Question

Surface area of a 4 dimensional surface.
Any of you guys though about this problem before?

Answer 1

If you consider the arc length of a curve, it generalizes into higher dimensions well, because the norm of a vector works in higher dimensions.\[
\int_C ds = \int ||\mathbf{r}'(t)||dt
\]But this isn't the case for surface area, because it uses the cross product which requires only three dimensions (apparently it works in 7 dimensions but I doubt that is useful.\[
\iint_S dS = \iint ||T_u\times T_v||dudv
\]In four dimensions, there is no single normal vector to a surface. In the same way that a curve has multiple normal vectors in three dimensions.

Answer 2

Tangent vectors will always be unique for curves regardless, so it's a good characterization.

In four dimensions, it might not even make sense to characterize surfaces by a normal vector. If we were to cut the surface into small pieces and to use the same parallelogram method, how would you find the area of a parallelogram made with 4D vectors... and would the approximation still hold?

Answer 3

If we go down a dimension into 2D, To find the area of a parallelogram made by vectors \(a\), \(b\) to be \(||a||\cdot ||b||\cdot \sin(\theta)\), where \(\theta\) is the angle between \(a\) and \(b\).

Answer 4

Perhaps using this, as opposed to the magnitude of the cross product, would create a good generalization. The only problem is we'd need to find \(\theta\) between two 4D vectors.

Answer 5

Suppose we use the dot product: \[
a\cdot b = ||a||\;||b||\;\cos(\theta)\implies \cos(\theta)=\frac{a\cdot b}{||a||\;||b||}\implies \sin(\theta) = \sqrt{1-\left(\frac{a\cdot b}{ ||a||\;||b||} \right)^2}
\]Then:\[
||a||\;||b||\;\sin(\theta) = ||a||\;||b||\; \sqrt{1-\left(\frac{a\cdot b}{ ||a||\;||b||} \right)^2}= \sqrt{(||a||\;||b||)^2-(a\cdot b)^2}
\]

Answer 6

If this works, then we have the general formula to be: \[
\iint_S dS = \iint \sqrt{(||T_u||\;||T_v||)^2-(T_u\cdot T_v)^2}dudv
\]It's be nice to do a few tests as a sanity check...
Anyone know any 4D surfaces, their area, and parameterization?

Answer 7

I don't think you can parameterize a surface in 4-D: http://en.wikipedia.org/wiki/Parametric_surface

At best, I think you'd have a parametric "solid," with three parameters, but there would still be a surface area, right? Interesting question.

Answer 8

Hmmm, well what sort of region would you get from: \[
\Phi(u,v) = (x(u,v),y(u,v),z(u,v),t(u,v))
\]My impression was that this would be a 4D parametric surface.

Answer 9

I would think that \[
\sigma(t) = (x(t),y(t),z(t),f(t))
\]Is a curve in 4D

Answer 10

Oh yes, you're right about that.
It appears to me that you'd have to introduce some topological reasoning. Does the notion of surface area exist in higher dimensions? If so, I'd be hard-pressed to find the surface area of this thing: http://en.wikipedia.org/wiki/Clifford_torus
Where would I even begin?

Answer 11

Well, is it a 4D creature? Because 4D creatures would be bounded by a 3D creature (solid?).

Answer 12

So it'd be like asking for the arc length of a sphere... does not compute.

Answer 13

I don't know what to tell you... This is over my head, but you may be on to something

Answer 14

It is above my head too... I only really understood the 4D hypercube and 4D hypersphere since they have very simple principles to them.

Answer 15

Maybe 'surface' is not the correct terminology. I'm talking about a 2D creature which exists in 4D environment. A 2D creature in a 2D environment stays flat. In 3D environment it can bend around. In 4D... it had even more freedom to move around. My intuition is that any 2D creature is parametrized with two input variables.

Answer 16

Okay let's just make the question simpler and ignore the whole surface thing...
Suppose you have two 4D vectors... \(a\) and \(b\).

What is the area of the parallelogram they make? They make a parallelogram, right?

For example, suppose \(a=\begin{bmatrix}1\\1\\0\\0 \end{bmatrix}\) and \(b=\begin{bmatrix}2\\1\\0\\1 \end{bmatrix}\)

Answer 17

Can we say there is an angle between these two? \[
a\cdot b = ||a||\;||b||\;\cos\theta
\]I'm wondering if this holds for higher dimensions.

