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December 7

Why aren't all houses made of brick?

Topic says it all. ScienceApe (talk) 00:26, 7 December 2012 (UTC)[reply]

let me guess...you just re read the three little pigs to your kid. Couldn't helpGeeBIGS (talk) 00:36, 7 December 2012 (UTC)[reply]

There are probably a great many reasons. Just one example — brick houses don't do well in earthquakes, so if you're building where earthquakes happen, brick is a bad choice. --Trovatore (talk) 00:37, 7 December 2012 (UTC)[reply]

Economics often has a lot to do with it. For example, the climate in the major Australian cities is not very different. However, in Perth double brick (brick outside, and rendered brick inside) construction completely dominates, but in other cities brick veneer (brick on the outside, frame&clad on the inside) dominates. The reason is an accident of history. In Perth a couple of clever chaps took advantage of local high quality clay deposits and devised their own factory machinery made from war surplus construction equipment - the result being that Perth was supplied with very cheap bricks. In other words in Perth double brick construction was cheaper but in other cities brick veneer was cheaper.

Double brick construction works very well in mediteranean climates such southern Australia, much of southern Europe. But in colder climates (ambient temperatures routinely below zero C) timber framed timber clad construction offers better thermal insulation, but is not maintenance free.

There are cultural aspects. For example, in many parts of Australia during the 1930's depression and at certain other times, people with low incomes got houses of timber frame asbestos/cement clad construction. Such houses were cheap to build and actually perform quite well in durability and in thermal properties, but ever since have had a cheap and nasty stigma. So, after the post- World War 2 ecomonic boom, anybody who didn't earn much usually stretched themselves and committed to an achievable dream - owning a brick house (even if that meant smaller rooms and paying off a loan for the next 30 years).

I imagine similar economic, environmental, and cultural considerations apply elsewhere in the world. Wickwack 124.182.154.215 (talk) 01:27, 7 December 2012 (UTC)[reply]

Relevant past discussion here. Evanh2008 ^{(talk|contribs)} 03:24, 7 December 2012 (UTC)[reply]

Traditionally houses were built in whatever building materials were available locally. I live in the UK and in my local area there are lots of sedimentary clays and sands and so, traditionally, the houses were built of brick. If you look at local maps from the 19th century there were brick fields and sandpits dotted all over the place. There is also some new red sandstone and some of the churches and public buildings are built of that. A few miles up the road you get into more rugged Millstone grit country and so all the houses, even basic workers' cottages, were built of stone. In other parts of the country, such as East Anglia houses were often built of flints set in lime mortar. As better and cheaper transport became available different building materials were used around the country and nowadays many new homes have timber frames with a brick veneer - quick to erect but still fit in with the traditionally built houses. The materials used today owe more to the speed and convenience of building rather than what is available locally. It is now very exensive to build using stone, whereas in some areas that was once the cheapest material. I don't know about other countries, but if you want to build a house in the UK it's generally easier to get planning permission if the building appears to fit in with the other buildings round about rather than being something that sticks out like a sore thumb. Also I think that most buyers prefer something that they expect to see in that area - unless they want to go for a big statement. Richerman (talk) 09:33, 8 December 2012 (UTC)[reply]

Good points. Here in Australia it is much the same with regard to getting the necessary Building Permits - Shires require that new dwellings and other buildings have an appearance that fits in with the existing buildings nearby. Some shires even specify what colours are to be used. Wickwack 120.145.28.146 (talk) 13:37, 8 December 2012 (UTC)[reply]

Interestingly (and embarrassingly) our article on building materials doesn't even mention bricks. Ghmyrtle (talk) 10:05, 8 December 2012 (UTC)[reply]

Venus in a small telescope

The other day I was looking at Venus with my 6.25" reflector. I was hoping to see it as a crescent. All I could see was a very bright spot, with all three eyepieces. Is Venus just too bright for that? Venustar84 (talk) 00:39, 7 December 2012 (UTC)[reply]

Estimate the current angular size: its diameter divided by its distance yields about 11 arc seconds. In principle, your telescope should be able to resolve about a half an arc-second in visible light. So, your minimum resolvable feature-size is good enough to image the structure of Venus' illumination pattern (its crescent); and because the planet appears bright enough that you should be able to magnify to fairly large size.

In practice, you're only about 10x above the resolution limit. So, any imperfections at all - atmospheric aberration, telescope alignment or mirror collimation issues, and so forth - will blur the image. How good is your scope's primary mirror, and how good are the optic(s) you're putting behind it? You can push it to its limits by measuring the circle of confusion of a bright star, who is well-approximated by a perfect point-source.

If you're collecting too much light - which happens when you're looking at planets - you can create an iris by cutting a hole of whatever size you want, out of cardboard or tinfoil, and shutter over some of your aperture. This dims the image and has the added bonus of improving your focus. Camera people call that process "stopping down the aperture." Telescope stores charge a lot of money for stops, but you can build one out of tinfoil or stiff card paper, without worrying too much about precision. Just try to be mostly circular, mostly centered, and mostly symmetrical. (Other experts recommend off-center stops, especially if your scope is Newtonian, for obvious reasons).

