# [bi|sur|in]jective coordinates

Discussion in 'Physics & Math' started by RJBeery, Sep 29, 2010.

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1. ### prometheusviva voce!Registered Senior Member

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Second year physics is a great year.

Wave mechanics and all it's confusion beckons. If you're really lucky you'll get to look at the Sommerfeld model as well. Good times.

3. ### RJBeeryNatural PhilosopherValued Senior Member

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Yeah, I'm really looking forward to anything related to wave mechanics and optics...

Anyway, the 2-sphere isn't a valid counter example because it can be described by a surjective mapping. Here's the difference (as I think of it...): the North Pole of a 2-sphere can be described as (r=constant, theta=90, phi={0, 1, 2, 99, -32.9...}), where phi is an infinite set of finite coordinates. The EH of a Black Hole, OTOH, actually requires the use of infinity as the Schwarzschild coordinate itself, meaning it is injective and to me "improper". Whether or not there are other coordinates that can describe the EH of the BH (but don't forget that infinite observer!) with surjectivity or bijectivity was at the core of our disagreement in the other thread.

5. ### przyksquishyValued Senior Member

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Polar coordinates on a sphere are problematic though. For example, in differential geometry there's a natural association between coordinate systems and basis vectors of the tangent space (ie. set of all vectors) at a point. For polar coordinates $(\theta,\,\varphi)$, call the associated vectors $\bar{u}_{\theta}$ and $\bar{u}_{\varphi}$. Intuitively, these are vectors that point along the gridlines defined by the coordinates. For example, $\bar{u}_{\theta}$ points in the direction of increasing $\theta$ with $\varphi$ kept constant. A problem at the poles is that these vectors become ill-defined. $\bar{u}_{\varphi}$ points nowhere (it becomes the null vector) while $\bar{u}_{\theta}$ becomes indeterminate (points in every direction). So if you're doing physics on a sphere, you already see you can't use polar coordinates everywhere: you couldn't express the components of vectors or vector fields at the poles. This sort of pathology is why differential geometers disallow these sorts of coordinate systems.

Last edited: Sep 30, 2010

7. ### Guest254Valued Senior Member

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1,056

What a shame, you don't seem to have understood the salient point of my first post. Particularly - any manifold can be mapped bijectively (i.e. both surjective and injective) to Euclidean space, since the sets have the same cardinality. But this mapping will not be, in general, a coordinate system. So your "theory" is more than a little daft, but this is not unexpected.

I would also suggest you stop using the phrase "coordinate mapping" until you've actually learnt what it means. Before you post next, I think it would be a good idea for you to read and digest the definition of coordinates given in the link.

8. ### QuarkHeadRemedial Math StudentValued Senior Member

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It has been mentioned to me, not unkindly, that my last post had rather limited validity. I believe I was aware of that, but thank my interlocutor nonetheless.

Lemme try and make amends, but first a caveat: in what follows I have, or the sake of brevity, cut an awful lot of corners which may make the true mathmen here cringe more than slightly.

Suppose $X,\,\,Y$ are topological spaces. One says that the map $f:X \to Y$ is continuous iff for $U \subseteq X$ that whenever $f(U)$ is open in $Y$ this implies that $U$ is open in $X$.

If $f^{-1}: Y \to X$ exists and is also continuous one calls $f:X \to Y$ a homeomorphism.

Let us suppose that spacetime is a 4-dimensional topological space, and define a set of local homeomorphisms $h_i:ST \to R^4$, the former being how I will denote spacetime. By local I mean that, for every $U_i \sub ST,\,\,\,h_k(U_k) \sub R^4 \ne h_k(U_j) \sub R^4,\,\,\,k\ne j$.

Now $R^4$ is naturally endowed with a global set of coordinates, so obviously each $h_i(U_i) \sub R^4$ is equipped with its own local coordinates. The continuous inverses $h_i^{-1}(h_i(U_i))$ "carry back" these local coordinates to local regions of spacetime. So spacetime can be "covered" by a set of local coordinate charts, which, unsurprisingly, is called an atlas

It is important to note that this is only possible because $h$ is a homeomorphism, so the assertion that there exists a single global set of coordinates for spacetime is equivalent to the assertion that $ST \simeq R^4$ globally.

Now every topological space comes endowed with a set of homeomorphic invariants, that is if two (or more) topological spaces are homeomorphic there is a list of properties that they must share, essentially by definition.

Let's select the property of compactness. Very loosely, this is the topological version of finiteness. I will give a (slightly naughty) definition in a mo, but first this:

there is a very famous and important theorem that states that the compact subsets of any $R^n$ are precisely those that are closed and bounded. So we can take as our definition of compactness to be this:

a topological space $X$ is compact iff every sequence in $X$ has a limit in $X$. This is easy enough to see for the closed and bounded sets, say, $[a,b]$. But equally easy to see is that $R^n$ itself is not closed (though it is bounded), therefore it is not compact.

