Wikipedia:Reference desk/Archives/Science/2010 December 4

= December 4 =

Camera error
Recently, after having not been used due to a low battery for a couple of months, my Canon PowerShot SD790 IS camera started to malfunction. After I charged the battery, upon attempting to use the camera, the pictures would sometimes come out like the first picture in the gallery below.

The error at that point was intermittent, as a few seconds later I was able to take a picture of the exact same area without a malfunction (the second picture). However, over the next few days, the malfunction kept getting worse, until six days later the third picture resulted from an attempt to take a picture of my television on a wooden table in a well-lit room. What could this camera malfunction be, and what are some ways that it could possibly be fixed?

PS: Note that this is not exactly an urgent or necessary inquiry, as I received a replacement camera as a birthday gift a couple days ago. A mere pointer to what the error is called and what causes it would be a suitable answer for me.

Thanks in advance, Ks0stm (T•C•G) 05:30, 4 December 2010 (UTC)


 * Maybe the (rechargable?) battery needs replacing. Side question - why is there a traditional red British phone box at the University of Oklahoma? Thanks 92.24.177.111 (talk) 12:33, 4 December 2010 (UTC)


 * As to the phone box, apparently it's just a phone box. What, did you think there wouldn't be an article on it? :-) 88.112.56.9 (talk) 14:20, 4 December 2010 (UTC)


 * Very good. I've taken the liberty of adding the middle photo above to that article's gallery, since it's mentioned in the text but not attributed. If the OP could take a nice close-up of the phone box, that would be even better. :) ←Baseball Bugs What's up, Doc? carrots→ 16:43, 4 December 2010 (UTC)


 * The first pic looks like the green component isn't being recorded. In the last pic, the same thing is true, but also the darkness seems to indicate that the other components are now recorded at reduced levels (it looks like only one line of red and one of blue is recorded for every 8 or so lines of data, so you have 1/8th the normal level of those colors). Does your camera have a gray-scale setting ?  If so, you might still be able to get a usable pic in a bright setting.  As for the cause, I don't know, perhaps some electrical contacts are broken and now have only an intermittent connection ?  Is it out of warranty ? StuRat (talk) 16:06, 4 December 2010 (UTC)


 * For a real mystery, where did the tree at the left, top of the 1st pic go, in the 2nd pic ? StuRat (talk) 16:21, 4 December 2010 (UTC)
 * The answer is that the middle picture was taken from maybe 20 or 30 feet farther to the right. Open two sessions and note the shift of background objects, such as the phone booth, and also that the small leafless tree in the foreground of the middle shot is the small then-leafed tree in the left picture. ←Baseball Bugs What's up, Doc? carrots→ 16:26, 4 December 2010 (UTC)
 * I don't think it is then-leaved, I think it is still unleaved in the picture on the left but then in front of the needled pine tree behind it. Wiki Dao  &#9775;  (talk)  16:34, 4 December 2010 (UTC)
 * You're right. I was confusing it with the tree that's right behind it in the line of sight. In any case, the middle picture was taken from a position farther to the right than the left picture was. Or to make it less confusing, the left picture was taken from farther left than was the middle picture. Another telltale sign is the driveway, clearly visible in the left picture, just a smidgen of it visible in the middle picture. ←Baseball Bugs What's up, Doc? carrots→ 16:36, 4 December 2010 (UTC)


 * If you look at those photos full-size, and especially if you zoom in, there's definitely a problem in the first photo, which looks as if it were taken through a screen door. Getting a new camera was probably more cost-effective than getting it repaired, though it might be worth it to take the camera and the evidence to a camera shop and see if they have any theories. ←Baseball Bugs What's up, Doc? carrots→ 16:49, 4 December 2010 (UTC)


 * Comment. The modern consumer digital-camera flow path consists of dozens, or even hundreds, of image-processing stages that convert the image from the physical world to JPEG format.  Without elaborate blueprints of the innards of the ASICs and software processes, we can only speculate what might be going wrong.  I suspect there is at least a mechanical shutter problem, or possibly a sensor problem, based on the varied exposure patterns in the corrupted images.  These have the hallmark of a CCD sensor's multi-field readout, with the later fields over-exposed (because the shutter never closed and new incident light kept coming in).  PowerShot SD790 (like most Canon cameras) uses a bayer-masked CCD; if mechanical shutter broke the exposure, the demosaicing algorithm could be misconfigured.  Similar field readout errors could be caused by insufficient power to the sensor, too.  If the image readout doesn't match the expected exposure level, subsequent inputs to image-processing stages would be poorly configured based on the disparity between intended and actual image characteristics, and would explain the color-clobbering and other weird glitches.  But, we can't know: there could be significant additional software- and hardware- image processing, filtering, and so on.  Any software or hardware error in these steps could corrupt the image.  Low battery may cause brownout; that could account for the shutter failure or electronics/software errors.  Nimur (talk) 22:18, 4 December 2010 (UTC)


 * I had an older model of Powershot that failed in a similar way. The picture took on a pink cast. It turned out to be a known problem with the internal wiring to the sensor. I sent it back to a Canon-approved dealer in the UK and they fixed it for free. --Heron (talk) 16:01, 6 December 2010 (UTC)

