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April 4

Black hole gravity

A black hole has gravity so strong that light can't escape. How does its gravity escape? 71.215.74.243 (talk) 03:49, 4 April 2012 (UTC)[reply]

Well, only objects with mass is trapped by the gravity well; grivity does not have mass. Plasmic Physics (talk) 03:55, 4 April 2012 (UTC)[reply]

Agreed. If gravitons exist, they must be massless. StuRat (talk) 04:32, 4 April 2012 (UTC)[reply]

Photons are also massless.--Modocc (talk) 04:48, 4 April 2012 (UTC)[reply]

as are synagogues.--188.157.191.113 (talk) 20:31, 4 April 2012 (UTC)[reply]

...although weighty topics are often discussed. StuRat (talk) 04:44, 6 April 2012 (UTC) [reply]

See gravitational wave. In M theory gravitons are not stuck to the brane (like particles with mass are) and so are free to radiate into higher dimensional space. SkyMachine ⁽⁺⁺⁾ 04:45, 4 April 2012 (UTC)[reply]

(ec) That doesn't help — photons are also massless, but cannot escape a black hole. It's well known that there are lots of difficulties reconciling quantum mechanics with general relativity; I wouldn't be surprised if this were one of them (or at least one problem with describing gravity in terms of virtual-graviton exchange).

Photons have no rest mass, but that seems rather an esoteric thought, as they are never at rest, and do have mass when in motion. By comparison, I believe that gravitons never have any mass. Any physicists here to verify this ? StuRat (talk) 04:51, 4 April 2012 (UTC)[reply]

Photons and gravitons both have zero invariant mass, which is what physicists usually mean by the word "mass" these days when not otherwise specified. I am reasonably sure that gravitons, assuming they exist, do indeed have nonzero mass in the sense of mass-energy — if they didn't, how could they affect anything else? You also skipped over my second paragraph, which shows that the same problem exists with photons and electric forces. --Trovatore (talk) 04:57, 4 April 2012 (UTC)[reply]

Note also that black holes can definitely have charge, and their charge can be felt via electrostatic forces, which according to QED are mediated by the exchange of virtual photons, even though real photons cannot escape the black hole. I don't know what the resolution of this problem is, or even whether there is an agreed resolution, but it does appear to show that the problem is not unique to gravity. --Trovatore (talk) 04:47, 4 April 2012 (UTC)[reply]

See General relativity. Gravity is not what you think it is. --Jayron32 04:25, 4 April 2012 (UTC)[reply]

For black holes to gravitate, their gravitons, if these exist (they have not yet been measured (perhaps due to extinction) and are only theoretical) would need to "escape"... in which case, the gravitons could not be interacting with themselves very often. Analogously, photons that get absorbed/emitted are able to form waves with minimal interactions with each other. --Modocc (talk) 04:45, 4 April 2012 (UTC)[reply]

Try Google for "how does gravity escape a black hole?" You get loads of links, like [1] [2] [3][4] etc. Some points from these:

How does the electric field escape a charged black hole?

Gravitons don't have to escape; the information needed is all present in the original collapsing star.

Virtual particles can do all sorts of unreasonable things, like escape the event horizon. They can't carry information, but they effectively mediate the force...

Wnt (talk) 05:50, 4 April 2012 (UTC)[reply]

You say virtual particles can't carry information (and I've heard it said before), but what does that actually mean? It is those virtual particles that tell us the mass and charge of the black hole - isn't that information? --Tango (talk) 06:06, 4 April 2012 (UTC)[reply]

Any information released would be random and unrelated to the original information of particles that fell into the black hole, appart from only a few sum total values such as total mass (which you infer from size of the event horizon), sum charge. SkyMachine ⁽⁺⁺⁾ 07:19, 4 April 2012 (UTC)[reply]

Quantum entanglement is handy of course, especially whenever information is lacking..., --Modocc (talk) 08:44, 4 April 2012 (UTC)[reply]

I like the stringball/fuzzball theory of black holes: where everything is crushed into a quivering, tiny non-zero yarn ball of superpositioned superstring where entanglement means nothing. Plasmic Physics (talk) 08:58, 4 April 2012 (UTC)[reply]

Gravity can't escape a black hole. Gravity is the curvature of spacetime. Small fluctuations of that curvature, i.e., gravitational waves, travel at the speed of light. Gravitational waves originating from inside a black hole can't escape the black hole any more than can light. For example, suppose an AM CVn star gets sucked into a supermassive black hole. To an outside observer, as the binary star approaches the event horizon, the frequency of the gravitational radiation from the binary star will slow down to doing nothing more than just producing a static increase in the black hole's Schwarzschild radius, very similar to how the light from the binary star will be observed to redshift down to nothing as it approaches the event horizon. An AM CVn star's orbital period is so short, and a supermassive black hole is so huge, that after the binary star has crossed the event horizon, it will still have enough time to produce a few more periods of gravitational radiation before it hits the gravitational singularity at the black hole's center. But those last few periods of gravitational radiation will never make it outside of the event horizon, just like the last bit of light produced won't. Red Act (talk) 15:30, 4 April 2012 (UTC)[reply]

