Tom Mattson

Those aren't paradoxes though...

Paradoxes of self-reference are notorious in logic.

"This sentence is false."

Another one is

"The sentence below is true.

The sentence above is false."

It's because the other 46 electrons are spin-paired. Each one of the inner electrons that is spin up has a partner that is spin down. Vectorially the angular momenta add up to zero. The only one that's left is that lonely little 47th electron.

Does this mean that the polarization is 2-dimensional,

Yes. More precisely, the space spanned by the photon's polarization vectors has dimension 2.

and hence a photon is like a spin 1/2 particle ?

No. First and most basic: 1/2 does not equal 1! The photon carries more spin angular momentum than say, an electron. And second, if the photon were spin 1/2 it would be a fermion. But the photon doesn't obey Fermi-Dirac statistics, it obeys Bose-Einstein statistics.

No, there are no known particles like that. However protons behave differently depending on the circumstances. At large distances the strong nuclear force doesn't bind particles together, so protons repel each other via their electric charges. However if two protons can be brought together close enough then they find there is a strong attraction between them. Also, neutrons aren't attracted to each other at large distances. But if you bring them in close together they are attracted to each other via the strong force.

They're not homo, but the proton appears to be bi-curious and the neutron is somewhat of a closet case.

Yes, if you change to spherical coordinates then not all of the components of the [tex]dx^{\mu}[/tex] have the same dimension. Consequently, not all of the components of the metric tensor do.

A metric is always a distance. Or do you mean metric tensor? I gather that's what you could mean by your reference to "components". If that is the case then all components of the metric tensor are dimensionless, as you can see from the equation of the invariant interval:

[math]ds^2=\eta_{\mu\nu}dx^{\mu}dx^{\nu}[/math]

[math]ds^2[/math] has dimension [math][L]^2[/math]. [math]dx^{\mu}[/math] and [math]dx^{\nu}[/math] have dimension [math][L][/math]. So [math]\eta_{\mu\nu}[/math] (and hence all of its components) are dimensionless.

I opened this week's New Scientist and was in for a bit of a shock when I saw a full-page ad title "No Black Holes (General Relativity Contradicts Itself)". The clincher:

 

 

 

Surprise!

 

You can read the full text of the ad here (we may leave a comment to let the author know we're discussing his ad). I'm naturally a tad skeptical (the argument doesn't make sense to me), but I thought I'd open this up to a typical SFN free-for-all.

 

What do you think? Another crackpot?

Smells like Zanket to me.

An electric field is actually the same thing as a magnetic field, that's why we call it an electromagnetic field.

An electric field is not the "same thing" as a magnetic field. For instance, a constant uniform electric field can accelerate a stationary charge in a straight line. A constant uniform magnetic field can't do that.

The way it works is that if you move through an electric field, you would call it a magnetic field.

No, you wouldn't. You would see an electric field and a magnetic field. There is no way to start in a frame in which [math]|\vec{E}|\neq 0[/math], [math]|\vec{B}|=0[/math] and (Lorentz) boost to a frame in which [math]|\vec{E}^{\prime}|=0[/math] and [math]|\vec{B}^{\prime}|\neq 0[/math].

so... it'd have to either be 100% efficient (not possible),

It's more specific than that. It would have to convert thermal energy into work with 100% efficiency. In that case you would have a violation of the 2nd law.

The other type of perpetual motion machine (PPM) would violate the first law. So if, say, the Earth were speeding up with no expenditure of energy, that would be such a PPM.

Back to the OP...

or is 'perpetual motion machine' more than just something that perpetually moves untill it breaks (i'm taking 'because it'll one day be destroyed' as a lame answre btw)

Yes, it something more than that. A perpetual motion machine of the first kind produces more energy than it consumes, which violates the first law of thermodynamics. A perpetual motion machine of the second kind spontaneously converts thermal energy into work, which violates the second law. The Earth does neither of these.

