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November 27

Absolute rest point

According to the book "Four Symphonies of an Oriental Universe" the author, Zhi-Zhong Cai, said we can locate an absolute rest point with some imagination. Let a point-source of light shine once at a point E, then the out going wave front will make a growing but thin out ball of light. Do you agree that the center of that growing ball of light is a fixed point as in the absolute rest universe? The reason is clear, because the speed of light is independent of the speed of the source of light so that the growing ball of light is growing from a fixed point in the whole universe or whole space. The only difficulty is that, unless the source of light stays shining like once every two seconds at that same event point E while the ball of light is growing, otherwise, we have no way to point out where is E. Because our technology is unable to show us the actual growing ball yet. It is the same for every event with a point-source of ligh shines once so that the location of that event is an absolute rest point that is secured by the growing ball of light, if we can see it, then we can try to locate its center point. That means, if we shine four point-source of lights at each of (0,0,0),(1,0,0),(0,1,0), and (0,0,1) of a Cartesian coordinate system roughly at the same time then we have an absolute rest frame. We cannot deny their existence just because we cannot figure out some way to use those growing balls of light to locate their centers. Please comment, thanks.Jh17710 (talk) 06:13, 27 November 2010 (UTC)[reply]

I don't understand why that reference frame would be anymore 'absolute' than any other reference frame or why it would be considered an 'absolute rest point' or even how it would be determined to be 'at rest' unless you just wanted to define it that way in your coordinate system. This may help. Sean.hoyland - talk 07:23, 27 November 2010 (UTC)[reply]

Thanks for the reference. The reason why it is an absolute rest point to any absolute rest frame is based on the character of light. Since the way light moves away from the souce point is independent from the speed of the source. That means, if just shine once, the center point of that growing ball of light is guarded by that thinning out and growing ball of light as an absolute rest point in the whole space; at least at time while the ball is recognizable. The problem is that one single growing ball of light will thin out as 1/(r^2) of density and the speed of light is too fast that scientists have no way to watch that growing ball so that even the growing ball can show them an absolute rest point, its center point, they have no way to mark it yet. But, there are rest points if we can figure out some way to see those growing balls to locate their center points.Jh17710 (talk) 22:30, 27 November 2010 (UTC)[reply]

Actually, all reference frames will agree that the expanding ball of light is spherical, and that its center is at rest. So you can't use it to pick out a preferred reference frame — all the frames will think they're the preferred one. --Trovatore (talk) 07:25, 27 November 2010 (UTC)[reply]

Trovatore, thanks for yuor kindness to show me the right way to describe that expanding ball. Let's look at the issue of spherical shap first. Because of the farther length looks shorter, the only perfect circle shape of a ball is when the ball is infront of us or when we turn our eye sight directly to it. That means even within the same reference frame, the spherical shap is not automatically maintained. Secondly, not all inertial frames can think they are the rest one and it will be very easy to figure out if, well, the if part is very difficult; if we can locate the expanding ball, then, instantly locate the center point, then, instantly decide if that center point is at REST in their systems.Jh17710 (talk) 22:30, 27 November 2010 (UTC)[reply]

OK, first of all, we need to distinguish between measurement in a reference frame, and visual appearance in a frame. There are complications involved in the visual appearance of objects at relativistic speeds (we surely have an article on that, but I'm not sure where it is).

But those complications don't interest us here. What we're concerned about is measurement of time/space coordinates. For example, you might set up a spherical shell of photodectors, all with synchronized clocks. When they detect the light, they send out a signal, time-stamped by what their clocks say. If all of those time stamps are the same, you conclude that the shell of light is centered at the center of the photodetectors.

