Your favorite mathematical theorems

In summary: The general case is not easy at all.In summary, we discussed various beautiful mathematical theorems such as the Cayley transformation, Noether's theorem, Liouville's theorem, the Schrödinger equation, the Pythagorean theorem, Stokes' theorem, the residue theorem, and the sum of consecutive numbers. These theorems have different qualities that make them beautiful, whether it's their elegance, usefulness, or unexpected results.
  • #36
atyy said:
But Euclid specified a point not as an abstract object, but as something physical "that which has no parts". Similarly, the Euclidean straight line is a model of a rigid body - the straight edge.
How do you figure these aren't abstractions? A thing with no parts has only one property left: location. The point of a point is to make a perfect specification of a location. If that location has any length, width, or height associated with it, it becomes ambiguous. A "part" is a fraction. A thing with no parts is a thing that can't be subdivided into any equal parts. Because, if it could, the location we're trying to specify would become ambiguous. If we could cut a point in half we would no longer be sure in which half we should locate the end of our line or the intersection of two lines, etc. Anyway, there is no physical object that conforms to that. It's an abstraction. And a straight line is a series of points with length only and no width or height. It, too, is abstract. There is no physical entity with length but without width and height.

I thought the "rigid body" of SR was required to have three dimensions and time. The whole point is to see how length and time are altered by great velocity relative to the observer.
 
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  • #37
zoobyshoe said:
How do you figure these aren't abstractions? A thing with no parts has only one property left: location. The point of a point is to make a perfect specification of a location. If that location has any length, width, or height associated with it, it becomes ambiguous. A "part" is a fraction. A thing with no parts is a thing that can't be subdivided into any equal parts. Because, if it could, the location we're trying to specify would become ambiguous. If we could cut a point in half we would no longer be sure in which half we should locate the end of our line or the intersection of two lines, etc. Anyway, there is no physical object that conforms to that. It's an abstraction. And a straight line is a series of points with length only and no width or height. It, too, is abstract. There is no physical entity with length but without width and height.

I thought the "rigid body" of SR was required to have three dimensions and time. The whole point is to see how length and time are altered by great velocity relative to the observer.

What you are saying makes sense, but the reason I think they aren't mathematical abstractions is that mathematicians seem not to accept these parts of the axioms any more. If I understand mathematicians correctly, one must always start with fundamental undefined objects and relations in the axioms that mutually define each other. Since Euclid never used "that which has no parts" in future derivations, here he must (according to me) have been doing physics, not maths. Or at least that would explain how even in a structure like GR in which Euclidean geometry is false, the Pythagorean theorem seems to remain a major motivation.

Here is the famous quote from Hilbert: "One must be able to say at all times (instead of points, lines, and planes) tables, chairs, and beer mugs." http://www3.canyons.edu/faculty/matsumotos/MathQuotes.htm
 
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  • #38
atyy said:
What you are saying makes sense, but the reason I think they aren't mathematical abstractions is that mathematicians seem not to accept these parts of the axioms any more. If I understand mathematicians correctly, one must always start with fundamental undefined objects and relations in the axioms that mutually define each other. Since Euclid never used "that which has no parts" in future derivations, here he must (according to me) have been doing physics, not maths. Or at least that would explain how even in a structure like GR in which Euclidean geometry is false, the Pythagorean theorem seems to remain a major motivation.

Here is the famous quote from Hilbert: "One must be able to say at all times (instead of points, lines, and planes) tables, chairs, and beer mugs." http://www3.canyons.edu/faculty/matsumotos/MathQuotes.htm
Okay, then:

''Mathematicians are like Frenchmen: whatever you say to them, they translate it into their own language and forthwith it is something entirely different.'' -Goethe
 
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  • #39
Jorriss said:
You need to look at better sources, Lagrangian mechanics has a rigorous framework.

micromass said:
Blatant to you maybe. I certainly know that it's rigorous enough, and I don't think I'm alone in this.

I'll show you how I see rigorous Lagrangian mechanics. Let's see what happens.