What kind of SIMUTANEITY make sense?
Currently there are three kinds of Simutaneity theories. The most popular one is that simutaneous events may lost their simutaneity if observed by observers in different inertial system when their relative speed is v>0. A famous case described it like followings, two lightnings hit two ends of a moving box at the same time, then, to the person at the center of that box, two events are not simutaneous. Logically speaking, this can happen; but, if we have a lot of observers in the moving system, not just one at the center of that box, then what will happen? Do you know a better case to explain this popular theory?Jh17710 (talk) 06:28, 4 December 2010 (UTC)
 * The second theory is the classic idea. It states, simutaneous events will keep their simutaneity to observers in any inertial system. The calssic case is, if two baseballs are hit at the same time to virtical and 45 degree angle but reach ground at the same time, then to observers in other inertial system, actual paths of two basealls may show different angles but two baseballs will always reach ground at the same time. When we have a lot of observers in any given inertial system, then, this theory will maintain physical laws in between inertial systems. We like physical laws be maintained in inertial systems, right?Jh17710 (talk) 06:28, 4 December 2010 (UTC)
 * The third theory is a visual version based on the second theory. Two simutaneous events will keep their simutaneity to observers keeping same distance from both event locations, starting the actual event time up to the recorded event time. The attached condition will explain the first case naturally, the observer moves to a location closer to the front event point before the observer record the event time so that the observed event times are different. The third theory also claims that two simutaneous events can be recorded different event times even in the same system if distances to two event locations are different.Jh17710 (talk) 06:28, 4 December 2010 (UTC)
 * Above analysis is to show that the simutaneity is distroyed by different distance, not by different speed.Jh17710 (talk) 06:28, 4 December 2010 (UTC)
 * No, only the first case is experimentally verfied; simultaneity is only preserved for observers in the same reference frames. If we have observer A and observer B, event C and event D.  If A and B are moving at different speeds, and C and D are at some distance from each other, if C and D are simultaneous in A's view, they will likely not be in B's system.  This is due to the relativity of simultaneity, which is a direct consequence of the experimentally verified theory of special relativity.  Your other two examples don't make any sense given what currently accepted theories explain about how the world works.  -- Jayron  32  07:08, 4 December 2010 (UTC)
 * Thanks for your time. Could you show me which experiment verified it? In your case of two observers A and B, could you provide more detail regarding what kind of tools they use to measure the event times of C and D? How they record each of the event times? By camcorders, by eyes and clocks, or what kind of device? If by eyes and clocks, our vision rely on lights to carry the picture of events C and D to our eyes so that we know what happen at what time, am I correct? If C and D are rest in A's frame, assume that all clocks at the location of C, D, and A are all synchronized, then, for observer Oc at event location of C, Pc, and Od at event location of D, Pd, will record different event time if the location of A, Pa, is different from the locations of C and D, am I correct?Jh17710 (talk) 15:35, 4 December 2010 (UTC)
 * Now go back to your claim {..if C and D are simultaneous in A's view, they will likely not be in B's system.}, but this time let we study it with distance in mind. If observer A is located at point Pa, we also assume that observers Oc and Od recorded same event time, Tc=Td, then C and D are simutaneous in A's view when distance PaPc=PaPd. If Tc=Td, but PaPc>PaPd, then A will record event C happens later than event D. For B, if B moves on the plan perpendiculer to the line segment PcPd at the middle point of it, then, B will always record that C and D are simutaneous no matter how B moves. Don't you think so?Jh17710 (talk) 15:35, 4 December 2010 (UTC)
 * Without going into the math in scrupulous detail; yes there are situations where two observers moving at different velocities can both record two events as simultaneous; however this is merely a geometry problem and not a physics one. It is possible to construct paths through spacetime whereby A and B are moving in different reference frames and both measure C and D as simultaneous.  Imagine a similar, related problem from the world of geometry.  If I place two points on a piece of paper, they are required to be colinear.  A third point can be placed that is sometime colinear with the other two; but does not have to be.  Imagine Event C and Event D as the two points.  If I place observer A on the paper on the line made by Event C and Event D, they will agree that they line up.  If I place observer B on the paper also somewhere on the same line, they will also agree that they line up.  However, if I place observer B on the paper somewhere NOT on the line, they will observe the two points (relative to B's position) to not be colinear.  They can extrapolate a line whereby C and D are colinear, but they will also recognize that they are NOT on this line.  It's the exact same situation in special relativity.  Imagine these lines happening in 4D spacetime rather than on a piece of paper; we call these lines worldlines, but they obey the same basic rules of geometry that all lines do.  We just have 4D spacetime as our paper.  Your postulates regarding perpendicular movement and the observation of similtaneity of events is an unsurprising consequence of basic geometry, and has no bearing on the reality of special relativity; in fact it confirms it rather than refutes it.  Your other two examples are still wrong.  -- Jayron  32  15:51, 4 December 2010 (UTC)
 * I know it will reach spacetime, sooner or later. The definition of the distance between an event and the origin point of the spacetime is man made. That definition is for the purpose of matching the equation of SR, t'=t/γ, period. You cannot prove SR by SR related theory. We have a lot of different way to define the distance of an event and the origin point of spacetime, if space and time can be put together to make a useful 4 dimensional structure. Whether our universe is Euclidean or not, is still a question. Within all kinds of force fields, it is hard to find a series of events moving at constant velocity, but that fact does not disprove an Euclidean universe.Jh17710 (talk) 16:40, 4 December 2010 (UTC)
 * Oh, and to answer your first question, the article Status of special relativity discusses these experiments in some detail. -- Jayron  32  15:58, 4 December 2010 (UTC)
 * There are some, but, could you briefly describe the one you think is the best, and the reason? Thanks.Jh17710 (talk) 16:40, 4 December 2010 (UTC)
 * Specifically the other two examples seem to rely upon some kind of universal time which transcends the reference frame ("same time", "recorded event time"), when the relativity of simultaneity is, in fact, an indication of (and consequence of) the lack of any universal time. --Mr.98 (talk) 15:07, 4 December 2010 (UTC)
 * You are absolutely correct about what SR claims, no universal time. However, that is just a claim, a hypothesis, isn't it? Theoretically, we have some way to calculate the absolute TIME PERIOD by counting the wave peaks from a known frequency souce of light or EM wave emitter at fixed distance. The formula is very simple, total wave peaks, N, times the wave length, L, then divides by the speed of light, c; time period between event C and D is Tcd=NL/c. That means, we can prove that time period is the same for rest and moving observers, theoretically. In his book published 9-1-2010, ZhiZhong Cai let a football game be broadcasting by AM waves at frequency of F and the twin brother B go 0.8c speed away from earth. In the leaving path, B will receive 20% of information so that the TV will show very slow motion of the front 20% portion of the broadcast and the remainder 80%, as well as the 100% section of the returning path, will show fast motion on TV so that when B meets A, B has watched the exactly same broadcast as A has watched, but in two kinds of motion-speeds. That is another way to explain Twin Paradox.Jh17710 (talk) 16:17, 4 December 2010 (UTC)
 * But we don't know how to match everybody's clock to show same TIME yet. The main difficulty is the starting time of our counting can be at any location of the wave, how far away from the wave peak is a kind of out of control. Once we figure out how to count the wave peaks exactly from a wave peak, then, we can start to figure out the unversal time, not just time period.Jh17710 (talk) 16:17, 4 December 2010 (UTC)