I have a 205 mm Newtonian scope, and lots of very good glass, a not-too-shabby set of cameras, and (formerly) a whole lot of human- and CPU-time to spend post-processing; but I have never successfully imaged any structure on Venus. I have had better luck with Jupiter clouds. Nimur (talk) 01:27, 7 December 2012 (UTC)[reply]

The crescent shape of Venus is visible using even moderately-good binoculars. Unless there's something seriously wrong with your telescope, you shouldn't have any difficulty resolving its shape. (Though if you do find it uncomfortably bright, you could partially obstruct the open end of your telescope aperture, or acquire a neutral density filter to take some of the edge off.)

That said, when exactly was 'the other day'? In recent weeks, Venus has been getting closer and closer to 'full' (see aspects of Venus for dates). Remember that Venus takes a lot longer than, say, the Moon to work through its phases. Based on this calculator widget, Venus was more than 80% illuminated on 1 November, and 88% illuminated on 1 December. Any time in November, Venus would only look a little bit out of round. Venus is also very nearly as far away from Earth as it can get (it is at its furthest when it is 'full', at superior conjunction on the opposite side of the Sun from Earth), and therefore very nearly at its smallest apparent size. At its furthest, its angular diameter is just under 10" (10 arc seconds); at its nearest, it swells to 66" across. TenOfAllTrades(talk) 01:35, 7 December 2012 (UTC)[reply]

As an amateur astronomer, I can clearly see Venus as a crescent (or some other phase) with my 60 mm (2.4 inch) telescope. I haven't tried recently, but if you're having trouble with your 6.25" reflector, and you're sure Venus wasn't near full that day, something is wrong. --140.180.249.232 (talk) 05:46, 7 December 2012 (UTC)[reply]

At 11", Venus must have been near-full (given that the angle for maximal distance (=full) is 10").

The theoretical limit of a 6" reflector is 0.75 arcsec (as a rule of thumb, it is 1 arcsec for the popular 4.5" tubes and inversely proportional to aperture.

Stopping the aperture is quite awful. It plays bloody hell with resolution. However, high-quality filters are a bit on the expensive side. If you want, you can make a simple cardboard stop or two (just cut a round hole and paint the cardboard black). You won't suffer much if you make a 2.25" (57mm) aperture - that's according to the rule above, still a 2-arcsec resolution, and if Venus is half or less lit, it'll be ~17 arcsec end to end.

From the data Nimur and Ten gave, it'll be about 9 months (calculated mentally) until Venus is down to half-lit. Hope that helps. - ¡Ouch! (^{hurt me} / _{more pain}) 09:56, 7 December 2012 (UTC)[reply]

Oh, and 6.25 is not "small." Those who laugh at it are brag-a-lots.

Wouldn't it be better to make a number of cutouts at the edge and keep your resolution? Filters for the moon and bright planets are rather cheap though, about $16 I think? Sagittarian Milky Way (talk) 15:30, 10 December 2012 (UTC)[reply]

As was pointed out in some TV segment I saw the other day, Galileo observed the different phases of Venus, whose size and shape demonstrated that Venus orbited the sun, not the earth. When it's at crescent phase it's closer to us and looks larger through a telescope. When it's at near-full phase it's farther away, so it looks small and roundish. But if you have doubts about the quality of your telescope, just aim it at some other objects, such as the moon, or one of the outer planets if they happen to be visible. Then you can tell if there's anything wrong with your telescope. ←Baseball Bugs ^{What's up, Doc?} carrots→ 14:00, 8 December 2012 (UTC)[reply]

Speed of light, etc.

All our articles on relativity seem to be a little sparse in this regard, so I hope no one minds me asking some dumb questions. What I am trying to work out is how exactly two objects can have independent velocities that sum to c or greater, without actually being measured by each other as accelerating to the speed of light. I believe the easiest way to put this into simple terms is the following:

Two objects (A and B) accelerate in the same direction at the same speed (for our purposes, let's peg it at 51% the speed of light) relative to an observer in an inertial reference frame. This observer sees two objects, both accelerating toward him at 51% the speed of light. Nothing weird so far, but then, B changes direction. While A continues to accelerate toward our observer, B reverses direction, now accelerating away from the observer at 51% the speed of light.

The observer (as far as I can tell) will not notice anything out of the ordinary. Both objects continue to move at the same speed relative to him/her; one has merely changed direction. Object A, however, has a problem -- according to our observer, B is now accelerating away from A at slightly over the speed of light. Because of special relativity, however, A measures the velocity of B at slightly under the speed of light (I think this is correct). Therefore, if asked to describe B's location in time and space, Object A and the observer will give two completely different answers. Issues like length contraction aside, the object they are observing possesses the same physical qualities, but the two will not agree on its location. How, if it all, is this resolved? Evanh2008 ^{(talk|contribs)} 04:08, 7 December 2012 (UTC)[reply]