We fix a dimension [itex]N=1,2,3,\ldots[/itex]. We fix some function [itex]L:\mathbb{R}^{1+2N}\to\mathbb{R}[/itex], and assume that all its second order partial derivatives exist, and are continuous. We agree on a custom, that the parameter of [itex]L[/itex] is often denoted as [itex](t,x,\dot{x})[/itex], where [itex]t\in\mathbb{R}[/itex], [itex]x\in\mathbb{R}^N[/itex] and [itex]\dot{x}\in\mathbb{R}^N[/itex]. We assume that spacetime points [itex](t_A,x_A)[/itex] and [itex](t_B,x_B)\in\mathbb{R}^{1+N}[/itex] are fixed. We define sets [itex]\mathcal{X}[/itex] and [itex]\mathcal{X}_0[/itex] in the following way:

[tex]
\mathcal{X}= \big\{x\in C^2([t_A,t_B],\mathbb{R}^N)\;\big|\; x(t_A)=x_A,\; x(t_B)=x_B\big\}
[/tex]

[tex]
\mathcal{X}_0 = \big\{x\in C^2([t_A,t_B],\mathbb{R}^N)\;\big|\; x(t_A)=0,\; x(t_B)=0\big\}
[/tex]

Now [itex]\mathcal{X}_0[/itex] is a vector space, while in most cases [itex]\mathcal{X}[/itex] will not be, at least not with the ordinary addition and scaling. By using the ordinary Euclidean norm in [itex]\mathbb{R}^N[/itex] we make [itex]\mathcal{X}[/itex] into a metric space by setting

[tex]
d(x,y) = \sqrt{\int\limits_{t_A}^{t_B} \|x(t)-y(t)\|^2dt} + \underset{t_A\leq t\leq t_B}{\textrm{max}} \|\dot{x}(t) - \dot{y}(t)\|
[/tex]

for all [itex]x,y\in\mathcal{X}[/itex]. We make [itex]\mathcal{X}_0[/itex] into a norm space by setting

[tex]
\|x\| = \sqrt{\int\limits_{t_A}^{t_B} \|x(t)\|^2dt} + \underset{t_A\leq t\leq t_B}{\textrm{max}} \|\dot{x}(t)\|
[/tex]

for all [itex]x\in\mathcal{X}_0[/itex]. Now these metric and norm spaces are related in an obvious way. For example [itex]d(x,x+h)=\|h\|[/itex] holds for all [itex]x\in\mathcal{X}[/itex] and [itex]h\in\mathcal{X}_0[/itex]. Then we define a mapping [itex]S:\mathcal{X}\to\mathbb{R}[/itex] by setting

[tex]
S(x) = \int\limits_{t_A}^{t_B} L\big(t,x(t),\dot{x}(t)\big)dt
[/tex]

Now [itex]S[/itex] is a continuous mapping. (A non-trivial claim! The continuity is considered with respect to the defined metric, of course.) For all [itex]x\in\mathcal{X}[/itex] there exists a continuous linear mapping [itex]DS(x):\mathcal{X}_0\to\mathbb{R}[/itex] and a real coefficient [itex]M(x)\in\mathbb{R}[/itex] such that

[tex]
\big|S(x+\eta) - S(x) - DS(x)\eta\big|\leq M(x)\|\eta\|^2
[/tex]

holds for all [itex]\eta\in\mathcal{X}_0[/itex] with the additional condition [itex]\|\eta\|\leq 1[/itex]. With a fixed [itex]x[/itex] this linear mapping is unique. (This means that the mentioned inequality does not hold with any other linear mapping even if the coefficient [itex]M(x)[/itex] was replaced with a new coefficient too.) If [itex]x[/itex] is a local minimum or a local maximum of [itex]S[/itex], then [itex]DS(x)=0[/itex]. (A non-trivial claim! The locality is considered with respect to the mentioned metric.) Also the relation

[tex]
DS(x)=0\quad\Longleftrightarrow\quad D_t\nabla_{\dot{x}} L\big(t,x(t),\dot{x}(t)\big) = \nabla_x L\big(t,x(t),\dot{x}(t)\big)\quad\quad\forall\; t_A\leq t\leq t_B
[/tex]

holds.

How do you like this theorem? Not very elegant, but I like this. I could mention this as one of my answers to the opening post's question.

I have few questions to those who claim that mainstream Lagrangian mechanics is rigorous. Firstly, have you seen anything like this anywhere ever? I know that this kind of mathematics can be found from some mathematics books, but they are not books on physics. I have never seen a book on physics or mechanics that would give results like this. Have you? Another thing is that do you think that I'm making things too complicated here? Do you think that Lagrangian mechanics can be made rigorous more easily, without these kind of function spaces? I would be interested to hear more.
 