Michelson–Morley experiment
(undent). No, the point is that, as an entire theory, Special Relativity is both mathematically consistent and experimentally consistent. That's why we can call it a "theory". You are confusing "theory" with "hypothesis", a common mistake. See Scientific theory. Things like the Michelson–Morley experiment confirmed it by proving that there is no "luminiferous ether" and thus there is no medium in which light travels. The fact that the speed of light is invarient regardless of what your own speed is is all that you actually need to prove. The rest of the theory results from the mathematics necessary to maintain an invarient light speed at all frames of reference. -- Jayron  32  16:57, 4 December 2010 (UTC)
 * Let us look at that experiment in more detail than the history I know. As I know, all related experiments started from ignoring the fact that two rays interfere each other at the detector are not split from a single ray except two times when the lab moves at around 135 and 315 degree relative to the direction of the incident ray. All the rest situations, two rays meet at detector are originated from two rays at the source of light. Looks like everyone just assumed that the difference between two source rays and single source ray is not a factor in the experiment because, as I know from http://en.wikipedia.org/wiki/Michelson%E2%80%93Morley_experiment, no one mentioned about that fact. Not even Paul Marmet when he pointed out clearly the two essential fundamental phenomena at http://www.newtonphysics.on.ca/michelson/index.html. He missed the fact of "there are two source of rays in most time of the experiment". Let me explain it. If we let the thickness of the semi-silvered mirror (SSM) be d and the length of each arm be D, and the portion of incident ray that refracts through the SSM reflects back from far mirror to the same SSM at a point P, then at 0 degree, the reflected portion of the incident ray will return to a point P0 a little bit, about 2.8dv/c, away from P to the far end of SSM as seeing from the source of light. At 90 degree, it will return to a point P90 about 2.8Dv/c away from P to the far end of SSM. At 180 degree, it will return to a point P180 about 2.8dv/c away from P to the near end of SSM and at 270 degree, the P270 is about 2.8Dv/c away from P to the near end of SSM. That means, only at about 135 degree when the distance changes from the far end to the near end and at about 315 degree when the distance changes from the near end to the far end the split two rays return back to the same point P. All the rest situations, the detector is comparing two rays from two incident rays, not one.Jh17710 (talk) 18:45, 5 December 2010 (UTC)
 * I saw above analysis from a paper about 5 years ago. I was so excited that I just wrote down the main idea as I provided here and I forgot to write down the website. I thought it should be easy to derive all details from the main idea. I tried to get the complete paper to post here, but I was unable to find it. Sorry about that. I then tried to figure out the detail of the numbers by myself. I found out at 0 degree, if the silver is applied on the front side of the SSM, then, I don't see the 2.8dv/c. If the silver is applied on the back side of the SSM, it is possible to find the distance of about 2.8dv/c between P and P0. To me, it is very difficult to follow Paul Marmet's idea for the 90 degree scenario. I think the distance is not related to D, most likely, it is just at the range of d. When I started from the beginning again, to examine the calculation of the time spent on two paths about 10 minutes ago, I found out, Paul Marmet is wrong. Originally, I thought his formulas for two essential fundamental phenomena are so beautiful but I knew his final calculations were double issuing of the factor so that he made a minor mistake there. Now, I know his beautiful analysis cannot apply to this experiment. Why? Because all three mirrors are NOT moving relative to the source of light! Since the single incident ray is not just light once, it is an ongoing ray, so that there will be time difference for the starting wave front but no difference in between following waves. Their device is unable to detect the difference if they just emit the light ray once. What they reported are all for the time difference in between following waves. So, there is no difference, for sure. I think I have solved the issue of Michelson–Morley experiment, about 15 minutes ago.Jh17710 (talk) 06:53, 8 December 2010 (UTC)

universal time
As for whether you could have universal time with SR; you can try, but it has no meaning. A lot of very smart people (e.g. Poincaré, Lorentz, etc.) attempted to integrate different conceptions of "local time" with "universal time." Einstein was in fact most radical in that he said (in a Machian line of thought) that you can just throw out the propositions which are not required or cannot be measured. We have no reason to suspect there is a "universal time". It is entirely unclear what that would mean physically or even philosophically. All we can measure is local time. Postulating universal time adds an extraneous and unsupported thing to a really quite simple theory that accords extraordinarily well with all experimental evidence so far, and does not get us anything for doing it. There is no philosophical or physical reason that designating one reference frame as "privileged" or "universal" makes any sense. You can do it just for kicks, if you want, but you don't learn anything from doing it. --Mr.98 (talk) 19:21, 4 December 2010 (UTC)
 * Please refer to above detail about the Michelson–Morley experiment. Other than it, please let me know what is the best experiment can support SR as you know, and briefly, about why you think so? Thanks.Jh17710 (talk) 18:56, 5 December 2010 (UTC)
 * There are lots of indications that SR stands up very well, certainly better than any aether theory, that do not rely on the specifics of the Michelson-Morley apparatus, if that is what is bugging you. But the larger issue is that there's simply no reason to posit universal time at all, no evidence for it whatsoever, no hypothetical way to measure it. So what's the point? What does it get us to assume one reference frame is "right" and the others are "wrong"? --Mr.98 (talk) 22:36, 6 December 2010 (UTC)
 * I just like to find out the truth. Could you please tell me the best experiment that verified SR except the Michelson–Morley experiment? Thanks.Jh17710 (talk) 07:05, 8 December 2010 (UTC)

