Slashdot editor says that Pioneer passed neptune on its way out of the solar system, Wikipedia claims it passed Pluto. Which is it? I think it's Neptune, can someone who is sure go fix the Wikipedia article?
thanks, you're a pal
Re:Whither the iMac?
on
RIP G4 PowerMac
·
· Score: 3, Informative
What are you talking about? When did anyone say that they were no longer shipping the G4 iMac?
They stopped production of the G4 powermac, but the G4 iMac is still around
It might not be as shiny as an Xserve cluster, but it is a heck of a lot cheaper.
Um, yeah, an Xserve is a piece of hardware, and a C-compiler is a piece of software. apples and oranges.
you could install distcc on your Xserve cluster, nnd it (the software) would still be cheap, and it (the hardware) would also be shiny.
Um, i guess the point is that if you want distributed compiling for MacOSX, heretofore the only option was Xserve cluster, and now linux/x86 cluster is also an option. ok ok....
i can't believe that someone puts forth such antiscientific falsehoods as fact, and gets modded up for interesting.
there is a fourth possibility that you missed: you don't know anything about cosmology.
the age of the universe has been pinned down at 13.7 Billion years. it is also extremely flat. we expect that even if the universe is closed, there are many many Hubble volumes, and so no matter how far we see, we will never see anything like an "edge". in fact, the Cosmological principle, which has significant experimental evidence, posits that the universe is homogeneous and isotropic. this explicitly rules out anything like an "edge".
nevertheless, an infinitely large flat universe is still possible, even with a finite age and a big bang. this is cosmology 101.
if you do not understand cosmology, that is fine, but you should instead make a question to try to understand better, instead of lying.
well, i guess my misconception is based on what i learned about the CMB: it is the light of last scattering at recombination, when the electrons and protons became hydrogen. that is the oldest light we can see, because before recombination, the universe was opaque due to the ionized hydrogen.
at least, this is the description of where the CMB comes from that i have heard, and it seems to be contradictory to what you and the article are saying.
or else maybe i am just badly confused.
In general, ionized and neutral gas are neither opaque or transparent. It depends on the density, temperature, the details of the radiation field, and also where an observer is located.
well sure, in general nothing can be 100% transparent, there are always some wavelengths that it can absorb. diatomic hydrogen can absorb some infrared wavelengths.
but the night sky is fairly transparent to visible light. i assume that this is because the universe is cold enough that the hydrogen is neutral. at a time when the hydrogen ionized, i would assume that most light would get absorbed.
the article says the opposite. do you agree with the article?
The newly discovered galaxy is likely to be a young galaxy shining during the end of the so - called "Dark Ages" - the period in cosmic history which ended with the first galaxies and quasars transforming opaque, molecular hydrogen into the transparent, ionized universe we see today.
ionized = transparent? i thought it was the other way around: neutral atomic hydrogen is transparent, ionized charged plasma is opaque. am i confused, or is it the article?
what is the status of GCC optimizations for the macintosh platform? why is the GCC team resisting? and can't apple just fork GCC if they need to make heavy modifications for the G5?
ut isn't it a trivial bundle, in two different ways, as you point out with reference to the example of a cylinder?
a bundle is a triple (E,M,p). E and M are manifolds, and p is a smooth map from E to M. until i specify a base space and a mapping, i don t have a bundle. if E is a product manifold, there are at least two easy choices to make it into a bundle, but there might be other choices. the point is, until i have made some choice, it is incorrect to say that M1xM2 is a bundle.
Anyway, most bundles are not trivial, and non-trivial bundles can't be simply "turned sideways" like a product can. I could be wrong, but it seems to me that in the topological category, and maybe in the smooth category as well, "most" bundles have a "unique" bundle structure map (in the sense that any two of them differ by a unique automorphism of the bundle space). On the other hand, spacetimes live in the category of smooth spaces with a tensor field which satisfies some conditions, etc... and in that category, maybe bundle structures are not unique?
the manifold M1xM2 can be a bundle in lots of ways, 2 of which are trivial. it is not a bundle until i choose such a way. your question about most bundles having a unique bundle structure doesn t quite make sense. do you want to know if most pairs of manifolds admit a unique mapping between them that makes them into a bundle? i think that is clearly not true. i don t even think it is true up to bundle morphisms.
i am aware of the difference between a trivial and a nontrivial bundle. i just am not sure why you want spacetime to be a bundle.
have to point out, however, that a 6-dimensional fiber bundle on a 4-dimensional space IS a 10-dimensional manifold!
yes, a 4 dimensional base space with a 6 dimensional fibre is a 10 dimensional manifold. but there are lots of other 10 dimensional manifolds that are not bundles, not even locally. i don t know why you want to restrict spacetime to be a bundle.
