Wikipedia:Reference desk/Archives/Science/2009 June 22

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June 22

Binary stars orbiting a planet

Would it be theoretically possible for a pair of binary stars to orbit a planet that sat between them? For that planet to be Earth-like? NeonMerlin 01:25, 22 June 2009 (UTC)

I don't think that would be stable. If the planet moved slightly towards one star it would accelerate towards it. There are several known stable solutions to the three-body problem, but I don't think that is one of them. It's very similar to L1, which isn't a stable equilibrium point. (It may be exactly like L1, I can't remember what the requirements for ratios of masses are...) --Tango (talk) 01:55, 22 June 2009 (UTC)

Something very close to what is being asked is possible. Although the L₁ Lagrangian point is nominally unstable, it’s possible to find stable periodic orbits around these points (called "halo" orbits), in the restricted three-body problem. Also, although the L₄ and L₅ Lagrangian points aren’t between the two stars, they are stable if the ratio of masses of the two stars is at least about 25, and would be reasonable good places for a planet to hang out without large variations in gravity or light. Red Act (talk) 02:12, 22 June 2009 (UTC)

According to [1], the answer is "sometimes" (specifically, 40% to 50% of observed binary systems). For the case of α Centauri, according to [2], the answer is yes. Someone42 (talk) 04:33, 22 June 2009 (UTC)

I don't believe that's the question being asked. Your links are about forming planets around one member of a binary pair. I believe the original question is whether a planet can exist at a stable point created by the combined gravity of both stars. Dragons flight (talk) 07:52, 22 June 2009 (UTC)

With a binary pair of stars - there must be stable lagrange points at which a planet could exist...in theory, it's a three body problem - and therefore, unstable - but providing the planet has negligable size compared to the stars, it's OK. But that's not quite what's being asked because the planet isn't "between" the two stars - it would be off to one side - and the stars aren't orbitting the planet. But given the right setup, I think the planet could possibly be somewhat earthlike. SteveBaker (talk) 15:21, 22 June 2009 (UTC)

If there is a big difference between the size of the stars (Red Act says above that the required ratio is at least 1:25, which rings a bell), then you could have the planet at L4 or L5 and there is no problem. If the stars are similar in size, which I think is what the OP is thinking of (although I may be wrong), then I'm not sure what stable solutions there are... I know of stable solutions for 3 similar sized bodies (there's a great one where they all follow a figure-of-eight), but not one for 2 similar sized large bodies and one small one. If there were a solution where the planet was just sat at the barycentre (which I think is what the OP is thinking of) then the planet could be Earth like, but it would be a little different due to constant daylight. If the planet rotated relative to the stars (which I actually suspect it wouldn't, I would expect it to be tidally locked) then you would get sunsets and sunrises, but there would be no night inbetween since when one sun sets the other rises. That would have some effect on the weather, but probably not anything that would completely rule out life as we know it. If it were tidally locked, then it is a little more difficult, but with the right kind of atmosphere the temperature variations wouldn't be too bad. --Tango (talk) 16:44, 22 June 2009 (UTC)

About that figure-eight orbit of three equal bodies – I haven't seen anything about how stable it is to small perturbations, or to differences between the masses; have you? —Tamfang (talk) 15:55, 26 June 2009 (UTC)

horsepower and cc

Is there any conversion factor between horsepower and cc of a motor car?? can these be relation in any such numerical figure?? or are these two different things?? —Preceding unsigned comment added by 119.153.21.132 (talk) 06:22, 22 June 2009 (UTC)

They are related, in that all other things being equal, a larger engine produces more power than a smaller one. However, many things are unequal, including bore, stroke, ignition, compression, fuel,... In general, modern engines with higher compression and direct injection get much more horsepower per volume of engine displacement than older, carburetor-based designs. --Stephan Schulz (talk) 06:53, 22 June 2009 (UTC)

There certainly isn't a direct conversion factor - there are too many clever ways of getting more HP from the same cc. To give you a concrete example: My wife's car (a MINI Cooper) has an identical engine (same cc's, same everything) as mine - but mine (A MINI Cooper'S) has a turbocharger, a free'er flowing exhaust system, and different settings on the engine management computer. I don't recall the exact numbers - but mine has about 50% more peak horsepower than hers. There is another version of the exact same car (the MINI One) that's sold only in Europe that's tuned for better fuel economy - it gets even less horsepower than my wife's car - and there are race-tuned versions that get 50% more horsepower than mine...so from the most extreme fuel-efficient version to the race-track version of the exact same car with the exact same engine - you get probably a 2:1 or maybe 3:1 difference in horsepower for the same number of cc's. SteveBaker (talk) 15:17, 22 June 2009 (UTC)

As Stephan and Steve said, there is no direct ratio of cc to hp. However, that doesn't stop people from measuring it. This guy, for example, has compiled an interesting table. It shows cc/hp ratios ranging from 1 (F1 engine) through about 12 to 15 for a production car, up to more than 40 for some unidentified industrial engines. I would be interested to know if there is a theoretical lower limit for this ratio. I guess it must be limited by the speed at which fuel can burn. --Heron (talk) 19:43, 22 June 2009 (UTC)

Is time finite or infinite?

Is time finite or infinite? Bus stop (talk) 07:41, 22 June 2009 (UTC)

The answer to your question isn't known. The solid consensus among scientists is that a Big Bang occurred, so time for sure has a finite beginning. However, there isn't a solid consensus regarding the ultimate fate of the universe. If it turns out that there will be a Big Crunch, then time also has a finite ending, so time is finite. But if there is a Big Freeze, then time will continue indefinitely, and time will be infinite. Red Act (talk) 08:06, 22 June 2009 (UTC)

I'm going to hedge my statement even further. There actually isn’t even a complete consensus as to whether the Big Bang marks the beginning of time. Although I don’t think the idea is widely favored, one possibility under consideration by some cosmologists is that the Big Bang was actually a Big Bounce. Red Act (talk) 08:15, 22 June 2009 (UTC)

I think a big crunch has been pretty much ruled out now. That would require the expansion to be slowly fairly quickly, it is actually measured to be speeding up. --Tango (talk) 14:40, 22 June 2009 (UTC)

Just to confuse things still further it may not even be 1 dimensional at the smallest scales though that's getting very speculative. See Spacetime and Multiple time dimensions. Dmcq (talk) 11:53, 22 June 2009 (UTC)