God's-eye view
I like to think of these questions as having to do with the "God's-eye view", i.e. if one could somehow be outside the universe looking in. When Einstein talks about "observers" he's often talking hypothetically, as if extremely fast events could somehow be observed by a normal human. ←Baseball Bugs What's up, Doc? carrots→ 16:52, 4 December 2010 (UTC)
 * Could you tell me more detail? Thanks.Jh17710 (talk) 18:56, 5 December 2010 (UTC)

time dilation
at what speed,as a fraction of c,does a moving clock tick at half the rate of an identical clock at rest? —Preceding unsigned comment added by 120.141.101.249 (talk) 12:47, 4 December 2010 (UTC)
 * See Time_dilation. -- Jayron  32  14:40, 4 December 2010 (UTC)
 * At about 86.6% of the speed of light, but the "moving clock" runs at normal rate in its own reference frame, of course. It only appears to be moving "slowly" when observed from an inertial frame moving at a different speed.  The effect is symmetrical because no one frame can be considered at absolute rest.    D b f i r s   15:45, 4 December 2010 (UTC)
 * If you look at the section 4 of Einstein's paper published 6-30-1905, he let the moving clock go circling and each time the moving clock meets the rest clock, he calimed that two clocks will read differently. I will say like in front of witness whoever read two clocks at that same time. How do you explain that the difference just {appears to be moving "slowly"} in front of that witness? Do you mean we should not let two clocks meet?Jh17710 (talk) 17:08, 4 December 2010 (UTC)
 * Your confusing accelerated motion with non-accelerated motion. Moving in a circle is an acceleration.  If two clocks are moving at a constant speed relative to one another, than the observer holding each clock thinks the other clock is moving too slow.  In order to get to the two clocks to exist in the same time and place, one or both of the observers will have to accelerate or decelerate to meet the other clock.  The exact path each observer takes in terms changing speeds will determine whose clock is ahead of whose.  For example, if the two observers came together to observe their clocks in such a way that each performed the exact same set of maneouvers and velocity changes in exact mirror image; then the two clocks would read the same time again.  If, however, one observer maintained the same speed, and the other observer changed their speed to "catch up" to the first, then their clocks would be different, specifically the observer who accelerated would have a clock which was ahead of the other person's clock. Moving in a circle is an acceleration, which is why the orbiting clock is ahead of the stationary clock.  -- Jayron  32  17:16, 4 December 2010 (UTC)
 * I just said what I understand from Einstein's paper about what Einstein thought at that time. Moving in a circle is an acceleration, but, I think Einstein knew it in year 1905, don't you think so? Or do you mean Einstein was confusing accelerated motion with non-accelerated motion? Not me, for sure. Now, let us use your explanation, but this time we let two clocks run circling at the same speed but different direction on two same size circles on one plane, meet at closest points, like 0.1 meter away from each other. In this situation, which clock will slow down? At the closest point, their relative speed is about 2v, so that the speed of their clocks should not be the same, but the situation is symmetric, which clock should run slower?Jh17710 (talk) 17:38, 4 December 2010 (UTC)
 * I think the two will each observe the other's clock to be varying in speed as it goes round the circle, but you need the mathematics of General relativity (a subject that I haven't much knowledge of) for this scenario.   D b f i r s   23:43, 4 December 2010 (UTC)
 * Thanks for your answer. I don't think GR has one final mathematical answer now. Physicists are still trying to decide which one is the best one. However, whatever the number will be, you see two clocks, 0.1 meter away from each other, showing a conflict scenario that each of two clocks is faster than the other clock. That is a paradox, isn't it?Jh17710 (talk) 19:27, 5 December 2010 (UTC)
 * To explain in a bit more detail; an observer moving with the clock you call the "moving" clock in your example will believe that the clock you call the "rest" clock is itself moving slower. That is, every observer believes himself to be at rest, and clocks not moving at his speed to be ticking slower.  That is because there is no universal rest frame.  -- Jayron  32  15:54, 4 December 2010 (UTC)
 * Whether there is universal rest frame or not, that is still a question. We are unable to locate it mainly because of lack of technology for now. One of possible ways is we can try to find a solution to watch the wave front of an expanding ball of light created by a single emission of a point source of light. The hypothesis of the speed, or we should say velocity, of light is independent to the speed, or velocity, of the source of light, provides us the best tool to understand the absolute rest universe that the speed of light, or the velocity of light, is relative to.Jh17710 (talk) 17:22, 4 December 2010 (UTC)
 * No, the speed of light is constant with respect to each local observer. There is no absolute rest frame anywhere in the universe.  We will never find one, however advanced our technology.  We can measure only speeds relative to other objects.  There is no fixed point in the universe.  I'm not sure that I really understand this, but Einstein seemed to have a good grasp of the idea.    D b f i r s   23:34, 4 December 2010 (UTC)
 * Now, let us look into your first sentence for more detail about the speed of light. The best way to find out the speed of light is by the formula that the speed of an EM wave is its wavelength times its frequency. Why? Because that character of light is wellknown and also because the measurement of distance and time period will always have errors. Even if we can make the error of measuring distance very tiny, the measurement of time is always questionable. Why? We know that the speed of atomic clock is faster in lower gravity field, then, there comes the ESSENTIAL QUESTION, under what strength of gravity should we measure the speed of light? For same method of measuring the speed of light, theoretically we will get different result on a mountain top and by a sea shore, then, which one is the real one?Jh17710 (talk) 20:54, 5 December 2010 (UTC)
 * Logically speaking, if we find out some way to count the number of wave peaks then we can make a final decision about that question. If we know the wavelength of an EM wave, for example, the red light or the calculated wave length for radio AM 1000, then, if we have some way to count the number of wave peaks we will be able to count the wave peaks to calculate the time period. The wave length will be constant if we keep same distance away from the source of the wave, no matter where is this setup located. On mountain top or at sea shore; the wave length will be the same, but, all kinds of clocks we know, so far, will have different speed respectively. That means, if we count the wave peaks to calculate the time period, then, guaranteed by the constant speed of light, the calculated time period will be the same on mountain top or at sea shore, so that we can use it to calibrate the speed of all kinds of clocks, theoretically.Jh17710 (talk) 20:54, 5 December 2010 (UTC)
 * Yes, you are correct, and I should have said that the speed of light in free space and in the absence of a gravitational field is constant for all observers. The effect of gravity between mountain top and deep mine is extremely small, but can be measured using atomic clocks.  Strictly speaking, though, isn't it time itself that is affected by gravity, not the actual speed of light.    D b f i r s   22:22, 5 December 2010 (UTC)
 * I think time is not affected by gravity. An atomic clock is affected by gravity does not mean time is affected by gravity, am I correct? Let us study this issue in more detail, if time is affected by gravity then what will happen? A clock with pendulum is affected by gravity but may be in different rate as an atomic clock is affected by gravity, then, we have quite the same ESSENTIAL QUESTION, should time be represented by the atomic clock or by the pendulum clock? If atomic clock is your choice, could you tell me which speed of light we should use? I mean, we should measure the speed of a red ray on top of a mountain or at a sea shore? Since the wavelength of red light is the same, should we have higher frequency of red light at sea shore compared with on top of mountain?Jh17710 (talk) 15:00, 6 December 2010 (UTC)
 * Urgh. Do you even read the articles in question, or are you just making stuff up?!?  Your first statement says that you think time is not affected by gravity.  This is very wrong.  See Gravitational time dilation which is also described in brief in the article time dilation.  If you start with a false premise, the proposals you create based on those false premises become invalid as well.  It has been experimentally verified that atomic clocks under different gravitational conditions move at different rates.  -- Jayron  32  15:20, 6 December 2010 (UTC)
 * Yes Jayron, I have read some of articles in question. Do you think we should measure the speed of a red ray on top of a mountain or at a sea shore? Since the wavelength of red lights is the same, if we measured a frequency of Fr for a red ray on top of a mountain, we will have a higher frequency for red light, Fr+, at sea shore. Do we take both of Fr and Fr+ as frequencies of red light?Jh17710 (talk) 18:26, 6 December 2010 (UTC)


