Wikipedia:Reference desk/Archives/Science/2008 August 20

= August 20 =

Black hole evaporation and time dilation
OK - so I have a question relating to the previous one. So I'm falling towards the black hole in by spaceship - my "mothership" is sitting a safe distance away observing my impending doom. From the point of view of the mothership, time for me slows down until I reach the event horizon and time stops dead. From my point of view, time in the outside universe (including the mothership) must therefore speed up. The closer I get to the event horizon, the faster time zips along. I see stars run out of fuel and die, I see the end of the universe sometime before I hit the event horizon.

But don't I also see the black hole evaporating? So doesn't the event horizon shrink as I approach it? Is it in fact the case that I can survive my death plunge because the event horizon always evaporates faster than I can go towards it? By the time I get there, the black hole weighs so little and is so tiny that I can escape it?

I'm sure there is a flaw in this - and in all likelyhood, the radiation from the rapidly (to me) decaying universe would cook me and spaghettification would have shredded me already. But this is a thought experiment.

SteveBaker (talk) 01:59, 20 August 2008 (UTC)
 * When the dude from Rush flew into the black hole it took long enough that they made a whole 'nother album. And at least a dozen more after that, so it must take a long time :) Franamax (talk) 02:30, 20 August 2008 (UTC)
 * The black hole is deeper in the gravitational field than you, so time for it is going even slower, not faster. Time further away from the black hole from you goes faster, time nearer goes slower. --Tango (talk) 02:54, 20 August 2008 (UTC)
 * OK - I'll buy that...so how come they evaporate at all when seen from the outside? From the point of view of someone a nice safe distance away, time has stopped at the event horizon so how can all of that virtual particle stuff happen? SteveBaker (talk) 03:36, 20 August 2008 (UTC)


 * It should be that the particle pair-production occurs outside the event horizon and one gets sucked in while the other flies away, that being the allegory usually used for Hawking radiation. So it takes finite time for an asymptotic observer to see the particle that escapes, though it should escape verrry slowly since its "formation". On the other hand, is its formation time well-defined since it's originally a virtual pair? SamuelRiv (talk) 05:13, 20 August 2008 (UTC)


 * I wouldn't say that "From the point of view of the mothership, time for me slows down". The mothership sees the light that you emit before crossing the event horizon redshifted into the far future. Redshift applies not only to individual wavelengths of light but to the overall duration of events, so the mothership sees you moving very slowly toward the horizon and never reaching it. But that's just a Doppler shift—it has nothing to do with the light as emitted by you or time as experienced by you. There is no reciprocal speedup, and you don't see the whole future of the universe. Here's a Penrose diagram of a classical (non-evaporating) black hole:

______*       \    /\     inside /  \ \/ outside \   /            \  /             \/


 * This doesn't show the "shape" of a black hole but it shows the causal structure: nothing can travel along a worldline that's closer to horizontal than the slashes / and \. The horizontal line is the singularity and the line dividing "inside" and "outside" is the event horizon. The vertex marked * is future infinity, which is where the outside spaceship ends up. So it's easy to see that the outside ship will never see anything from inside the horizon. But you won't see the whole future of the universe either; no matter where you hit the singularity, your past light cone will exclude some neighborhood of the point *.


 * Here's a Penrose diagram of an evaporating black hole, taken from Hawking's 1975 paper:

*             |\              | \          ____|  \          |  /   /     inside /   / |/ outside |  /          |  /          | /          |/


 * This uses a polar coordinate system and the vertical lines are r=0; nothing special happens there. The horizontal line is again the singularity. From this diagram you can conclude that (a) the black hole won't evaporate before you fall in; (b) you won't see the whole future of the universe before you hit the singularity; (c) the ship outside will see you fall in at the same moment it sees the black hole evaporate (in contrast to the non-evaporating case, where it has to wait forever to see you fall in).


 * However, it's not known if this diagram is correct, and many people suspect it isn't. So in truth nobody knows. -- BenRG (talk) 13:32, 20 August 2008 (UTC)

Using sig figs in calculations with both addition and multiplication
How do you use sig figs in a calculation when using both addition and multiplication?

For example: Find the molar mass of CO2

In this case, we would do 12.01g + 2*16.00g to find the molar mass.

Do I first round to the correct sig figs for each of the values (12.01g and 2*16.00g) and then add, using the correct decimal places, or should I find the result of (12.01g + 2*16.00g) then round using 4 sig figs?

Although doing it either way produces the same result in this example, in other cases that I've tried, rounding at different points produces different results. What is the correct way, and/or what method is most commonly used? —Preceding unsigned comment added by 68.111.75.89 (talk • contribs) 13:58, 20 August 2008


 * It's generally recommended to keep things as exact as possible until the very last step, since round-off error has a tendency to accumulate (i.e. the more often you round, the more information you lose, and hence the more likely it is that your final answer will be way off the mark). Confusing Manifestation (Say hi!) 04:01, 20 August 2008 (UTC)


 * You shouldn't report the final answer to 4sf at all, since you can't be sure it's accurate. Consider if the real values were 12.014 and 16.004 (which round to the values you gave), then the correct answer would be 44.022, which to 4sf is 44.02, however using the figures you started with you would get 44.01. Every time you do a calculation using an approximate value you lose some precision, so you should always report the final answer to a lower precision that you started with (I'm not sure how much lower, I'm sure there's some rule for it, though). --Tango (talk) 17:24, 20 August 2008 (UTC)


 * Well - be a little careful. You can often start off with a large quantity of inaccurate data and compute an average that has greater precision than any of the individual numbers in the input data.  Providing the errors in the inaccurate data are random of course.  SteveBaker (talk) 18:48, 20 August 2008 (UTC)


 * That's pretty much the only exception to the rule (that I can think of, anyway), but yes, you're absolutely right. --Tango (talk) 23:37, 20 August 2008 (UTC)


 * Hmmm - what happens to precision when you do stuff like squaring a number or calculating a square root?  Let's not think in terms of "significant digits" - let's think about the actual worst-case error:


 * If I have a square, whose side I'm going to explicitly state has a length of 4cm plus or minus 0.5cm...I'm saying that there is a plus or minus 12.5% error. Now I calculate the area of that square: I end up with 16cm2 - with some kind of error.  We know that the "true" size of my square was definitely somewhere between 3.5 and 4.5 cm so the the area can really only lie between 12.25 and 20.25 cm2 - so I should really say that the result is halfway between those answers: 16.25 plus or minus 4.0cm2.  The error is now plus or minus almost 25%.  So squaring this number almost doubled the error.


 * BUT: Suppose we reverse the operation. If you told me that the area of a square is 16.25cm2 plus or minus 4.0 and asked me to calculate the length of the side, I could (if I was being sufficiently anal about it) say that the true area lies between 12.25 and 20.25 - so the result lies between the square roots of those two numbers - which is 3.5 and 4.5 respectively - so I can reliably say that the result is 4.0cm plus or minus 0.5cm.  So taking the square root of a number with a 25% error bar produced an answer with a 12.5% error bar.  Hence, taking the square root of a number improves the precision of the result!  Taking the cube root improves it even more.


 * In general - there are many situations where a small error in the input to a calculation can produce ENORMOUS errors in the output...and reversing those calculations has the opposite effect.


 * So no - it's not just averaging. The idea of teaching people to use "significant digits" as if it were a religious matter is bogus - it's a very rough rule of thumb when you need a rough idea of how much precision to put into your answer.  Those rules cannot be applied rigorously.  We should teach people to use error limits instead.  In a bygone era when crunching the numbers was horribly time-consuming, then an approximate way to estimate error was an acceptable way to go.  But these days, we can do much better.


 * SteveBaker (talk) 14:57, 21 August 2008 (UTC)

Earth's rotation
If the Earth is rotating so fast, why don't we feel anything?