These sort of apparent paradoxes show up all the time in special relativity problems. See the Ladder paradox for another one. The answer to your question is that both Object A and our observer are perfectly capable of calculating the location and velocity of B and both will arrive at the same answer so long as both work from the same set of data. If they don't get the same answers, then they have left something out of their calculations. It's the same basic problem with observing any far away object. Where we see a star now in the sky isn't where the star is now. An observer on the other side of the galaxy will see the same star in a different location, but given the right information about the star's position and velocity we can both calculate the star's actual position and come up with the same answer. --Jayron32 04:17, 7 December 2012 (UTC)[reply]

You need to think a lot more carefully about how your "observers" are measuring these things. With a radar gun? Then how do you convert a redshift into meters per second, and why? Or by timing how long it takes to go a given distance? Then how are you determining the elapsed time and the distance? The reason no one understood special relativity before Einstein's 1905 paper (even though all the math had been worked out before) is that they imagined that measuring speeds was something you could just do, and the details were irrelevant. Einstein carefully considered what it actually means to measure a speed, and showed that the specialness of the speed of light is quite natural. Unfortunately, most people still don't get it. They talk about special relativity in terms of "observers" who somehow just know the length and speed of everything (though apparently they don't really, since they all disagree with each other). Special relativity seems mysterious when described in this way. It isn't really.

You might want to read this translation of Einstein's paper. It's still one of the best introductions to special relativity. Note that what he calls "observers" in the paper are just scientists who look at things. The strange beast that's now called an "observer" is a later invention. -- BenRG (talk) 04:42, 7 December 2012 (UTC)[reply]

I have always taken that latter usage of "observer" to mean the God's eye view of things, on the assumption that God is not bound by the laws of physics and hence can observe anything. In the narrower intention that Einstein evidently meant, the "observer" would be more like a "measurer". Michelson, for example, didn't literally "observe" light traveling at the speed of light, but he invented a way to measure it. ←Baseball Bugs ^{What's up, Doc?} carrots→ 13:53, 8 December 2012 (UTC)[reply]

Here are some diagrams that I've been meaning to make since forever, showing the Euclidean versions of velocity addition, length contraction, and (since someone mentioned it above) the barn-pole paradox.

Velocities are slopes (distance over time) and I hope the first diagram convinces you that there's no reason to expect simply adding the slopes to give the right answer. You can just add the angles, though (as long as the lines are in a common plane). The spacetime version of an angle is called rapidity and it's related to ordinary velocity v by v = tanh α (compare the Euclidean version: v = tan θ). In the length contraction case, the main point is that the two "observers" are making two different measurements, and it shouldn't be surprising that they get different results. It doesn't mean that reality is different for different people. The barn-pole paradox just illustrates a common beginner's mistake in applying the Lorentz transformation (which is a spacetime rotation).

I would like to add these diagrams, and the twin paradox one that I mentioned in another thread, to the articles someday, but I'm not sure how to deal with the likely claim of original research. It's hard for me to believe that no textbook in the last 107 years has used similar diagrams, but I haven't found one yet. -- BenRG (talk) 05:13, 10 December 2012 (UTC)[reply]

See the article on velocity addition (which is what the OP is attempting to apply here). Modocc (talk) 15:43, 8 December 2012 (UTC)[reply]

How much, or, In what ways, except from hormonal action, does Aging related to the Nervous system in compare to Genetic contorl?

In other words, How much does Aging affected by the Nervous system (and the endocrine system that it governs), In compare to Genetic control? — Preceding unsigned comment added by 109.66.11.190 (talk) 06:42, 7 December 2012 (UTC)[reply]

Genetics determines the form of the nervous system to a very detailed level. Aging only causes some minor deterioration, except in the case of diseases such as Alzheimer's disease. Looie496 (talk) 07:16, 7 December 2012 (UTC)[reply]

Hi. please note that I'M not asking how aging affects the NS. 109.66.11.190 (talk) 14:28, 7 December 2012 (UTC)[reply]

With all respect, the grammar of your question is so bad that it's almost impossible to understand what you are asking. I made a guess; apparently it was wrong. Can you try to make the question clearer? Looie496 (talk) 15:33, 7 December 2012 (UTC)[reply]

I'll try to translate:

In what manner and to what degree is aging related to the nervous system, as opposed to hormonal and genetic factors? In other words, how much is aging affected by the nervous system (and the endocrine system that it governs), in comparison to genetic control?

Or something like that. Evanh2008 ^{(talk|contribs)} 15:40, 7 December 2012 (UTC)[reply]

That's like asking how much driving is affected by cars, in comparison to roads. How could you answer a question like that? Looie496 (talk) 15:51, 7 December 2012 (UTC)[reply]

About not eating for 4 days straight, and subsisting only on fluids...

closed, weird question about washlets, asking for personal advise

The following discussion has been closed. Please do not modify it.

I only ask for nutritional advice, nothing medical, as no one is allowed to here.

Hello,

I am due to go to a ski resort in Monarch, Colorado on around January 2nd, 2013 and be on the whole trip for 4-5 days.

