https://www.youtube.com/watch?v=M9sbdrPVfOQ :)
But I never understand I think we have to have a brain in 4D to understand.
You can also see that :
A conference really interesting Here
And a short video Here
$endgroup$
add a comment |
$begingroup$
This man can ! https://www.youtube.com/watch?v=M9sbdrPVfOQ :)
But I never understand I think we have to have a brain in 4D to understand.
You can also see that :
A conference really interesting Here
And a short video Here
$endgroup$
This man can ! https://www.youtube.com/watch?v=M9sbdrPVfOQ :)
But I never understand I think we have to have a brain in 4D to understand.
You can also see that :
A conference really interesting Here
And a short video Here
answered Apr 25 '15 at 22:19
Hexacoordinate-CHexacoordinate-C
720829
720829
add a comment |
add a comment |
protected by Zev Chonoles May 15 '15 at 15:52
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
1
$begingroup$
Note that, in a very literal sense, four dimensions don't fit in the world that you're used to. A fourth dimension would require adding quite a lot of space to the universe. (Inb4 strong theorists: I'm trying to keep this simple.)
$endgroup$
– Akiva Weinberger
Apr 26 '15 at 0:51
17
$begingroup$
I think the book Flatland tries to wrap your head around that.
$endgroup$
– Akiva Weinberger
Apr 26 '15 at 0:53
1
$begingroup$
*string. Whoops.
$endgroup$
– Akiva Weinberger
Apr 26 '15 at 4:08
8
$begingroup$
That's simple. For $mathbb R^n$ and let $n=4$. ;-)
$endgroup$
– Vim
Apr 26 '15 at 7:52
5
$begingroup$
@Vim Well that answers one of OP's questions, as he thinks: "Yep, should've asked on physics.SE" ;)
$endgroup$
– OJFord
Apr 26 '15 at 12:32