Is the Distance Formula applicable for 4-Dimensional Coordinate System? If yes, what is its physical significance?

Question

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?

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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.

Answer 1

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.

Answer 2

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.