 * Is this a homework question? Looie496 (talk) 17:55, 4 December 2010 (UTC)
 * Originally I thought so. But from the history Reference desk/Archives/Science/2010 November 19 and the others who understand it better then me I think the answer is no, it's closer to OR (to put it mildly) Nil Einne (talk) 18:17, 4 December 2010 (UTC)

on p block elements
why p block elements consists metals,non-metals —Preceding unsigned comment added by 117.197.60.188 (talk) 15:57, 4 December 2010 (UTC)
 * It has to do with effective nuclear charge; whether an element is a metal or a nonmetal is dependant upon how much "pull" the valence electron level "feels" from the nucleus. Larger atoms have more electron shielding, due in large part to the presence of the core "d" and "f" electrons, than smaller atoms.  This additional shielding explains why nitrogen and phosphorus are nonmetals, while bismuth is a metal, though they are both in the same group.  The electrons in the outer shell of nitrogen experience greater attractive pull from the nucleus; so it has more non-metal character (higher first ionization energy, greater electronegativity) than that of Bismuth, whose valence electrons are "shielded" by all of those extra electrons, making Bismuth more metalic in character.  -- Jayron  32  16:04, 4 December 2010 (UTC)

Time travel question
Hello science volunteers. This is a great service you have here. I hope you can help me.

I have read Earth's Rotation and Earth's Orbit and Time Travel. I understand that theoretically, if a person travelled in time only, they would not end up on the same place on earth at a different time but would end up out in space in a different time - because the earth's movement in its orbit plus the sun's movement around the milky way and the milky way's movement through space would all move the earth away from their starting spot.

So for someone travelling a couple of hundred years through time, I understand they'd end up way out in space.

What I'd like to understnad is what might happen for a person who travelled just one second or even a fraction of a second. To the point where they ended up say at about the height of a tall building. Or buried a short way underground. How can I calculate this?

1. I read in the articles that the speed of the earth's rotation at the equator is 465 m/s. How do I calculate this speed for a different latitude?

2. I'm having trouble visualizing - if a person moved a short fraction in time, would the earth's curvature put them underground or up in the air? Or neither? Is the curvature a factor at all?

3. The article says the earth moves in its orbit at 300,000 m/s. So If a person travelled in time a thousandth of a second, they should end up 300 m away from the starting point. But how do I figure out whether this would be up in the air or underground? I imagine they'd have to start out at either dawn or sunset, but I can't figure out which would give which.