I agree with everything that's been said already - but I'd say that on balance, the smart money is on time having a finite beginning at the moment of the big bang - and no ending (ie it's infinite) - although when the overall entropy of the universe gets small enough, nothing will be "happening" so it would be hard to measure the progress of time. But there are other possibilities that have not yet been entirely discounted. 15:10, 22 June 2009 (UTC)

Don't you mean when entropy (absence of ability to complete work) gets large enough? Or am I missing something? —Anonymous Dissident^Talk 15:18, 22 June 2009 (UTC)

Actually, I'd suggest that many physicists believe that some form of time and space existed before the Bang, but that our theories and experiments are simply inadequate to understand what came before the apparent singularity. Quantum gravity, whatever form that ultimately takes, will almost certainly eliminate true singularities and create a path to considering space-time before the Bang (though that space-time may be structurally very different than our present reality). Dragons flight (talk) 16:01, 22 June 2009 (UTC)

I think most physicists consider "what happened before the big bang" to be an unscientific question since it is impossible for us to know. The big bang wipes the universe clean, so to speak, so there is no information left from that time for us to observe. If we can't observe something then, from the point of view of science, it doesn't exist. --Tango (talk) 16:48, 22 June 2009 (UTC)

There is a difference between an unanswered question and an unscientific question. No one has evidence of extraterrestrial life, and yet there are many theories for what could exist out there based on our understanding of chemistry and terrestrial life, and someday those theories may be tested. Similarly, there are theories (e.g. Big Bounce, brane collisions, "true" vacuum) for what might have come before the Big Bang. We don't know how to test those ideas today, but that doesn't mean that they will never be tested. For example, most quantum gravity proposals lead necessarily to the conclusion that there was a "before the Bang". If we ever validate one of those proposals as likely correct for the modern universe, then it would be indirect evidence of what the universe may have been like before the Big Bang. Nor is it necessarily true that the Big Bang wipes the universe clean. In the speculative brane collision cosmologies, there could be remnants of before the Bang imprinted on the distribution of dark matter. (That no one has found evidence of this doesn't make it uninteresting to look.) As a physicist, I would suggest that most other physicists see it as a largely intractable question, but not an intrinsically unscientific one. Dragons flight (talk) 17:24, 22 June 2009 (UTC)

Would a better question be "is time as we know it finite or infinite?" A different version of "time" could exist before the big bang and after the big crunch. --Navstar (talk) 00:45, 23 June 2009 (UTC)

Or perhaps just try to figure out how long it took to read this sentence? See Xeno's paradoxes :-). --SB_Johnny | ^talk 01:05, 24 June 2009 (UTC)

Inverted vertical tube manometer

Why the pressures within an inverted vertical tube manometer is being substracted..? —Preceding unsigned comment added by Sreekanth awh (talk • contribs) 08:26, 22 June 2009 (UTC)

Beam of light

Is there an angle that you could cast light at (in a beam) where the circle iluminated would appear the same size at all distances? Say if you shon it at a wall a foot away from you and held a ruler a couple of inches away from your eye the circle of light would appear 4" wide, and it would also appear that if you stood 10 feet away from the wall, due to perspective. Gunrun (talk) 08:43, 22 June 2009 (UTC)

I would think so. The apparent radius of a circle goes like 1/d with d the distance between you and the circle. The actual radius of the illuminated circle is proportional to d. You might want to have a look at http://en.wikipedia.org/wiki/Gaussian_beam

You'd need a "collimation" lens or mirror which makes the incoming light exit in a parallel beam. SteveBaker (talk) 15:06, 22 June 2009 (UTC)

Help with project

My friends and i are trying to do a project in the summer, which would keep us engaged, and at the same time learn something new. We are all first year undergraduate students. The basic idea is this: You have a (computer) mouse which eats light, that is, goes to the brightest area it can. It can do cool things like following the light from a torch, etc. In addition, whenever it collides with an obstacle, it veers back a couple of meters and then zooms off in some other direction. It isn't wired. This is to be done with basic electronics. We are following this book, and you can find the circuit diagram and other details in the link. But the thing is, we are facing multiple problems, and in need of help. The first problem is the IR emitter thingy. The book says the IR emitters (removed from an old mouse) will sense the incoming light, and hence act like the "eyes" of the mouse. It has given clear instructions on how to desolder the emitters and all. My question is, how can IR emitters possibly detect light? Shouldn't you use detectors? This can't be a printing mistake, he's used it several times. Then comes another problem. According to his illustrations, his emitters have 2 terminals, whereas our emitters have 3. What do we do? Worse, our detectors have four! Are there various types of emitters/detectors used in a mouse pcb,and are we stuck with a different kind? If that is the case, would the correct IR emitters/detectors be available in any electronics shop (The major electronics hub in the city is quite some distance away)? I think we are seriously in need of help... Rkr1991 (talk) 09:13, 22 June 2009 (UTC)

You really want to talk with some robot building enthusiasts, there's lots of them around on the web and they're very happy to help with newcomers. You might even find a club nearby if you're lucky. Just google for things like robots builders enthusiasts. Dmcq (talk) 12:13, 22 June 2009 (UTC)

It would be nice if someone could give me a direct link of one such site that's really good... Rkr1991 (talk) 13:07, 22 June 2009 (UTC)

I guess from what you are saying that your university does not have a robotics course or club, that would be best but you can always found one. You really do need to get to grips with searching the web, for instance you have to look at electronics catalogues like Maplin if you are going to strip things down and use the bits. Did you try a google and look at the top few entries for clubs? Wikipedia has articles on robotics from Portal:Robotics but it tends to be more institutional stuff rather than how-to and friendly. You have to suck it and see and sometimes just put the blue smoke down to a learning experience. Dmcq (talk) 14:14, 22 June 2009 (UTC)

Well, there are two kinds of mice:

1. Mechanical (the kind with a ball underneath)
2. Optical (the kind with LED's underneath)

• The mechanical kind do have light sensors in them - but they'd be hard to use. If you pull the mouse apart, you'll see that the ball rolls against a pair of rollers - and if you dismantle the mouse enough - you'll see that each roller connects to a slotted disk that sits inside a U-shaped black box. That box has an LED light and a light sensor - one on each side of the disk that count the number of times the disk interrupts the light beam. But to use the light sensor in your application, you'd have to somehow get it out of the black plastic enclosure without damaging it. Maybe you can do that - but it's not something I've ever tried.
• The optical kind uses a camera that looks down at your desktop and watches it slide past - measuring the speed that it moves. This is a very low-resolution camera (I vaguely recall it's a monochrome 64x64 pixel gizmo) - and the lens it looks through is focussed to produce a sharp image at a distance roughly equal to the thickness of the little rubber feet on the mouse! So you aren't going to be taking any photo's with it! However, technically - I guess you could remove the lens and let it see an amazingly blurry view of the world - which would certainly give you an overall idea of the light level. However, interfacing to the camera might take some fancy electronics...and I don't know how fancy you want to get!