If we were to jump 100m above the ground, should the Earth which is constantly rotating move under our feet so that we land somewhere else? —Preceding unsigned comment added by 68.111.75.89 (talk) 04:05, 20 August 2008 (UTC)


 * The rotation isn't very fast -- it takes a whole day to complete one revolution, and that's why the sensation of turning is far too little to feel. If you were able to jump say 100 km off the ground so that it took you several minutes to fall back down, then the Earth's rotation would accumulate enough for you to notice it.


 * You think the Earth's rotation is fast because of the linear speed of its surface motion. This is indeed fairly large: at the Equator just over 1,000 km/h or about 650 mph, in the mid-latitudes about 700 km/h or 450 mph.  But you are moving along with it at the same speed, so you don't notice anything.  It's just like when you walk around on a ship, train, bus, or airplane: you share the vehicle's speed, and you don't stop sharing it when you jump off the floor.


 * --Anonymous, edited 05:25 UTC, August 20/08.


 * When you jump into the air, you start off moving sideways at the exact same speed as the earth is moving - the air around you is also moving at the same speed...so there is nothing to "slow you down" - so you and the earth keep moving together until you come back down again. The only slight deviation is if you were to move very fast to the north or south when the "Coriolis effect" would cause you to be deflected off to one side a tiny bit.  But the force is very small - you really don't notice it until you've moved a long distance in a relatively short time. Hurricanes feel the rotation - but that's because they are big enough.  That's what starts them off spinning - and it explains why they spin clockwise in the southern hemisphere and anticlockwise in the north.
 * One way to really see proof that the earth is rotating is to make yourself a Foucault pendulum. This is a big, heavy pendulum that can easily swing in any direction and which has enough energy stored in it to continue to swing freely for many hours.  If you imagine setting the pendulum up at the North pole - then as the earth rotated - the pendulum would continue to swing in the same straight line.  From the perspective of someone slowly rotating with the Earth, it would appear that the pendulum was slowly changing its swing - moving at about half the speed of the hour hand on a clock so it would complete an entire revolution in 24 hours.  But the experiment works even if you aren't at the North pole (although it works better the further north or south of the equator you go. SteveBaker (talk) 06:24, 20 August 2008 (UTC)
 * "... so you and the earth keep moving together until you come back down again" - I figured as much (not that I'm not the OP). :)
 * This should be equal to jumping in a train (moving at a constant speed), right? (For those who haven't tried, you don't slam into the car's back wall!)-- Aeluwas (talk) 08:55, 20 August 2008 (UTC)
 * Not exactly equal, but good enough. A person who jumps up initially moves horizontally with the Earth, but the Earth curves away while the person continues to travel in a straight line (Newton's first law of motion).  If Earth were perfectly round and humans were capable of jumping straight up, the person would indeed land in a different place  --Bowlhover (talk) 11:25, 20 August 2008 (UTC)
 * Surely the person doesn't travel in a straight line as they continue to be affected by gravity. AlmostReadytoFly (talk) 13:27, 20 August 2008 (UTC)
 * They would tend to travel in a straight line, not with the Earth (in a circle). --Bowlhover (talk) 20:37, 20 August 2008 (UTC)
 * Some quantitative calculations:
 * Because Earth rotates faster at the equator, the maximum centrifugal force an object can experience is felt if the object is at the equator. The force is given by F=mv^2/r, where m is the object's mass, v is Earth's tangential speed of rotation, and r is Earth's radius.  r=6378 100 m and v=463.8 m/s at the equator, so every kilogram of mass feels 0.034 N of force.  A 75 kg human feels 2.5 N which is about the weight of a 0.26 kg mass.  However, the only effect this force has at the equator is weight reduction, and 0.26 kg is not very impressive.
 * Assuming I've done the math correctly, the maximum horizontal force the centrifugal force can supply is half the force at the equator. At 45 degrees of latitude where the horizontal force is the strongest, a 75 kg human feels a force of 1.25 N pushing towards the south.  1.25 N is small compared to the forces involved in everyday life, so it's not noticed.  --Bowlhover (talk) 11:08, 20 August 2008 (UTC)


 * Two corrections: (i) centrifugal force is a fictitious force - the forces on a object at the equator do not have to balance because the object is not in equilibrium - it is travelling in a circle (well, in a much more complex path if you take into account the Earth's motion round the Sun etc., but we can approximate the motion locally as circular motion); (ii) objects at 45 degrees of latitude do not experience a "horizontal" force because horizontal and vertical are defined relative to local perceived direction of gravity. In other words, a force even as small as 1% of the weight of an object would cause a snooker ball on a smooth table to roll - but we would attribute this motion to the table being slightly tilted, not to a mysterious horizontal force. Gandalf61 (talk) 12:37, 20 August 2008 (UTC)


 * Oh...Now you've gone and done it! Now I have to rant about people who complain about use of the term "centrifugal force". (a) It's a very handy shorthand that makes a lot of calculations easier to do in a rotating frame of reference.  Anyone who gives a damn is fully aware of it's "fictitious" nature - so we don't need to be corrected - Mmmm'k?  Thanksmuch. (b) Nearly every force we use in physics is "fictitious" in some way or other.  You rest a brick on a table and Newton tells us that the table produces an equal and opposite "force" that opposes gravity...but really, that's the nuclear forces acting between all of the molecules of the table and brick.  Gravity is considered a "force" - but really it's only an aspect of the curvature of space-time.  Friction, drag, lift (as in airplanes), coriolis, air pressure...are all "forces" that don't really exist.  If we have to expand every explanation down to the "fundamental" level - we'll never get any work done!  Why does everyone pick on poor old super-handy centrifugal force for making nit-picky academic points?  Bah!  I intend to continue to use the term whenever and where ever it's convenient - I very much doubt that anyone will be even slightly confused. SteveBaker (talk) 16:24, 20 August 2008 (UTC)


 * ...and the real reason you don't feel that lateral force is because the earth isn't a sphere. It's distorted into an oblate spheroid precisely because of the effect of (yes, Gandalf61: I'm going to say it) centrifugal force.  Hence "horizontal" as defined as "tangential to the mean sea level surface of the earth" is handily orthogonal to the net local gravity+centrifugal vector...so you don't feel any lateral force - no matter how small.  If you did - then the water in the ocean would flow sideways and thereby automatically adjust. "Mean sea level" would (as if by magic) correct for it.  Hence, no lateral force. SteveBaker (talk) 16:31, 20 August 2008 (UTC)


 * Well, I wouldn't use sea level to define horizontal and vertical - I would use a spirit level and/or a plumb line. But we reach the same conclusion - there is no "horizontal" force. Any chance you might stop ranting at me now ? Gandalf61 (talk) 16:41, 20 August 2008 (UTC)
 * The ocean is a VERY big spirit level - it'll do just fine! Why would you trust a tiny little amount of liquid sloshing in a tube when you can use an entire ocean?  As for the rant...er, sorry - but sadly, I just started another rant (below) about your incorrect ideas about friction!  It's nothing personal...honest! :-P SteveBaker (talk) 17:31, 20 August 2008 (UTC)


 * I can't believe I forgot "down" is not necessarily "towards Earth's centre". Anyways, to SteveBaker: your rants, particularly those concerning angle of attack and Bernoulli's principle, have become famous on the science Reference Desk.  See http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Archives/Science/2008_August_4#Why_can.27t_bullets_fly_.3F, for example.  --Bowlhover (talk) 22:07, 20 August 2008 (UTC)


