Measuring distances: why the Pythagoras formula

In summary: because in a vector space, the square root of a vector is just the length of the vector divided by the vector's magnitude (or cross product).
  • #1
krishna mohan
117
0
hi...

We generally use the Pythagoras formula for distance between two points in 2D, when the Cartesian co-ordinates are given...

One directly extends it to 3D..having the distance going as [tex]\sqrt{x^{2}+y^{2}+z^{2}}[/tex]...


For a curved co-ordinate system, we have distances measured by something like

[tex]ds^{2}=g^{\mu\nu}dx_{\mu}dx_{\nu}[/tex]...

I was wondering why it is all about squares and square roots..and not, say cubes and cube roots or fourth roots or something...
 
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  • #2
krishna mohan said:
hi...

We generally use the Pythagoras formula for distance between two points in 2D, when the Cartesian co-ordinates are given...

One directly extends it to 3D..having the distance going as [tex]\sqrt{x^{2}+y^{2}+z^{2}}[/tex]...

For a curved co-ordinate system, we have distances measured by something like

[tex]ds^{2}=g^{\mu\nu}dx_{\mu}dx_{\nu}[/tex]...

I was wondering why it is all about squares and square roots..and not, say cubes and cube roots or fourth roots or something...
There is general theory: theory of metric spaces, where You may assume that metric, that is the mathematical form used for measuring distances is general, function of Your choice.
Function may be treated as metric if has following properties:

[tex]
\rho(A,B)>0
[/tex]

[tex]
\rho(A,B) = 0 \iff A =B
[/tex]

[tex]
\rho(A,B) = \rho(B,A)
[/tex]

[tex]
\rho(A,B) \leq \rho(A,C) + \rho(C,B)
[/tex]

[tex] A,B,C[/tex] are points of space. Points may mean whatever You like! For example it may be points in Euclidean space, functions, vectors, or even geometrical shapes, or colors, as long as You provide valid [tex]\rho[/tex] You will end with space with metric measuring some kind of distances.
For typical, Euclidean space, there are non standard examples of metric ( You provided us with standard one):

[tex]
\sum \left|x_i -y_i\right|
[/tex]

where this sum is over space dimension indexes. This is so called taxi metric. Other examples You may find in wikipedia http://en.wikipedia.org/wiki/Metric_(mathematics)

Important types of metric are defined in spaces of functions ( general in vector spaces) where this structure provides us an ability to perform for example analysis in general functional operators etc. This are Banach and Hilbert spaces with metric which is given by the so called norm.

So in fact there are norms for which You use squares etc in physical and mathematical practice in functional spaces. For example norm ([tex] \left\| f \right\| = \rho(0,f) [/tex]) for function [tex]f[/tex] may be given by:

[tex]\left\| f \right\|_p = ( \int \left| f(x) \right| ^p dx) ^\frac{1}{p}[/tex]

It is called [tex]L^p[/tex] norm and space of functions for which it is well defined is called [tex]L^p[/tex] space. For [tex] p =2 [/tex] You will find important case 2-norm, which is used in Quantum mechanic Hilbert spaces called [tex]L^2[/tex] .
 
  • #3
Thank you...that was very illuminating...


But I still wonder...why the general distance formula is the one with the squares and square roots? Why did nature not choose, say, the taxicab metric? Is there something that tells me it has to be this way?
 
  • #4
krishna mohan said:
Thank you...that was very illuminating...

But I still wonder...why the general distance formula is the one with the squares and square roots? Why did nature not choose, say, the taxicab metric? Is there something that tells me it has to be this way?
Ha! This is the Question! If You think about curved spacetime in GR Theory, Then You will see that Euclidean metric is flat. It is not by taking other spaces/metrics relative to Euclidean, but it is somehow real thing: differential of [tex]x^2 [/tex] is linear function with constant coefficients ( independent od x-coordinates). This is important thong in my opinion which is not true for other metrics.
 