I've built a robot just like you're describing - but I cheated and used Lego Technics - which lets you build such a thing in about an hour - and get the software working in a couple of hours. It can certainly be done. My son and I actually built two of them - put IR LED's on the top of each one and programmed them to play "Tag" - one robot is programmed to seek light and the other to avoid it - they each have a 'skirt' with microswitches as a 'bump' sensor - when one robot hits the other, it changes from seeking light to avoiding it and vice versa. The one that's set to seek light doesn't start moving for 10 seconds in order to give the other one time to escape - and (theoretically) they change sides. It's actually a little bit more complicated than that - they send each other messages via IR to be sure it was a legal "Tag" and not a collision with the wall or something. Anyway - it's hilarious to watch - and an interesting programming challenge to have two people each program their own bot and run it as a competition. SteveBaker (talk) 15:04, 22 June 2009 (UTC)

By the way my favourite book on this sort of thing is Hardware Hacking Projects for Geeks by Scott Fullam. Rather self identifies I guess :) With projects like for instance how to hack a furby (rather a difficult one that). or turn yourself into a Borg. Dmcq (talk) 21:04, 22 June 2009 (UTC)

We have dismantled a mechanical mouse, and got the two IR emitters and detectors out of the PCBs safely. Now, thanks for all your help, but it would be nice if someone could directly address the two questions i asked, one whether if we should use emitters or detectors, and the other, their availability and terminals (please refer to my posts above). In the meantime, I'll just hunt for some Robotics sites and clubs.... Thanks for your help everyone :-) Rkr1991 (talk) 05:14, 23 June 2009 (UTC)

Where are you, Spinningspark ? I thought you were the resident electronics specialist... Rkr1991 (talk) 05:22, 23 June 2009 (UTC)

Anyway to your question, I'd have expected four IR emitters and detectors in a mouse two for each wheel. The emitters are LEDs that emit the IR and I'd expect them to be clear and only have two legs each. If they have put two detectors in a single package then that may explain why you've found only two detectors and they have three legs. Some detectors actually do have three legs but that would be overkill for a mouse. The detectors are probably red to filter out other light. The detectors are what you want to detect the light but you probably want both the emitters and detectors, it can be quite hard enough working with the changing ambient light never mind the occasional TV controller. It is common to modulate the emitter and detect a modulated light at the detector to cut out such interference - also LEDs can be run brighter being pulsed. Dmcq (talk) 09:03, 23 June 2009 (UTC)

By the way I'd guess an IR emitter probably can be used as a detector but it sounds rather a strange idea to me when you've already got good detectors designed for the job. Dmcq (talk) 09:08, 23 June 2009 (UTC)

No - you can't use an LED as a light detector! The IR emitters in a mouse don't look like regular LED's - they are encapsulated...but if you need to detect light - the emitters aren't really of any interest. But if you have the mouse's light detectors separated out without damaging them - then you should be able to use them to detect sunlight and things like your TV remote...but many other kinds of light source don't produce enough IR to be detectable - so your robot might not be interested in seeking things like flashlights. Also, I have no idea what the sensitivity of the mouse's detectors are - they might require quite a lot of light to make them trigger because they're designed to operate with an IR LED about 5mm away - and they are also working hard to exclude the possibility of sunlight or other stray light sources from triggering the thing accidentally - so it may be designed to make sure that what you're trying to do won't work! Finding the manufacturer's data sheet for the part you're using would help an awful lot! If it has any serial numbers or anything on it - then typing that number plus "mouse" into Google would probably turn up what you need to know. SteveBaker (talk) 18:35, 23 June 2009 (UTC)

OK - let's try to answer the actual questions here.

• There is no way a light emitter can be used as a detector. Your book is wrong. If it's not a misprint then the author is an idiot...which is actually possible - many of these books are written by enthusiastic amateurs who are sloppy with scientific terms.
• So what you want is the detector. The two slotted disks in the mouse are used by the electronics to figure out by how much the wheel has turned - and in which direction. With a simple IR emitter and detector pair - one on either side of the wheel - you could count the number of dark/light transitions are picked up by the detector and know how many slots have passed by - and therefore, by how much the mouse has moved in that axis. However, that's not enough - you've also got to figure out the DIRECTION it's moving (forwards/right or left/backwards) - and for that you need a second detector positioned a little off to one side of the first one. By noting which detector triggers the dark/light transition first - you can figure out the direction as well as the distance.
• Hence, each axis of the mouse needs TWO detectors - but you only really need one IR light source. So the black block on the side of the slotted disk that contain the emitter will have two pins - power and ground - probably just like a regular IR LED - and the black block on the opposite side which contains the detector will have three pins - probably a common ground wire and two pins where the light brightness value appears as a small voltage...or possibly as a variable resistance to that common ground.
• You'll have to figure out which pin of the detector is ground. If you can trace where the three copper tracks on the circuit board went - then probably you'll find that the ground pin connects to a dozen places in the circuit - but the two other outputs each go to just one or two places. Try to trace where they go - see if there are any resistors or anything connected up along the way.
• So I guess you could try to get a reading on a multimeter on the most sensitive range - see if putting the detector in bright sunlight produces a voltage or a change in resistance...that'll give you an idea of what circuits to use to interface to this beast.

But still, the best advice is to try to find a data sheet for the device...there is no substitute for knowing what the heck the thing does in detail.