With me so far? Here's the surprising thing: No matter what reference frame you happen to be in, you can set up a concentric system of such shells, according to which you will conclude that the center of the shell of light is at rest. The why of that is a little involved to explain in detail, but it all works out. --Trovatore (talk) 22:45, 27 November 2010 (UTC)[reply]

Trovatore, I mean, it is really hard to figure out if a shape is spherical at far away, because our tools are very limited; it is even more difficult to measure from another inertial system, no tool at all. But you are right, that is not the main point. On the shell of photodetectors, we will need four wave peak counters at tips of the largerest internal triangular pyramid, let the center send out fixed frequency signals, then, if all four frequencies remained the same for a little while, then, yes we are lucky to witness an absolute rest point at that little while. I don't trust clock.Jh17710 (talk) 00:40, 28 November 2010 (UTC)[reply]

I made a wrong statement above. I will say it at the end of this paragraph. I found out it is not easy to synchronize clocks. If we think a little bit deeper, the four clocks on the spherical shell cannot have the same gravity so that their speeds will be different! Even if we can adjust their speeds, the mistake I made is that, all wave peak counters will count the same frequency within one minute of time, since we assumed that their distances from the source of light are all fixed. The first wave front may reach four clocks at different time, but, since all the following waves are all have the same differences in between four clocks so that the wave peaks count at each of four clocks will be the same in the same period of time. Sorry for the mistake.Jh17710 (talk) 15:43, 28 November 2010 (UTC)[reply]

The problem of detecting the wave front of that SINGLE expanding ball of light is our technology is way behind on two issues relative to measurement, one is to adjust the speed of four clocks so that they are the same against to gravity and temparature or other factors, and another one is to match the numbers showing on four clocks. Until we can control the synchronization of four clocks in this discussion, we have no way to find out if the SINGLE expanding ball of light reaches four clocks at exactly the same time.Jh17710 (talk) 15:43, 28 November 2010 (UTC)[reply]

Well, there is no "exactly the same time" (or "exactly" anything) in engineering. But you don't need it. You put the detectors in free space. They keep their relative positions constant, and their clocks synchronized, by sending laser messages to each other with timestamps, and making corrections with little hydrazine rockets when necessary. It's not going to be perfect, but it can narrow down the center of the expanding ball of light within some reasonably small margin of error.

And you can put several larger collections of these assemblies of photodetectors at larger distances, and decide whether the center of the ball of light is moving (again, within some margin of error) in your reference frame.

Then the surprising, counterintuitive thing is this: No matter what reference frame you are in, you will determine that the center of the ball of light is at rest, within your experimental error. --Trovatore (talk) 20:59, 28 November 2010 (UTC)[reply]

Trovatore, there are quite a few issues in your last comment. First of all, if you like to let four detectors keep constant relative positions do you think it may be better if we use one solid frame to install them? The second issue is laser messages with timestamps, since they are at four tips of a triangular pyramid, that can be a good idea to synchronize their clocks; but, the top one with different gravity potiential may adjust its speed a little bit to match other 3 clocks at about the same altitude. Then the third issue is if we have two sets of this kind of shell, one is rest and the other one is moving directly away from the rest set at any speed then back and forth at any other speed, we should find out, logically speaking, all of eight devices will count the same total amount of wave peaks within same period of time. Do you expect this result? The fourth issue is that above result does not mean both center points are absolute rest points. The key point is the SINGLE expanding ball of light, does it reach four devices at the same time? Or, we can arrange the experiment in one dimensional space to get more chance of being rest and to make things easier, just use a section of train. Logically speaking, if we fire two bullets at the same time at the center of that section of train to two ends, then, no matter how that train moves, bullets reach ends at the same time, but, not if we fire two rays of light. Is that right?Jh17710 (talk) 02:22, 29 November 2010 (UTC)[reply]

Let me just take your last question, because it's the one that treats the key issue.