As in electromagnetism, the gravitational field depends on a retarded potential. The source of the retarded potential outside a black hole is the collapsing matter before it became a black hole. It isn't the mass inside the black hole (if it were, that would imply that light could escape the black hole). -- BenRG (talk) 17:21, 4 April 2012 (UTC)[reply]

Asking how a graviton could escape gravity seems somewhat similar to me as asking how an electron could avoid getting electrocuted... --Mr.98 (talk) 01:49, 5 April 2012 (UTC)[reply]

What's the right word? gravitate? - do gravitons gravitate? Plasmic Physics (talk) 03:32, 5 April 2012 (UTC)[reply]

Before deciding what they do, we should probably have some evidence they exist, and since a) there is yet no evidence that they do and b) there are perfectly workable theories of gravity that don't need them too, it doesn't really matter for this discussion. General relativity provides a very good, mathematically rigorous and very consistent with experimental evidence model which doesn't require us to invent properties for a particle no one has yet found and which no one can even agree on what it should behave like enough for us to know how to look for it. --Jayron32 03:40, 5 April 2012 (UTC)[reply]

Everyone agrees on what a graviton is: it's a quantized vibration of the gravitational field. It's hard to imagine a quantum theory of gravity that wouldn't have that. Part of the problem here is the idea that quantum forces are transmitted by particles, whereas classical forces are transmitted by fields, and therefore you need to throw away all of your classical knowledge and approach things in a completely different way when you go quantum. That's not true. Forces, whether quantum or classical, are transmitted by fields. There's also a virtual-particle picture of forces, both quantum and classical. It's often useful for calculation in the quantum case, and not so usefully classically, but if you're just trying to understand the nature of forces, then that shouldn't matter.

Anyway, gravity does gravitate. In the virtual-particle picture that shows up as interaction vertices involving only gravitons (whereas there are no interaction vertices involving only photons). It makes sense to ask whether a gravitational wave can escape a black hole, and in particular it makes sense to ask whether a real graviton (the weakest possible gravitational wave) can escape a black hole. I don't know if it makes sense to ask if a virtual graviton can escape a black hole. Probably it doesn't make sense. -- BenRG (talk) 07:15, 5 April 2012 (UTC)[reply]

A black hole at the distance farther than event horizon is any star,"the effect of supernova in our world is much much more than black hole"suppose that our sun comes to be any black hole all solar system family will remain in their placement ,the gravity fields are equal and wonderful for you is that orbitals will remain same.--Akbarmohammadzade (talk) 14:34, 5 April 2012 (UTC)

 the wonderful subject which have not been worked on is reverse
world creating in such planetary system. the observatories in thus field
of study might see world reverse.--Akbarmohammadzade (talk) 14:34, 5 April 2012 (UTC)[reply]

In fact I love supernovae ,for their light celebrating at start , their palpitating as our heart , for their expanding as the universe and their role in our existence with sending heavy elements here--Akbarmohammadzade (talk) 15:33, 5 April 2012 (UTC)Italic text[reply]

orce carrier particles in a black hole

If the fundamental forces of nature are carried in particles/waves, as quantum physics maintains that they are, then how are gravitrons (for mass) and charge carriers able to exert influence on the outside world, seeing as they cannot escape the black hole? Thus, a black hole should not be able to exert the full amount of force due from its mass nor its charge. Magog the Ogre (t • c) 15:45, 2 March 2013 (UTC)

   A (theoretical) graviton is itself massless, so is not affected by gravity. How can it "carry mass" yet itself be massless ? Think of it as carrying information, like if you carry a bankbook with your bank balance written on it, but that doesn't mean you have the actual cash on you. StuRat (talk) 16:38, 2 March 2013 (UTC)

       StuRat, sometimes the answers you give are very useful, but I wish you would stop answering questions where you don't have the slightest idea what you are talking about. Massless particles are indeed affected by gravity, for example photons are redshifted when they pass through a gravitational field. Regarding Magog's question, there is no accepted theory of quantum gravity currently, but if one is ever developed, it will probably have the gravitons created in the region outside the event horizon. The story is clearer for charge carriers. Quantum field theory allows for the creation of an electron-positron pair out of the vacuum, with one of them falling into the black hole and the other escaping to infinity -- this mechanism allows electric field propagation without anything actually escaping across the event horizon. Looie496 (talk) 16:53, 2 March 2013 (UTC)

               To clarify, you mean particles with no rest mass can be affected by gravity, due to the mass they gain due to relativity effects at high speeds, right ? Yes, I neglected to include that (and our article on gravitons really should say it has 0 "rest mass", not just 0 "mass", but I can't figure out how to change it). StuRat (talk) 17:26, 2 March 2013 (UTC)

                   They don't gain mass, but they are affected by gravity because space time is curved. Light has a mass of exactly 0, but it cannot escape a black hole due to the curving of space time. Magog the Ogre (t • c) 17:38, 2 March 2013 (UTC)

                       But, again, photons have zero rest mass, but do have relativistic mass. StuRat (talk) 20:55, 2 March 2013 (UTC)