  • #5
to understand why it's all about squares and square roots , you need to think about the space as a vector space , i will illustrate on a 2D Euclidean Space with Cartesian coordinates for simplification but you can generalize it after ...
suppose that we need to measure the distance between two points [tex]P_1,P_2[/tex] , by thinking about the space as a vector space , we know that we can define the two points by two position vectors [tex]\vec{r_1},\vec{r_2}[/tex] ... using these two position vectors we can construct a new vector [tex]d\vec{r}[/tex] which is :
[tex]d\vec{r}=\vec{r_2}-\vec{r_1}[/tex]
this new vector is a vector that connects both [tex]P_1[/tex] and [tex]P_2[/tex] and its length (norm) is the distance between the two points .
we are working on a 2D vector space , so each vector can be represented by two coordinates and two unit vectors , so :
[tex]\vec{r_1}=x_1\bold{i}+y_1\bold{j}[/tex]
[tex]\vec{r_2}=x_2\bold{i}+y_2\bold{j}[/tex]
therefore , we can state the vector [tex]d\vec{r}[/tex] as :
[tex]d\vec{r}=(x_2 - x_1)\bold{i}+(y_2 - y_1)\bold{j} [/tex]
[tex]\therefore d\vec{r}=dx\bold{i}+dy\bold{j} [/tex]
[tex]where : dx=x_2-x_1 , dy = y_2-y_1[/tex]
the length (norm) a vector is the square root of the dot product of the vector by it self , means that :
[tex]\abs{d\vec{r}}=ds=\sqrt{d\vec{r}.d\vec{r}}[/tex]
or : [tex] ds^2=d\vec{r}.d\vec{r}[/tex]
[tex]\therefore ds^2=(dx\bold{i}+dy\bold{j})^2 = dx^2 + dxdy\bold{ij} + dy^2 [/tex]
and the second term will be canceled because of the orthogonality of the coordinate system leaving us with :
[tex]ds^2 = dx^2 + dy^2[/tex]
and that's why it's all about squares and square roots .. !
 
  • #6
hi krishna! :smile:
krishna mohan said:
… I was wondering why it is all about squares and square roots..and not, say cubes and cube roots or fourth roots or something...

I think it's basically because you need an inner product, and that involves two vectors, or two copies of the same vector.

One vector is in the original space, and the other is in the dual space (the space of functionals, or scalar-valued functions).

It's difficult to see how you could define a third space (or fourth etc). :wink:
 
  • #7
krishna mohan said:
Thank you...that was very illuminating...


But I still wonder...why the general distance formula is the one with the squares and square roots? Why did nature not choose, say, the taxicab metric? Is there something that tells me it has to be this way?
Nature didn't "choose" anything. Distance is purely a human invention. As kakaz said, you could as easily define distance to be the sum of the absolute values of the coordinate distances or the largest of those absolute values. The "standard distance formula" happens to be easy to use. Without at least an even power, you have to have absolute values to avoid "negative" distances. And, absolute value, not being 'smooth', is awkward to work with. And, while any even power would do, "2" is the smallest and so the simplest.

I still remember the shock I felt when, while reading Eddington's "The Mathematical Theory of Relativity", where he was calculating the gravitational "pull" between two bodies, he remarked that we have to decide, arbitrarily, which of several values to use as the "distance" between them.
 
  • #8
HallsofIvy said:
Nature didn't "choose" anything.
I will not agree. Nature choose some kind of space-time as flat. When there is no matter, no energy, no other disturbances, then there is an Euclidean metric - and obviously that is the choice. Why? I do not know: but I assume that answer would be interesting...

In mathematics Euclidean metric is limiting case for hyperbolic or elliptic geometries which are curved, whilst Euclidean one is flat.

It is interesting coincidence that locally our space is Euclidean...
 
  • #9
krishna mohan said:
I was wondering why it is all about squares and square roots..and not, say cubes and cube roots or fourth roots or something...

Think about how the pythagorean theorem combines with the coordinates of a point on a circle of radius r, centered at the origin: (r * cos t, r * sin t).

[tex]\sqrt{ (r * \cos t)^2 + (r * \sin t)^2} = \sqrt {r^2 \cos^2 t + r^2 \sin^2 t} =\sqrt {r^2 (\cos^2 t + \sin^2 t)} = \sqrt {r^2 (1)} = r[/tex]
 
  • #10
krishna mohan said:
But I still wonder...why the general distance formula is the one with the squares and square roots? Why did nature not choose, say, the taxicab metric? Is there something that tells me it has to be this way?
It had to pick something!