SteveBaker (talk) 00:43, 24 June 2009 (UTC)

The dismantled mouse in question is more than 20 years old, there's no way I can get a data sheet for that old thing ! I'll try bread boarding the circuit now with Steve's advice using the detectors, and I'll get back if I have any doubts. And thanks for the help, everyone... Rkr1991 (talk) 08:03, 24 June 2009 (UTC)

Steve, our article LED as light sensor and the internet disagree with you. 76.11.230.33 (talk) 05:03, 25 June 2009 (UTC)

Well I just took a red LED out and tried it to see if it could be used as a detector, and I think it pretty definitely can't. I thought there was a chance it would work but no it doesn't look that way. There's nothing like checking with the real world to get yes/no answers. Dmcq (talk) 10:01, 24 June 2009 (UTC)

I've tried using an LED as a detector, and it does work. However, it's not very sensitive at all. Using just a multimeter, it could differentiate between blindingly brilliant light, and merely dazzling. Bunthorne (talk) 04:48, 26 June 2009 (UTC)

I think you're all making a mistake. The emitters maybe be LEDs, but that doesn't make them equivalent to normal red LEds. These IR LEDs might detect light with a bit more precision. I'll check it out and get back to you. Rkr1991 (talk) 12:39, 27 June 2009 (UTC)

Wisdom Tooth Extraction After effects

This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis, prognosis, or treatment recommendations. For such advice, please see a qualified professional. If you don't believe this is such a request, please explain what you meant to ask, either here or on the Reference Desk's talk page. Algebraist 09:47, 22 June 2009 (UTC)

This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis or prognosis, or treatment recommendations. For such advice, please see a qualified professional. If you don't believe this is such a request, please explain what you meant to ask, either here or on the Reference Desk's talk page. Algebraist 09:47, 22 June 2009 (UTC)--~~~~

Reaching the lowest possible temperature

How can we reach the lowest possible temperature? If something is 0 C, I can put something -5C and cool it a little. But since the absolute zero temperature does not exist, how can be cool something until the lowest existent temperature, there is nothing cooler than it...--Mr.K. (talk) 09:56, 22 June 2009 (UTC)

Bringing an object in contact with a cooler object is not the only way of cooling it. see for example http://en.wikipedia.org/wiki/Laser_cooling

Or even have a look at fridge, have you wondered how that works? Dmcq (talk) 12:25, 22 June 2009 (UTC)

For all that though, an object reaching absolute zero is not possible, because the laws of thermodynamics will insist that some energy enter your sample, which will raise the temperature above 0K. You would need a way to perfectly isolate your sample, and that is not possible. How close you get to 0K will depend on that factor. 65.121.141.34 (talk) 14:01, 22 June 2009 (UTC)

Although we have got remarkably close to it; see absolute zero.--Shantavira|^{feed me} 15:28, 22 June 2009 (UTC)

Even more remarkably, using the technical definition of temperature given by statistical thermodynamics, it is actually possible to have objects with negative absolute temperatures -- and such things have even been created in the laboratory. The way it works out, though, is that negative temperatures are hotter than infinity, and such a system will always lose heat to any system at a positive temperature that it is in thermal contact with. It remains true that no system can be at perfect absolute zero -- systems at negative temperatures cool by dropping to negative infinity and then dropping from positive infinity. Looie496 (talk) 16:58, 22 June 2009 (UTC)

Telescope on the moon

OK, forgive my lack of comprehension of science. As I understand it, because light moves extremely fast, but not infinitely quickly, whatever you see took place a fraction before you actually noticed it. And furthermore, the further away the incident is, logically, the longer ago the incident took place. So I wondered, (Q1) what happened if someone had an improbably powerful telescope and was on the moon, looking at Earth. How long ago would whatever he saw actually have taken place?

Q2 What about if the person with the telescope was even further away, say in some high-tech capsule that allowed him to survive and also get a good view from Alpha Centauri?

Q3 How far away would you need to be to see, say, the Romans invading Britain?

If there's a flaw in my logic, please point it out. --Dweller (talk) 10:13, 22 June 2009 (UTC)

The moon is about one light-second away, so an observer sitting there "now" (scare quoted because that doesn't quite make sense: see relativity of simultaneity) would see events on Earth that took place about a second ago. Similarly, Alpha Centauri is about four and a half light-years away, so an observer there would see events four and a half years ago. To see the Romans invading Britain, you'd have to be about two thousand light-years away (more or less depending on which invasion you want), perhaps somewhere near NGC 6826. Algebraist 10:20, 22 June 2009 (UTC)

Thanks for that (and that's a very pretty picture in your link). Is there some kind of 2D map of [our bit of?] the universe that I can look at that'd show me where we're talking about, to contextualise with other objects even I might have heard of? And, if I might throw yet another question, can I assume that an object c.2,000 light years away is about 500 times further away than something about 4 light years away, or is it some kind of non-linear scale? I'd assume it is, but one thing I've learned is not to make too many assumptions in ignorance. --Dweller (talk) 10:39, 22 June 2009 (UTC)

As the article states, 'light-year' is just a fancy (and shorter) way of saying '9,460,730,472,580.8km', so it's a perfectly linear scale (up to the usual quibbles involving relativity). The best map of the universe I know is this (but note that it is not linearly scaled). I can't find a decent linearly-scaled map of nearby space at present. Algebraist 10:50, 22 June 2009 (UTC)

Interesting, thanks. I can't really understand the map properly (I get that it's a log scale) - why is it rendered as a column, not (as I expected) a square? Presumably, the real universe is roughly spherical from whatever its central point is... so I expected a 2D rendering to resemble the 2D renderings we do of Earth. --Dweller (talk) 11:17, 22 June 2009 (UTC)

I think it's that shape largely so it makes a good wallchart, though the paper may contain other reasons (I read it a while ago and can't remember). The question of the Shape of the Universe is a complicated one on which I am not competent to pronounce. Algebraist 11:43, 22 June 2009 (UTC)

Maps of the Earth's surface usually only attempt to show the two dimensions of longitude and latitude. Drawing a map of the universe presents the challenge of presenting the three dimension of right ascension, declination (the astronomical equivalents of longitude and latitude) and radial distance in a two-dimensional format. The solution adopted in the case of the "column" map is to simply throw away one dimension - it does not attempt to represent declination. The vertical axis is, as you say, a logarithmic distance scale and the horizontal axis is right ascension (there is a further simplification for objects within the solar system which do not have a constant right ascension - they are placed at an arbitrary horizontal co-ordinate a horizontal co-ordinate that corresponds to their right ascension at a specific date and time). So, for example, the Andromeda Galaxy (M31) is shown almost directly above the Small Magellanic Cloud (SMC) because they have similar right ascensions, even though their declinations are very different - the SMC at a declination of -72^o is a southern sky object, whereas M31, at a declination of 41^o, is a northern sky object. To see how the universe is mapped onto the rectangular column, imagine that all the objects shown are projected onto a disc around the celestial equator centered on the Earth; cut a small hole in the middle of the disc and make another cut from this hole out to the edge of the disc; then fold the sides of the cut disc together (like closing a fan), until you have the column shown. Gandalf61 (talk) 12:06, 22 June 2009 (UTC)

They're not shown at arbitrary right ascension, they're shown as they were at 0448 UT on the 12th of August 2003. Algebraist 12:17, 22 June 2009 (UTC)

Ah, yes - thank you - fixed in my post above. Gandalf61 (talk) 12:28, 22 June 2009 (UTC)