Garsk (talk) 16:12, 4 December 2010 (UTC)


 * It's a very tricky question. There are two major problems, and we'll ignore the first for the sake of answering your question.  (the first being that time travel simply doesn't work, excepting the sort of time travel we all do, moving forward through time).  The second problem, if we assume someone can "travel" backwards through time, the space problem isn't even as simple as you make it out to be.  Because of the implications of the relativity theories (special relativity and general relativity) there is no privileged "frame of reference".  That is, your question assumes one can remain "stationary" when one jumps through time.  The problem is, there is no absolute stationary.  That's what "relativity" means; all measurements are relative to each other, and no set of conditions can be defined as "at rest" execpt arbitrarily.  That is, we call where we are now "at rest"; however thats for the convenience of the math and not representative of any reality.  I suppose, if one wanted to make some assumptions here, you could calculate your trajectory, and based on inertia, calculate where you would have been relative to your old position had you stayed at the same time, and then find out where that was at the new time.  The problem is defining your trajectory.  The earth spins, it moves around the sun, our galaxy rotates, the local group of galaxies is moving, its a messy issue to work out.  Of course, this is all moot since time travel in the Back to the Future sense is impossible.  -- Jayron  32  16:47, 4 December 2010 (UTC)
 * You always say that, but Time travel doesn't say that. I don't mean that it would involve DeLoreans and blazing tire tracks, or be feasible for humans or in any way useful, but issuing a decree that time travel is limited to forwards travel at one minute per minute seems premature. 81.131.27.188 (talk) 17:22, 4 December 2010 (UTC)
 * There is no known evidence that backwards time travel has ever occurred, nor does it make logical sense that it could occur. The answers to the paradoxes that could result are "explained" by looney theories about the universe magically splitting in two. It makes for interesting fiction, certainly. Like a math teacher of mine once said, "If you start out with incorrect assumptions, you're liable to get interesting results." ←Baseball Bugs What's up, Doc? carrots→ 17:31, 4 December 2010 (UTC)
 * Heh. Well, it's all about the Many-worlds interpretation, innit. Some people find it repellent, others find it attractive. (Some of those are awful cranks, but that's not the fault of the idea.) If we're being scientific, we all ought to rein those feelings in. It's not a fringe theory, anyway. 81.131.27.188 (talk) 17:55, 4 December 2010 (UTC)
 * Bugs, are you really qualified to pass judgment on quantum mechanics and general relativity? I'm not, and I suspect I have far more knowledge of its details and implications than you do. The truly wise know when to hold their tongue. --Mr.98 (talk) 19:00, 4 December 2010 (UTC)
 * If there is any evidence that actual backwards time travel has ever occurred, please place it here on the desk. :) ←Baseball Bugs What's up, Doc? carrots→ 19:35, 4 December 2010 (UTC)
 * That's a non sequitur. I'm not sure you realize that, though, which is why you are not exactly qualified to provide an adequate answer to any part of this question. --Mr.98 (talk) 19:46, 4 December 2010 (UTC)
 * No it's not. You can invent hypotheses all day long, but unless you can find evidence that support those hypotheses, they remain in the realm of fantasy. ←Baseball Bugs What's up, Doc? carrots→ 19:49, 4 December 2010 (UTC)


 * EC Sure, no problem. Time Traveler Caught on Film :) A Quest For Knowledge (talk) 19:48, 4 December 2010 (UTC)
 * And actually, I'm just going by what Stephen Hawking said some years back, about time being a vector, a one-way arrow. But I'm sure you have far more knowledge of the details and implications than he does. :) ←Baseball Bugs What's up, Doc? carrots→ 19:39, 4 December 2010 (UTC)
 * Personally, no, but Hawking is not the only guy in physics, nor universally agreed with by professional physicists, and I know that much. I certainly know when I am out of my depth, which you apparently do not. --Mr.98 (talk) 19:47, 4 December 2010 (UTC)
 * Kurt Gödel wasn't exactly a crank: closed timelike curve. See also Novikov self-consistency principle, on avoiding the grandfather paradox.  It's no crazier than the rest of general relativity, really. TenOfAllTrades(talk) 19:17, 4 December 2010 (UTC)
 * Time travel in the sense of the OP, or as in Back to the Future, involve the time traveler having a discontinuous world line. There is no evidence that suggests that time travel in that sense might be possible.  The kinds of time travel discussed in Time travel that are known to actually physically exist (time dilation) or haven't yet been ruled out as conflicting with our current understanding of physics (closed timelike curves) involve continuous world lines.  Red Act (talk) 20:33, 4 December 2010 (UTC)

Thank you for the background. I'm sorry I didn't understand all of it, but I get the idea you can't answer my question because you say time travel is impossible anyway. Could you pretend that it is for the sake of the calculations I have asked about? How would I calculate where I was a thousandths of a second ago or where I will be a thousandths of a second in the future, with respect just to the earth's motion around the sun and its own rotation? Is that more answerable? And thanks again to everyone who tried. Garsk (talk) 19:33, 4 December 2010 (UTC)
 * Forward time travel is not only possible, we're all doing it right now. For the calculations, though, you first have to define a frame of reference. For example, if you knew where the "center" of the universe is, you could travel in the direction of the line defined by its point and the point where you are, in which case you would be best off to be on the lee side of the earth from it. But you can use that same principle for any frame of reference, for example you could travel along the line connecting the center of the sun with the center of Polaris. Again, make sure you're going "up" rather than "down" at that moment, or you might find yourself in an early grave. ←Baseball Bugs What's up, Doc? carrots→ 19:43, 4 December 2010 (UTC)


 * General relativity doesn't give any simple answer. In order to know "where" you end up, you also have to know "how" you traveled in time.  Since we don't know of any concrete way to travel in time, there is no way to say anything about where you could end up.  Dragons flight (talk) 20:03, 4 December 2010 (UTC)

I guess the problem is I really don't understand why my question can't be answered on an imaginary basis. Probably I don't get it because I don't understand the "real-world" physics of time travel at all. Apologies for that, because I do appreciate you are all trying to help.

However, the Time Travel article says that this idea has been used by science fiction writers. I know fiction isn't the real world but for this question I don't really care; I just want to be able to imagine what was supposedly going on in such stories.

Can it be answered, with respect to the earth's motion around the sun, where the point in space is that a person on a particular spot on the surface of the earth now would reach in one-thousandth of a second, relative to that spot on the earth now? Baseball Bugs is getting close to what I am trying to ask - I want to know if this spot would be up or down and which part of the earth is the lee side or trailing edge and which part the leading edge.