I'm afraid that at the ski resort, there may be no washlet / bidet-like devices and the resort staff would likely reject my donation of a BioBidet attachment device to install on the toilet of the accommodation I'll stay in. (I'll still bring one in the remote off-chance that they consider my digestive needs and accept it after all.)

Being faced with undue hardship without access to a bidet, I would much rather subsist on fluids during the whole four days. Weighing 206 pounds currently, I would have 25 pounds to lose until reaching the normal-weight threshold anyway.

However, if the resultant hunger makes me ravenous, I may take yogurt in hopes that it's only urinated out, as yogurt is still more liquid than solid.

Therefore, the questions:

1. After explaining my digestive concerns for needing a bidet, would they let me install the bidet-attachment on their toilet in the room I'm in? (It's a donation that I'd like to have remain there after I return home with my ministry group.)

1a. Why might they not allow me to, when bidets clearly have better results than the "old way?" Wouldn't it be like rejecting the donation of a shower stall 100 years ago?

2. If I subsist strictly on fluids for the 4-5 days I'm away from home, how many pounds would I stand to lose? Also, what unpleasant nutritional side-effects would be bound to happen?

3. If the only things I eat are no tougher than yogurt, will I still be compelled to take the #2 before the 4 days, or will only the #1 happen?

4. What (else) can I eat that will not force me to take the #2, no matter how much of it I have? (Someone up to no good from AOLAnswers answered, "Hemlock. Try it." Needless to say, I have thus far, found no non-poisonous forms of hemlock, so it's seemingly not like mushrooms.)

I hope the answers prepare me well for this trip. --Mount Zynar (talk) 07:26, 7 December 2012 (UTC)[reply]

You're approaching this from a totally wrong angle. You should be asking what you can use in lieu of a bidet, and the answer is quite simple. Pop in the shower and wash yourself off down there. Like you, I like my nether regions to be clean, and when at home, I always take a partial shower afterwards. Just takes a minute or two, and it's a lot easier, and healthier, than starving yourself. Dominus Vobisdu (talk) 07:46, 7 December 2012 (UTC)[reply]

Do you want to step on what others leave behind (especially that which comes from "up there?") And do you ever want to stick your foot in the toilet bowl, even after it's flushed clean? Think about it. That's why I would not use that particular shower-option in place of bidets. --Mount Zynar (talk) 08:56, 7 December 2012 (UTC)[reply]

You could clean it, with bleach, before you leave. StuRat (talk) 09:16, 7 December 2012 (UTC)[reply]

Along those lines, request a handicapped access room, as they often have a movable shower hose, which would be a lot easier to use as a bidet. You could also ask if they have any rooms with bidets. If they have international tourist there, it's possible, although those rooms might cost more. StuRat (talk) 07:56, 7 December 2012 (UTC)[reply]

And, if forced to use a toilet, a bit of moisturizer squeezed onto the toilet paper can both assist in cleaning and prevent irritation. You could also just attach the device without asking them. You might want to remove it before the maid arrives each day, as there's a slight chance she might complain to the hotel management. But, most likely, they would just ignore it. I'd remove it when you go, though, or they might very well complain. I myself have made little fixes to hotel rooms I've stayed in, from putting bright CFLs in place of dim incandescent bulbs, to fixing a rattling A/C unit and a toilet that had a lever which stuck down. I never asked permission. StuRat (talk) 07:54, 7 December 2012 (UTC)[reply]

That all liquid diet might also result in diarrhea, so I'd avoid that. But, if you insist, all least do a dry run first (or is it a wet run ?), so you discover any problems it causes while you still have access to a bidet. StuRat (talk) 07:55, 7 December 2012 (UTC)[reply]

Stu, how are we sure that an all-liquid diet won't just cause frequent urination? --Mount Zynar (talk) 08:56, 7 December 2012 (UTC)[reply]

With pure water, you might be right. But other liquids are likely to contain some indigestable material. For example, pulp in orange juice. StuRat (talk) 09:14, 7 December 2012 (UTC)[reply]

You also misunderstand what faeces consist of: it is entirely possible to pass a stool without eating any solid food because of the dead cells the body has to eliminate, which it does in faecal matter. --TammyMoet (talk) 08:23, 7 December 2012 (UTC)[reply]

How long would that take to happen? On the night of New Years, I plan to take an hour or two committing multiple "enema washes" to empty out the bowels as much as I possibly can. If the process, involving only dead cells, would take more than the 4-5 days the trip lasts, and I don't eat any solids, then I don't think it's a problem. So how many days would it take? --Mount Zynar (talk) 08:56, 7 December 2012 (UTC)[reply]

Is it just me who finds this question weird? And although this is no medical question per se, doing outdoor sports like ski and not eating anything solid could have some health implications. OsmanRF34 (talk) 14:44, 7 December 2012 (UTC)[reply]

This is at least the fourth or fifth time the Refdesk has gotten hit with washlet-related spam. The idea that a ski resort shower won't work as a substitute for a final wash seems ridiculous. (Even if someone found that repulsive, the person before him didn't... unless he's avoiding the shower too) Note this is a brand new account. Prolonged fasting verges on a medical issue, especially in someone not accustomed to the practice who says he's going skiing, and enemas are definitely not nutritional. Even I can accept culling this one. Wnt (talk) 14:52, 7 December 2012 (UTC)[reply]