No, in fact, that's not right. If you fire two rays of light forward and backward from the center of a train moving at a constant speed, they will (as measured in the train's frame) reach the ends of the train at the same time, no matter how fast the train is moving. --Trovatore (talk) 06:27, 29 November 2010 (UTC)[reply]

That is the key issue. Looks like your main point is on the "measured in the moving system". That means, if measured in the rest system and if when v=o, both lights hit ends at the same time then when v>0, the light will hit the rear end first. Is that what you mean?Jh17710 (talk) 03:02, 30 November 2010 (UTC)[reply]

No, it isn't. I meant precisely what I said. If it's hard to make it square with your intuition, that's fine; that's because your intuition is wrong. You need to get a new intuition, one that's right. The way to do that is to work through why what I said is right. --Trovatore (talk) 03:12, 30 November 2010 (UTC)[reply]

Oops, sorry. I didn't see your "measured in the rest system". Yes, you're right. --Trovatore (talk) 03:49, 30 November 2010 (UTC)[reply]

Thanks. Now I have to study about this issue for a while.Jh17710 (talk) 04:14, 30 November 2010 (UTC)[reply]

I think it is not easy to explain this issue. Could I ask two more questions? The first one is, since you think both ray will reach ends at the same time no matter how fast the train moves, do you think the period that rays take to travel half of a section of train will remain the same or shorter? The second one is if we replace rays by bullets, then, do you think bullets will reach ends at the same time to rest observers? I think the answer for the second question is yes, but, I don't know your answer for the first one.Jh17710 (talk) 14:13, 1 December 2010 (UTC)[reply]

Let's ignore the first question for now. About the second question, according to Red Act, we should have a lot of clocks at the rest system, so that when bullets reach ends at the same time, there are observers at two locations of contacting point where bullet reach end wall so that we expect bullets reach ends at the same time as measured in the rest system. Compare this with the event of rays, the rest system observed different results but the moving system observed same result; that means {physical law regarding emission of bullets and rays} is not maintained in between inertial systems. We may wait until our technology allow us to actually measure the time so that this issue can be verified directly, or, follow the expectation of relativity principle, either assume rays reach ends at the same time to observers of the rest system (That is the hypothesis of emission theory of light.) or rays reach ends at different time to observers of the moving system.Jh17710 (talk) 03:55, 2 December 2010 (UTC)[reply]

Simultaneity

The problem boils down to relativity of simultaneity. The center of a growing light ball is the spatial center of a set of events which occur at the same time. But observers disagree about what "at the same time" means, which makes them disagree as to where the "center" of the growing light ball is. Red Act (talk) 08:21, 27 November 2010 (UTC)[reply]

Red Act, yes, the center of an expanding ball of light is the spatial center of a set of events, like when the radius of that ball is at 10', 20', 30',...but I don't see you point of "at the same time". Could you explain it a little bit? Thanks.Jh17710 (talk) 22:30, 27 November 2010 (UTC)[reply]

I assume you're familiar with conic sections. If you slice a cone along parallel planes at the correct angle, you get circles; the centers of those circles lie on a line through the cone. If you slice the cone along parallel planes that are inclined to those other planes by less than 45°, you get ellipses. The centers of the ellipses (for any particular set of parallel planes) also lie on a line, but it's a different line. The slope of that line is the velocity of the inertial reference frame whose planes of simultaneity are the parallel planes that you used. Thus, depending on which reference frame you use, you get a different "center". That center is always at rest with respect to the reference frame that you used. -- BenRG (talk) 01:14, 28 November 2010 (UTC)[reply]

BenRG, for events happen at points, lines, or even plans, we may have problem of "at the same time" or different time, but, this will not happen for an expanding BALL like this case. They will never happen at the same time from any observers. Am I correct?Jh17710 (talk) 03:25, 28 November 2010 (UTC)[reply]

Absolute time

According to the book "Four Symphonies of an Oriental Universe" the author, Zhi-Zhong Cai, said we can have a better clock than current atomic clock. If I have some device can count wave peaks then just select a source of light with fixed frequency which is moving at constant speed v relative to me along the line connecting that source and me, and we are ready to use that device to measure time. According to Doppler effect, no matter the moving party is the source or me and no matter what is the value of v, the frequency and the wave length of light may change but the speed of light will remain the same, so that my time can be calculated by (the total wave peaks I count by the device x the wave length)/(the speed of light). That means, the time period calculated at different relative speed will be the same. Could we use this idea to make a clock to measure time period? If yes, then, a moving clock will not slow down. If the device can be made very small, then we can put the device and the source of light in a small vacuum box and use it as a clock, that will be v=0, and since the speed of light is independent of the change of gravity, it will be a better clock than atomic clocks. Isn't it?Jh17710 (talk) 07:01, 27 November 2010 (UTC)[reply]

