## Recommended Posts

In another thread on this subject it was stated that “we can establish a one-to-one mapping between any Rm and any Rn.”

There is a relationship that occurs in R3 that I believe cannot be mapped to R2.

The relationship exists between the sphere center and the surface curvature.  I believe that because R2 has no third orthogonal plane or axis that this relationship does not exist in R2.

This is best described as a relationship that exists between the tangent to a small circle on the surface of a ball and the gross position of the ball in 3-space.

The illustrations show how this relationship can be seen.

Mathematically, it is expressed as an identity:

$(\cos\frac{\phi}{2}\:\sin\frac{\lambda}{2})^2 +(\cot\frac{\phi}{2}+\cos\alpha)^2=1$

Does the fact that this mathematical identity exists and is not represented in the current mathematical approach to modeling spacetime have any significance?  To me it seems that this isn’t an issue unless the spacetime is curved, at which point it appears that the whole idea of using Rn-1 breaks down. Why isn’t this the case?

##### Share on other sites
7 hours ago, steveupson said:

In another thread on this subject it was stated that “we can establish a one-to-one mapping between any Rm and any Rn.”

Since I expect that comment was mine here is some backup.

Any Rn has the cardinality (loosely same number of points) as R and so

Rn has the same cardinality as Rm.

Note that this does not say that the mapping is unique.

The non uniqueness is brought out in the link.

Here is a finite example.

Consider two (infinite) sets : {1,3,5...} and {0,2,4...}

You can make the obvious pairing

10,32,54...

But others are available

$1 \to 2,3 \to 4,5 \to 6...$

Edited by studiot

##### Share on other sites
7 hours ago, steveupson said:

In another thread on this subject it was stated that “we can establish a one-to-one mapping between any Rm and any Rn.”

There is a relationship that occurs in R3 that I believe cannot be mapped to R2.

Mapping the members of the set doesn't necessarily imply that relationships within a set can be mapped to the other.

##### Share on other sites
5 minutes ago, Strange said:

Mapping the members of the set doesn't necessarily imply that relationships within a set can be mapped to the other.

Suppose you are talking about R3  do the relationships between the members of the set equate to all the points in a 3D  space?

So (1,6,7) could be mapped to(1,6) in R2 and so on?

(2,8,512) could be mapped  to(4,16)? Or would it have to be(4,64)? Either would work?

Trying to familiarise myself with the "set" terrain as applied to dimensional spaces in general.

##### Share on other sites
1 hour ago, Strange said:

Mapping the members of the set doesn't necessarily imply that relationships within a set can be mapped to the other.

That was the simpler version I was looking for.

Thank you Strange, +1

1 hour ago, geordief said:

Suppose you are talking about R3  do the relationships between the members of the set equate to all the points in a 3D  space?

So (1,6,7) could be mapped to(1,6) in R2 and so on?

(2,8,512) could be mapped  to(4,16)? Or would it have to be(4,64)? Either would work?

Trying to familiarise myself with the "set" terrain as applied to dimensional spaces in general.

1) Yes

2) Yes, there are many types of mapping with fancy words to describe them. This one can't be one to one.

3) Either could be done, but see not (2).

An ismorphism is not only one-to one it also preserves the structure of the set, eg to order of the elements (if that is important)

Edited by studiot

##### Share on other sites
1 hour ago, Strange said:

Mapping the members of the set doesn't necessarily imply that relationships within a set can be mapped to the other.

It's hard for me to imagine the usefulness of a mapping that loses so much information about position (in 3D) when we are trying to understand position (in 4D.)

Why would a mapping that doesn't contain this relationship be relevant to physics?

edited to add>  I can understand that it may not be relevant to Galilean spacetime but it doesn't make sense to me that it isn't relevant to relativistic spacetime.

Edited by steveupson

##### Share on other sites
1 minute ago, steveupson said:

Why would a mapping that doesn't contain this relationship be relevant to physics?

Who says it would be?

##### Share on other sites
5 minutes ago, Strange said:

Who says it would be?

Everyone except me.

##### Share on other sites
2 minutes ago, steveupson said:

Everyone except me.

Citation needed.

##### Share on other sites
5 minutes ago, Strange said:

Citation needed.

The Lorentz transformation doesn't maintain this relationship.   I seem to be the only one who sees this or thinks it might be important.

##### Share on other sites
19 minutes ago, steveupson said:

It's hard for me to imagine the usefulness of a mapping that loses so much information about position (in 3D) when we are trying to understand position (in 4D.)

Why would a mapping that doesn't contain this relationship be relevant to physics?

edited to add>  I can understand that it may not be relevant to Galilean spacetime but it doesn't make sense to me that it isn't relevant to relativistic spacetime.

The whole point about these mappings, from the point of view of Physics, is that each different model of Rn leads to a different geometry, characterised by it intrinsic curvature.