It's sort of tautologous -- if nature had "chosen" the taxicab metric, you would be here asking why that metric was chosen rather than one based on squares and square roots!
 
  • #11
Truth_Seeker said:
the length (norm) a vector is the square root of the dot product of the vector by it self , means that :
[tex]\abs{d\vec{r}}=ds=\sqrt{d\vec{r}.d\vec{r}}[/tex]

I believe the concept of a norm of a vector is an abstraction from our usual distance formula for the distance between two points... thus it does not seem appropriate to use it here..
 
  • #12
tiny-tim said:
hi krishna! :smile:


It's difficult to see how you could define a third space (or fourth etc). :wink:


Yep..that was something I also thought about..but I do not see what prevents me from defining a dot product between, say, three vectors..and then use it to define my distance or the norm of a vector...
 
  • #13
HallsofIvy said:
Nature didn't "choose" anything. Distance is purely a human invention.

I still remember the shock I felt when, while reading Eddington's "The Mathematical Theory of Relativity", where he was calculating the gravitational "pull" between two bodies, he remarked that we have to decide, arbitrarily, which of several values to use as the "distance" between them.

Hey..thats more like what I was looking for..do you have any more references?
 
  • #14
kakaz said:
I will not agree. Nature choose some kind of space-time as flat. When there is no matter, no energy, no other disturbances, then there is an Euclidean metric - and obviously that is the choice. Why? I do not know...

This is a very interesting issue but at the same time it confuses me.
Is it actually so safe to say that our Universe in its very essence is Euclidean?
 
  • #15
krishna mohan said:
Yep..that was something I also thought about..but I do not see what prevents me from defining a dot product between, say, three vectors..and then use it to define my distance or the norm of a vector...

yes, but you'd need a 3-dot product of a vector with itself

you'd need three copies of the vector in three different spaces

(like, for the standard dot product, you have two copies, one in the original space, one in the dual space).

You still need 3 spaces. :wink:
 
  • #16
mnb96 said:
This is a very interesting issue but at the same time it confuses me.
Is it actually so safe to say that our Universe in its very essence is Euclidean?
That is good point!
At least locally: YES. So for small distances and not very large mases and energies, our universe is Euclidean.

But there is other one point of view we may say the same: at cosmological scales, there is doubt if our universe is expanding ( hyperbolic geometry) or collapsing ( elliptic one), so we may say, that up do our measurements Universe as a whole is very near Euclidean solutions of gravitational equations. And this is even more difficult to explain!
 
  • #17
tiny-tim said:
y

You still need 3 spaces. :wink:

Yes...but is there any difficulty in defining three spaces?
 
  • #18
kakaz said:
That is good point!
at cosmological scales, there is doubt if our universe is expanding ( hyperbolic geometry) or collapsing ( elliptic one), so we may say, that up do our measurements Universe as a whole is very near Euclidean solutions of gravitational equations. And this is even more difficult to explain!

In fact, I was attending a talk yesterday...and the cosmologists showed some graphs and some equations and stated that the known universe has intrinsic curvature very near zero...I guess that is the same as saying the space is Euclidean...
 
  • #19
krishna mohan said:
Yes...but is there any difficulty in defining three spaces?

One will be the original space, one can be the dual space, what can the third one be? :confused:
 
  • #20
krishna mohan said:
In fact, I was attending a talk yesterday...and the cosmologists showed some graphs and some equations and stated that the known universe has intrinsic curvature very near zero...I guess that is the same as saying the space is Euclidean...
Yes exactly - it is the same. So it is vary strange, because such situation, requires some very precise relation for mater and energy of Universe to follow. It is amazing coincidence to be seen in reality. So it is very strange...
 
  • #21
tiny-tim said:
hi krishna! :smile:


I think it's basically because you need an inner product, and that involves two vectors, or two copies of the same vector.

One vector is in the original space, and the other is in the dual space (the space of functionals, or scalar-valued functions).