Yes, you can make a clock that measures a time coordinate independently of your motion if you use an external reference. A more familiar example is a cell/mobile phone that gets its clock setting from the cell tower. Another theoretical example would be a clock that measures the temperature of the cosmic microwave background, though that's cooling so slowly that it wouldn't give you a very precise time in practice. There are also inertial navigation systems that, in certain restricted cases, can keep time independent of your motion without any external reference. But this has nothing to do with the accuracy of atomic clocks. They do a very good job of what they are designed to do, which is count very nearly equal intervals of proper time. The clocks I described before are measuring some form of coordinate time, which can also be useful, but it's not the same thing.

I'm not familiar with the book you mention, but I would not use it as a way to learn physics. It sounds like the author is pretty clueless. A lot of people like that write "physics" books, for some reason. -- BenRG (talk) 09:43, 27 November 2010 (UTC)[reply]

Thanks. That book just published 9-1-10 at Beijing, it is in Chinese.Jh17710 (talk) 03:26, 28 November 2010 (UTC)[reply]

The article Comoving_distance#Comoving_coordinates covers comoving time in a bit of detail. --Jayron32 13:48, 27 November 2010 (UTC)[reply]

I think there are at least two issues. First, the speed of light does change in a gravitational field. Second, and more importantly, the method depends on the distance between the two objects remaining constant to an extremely high degree of precision. This means that any vibration, such as caused by sound waves, will cause a wobble in the distances. I don't see how it would be possible to solve that problem. Looie496 (talk) 17:05, 27 November 2010 (UTC)[reply]

Looie496, the frist issue is still unknown, we don't even know if the gravity will change the direction of light; the change might be caused by the different density of air around the sun. The second issue is very true. It is even worse for a constant velocity v in between two objects. However, the spirit in the idea is that, theoretically speaking, in front of a light source, we let a counting device move under a relative constant velocity v, then, if we use (total wave peaks x the wave length)/(the speed of light) to define time, that kind of time will be independent of v. What do you think?Jh17710 (talk) 23:13, 27 November 2010 (UTC)[reply]

The GR prediction of gravitational lensing has been experimentally confirmed to very high precision. Maybe 90 years ago you could have argued that it was refraction in the Sun's atmosphere, but not today. See Tests of general relativity.

The questions you ask in this thread about time in special relativity could be rephrased as questions about position in Newtonian physics. Can you build a device that determines its absolute position (relative to Earth's surface) by triangulating with cell towers? Yes. Can you put the cell towers and the receiver together in a box to get a device that knows its absolute position without an external reference? No. Can you use inertial navigation to keep track of your absolute position without an external reference? Yes (subject to certain caveats). In special relativity the same things are true of the time coordinate, for essentially the same reasons. -- BenRG (talk) 00:36, 28 November 2010 (UTC)[reply]

That way of counting time is just an idea. Like Looie496 said, it is almost impossible to make a device to count wave peaks due to the difficulty of keeping same distance. However, even if gravity can change the direction of light, the speed is the same so that the idea of that book still works, right? About a device to locate an absolute rest point, it is even more difficult as discussed in the thread above this one.Jh17710 (talk) 01:58, 28 November 2010 (UTC)[reply]

Pulsars are also better "clocks" than nuclear decay. ~A H 1^(TCU) 03:27, 28 November 2010 (UTC)[reply]

If we have source and device

Even if you had such a light source, and an object capable of counting the light pulses, that still won't tell you how much time passed. You'd have to know the frequency of the light being emitted. You'd need another clock to figure out how fast that one goes, at which point you might as well use the other clock. — DanielLC 05:52, 28 November 2010 (UTC)[reply]