                           Gravity affects everything equally (equivalence principle); the mass really doesn't matter, whether it's rest mass or relativistic mass. Also, relativistic mass has never been a very useful concept, and isn't usually taught any more. It leads people to think that rapidly moving objects should collapse into black holes, for example, which they certainly don't. -- BenRG (talk) 23:30, 2 March 2013 (UTC)

           So, in theory then, a black hole could have a hugely positive charge, but only the content which lay on the event horizon would be able to affect the outside world in terms of charge? (I realize that incoming material does not fall into a black hole but forever stays on the event horizon... but what about material that was present when the black hole expanded, or material that was at the center of the original star, say a magnetar, when the black hole formed) And what does this say about the speed of gravity? Magog the Ogre (t • c) 17:08, 2 March 2013 (UTC)

               Matter does cross the event horizon. You can't see it cross from outside, because no light from the crossing can ever reach you, but it does cross. The electromagnetic field of a black hole has the same strength as the total field of the matter than went into the black hole, but it doesn't come from inside; it is a "fossil field" (retarded potential) from before the matter crossed the event horizon. The same is true of the gravitational field. This doesn't change (as far as anyone knows) in quantum mechanics; nothing that happens inside the event horizon affects anything outside. As for the virtual particle picture, I think this is just an example of a case where it isn't very helpful (as Count Iblis sort of says below). -- BenRG (talk) 23:30, 2 March 2013 (UTC)

Virtual particles are just mathematical tools to do computations, they are unphysical and don't stick to the equations of motions of real particles. E.g. in case of Compton scattering, two Feynman diagrams are needed to compute the amplitude. In one of these diagrams, the emitted photon leaves the electron before the photon that is going to be absorbed has arrived. Count Iblis (talk) 18:45, 2 March 2013 (UTC)

exact che soale mahi .... absolute favorite question!!

Why is pressure a scalar?

Consider this figure, which shows the forces due to pressure on an infinitesimal quantity of fluid. It's clear that p_y(y)=p_y(y+dy)+ (infinitesimal), or else there'd be an infinite acceleration in the y direction. Similarly, p_x(x)=p_x(x+dx) + (infinitesimal). But pressure is treated as a scalar, which implies/assumes that p_x=p_y. What's the justification for this assumption? It's not surprising, physically, that this would be the case if the fluid were incompressible, but is there a more rigorous justification for the claim that pressure is a scalar? 65.92.5.132 (talk) 04:21, 4 April 2012 (UTC)[reply]

In the general case, you have a stress tensor. Intuitively, I think it's the case that the stress tensor has to reduce to a single number, the pressure, in equilibrium conditions in a material that has zero tensile strength, but I can't say I'm sure about that. --Trovatore (talk) 04:26, 4 April 2012 (UTC)[reply]

Almost. In the most general case (nonzero viscosity and/or nonequilibrium system), you can have shear stresses, too. In a static system and the case of zero shear modulus, shear stresses must vanish and the stress tensor has to be diagonal in all reference frames. The only rank-2 tensor that is diagonal in all reference frames is a multiple of Kronecker delta (a tensor with all diagonal components equal to each other),

p\delta _{ij}

where p is pressure. This fact is known as Pascal's law (though he obviously discovered it experimentally long before people knew about tensors or shear moduli.)--Itinerant1 (talk) 05:35, 4 April 2012 (UTC)[reply]

And to put this into more intuitive terms, consider the figure from the original question, and draw a plane at a 45 degree angle through the box. If

p_{y}(y)\neq p_{x}(x)

, you can calculate that the force exerted by the fluid on one side of the plane on the fluid on the other side is not perpendicular to the plane. --Itinerant1 (talk) 05:43, 4 April 2012 (UTC)[reply]

This is why I visualise pressure as equivalent to energy density, which is a scalar quantity. I used to think that since force is a vector quatity, pressure must also be a vector quantity; that pressure is the average quantity of force applied at a right angle a surface. Plasmic Physics (talk) 06:19, 4 April 2012 (UTC)[reply]

Okay, so what I've understood so far is that 1) in general, the forces on a small fluid element won't be represented by a scalar, but by a tensor, so that the force on the top of a small cube of fluid will not in general be the same as the force on the side, and 2) in the case of either no shear forces or static fluid, it can be shown that the tensor reduces to $p\delta _{ij}$ , which means that we can talk about pressure as a scalar (the proof of this is clear when there are no shear forces, but why does this hold in a static fluid?).

