  # Tom Mattson

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## Everything posted by Tom Mattson

1. 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."
2. 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.
3. Yes. More precisely, the space spanned by the photon's polarization vectors has dimension 2. 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.
4. 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.
5. Yes, if you change to spherical coordinates then not all of the components of the $$dx^{\mu}$$ have the same dimension. Consequently, not all of the components of the metric tensor do.
6. 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: $ds^2=\eta_{\mu\nu}dx^{\mu}dx^{\nu}$ $ds^2$ has dimension $[L]^2$. $dx^{\mu}$ and $dx^{\nu}$ have dimension $[L]$. So $\eta_{\mu\nu}$ (and hence all of its components) are dimensionless.
7. ## How is it possible that the magnetic fields is perpendicular to the electric 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. 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 $|\vec{E}|\neq 0$, $|\vec{B}|=0$ and (Lorentz) boost to a frame in which $|\vec{E}^{\prime}|=0$ and $|\vec{B}^{\prime}|\neq 0$.
8. 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.
9. Back to the OP... 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.
10. 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.
11. 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. Then why did you quote the SR result?
12. 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? Well, yes, using SR to draw inferences about gravitational time dilation isn't exactly good science.
13. 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 $P\subset\mathbb{Z}$, no? But no matter, because as you point out... 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... Who said that? Not me. Yes and then again, one mightn't say it. The definition I cited is certainly sensitive to the set S. That is not how I defined a binary operation. I defined it as a map $b:S\times S \rightarrow S$. My definition necessarily entails closure of S under b, while yours doesn't.
14. 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. It's a property of + acting on P. P is just a set, and set theory knows nothing of +. 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. 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: 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... $f:\mathbb{R} \rightarrow [0,1]$, given by $f(x)=\sin(x)$ is not the same function as $g:\mathbb{Q} \rightarrow [0,1]$, given by $g(x)=\sin(x)$ So if we look at + as a function, then under this interpretation of functions $+_{\mathbb{Z}}:\mathbb{Z}\times\mathbb{Z} \rightarrow \mathbb{Z}$ is not the same as $+_P:P\times P\rightarrow P$. hmmmmm....(scratches head)
15. 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. 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.
16. Why yes, I posted a complete proof of that just one month ago! http://www.scienceforums.net/forum/showthread.php?t=26984
17. 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.
18. 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.
19. 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.
20. 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?
21. So you're saying that your drawing of a dog can fetch your slippers?
22. 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"?
23. What does the "o" represent? Surely it isn't zero. And what is "dWt"?
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