I still don't know what you mean by "equivalence", but it certainly isn't the normal definition. Two statements are "equivalent" if, under a given set of circumstances, either they both hold, or neither of them holds. SR time dilation and gravitational time dilation obviously do not enjoy the status of "equivalence", under that definition.

Tom: there is a deep equivalence between special and general relativity with respect to time dilation.

I don't know what you mean by that, but it is clear that the equations for SR and gravitational time dilation are not equivalent.

I can explain what gravitational time dilation actually is.

Then why did you quote the SR result?

It's to do with time dilation, Martin. We know that time dilation is programmed into GPS. It's real. And we know that the more gravity there is, the more time dilation there is. Consider a supernova, a collapsing star that has formed a black hole. At the event horizon, time dilation goes infinite:

 

 

The time dilation is infinite.

That's the SR time dilation formula. What makes you think it applies to gravitational time dilation? I don't know much about GR, but Mortimer at Physics Forums presented the gravitational time dilation formula thusly:

http://www.physicsforums.com/showthread.php?t=72898

It clearly does not diverge to infinity. Does anyone know if this equation is correct for black holes?

I know why there are no black hole singularities. It's crushingly simple. But the explanation is deemed to be pseudoscience, so I can't share it here.

Well, yes, using SR to draw inferences about gravitational time dilation isn't exactly good science.

Umm. So, if P is "just a set", and set theory knows nothing of +, how can you then assert that [edit]"It's a property of + acting on P"[/edit]?

Because operations can be defined on sets? I really don't see why my remarks are so bewildering. If you're only talking about some set P, then surely the only "properties" of P that exist without introducing an operation such as + are set theoretic properties such as [math]P\subset\mathbb{Z}[/math], no? But no matter, because as you point out...

Recall I suggested the notion of an "algebraic set", i.e one with a binary operation.

By the definition of a binary operation that I cited, P is not an algebraic set, because as I said, "Right, because + is not a binary operation on P, "to which you responded...

Oh? 3 +5 is undefined? Surely not.

Who said that? Not me.

I completely agree, but not for the reasons you gave. Binary operations, one might say, are insensitive to the space they operate on.

Yes and then again, one mightn't say it. The definition I cited is certainly sensitive to the set S.

As you did; +: X × X → Y, ×: Y × Y → Z, where possibly X = Y, Y = Z, I'm reasonably sure all other imaginable operations can be derived from these.

That is not how I defined a binary operation. I defined it as a map [math]b:S\times S \rightarrow S[/math]. My definition necessarily entails closure of S under b, while yours doesn't.

Umm, well, I still don't see your point. Every group G has at least one subgroup, that is G itself, by the definition. How does this comment work here?

If H=G itself, then the requirement that H be closed under * is redundant. It's only not redundant if H is a proper subset of G.

But this is a property of P, not the operation +.

It's a property of + acting on P. P is just a set, and set theory knows nothing of +.

Sorry to sound assertive, but; +: Z × Z → Z defines an abelian group (let's say) whereas +: P × P -/-> P, rather +: P × P → Z, hence P is not closed under +, and therefore P is not a group, therefore not a subgroup of Z.

Right, because + is not a binary operation on P. So in the formalism I learned, we don't even get to the group axioms. <P,+> isn't eligible for consideration as a group.

But for certain sure, P is a proper algebraic subset of the algebraic set Z. Are you suggesting there are two "varieties" of +?

No. I'm suggesting that + is a binary operation on some sets, and it is not a binary operation on some other sets.

Edited to add:

How do you define binary operation, Xerxes? The definition I cited is the only one I've ever seen.

Edited to add again:

Xerxes: Are you suggesting there are two "varieties" of +?

 

Tom: No.

Actually, come to think of it, maybe I am! In the spring I took a second algebra course, and in that course the professor was very specific about what he meant by a function. To define a function, one must specify the domain, the codomain (a superset of the range), and a rule for assigning each element of the former set to a unique element of the latter set. He maintained the point of view that changing any of these three items changes the function. For instance...