A product is a specific example of a fiber bundle---a trivial example, in the technical sense of "trivial"... So why can't space-time be a nontrivial bundle?
the property of being a bundle is extra structure that a manifold can have. not all manifolds have this structure. of course, if spacetime is M4xCY3, then it can be made into a trivial bundle rather obviously, but the point is, there is no reason to do this. a product of manifolds is just a product of manifolds. not a bundle.
Example: a cylinder in R^3 is a bundle of lines over a circle. It's trivial---a cylinder is the product of a line and a circle.
yes, but you can also make the line the base space of the cylinder, and let each fibre be the circle. this is a totally different bundle, but the same manifold. furthermore, there is no canonical "best choice" of which of these two bundles i should choose, if i want to start calling this cylinder a bundle. since i am only interested in what kind of topology and geometry spacetime has, it makes no sense to choose some bundle structure on it.
so let s stop talking about bundles. i guess you still want to ask the question why is spacetime M4xCY3 instead of having M4 embedded in the 10 dimensional spacetime in some nontrivial way. i think that s a good question. i donno
I don't know why they take a product $M \times V$
because they want it to describe our universe. so some part of spacetime has to be 4 dimensional minkowski space. that s M. the rest is V.
why not allow a more general bundle of C-Y 3-folds?
spacetime is not a bundle. it is just a manifold.
The imaginary version of string theory which exists only in my mind has the universe as a bundle of Calabi-Yau 3-folds over $M$ (a real 4-manifold with a Minkowski metric, or something like that... anyway, a $(3,1)$ form --- that's a Minkowski metric, right???)
it is not a bundle! the spacetime is a 10 dimensional manifold, not a 4 dimensional base space with a 6 dimensional fiber.
by the way, a metric is a (0,2) tensor, not a (3,1) tensor, and not a form at all (since it is a symmetric tensor, and forms are necessarily antisymmetric). you are thinking of the signature. the signature of the metric is the number of positive diagonal elements minus the number of negative diagonal elements, in some diagonal basis. for minkowski space, or any lorentzian metric, the signature is 3-1.
You do know that that's talking about the OLD lisence right?
yeah
That said I think some of these issues may still be valid
well yeah, that was my question.
the FSF has commentary on a variety of open source lisences. according to them, the plan 9 license did not qualify as a free software license, for a variety of reasons, the worst of which is a clause allowing Bell Labs to restrict and revoke your license under certain unreasonable conditions. see this
i wonder if this new revised license has fixed any of those problems?
here is the statement from RMS.
When I saw the announcement that the Plan 9 software had been
released as "open source", I wondered whether it might be free
software as well. After studying the license, my conclusion was that
it is not free; the license contains several restrictions that are
totally unacceptable for the Free Software Movement. (See
http://www.gnu.org/philosophy/free-sw.html.)
I am not a supporter of the Open Source Movement, but I was glad when
one of their leaders told me they don't consider the license
acceptable either. When the developers of Plan 9 describe it as
"open source", they are altering the meaning of that term and thus
spreading confusion. (The term "open source" is widely misunderstood;
see http://www.gnu.org/gnu/philosophy/free-software-fo r-freedom.html
Here is a list of the problems that I found in the Plan 9 license.
Some provisions restrict the Plan 9 software so that it is clearly
non-free; others are just extremely obnoxious.
First, here are the provisions that make the software non-free.
You agree to provide the Original Contributor, at its request, with a
copy of the complete Source Code version, Object Code version and
related documentation for Modifications created or contributed to by
You if used for any purpose.
This prohibits modifications for private use, denying the users a
basic right.
and may, at Your option, include a reasonable charge for
the cost of any media.
This seems to limit the price that may be charged for an initial
distribution, prohibiting selling copies for a profit.
Distribution
of Licensed Software to third parties pursuant to this grant shall be
subject to the same terms and conditions as set forth in this
Agreement,
This seems to say that when you redistribute you must insist on a contract
with the recipients, just as Lucent demands when you download it.
1. The licenses and rights granted under this Agreement shall
terminate automatically if (i) You fail to comply with all of the
terms and conditions herein; or (ii) You initiate or participate
in any intellectual property action against Original Contributor
and/or another Contributor.
This seemed reasonable to me at first glance, but later I realized
that it goes too far. A retaliation clause like this would be
legitimate if it were limited to patents, but this one is not. It
would mean that if Lucent or some other contributor violates the
license of your GPL-covered free software package, and you try to
enforce that license, you would lose the right to use the Plan 9 code.
You agree that, if you export or
re-export the Licensed Software or any modifications to it, You are
responsible for compliance with the United States Export
Administration Regulations and hereby indemnify the Original
Contributor and all other Contributors for any liability incurred as a
result.