There are no shear forces in static fluid, because the shear modulus of a fluid is virtually zero, so the only possible source of shear forces is viscosity, and those are zero as well if there fluid is static.--Itinerant1 (talk) 18:14, 4 April 2012 (UTC)[reply]

But, I still have a problem. Even when the forces are treated as a tensor, pressure is still treated as a scalar. For example, in a Newtonian fluid, the diagonal components of the stress tensor $T_{ii}$ is equal to $-p+2\mu {\frac {\partial u_{i}}{\partial x_{i}}}$ , where μ is the kinematic viscosity and p is the pressure, clearly treated as a scalar. So, what gives? Is pressure being defined as a scalar? 65.92.5.132 (talk) 15:03, 4 April 2012 (UTC)[reply]

According to this, yes. It is defined as 1/3 the trace of the stress tensor.--Itinerant1 (talk) 18:26, 4 April 2012 (UTC)[reply]

Thanks. 65.92.5.132 (talk) 01:16, 5 April 2012 (UTC)[reply]

there are 4 kinds of topology in Set{1,2},and 29 in Set{1,2,3}, is exist a Formula in a Set {1,2,3,4,5,6...n}?

there are 4 kinds of topology in Set{1,2},and 29 in Set{1,2,3}, is exist a Formula in a Set {1,2,3,4,5,6...n}? — Preceding unsigned comment added by Cjsh716 (talk • contribs) 04:34, 4 April 2012 (UTC)[reply]

A topology τ on a set X is a subset of X which satisfies a number of rules. Thus any valid τ must be an element of the power set of X, and thus the number of topologies must be less than or equal than 2ⁿ where n = |X|. So how can there be 29 possible topologies of the set {1,2,3} as 29 > 2³ = 8? (Or am I misunderstanding something about your notation?) -- ToE 06:37, 4 April 2012 (UTC)[reply]

The 29 appears in our article Finite topological space#Number of topologies on a finite set, so I struck my comment above and will withdraw from this discussion until I better understand what is going on. -- ToE 06:42, 4 April 2012 (UTC)[reply]

A topology is a set of elements of the power set, not just an element of it, so the number of topologies is bounded by 2^(2^n), not 2^n. AndrewWTaylor (talk) 08:19, 4 April 2012 (UTC)[reply]

29 doesn't look much like it would be 2^(2^n) of anything. I guess 2^(2^3-3)-3....--188.157.46.184 (talk) 08:28, 4 April 2012 (UTC)[reply]

try http://oeis.org/A000798 --Digrpat (talk) 09:13, 4 April 2012 (UTC)[reply]

~~In most cases n~~Not every subset of pow(X) is a topology, so 2^(2^n) is an upper bound, not the actual number of topologies. AndrewWTaylor (talk) 09:32, 4 April 2012 (UTC)[reply]

To quote the article: "There is no known simple formula to compute T(n) for arbitrary n." You can write a program to enumerate all subsets and count the ones which are valid topologies. There are probably more efficient algorithms, but I don't know what they are. Dcoetzee 10:04, 4 April 2012 (UTC)[reply]

There's an efficient algorithm here: Posets on up to 16 points--Itinerant1 (talk) 10:11, 4 April 2012 (UTC)[reply]

How is heat energy converted to mass?

Mass is converted to energy in nuclear fusion and fission. Energy can be converted to mass, too. I see that Mass–energy equivalence#Practical examples says that each kilogram heated 1°C gains 1.5 picograms of mass. Where and what is that new mass from heat? 71.215.74.243 (talk) 08:09, 4 April 2012 (UTC)[reply]

It is in the motion of the particles, because heat energy is (mostly random) motion. A fast-moving particle simply has more relativistic mass than a slow-moving one. There are no additional particles making up the mass difference. 157.193.175.207 (talk) 09:09, 4 April 2012 (UTC)[reply]

Mass is the energy of an object at rest. If you add heat to an object, it will gain mass, simply because mass and rest energy are synonymous. The equation E = m c^2 has to be understood as a trivial statement where the factor c^2 does nothing more than convert between conventional units we use for mass and energy. Count Iblis (talk) 15:02, 4 April 2012 (UTC)[reply]

~~You means 'at rest' in your frame of ref. In someone elses frame moving relative to yours, the object would have energy. Yes?~~ Struck edit by banned user. Franamax (talk) 16:06, 4 April 2012 (UTC) — Preceding unsigned comment added by 92.28.95.250 (talk) 15:12, 4 April 2012 (UTC)[reply]

Yes, then the object would have more energy in addition to the rest energy. That additional energy is called "kinetic energy". Count Iblis (talk) 15:49, 4 April 2012 (UTC)[reply]

can some women not orgasm from intercourse, but only clitorally

or if they are unable to do the former, is it considered a dysfunction, what is the cure or can be done (other than obviously wearing socks). --188.157.46.184 (talk) 09:29, 4 April 2012 (UTC)[reply]

It amuses me that some men seem to think that the only acceptable orgasm is a vaginal one. Oh, and the reference to socks. --TammyMoet (talk) 09:31, 4 April 2012 (UTC)[reply]

It's an easy mistake to make - after all, reproduction is through intercourse and men orgasm that way. Having just read http://www.scarleteen.com/article/advice/the_great_no_orgasm_from_intercourse_conundrum however ,I must apologize Tammy, I didn't know this fact. Specifically, my partner thought of herself as having a 'dysfunction' in this way, causing emotional problems like she was 'broken' in some way. To me, I love getting her off orally, and after she has come intercourse is also good for her, but she is upset that she can't come from that. Actually I just learned this would put her in the majority. What do you recommend, if she has come orally and we are having intercourse? Should I touch her as well, or have her touch herself, or use a vibrator or what.. Or is intercourse just a chore in reality for her, despite what she might say about liking it (even if she doesn't get off from it). 188.157.46.184 (talk) 09:37, 4 April 2012 (UTC)[reply]