DanielLC, the key issue of a light source is having stable frequency. We have some known natural frequencies and radar will emit EM waves with almost fixed frequencies. So long as the souce and the device are keeping same distance or one of them is moving away directly from each other at constant speed (I will say this is impossible in real world, but, theoretically we should include the relative moving situation.) then the device will receive EM waves with fixed frequence. If we do have a known source, like red light, and a device, then we don't need any clock. That source and device make a perfect clock themselves so long as we can find some way to keep a constant distance between them. With the known frequency, we can calculate the time by the total count of wave peaks. Right?Jh17710 (talk) 17:22, 28 November 2010 (UTC)[reply]

Force at a distance without any connection

How does the force from gravity or magnetism communicate itself to distant objects, when nothing connects or passes between them and the attractor? And there's no aether either.

I'm aware of the teaching model that looks like ball-bearings on a rubber drum, but what is this curved surface and the third-object gravity embodied in in reality? 92.15.11.45 (talk) 14:18, 27 November 2010 (UTC)[reply]

There isn't any need for solid objects to be involved. In fact, what we see as solid objects ... aren't. The electromagnetic repulsion between electrons is what keeps the things we see as solid about an angstrom away from each other instead of passing right through each other. It's more like the whole universe is made of infinitely-small particles than of things that can actually touch each other. Paul (Stansifer) 15:02, 27 November 2010 (UTC)[reply]

I'm already aware that things are made of atoms and smaller things. That does not explain how for example a star pulls on the planets orbiting it, or how a magnet attracts things from a distance. 92.15.11.45 (talk) 15:28, 27 November 2010 (UTC)[reply]

Just about everything attracts or repels anything else at a distance. --Chemicalinterest (talk) 15:37, 27 November 2010 (UTC)[reply]

I'm asking how this happens. 92.15.11.45 (talk) 15:45, 27 November 2010 (UTC)[reply]

When I first read your question I found the article Action at a distance (physics). This didn't seem much help and I didn't understand it, so I didn't mention it, but on reflection, maybe it's helpful to you, so there you are. There seems to be some debate about how it happens and what it means. 81.131.46.126 (talk) 16:32, 27 November 2010 (UTC)[reply]

This is a good question, and has been answered on the desk many times. Action at a distance was one of the most plaguing problems of 20th century physics, but for the most part it has been resolved. There are a few critical mechanisms that apply to different kinds of fundamental interactions:

Forces like electromagnetism and the strong nuclear force are mediated by a particle, the photon or gluon. These forces "propagate" at the speed of their carrier-particle, and in the case of the photon, that particle also conveys energy and momentum. The mathematics to ensure energy-conservation are explained by treating force-carriers as virtual particles until they undergo an interaction.
The special case of gravity is explained by general relativity - there is currently no known force-carrying particle for gravity. Instead, Einstein's theory of gravity explains that all mass perturbs the geometry of space - so simply by existing, massive objects cause other objects to change their trajectories. Gravity therefore changes the shape of the inertial trajectory by warping the geodesic of space. General relativity mathematically describes the relationship between mass and the resulting warping of space-time geometry - in other words, allowing you to calculate the effective force. It also can describe how quickly gravitational information propagates - that is, when you move a massive object, how soon other objects in the "neighborhood" realize that something has changed.
The last case, quantum entanglement, applies to certain quantum properties, but in general does not convey energy or information.

Hopefully this explains the various ways things can "act at a distance." Nimur (talk) 16:35, 27 November 2010 (UTC)[reply]

Note that (virtual) gravitons are hypothesized as the force carrier particles for gravity. And if you're having a hard time understanding how gravity can both be transferred by a particle and by curvature in space-time, you're not alone. Reconciling the two is one of the key stumbling blocks in constructing a theory of everything which combines quantum mechanics (forces-as-virtual-particles) and general relativity. -- 174.24.198.158 (talk) 19:10, 27 November 2010 (UTC)[reply]

you might be intested in this answer. 62.54.13.205 (talk) 23:04, 27 November 2010 (UTC)[reply]