The galaxies from the Sloan Digital Sky Survey are actually close to the equatorial plane, so the top part of the diagram is a slice rather than a projection. As for why it's a column instead of a disc: if you just plot a scale model of the equatorial plane then a lot of small-scale structure is invisible because each pixel of the image is millions of light years across. If you logarithmically scale the radial distance then you can fit in small-scale and large-scale structure, but shapes are distorted. To avoid distorting shapes you have to rescale the angular direction also, and it turns out the rescaling that's needed is to make it a fixed width at all distances. This is familiar as the fact that the complex logarithm (which is conformal, i.e. "shape-preserving") repeats at a fixed interval of 2π in the imaginary direction. The mapping they used is actually the complex logarithm with the imaginary part plotted horizontally. The authors compare it to this famous New Yorker cover. -- BenRG (talk) 13:57, 22 June 2009 (UTC)

It's probably impossible to see the romans invade britain now though, unless you're at the planet already, because to get there to see it you'd have to travel faster than than the speed of light. The invasion of the romans is moving away from our planet at light speed! Gunrun (talk) 10:46, 22 June 2009 (UTC)

Yes, I got that. If you could travel faster than light, I suppose you wouldn't need to travel to NGC 6826, you could just go to Kent! --Dweller (talk) 11:14, 22 June 2009 (UTC)

Assuming the age of Earth to be 5 billion years, at what distance in the universe would one have to be to see Earth being "born," and is there enough distance in the universe to get far enough away from the Earth to get this view, or is the universe just too small for this? And if the universe is too small to find a sufficiently distant vantage point from which to see the Earth being born, where does the light go to when it runs out of "universe" in which to travel? In fact, if I can ask one more question, where in general does light go when it reaches the outer limits of the universe? 12:07, 22 June 2009 (UTC)

Now we're into large enough scales that general relativity becomes unavoidable. To be looking at an event here 5Gyears ago, an observer would have to be five billion light years away in terms of light travel distance, but not in terms of more commonly used measures such as comoving distance or cosmological proper distance. See Distance measures (cosmology) for a discussion. Light does not reach the outer limits of the universe, because no such limits exist. Algebraist 12:13, 22 June 2009 (UTC)

You'd also need a pretty big telescope to see these things. To see the Romans invade Britain at 25 frames per second and be able to see individual Romans, the width of the mirror would have to be um well I haven't worked it out but I'd guess quite a bit bigger than the earth. Dmcq (talk) 12:34, 22 June 2009 (UTC)

We need Stephen Hawking here. Is Stephen Hawking in the house? Bus stop (talk) 12:37, 22 June 2009 (UTC)

A back-of-an-envelope calculation, ignoring all possible technicalities, suggests we're talking about an optical telescope about the size of the orbit of Pluto. Algebraist 12:40, 22 June 2009 (UTC)

Perhaps rather than one stationary large mirror we could have a series of smaller, rapidly moving mirrors that gather up light in the course of their journeys, interpolating data at convenient intervals. Bus stop (talk) 12:52, 22 June 2009 (UTC)

The light that reached that galaxy would also be a different shape, right? They'd have to decode it in some complicated way as well, right? ~ R.T.G 13:14, 22 June 2009 (UTC)

Chaps, you sound like a group of ancient scientists laughing about how many horses you'd need to tether together to get an object to move fast enough to fly. Any technology good enough to be able to get a clear view from the surface of a star is probably capable of making a telescope that doesn't depend on massive mirrors. --Dweller (talk) 13:20, 22 June 2009 (UTC)

Well, yes, but that wouldn't be nearly as much fun. Algebraist 13:51, 22 June 2009 (UTC)

That's big. But it is an opportunity. Any sufficiently advanced civilization then that was interested in what was going on round them would build telescopes the size of the orbit of Pluto. It must be possible to spot one of those I'd have thought :) Dmcq (talk) 13:48, 22 June 2009 (UTC)

One way to avoid improbably large lenses and/or mirrors would be to use gravitational lensing. Since gravity bends light - if you position your camera on the opposite side of (say) some suitably gigantic black hole - then you can use it as a lens. Of course you've just exchanged the problem of building a lens the size of the orbit of pluto for the inconvenience of having to fly your camera out to the right place (which might take a few thousand years) - and wait another few thousand for the photos to get back...but it would certainly be cheaper if you're not in a hurry. SteveBaker (talk) 14:47, 22 June 2009 (UTC)

Even cheaper if you don't pay for one hour processing. NotAHen (talk) 22:35, 22 June 2009 (UTC)

re a map of the universe, it seems mandatory that this should be linked from here. Jørgen (talk) 15:38, 22 June 2009 (UTC)

As the Pluto-orbit-sized telescope would be at NGC 6826, we would need to have another 'here' in order to be able to spot it, wouldn't we? If it were to replace the orbit of Pluto, then "they" would be using it for something other than observing Roman invasions. - KoolerStill (talk) 05:43, 23 June 2009 (UTC)

And there would be all kinds of problems with a telescope that size. For one, you would have to aim it at England, taking into consideration that England is spinning around the earth's axis at about 1,000 feet per second, the earth is orbiting the sun at about 120,000 feet per second, and that is only the earth moving, your telescope is probably moving as well. 65.121.141.34 (talk) 13:19, 23 June 2009 (UTC)

Apparent contradiction?

According to the article on Thymine, the molecular formula of this nucleoside is C₅H₆N₂O₂. However, in the image to the right I can only see two hydrogens and no carbons at all. Where are the four missing hydrogens and the five missing carbons? --83.34.187.22 (talk) 12:00, 22 June 2009 (UTC)

See skeletal formula. Algebraist 12:04, 22 June 2009 (UTC)

Rain in the East Coast, USA

Is there a scientific reason why the East Coast is getting so much rain in the past few weeks? --Reticuli88 (talk) 13:23, 22 June 2009 (UTC)

The Gulf Stream? El Nino? Weather patterns converging near a subtropical low and jet stream? ~AH1^(TCU) 00:40, 26 June 2009 (UTC)

Farthest blue-shifted object?