 * The answer here is strikingly simple. The real question, buried in this specific example, is what approximations are valid in this scenario.  If you jump 100 meters, the scale involved is sufficiently small that gravity can be considered a constant force, the earth can be considered flat, and so forth; and you have a purely ballistic trajectory.  This is a very easy math problem and is often taught in algebra-level introductory physics.  If you wanted to jump, say, 100 kilometers, the scales would be quite different and the approximations would be very inapplicable, and you would have to use orbital mechanics to determine the new trajectory.  There are some "universal" details you can still count on - momentum will be conserved, inertia still exists, energy is still conserved - only, your equations will be a little messier.  Nimur (talk) 17:01, 20 August 2008 (UTC)

How a redshift affects the appearance of the moon on the horizon
A recent disscussion a got involved in has me curious and I need help in sttling an argument and that is wether or not the earth's gases are more condensed at the equator when the moon rises in the evening causing it to look oblong or just merely bigger than say half an hour after the moon rises. I know that the moon is only one size but is there certain times of the year (eg. summer equinox to fall equinox) when the redshift seems to make the moon appear more oblong or oval shape than other times of the year? —Preceding unsigned comment added by 69.31.224.52 (talk) 04:33, 20 August 2008 (UTC)


 * There are really three parts to this answer:


 * Red-shift doesn't apply to the moon to any measurable degree. Red-shift happens due to the doppler effect when an object is moving away from you at some significant fraction of the speed of light.  For objects that are an ENORMOUS distance away, the expansion of the universe makes them seem to move away from us faster the further away they are - so they red-shift.  But the moon stays pretty much the same distance from us all the time and it's not far enough away for the expansion of the universe to be measurable - so definitely no red-shift for the moon.  The appearance of a red-shifted object are that it's colours become more reddish - it has nothing to do with the apparent size of the object.


 * The moon sometimes looks distorted when it's close to the horizon because of heat trapped close to the ground - hot air diffracts light differently from cold air. That explains the oblong/oval appearance - and it's more obvious in the summer when the ground is hot and more obvious still at the equator where the ground is hottest of all.


 * The apparent change in size of the moon is an optical illusion. That illusion happens just as much in winter as in summer - so it's not related to the layers of hot air thing.  We have an article about that illusion here: Moon illusion (of course we do!)


 * I think that clears up everything.
 * SteveBaker (talk) 06:08, 20 August 2008 (UTC)


 * And just for fun, the moon does periodically appear larger and smaller, but that's an artifact of its elliptical orbit and it's not of a magnitude remarked on by the casual observer. &mdash; Lomn 13:03, 20 August 2008 (UTC)


 * "The appearance of a red-shifted object are that it's colours become more reddish..." Not quite. redshift makes the wavelengths all increase, so red becomes closer to infrared, green becomes closer to red, blue becomes closer to green, violet becomes closer to blue, and ultraviolet becomes closer to violet. — DanielLC 15:51, 20 August 2008 (UTC)

size of earth
What is the size of earth? —Preceding unsigned comment added by 68.111.75.89 (talk) 04:46, 20 August 2008 (UTC)


 * You may want to check out our article for Earth, which, among other things, includes this information. Specifically, you want to look at "physical characteristics" in the sidebar. -- Captain Disdain (talk) 05:35, 20 August 2008 (UTC)


 * There's an entire field devoted to measuring the Earth: see geodesy. --Bowlhover (talk) 06:05, 22 August 2008 (UTC)

Banked Corner
Why does a car going around a banked/sloped corner slip down if it is not going fast enough?

125.238.173.197 (talk) 07:37, 20 August 2008 (UTC)
 * The same reason anything slides down a slope: because of gravity. But if a car is going fast enough on a banked turn, then its effective centrifugal force forces it away from the centre of the curve, keeping it up the slope. AlmostReadytoFly (talk) 08:21, 20 August 2008 (UTC)


 * Does it? It would be a very poorly designed road (and tyres) if a car "slipped down" because it wasn't "going fast enough". Vehicles need to be able to stop safely on any road. Even in icy conditions the camber is unlikely to be so great that this would happen.--Shantavira|feed me 12:55, 20 August 2008 (UTC)
 * Oh, real roads. For the sake of the problem, I assumed frictionless roads. ;-) AlmostReadytoFly (talk) 13:23, 20 August 2008 (UTC)


 * While it's always dangerous to assume things, here in New England, winter icing occasionally produces conditions that are virtually friction-free and cars do, indeed, sometimes slip off of roads (and driveways!) in surprising directions including down a superelevated turn.


 * Atlant (talk) 14:46, 20 August 2008 (UTC)


 * Depends on the road and the anticipated speed. On the U.S. interstate, curves rarely exceed a one degree curvature and are banked to allow vehicles to negotiate them safely at 70 mph.  At that speed, you don't need a whole lot of banking, but there are those who end up testing (and exceeding) the limits. By contrast, Phoenix International Raceway has curves with 9 degree and 11 degree banking.  Record qualifying times are around 135 mph.  — OtherDave (talk) 13:27, 20 August 2008 (UTC)


 * Are you perhaps asking about banked race-tracks? The Talladega Superspeedway Nascar track has a 33 degree banked corner.  So if you parked your car on a 33 degree slope - would it slide down?  Well, if it was icey, it certainly would.  I know my car wouldn't slip because I have a lateral accellerometer (for measuring cornering forces in Autocross driving) - and I can exert almost 1g laterally before the car will slip - and that's when the wheels are rolling...you get better grip when parked.  A 33 degree slope doesn't come close to producing a 1g lateral force.  But a heavier car with narrower and worn-out tyres and poorly designed suspension might maybe slide on a 33 degree slope...it's hard to get lateral g-force data so I don't see how we could know for sure.  SteveBaker (talk) 16:03, 20 August 2008 (UTC)


 * Umm .. isn't the coefficient of friction between tyres and road surface the only important factor as regards sliding down the slope when stationary ? Surely weight of car and width of tyres are irrelevant here ? Physics 101 says car can sit on 33 degree lateral slope without sliding as long as coefficient of friction is at least tan 33 degrees, which is about 0.65 (of course if its centre of gravity is too high, it might tip before it slips). Gandalf61 (talk) 16:23, 20 August 2008 (UTC)


 * This has been discussed before and in far greater detail (as well as by people who understand it far better than I), but friction is an excellent example of where a 101-abstraction maps incredibly poorly to physical reality. &mdash; Lomn 16:50, 20 August 2008 (UTC)


 * Yep - that "friction doesn't depend on surface area" thing is complete nonsense out here in the real world. Take a look at a Formula I race car...or a dragster...what do you notice about the tyres?  Kinda wide aren't they!  If narrow tyres would produce just as much friction, why don't they use super-skinny tyres to save weight?   Huh...maybe there's a reason for that?  I drive Autocross events where grip is everything.  Wide tyres make a SPECTACULAR amount of difference to cornering grip.  Feynman's lectures on physics (which you certainly should have read in Physics 101) has a delightful rant about the "laws" of friction.
 * From our article Friction:
 * "This approximation mathematically follows from the assumptions that surfaces are in atomically close contact only over a small fraction of their overall area, that this contact area is proportional to the normal force (until saturation, which takes place when all area is in atomic contact), and that frictional force is proportional to the applied normal force, independently of the contact area (you can see the experiments on friction from Leonardo Da Vinci). Such reasoning aside, however, the approximation is fundamentally an empirical construction. It is a rule of thumb describing the approximate outcome of an extremely complicated physical interaction. The strength of the approximation is its simplicity and versatility – though in general the relationship between normal force and frictional force is not exactly linear (and so the frictional force is not entirely independent of the contact area of the surfaces), the Coulomb approximation is an adequate representation of friction for the analysis of many physical systems."
 * SteveBaker (talk) 17:25, 20 August 2008 (UTC)