Answer 18

I'm having a hard time picturing a 4D surface, let alone a 4D vector... But supposing they are a parallelogram, isn't the area \(a\times b\)? But the cross product doesn't seem to work here...

As for your angle question: orthogonality (perpendicularity) is defined for higher dimensional vector spaces, so I would think so. http://www.wolframalpha.com/input/?i=angle+between+%281%2C1%2C0%2C0%29+and+%282%2C1%2C0%2C1%29

Answer 19

Well, think about what area of a parallelogram in 2D, when you have two vectors. You can still find its area.

Answer 20

In both the 2D and 3D case, the area of the parallelogram is \[
||a||\;||b||\;\sin\theta
\]

Answer 21

When they're completely orthogonal, \(\sin90^\circ = 1\)

Answer 22

Oh right, I forgot about that equation for cross product. I was thinking of the determinant method. So you have \(a\times b=||a||\;||b||\sin\theta=\sqrt3,\) since the angle was found to be \(\frac{\pi}{6}\).

Answer 23

I use the fact that \(\sin\theta = \sqrt{1-\cos^2\theta}\) up above to get: \[
||a||\;||b||\;\sin\theta = \sqrt{(||a||\;||b||)^2-(a\cdot b)^2}
\]

Answer 24

So then, for the sake of generalization, I replace
the \(||T_u\times T_v||\) with \(\sqrt{(||T_u||\;||T_v||)^2+(T_u\cdot T_v)^2}\)
in the surface integral equation to get:\[
\iint_S dS = \iint \sqrt{(||T_u||\;||T_v||)^2-(T_u\cdot T_v)^2}dudv
\]But I just have no idea how to test this.

Answer 25

Use grassman or exterior algebra which is used to measure areas, volumes etc. Take 2 vectors in 4d space - the e's are the bases vectors like i,j,k - except there's 4 of them.
\[u = u^{1}e_{1} + u^{2}e_{2} + u^{3}e_{3} + u^{4}e_{4}\]\[v = v^{1}e_{1} + v^{2}e_{2} + v^{3}e_{3} + v^{4}e_{4}\] Then the wedge product of the two is \[u \Lambda v = ( u^{1}e_{1} + u^{2}e_{2} + u^{3}e_{3} + u^{4}e_{4}) \Lambda ( v^{1}e_{1} + v^{2}e_{2} + v^{3}e_{4} + v^{3}e_{4} )\] I'm using lambda to represent the wedge operator. The wedge product reduces to:
\[(u^{1}v^{2}-u^{2}v^{1}) e_{1}\Lambda e_{2} +(u^{1}v^{3}-u^{3}v^{1}) e_{1}\Lambda e_{3} +(u^{1}v^{4}-u^{4}v^{1}) e_{1}\Lambda e_{4} +\]\[(u^{2}v^{3}-u^{3}v^{2}) e_{2}\Lambda e_{3} +(u^{2}v^{4}-u^{4}v^{2}) e_{2}\Lambda e_{4} +(u^{3}v^{4}-u^{4}v^{3}) e_{3}\Lambda e_{4} \]Funny, this is a vector in a 6d space (a subspace of the 4x4 bivector space)- where ei^ej are bases vectors (ei^ei is zero). Just take the square root of the 6 co-efficients squared to get the length of this vector and that's the area. Its just algebra.

Now lets make it more intuitive.

To get the length of a line in 3D, project the line on to the x-axis, y axis, z-axis and then take the square root of the square of those (3) co-efficients.

To get the area of a parallelogram in 3D, project the parallelogram onto the xy plane, yz plane and zx plane and take the square root of the square of the areas of those 3 projected parallelograms.

To get the area of a parallelogram in 4D, project the parallelogram onto the wx,wy,wz, xy,xz, yz and take the square root of the square of the areas of those 6 projected parallelograms.

Grassman algebra is just and algebra for doing just this. Type "exterior algebra" in wiki for more info.

Answer 26

Also to take the 2 vectors from the example above (1,1,0,0) and (2,1,0,1) we have\[(1\times1-1\times2,1\times0-0\times2,1\times1-0\times2,\]\[1\times0-0\times1,1\times1-0\times1,0\times1-0\times0)\] = \[(-1,0,1,0,1,0)\]This (6d) vector has a magnitude of \[\sqrt{3}\] This is the area of the parallelogram spanned by the 2 vectors (if my arithmetic is correct)