Large-scale movement of the universe is dominated by the metric expansion of space, making further objects recede faster. But what is the farthest object that does actually travel towards us? More generally, what it the object with the largest difference between expected and observed red shift (either way)? --Stephan Schulz (talk) 14:59, 22 June 2009 (UTC)

I don't think anything outside the Virgo Supercluster (or Local Supercluster) is blue-shifted. If it is, then it's just a fluke. According to our article on the Virgo Cluster (which is the largest cluster in the local supercluster), its gravity only slows our recession from it by 10%, so nowhere near enough for us to be moving toward it. So I suspect the farthest blue-shifted galaxy will either be in the Local Group (in which there are plenty) or in a group very nearby in the Local Supercluster. --Tango (talk) 16:54, 22 June 2009 (UTC)

We have several blue-shifted galaxies listed in the M81 Group. I can't find any blue-shifted galaxies in other nearby groups, so I'm going to put my money on the furthest away one in that group, which I can't determine (the distances don't seem to be listed anywhere, maybe because they aren't known very precisely). --Tango (talk) 17:01, 22 June 2009 (UTC)

The NASA/IPAC Extragalactic Database (NED) is searchable, but I'm not familiar with all the options. Using the "Advanced All-Sky" search I was able to get a list of 7227 blue-shifted objects (takes a while to load), but I don't see a way to get it to include distances to each object. Is anyone familiar enough with this database to tease out distances to each object? 152.16.16.75 (talk) 01:49, 27 June 2009 (UTC)

Colours of night sky objects?

Is there an online database that lists the apparent or observed colours of night sky objects to the human eye using some sort of scientific scale? For example, Mars and Betelgeuse are redder. I know Rigel is blue in absolute terms, but is it blue to the naked eye from a human perspective? I want a version of the Hertzsprung-Russell_diagram using observed data from the human eye on Earth rather than absolute data out there in space. --Sonjaaa (talk) 16:53, 22 June 2009 (UTC)

I saw something like this a while ago but I can't find it now. If you want to search around a little, try looking up the photography technique using long exposures and movement so that stars appear as smears which makes their apparent color much more noticeable than when they are just points. Sifaka ^talk 18:37, 22 June 2009 (UTC)

The Cambridge Star Atlas [3] classifies stars by their visual appearance and presents them in color right on the charts. Will Tirion also did Sky Atlas 2000.0 [4]. It's open to interpretation as different eyes respond differently to very low levels of light, but they're the best I know of. Acroterion _(talk) 21:19, 22 June 2009 (UTC)

Does a star or planet's B-V colour index represent its apparent colour to a human observer on Earth?--Sonjaaa (talk) 00:33, 23 June 2009 (UTC)

Only in the broadest terms. Rigel and Betelgeuse are useful comparisons, both being in Orion (constellation), with Rigel (to me, anyway) seeming bluish-white and Betelgeuse rosy. Acroterion _(talk) 03:46, 23 June 2009 (UTC)

Mold on mushrooms

I often make spaghetti with different kinds of spaghetti sauces, and it seems that the mushroom sauce gets mold much more often than the tomato-only sauce. I can't imagine why, since both are the same brand, bought in the same amounts, stored in the same type of jar in the same place in the same refrigerator; the only difference between the types is the presence of cooked chunks of mushrooms. Obviously the mushrooms and the mold are different species, so it's not as if the mushrooms are manifesting themselves as mold, but realistically is there any possibility that the mold grows more readily on the mushrooms themselves, and/or that the presence of the mushrooms makes the sauce a better host for mold? Nyttend (talk) 17:49, 22 June 2009 (UTC)

My guess would be that the acidity of the tomato sauce retards mold growth. The mushrooms floating at top might provide a low-acidity area for mold growth to start. It could also be that the recipes are different regarding the sauce itself, for example one saltier than the other. Or, well, who knows? Looie496 (talk) 18:07, 22 June 2009 (UTC)

(Post e.c.)There are quite a few fungal species that specifically parasitize other fungi including mushrooms, the genera of which are escaping my memory right now; however, I strongly doubt that those are the fuzzy culprits in this case. Most likely what is infesting your sauce is similar to what infests your fruit, bread, cheese, or other items. My intuition is that your mushroom sauce has a more favorable environment for growth. The process of preparation or different ingredients may change the pH or solute concentration of the sauce to a more favorable range for instance or if it includes cooking may deactivate some of the natural preservative compounds in the tomato sauce. It could also very well be that mushrooms may be a good host for mold the same way that certain items tend to fuzz over before others do. Hope this helps. Sifaka ^talk 18:26, 22 June 2009 (UTC)

I can confirm that the mold on the mushroom sauce looks the same as the mold on the non-mushroom sauce. What I meant by "cooking" was the preparation that Hunt's puts into before I buy it — between opening the can and actually preparing the pasta, the only thing I do with the sauce it putting it into the jar. Nyttend (talk) 21:16, 22 June 2009 (UTC)

chicken eggs

Are chicken eggs sterile (free of bacteria) before the shell is opened? 65.121.141.34 (talk) 18:31, 22 June 2009 (UTC)

There are bacteria all over the surface of the shell which are there before the egg is opened. According to Egg (food)#Contamination, if the egg is laid by a healthy chicken then yes they are sterile inside, but according to this source, some very few eggs do have bacteria in the interior. Sifaka ^talk 18:48, 22 June 2009 (UTC)

Oops accidentally posted here.. removed!--87.113.12.133 (talk) 20:00, 22 June 2009 (UTC)

I see, the article says that about 1/30,000 chicken eggs in the US is contaminated with salmonella. Now I can accept the risk of eating raw cookie dough (homemade, not sold in the store, because who know how many eggs are in 1 batch). 65.121.141.34 (talk) 20:09, 22 June 2009 (UTC)

As a matter of fact - there has just been a massive recall of raw cookie dough for precisely this reason. SteveBaker (talk) 00:19, 23 June 2009 (UTC)

The recall was due to E. coli O157:H7, which is a little surprising. (See, for example, this discussion.) -- Coneslayer (talk) 11:31, 23 June 2009 (UTC)

Eradicated bacteria on a space station

If we were to create a substantial population in a space station (something like Babylon 5) and ensured that the construction and all people sent there were free of all bacteria/viruses would it remain sterile indefinitely or would the human flora be likely to eventually mutate into forms that cause illness? Thanks :-) --87.113.12.133 (talk) 19:58, 22 June 2009 (UTC)

Not quite addressing the original question here, but I would like to point out that the people who lived on such a space station long term, and eventually left would probably have the equivalent of a compromised immune system, and it would not be advisable for them to leave. 65.121.141.34 (talk) 20:06, 22 June 2009 (UTC)

Human gut flora already causes illness if it colonises places other than the gut. Always wash your hands - especially in space. 86.4.190.83 (talk) 20:48, 22 June 2009 (UTC)