It is unacceptable for a license to require compliance with US export
control regulations. Laws being what they are, these regulations
apply in certain situations regardless of whether they are mentioned
in a license; however, requiring them as a license condition can
extend their reach to people and activities outside the US
government's jurisdiction, and that is definitely wro
Slashdot Mirror
thanks, you're a pal
They stopped production of the G4 powermac, but the G4 iMac is still around
Um, yeah, an Xserve is a piece of hardware, and a C-compiler is a piece of software. apples and oranges.
you could install distcc on your Xserve cluster, nnd it (the software) would still be cheap, and it (the hardware) would also be shiny.
Um, i guess the point is that if you want distributed compiling for MacOSX, heretofore the only option was Xserve cluster, and now linux/x86 cluster is also an option. ok ok....
of course, the age of the universe is a theory based on data. however, it is quite a significant amount of data.
i can't believe that someone puts forth such antiscientific falsehoods as fact, and gets modded up for interesting. there is a fourth possibility that you missed: you don't know anything about cosmology. the age of the universe has been pinned down at 13.7 Billion years. it is also extremely flat. we expect that even if the universe is closed, there are many many Hubble volumes, and so no matter how far we see, we will never see anything like an "edge". in fact, the Cosmological principle, which has significant experimental evidence, posits that the universe is homogeneous and isotropic. this explicitly rules out anything like an "edge". nevertheless, an infinitely large flat universe is still possible, even with a finite age and a big bang. this is cosmology 101. if you do not understand cosmology, that is fine, but you should instead make a question to try to understand better, instead of lying.
your sig.... that is a quote from one of the 2001 installments from Clarke, not Pratchett, i think
well, i guess my misconception is based on what i learned about the CMB: it is the light of last scattering at recombination, when the electrons and protons became hydrogen. that is the oldest light we can see, because before recombination, the universe was opaque due to the ionized hydrogen. at least, this is the description of where the CMB comes from that i have heard, and it seems to be contradictory to what you and the article are saying. or else maybe i am just badly confused.
well sure, in general nothing can be 100% transparent, there are always some wavelengths that it can absorb. diatomic hydrogen can absorb some infrared wavelengths.
but the night sky is fairly transparent to visible light. i assume that this is because the universe is cold enough that the hydrogen is neutral. at a time when the hydrogen ionized, i would assume that most light would get absorbed.
the article says the opposite. do you agree with the article?
and molecular hydrogen is opaque?
only 1 pci slot? my motherboard has 6! so does every motherboard i have ever seen. perhaps you have only one AGP slot?
ionized = transparent? i thought it was the other way around: neutral atomic hydrogen is transparent, ionized charged plasma is opaque. am i confused, or is it the article?
what is the status of GCC optimizations for the macintosh platform? why is the GCC team resisting? and can't apple just fork GCC if they need to make heavy modifications for the G5?
ut isn't it a trivial bundle, in two different ways, as you point out with reference to the example of a cylinder?
a bundle is a triple (E,M,p). E and M are manifolds, and p is a smooth map from E to M. until i specify a base space and a mapping, i don t have a bundle. if E is a product manifold, there are at least two easy choices to make it into a bundle, but there might be other choices. the point is, until i have made some choice, it is incorrect to say that M1xM2 is a bundle.
Anyway, most bundles are not trivial, and non-trivial bundles can't be simply "turned sideways" like a product can. I could be wrong, but it seems to me that in the topological category, and maybe in the smooth category as well, "most" bundles have a "unique" bundle structure map (in the sense that any two of them differ by a unique automorphism of the bundle space). On the other hand, spacetimes live in the category of smooth spaces with a tensor field which satisfies some conditions, etc... and in that category, maybe bundle structures are not unique?
the manifold M1xM2 can be a bundle in lots of ways, 2 of which are trivial. it is not a bundle until i choose such a way. your question about most bundles having a unique bundle structure doesn t quite make sense. do you want to know if most pairs of manifolds admit a unique mapping between them that makes them into a bundle? i think that is clearly not true. i don t even think it is true up to bundle morphisms.
i am aware of the difference between a trivial and a nontrivial bundle. i just am not sure why you want spacetime to be a bundle.
have to point out, however, that a 6-dimensional fiber bundle on a 4-dimensional space IS a 10-dimensional manifold!
yes, a 4 dimensional base space with a 6 dimensional fibre is a 10 dimensional manifold. but there are lots of other 10 dimensional manifolds that are not bundles, not even locally. i don t know why you want to restrict spacetime to be a bundle.
A product is a specific example of a fiber bundle---a trivial example, in the technical sense of "trivial"... So why can't space-time be a nontrivial bundle?
the property of being a bundle is extra structure that a manifold can have. not all manifolds have this structure. of course, if spacetime is M4xCY3, then it can be made into a trivial bundle rather obviously, but the point is, there is no reason to do this. a product of manifolds is just a product of manifolds. not a bundle.