Cunnilingus aint so bad. Some women have their orgasm hang ups too. But you should probably address this question to Dan Savage if you want the expertize on this, or atleast subscribe to his Savage Love podcast. SkyMachine ⁽⁺⁺⁾ 09:44, 4 April 2012 (UTC)[reply]

Um, thanks for the recommendation, but I think this place can answer the question fine. I do love going down on her, she comes very hard like that, then asks me to put it in, and after a while she is upset that she isn't coming that way too. What do you recommend? One thing would be, we could just stop after a couple of minutes of intercourse... or, use a vibrator/finger her clit as well during it. I don't know, nothing works especially great, and more importantly she has hangups about it and will go as far as to start crying. But as I've just found out, this is perfectly normal (and doesn't bother me at all - the fact that she comes clitorally). Recommendations? 188.157.46.184 (talk) 09:48, 4 April 2012 (UTC)[reply]

In short: yes, there are women who cannot orgasm from vaginal penetration. This is not a dysfunction, but an individual variation. If she's not able to handle penetration for whatever reason - and it sounds like she's really not able to right now, if she can't bear it even with added clitoral stimulation - it's probably best to avoid it in favor of other things. If this frustrates you, just let her know and I'm sure she'll do her best to find a way to make sure you're enjoying yourself too. Most importantly make sure she doesn't feel broken or messed up, because she isn't, just different. Dcoetzee 09:59, 4 April 2012 (UTC)[reply]

You can digitally (with fingers) or with a vibrator stimulate the G Spot to arouse the underside of the female prostate. This is much harder to do in regular intercourse. SkyMachine ⁽⁺⁺⁾ 10:07, 4 April 2012 (UTC)[reply]

this is a good recommendation. are there any positions incidentlaly that are particularly good at stimulating the g-spot. I don't care if I put it in or not, I just want her to have fun. She's the one that asks me to. 188.156.25.146 (talk) 10:13, 4 April 2012 (UTC)[reply]

Well digital stimulation it doesn't matter, so long as she is comfortable. The roman way is your best bet for G Spot stimulation during intercourse. SkyMachine ⁽⁺⁺⁾ 10:24, 4 April 2012 (UTC)[reply]

Tammymoet, I don't know why you find my socks reference weird, just google "socks orgasm" -'When they gave the couples socks to wear, about 80% of the couples were able to achieve orgasm compared with 50% previously in this'. given that you can't come vaginally (not specifically at you, at anyone) is vaginal intercourse still 'fun'? or should we not be doing it at all. 188.156.115.31 (talk) 11:44, 4 April 2012 (UTC)[reply]

Well I also wondered what the socks referred to (did Tammymoet say it was weird - I don't think so) because in my 50 plus years of a pretty open life I had never heard about this unlikely use of socks. Placebos eh? don't you just luvvum! Caesar's Daddy (talk) 13:20, 4 April 2012 (UTC)[reply]

Just so long as it isn't some veiled message alluding to the perpetrating of sock puppetry, which it very easily could be read as. SkyMachine ⁽⁺⁺⁾ 13:41, 4 April 2012 (UTC)[reply]

This sounds cool (news link [5]) - but come on, it was a study of 13 people. That's 4 out of 13 people different, in a study which can't have a very good control ... I'm not even hunting for this one in the literature. The orgasm may blow your socks off, but the study? Not so much. Wnt (talk) 19:21, 4 April 2012 (UTC)[reply]

I'll just say that the best positions for stimulation for the woman are with her on top, or from behind on the side (according to Alex Comfort's Joy of Sex). Actually, why don't you get a copy of "The Joy of Sex" and work your way through it together? (by the way I'm still laughing at the sox...) --TammyMoet (talk) 18:29, 4 April 2012 (UTC)[reply]

Yep, they're a joke this season. A Quest For Knowledge (talk) 18:40, 4 April 2012 (UTC)[reply]

Live engineering status for UK TV and radio broadcast towers

I'm currently unable to receive digital radio; as two unrelated receivers experience the problem, I expect that the transmitter is currently off or operating on reduced power. Here in Yorkshire we've had some late snow and high winds and some places have experienced power outages and unplanned voltage excursions, so that's understandable. I think the DAB signal comes from Bilsdale transmitting station - I'm getting DVB-T okay, so it's possible that the radio transmitters alone are off (perhaps Crown Castle, in their infinite wisdom, have chosen to prioritise Cash in the Attic over The World at One (grrr)). I'd hoped I could find a website that shows the current status of the various transmitters on that (and in general other) towers. While it's mostly of academic interest to me, I'm sure it's vital for professional TV and radio installers (who don't want to waste time investigating a customer's reception when the problem is a failure at the broadcast tower). I can't find such a site, either for planned or exceptional outages - does such a resource exist? 46.208.221.86 (talk) 12:58, 4 April 2012 (UTC)[reply]