It's difficult to see how you could define a third space (or fourth etc). :wink:

Hi,
I'm new here. I'm also new to mathematics. I have finished an introductory university course in linear algebra and am very familiar with vector spaces. However, I have never heard of a dual space nor am I familiar with its role regarding inner products. Will you explain further, please?
 
  • #22
Welcome to PF!

Hi Bilbo Baggins! Welcome to PF! :smile:

From http://en.wikipedia.org/wiki/Dual_space" …

Given any vector space, V, over some field, F, the dual space, V*, is defined as the set of all linear functionals on V, i.e., scalar-valued linear maps on V (in this context, a "scalar" is a member of the base-field F). V* itself becomes a vector space over F under the following definition of addition and scalar multiplication:

for all φ,ψ in V*, a in F and x in V. Elements of the algebraic dual space V* are sometimes called covectors or one-forms.

So the dual of the vector a is the function (or covector or one-form) which sends any (ordinary) vector b to the scalar a.b :wink:
 
Last edited by a moderator:
  • #23
krishna mohan said:
In fact, I was attending a talk yesterday...and the cosmologists showed some graphs and some equations and stated that the known universe has intrinsic curvature very near zero...I guess that is the same as saying the space is Euclidean...
Actually that is saying that space is NOT Euclidean- but very close to it. And, by the way, I feel sure he said that the average curvature is very near zero. The curvature close to any massive body is definitely not 0.
 
  • #24
Yes...he talked about the intrinsic curvature...that must mean curvature without any masses...
 
  • #25
krishna mohan said:
hi...

We generally use the Pythagoras formula for distance between two points in 2D, when the Cartesian co-ordinates are given...

One directly extends it to 3D..having the distance going as [tex]\sqrt{x^{2}+y^{2}+z^{2}}[/tex]...For a curved co-ordinate system, we have distances measured by something like

[tex]ds^{2}=g^{\mu\nu}dx_{\mu}dx_{\nu}[/tex]...

I was wondering why it is all about squares and square roots..and not, say cubes and cube roots or fourth roots or something...

The metrics that use cubes or some other real number than 2 do no have a notion of inner product - that is they do not have an idea of an angle between two vectors. The Pythagorean metric is more than just a distance measure because it does have an idea of angle. this is why it is used preferentially.

In Physics inner products are used to compute many important quantities such as work and flux. The inner product allows you to find the component of a field in the direction of motion of a particle that it is acting on. So inner products are natural and essential to Physics.

The basic concept here is the inner product not the Pythagorean distance measure.
 
  • #26
wofsy said:
The metrics that use cubes or some other real number than 2 do no have a notion of inner product - that is they do not have an idea of an angle between two vectors.

Could you give me a reference please?
 
  • #27
krishna mohan said:
Could you give me a reference please?

I'll look for one. But is a simple excercise to show that an inner product determines a quadratic metric and visa versa.
 

1. What is the Pythagoras formula?

The Pythagoras formula is a mathematical equation used to calculate the length of the hypotenuse of a right triangle. It states that the square of the length of the hypotenuse (c) is equal to the sum of the squares of the other two sides (a and b), or c² = a² + b².

2. Why is the Pythagoras formula important?

The Pythagoras formula is important because it allows us to calculate distances and lengths in two-dimensional space. It is also the basis for many other mathematical concepts and is used in various fields such as engineering, architecture, and physics.

3. How is the Pythagoras formula used to measure distances?

The Pythagoras formula is used to measure distances by finding the length of the hypotenuse of a right triangle, which can represent the distance between two points in two-dimensional space. By using the formula, we can calculate the distance between any two points on a coordinate plane.

4. Can the Pythagoras formula be applied to three-dimensional space?

Yes, the Pythagoras formula can also be applied to three-dimensional space. In this case, it is known as the 3D Pythagoras formula, and it is used to find the length of the diagonal of a cuboid or rectangular prism.

5. Are there any real-life applications of the Pythagoras formula?

Yes, there are many real-life applications of the Pythagoras formula. For example, it is used in construction to determine the length of diagonal beams or to calculate the distance between two points on a map. It is also used in navigation to find the shortest distance between two points, and in sports to calculate the distance of a throw or kick.

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