Agreed, furthermore, "Committing multiple enemas" sounds illegal in this context.165.212.189.187 (talk) 16:20, 7 December 2012 (UTC)[reply]

Wow, given that post-game analysis I am almost tempted to click "show" to see what I missed. μηδείς (talk) 17:09, 7 December 2012 (UTC)[reply]

Lung shot

Does getting shot in the lung feel like drowning?--Wrk678 (talk) 09:38, 7 December 2012 (UTC)[reply]

You have two lungs, and if only one fills with blood, you can still breathe, so it wouldn't be as bad as drowning. You would need to remain calm, though, as you might use more oxygen in an excited state than one lung can supply. StuRat (talk) 09:54, 7 December 2012 (UTC)[reply]

What you experience would depend on whether you are bleeding into the lung, which is not guaranteed even by being shot through it. If there is little bleeding and you've simply open an air passage between the lung and the chest cavity, our hypothetical victim has a pneumothorax. As you can read from that article, the symptoms don't sound anywhere close to the feeling of drowning. In the case of significant bleeding into the lungs, it would be called hemothorax, but the mortality caused by hemothorax is typically related to the loss of circulating blood rather than suffocation (although shortness of breath may be present). So in both cases, I suspect the answer is no. Someguy1221 (talk) 10:01, 7 December 2012 (UTC)[reply]

Only one example, but this 15-year-old said, "Do you ever get a side ache so bad you just want to lay down and give up?" Clarityfiend (talk) 21:19, 7 December 2012 (UTC)[reply]

Are German Turks darker than normal Turks?

I got this impression, but does not have a source confirming or denying it. OsmanRF34 (talk) 14:28, 7 December 2012 (UTC)[reply]

Could it be due to acquired Schadenfreude? μηδείς (talk) 17:07, 7 December 2012 (UTC)[reply]

I really meant it as a serious question. I have the impression that they immigrated from a region from Turkey where people are darker, and they didn't mixed too much with local Germans, who obviously have a fairer skin tone. OsmanRF34 (talk) 22:38, 7 December 2012 (UTC)[reply]

The people you are calling German Turks may be another ethnic group, such as the Romani people. StuRat (talk) 23:40, 7 December 2012 (UTC)[reply]

I have tutored two German Turks who emigrated to the US, and neither was particularly dark. I have met Spaniards and Italians darker than them. Both were also strongly Turk-nationalist, which makes me doubt they were of some minority ethnicity such as Gypsy. (I have also met blond Armenians from Turkey one would think were Poles.) μηδείς (talk) 02:45, 8 December 2012 (UTC)[reply]

The skin colour may vary due to regional differences in Turkey. The Turks in Germany came here mostly in the 1960s and 1970s to work in German industry as "Gastarbeiter" Italy, Spain and Greece were other places were workers were allowed to come to Germany. I do not have a chart from which places the most of the Turks came but I guess it was from the poor part which is the anatholian montains in the east. I look for a chart to ducument this. --Stone (talk) 13:51, 8 December 2012 (UTC)[reply]

In what ways, except from hormonal action, does Aging related to the Nervous system in compare to Genetic contorl?

Because this last question of mine became a Cynic charade, i decided to ask it again. Ben-Natan (talk) 16:44, 7 December 2012 (UTC)[reply]

So let's see if you understand this time. Your question, as you ask it, makes no sense. Do you mean "How is the Nervous system affected by the aging process, and how does this compare to the effects of genetic controls?" --TammyMoet (talk) 18:11, 7 December 2012 (UTC)[reply]

No man. is there anything extra, besides the neurologically-governed hormonal system, and genetic control, that affects aging?. 109.66.11.190 (talk) 19:07, 7 December 2012 (UTC)[reply]

Thank you. Why didn't you say that the first time? Anyway, diet, exercise and smoking all affect the aging process in humans. --TammyMoet (talk) 20:39, 7 December 2012 (UTC)[reply]

UV rays too, but that’s partly due to the genetic issues they tend to cause. Evanh2008 ^{(talk|contribs)} 04:15, 8 December 2012 (UTC)[reply]

in what existing\potential methods we govern genetic control? (it seems to be the most affecting factor on aging).

I have absolutely no idea what you just said. Evanh2008 ^{(talk|contribs)} 05:48, 10 December 2012 (UTC)[reply]

Nullification of time dilation effects via deceleration?

I wonder if time dilation effects might be nullified (or otherwise altered) by a deceleration equivalent to the rate of acceleration or some such? Or does relativity already preclude such things? — Preceding unsigned comment added by 66.87.127.23 (talk) 17:03, 7 December 2012 (UTC)[reply]

Deceleration is a kind of acceleration. By the principle of equivalence they're indistinguishable. 87.114.106.165 (talk) 17:07, 7 December 2012 (UTC)[reply]

I see, thanks!