I've noticed that such info is rarely made publicly available. I suspect the reason is that they don't want advertisers (or customers, in the case of publicly supported stations) to get a true picture of how unreliable they really are, as this could adversely affect their income. As for the installers, I would expect them to carry a portable antenna with them, so they can verify that there is a signal, before working on the customer's equipment. This would also allow them to determine the optimal location and direction for the antenna, before they begin installation. StuRat (talk) 15:03, 4 April 2012 (UTC)[reply]

There used to be a weekly program on British TV aimed solely at installers. It gave details of new transmitters or relays, the frequency bands that the handful of analog stations were broadcast on, and the polarity of the antenna. This also listed planned outages and low power events. This was usually on between Carole Hersee and Pages from Ceefax; I've not seen it for 15 years at least. -- Finlay McWalterჷTalk 15:44, 4 April 2012 (UTC)[reply]

The BBC has a 'Reception Problems' tool at https://faq.external.bbc.co.uk/templates/bbcfaqs/emailstatic/interferencePage?entryID=&id=POKHPGEVQP0B1IPKD0SE2UO6DM&moduleID=&reset=true. You can enter your postcode and find out what's going on in your area. Without knowing your postcode I can't really tell what the problem might be, but just for the hell of it I made up a Bradford postcode, and found out that there's some -engineering work going on on Ilkley Moor (hopefully they've remembered to wear a hat) and the Keighley transmitter is operating at reduced power. - Cucumber Mike (talk) 15:36, 4 April 2012 (UTC)[reply]

Using a Helmsley postcode, I have established that the Bilsdale transmitter was off air until 13:41, so although you will have missed The World At One, hopefully you were back up and running in time to catch The Archers. Jill is overruled. Meanwhile Tony is keen to help. Sounds riveting... - Cucumber Mike (talk) 15:42, 4 April 2012 (UTC)[reply]

Thanks Mike, that's exactly what I was looking for - and it's entirely consistent with the pattern in which the radio stations came back to life. I doubt anyone would forget their hat 200m above Wharfedale, especially today. 46.208.224.194 (talk) 23:02, 4 April 2012 (UTC)[reply]

They might if they'd been "a-courting" Mary Jane. {The poster formerly known as 87.81l230.195} — Preceding unsigned comment added by 90.197.66.205 (talk) 17:27, 5 April 2012 (UTC)[reply]

empty space

if we were able to go out deep into empty space and encapsulate a "volume" of empty space. would that capsule be able to then either compress to a zero space or expand to a very large space without affecting/being affected by the interior empty space? — Preceding unsigned comment added by 165.212.189.187 (talk) 14:55, 4 April 2012 (UTC)[reply]

Pretty much, yes. However, since it's not a complete vacuum, there would be a limit to how far you could compress it. StuRat (talk) 14:59, 4 April 2012 (UTC)[reply]

How about expanding it? would you be "creating" additional space in the universe or "transferring" it from the outside of the capsule to the inside? — Preceding unsigned comment added by 165.212.189.187 (talk) 15:17, 4 April 2012 (UTC)[reply]

If you enlarge or shrink a container in a vacuum, you aren't enlarging, shrinking, creating or destroying the space within the container; all you're doing is moving the container's walls within space. Red Act (talk) 15:39, 4 April 2012 (UTC)[reply]

So, yes, you are "moving space from outside to inside" when you expand the vacuum, even if it's completely enclosed, and without affecting whatever empty space is inside. 188.36.162.23 (talk) 16:00, 4 April 2012 (UTC)[reply]

You are treading dangerously close to that line where you make so many impossible assumptions that science can't give you an answer. There is no such thing as an impermeable container that you can change the enclosed volume at will. I think you might be confused by the idea of a piston filled with fluid; when you pull it up, the pressure inside decreases. But this pressure is only caused by molecules bumping into each other. Nothing is being done to the actual space within the cylinder (whatever that means). If you had instead somehow had a perfect vacuum (which, I must reiterate, is impossible) and pulled up or pushed down on the piston, nothing special would happen. -RunningOnBrains^(talk) 16:28, 4 April 2012 (UTC)[reply]

You aren't "moving space from outside to inside". Space doesn't have a state of motion. When you imagine space going into your box, you are imagining the space as being at rest relative to some lab frame, through which the side of the box is moving. It doesn't work that way. -- BenRG (talk) 17:36, 4 April 2012 (UTC)[reply]

What doesn't work "that way"?; Then how does it work?165.212.189.187 (talk) 17:45, 4 April 2012 (UTC)[reply]

if the box has the mechanisms to expand while keeping what is out out and in in then what would happen?165.212.189.187 (talk) 17:52, 4 April 2012 (UTC)[reply]

You can't move quantities of space around. Space isn't a substance. When the box expands, the amount of volume inside the box is increasing and the amount of volume outside the box is decreasing, but that's exactly what it means for the box to expand. You can't divorce the two, because they're just two ways of describing the same concept. The only way for the box not to increase the volume inside is by not expanding. Rckrone (talk) 03:24, 5 April 2012 (UTC)[reply]

Hmmm, maybe Casimir_effect#Measurement applies? I don't know - is restricting virtual particles like "space" being compressed? - I never even really understood why we have to believe in vacuum fluctuations rather than just minute fluctuations of charge within the conducting plates, like London dispersion force. Wnt (talk) 17:22, 4 April 2012 (UTC)[reply]