So in R2, the plane and the surface of the sphere serve as suitable models.

The flatlanders could determine whether their universe was planar or spherical by measurement made purely within their universe (ie in 2D).

##### Share on other sites
5 minutes ago, steveupson said:

The Lorentz transformation doesn't maintain this relationship.

Can you demonstrate that the Lorentz transform "is a relationship that occurs in R3 that ... cannot be mapped to R2"

##### Share on other sites
17 minutes ago, studiot said:

The whole point about these mappings, from the point of view of Physics, is that each different model of Rn leads to a different geometry, characterised by it intrinsic curvature.

So in R2, the plane and the surface of the sphere serve as suitable models.

The flatlanders could determine whether their universe was planar or spherical by measurement made purely within their universe (ie in 2D).

The point of this thread is to point out that the intrinsic curvature is not accurately captured by the current methods.

Trust that I understand the current method and that you don't really understand the identity above or why it shows that the curvature is more than what flatlanders can see.

21 minutes ago, Strange said:

Can you demonstrate that the Lorentz transform "is a relationship that occurs in R3 that ... cannot be mapped to R2"

First of all, I'm not sure what it is that you want a demonstration of.  And secondly, my ability (or inability) to demonstrate something other than what I am claiming has nothing to do with whether or not my claim is accurate.  It’s math, it’s either correct or it isn’t.

I appreciate that you have at least tried to work the math yourself.  It would greatly facilitate this discussion if someone other than me would try and verbalize what they think the identity above means.  That way it might be easier to assess where to focus the conversation. If the identity has no meaning to anyone then I’m pretty sure that nothing that I can say will make any sense at all.

##### Share on other sites
1 minute ago, steveupson said:

The point of this thread is to point out that the intrinsic curvature is not accurately captured by the current methods.

Trust that I understand the current method

Given the lack of mathematical knowledge you have admitted to before (you have struggled with simple trigonometry) I find it rather implausible that you have a good grasp on differential geometry and pseudo-Riemannian manifolds.

3 minutes ago, steveupson said:

First of all, I'm not sure what it is that you want a demonstration of.

Well, for one thing, the Lorentz transform I am aware of only applies to one dimension. So perhaps you could explain how it is applied to 2 or 3 dimensions and what sort of mapping it provides between these.

Giving your statement immediately above, perhaps you could explain how the Lorentz transform relates to the intrinsic curvature of pseudo-Riemannian manifolds?

4 minutes ago, steveupson said:

And secondly, my ability (or inability) to demonstrate something other than what I am claiming has nothing to do with whether or not my claim is accurate.  It’s math, it’s either correct or it isn’t.

And, as you are making the claim, it is up to you to show that it is correct.

6 minutes ago, steveupson said:

It would greatly facilitate this discussion if someone other than me would try and verbalize what they think the identity above means.

As you are the one who is making assertions about some such relationship, it is necessary for you to identify the specific relationship you are concerned about (and what it means).

##### Share on other sites

I have never made the admission that you attribute to me.  Yes, I am one of those self-taught geniuses that has never had any use for algebra or any of the other standard bookeeping methods.  I generally use my own shorthand, which does complicate things a great deal when trying to explain this problem.

Once again, the math is either correct or it isn't.  Its accuracy has nothing at all to do with my understanding of trig or algebra or anything else.  It has nothing to do with me at all.

The one dimension that you're transforming isn't what you think it is.  It's either no longer orthogonal to anything or its metric has changed.  Do the math.  If you have questions about the identity above then please ask.

Edited by steveupson
syntax, spelling

##### Share on other sites
12 minutes ago, steveupson said:

Yes, I am one of those self-taught geniuses

Of course you are.

12 minutes ago, steveupson said:

Once again, the math is either correct or it isn't.

OK. Then you should be able to demonstrate this, rather than us having to rely on your assertions (after all, it's not about you or what you believe).

##### Share on other sites
1 hour ago, steveupson said:

The point of this thread is to point out that the intrinsic curvature is not accurately captured by the current methods.

Trust that I understand the current method and that you don't really understand the identity above or why it shows that the curvature is more than what flatlanders can see

Perhaps if you cleaned and polished your genius spectacles beofre you read my posts you would refrain from posting such rubbish.

Of course it is distinguished and measured, if that is what you mean by capture (how do you capture curvature?) within the manifold.

That is the definition of intrinsic curvature.

There are indeed more esoteric curvatures that can't be so determined.

Again the point is that in Physics we are seeking reasonable manifolds that do not exhibit pathogenic properties in order to model reality.

The Lorenz relationship is an additional constraint, as is the invariant s2 = x2 + y2 + z2 + (ict)2
The  imaginary i provides the nencessary orthogonal rotation and leads to the negative sign usually seen in the formula.
This invariant is commonly known as the Minkowski interval.