 * The design of F1 tyres is governed by many engineering factors. Skinny tyres would have to be inflated at higher pressures. Higher pressures require thicker tyre walls and increase tyre wear. Skinny tyres would also deform more under high downforces and when cornering, when they must communicate lateral forces to the axles and chassis. Wider tyres are less affected by unevenness and small patches of oil on the track. Bicycles and motorcycles manage very well on skinny tyres. I suggest that the idea that lateral friction increases with tyre width is not as obvious as you say. Gandalf61 (talk) 22:34, 20 August 2008 (UTC)


 * We had this EXACT same argument on this very page two years ago - and you were wrong then too, Read Feynman's Lectures on Physics - then tell me that someone with a nobel prize in physics didn't understand friction as well as you do. SteveBaker (talk) 13:54, 21 August 2008 (UTC)


 * Steve - as I said before on your talk page, the tone that you are using in responses like this is somewhat offensive and is inapproriate for the RDs. Perhaps you could tell us exactly what Feynman says on the subject or provide a relevant link ? I think that would be more instructive than taking cheap shots at me - although possibly less satisfying for yourself and less entertaining for other readers. Gandalf61 (talk) 14:28, 21 August 2008 (UTC)


 * The last time (of several times) I discussed this here was: Spherical wheels on cars - I refer you to that discussion. SteveBaker (talk) 17:54, 21 August 2008 (UTC)


 * (unindent) Okay, I have reviewed that thread. I see that Feynman says the standard model of friction is a "good empirical rule" that is correct in "certain practical or engineering circumstances". The quote from our friction article that you gave above says it is "an adequate representation of friction for the analysis of many physical systems". One further source - Hyperphysics, Georgia State University says

Part of the standard model of surface friction is the assumption that the frictional resistance force between two surfaces is independent of the area of contact. While exceptions exist, the assumption has enough validity to be useful in many circumstances. For example, it implies that wide tires will not in general give better traction than narrow tires, and will not change the normal braking distance for a car. Better traction can be obtained with wide tires, or tires with lower pressure, if the pressure changes the coefficient of friction, as on a surface of snow.


 * So all these sources agree that the standard model is an empirical law that has limitations, but is valid in many scenarios. I see no reliable sources that support your bald assertion that it is "complete nonsense out here in the real world". Gandalf61 (talk) 20:38, 21 August 2008 (UTC)


 * What use is it if it has all of these exceptions? We shouldn't be calling it a "law" - that's dangerously deceptive - at best it's a rule of thumb - but not really even that because it's stating that the contact area doesn't matter (which is a qualitative statement) when in fact it should only ever be a quantitative statement: "under a lot of circumstances the effect of the contact area is negligable".  Teaching it the way they do in schools elevates a hackish approximation that only works over some limited set of conditions - to the status of a Law like the laws of motion.  There doesn't seem to be any well-defined set of circumstances when it works and when it doesn't.  At least when Newtons laws of motion break down, we can say "they are accurate at speeds that are a small percentage of the speed of light".  There is no such clear guideline for the applicability of F=un - just some vague hand-waving that says that it sometimes works and sometimes doesn't!  That's hardly useful science!  I have no idea when it's going to work or when it's a case when it's wildly incorrect (as is CERTAINLY the case with car tyres).  It even fails to take account of the wildly different numbers you get for static and dynamic friction (a critical factor in answering this very question in fact).


 * Empirical rules that produce only small errors - or which work well - but only over a carefully specified range of conditions - are useful. Rules that just randomly fail without warning are useless!  This is one of those.  I drive an autocross car - I have an accelerometer in there with me so I can measure the force exerted by the tyres on the road by turning off the ABS, getting the car up to (say) 30mph, standing on the brakes and watching the accelleration.  I tune things like braking distances by doing careful experiments.  There are a lot of variables here - but I can control for most of them - the exception being the tread pattern...but we can figure that into the area calculations.  The wide tyres give me about 30 to 40% more deceleration than the skinny ones made from the same rubber compound - and they have a "contact patch" which is about twice what the skinny ones have.  That says that for rubber on asphalt at the pressures you get with a car, the frictional force is almost proportional to the contact area...that's a lot different from "negligable".


 * F=un says nothing about whether the surfaces are moving or static - and that's crucial to how wheels work without sliding about all over the place. The reason cars have ABS braking these days is 100% because the "Law" of friction doesn't work. SteveBaker (talk) 00:53, 22 August 2008 (UTC)

Brass shell casings corrosion rates in soil
I am trying to find out what exactly would be the effect on a 9mm brass shell casing after sitting  in Frelsburg clay soil  for  15 years  virtually undisturbed. The casings were supposedly ejected from an undetermined semi-automatic weapon and therfore would not have been buried too deep in the backyard of a house in Burton Texas.

What type of corrosion would one expect to see on this fired brass cartridge casing which is 70 % copper and 30 % zinc after sitting the the clay for 15 years?

Should one expect just/only some discoloration, or would the metal actually be compromised to the point of seeing actual physical damage like  decomposition,  or just plain disintegrated ? These casings were not more than 1 to 6 inches beneath the ground's surface and were found with a hand held metal detector.

thanks ! I am trying to determine if some evidence in my friends murder trial was planted by the Harris County Sheriff's Cold Case Squad.

No one bothered question the integrity of the so-called evidence. —Preceding unsigned comment added by Pamfrancis (talk • contribs) 08:33, 20 August 2008 (UTC)


 * While I am skeptical of the value of using Wikipedia to refute police evidence, various references suggest that there would be little corrosion in the environment you describe: [CuZn30, CuZn33, CuZn36 and CuZn37 tarnish slowly to a uniform dull bronze colour with no pitting or localised attack. Hope that helps &mdash; [[User talk:Lomn|Lomn]] 12:59, 20 August 2008 (UTC)

Lomm - the webpage you referenced above is from the metal expert /metallurgist.that we have hired - David Hendrix. The issue is that the shell casings that were extracted from the yard 13 years after Steve lived there( They were retrived in 2001) look pretty good. Other shell casings that did not match the crime scene casings through ballistics, extracted from the same yard - were too corroded for the Sheriff's office balistic expert to even test.  Only  the ones that matched balistically were in good condition, everything else was of no value ( looked like hell - as they should after 13 years) So why would these magic matching shell casing be in a state that he could use to examine match and the others were not - same yard, same amount of time, same clay soil. Also the Sheriff's didn't bother to document said extraction with photos, a diagram, a video, any witness, nothing. NO law enforcement person would skip these vital steps, especially since the location and the legitimacy of this extraction screams the need of this kind of documentation to prove that they got them from where they said they did. Pamfrancis (talk) 22:21, 21 August 2008 (UTC)


 * Pamfrancis, what you really want to do is talk to your friend's lawyer and a forensics expert, not a bunch of people on the internet. There are undoubtedly a lot of factors that are going to affect this whole thing that you either aren't aware of or haven't thought to mention, and anything we say is going to be a really crappy basis for any decisions you make. If this was for a hobby, that'd be one thing, but if this is really a murder trial, you don't want to take chances. (That said, though, I'm pretty skeptical that cops working on a cold case investigation would plant evidence unless they had some really, really strong motivation to do so. I mean, sure, some cops are just plain bad, but even then they don't really do this kind of stuff just because. It might be one thing to close an active case and get that stat, but people working on 15-year-old cold cases aren't generally expected to crack them, since they're, well, cold cases.) -- Captain Disdain (talk) 13:17, 20 August 2008 (UTC)


 * I agree that you should seek an expert - but I've gotta say that it doesn't look good. Brass is pretty corrosion-resistant.  Metal detector enthusiasts frequently find brass artifacts that have been buried for a hundred years or more and they pretty much look new. This for example.  It's going to turn a dull brown - but you aren't looking at the sort of corrosion you'd expect from iron or something.  I really doubt you'll prove anything this way...particularly if you're trying to prove something as controversial as "planted evidence".  But find a lawyer and an expert...don't take my word for it. SteveBaker (talk) 17:11, 20 August 2008 (UTC)