[math]f:\mathbb{R} \rightarrow [0,1][/math], given by [math]f(x)=\sin(x)[/math]

is not the same function as

[math]g:\mathbb{Q} \rightarrow [0,1][/math], given by [math]g(x)=\sin(x)[/math]

So if we look at + as a function, then under this interpretation of functions [math]+_{\mathbb{Z}}:\mathbb{Z}\times\mathbb{Z} \rightarrow \mathbb{Z}[/math] is not the same as [math]+_P:P\times P\rightarrow P[/math].

hmmmmm....(scratches head)

You surely must have closure as an additional axiom?

Not for groups, but for subgroups. If you take the definition of binary operation that I was taught, then later on down the line we have to have the following theorem.

Let <G,*> be a group and let H be a nonempty subset G. <H,*> is a subgroup of <G,*> iff it is closed both under * and the taking of inverses.

I cannot see in what sense the + on Z is said to be a "complete" operation, whereas the + on P (it's subset) is partial. Did your prof give any more info?

No, but I think he pretty much said it all with what I quoted. + on Z is a map from ZxZ-->Z, and it's not a map from PxP-->P. There's the difference right there.

Why yes, I posted a complete proof of that just one month ago!

http://www.scienceforums.net/forum/showthread.php?t=26984

EDIT:@Tom: [smart-ass]Afaik, closure is a group-axiom.[/smart-ass]

That's what I thought too, until last fall. I took a graduate course in algebra, and the professor defined a group as a set with binary operation that satisfies the axioms for associativity, identity element, and inverses. That day I was the smart-ass, and I "reminded" him of closure. He rolled his eyes and said that that comes with "binary operation". He said that without closure the operation is just a "partial operation".

But whatever, you can certainly list closure a second time if you want. The wiki page on Groups does just that.

Although IIRC associativity, identity element et cetera et cetera are axioms of addition if not of binary operations generally, no?

No, they aren't. As river_rat has already pointed out, addition is defined on the naturals, which don't include zero or negative numbers (so no additive identity or inverses). The definition of a binary operation b on a set S implies closure of S under b, but it does not imply not any of the group axioms for S.

Addition is a GROUP operation.

No, addition doesn't "belong" to any particular algebraic structure. Addition is a binary operation, and the definition of a binary operation is made prior to the definition of a group, monoid, or even semigroup.

And what is the definition of a binary operation, you might ask?

A binary operation b on a set S is a map b:SxS-->S. That's it! There is nothing more implied by this. Not associativity, not an identity element, and not inverses. Of course, mathematicians find it perverse to use the "+" symbol to indicate a noncommutative binary operation, so they almost universally never do it. But that's just an aesthetic choice.

One day I believe humans will understand every event within the human brain down to all X quintillion atoms. Not surprisingly, "subjectivity" will no longer be applied to the human experience,

Why not? Say a man who is born deaf is the one who comes up with the stroke of genius that maps out all brain functions in terms of atomic states. What makes you believe that he could ever possibly know what it is like to hear the sound of his own voice?

But when I draw one on a piece of paper, my dog will ostensibly look different from your dog. Sure they have some of the same characteristics, but they are certainly different. Just like my representaiton of Dirac matrices look different from yours---that's ok, they still DO the same thing, they still ACT on the same space, they just LOOK different. He wasn't so amused.

So you're saying that your drawing of a dog can fetch your slippers?

Do you not see what I'm getting at? Even the computer is still your own perception. The GM counter is still your own perception. There is no method to ever avoid this. You must take the leap of faith that those perceptions are always accurate, even when we know that not to be the case.

We assume that our waking, sober world is perceived without substantial error. That is the assumption science must make.' And it is an assumption that may turn out to be either right or wrong.

Why must one take that leap of faith? Why must one make that assumption?

Why not simply regard science as the discipline which seeks to determine the laws that describe what we observe, rather than what is "really there"?

What does the "o" represent? Surely it isn't zero. And what is "dWt"?