Midnight sun question

Because of possible changes to the earth's tilt on its axis and to the earth's orbital path around the sun, is it possible for a place right around or near the Arctic Circle to barely experience the midnight sun on one year during the summer solstice night, but on another year, not experience the midnight sun let's say, for 10 minutes during the summer solstice night? Willminator (talk) 15:43, 27 November 2010 (UTC)[reply]

The arctic circle, by definition, shifts along with every change, no matter how small, in the obliquity of the ecliptic. The arctic circle is not defined by maps, and is not fixed. It is, literally, equivalent to the obliquity of the ecliptic (the latter, expressed in latitude degrees from the pole, equals the arctic circles). The answer is flat-out NO, by definition. 63.17.43.56 (talk) 11:48, 1 December 2010 (UTC)[reply]

Well, the largest nutation is only some 20 sec of arc. On the Earth's surface that's about 2000ft of distance. The sun takes only 2 minutes to cover its own diameter. So on a very flat piece of land at approx 66½ degrees north (or south, six months later) you might just be able to catch a 'momentary' extension (or reduction) in night-time, as measured against the previous year. The sun would cover the 2000 ft in about a second or so I would think. This is a bit rough and ready, so I think it needs someone else to double check the assumptions I'm making.--Aspro (talk) 16:47, 27 November 2010 (UTC)[reply]

~~I don't see where you get the 2000 ft from (20 seconds of arc north/south on the Earth would be 1215 ft)~~(no, you were right about the distance), and while the Sun does indeed cover that apparent angle in less than a second, it's not what matters here, since the direction of the Sun's apparent motion is very close to east/west in the case of the midnight sun. What cannot happen is that the Sun never touches the horizon in one year but sets completely below the horizon in the next year (or the year after that), because the Sun's apparent size is roughly 30 minutes of arc. And if you consider not nutation but the fact that days and years don't align (i.e. if the exact time of solstice is at midnight for one location, it will be roughly at noon for that same location 2 years later, since 1 year is approximately 365 and a quarter days), then the difference will be only 3.3 seconds of arc (the formula is arccos(sin(axial tilt)*cos(angle that Earth travels in its orbit in half a day)) - arccos(sin(axial tilt))). Icek (talk) 23:43, 27 November 2010 (UTC)[reply]

Good questions. To do it in my head, I assumed a nautical mile is about 6080 ft and the illuminated parabola as a simple triangle with a 'lateral' base (therefore) of about 2000 ft west to east (not east-to-west as you put it because that is not the way the sun orbits), and about 2000 ft in the NS axis to keep things simple. I wanted to avoid discussions about twilight because it would be obvious to all that the effect would not be great enough to go into proper night. Likewise, whether 'day' starts at first rays or at the full disc becoming visible, is more of a academic question than what the OP is really asking. So I am assuming sun's rays. Perhaps the OP can comment on whether s/he meant this. The point about 'two years' is misleading pedantism in regards to the spirit behind the question. For we are talking about the apparent altitude extreme at the solstice, not a date on somebodies calendar. For it to be noticeable to the eye, the ground would need to be very flat, or you would need to assume the exact same position each year. Yet, if one used say a sextant, the difference in altitude -I expect - would be just discernible. Be on guard though. In such cold climes atmospheric 'tunnelling' may cause the sun to be visible when it is actually below the visible horizon. --Aspro (talk) 01:05, 28 November 2010 (UTC)[reply]

By midnight sun I mean the sun's disk being above the horizon at night in the summer solstice. A place that gets white nights is where the sun's rays are visible in the middle of the night, but the sun stays below the horizon around summer solstice. What I'm talking about is about a place where at least half of the sun's disk is above the horizon right in the summer solstice night, which happens around the Arctic Circle line. What I'm asking is if there a year around the Arctic Circle line where at least part of the sun's disk is exposed in the middle of the summer solstic night, but on another on another year in the summer solstice night it's not, or will the sun's disk be always partly exposed at that such location every year on the summer solstice night. I hope I've clarified what I tried to ask. Willminator (talk) 03:04, 28 November 2010 (UTC)[reply]