E.coli which is part of normal flora has several serotypes which are major causes of food poisoning. I'm not 100% about what that actually means, but I think it convinces me that even if those serotypes didn't make it onto your station, it wouldn't remain sterile indefinitely . The debate however is for how long, if you really managed to get a completely sterile human population it could possibly be centuries or millennia before anything mutated significantly enough to cause illness. To get a bit more sci fi on the idea, on your station you'd also have to include something to sustain a human population for generations. This means you'd probably need many other organisms to take care of nutrition and decomposition. I find it hard to imagine an environment which is can sustain a human population for generations but is completely sterile. I think you'd quicker develop a population with a very poor immune system, if a bug finally did mutate or get on board, it would wipe you out most of your population.Vespine (talk) 23:15, 22 June 2009 (UTC)

"Normal human [gut] flora" are bacteria, so your premis is inconsistent: you cannot be bacteria-free and also have normal human flora. Separate;y from that, I'm not clear on how you would eliminate all bacteria and virii from a small human population in the first place. But let's assume you magically eliminate all bacteria and virii, and then re-introduce a completely-controlled set of flora as would be needed for human health. Then, you will probably get evolved pathogens as the flora mutate (as in the earlier answer.) You might also get "spontaneous" virii, as the human genome incorporates virus DNA that has been inserted by retrovirii, and it is possible that mutation or some other replication error would "liberate" a virus particle. It only takes one. -Arch dude (talk) 01:34, 23 June 2009 (UTC)

Viruses, with respect, and confirmed by Collins English Dictionary, The Oxford English Dictionary and Dictionary.com. 86.4.190.83 (talk) 07:09, 23 June 2009 (UTC)

virii would be the plural of virius (assuming that virius is o-stem masculine). —Tamfang (talk) 16:15, 26 June 2009 (UTC)

Thanks all. I'm particularly interested in the fact DNA can spontaneously create a virus, I'll have a further look at that --87.113.128.46 (talk) 22:15, 23 June 2009 (UTC)

star diameter

Do we know of any stars whose diameter is larger then the orbit Pluto would make (at maximum distance, assuming its eccentricity was 0.0000)? What would be the angle we would be able to see if it was at the distance of Alpha Centari (sp?)?65.121.141.34 (talk) 20:53, 22 June 2009 (UTC)

Guess: No. Too big. No star could be that big. (Just a guess.) Bus stop (talk) 20:57, 22 June 2009 (UTC)

Any star even remotely that big would have long since become a black hole. To be that big and not be a black hole, you'd have to have a density of less than 3 kilograms per cubic metre (about twice that of air), and I'm pretty sure you couldn't make a star with such low density. Algebraist 21:05, 22 June 2009 (UTC)

And for the easy part: an object that big at the distance of Alpha Centauri would have an angular size of about 1.25 arcminutes, twelve times smaller than the full moon. Algebraist 21:07, 22 June 2009 (UTC)

Interesting, I got 0.98', did you use Pluto's aphelion distance or something? --Tango (talk) 21:15, 22 June 2009 (UTC) No, I lost a factor of two, it's more like 0.49'. --Tango (talk) 21:18, 22 June 2009 (UTC)

I used the aphelion, as suggested by the OP's 'at maximum distance'. Pluto's aphelion is about 50AU, so the diameter is 100AU. Alpha Centuari is about 4.4 light years away, which according to Google is about 278255AU, so the angle is 100/278255 radians, which (according to Google again) is about one and a quarter arcminutes. Algebraist 21:23, 22 June 2009 (UTC)

I was right the first time, I actually lost two factors of two in different directions. The difference is simply because I used the semi-major axis. I didn't read the question carefully enough! --Tango (talk) 21:26, 22 June 2009 (UTC)

What were you doing with another factor of two anyway? Algebraist 21:28, 22 June 2009 (UTC)

I guess strictly speaking it was the same factor of two twice. I managed to get very confused about whether I was using the radius or diameter of Pluto's orbit. --Tango (talk) 21:38, 22 June 2009 (UTC)

(ec) I believe Antares is one of the largest stars we know of and its diameter is between those of the orbits of Mars and Jupiter, so quite a long way short of Pluto. I don't know if it is impossible for a star to be much bigger than that, but I would be surprised if any were a big as Pluto's orbit. Pluto's orbit at the distance of Alpha Centari would span just under 1 arcminute, if my calculations are correct. That's about the angular diameter of Venus when it is at its closest. I've just fixed an error, I get about half an arcminute. --Tango (talk) 21:15, 22 June 2009 (UTC)

We have List of largest known stars, of course. If the data is correct, VY Canis Majoris's diameter is twice that of Betelgeuse and four times that of Antares, and it would extend out to about the orbit of Saturn. Still a way to go to Pluto, though (and the more boring theory puts it at only 600 solar radii). --Stephan Schulz (talk) 21:34, 22 June 2009 (UTC)

I took a look at Pluto, Stellar classification, and List of largest known stars and found some cool stuff:

1. Star diameter is measured in solar radii (R_☉) (1 R_☉ = 69,550 km = 0.004652 AU)
2. Plutos orbit: 4,436,824,613 km (about 30 AU) at perihelion and 7,375,927,931 km (about 49 AU) at aphelion
3. A star would therefore have have to be 6,379 R_☉ to extend to the closest point to the sun in Pluto's orbit, and 10605 R_☉ to extend to its furthest.
4. The largest known star (though that size is disputed), is the hypergiant VY Canis Majoris, at between 1,800 R_☉ and 2,100 R_☉, not even close to the orbit of Pluto, but extending to somewhere around the orbit of Saturn.

Stuff I didn't know yesterday. Hope that helps! – ClockworkSoul 21:59, 22 June 2009 (UTC)

Small stars/planets

In reading about small stars (due to an earlier question here), I found this quote: "The relatively puny body weighed in at 96 times Jupiter's mass - above the threshold of 75 Jupiter-masses required for a bona fide star, which must also burn hydrogen." So, are there hydrogen-burning planets, considered planets simply because they are less than 75 Jupiter-masses? -- kainaw™ 22:19, 22 June 2009 (UTC)

Objects around the boundary between stars and planets are called brown dwarfs. There is no generally agreed upon definition of where to draw the line between planets and brown dwarfs and between brown dwarfs and stars. What elements they fuse is a key factor in most proposed definitions, but it doesn't seem to be quite that simple. --Tango (talk) 22:21, 22 June 2009 (UTC)

Also, once an object begins to fuse hydrogen, its radius expands from the outward pressure. Take a look at brown dwarf, in particular the sections "Distinguishing high mass brown dwarfs from low mass stars" and "Distinguishing low mass brown dwarfs from high mass planets". – ClockworkSoul 22:24, 22 June 2009 (UTC)