66.87.127.23 (talk) 17:44, 7 December 2012 (UTC)[reply]

Besides that, acceleration does not cause time dilation. Dauto (talk) 20:00, 7 December 2012 (UTC)[reply]

Doesn't it? Dbfirs 23:18, 7 December 2012 (UTC)[reply]

Nope... Dauto (talk) 01:18, 8 December 2012 (UTC)[reply]

Well, OK, not directly, I suppose ... but different acceleration histories will result in a difference in time. I agree that it's not quite the same as time dilation being "caused" by acceleration. Dbfirs 07:46, 8 December 2012 (UTC)[reply]

It's perfectly reasonable and accurate to say that time dilation is a direct consequence of acceleration. Consider the twin paradox, for example. The twin that ages least is the one that experiences the most acceleration. The way Einstein put it, at any given instant of acceleration the twin is momentarily in a different inertial reference frame (and thus stationary). In other words, the twin with the highest surplus of delta V stays the youngest.

Sebastian Garth (talk) 15:29, 8 December 2012 (UTC)[reply]

No, it doesn't work that way. All the dilation is accountable from the integrated dilation due to relative motion. There is no correction factor or extra term due to acceleration. For instance, if one twin travels at speed v along a circle of radius R and the other twin travels at speed 2*v along a much wider circle (say a radius of 10*R),after the faster twin completes a turn (and the slower one completes 5 turns), the faster moving twin will have aged more slowly than the slow moving one even though his acceleration was smaller. Dauto (talk) 16:46, 8 December 2012 (UTC)[reply]

In that case, the faster object would initially be more dilated but would eventually be "overtaken" by the slower object due to the the cumulative effect of acceleration on the latter. Integrating relative motion is meaningless here because the process is asymmetrical. Otherwise, how could we differentiate one twin from the other in terms of say the classical (rectilinear) experiment?

Sebastian Garth (talk) 17:37, 8 December 2012 (UTC)[reply]

Your understanding of the circular motion case is incorrect. No "overtaking" due to acceleration takes place. The faster moving twin becomes younger (By that I mean he ages more slowly) regardless of the amount of acceleration it has been subjected too. Note that in the rectilinear case the twin that accelerates must take into account the change in simultaneity that happens due to him changing reference frames when he accelerates. This is a separate effect from the time dilation effect. That's why it is always a good idea to have at least one observer remain un-accelerated and calculate all time dilation factors with respect to him. Dauto (talk) 19:57, 8 December 2012 (UTC)[reply]

Let's go back to the original thought experiment. Why does the traveling twin experience time dilation then?

Sebastian Garth (talk) 22:18, 8 December 2012 (UTC)[reply]

Alright, may be an example is in order here. Let's call the twin that remains on Earth "Earl" and let's call the one that travels "Travis". Let's also say that they were both 20 years old when Travis started traveling to planet-X located 8 light-years away from Earth. In other words the proper distance between Earth and planet-X is $L_{0}=8cy$ , where cy stands for light-year. Lets split Travis motion into 5 steps

Step 1: Travis accelerates instantaneously to the speed $v_{1}=v={\frac {4}{5}}c$ . There is no time dilation associated with that acceleration. That acceleration is (off course) not realistic, but it makes the math simpler. We shall label that reference frame with a prime " $\,S'$ ", and Earl's reference frame is not primed " $\,S$ ". The Lorentz factor for that speed is $\gamma _{v}={\frac {1}{(1-(v/c)^{2})^{1/2}}}={\frac {5}{3}}$ .

Step 2: Travis remains at that speed for 10 years - Earl's time - that is $t_{1}=10y$ , where y stands for years. That's just enough time for Travis to reach planet-X: $L_{0}=vt_{1}=({\frac {4}{5}}c)(10y)=8cy$ . Due to time dilation Earl observes Travis age only 6 years - that is $t_{1}'={\frac {t_{1}}{\gamma _{v}}}=6y$

Step 3: Travis accelerates (again instantaneously) to the speed $v_{2}=-v=-{\frac {4}{5}}c$ . Again, no time dilation applies here. We shall label that reference frame with a double prime " $''$ " " $\,S''$ ".

Step 4: Travis remains at that speed for 10 more years - Earl's time - that is $t_{2}=10y$ , which is enough time to bring Travis back to Earth: $L_{0}=vt_{2}=({\frac {4}{5}}c)(10y)=8cy$ . Again, due to time dilation Earl observes Travis age only 6 years - that is $t_{2}''={\frac {t_{2}}{\gamma _{v}}}=6y$

Step 5: Travis accelerates instantaneously once more to Earth's rest frame without any time dilation.

Total travel time for Earl: $t_{Earl}=t_{1}+t_{2}=10y+10y=20y$ .

Total travel time for Travis: $t_{Travis}=t_{1}'+t_{2}''=6y+6y=12y$ .

That's what happened from Earl's point of view. Let's explore what happened from Travis point of view.

Step 1: Travis accelerates instantaneously to the speed $v_{1}=v={\frac {4}{5}}c$ . The distance between Earth and planet-X is Lorentz contracted to $L'={\frac {L_{0}}{\gamma _{v}}}={\frac {8cy}{5/3}}={\frac {24}{5}}cy=4.8cy$ .