Garsk (talk) 20:52, 4 December 2010 (UTC)
 * It cannot be answered because there is no unique way of identifying "here" now with "here" some time later. Take an example of real travel. Say, at 9am you sit down in your seat ("here") in a lecture hall of the University of A. Two hours later, lecture finished, you're still in your seat in the lecture hall of the University of A. On the other hand, imagine you take a train, at 9am you're sitting in your seat (here) in the train at the station of A. Two hours later the train arrives at B. Now, "here" is your seat in the train at the station of B. What "here" at 11am means actually depends on which reference system you choose. If you define "here" with respect to a reference system that is fixed with respect to your seat and the train, then you are "here" both at 9am and at 11am. If you define "here" with respect to a reference system that is fixed with respect to the surface of the earth, then "here" is the station of A at both 9am and 11am. You are not "here" any more, you have travelled "there". There is no such thing as "travelling just in time", this depends on which reference system you choose.
 * Having gotten that out of the way, we can just choose one. Say we choose one which is fixed with respect to the centre of the sun and which is not rotating. Then, yes, you can answer the question of where you will be a millisecond from now if you stay fixed with respect to that reference system which we have defined just now. The answer depends on what time of day you choose to stay fixed. At 6am you'll hop 30 metres underground (the earth's orbital velocity is 30,000 m/s, not 300,000), at 6pm you'll hop 30 metres in the air (whether straight up or down or at an angle depends on your latitude and the time of year). Does that satisfy you?
 * With respect to fiction, there is of course nothing that can be said. If a writer chooses to allow time travel they are also free to specify the details of how it's supposed to work because just changing that one detail actually throws out all of our theories of physics which do not allow time travel. H.G.Wells lets his time machine follow all the movements of the earth so that it stays fixed with respect to the surface of our fair planet. Others may follow an interpretation like the one that you suggest. --Wrongfilter (talk) 21:26, 4 December 2010 (UTC)
 * I just also want to point out that only taking into account relative motion vise a vise the Earth or even the Sun omits the relative motion that you get from the rotation of the Milky Way (which is on the order of 220 km/s or something like that). Just to illustrate part of the problem here. --Mr.98 (talk) 22:27, 4 December 2010 (UTC)
 * You should also take into account the motion of Dark flow if you believe that it exists. Science fiction time travel usually just assumes that the device includes some form of stabilisation with respect to local space and objects, though some stories assume a fixed distance from the centre of the earth and so time travellers end up buried etc.  In fact no reference frame exists in an absolute sense, so there is no fixed point in the universe from which to measure displacement.  See also Metric expansion of space.    D b f i r s   23:27, 4 December 2010 (UTC)

Calculations
OP, you just want something like this for eg. your first question ("How do I calculate this speed for a different latitude?"): $$\mathrm{v}_\perp=r\,\frac{d\phi}{dt}=r\,\omega$$ from the Angular velocity article... right? Now say R is the radius of the Earth (use the approximation of 6371km), and r is the radius of rotation (ie. distance perpendicularly to the axis of rotation) at a given latitude (say 39°): $$r=R\,\cos(Latitude)=6371km*\cos(39^\circ)$$ so that, with your value of 465m/s tangential velocity at the equator: $$\frac{\mathrm{v}_{\perp39^\circ}}{\mathrm{v}_{\perp0^\circ}}=\frac{r\,\omega}{R\,\omega}=\frac{r}{R}$$ or: $$\mathrm{v}_{\perp39^\circ}=\mathrm{v}_{\perp0^\circ}\,\frac{r}{R}=465m/s*\cos(39^\circ)\approx361m/s$$ Does that make sense? This is what you are asking for, isn't it? Wiki Dao &#9775;  (talk)  22:02, 4 December 2010 (UTC)

Thank you all. Very much. Thats is what I was looking for. Thanks for persevering with me to get there. Plus all the extras.

(WikiDao I had to laugh - knowing nothing about math I thought it would be just numbers. I guess it is just too complicated for a nonmathematician! Thanks for taking the trouble to write and calculate it all out.)

Cheers all. Garsk (talk) 00:24, 5 December 2010 (UTC)


 * NP :) Btw: re. 2) the change due to the rotation of the Earth would be "toward the horizon" ie. along the ground (for short times), and re. 3) yes, whether the person goes "up" or "down" or "sideways" in this scenario would depend on the time of day (dawn→down, dusk→up, noon→east, midnight→west if I understand what you are imagining correctly, though there are all kinds of conservation-of-momentum issues and so on, not to mention all the other "time travel" issues mentioned in the comments above...). Wiki Dao  &#9775;  (talk)  02:04, 5 December 2010 (UTC)

Cars/trucks with fixed axles
Did early cars or trucks have a fixed rear axle, so both wheels had to turn the same speed ? This would obviously cause skittering in turns, where the wheels need to turn at different speeds, so would have been far from ideal. If they did have those, what's the last vehicle to feature such a fixed axle ? StuRat (talk) 19:39, 4 December 2010 (UTC)


 * The differential was widespread even before the automobile - it was used in carts, horse-carriages, and chariots since ancient times. Regarding the "last vehicle" to use a fixed-axle, well - these still exist today, in the form of low-cost carts, tuktuks, wheelbarrows, and so on.  Nimur (talk) 20:05, 4 December 2010 (UTC)


 * Thanks. I understand that trains still use solid axles, so how do they deal with turning ? (Obviously they don't make sharp turns, but it's still got to be an issue even for gradual turns). StuRat (talk) 22:33, 4 December 2010 (UTC)


 * A brief explanation is given in our article: Rail adhesion. The part of the wheel in contact with the rail is not quite flat, and this slight tapering of the tread ensures that "when the train encounters a bend, the wheelset displaces laterally slightly, so that the outer wheel speeds up (linearly) and the inner wheel slows down, causing the train to turn the corner".  Perhaps a rail expert can give a fuller explanation?    D b f i r s   23:14, 4 December 2010 (UTC)


 * The trick is that the bit of the left wheel that's touching the rail isn't at the same distance from the axle as the bit of the right wheel that's touching the rail. One of them has to travel further because it's on the outer rail of the curve, but it does travel further because it's also further from the axle.  --Anonymous, 04:44 UTC, December 5, 2010.