In that case the answer is no. The sun's angular diameter is about 1920 arcseconds and we are only talking about a change of about 20 arcseconds in altitude. As this would only represent the edge of the disc showing or not showing, it would be unlikely to cast a shadow bright enough to be seen above the twilight. However, with the aid of a post driven deep enough into the ground to prevent frost movements' so as to provide a fixed sight-line to the horizon, it should be able to indicate that in some years a few rays of midnight sun are still observable and in other years there is a brief moment of no disc to be seen at all. The effect though is very small and not something one would notice unless one took the trouble to observe it carefully. Even then, as I have already mentioned, the closer to the horizon that one observes events, the more the atmosphere can distort the true position of an object and make those observations unreliable. So the effect is of academic interest only.--Aspro (talk) 11:19, 28 November 2010 (UTC)[reply]

If you want to take a long-term perspective, the Earth's obliquity periodically changes from 22.1 to 24.5 degrees with a ~40,000 timescale due to interactions with other planets. So, if a hypothetical observer that could stay around 20000 years, then they would notice a quite substantial variation in the range of the midnight sun, even though these changes are pretty negligible on the scale of human lifetimes. Dragons flight (talk) 11:43, 28 November 2010 (UTC)[reply]

Aspro, which illuminated parabola are you referring to?
Why do you think that my 2 years have anything to do with the different human-made calendars? All I wanted to say is that the exact time of the solstice - the point in time when the angle between Earth's axis and a line connecting Earth and Sun reaches a minimum - may be at midnight for one particular location and one particular year. In the following year the exact time of solstice will be shifted by about 6 hours in local time of day, and in the year after that by about 12 hours. Obviously the distance in time to the next midnight cannot be more than 12 hours. So for a midnight 12 hours distant in time from the exact time of solstice the Sun's apparent position is 3.3 seconds of arc lower than for a midnight at the exact time of solstice (if there was no change in axial tilt). For 6 hours it would be 0.8 seconds of arc. Icek (talk) 11:55, 28 November 2010 (UTC)[reply]

The places you can see the midnight sun are those north of the polar circle. The arctic one is moving north by 14.2 metres per year. make of that what you like :-) EverGreg (talk) 20:48, 28 November 2010 (UTC)[reply]

Aspro, explain to me if I misunderstood anything in this discussion. If the partial midnight sun, or where part of the sun's disk is above the horizon at the sun's lowest point in the middle of a summer night, happens only once a year always every single year on the summer solstice night, which is possible on a location at or near one of the polar circles, like somewhere around Fort Yukon, Alaska; why do solar and lunar eclipes happen once in a while? It seems as if the earth's tilt axis and the earth's eliptical, orbital path around the sun change is not the same it was the previous year. Willminator (talk) 20:10, 29 November 2010 (UTC)[reply]

If I may answer as long as Aspro doesn't: Eclipses - solar or lunar - depend on the orbit of the Moon and are unrelated to what we were talking about. You could ask about the visibility of the Moon.

Having said that, there is a Moon-related effect on the apparent altitude of the Sun: Due to the motion of Earth around the common center of gravity with the Moon, Earth's center will be above and later below the ecliptic plane by a few hundred kilometers every lunar orbit. This results in a shift of the apparent position of the Sun by about 1.7 seconds of arc - also a very tiny amount. In fact, one often makes the assumption that the Sun's apparent position in the sky relative to the stars is the same everywhere on Earth, but there is indeed a variation larger than the one having to do with the Moon's orbit, about 18 seconds of arc from one side of the Earth to the other (e.g. between north pole and south pole on equinox; this may be called parallax). Icek (talk) 14:46, 30 November 2010 (UTC)[reply]

What does unranked mean in wikipedia's Scientific classification ?