So the assertion that there is a global coordinate chart for all spacetime implies that spacetime is globally homeomorphic to $R^4$ which implies that spacetime is not compact which implies that there are sequences in spacetime that have no limit there.

Anyone who thinks this is NOT absurd, needs to tell us where these limits might lie.

Hint

9. ### RJBeeryNatural PhilosopherValued Senior Member

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You're right Guest; I'm wrong to say that spherical coordinates are "valid everywhere", according to the definition of coordinate charts.
In fairness, though, I feel that precise phrase was "put in my mouth":
**It is here that AlphaNumeric suggests that an area of indeterminability in a coordinate system, which is what we were discussing, means that it is invalid. My retort was that a sphere had no such area and therefore presumed that his label of "invalid" did not apply.
Point being, I was wrong in my use of terminology but not my meaning. I tried to make this clear when I wrote
Here, it should be obvious that what I mean by "valid" is the English definition of "appropriate" (as contrasted by the word "fails").

I would think that non-standard usage of terms is excusable given the fact that I usually preface my threads with an apology for my layman's terminology AND I try to define the terms as I use them. If you'd like to play a game of "gotcha" I've seen some pretty epic ones but I really don't see how that benefits the forum. Temur not only had no problem understanding my posts, he does a very nice job responding in a way that is readable to us "laymen"...if you are unable to do so, what does Temur have that you do not?

10. ### Guest254Valued Senior Member

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Well it only took two threads and umpteen posts. I'm glad you've learnt something.

AlphaNumeric was perfectly accurate with his use of the phrase "valid coordinates" - the standard coordinates used in the study of the Schwarzschild solution to the Einstein equations are not globally valid. He can not be held accountable for your lack of knowledge and understanding.

11. ### QuarkHeadRemedial Math StudentValued Senior Member

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So it seems I wasted my time! This is not unusual here, but dispiriting all the same. But here is something I don't quite get
Why are you a layman in mathematical terms? My Physics education stopped at what we call here A-level, say at 17 or 18 years old, but I did learn that you cannot do physics without knowing a fair bit of mathematics.

I find it very hard to believe that you are in year 2 at a reputable university studying physics, and yet have so little knowledge of mathematics. How can this be?

12. ### RJBeeryNatural PhilosopherValued Senior Member

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Frankly it's this kind of post that makes me say you offer very little to the conversation. If you think I contend that the Schwarzschild coordinates are globally valid in the study of BH's then you are lost in this conversation or just posting randomly.

QuarkHead's post, OTOH, appears to have substance and I want to read and reply to it after lunch...

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14. ### funkstarratsknufValued Senior Member

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QuarkHead: I'm not following. Why should it be absurd that spacetime is not compact? I know next to nothing about GR spacetimes, but I do remember enough topology that, intuitively, I would be more surprised by finding that spacetime is compact, than not. Can you elaborate?

15. ### Guest254Valued Senior Member

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It's anything but absurd. If spacetime were to be compact, then a natural consequence is the existence of closed timelike curves.

16. ### RJBeeryNatural PhilosopherValued Senior Member

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QuarkHead: I don't have time to fully digest what you've written (the last math course I took was back in '97; shaking these cobwebs off isn't easy but I'm trying!), but I wanted to respond with something so you don't feel like you're lacking attention before I run out the door. Take these comments with my sincere humility.
...OK aren't you making a presumption here? I've always considered the mathematical world to be on an "informational plane" outside of Reality, if you will, and Physics is involved with picking and choosing what aspects of that imaginary world seem to represent what we observe. Would you make this same statement if time and space were finite? I frequently think about the divide between the mathematical world and Reality, and how the two interact. I've always had a problem with the idea that infinity has any physical significance, as I said, which is the motivation behind my assertion that objects unable to be described other than with an injective mapping are not physical. Also, when I read homeomorphism I think bijection, which makes me think that you are raising the bar on my original theory (recall the 2-sphere, etc)...I could very well be wrong on this point, I'm just trying to finish my thought before leaving...anyway, friend, NOT that I don't appreciate the lesson, but if you think what I'm writing is complete garbage it would save you a lot of work if you could just name a physically undeniable object that can only be described by an injective coordinate mapping.

I'm relying here on your intellectual honesty and capability to comprehend what it is that I'm asking for, rather than feigning confusion because of my inability to use the precise and succinct terminology. You seem like a clever guy and I bet you'll be able to translate to and from layman...

17. ### quadraphonicsBloodthirsty BarbarianValued Senior Member

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Not too be too much of a nit-picky derailer here, but people seem to be using the simple term "manifold" in this thread when they are actually talking about certain specific types of manifolds. Remember that Euclidean space is itself a manifold to begin with. If you mean to refer to a non-trivial/differentiable/Riemannian/whatever manifold, then please say so. I realize that such are mostly what people are interested in when they get into this area, but the failure of clarity makes it more difficult for people trying to get their bearings and learn new material.