You don't have to, and shouldn't. As explained in hep-th/0503158, the Casimir force is just a relativistic quantum correction to the electromagnetic force. -- BenRG (talk) 17:36, 4 April 2012 (UTC)[reply]

That plugs a hole ... thanks!!! But the Casimir effect article is written entirely in terms of virtual particles. I added quotes and a link to the reference, not wanting to make a major revision to a basic physics article I scarcely understand on my own, but somebody should increase the prominence of this saner-sounding interpretation. (I mean, it's like we were describing electric power transmission in terms of holes...). Wnt (talk) 19:13, 4 April 2012 (UTC)[reply]

I don't want to edit it because I think I would be opposed on the grounds that Jaffe's paper isn't mainstream. Most papers still get this wrong. Physicists who read Jaffe's paper change their minds, but nonphysicists have nothing to go on but numbers. I've been waiting for more physicists to come around, but it's taking a long time. -- BenRG (talk) 20:40, 4 April 2012 (UTC)[reply]

Not mainstream? He's writing from the theoretical physics department at MIT. Can you get more mainstream than that? Wnt (talk) 22:39, 4 April 2012 (UTC)[reply]

Sure you can. I don't feel like digging up references, but it's easy to find distinguished particle physicists effectively contradicting Jaffe in published papers. I think there are examples in Jaffe's paper. On the other hand, I'm pretty sure that none of them cite Jaffe. I could try to make a case for Jaffe on the grounds that, seven years later, no one who has read the paper has said that it's wrong. I'll think about it. -- BenRG (talk) 00:25, 5 April 2012 (UTC)[reply]

the OP is basically asking, if you have a collapsible structure with webbing, for example, in space, and you open the structure so that instead of containing 10 cubic centimenters it contains 100 times as much volume. Anything weird or strange happen? No. Where does the extra 90 cubic centimeters of volume come from? Outside of the original confines of the box. This is weird to the OP because if it were on Earth, and an enclosed container contained 10 cc of air, and then 100 cc of air when opened, then it must have a leak. In space no leak is required. (The OP is almost thinking in terms of ether, I think, and my answers were along those lines). In this way of thinking: when you open the webbed box ether magically moves from outside to inside, penetrating the wall. In fact since space is empty, nothing needs to move. I guess if it were a theoretically really impermeable membrane then the few particles per million that space has since it's not a perfect vacuum would be reduced to a lower ratio. (this would especially apply if you're in a part of space with a low density cloud of particles) But the OP is asking about the vacuum part, not the paricles part, and isn't asking or thinking about space with a cloud of particles in it, but mostly empty deep space. 188.157.140.201 (talk) 18:15, 4 April 2012 (UTC)[reply]

What's wierd to me is that you presumed that that was wierd to me. — Preceding unsigned comment added by 165.212.189.187 (talk) 18:57, 4 April 2012 (UTC)[reply]

I don't get it, why even post the question? Something had to be weird to you, as it's a completely straightforward question... what made you ask it? 188.157.140.201 (talk) 19:15, 4 April 2012 (UTC)[reply]

I don't know, I guess I'm arguing for the ether theory. I thought that there would be an upper limit to the expansion of the container besides its obvious physical limitations of having enough matter. Doesn't the energy increase as the container expands? Isn't the space inside sequestered from the space outside? If there was even one atom inside the container would it eventually rip apart or will the atom stop the expansion? — Preceding unsigned comment added by 165.212.189.187 (talk) 19:42, 4 April 2012 (UTC)[reply]

Well, the point of empty space is stuff can move through it freely. If that weren't so... it wouldn't be empty space. But if there are atoms in it (and there are - outer space (the interstellar medium) is not literally empty space - then you could eventually concentrate them to something substantial. See Bussard ramjet, for example. Wnt (talk) 20:03, 4 April 2012 (UTC)[reply]

(edit conflict)Ah here we go; I think I get what you don't get now. You're asking about some ideal situation where you somehow got a "true vacuum" (aka nothing), which really is physically impossible, but I'll try to go on. You should re-read my explanation above. It seems you have a misunderstanding about how pressure works. You are not doing anything to "space" when you, say, pull up on a piston. When you put something, like say a marshmallow, in a vacuum, it expands. But this is not due to some negative force that pulls it apart: this is due to the air bubbles trapped inside (since it was made at atmospheric pressure) exerting an outward force on the marshmallow. Likewise, at sea level, if you pull up on a piston to lower the pressure inside, the resisting force you feel is not that the air or space or anything within the piston is resisting you. The force that is resisting is the atmospheric pressure pushing down on the piston. We don't notice air pressure in our every day experiences because it's essentially the same all around us; there are no pressure differences, or gradients. But atmospheric pressure is quite strong: about 10 kPa, or 10,000 Newtons per square meter. To put this in common units, on every 1 square centimetre (0.16 in²) of surface, the air is pushing down with 1 newton (0.22 lbf) of force, the weight of 0.1 kilograms (0.22 lb) (these units seem small, but even for a decent-sized surface they add up quickly!). If the pressure is the same in the piston as out, there is no net force. But when you pull up, the pressure inside the cylinder decreases, so you must apply the same force as was previously applied by the inner pressure.