  • #40
http://d2tq98mqfjyz2l.cloudfront.net/image_cache/137408946864354.jpg​

More seriously- the Divergence theorem.
(Stokes' is taken)
 
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  • #41
Particularly beautiful? How about de Rham's theorem?
 
  • #42
Stone-Weierstrass theorem(s)
 
  • #43
Enigman said:
More seriously- the Divergence theorem.
(Stokes' is taken)

I think most people included the Divergence theorem when they said Stokes's theorem. I did. I'd be interested to know if that was indeed what people meant (probably you too, but you were joking:)

Here is an interesting blog post which talks about the Stokes's theorem in a way I think about it - the way in Spivak's Calculus on Manifolds: http://owl-sowa.blogspot.com/search/label/Endre Szemerédi: "Let me clarify how I understand the term “conceptual”. A theory is conceptual if most of the difficulties were moved from proofs to definitions (i.e. to concepts), or they are there from the very beginning (which may happen only inside of an already conceptual theory). The definitions may be difficult to digest at the first encounter, but the proofs are straightforward. A very good and elementary example is provided by the modern form of the Stokes theorem. In 19th century we had the fundamental theorem of calculus and 3 theorems, respectively due to Gauss-Ostrogradsky, Green, and Stokes, dealing with more complicated integrals. Now we have only one theorem, usually called Stokes theorem, valid for all dimensions. After all definitions are put in place, its proof is trivial. M. Spivak nicely explains this in the preface to his classics, “Calculus on manifolds”."

The funny thing is at the end he says Euclidean geometry is dead. But as were discussing, although I usually think of geometry as Riemannian geometry just out of naive physics habit, it doesn't seem possible to motivate Riemannian geometry without Euclidean geonetry. So in that sense, how could Euclidean geometry be dead, even though it is true that metric geometry is the form which seems more alive now. What do others think?
 
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  • #44
The proof that there are more reals than integers, by Cantor diagonalization.
 
  • #45
jostpuur said:
Well I have never seen rigor Lagrangian mechanics either, except in my own notes. The physicists talk about minimizing the Lagrangian, but they have no clue of what kind of function space the domain is, or what metric it would have. Of course the domain space must have some metric, because otherwise the Lagrangian couldn't have local minima. I mean how could a mapping

[tex]
L:?\to\mathbb{R}
[/tex]

have a local minima, if nobody knows what the domain is, or what metric it would have?

Hmm. I believe pages 4-5 of Quantum Mechanics for Mathematicians provides the necessary rigor. Walter Thirring's Classical Mathematical Physics also goes through the requisite differential geometry. The formalism is that of finding the stationary point of a functional using generalized derivatives, such as the Gateaux and Frechet derivatives, so a tiny bit of functional analysis is required.

As for myself, I always loved the generalized Stokes Theorem, as many others have already mentioned.
 
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  • #46
Although not theorems, an engineer I've always been blown away by Fourier series & transform. So incredibly useful for a wide range of applications. While it seems obvious that you should be able to use different function spaces to simplify problems in this case it just works out so elegantly.

As for real theorems, I'm going to go with Residue theorem or Stokes theorem.
 
  • #47
Demystifier said:
The proof that there are more reals than integers, by Cantor diagonalization.

When I first read the proof the Goedel-Rosser incompleteness theorem, I read a proof with an explicit costruction of an undecidable statement, and it seemed very mysterious. Later, I learned that there was a later proof, by Tarski, based on Cantor's diagonalization.

I'm still trying to understand this method based on diagonalization. Dan Hathway has a write-up here based on Hinman's book: http://danthemanhathaway.com/MathNotes/Incompleteness.pdf.
 
  • #48
atyy said:
When I first read the proof the Goedel-Rosser incompleteness theorem, I read a proof with an explicit costruction of an undecidable statement, and it seemed very mysterious. Later, I learned that there was a later proof, by Tarski, based on Cantor's diagonalization.

I'm still trying to understand this method based on diagonalization. Dan Hathway has a write-up here based on Hinman's book: http://danthemanhathaway.com/MathNotes/Incompleteness.pdf.
I think the original Goedel's proof is also based on Cantor's diagonalization, as is any proof in abstract logic based on self-reference.
 

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