Edited by studiot

##### Share on other sites
5 hours ago, studiot said:

1) Yes

2) Yes, there are many types of mapping with fancy words to describe them. This one can't be one to one.

3) Either could be done, but see not (2).

An ismorphism is not only one-to one it also preserves the structure of the set, eg to order of the elements (if that is important)

Hope I am  keeping to the topic but would what is described as a 3D television screen be an example of a one to one mapping of 4D spacetime to a 2D(+1) surface?

All the members of the set of the scene being broadcast are projected onto the screen and each event being recorded has its own unique place on the screen.

Is this holograph territory?

##### Share on other sites
7 hours ago, studiot said:

Perhaps if you cleaned and polished your genius spectacles beofre you read my posts you would refrain from posting such rubbish.

Of course it is distinguished and measured, if that is what you mean by capture (how do you capture curvature?) within the manifold.

That is the definition of intrinsic curvature.

There are indeed more esoteric curvatures that can't be so determined.

Again the point is that in Physics we are seeking reasonable manifolds that do not exhibit pathogenic properties in order to model reality.

The Lorenz relationship is an additional constraint, as is the invariant s2 = x2 + y2 + z2 + (ict)2
The  imaginary i provides the nencessary orthogonal rotation and leads to the negative sign usually seen in the formula.
This invariant is commonly known as the Minkowski interval.

There are fundamental rules (Pythagoras and plane trigonometry) that are associated with 2 dimensions.  Moving from R2 to R3 simply adds an additional set of these exact same rules, arranged orthogonality,  to the single set of rules that already exists.  Do you understand this?

Also, it can be and has been shown mathematically that a completely new set of rules applies to R3 that cannot exist in R2.  This set of rules are in addition to Pythagoras and plane trigonometry.  Do you understand this?

I'm trying to drill down on some way to explain this to you.  Either you follow me this far, or you don't and I'll have to back up even further.

4 hours ago, geordief said:

Hope I am  keeping to the topic but would what is described as a 3D television screen be an example of a one to one mapping of 4D spacetime to a 2D(+1) surface?

All the members of the set of the scene being broadcast are projected onto the screen and each event being recorded has its own unique place on the screen.

Is this holograph territory?

##### Share on other sites
1 hour ago, steveupson said:

I am not clear where you say I am transiting to "Euclidean 3-space " from.

As I understand my scenario the "transition" is from the world observed by the camera(3D+1 or 4D) to the surface of the viewing screen (2D or 2D+1 if I include time.)

So ,with time  being accounted for I am going from 4D to 3D and otherwise it is simply 3D to 2D.(eg taking a photo)

##### Share on other sites

The way that I understand it, when time is removed from spacetime we are left with Euclidean 3-space.  What we are talking about when we use state-of-the-art math is really a model where we have x-y, x-z, and y-z planes arranged orthogonal to one another.

The geometric rules that apply in any plane are the same as with any other plane, and these geometric rules are consistent with trigonometry and Pythagoras.  A much better nomenclature for this type of construction would be 2D3, or 2D x 2D x 2D.  This is what yields our octants, this 23 relationship. As this is continued further then we get 2D4 or what I believe is called a 4-dimensional manifold, or what you might call 3D+1 or what others might call R4.  This further doubling of the divisions is what causes the need for imaginary numbers, as was pointed out by another member above.

We assign time to this last "2D construction" (clarification to follow) that is added, but we have to treat time differently than we do the other "dimensions" because time is a different quantity than distance or length, which was the only quantity that was contained in the first three 2D constructions or planes.

The reason I say “dimensions” in quotes is because in physics the term can have a completely different meaning.  It can also be used to distinguish between different quantities such as distance or time or temperature or mass, etc.  There are currently seven of these base quantities that can be mixed and matched in certain ways and that fall under a basic branch of math/physics called dimensional analysis.  In this context each different base quantity can be called a dimension.

Where it gets really complicated is when we need more than one instance of any particular base quantity in order to represent a physical phenomenon.  For instance, in the current state-of-the-art model of spacetime we need three instances of length and one instance of time. This is where we can clarify the reference above that talked about time being a "2D construct."  Obviously, since time only exists as on instance of the quantity in spacetime, it doesn't need to be combined with itself like distance does.  Therefore it differs both mathematically and conceptually from the way distance is considered.  That’s probably why time has been selected in the mathematical treatment to get the imaginary coordinates rather than any of the other length “dimensions.”

Being aware of this reality, it is fairly difficult to stay on track with our terminology.  I am currently trying to stay in context with the terminology that was introduced in the previous thread about surfaces in a higher dimension.

Try and see the symmetry of what I’m talking about.  This symmetry is a very fundamental connection, far deeper than the few associations that are necessary in order to add or remove orthogonal planes from a particular geometry as is done with Rn+1 or Rn-1 and so forth.

