3
$\begingroup$

The distance ($D$) between two points $(x_1,y_1)$ and $(x_2,y_2)$ on a plane is given by $$D=\sqrt {(x_2-x_1)^2+(y_2-y_1)^2}$$

The distance ($D$) between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ in space is given by $$D=\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$

The above two formulas are similar in many ways. We can just consider the two dimensional case as a special case of the three dimensional case when the $z$ coordinates are the same. Constructing a similar formula for a 4-Dimensional Space, the distance (D) between the coordinates $(x_1,y_1,z_1,a_1)$ and $(x_2,y_2,z_2,a_2)$ in 4-dimensional space is given by $$D=\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2+(a_2-a_1)^2}$$

Is this formula correct/applicable to 4-dimensional cartesian coordinate system? If yes, what is its significance? What is the significance/meaning of a point in 4-dimensions?


A little bit of Mathematical Background: I am a high school student, who has studied only till 3-dimensional geometry. I just improved upon this to 4-dimensions by seeing the pattern.

$\endgroup$
4
$\begingroup$

What you have defined is called a metric on $\mathbb{R}^4$ - a notion of a distance function for points in four-dimensional real space. Mathematicians often work with the concept of metric spaces, formally defined as a pair $(E,d)$, where $E$ is a set and $d:E\times E\to\mathbb{R}$ satisfying a particular set of identities (where $x,y\in E$):

  1. $d(x,y)\geq0$
  2. $d(x,y)=0\iff x=y$
  3. $d(x,y)=d(y,x)$
  4. $d(x,z)\leq d(x,y)+d(y,z)$

So, in your case, what you have conjectured is that the function $d:\mathbb{R}^4\times\mathbb{R}^4\to\mathbb{R}$ satisfying: $$d\big((x_1,x_2,x_3,x_4),(y_1,y_2,y_3,y_4)\big)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2+(x_4-y_4)^2}$$ is a metric on $\mathbb{R}^4$. Try proving it!

But there are other metrics on $\mathbb{R}^4$ other than just the Euclidean one. For example, there is the taxicab metric: $$d\big((x_1,x_2,x_3,x_4),(y_1,y_2,y_3,y_4)\big)=\sum_{i=1}^4|x_i-y_i|$$ And also the discrete metric: $$d(\vec x,\vec y)= \begin{cases} 1, & \vec x\neq\vec y\\ 0, & \vec x=\vec y \\ \end{cases}$$ It is also a good exercise to prove that these are also metrics on $\mathbb{R}^4$, and hence also perfectly valid notions of distance according to our definition. Your extension of the Euclidean metric to $\mathbb{R}^4$ preserves some nice qualities that we are used to - namely, translation and rotation invariance - but it is good to know that it is not the only way to define distance in $\mathbb{R}^4$.

As for how to interpret points in $\mathbb{R}^4$, think about how you view points in lower dimensions first. In $\mathbb{R}$, points lie on a number line. In $\mathbb{R}^2$, points lie on a plane. In $\mathbb{R}^3$, points lie in a larger three-dimensional space. In each case, we add a new dimension to where we can consider a point lying, but fundamentally nothing much changes about how we think about and consider points. You could, if you wanted to, just think about a point in $n$-dimensional real space as an ordered $n$-tuple of coordinates, each lying somewhere on the real line $\mathbb{R}$. So taking the case $n=4$, we can think of a point in $\mathbb{R}^4$ as being an ordered $4$-tuple of coordinates $(x_1,x_2,x_3,x_4)$ each lying on the real number line somewhere.

$\endgroup$
  • $\begingroup$ Thank you for your answer. Could you please tell whether $d(x,y)$ is a distance function, since many properties of distances match with the four identities mentioned in your answer. $\endgroup$ – Intellex Oct 18 at 4:02
  • 1
    $\begingroup$ Yes, I am denoting $d(x,y)$ as a distance function. Any metric/distance function must satisfy those identities that I listed above; those are in a way the defining properties of a notion of distance. $\endgroup$ – csch2 Oct 18 at 4:04
  • $\begingroup$ It was easy to prove identities 1,2, and 3. I think the fourth identity is related to "sum of any two sides of a triangle is greater than or equal to (in the degenerate case) the other side". Any other way to prove this? Could you please give a small hint. $\endgroup$ – Intellex Oct 18 at 4:09
  • 1
    $\begingroup$ The discrete metric is still defined on $\mathbb{R}$. For $x,y\in\mathbb{R}$, we say $d(x,y)=1$ if $x\neq y$ and $d(x,y)=0$ otherwise. The first three rules are easy to see, and the triangle inequality is easy to check as well - just reduce to a few cases. It's similarly easy to check that the discrete metric is a metric on $\mathbb{R}^4$ - again, the first three rules follow from the definition of the discrete metric and the triangle inequality follows from case analysis. $\endgroup$ – csch2 Oct 18 at 4:27
  • 1
    $\begingroup$ Remark that's relevant WRT “physical significance”: in relativistic mechanics, physicist often talk about about a “metric” in 4D space that's actually not a mathematical metric, but rather a bilinear form. It looks similar to the Euclidean distance on paper, but actually behaves quite different. en.wikipedia.org/wiki/Minkowski_space $\endgroup$ – leftaroundabout Oct 18 at 10:45
3
$\begingroup$

Mathematically we can consider any dimension of Euclidean space we want. You just use enough coordinates to specify each point. The distance formula is just like you say with one term per dimension. We can then do geometry, such as counting the number of regular hypersolids, computing the hypervolume of an $n-$dimensional ball, considering the dimensionality of subspaces, etc.

$\endgroup$
  • $\begingroup$ Thank you for your answer. May I know what does distance mean in $4$ dimensions. For $2$ and $3$ dimensions, it seems obvious. Or do you recommend me to revisit this topic once I get into college? $\endgroup$ – Intellex Oct 18 at 3:46
  • 2
    $\begingroup$ It means the same thing in four dimensions as in two or three. It is the distance between two points, the length of the line segment connecting them. It is hard to imagine four dimensions, but analogies with the step from two to three can be helpful (and also misleading). You have a fourth coordinate axis that is perpendicular to the other three, so it takes four coordinates for each point. $\endgroup$ – Ross Millikan Oct 18 at 3:58
  • 1
    $\begingroup$ It seems counter-intuitive to think of a line perpendicular to the three dimensional coordinate axes at the same time. $\endgroup$ – Intellex Oct 18 at 4:14
  • 2
    $\begingroup$ That is because your brain is stuck in three dimensions. It is hard to imagine, so you get forced to rely on formal arguments with less intuition. $\endgroup$ – Ross Millikan Oct 18 at 4:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.