18. ### AlphaNumericFully ionizedRegistered Senior Member

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Coming from the guy who then states his completely unsupported and vague belief of "I've always considered the mathematical world to be on an "informational plane" outside of Reality, if you will, and Physics is involved with picking and choosing what aspects of that imaginary world seem to represent what we observe."

The technical details of manifolds etc are not something people just picked randomly, they are there to make sure the foundations upon which calculus on manifolds is built are solid. If you remove axiomatic properties like being a topological space or not needing bijective maps then many results which are used in physics become flawed as they rely on those properties.

And without actually knowing anything about any relevant mathematics you do this how? You just admitted "the last math course I took was back in '97" so clearly you aren't using any mathematics or anything related to mathematics to consider " the divide between the mathematical world and Reality", which is a contradiction.

Perhaps you should find out how its used in physics and mathematics before making your conclusions?

You cannot define calculus in terms of coordinates which are not 'nice'. For instance, suppose you have a function on a sphere which depends on the polar angles, $f(\theta,\phi)$. You can consider how this quantity varies as you vary the angles, ie you compute $\frac{\partial f}{\partial \theta}$ and $\frac{\partial f}{\partial \phi}$. These are generally ill defined at the poles, where $\theta = 0,\pi$ and $\phi$ can be anything. If $\phi$ can be anything then you can not talk about 'smoothly varying' something in terms of it as it can have a discontinuity.

For instance, if you went from London to the North Pole in a straight line (parametrised by $\lambda$) you'd be moving along the path $(\theta,\phi) = (\lambda,0)$. The coordinates smoothly vary with $\lambda$. At the North Pole you turn on the spot by any angle not equal to $\pi$ and carry on walking (now heading to the South Pole). Your path is continuous as you don't 'jump' from any one spot to another yet you'll find that at $\theta = 0$ there is a jump in $\phi$ such that $\frac{\partial f}{\partial \phi}$ is infinite. Clearly this doesn't reflect the physical reality, as you've walked in a continuous path and thus any quantity you're measuring varies continuously too.

Thus the pole in the $(\theta,\phi)$ set of coordinates is 'bad' and you cannot describe consistently any function/object which passes through that point in those coordinates. Hence you need a second set of coordinates. Its only that one point, as removing it makes the sphere equivalent to the flat plane via stereographic projection, but its enough to see an calculus based work defined on a sphere can't be described correctly using only one coordinate set.

19. ### RJBeeryNatural PhilosopherValued Senior Member

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If you're asking me the secret to creative insight I cannot help you...maybe take that stick out of your "arse" and smoke a :m:joint:m

haha OK bro just kidding, settle down.

It is an important question you ask, though. Do you disagree that we can have fundamental guiding physical principles that AREN'T rooted in mathematics? Of course math helps guide our understanding but to me the difference is in how much credence we lend those mathematical descriptions. I'm not claiming the math is irrelevant, but it also isn't REALITY (and I think that's why we have a disconnect on black holes; you accept them because the math permits it regardless of the implications). How do you, for example, mathematically quantify the aesthetics of a QM interpretation? You cannot, which is why, in my opinion, many Physicists have no opinion on the matter! "It is what it is! Shut up and calculate!" They [you] get so caught up in the math of it all that they [you] lose sight of the big picture...

20. ### NeverflyBannedBanned

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R.J. Berry, I have a question for you...

Why does light travel slower through a medium?

Research the answer, then shoot me a P.M. and explain the entire process, please.
Since it's not really relevant to this thread- that is why I'd like the answer in a P.M.

However, the reason I am I'm confronting you in this thread, now, about that particular topic IS very relevant to why you have gotten some of the responses you have gotten.

21. ### CptBorkValued Senior Member

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Everything in math is relevant to reality. For instance, I can guarantee you that if you have a properly working calculator, pick a prime number $p$, an integer $a$ not divisible by $p$, and punch in $a^{p-1}-1$, the result will always be divisible by $p$. So at a minimum, I can tell you what your calculator is going to do- predictive power, my friends.

22. ### Guest254Valued Senior Member

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I'm really, really disappointed that you've failed to understand this point. Any manifold can be mapped bijectively to a Euclidean space. This statement is clear enough, isn't it? So the sphere, the plane, the Klein bottle, all the pseudo-Riemannian manifolds corresponding to solutions to the Einstein equations.... can be mapped bijectively to Euclidean space.

So every manifold is "physical" according to your definition, which is why I told you it was more than a little daft. I feel bad for having to explain this again.

23. ### RJBeeryNatural PhilosopherValued Senior Member

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Well, frankly Guest I have little faith that any miscommunication between us isn't intentional feigned confusion on your part due to some technically inappropriate but otherwise understandable term usage on my part. Either that or, as I said before, there are other posters here that have the ability to translate to-and-from layman that you are lacking.

Hell, I'll give you the benefit of the doubt...so what is the function that describes the 2-sphere bijectively?