Step 2: Travis remains at that speed for 6 years - that is $t_{1}'=6y$ . That's just enough time for Travis to reach planet-X: $L'=vt_{1}'=({\frac {4}{5}}c)(6y)=4.8cy$ . Due to time dilation Travis observes Earl age only 3.6 years - that is $t_{1}={\frac {t_{1}'}{\gamma _{v}}}={\frac {6y}{5/3}}=3.6y$

Step 3: Travis accelerates instantaneously from $v_{1}=v={\frac {4}{5}}c$ to $v_{2}=-v=-{\frac {4}{5}}c$ . There is no time dilation here but there is shift in simultaneity. There is no shift in step 1 (or step 5) because the distance between the twins is zero for those steps. The simultaneity shift is given by $\Delta t''={\frac {\gamma _{u}uL'}{c^{2}}}$ where $\,u$ is the relative speed between the reference frames $\,S'$ and $\,S''$ which is given by the Velocity-addition formula: $u={\frac {v_{1}-v_{2}}{1-v_{1}v_{2}/c^{2}}}={\frac {v+v}{1+(v/c)^{2}}}={\frac {(8/5)c}{1+16/25}}={\frac {40}{41}}c$ . The Lorentz factor for that speed is $\gamma _{u}={\frac {1}{(1-(u/c)^{2})^{1/2}}}={\frac {1}{(1-(40/41)^{2})^{1/2}}}={\frac {41}{9}}$ . And we get the simultaneity shift $\Delta t''={\frac {\gamma _{u}uL'}{c^{2}}}={\frac {{\frac {41}{9}}{\frac {40}{41}}c{\frac {24}{5}}cy}{c^{2}}}={\frac {64}{3}}y$ . That means that in reference frame $\,S'$ Travis turns 26 years old simultaneously with Earl turning 23.6 years old, but in reference frame $\,S''$ the former happens ${\frac {64}{3}}$ years after the latter, which means that (taking into account time dilation) Earl has aged an extra $\Delta t={\frac {\Delta t''}{\gamma _{v}}}={\frac {(64/3)y}{5/3}}={\frac {64}{5}}y=12.8y$

Step 4: Like in step 2 $t_{2}''=6y$ and $t_{2}=3.6y$

Step 5: No time elapses in step 5 for either Earl or Travis.

Total travel time for Travis: $t_{Travis}=t_{1}'+t_{2}''=6y+6y=12y$ .

Total travel time for Earl: $t_{Earl}=t_{1}+\Delta t+t_{2}=3.6y+12.8y+3.6y=20y$ .

Dauto (talk) 18:36, 9 December 2012 (UTC)[reply]

Could you explain (just for my peace of mind) how your mathematics, and the Gruber and Price examples, square with "This is because gravitational time dilation is manifested in accelerated frames of reference" in the article Gravitational time dilation, and how they square with the Equivalence principle? (I think I need to do a lot more reading on this before I understand it!) Dbfirs 21:22, 9 December 2012 (UTC)[reply]

That statement is misleading at best. The concept of gravitational time dilation itself is misleading and would be more aptly described as gravitational Doppler shift. If you send a light signal from the front end of a rocket to its back end, by the time the signal reaches its destination the rocket will have accelerated a bit, so the source is perceived to be moving towards the destination and a blue shift is observed. If the signal is going in the opposite direction, a redshift is observed instead. By the equivalence principle a similar Doppler shift must be observed by some one in the presence of a gravitational field. That shift is often called a gravitational time dilation but that causes a lot of confusion in my opinion. I prefer the term gravitational (Doppler) shift. Dauto (talk) 01:01, 10 December 2012 (UTC)[reply]

... so does "gravitational (Doppler) shift" explain why clocks at the top of a tower run faster (let's assume the tower is on a non-rotating planet to filter out differences in speed and centripetal acceleration)? (I suspect that if I understood the metric tensor of general relativity properly, then I wouldn't need to ask!) Dbfirs 12:35, 10 December 2012 (UTC)[reply]

Yes, Gravitational Doppler shift explains why a clock at the top of a tower moves faster. Dauto (talk) 14:52, 10 December 2012 (UTC)[reply]

Look, the twin paradox/effect is really very simple. It's closely analogous to the fact that a straight line is the shortest distance between two points. In spacetime it comes out the other way—the straight line is the longest distance (elapsed time) because of a sign change, but it's essentially the same thing. A straight line is a nonaccelerating worldline in flat spacetime. A curved path is an accelerating worldline. So it's true that if one twin accelerates and the other doesn't, the one that accelerates will end up younger. But it's a bit silly to say that the acceleration (bending) "causes" the extra distance, and ridiculous to say that the extra distance happens during the acceleration. The extra distance isn't in any particular place; it's just that the curve overall is longer. I could reproduce Dauto's long calculation in Euclidean geometry, and it would be almost the same aside from some sign changes. But there's no point; a simple diagram is sufficient.