I am going to answer both Captain Disdain and Steve B here -  The fact that only the matching shell casings look good and were able to be tested by the sheriff's ballistic expert is what makes this so questionable- I have hired a lawyer, I have all sort of experts lined up and ready to go now that we have the courts permission to view/test the evidence shell casings introduced at his trial in 2004. In the beginning, they couldn't find them. That was the first 2 years, then they found them, Now the judge in the case has given us his OK. Remember this is Texas, if you are indicted, your guilty. The Sheriff's don't care, that's there job. Both the DAs and the law enforcement believe it is their job to get an indictment - it is the courts job to throw out the evidence, or the defense lawyers job to show reasonable doubt. Neither happened - and i won't even go there - Remember I have been workin on this since the day of his conviction 4.5 years ago. This evidence was collected by the newly formed Cold Case Squad. It is the only "evidence' they could get on Steve who was never even a suspect in the oriiginal investigation. thanks for the help! Information is information where ever I get it, even on the internet. :)Pamfrancis (talk) 22:16, 21 August 2008 (UTC)

Plus I have a metal detector on my porch that I am going to use to go to the same yard to see if I can find some more shell casings, that look like they should - OLD - Pamfrancis (talk) 22:21, 21 August 2008 (UTC)

Brightness of the Sun and other stars
If the Sun were compared to a light bulb, what is its total output in lumens? Is the lumen output of a star proportional to the 'luminosity' as on http://en.wikipedia.org/wiki/Main_sequence_stars, or is it different due to being weighted to a different spectral response curve? If it is different, what is the approximate lumen output of each class of main sequence star?

-User: Nightvid (unregistered) —Preceding unsigned comment added by 128.8.238.179 (talk) 13:54, 20 August 2008 (UTC)
 * I'm not sure about light bulbs but this source says that "The ratio of sunlight to candlelight is about 259,000:1". —Cyclonenim (talk · contribs) 14:14, 20 August 2008 (UTC)


 * That's just meaningless! At what distance from the candle?  At what distance from the sun?  If you are 10 feet away from the sun - it's a heck of a lot more than 259,000 times brighter than being 10 feet from a candle.  If you are a billion miles from the sun and only one centimeter from the candle - then the candle is considerably brighter.  You're quoting this number accurate to better than 1% - but there is considerably more than 1% difference between the brightness of the brightest and dimmest candles.  This is the kind of crappy "factoid" that gets the world in trouble!  What the article actually SAYS is that this is at 3 feet from the candle and at the surface of the earth - so at 93 million miles from the sun.  (There is still far too much precision in the answer though.) SteveBaker (talk) 15:36, 20 August 2008 (UTC)
 * Well, the issue is how to measure "brightness", and the answer is measure one of the following "more pure" quantities:
 * total power output
 * power received at fixed distance per unit-area
 * Furthermore, you can adjust the "total power" to restrict only to "power emitted in visible light spectrum", or to normalize the power-vs-frequency profile based on an approximation of human vision, or any other contrived scheme. We have empirical units such as watts, candela, lumen, etc...  Proper usage of these units requires understanding of what you are actually measuring.  Nimur (talk) 17:07, 20 August 2008 (UTC)
 * That's as perceived on earth, though, right? I think the OP wants to compare actual light output. To the OP: yes, luminosity of stars is measured in terms of the raw power output, while the definition of the lumen is weighted towards wavelengths the human eye perceives strongly, so there won't be a direct relationship. I don't know how big a difference this will make or the answer to your last question. Algebraist 14:22, 20 August 2008 (UTC)


 * In lumens: A candle (nominally 1 candela) outputs 4 x pi lumens.  The sun outputs 3.75 x 1028 lumens. So the sun produces about as many lumens as 3x1027 candles...3,000,000,000,000,000,000,000,000,000 !  Because we're talking lumens (which are adjusted to allow for the nature of human vision) and candle flames vary a bit in color and intensity, this is a very inexact comparison. SteveBaker (talk) 15:36, 20 August 2008 (UTC)


 * Luminosity of stars is generally stated in terms of Absolute magnitude - which uses the sun as a reference. The trouble is that this term derives from a time in history where someone just kinda eyeballed a star and guesstimated how bright it was.  Nowadays there is a mess of derived terms - "Absolute Visual Magnitude", "Absolute photographic magnitude", "Absolute bolometric magnitude" - and they're all different.  But Absolute Visual Magnitude could be roughly compared to a scale of lumens because they both talk about total energy output in the range of the human visual system.  Remember though that magnitude is a logarithmic scale.  SteveBaker (talk) 15:43, 20 August 2008 (UTC)


 * So using the sun's 3.75 x 1028 lumens as a reference, the lumen output of other stars can be calculated by converting the "Absolute Visual Magnitude" to lumens, although this isn't exact due to magnitudes based on "eyeballing", right? Then my last question is equivalent to (short of a conversion factor) "What is the Absolute Visual Magnitude of each class of main sequence star?".
 * -Nightvid —Preceding unsigned comment added by 128.8.238.179 (talk) 21:22, 20 August 2008 (UTC)


 * But it's not that simple. Stellar magnitudes are represented on a logarithmic scale and most of them are negative numbers.  A magnitude -3 star is 2.512 times brighter than a magnitude -4 star which is 2.512 times brighter than a magnitude -5 star.  So you CAN'T simply say the conversion factor from magnitude to lumens is the sun's brightness in lumens divided by the brightness in absolute magnitude.  You need some kind of equation which is going to have logarithms and stuff in it.   I'm too tired to figure it out right now - but I'm sure someone here will do it. SteveBaker (talk) 03:22, 21 August 2008 (UTC)


 * (to Nightvid) I think you misunderstood slightly. Steve was saying that the visual magnitude system was based on eyeballing for most of its history, which started during the time of Hellenistic Greece.  Since the human eye is not very accurate, cameras were later used to measure brightness, but early films were much more sensitive to blue light than red light.  Hence, blue stars appeared much brighter to cameras than to the human eye, while red stars seemed to be much dimmer.  The photographic magnitude system was invented as a result.  Bolometric magnitude was later used to express a star's total luminosity for all of the electromagnetic spectrum, not just the visible band.
 * The measurement of brightness in astronomy is called photometry; if you read the article, you'll find it's a very exact science using expensive CCDs and software. The "V" band corresponds to the visual band, so this is the band to search for on Google for data on precise visual magnitudes.
 * In the magnitude system, every five-magnitude increase represents a dimming by a factor of exactly 100. Hence, a magnitude-0 star is exactly 100 times as bright as a magnitude-5 star.  Since this system is logarithmetic, each magnitude represents an increase or decrease in brightness by a factor of 2.512 (the fifth root of 100).
 * The spectral classification article has data on the luminosities of stars in each class. The Sun has a bolometric magnitude of 4.75, so you can easily calculate the absolute magnitudes represented by the data.  An O-type, for example, outputs ~1 400 000 times as much power as the Sun and has a magnitude lower by log (base 2.511) 1 400 000.  That's about mag -10.62.  If you want figures for visual brightness, use this conversion table.  The data in the spectral classification table is so approximate that this is not necessary, but it's important for specific stars where exact photometric data is available.  --Bowlhover (talk) 08:33, 22 August 2008 (UTC)