Sorry, I've gotten on a bit of a tangent. The point that I think you need to understand is this: pressure is caused by the force of collisions of fluid molecules against a surface. Say you have a box that has a certain pressure of gas inside. If you expand that box, so now the gas has a greater volume and a lower pressure, you have not applied some magical negative force on the molecules, you have simply allowed them more room to move, which means fewer collisions with the side of the box on average, so a lower pressure. "Space" is not a thing in the classical sense, it is the absence of things, and so your container is not interacting with it. Saying that after expansion, there is "more space" in the box, so "where did it come from" is not a valid question, as space is nothing. I hope I have answered your question. -RunningOnBrains^(talk) 20:17, 4 April 2012 (UTC)[reply]

Yes, I was misinterpreting the "negative pressure" idea. can you have infinite negative pressure? — Preceding unsigned comment added by 165.212.189.187 (talk) 20:25, 4 April 2012 (UTC)[reply]

Well, there's the Big Rip idea. If I remember correctly, dark energy = a cosmological constant. The more dark energy, the higher the negative pressure on the universe, which somehow causes space itself to expand. With infinite negative pressure, space expands infinitely. Unless I got the sign wrong. I never really understood this stuff ... it seems as bizarre as some of the talk above to me. How can pressure cause space itself to expand/contract? Wnt (talk) 20:49, 4 April 2012 (UTC)[reply]

"Pressure" itself has no effect on space. With regard to dark energy, "negative pressure" is shorthand for the effect dark energy is having on spacetime. It's not actual pressure in the barometric sense, but a repulsive force. — The Hand That Feeds You:^Bite 21:37, 4 April 2012 (UTC)[reply]

No, it is honest to goodness negative pressure. A more familiar name for "negative pressure" is "tension". I'm not sure where the "negative pressure" terminology came from—it seems pointlessly obscure.

Pressure and tension do gravitate, and tension gravitates negatively. This is a gravitomagnetic effect—pressure is like a current of mass (more precisely, of momentum). -- BenRG (talk) 23:23, 4 April 2012 (UTC)[reply]

Didn't someone just ask why pressure is scalar (has no direction)? Thus isn't negative pressure quite different from tension, as tension most definitely has a direction!! (just pull a boat from far away and it will move toward you - quite different from trying to pull up on it from a helicopter. So negative pressure != tension! 134.255.48.97 (talk) 23:44, 4 April 2012 (UTC)[reply]

Tension can be omnidirectional, though it's harder to think of realistic examples (a gas can't be under tension). One example is a sphere of metal that shrinks when cooled surrounded by a spherical shell of metal that doesn't shrink (or shrinks less). When it's cooled, the sphere should be under roughly constant tension. Another example is a boat with a bunch of ropes attached, pulling in all directions. Pressure can also be unidirectional (e.g., water pressure in a pipe, or pushing a boat). -- BenRG (talk) 00:25, 5 April 2012 (UTC)[reply]

Water pressure in a pipe isn't unidirectional, it's just being resisted in some directions and not in others. The pressure is still there in the directions towards the walls of the pipe. --Tango (talk) 05:03, 5 April 2012 (UTC)[reply]

hydrocortisone

If someone uses a ear drop that has 1% hydrocortisone in it, how much hydrocortisone is absorbed into the bloodstream? — Preceding unsigned comment added by 64.38.226.84 (talk) 21:08, 4 April 2012 (UTC)[reply]

It's very difficult to say without a physical examination. It depends on things like earwax, age, and liver metabolism. 71.215.74.243 (talk) 21:15, 4 April 2012 (UTC)[reply]

Yeah, there's no direct answer for this. It's going to depend on the individual's body chemistry, as well as how much of the drops are used. — The Hand That Feeds You:^Bite 21:40, 4 April 2012 (UTC)[reply]

And whether they subsequently lie on their side, and if so, which side. --Colapeninsula (talk) 09:14, 5 April 2012 (UTC)[reply]

A clinically insignificant amount. Do the math. An ml of 1% hydrocortisone will have 10 mg of hydrocortisone. It is roughly 20 drops. Typical dosing might be 4 drops tid, or about 6 mg a day. When given by mouth, 5 mg a day may be full replacement/full suppression for a 2-5 year old child, but only a small amount of the otic hydrocortisone gets absorbed, much less than 6 mg. refalteripse

Frank Sherwood Rowland

You have put in the wrong portrait. (4 April 2012 at 3:13 PDT) — Preceding unsigned comment added by 66.241.92.137 (talk) 22:13, 4 April 2012 (UTC)[reply]

You should post this to the talk page for that article, along with any proof you have (preferably the correct pic). StuRat (talk) 22:16, 4 April 2012 (UTC)[reply]

Looks an awful lot like the guy in the Telegraph obituary photo. Clarityfiend (talk) 02:53, 5 April 2012 (UTC)[reply]

Header linked for convenience. Richard Avery (talk) 07:21, 5 April 2012 (UTC)[reply]