In general relativity it's the same except that instead of a flat piece of paper you're drawing lines on a warped piece of paper (maybe it got wet and you dried it out). -- BenRG (talk) 03:03, 10 December 2012 (UTC)[reply]

On the right is a diagram I made ages ago showing the Euclidean version of a common presentation of the twin paradox. This diagram is far more complicated than the simple geometry of the triangle requires. There's no reason to slice the triangle horizontally or diagonally. Nobody would do it teaching Euclidean geometry. For some reason they do it when teaching the twin paradox. -- BenRG (talk) 03:09, 10 December 2012 (UTC)[reply]

You might also be interested in the diagrams I just added to another recent thread. -- BenRG (talk) 05:28, 10 December 2012 (UTC)[reply]

What was the format of a single sample of audio recorded on a cassette tape?

Say I hit record on a tape recorder and record for a fraction of a second the sound of me saying "AAHH." The smallest frame of the sampling rate, corresponding to one momentary sample (I assume signifying what the voltage caused by the fluctuating mirophone at that moment was), what format was THAT encoded in on the tape? Analog, I know, but how was the amplitude of the microphone at one moment encoded on the tape? Was the voltage fluctuation range of all microphones standard or did the recorder hardware somehow sense the microphone's range and normalize it to the standard range? So if the amplitude to encode onto the tape were, say, between -X and +X volts, how was A.BCD volts encoded on the tape? What was the format for encoding an analog value? 20.137.2.50 (talk) 19:56, 7 December 2012 (UTC)[reply]

You're overcomplicating things, perhaps. The degree of magnetization is proportional to the signal. There are no "frames", there is no "sampling rate", there are no "samples", there is no "encoding". There are no discrete "moments". --jpgordon^{::==( o )} 16:10, 8 December 2012 (UTC)[reply]

Then please expand upon what exactly "degree of magnetization means" as if I didn't understand at all what you meant by that. I thought a single spot on magnetic tape was either '1' or '0'.67.163.109.173 (talk) 16:17, 8 December 2012 (UTC)[reply]

Not if you're referring to analog sound recording. --jpgordon^{::==( o )} 16:28, 8 December 2012 (UTC)[reply]

If it was just 1s and 0s, it would be a digital recording. As for the "degree of magnification", I'm going to assume that what jpgordon refers to is the strength of the magnetic field on the tape as left by the recording head. WegianWarrior (talk) 16:39, 8 December 2012 (UTC)[reply]

OK, then maybe what I should be asking is for a link to a physical specification as to the corresponding between minimum and maximum magnetization for standard cassette tape and minimum and maximum microphone/speaker signal. 67.163.109.173 (talk) 17:02, 8 December 2012 (UTC)[reply]

Orders of magnitude (magnetic field) gives the magnetic field of a tape head (I think it means during writing) at 24 µT (240 mG). The write head is given a current calibrated so that the highest amplitudes will produce a field strength on that head of say 240 mG (and -240mG for the opposite peak). This writes an analog signal onto the tape, such that if one subsequently passes the read head over the tape, the magnetism of the tape induces a current in the coil around the read head (probably about an order of magnitude less than the write current) which is amplified to produce a listenable analog electronic signal. The designers of a given tape device know the performance of their read and write electromagnets and so calibrate their circuits so that the signal level they use in their electrical circuits maps to a full (but not overdriven) use of the available magnetic response of the tape. In an idealised view the tape is truly analog in both the imaging of the amplitude and the time resolution, and so the input signal is simply transduced as a magnetic image. But in reality the the oxide on the tape forms a series of tiny magnetic domains, which limits the spacial (and thus temporal) resolution of the tape (e.g. if you had a tape with very chunky domains, it would image high frequency sounds poorly). Also the tape head does not produce a perfectly point-sized magnetic field, which means the magnet is writing to a few adjacent domains simultaneously. Tim Hunkin has a lovely demonstration of how this works (it used to be on his website, but I think he's taken it down) where he makes magnetic tape by putting ground rust on some Sellotape, and then rubs it past a simple electromagnet driven off a microphone (into which he yells) - then he drags the tape back across the electromagnet and you can (just) hear his voice back from the microphone (which functions like a little speaker). Back to your plot, above - if you had a really teeny tiny magnetic compass, and you moved it slowly down the tape, its needle would be defected back and forth - the extent of its deflection would correspond to the extent of deflection in your plot from the zero centreline. -- Finlay McWalterჷTalk 17:16, 8 December 2012 (UTC)[reply]

Hunkin's magnetic-tape demo is on an episode of The Secret Life of Machines. DMacks (talk) 08:37, 9 December 2012 (UTC)[reply]

Unfortunately this is one of those subjects where it is necessary to overcomplicate things. The magnetization on the tape is not a straightforward monotonic function of the signal, because of tape bias, especially AC bias. AC bias is used in audiocassette recording to overcome the hysteresis and nonlinearity of the recording medium.--Heron (talk) 03:40, 10 December 2012 (UTC)[reply]

There also exist Digital Audio cassettes whose standard is described in the linked article. SkylonS (talk) 09:59, 10 December 2012 (UTC)[reply]