What does unranked mean in wikipedia's Scientific classification ? In the horticulture class I am taking they dicuss Plant Hierarchical Classification. Kingdom,Phylum,Class,Order,Family, Genus, Species. There is no "unranked". Does unranked refer to the plant or the subdivision of the classification. Is wikipedia using scientific classification or something else? —Preceding unsigned comment added by Zoztrog (talk • contribs) 17:12, 27 November 2010 (UTC)[reply]

It is impossible to cleanly fit the Linnaean categories to the evolutionary trees produced by modern evolutionary biology -- there aren't enough levels, and the Linnaean levels don't have any basic underlying biological reality anyhow. "Unranked" means a grouping of species that does not match up with any of Linnaeus's levels. Looie496 (talk) 18:13, 27 November 2010 (UTC)[reply]

As a footnote, my experience is that textbooks tend to run about 10 years behind the leading edge of science. I expect that the textbook presentations will gradually change over the coming decade. In current biology, the Linnaean scheme is definitely on its way out. We can expect the concepts of genus and species to stick around, but above that, pretty much every Linnaean grouping is useless. Looie496 (talk) 18:19, 27 November 2010 (UTC)[reply]

Agreed. While I think that the lagging of textbooks behind the leading edge of science has its upsides (some call it the "bleeding edge" for a reason), what disturbs me is the way in which clearly-dated dogma is still being taught in classrooms. The Linnean hierarchy is still being drilled by rote in many universities, rather than using it as a shared construct with a tangible (if flawed) basis in observation. I hope that university biology faculty take the leading edge into account when they teach, and avoid having students memorize things that will have no utility in a few years. It's hard to teach concepts, but it's rewarding. -- Scray (talk) 21:06, 27 November 2010 (UTC)[reply]

If they're still teaching Linnean classifications, they're probably more than ten years past the leading edge then. That stuff was already well on the way out fifteen-plus years ago when I was at university and it's not like I was reading cutting edge bio stuff at the time. Matt Deres (talk) 14:41, 29 November 2010 (UTC)[reply]

CLONE

THE PROCESS OF CLONING,IS SEXUAL OR ASEXUAL REPRODUCTION —Preceding unsigned comment added by 41.211.232.70 (talk) 17:19, 27 November 2010 (UTC)[reply]

Our cloning and asexual reproduction articles should clear this up for you. Is this homework by any chance? -- Scray (talk) 17:57, 27 November 2010 (UTC)[reply]

See Dolly (sheep). (and please don't write in ALLCAPS) ~A H 1^(TCU) 03:22, 28 November 2010 (UTC)[reply]

Growth of trees

I've searched and maybe I just don't know how to use it well, but I'd like to know, If I make a notch/ scratch/ marking in a tree, will the marking rise as the tree grows? or will it stay the same distance from the ground? This is for research for a book I am writing. thank you!

46.117.94.124 (talk) 21:18, 27 November 2010 (UTC)[reply]

The only part of the tree trunk that is living or growing is the cambium. So the tree trunk just gets fatter with new rings, whilst only the new shoots grow upwards from the top. In other words, a blaze will stay at the hight that it was made. Does that help? --Aspro (talk) 21:33, 27 November 2010 (UTC)[reply]

A tree (and almost all plants) grows from the top, so a mark will stay at the same height. However grass is unusual - it grows from the bottom. Ariel. (talk) 23:09, 27 November 2010 (UTC)[reply]

You can also see this with signs, e.g. street signs, that are put on trees. The signs will remain at the same height above the ground. The tree may eventually even grow around the sign if it's there long enough. There is an example of this not far from the house I'm staying in now. The bark is slowly enveloping the signs. Dismas|^(talk) 00:50, 28 November 2010 (UTC)[reply]

Pics or it didn't happen. -- BenRG (talk) 02:30, 29 November 2010 (UTC)[reply]

I was actually going to ask the same question! I recently moved into a house that has a data cable which is secured to a tree between the street and my house. The anchor is barely half way up the tree and I was wondering if I would have to have it moved at some time in the future. I did suspect the answers above but glad to have it confirmed. Vespine (talk) 23:15, 28 November 2010 (UTC)[reply]