Scientific/medical reasons for Heartbreak
What are the scientific/medical reasons for experiencing heartbreak? What kind of person or what condition does a person have that cannot experience heartbreak? --12.33.211.29 (talk) 18:03, 20 August 2008 (UTC)
 * We have an article: Broken heart (also see grief as heartbreak is considered to be a form of grief-reaction). As for those who cannot experience heartbreak, well, that could only happen if they didn't care in the first place i.e. sociopaths, and psychopaths but even these may feel selfishly-motivated grief. Fribbler (talk) 18:07, 20 August 2008 (UTC)
 * Also find Broken Heart Syndrome for the physical effect of negative emotional trauma. Julia Rossi (talk) 06:26, 21 August 2008 (UTC)

Tea
Tea is made from the leaves of Camellia sinensis. Can a similar brew be made from other species of Camellia? DuncanHill (talk) 18:29, 20 August 2008 (UTC)
 * You could brew/seep just about any plant matter in water to make a at least a mildly flavored drink. It's a question of whether it tastes good enough to drink. I know, for example, Slashfood (a cooking and food blog) discusses "backyard teas" a few times. These are teas made not from Camellia sinensis, but from plants you would find in your backyard. Rooibos is a popular tea alternative as well. I assume tea could be made from other species of Camellia, but I have not found any evidence of any commercial production of such teas. --Russoc4 (talk) 00:36, 21 August 2008 (UTC)


 * You might want to be a bit more careful than that though - lots of plants are poisonous or have very serious effects on your health. I'd want to check before making drinks from any old random plant! SteveBaker (talk) 03:14, 21 August 2008 (UTC)


 * Under tomato Wikipedia says, "tomato leaves and stems actually contain poisonous glycoalkaloids, but the fruit is safe". A rather common flowering shrub planted in gardens is the oleander. Here is what Wikipedia has to say about this plant. "Oleander is one of the most poisonous plants and contains numerous toxic compounds, many of which can be deadly to people, especially young children. The toxicity of Oleander is considered extremely high and it has been reported that in some cases only a small amount had lethal or near lethal effects (Goetz 1998)." Andme2 (talk) 03:27, 21 August 2008 (UTC)


 * Well, yes, obviously you can't use anything you find. I thought common sense would dictate that. That's also why I only provided the link for some safe samples. Apparently it's not enough to help answer a person's question. --Russoc4 (talk) 15:01, 21 August 2008 (UTC)


 * Another note, the article Herbal tea suggests that "An herbal tea, tisane, or ptisan is an herbal infusion made from anything other than the leaves of the tea bush (Camellia sinensis)... In some countries (but not in the United States) the use of the word tea is legally restricted to infusions of Camellia sinensis (the tea plant)." These are not cited though. --Russoc4 (talk) 15:14, 21 August 2008 (UTC)
 * I've been looking, but I have not found any discussion on caffeine in Camellia plants other than Camellia sinensis. --Russoc4 (talk) 15:26, 21 August 2008 (UTC)

A lot of fruit plant leaves can be used for tea, and lemon, orange, grapefruit leaves even contain a bit caffeine. Tasteful too are strawberry, raspberry and the currants. --Ayacop (talk) 16:26, 21 August 2008 (UTC)


 * I was really just asking about Camellia. An estate in Cornwall has started growing camellia sinensis, and producing tea commercially from it. I was wondering if any of the other camellias which flourish in Cornwall could be used to make a drink like tea (that is, actually tasting like tea, not like raspberries, or chamomile, or suchlike). DuncanHill (talk) 16:56, 21 August 2008 (UTC)
 * Again, it's theoretically possible, but there does not seem to be any commercial value to Camellia plants other than the one. --Russoc4 (talk) 19:42, 21 August 2008 (UTC)

Black holes again: interior
It was stated previously that the mass of a black hole would be concentrated within a single point of the interior. I doubt this (as a general case, at least) based on the following: think of the earth and the surrounding galaxy, think of the interstellar space and the neighboring galaxies. The space there has a density, widely varying, but none-zero and (?) above a certain ground level (the cosmic background radiation, at least). Now think of a sphere with radius R around the earth (or any other point you like). Think R grow bigger and bigger and consider the mass contained. As the Schwartzschild radius of any mass is growing linearly with mass, but at constant (minimal) density, the mass grows cubic with radius, at some radius the mass contained within will will be contained entirely inside its Schwarzschild radius --- it will be a black hole, if looked at from outside. To cast it into a question: are we inside a black hole, if looked at from far enough away? 93.132.147.52 (talk) 18:54, 20 August 2008 (UTC)
 * I think you're describing the cosmological horizon. While it's a form of an event horizon, it's not classically understood as a black hole. &mdash; Lomn 19:02, 20 August 2008 (UTC)
 * I don't think that's what he or she is referring to. Imagine an infinite, homogeneous medium.  Pick a point in that medium, and imagine a sphere centered on that point, of radius r.  As you increase r, the enclosed mass will eventually exceed c²r/2G.  The OP seems to believe that, at this point, r will be an event horizon, because it is the Schwarzschild radius for the enclosed mass.  But the enclosed mass will not (necessarily) collapse to a singularity, and it's the singularity that results in the event horizon.  A singularity severely warps spacetime; a homogeneous medium does not.  (The OP actually allowed for an inhomogeneous medium, like the real universe, but for moderate clumpiness that's not forming singularities, you can just imagine a homogeneous medium.) -- Coneslayer (talk) 19:56, 20 August 2008 (UTC)
 * It depends on the shape of the universe, but it's quite possible that the (observable) universe is a black hole. If so, it just hasn't finished collapsing yet.  --Tardis (talk) 20:02, 20 August 2008 (UTC)


 * The Schwarzschild solution and hence the Schwarzschild radius, implicitly assume that expansion is neglible in the local neighborhood of the black hole. As we know the expansion of the universe is not neglible when applied to the universe as a whole, these solutions are not applicable whem considering the universe as whole.  Nonetheless, it is still possible that the universe as a whole does constitute a type of black hole (though recent measurements make this increasingly unlikely).  If so, the singularity for the universe would occur in the future via the Big Crunch.  More generally, all the mass inside a black hole must reach a singularity in finite proper time (time as measured by a local observer).  So it is possible for transient inhomogenuities in mass to exist within a black hole, but they would have to be eliminated eventually.  Dragons flight (talk) 20:14, 20 August 2008 (UTC)


 * This stuff is regularly fantastic. It's impossible to tell when you people are kidding.  Wanderer57 (talk) 20:17, 20 August 2008 (UTC)
 * I've run into editing conflicts now several times. I'm not sure if I should get annoyed that I have to retype so often or to feel happy that the topic has caught so much attention --- I'm inclined to the later. As I can tell for kidding, I'm pretty sure for the OP and quite sure for the answers that the subject is treated seriously. 93.132.147.52 (talk) 20:26, 20 August 2008 (UTC)
 * OFF TOPIC TIP: If you get hit by an edit conflict, hit the "BACK" button on your browser - it should take you back to the editing window. Use the mouse to select the text you typed and use Ctrl-C to copy it into your clipboard.  Now you can navigate to the newly edited version of the page (I usually hit the "BACK" button again then hit "RELOAD" to refresh it).  Then you can hit the [EDIT] widget again and use Ctrl-V to paste your text back in again.  Then you can hit "Save page" again - and this time, so little time will have elapsed between hitting EDIT and SAVE that it's very unlikely anyone will get in ahead of you. SteveBaker (talk) 03:12, 21 August 2008 (UTC)
 * You've clearly never had to write a reply on Keeper76's talk page during a busy period. Crickey. —Cyclonenim (talk · contribs) 09:07, 21 August 2008 (UTC)

I was not thinking on the effects of totally (that's to say, sufficiently for the effect) homogeneous medium outside the black hole. That splits my question into two: what if the density would drop just outside the schwarzshield radius, leaving uncounted galaxies, unwilling to collapse to a singularity right now, inside? And what if two black holes of equal mass would decide not to drop into each other but to orbit around the common barycenter? And more, if there where three BHs, would it be possible for one of them to bend spacetime as much that it could be expelled? 93.132.147.52 (talk) 20:49, 20 August 2008 (UTC)