Example: a cylinder in R^3 is a bundle of lines over a circle. It's trivial---a cylinder is the product of a line and a circle.
yes, but you can also make the line the base space of the cylinder, and let each fibre be the circle. this is a totally different bundle, but the same manifold. furthermore, there is no canonical "best choice" of which of these two bundles i should choose, if i want to start calling this cylinder a bundle. since i am only interested in what kind of topology and geometry spacetime has, it makes no sense to choose some bundle structure on it.
so let s stop talking about bundles. i guess you still want to ask the question why is spacetime M4xCY3 instead of having M4 embedded in the 10 dimensional spacetime in some nontrivial way. i think that s a good question. i donno
because they want it to describe our universe. so some part of spacetime has to be 4 dimensional minkowski space. that s M. the rest is V.
why not allow a more general bundle of C-Y 3-folds?
spacetime is not a bundle. it is just a manifold.
The imaginary version of string theory which exists only in my mind has the universe as a bundle of Calabi-Yau 3-folds over $M$ (a real 4-manifold with a Minkowski metric, or something like that... anyway, a $(3,1)$ form --- that's a Minkowski metric, right???)
it is not a bundle! the spacetime is a 10 dimensional manifold, not a 4 dimensional base space with a 6 dimensional fiber.
by the way, a metric is a (0,2) tensor, not a (3,1) tensor, and not a form at all (since it is a symmetric tensor, and forms are necessarily antisymmetric). you are thinking of the signature. the signature of the metric is the number of positive diagonal elements minus the number of negative diagonal elements, in some diagonal basis. for minkowski space, or any lorentzian metric, the signature is 3-1.
working in momentum space instead of real space is just a convenient choice of coordinates. it does not make the results any less real.
wolve is not a word
7.0? is that panther? where did you get it?
from the article: In May, Novell Inc. claimed that it, and not IBM, had the rights to the Unix source code -- a claim it later retracted.
Novell retracted their claim? i never heard that. why did slashdot not tell me?
shit, i fucked up that link. now my joke sucks. see below.
MAC ain't gonna do it... they're too desktop focused.
Guy About Had It With People Who Confuse "Apple," "Mac."
MAC ain't gonna do it... they're too desktop focused.
Guy About Had It With People Who Confuse "Apple," "Mac."
the article says the new one was accepted by OSI. that doesn t mean it s free, not by a long shot. open source software != free software
You do know that that's talking about the OLD lisence right? yeah That said I think some of these issues may still be valid well yeah, that was my question.
i wonder if this new revised license has fixed any of those problems?
here is the statement from RMS.
When I saw the announcement that the Plan 9 software had been released as "open source", I wondered whether it might be free software as well. After studying the license, my conclusion was that it is not free; the license contains several restrictions that are totally unacceptable for the Free Software Movement. (See http://www.gnu.org/philosophy/free-sw.html.)
I am not a supporter of the Open Source Movement, but I was glad when one of their leaders told me they don't consider the license acceptable either. When the developers of Plan 9 describe it as "open source", they are altering the meaning of that term and thus spreading confusion. (The term "open source" is widely misunderstood; see http://www.gnu.org/gnu/philosophy/free-software-fo r-freedom.html
Here is a list of the problems that I found in the Plan 9 license. Some provisions restrict the Plan 9 software so that it is clearly non-free; others are just extremely obnoxious.
First, here are the provisions that make the software non-free.
You agree to provide the Original Contributor, at its request, with a copy of the complete Source Code version, Object Code version and related documentation for Modifications created or contributed to by You if used for any purpose.
This prohibits modifications for private use, denying the users a basic right.
and may, at Your option, include a reasonable charge for the cost of any media.
This seems to limit the price that may be charged for an initial distribution, prohibiting selling copies for a profit.
Distribution of Licensed Software to third parties pursuant to this grant shall be subject to the same terms and conditions as set forth in this Agreement,
This seems to say that when you redistribute you must insist on a contract with the recipients, just as Lucent demands when you download it.
1. The licenses and rights granted under this Agreement shall terminate automatically if (i) You fail to comply with all of the terms and conditions herein; or (ii) You initiate or participate in any intellectual property action against Original Contributor and/or another Contributor.
This seemed reasonable to me at first glance, but later I realized that it goes too far. A retaliation clause like this would be legitimate if it were limited to patents, but this one is not. It would mean that if Lucent or some other contributor violates the license of your GPL-covered free software package, and you try to enforce that license, you would lose the right to use the Plan 9 code.
You agree that, if you export or re-export the Licensed Software or any modifications to it, You are responsible for compliance with the United States Export Administration Regulations and hereby indemnify the Original Contributor and all other Contributors for any liability incurred as a result.
It is unacceptable for a license to require compliance with US export control regulations. Laws being what they are, these regulations apply in certain situations regardless of whether they are mentioned in a license; however, requiring them as a license condition can extend their reach to people and activities outside the US government's jurisdiction, and that is definitely wro