In the Cartesian system we can begin with three axes and then divide each axis into identical units.  This does not get us completely to where we want to be.  We have to make it axiomatic that the three axes are at right angles to one another.  Direction is an axiom and distance is a quantity, or what we call the metric.

Consider the opposite arrangement.  We have three axes and we connect a point on each axis with a point on the others and we make the angles between them identical angles. By this I mean that the angle formed by each axis with each connecting line is the same. This arrangement is perfectly symmetrical to the identical units in the first example.  We still must make something else axiomatic in order to achieve orthogonality.  This time we must make all these connected points equal distance from the origin.  Instead of equal right angles as in the first case, we have to assign equal lengths to each axis.

In either case, the expressed quantities are only one half of the picture or reference frame while the other half of the picture, the given or axiomatic quantity, is already baked into the system.

It is possible, in Euclidean 3-space, to construct a reference frame that utilizes direction or orientation as the expressed quantity.  In this symmetrical system, length will be axiomatic and without orientation.  Think diameter of a sphere.  It is a length without direction.  In this additional view of Euclidean 3-space, a vector becomes the symmetrical opposite of what we are used to.  The direction is the magnitude or scalar value while the vector becomes the magnitude and length.

I know it sounds extremely weird, but it’s only math.  Try the math.  See that it works. If I had the algebra skills necessary to write this all down algebraically I would continue to do the math and take it to the next level.  But I’m pretty sure that I’ve gone as far as I can without more help. The mathematical identity is the key to understanding all of this.  This identity is something that we just stumbled across while trying to solve some different problems.  We were able to recognize why it is so significant and precisely why it differs from all of the other known geometric identities.

The first step in problem solving is recognizing that there is a problem. It should be clear to everyone that our current understanding of spacetime has some issues.  I believe the math that has been done so far shows that the problem is with how we are doing the geometry.  The standard methods overlook some very basic things that must be included in order to represent a curved spacetime model more accurately.  As a matter of fact, it was once argued by some members that there was no underlying function to the new model, but that argument was dropped when the algebraic expression for the function was produced.  Then it was argued that there was no new identity, everybody move along, nothing to see here.  Just high school math that could be accomplished in an afternoon. Now it is being argued that the new identity is not significant to relativistic transformations, even though this was the original claim. At some point the light will come on for someone.

The identity expressed in the OP must be included as part of any accurate representation of Euclidean 3-space.  Until now, omitting this fundamental element of Euclidean 3-space from our models has never been much of an issue. It’s only when we try and take a close look at curved spacetime that the problems with maintaining this identity become realized.

Maintaining this identity in any more accurate model of spacetime is not an easy undertaking.  It’s a daunting project.  But it must be done in order to make the math work out correctly.

##### Share on other sites
7 hours ago, steveupson said:

The way that I understand it, when time is removed from spacetime we are left with Euclidean 3-space.

Not necessarily. Space doesn't have to be Euclidean, and isn't in the presence of mass.

##### Share on other sites
On 30/11/2017 at 3:02 AM, steveupson said:

Mathematically, it is expressed as an identity:

${\left( {\cos \frac{\varphi }{2}\sin \frac{\lambda }{2}} \right)^2} + {\left( {\cot \frac{\varphi }{2} + \cos \alpha } \right)^2} = 1$

OK so if it is an identity the equation asserts that this is true for any value of phi, lambda and alpha so let us try some

${\left( {\cos \frac{\varphi }{2}\sin \frac{\lambda }{2}} \right)^2} + {\left( {\cot \frac{\varphi }{2} + \cos \alpha } \right)^2} = 1$

Put

$\varphi = \frac{\pi }{2}\quad \lambda = 0\quad \alpha = \pi$

This computes to

${\left( {\sqrt {\frac{2}{4}} *0} \right)^2} + {\left( {1 + \left( { - 1} \right)} \right)^2} = 0 \ne 1$

OOps so the equation is not an indentity.

##### Share on other sites
5 hours ago, studiot said:

OK so if it is an identity the equation asserts that this is true for any value of phi, lambda and alpha so let us try some

(cosφ2sinλ2)2+(cotφ2+cosα)2=1

Put

φ=π2λ=0α=π

This computes to

(240)2+(1+(1))2=01

OOps so the equation is not an indentity.

Correct.

It is an identity in three dimensions.

It doesn't exist in two dimensions.

6 hours ago, Strange said:

Not necessarily. Space doesn't have to be Euclidean, and isn't in the presence of mass.

Correct.

##### Share on other sites
11 minutes ago, steveupson said:

Correct.

It is an identity in three dimensions.

It doesn't exist in two dimensions.

Are you saying that there is a difference between substituting the values φ=π2λ=0α=π in three dimensions and two dimensions?