 * Anything inside a blackhole must collapse into a single singularity within finite time. General relativity does not permit any permanently stable orbits in the interior of a black hole.  Achieving a stable orbit inside the event horizon, relative to the singularity, is functionally equivalent to going faster than the speed of light, i.e. it's impossible.  Dragons flight (talk) 21:16, 20 August 2008 (UTC)
 * Thats common talk, but just why? From inside, there doesn't seem to be a need for it. And finite time for what kind of observer? Inside or outside? By the way, I've never been convinced that faster then light is impossible. 93.132.147.52 (talk) 21:30, 20 August 2008 (UTC)
 * Finite "proper time", which means time as measured by the object itself (the one that's falling in). From the point of view of an external observer, it takes an infinite amount of time just to reach the event horizon. If you don't accept the speed of light as a fundamental limit, then you have to reject the whole of relativity which includes the whole existence of black holes (some other theory may have something similar, but it won't be the same as what we know as a black hole), so the question becomes moot. If you accept the basic principles of GR, then it the maths is quite clear that objects that cross the event horizon have to hit the singularity in finite proper time. I'm not sure how to explain it without the maths, hopefully someone else can. There is a special case which is easy to explain, though - the case of an inert falling object, one without any kind of engine. If it's not going to hit the singularity it needs to either go into orbit, or leave the black hole entirely, and orbital velocity and escape velocity can both be calculated using simple Newtonian mechanics and you can see from their formulae that they would be greater than the speed of light once you cross the event horizon (actually, orbital velocity is greater from a even greater distance). --Tango (talk) 23:52, 20 August 2008 (UTC)
 * I feel slightly misunderstood. For the speed of light, I see that it's impossible to accelerate as much as to (exactly) reach it or exceed it, but I can't see that it would be ruled out that something could be faster than light (space like) right from the start.
 * Your "Anything inside a blackhole must collapse into a single singularity within finite time" holds in the simplest possible case (Deriving_the_Schwarzschild_solution) but I don't see or have references if this holds for more complex situations, too. Although, an object moving in this metric must be small enough not to disturb the metric 93.132.168.99 (talk) 08:14, 21 August 2008 (UTC)


 * User:93.132.168.99 - you aren't misunderstood. I perfectly understand what you are saying - but it's still wrong.  Objects moving faster than light ("tachyons") are completely forbidden by the very same math that prevents a normal object from attaining light speed.  It's not like the situation with photons which can only move at the speed of light because they started off moving at that speed and have zero rest-mass.  When you calculate the mass or length or time-dilation factor for tachyons using the lorentz transform, you end up having to calculat the square root of (1-v2/c2).  If v (the velocity of the object) is greater than c (the speed of light) then v2/c2 is bigger than 1.0 - which makes one-minus that be a negative number.  Stick a negative number into your pocket calculator and hit the square root button!  It says "ERROR"!  When you take the square root of a negative number you get a 'complex number'.  That would leave mass/length/time for a tachyon to be a complex number.  But it's a rule that no matter what, no "real world" result can ever be a complex number - they are merely a convenience for doing calculations - in absolutely ALL real world calculations, the complex number ends up dropping out of the calculations before you get the answer.  That wouldn't happen for tachyons.   Therefore it's very safe to assume that they don't/can't exist.  You can also show that such objects would require an infinite amount of energy in order to slow DOWN to anything less than infinite velocity.  Something that has infinite velocity can't be at any place in space where you might observe it because it's gone already.  Tachyons are *SO* not possible! SteveBaker (talk) 13:47, 21 August 2008 (UTC)


 * Don't think I agree with you there. You might as well say that spacelike intervals don't exist because $$\sqrt{dt^2 - dx^2 - dy^2 - dz^2}$$ is imaginary. I would say that tachyons have "spacelike mass", or negative mass-squared. It's a perfectly well defined concept. In fact it's a key part of the Higgs mechanism in the Standard Model. There's also no reason you couldn't detect a particle moving with infinite speed—it's only necessary that the particle's worldline intersect the detector's worldline, which can still happen if one of the worldlines happens to be sideways. The real problem with tachyons is that, classically at least, they wreak havoc with causality. -- BenRG (talk) 15:24, 21 August 2008 (UTC)

I don't think I believe that the mass of a black hole is concentrated at the singularity. The Schwarzschild geometry is a vacuum geometry. But the singularity of a Schwarzschild black hole is in the future (as in the Penrose diagram I drew above). So, if there's mass there, where was it at earlier times? As far as I can tell, it was in the gravitational field. Gravitational fields can carry mass-energy but regions with a strong gravitational field are still called vacuum. So I would say that the mass of the Schwarzschild black hole is everywhere, but in a form that's missed by some accounting systems.

I know almost nothing about more realistic black hole models, but in the collapsing and evaporating black hole from Hawking's paper (the other Penrose diagram above) the singularity is still spacelike and times after it forms are not defined. All of the mass of the collapsing star ends up at the singularity, but there's no later time when it's "still there". I've never seen a realistic model with a timelike singularity, but I suppose they exist. Even in that case, though, time and space are so mixed up in black holes that I don't think you can ever say that the mass has "already" reached the singularity. If you're outside a black hole then by definition you're not in the causal future of any point in the hole, so every point in the hole is either causally disconnected from you or in your causal future. So it's perfectly consistent to take the position that nothing inside the event horizon has happened yet; you can even say that the event horizon hasn't even formed yet.

In the cosmological scenarios it starts to matter a lot how you define "black hole". -- BenRG (talk) 15:09, 21 August 2008 (UTC) Just thinking (being a non mathematics person)Relativity & Quantum cannot explain a Black hole due to the answer always ending in infinity; am i wrong in thinking that the total mass in the universe of everything after the big bang we could say is the the total amount it takes to fill a black whole, when it is then full it destabilizes and explodes; thus creating a big bang...(probs just had to much to drink last night but makes sense to me :)
 * People often say that a black hole is a region of spacetime from which not even light can escape. In that case the whole universe is a black hole, since light can't escape from the universe. But also the causal future of any point in spacetime (a future light cone) is a black hole, so black holes are forming all around us and engulfing us all the time.
 * A better definition is that a black hole is a region in which you'll inevitably hit a gravitational singularity in the future. That doesn't work in a Big Crunch universe, where everything eventually hits a gravitational singularity. In fact there's no way to unambiguously distinguish the Big Crunch singularity from a black hole singularity. Black holes are just parts of the Big Crunch that happened early (except that, as I said, they haven't really happened yet). If the universe exists forever, as it does in the ΛCDM model, then this definition implies that no large region of the cosmos constitutes a black hole because (by assumption) it won't end up as a singularity. This definition also doesn't work with exotic wormhole geometries where you can avoid the singularity.
 * An even better definition, which avoids mentioning the unobservable interior of the hole, is that a black hole is a region from which you can't get to future infinity. Again that doesn't work in a Big Crunch universe. It gets interesting in a universe that becomes de Sitter in the future (as ΛCDM says ours does), since future infinity in a de Sitter universe is a 3D surface instead of a point, reflecting the fact that different parts of the universe become causally disconnected in the future. So you could, if you liked, declare a "black hole" to be a region from which you can't get to our future infinity, i.e. the point the Local Supercluster is headed for. In that case we're surrounded by a black hole with an event horizon that's about 15–16 billion light years away in every direction ($$\int_{\text{now}}^\infty \frac{a(\text{now})}{a(t)} \, c \, dt$$). In the future that distance will increase to an asymptotic value of about 16–17 billion light years ($$\frac{c}{\sqrt{\Omega_v}H_o}$$). It's an inside-out black hole, i.e. the interior of the hole is the part outside the sphere. Phenomenologically it's very much like an ordinary black hole. The effective surface gravity at the horizon goes to infinity; objects falling through redshift out of visibility; and I think it Hawking radiates, but don't quote me on that. I don't think it would normally be classified as a black hole, however.