 * It's nice to listen to Feynman explain: http://www.youtube.com/watch?v=y7h4OtFDnYE —Preceding unsigned comment added by 99.226.188.46 (talk) 04:49, 5 December 2010 (UTC)


 * Thanks, everyone. StuRat (talk) 08:13, 9 December 2010 (UTC)

problems with WWVB NIST time radio broadcast
Are there any problems with WWVB NIST time radio broadcast or atmospheric conditions interrupting the signal?

My father and I both live on the east coast of the US, 14 miles apart. For years both of us have had a clock updated by WWVB. For several days both of them have been way off. Usually putting them in a west window overnight makes them reset, but this has not happened. I changed the batteries and they still have not reset themselves from the WWVB signal. Is there some sort of broadcast problem or interference? Or did both clocks happen to go bad about the same time? Bubba73 You talkin' to me? 20:40, 4 December 2010 (UTC)
 * The official WWVB monitoring station reports good signal quality on the east-coast for the last few days, and there has not been a scheduled broadcast outage since October 2010. More official info from NIST: WWVB Radio Controlled Clocks, including estimated reception quality vs. time of day (this changes as the ionosphere's E-region forms and dissipates based on day/night sunlight incidence, affecting the skywave pattern).  Nimur (talk) 21:38, 4 December 2010 (UTC)


 * Thanks, I didn't know about that information. Both clocks were off by several hours and an odd number of minutes (so it isn't a time zone issue). Bubba73 You talkin' to me? 22:07, 4 December 2010 (UTC)


 * My clock would never reset sitting in the window. Tonight I took it outside, took out the batteries to reset it, and in a few hours it picked up the time.  Bubba73 You talkin' to me? 07:18, 5 December 2010 (UTC)


 * Except that it is 39 or 40 seconds slow. Bubba73 You talkin' to me? 07:21, 5 December 2010 (UTC)

Step-by-step method for writing the electron configuration of an arbitrary atom
Can someone please give me a step-by-step method of how to write the electron configuration of any atom; one way according that doesn't follow the Aufbau principle and another that does. I know that the specific method I have to use takes into account the position of Lanthanum and Actinium. Also, please keep in mind that I'm in the 11th grade so the notation should not take into account any Aufbau exceptions of consequences of relativistic motion of the electrons. --Melab±1 &#9742; 22:34, 4 December 2010 (UTC)


 * Are you asking how to record an electron configuration which is already known, or how to determine the configuration for a given atom ? StuRat (talk) 16:41, 5 December 2010 (UTC)


 * How to determine the configuration of a given atom. --Melab±1 &#9742; 19:15, 5 December 2010 (UTC)


 * OK, and what's wrong with the method they taught you in school ? StuRat (talk) 19:22, 5 December 2010 (UTC)

The shortcut method for writing the electron configuration can be done 2 ways for high school students: Those are the two methods taught to high schoolers, which ignores the exceptions. Hope that helps. Its easier to explain in person and with spoken words, but that's the best I can do in text. -- Jayron  32  19:40, 5 December 2010 (UTC)
 * Method A: The Aufbau method. You need a convenient way to memorize the filling order for orbitals.  The easiest method is called the "diagonal rule", and this search of Google Images gives you lots of good examples.  Once you have the correct order of the orbitals, and also know the number of electrons in each orbital (s=2, p=6, d=10, f=14: It's the odd numbers doubled, by the way, so if you know your odd numbers, you know this too), you just fill them up in order (which is exactly what Aufbau means).  Thus, take Sodium, atomic #11.  We know we have 11 electrons to fill in.  Using the diagonal rule, (see the pictures I provided) we fill up the orbitals with electrons until we run out.  Start with 1s2.  Now we have 9 left.  The next to fill up is 2s2.  Now we have 7 left.  The next to fill up is 2p6.  That leaves us with one left.  The next orbital is 3s, which can hold up to two electrons.  Since we only have 1 left, we write: 3s1.  That gives us a final electron configuration of 1s22s22p63s1.
 * Method B: The periodic table method. The period table is basically a giant grid, and the coordinates of that grid tell you the electron configuration.  Each "block" on the periodic table corresponds to a type of orbital, so the leftmost block, with 2 columns is the "s" block, the rightmost block, with 6 columns is the "p" block, the center block with 10 columns is the "d" block, and the bottom "detached" section, with 14 columns, is the "f" block.  Each column within the block is the number of electrons in that orbital, and each row of the period table is the energy level for that orbital.  Lets take Phosphorus as an example.  You can "count up" to phosphorus through the period table, you go through the first row, which only has s block electrons, so you start 1s2.  The second horizontal row has 2 s block elements and 6 p block elements, so you write 2s22p6.  The third row, which contains phosphorus, has 2 "s" block elements, and phosphorus is the third p block element on the third row.  So that gives us 3s23p3.  Done.  If you have to count through the "d" block or "f" block elements, remember that the energy level for the d block is one lower than the row you are on, so the electron configuration for vanadium, the third element in the 4th row in the d block, is ...3d3 (subtract one from the row number for "d" block elements) and for "f" block elements, you subtract 2.  So uranium's last entry in its configuration is ...5f3, being on the 7th horizontal row, subtract 2 to get the energy level of "f" electrons.


 * Thank you very much! I'll be sure to put this in my notes. --Melab±1 &#9742; 20:46, 5 December 2010 (UTC)

StuRat (talk) 02:33, 6 December 2010 (UTC)