21 Hockey Pucks
I work at a local ice arena, and just before I left for home, some guy walks in and asks to buy 21 pucks. While I'm gathering the pucks I, of course, ask him what he needs 21 pucks for. He says he needs puck for "practice, or something like that" and 21 "seems like a good number". I've worked at the arena for years now, and play hockey myself, so I know this kid doesn't play, and he couldn't answer a few basic questions about the sport. Can anyone think of any possible use for 21 pucks? That's between 115.5 and 126 ounces of vulcanized rubber. --Willworkforicecream (talk) 21:57, 20 August 2008 (UTC)


 * Some act of violence comes to mind. Then again, 21 is an usual number. Perhaps he's inventing some idiotic game for the next time he and his buddies are drunk.--El aprendelenguas (talk) 22:05, 20 August 2008 (U


 * A google search reveals a few crafts projects you might do with a hockey puck, but nothing that specifically asks for twenty-one of them. I've heard of them being used as feet for things, perhaps he has four tables that need to be an inch taller?
 * I can't think of any acts of violence that wouldn't be better and more cheaply accomplished by a simple rock or brick.
 * That's what I figured, at $2.00 a puck, rocks seem like the way to go. --Willworkforicecream (talk) 03:39, 21 August 2008 (UTC)
 * If you're sure he doesn't play hockey, I'll bet he's using them as raw material in some basic project, Perhaps as weights or rollers or something. APL (talk) 23:15, 20 August 2008 (UTC)
 * Incidentally, do you always grill customers on the basic facts of hockey before selling them pucks? APL (talk) 23:23, 20 August 2008 (UTC)
 * I do if they're buying such large quantities of pucks.--Willworkforicecream (talk) 03:39, 21 August 2008 (UTC)


 * Maybe he was hired to test out the customer service of your store, a test which unfortunately you appear to have failed? Nil Einne (talk) 11:51, 21 August 2008 (UTC)
 * So, what potential bad scenario do you generally forsee when you encounter a non-hockey-player trying to buy pucks in volume? That the guy is going to start whanging them at people, windows and/or passing cars? --Kurt Shaped Box (talk) 12:06, 21 August 2008 (UTC)


 * To honor a visiting dignitary or head of state? TenOfAllTrades(talk) 23:42, 20 August 2008 (UTC)


 * That would presumably be quite painful. --Bowlhover (talk) 08:58, 21 August 2008 (UTC)


 * There are perhaps a range of other uses for hockey pucks - but why would he lie about his reasons for needing them? Maybe there is a school who are starting hockey practice - he might maybe work for the school or be something to do with a parent's group and have been sent out to buy the pucks on behalf of the team - although he might personally have little or nothing to do with the team.  That's why he'd be vague about the precise reason they're needed - and he might easily not know anything about the sport. SteveBaker (talk) 03:07, 21 August 2008 (UTC)
 * I doubt it is another school starting a team, seeing as there are three school teams in the valley, and I have played on all three teams.--Willworkforicecream (talk) 03:39, 21 August 2008 (UTC)


 * It's not a good idea to be suspicious because some people who are simply shy. I used to buy equipment for various reasons ranging from science to toad keeping, but I rarely had non-innocuous motives.  Nevertheless, I tried my best to hide my actual intentions for fear of being perceived as "unusual".  It was due to my introversion and nothing else.  --Bowlhover (talk) 05:47, 21 August 2008 (UTC)


 * I'm going agree with Bowlhover 100%. Honest, but slightly unusual answers like "I'm building a habitat for a pet tortoise." just invites a conversation I'm just not interested in having with some random stranger. Especially not if I'm in a hurry.APL (talk) 13:21, 21 August 2008 (UTC)


 * For backyard (air?) pistol shooting targets, maybe? Perhaps he only had enough money with him to buy 21 pucks? --Kurt Shaped Box (talk) 11:57, 21 August 2008 (UTC)
 * Inventor. --jpgordon&#8711;&#8710;&#8711;&#8710; 15:24, 21 August 2008 (UTC)
 * My brother used hockey pucks to put a body lift into his Jeep Wrangler. He's pretty cheap and didn't want to cough up for the commercial kits.  Have seen this use on the internet in a number of places although I am too lazy to google it right now. 142.36.194.253 (talk) 22:17, 21 August 2008 (UTC)
 * That's plausible - but why 21 of them? You'd think he'd have gone with a round number - like maybe 20!   But I could imagine there being 21 kids in a class of some kind taking hockey lessons. SteveBaker (talk) 00:20, 22 August 2008 (UTC)
 * I'm not sure that logic is right. If I needed some number of hockey pucks for a construction project then it would be the most natural thing in the world to pick up a spare in case I mucked one up, but it would seem weird to buy hockey pucks for a class and get exactly enough for the students. APL (talk) 05:08, 22 August 2008 (UTC)
 * Or "I need 3 points of support for my engine/whatever, and 7 pucks each should be just right, and 3*7=21." jeffjon (talk) 17:49, 25 August 2008 (UTC)

Moving toilet water
I've been experiencing rain from Tropical Storm Fay all of yesterday and today, and noticed that the water in my toilet bowl keeps shaking prior to usage and flushing. I've just lifted the lid and the water slightly ebbs and flows, sometimes more noticeable than others. I've even flushed, waited 5 minutes with the lid open, and still the water is shaking. I haven't noticed this before the storm, so it makes me wonder if this is caused by low pressure or something? Or is my toilet just possessed? Thanks!--El aprendelenguas (talk) 22:00, 20 August 2008 (UTC)


 * How windy is it outdoors? A gusty and irregular wind can induce surprisingly large pressure differences in the vent pipe that goes up through the roof; I've observed one good blast can make the water go up, then down by half an inch or more.  (For other reasons, once it goes down it doesn't come back up all the way, either.)
 * It's not inconceivable that the right kind of blowing can set up something like a standing wave in the vent that in turn causes vibration and rippling in the water. I can't prove that's the case here, but it's the best explanation I've been able to come up with for myself over the years :-)   --Danh, 70.59.119.73 (talk) 23:44, 20 August 2008 (UTC)
 * I agree. I've known the toilet bowl dry out completely in high winds, presumably by some mechanism similar to what you describe. Once the water gets blown past the U-bend, it doesn't come back. --Tango (talk) 23:55, 20 August 2008 (UTC)


 * Are you sure you don't just have a leaky flap valve? When that big rubber thing at the bottom of the cistern leaks, you get a slow trickle of water running down the side of the bowl which is sometimes not at all noticeable - except that it causes ripples in the water at the bottom.  Low pressure shouldn't make a difference - it could just be a coincidence that the flap valve developed it's leak at that moment.  Not all toilets have flap-valves (the ones here in the US do - the ones in the UK use a syphon mechanism which develops different problems) - but anything that produces a constant tiny trickle of water would do that. SteveBaker (talk) 03:02, 21 August 2008 (UTC)


 * I can confirm that the water level in my toilet fell by about half and inch during a squall a few days ago. I was actually meaning to question it here... --Kurt Shaped Box (talk) 00:56, 22 August 2008 (UTC)


 * If you live in a high-rise building, it is normal for the building itself to shake or rock slightly in a high wind, which might agitate the water perceptibly. --Anonymous, 04:35 UTC, August 21, 2008.

Thanks for the answers. I think it is the wind and the air vent, like many of you suggested. Anyway, the storm has passed for the most part, and the water is still.--El aprendelenguas (talk) 19:57, 21 August 2008 (UTC)