Thanks for that article, Kainaw. Bus stop (talk) 22:26, 22 June 2009 (UTC)

In a sense, this is the wrong place to ask this question because it is another one of those problems that has more to do with the English language than with science. Just as it was highly contentious when Pluto was "demoted" from plantary status - it might be just as bad to promote a super-Jupiter exoplanet to "star" status (although fans of the 2010 sequel to 2001 might think otherwise!). It seems kinda wrong to call the second largest object within its solar system a "dwarf" and something that's probably a rather pretty pinkish-purple color "brown". But it's just linguistics - there is certainly a continuum of objects out there from things that are clearly NOT stars to things that clearly ARE - somewhere between those two extremes there is something which will allow debate on the subject to rage on for years! SteveBaker (talk) 00:14, 23 June 2009 (UTC)

In my opinion, this is a question about scientific terminology, so it belongs in science as much as language. We get many questions about scientific terminology and I feel that pushing them into a desk that, as a whole, is not expected to be watched by many people with a scientific background will cause the questions to go unanswered. -- kainaw™ 11:27, 23 June 2009 (UTC)

I don't disagree - which is why I said "In a sense..." - rather than "This question doesn't belong here - I've moved it to the language desk for you". But the answer is the same - it's a completely arbitary choice made by people who used to think that there was such an obvious distinction between a "star" and a "planet" (or between a "planet" and a "rock") that no detailed definition was required. But as astronomers find all of these exotic objects out there that blur the boundary - they are faced with a linguistic nightmare. Hence they erect arbitary 'bright line' definitions that are never entirely satisfactory - but at least give you an idea of where to look up information and how to speak clearly to some particular audience. However, there is rarely any actual science involved in making these distinctions - it's mostly about how to define the term clearly without having to rewrite half a million textbooks. (Sadly - that's where they went wrong with Pluto.) So there isn't a scientific answer to your question - the answer is "well, that's how the words are defined"...it's unsatisfying - but true. SteveBaker (talk) 13:06, 23 June 2009 (UTC)

There wasn't really a good solution with regards to Pluto, though. The textbooks needed to be rewritten to remove Pluto as a planet, rewritten to add several other objects as planets or left inconsistent. Any of those options would have really annoyed a large group of people. --Tango (talk) 15:18, 23 June 2009 (UTC)

Talk:Liquid crystal display

There are accumulating unreplied remarks on that page and, of course, there is probably a bunch of you who could answer them. (is mainstream isn't it?) ~ R.T.G 22:32, 22 June 2009 (UTC)

Feel free to go there and recommend that people direct questions that are unrelated to the business of writing that article to us here at the science ref desk. But we shouldn't create little side-branches of the Ref Desk in odd little places like that. Article talk pages are intended to be forums where you talk about the business of writing about the article - they aren't really places for answering questions about the subject matter of the article. SteveBaker (talk) 22:58, 22 June 2009 (UTC)

I think they are mostly concerning the content of the article but they are also very technical posts and unanswered for some time (stuff like Mother glass and "Wasn't that invented by so and so?"). ~ R.T.G 00:14, 23 June 2009 (UTC)

A better place to note stuff like this about an article is probably the discussion page of the WikiProject that encompasses that article. Tempshill (talk) 01:31, 23 June 2009 (UTC)

The OP doesn't mention their own (speculative) question that has gone unanswered at Talk:Liquid crystal display. I have added a reply that I hope will satisfy. Cuddlyable3 (talk) 14:41, 23 June 2009 (UTC)

I think the lead was very technical and even wrong when I asked that question. Very good, Cuddlyable3, thank you. I believe that is the first response to a post in a year. ~ R.T.G 16:37, 23 June 2009 (UTC)

Special relativity and Emission theory

Does an observer move relative to the light cone? Special relativity says "Special relativity incorporates the principle that the speed of light is the same for all inertial observers regardless of the state of motion of the source.", and if an observer moves relative to the light cone, the speed of light is not the same for the observer. And as, if we slide the axis of the path of the observer to enable to regard the observer not moving, then the axis of the light cone is slided accordingly, and it turns out that the speed of light is fixed to the source (like in emission theory). But if an observer does not move relative to the light cone, the plane sharing simultaneity does not seem to tilt. Like sushi (talk) 23:18, 22 June 2009 (UTC)

The question doesn't make sense. A light cone is only defined for a specific event, that is a fixed point in space and time. If you are moving (or even if you're not - you'll at least be moving through time) then at each point you have a different light cone. It doesn't make sense to talk about moving relative to a light cone or not. --Tango (talk) 23:41, 22 June 2009 (UTC)

Maybe the use of the term is wrong. I meant by "light cone", the cone made by light from a source. And if there are numbers of observers in the same reference frame, the motion of the observers (the refrence frame) is notable.

Like sushi (talk) 00:18, 23 June 2009 (UTC)

I mean by "moving relatively to the light cone", that the axis of the path of the observer is not parallel to the axis of the cone.

Like sushi (talk) 00:37, 23 June 2009 (UTC)

So you don't mean the technical term? A "light cone" (well, a "future light cone", which is probably what is relevant here) in relativity is the region of spacetime that light emitted from an event can fill (it's not really a cone, it's the 3D analogue of a usual 2D cone). If that's not what you mean, you are going to need to be more precise about what you do mean. For example, are you talking about a cone is 3D space or 4D spacetime? --Tango (talk) 01:20, 23 June 2009 (UTC)

What I meant is a 4D one, though it can serve, in this case, even if it was 2D + time, or even 1D + time. And it is not "the region of spacetime that light emitted from an event can fill" but "actually is going to fill". And it is (as I thought you might be taking it) not from a point the observer goes through, but from a source apart from him.

And I repeat. I mean by "moving relatively to the light cone", that the axis of the path of the observer is not parallel to the axis of the cone.

Like sushi 01:34, 23 June 2009 (UTC)

In special relativity, the light cone does not have a unique axis. One of the consequences of relativity theory is that the light cone emanating from a given point in space-time is the same regardless of the velocity of the object that emits the light. No matter what the velocity, relativistic time dilation causes the light to look from the emitter's point of view like it's traveling away at constant velocity in all directions. Looie496 (talk) 04:32, 23 June 2009 (UTC)

(Edit conflict) (indent) The problem is that you can’t just ask if something is "moving", or talk about "the axis" of a light cone, without specifying which inertial frame of reference "motion" is defined relative to, and "the axis" of a light cone is defined in. To use a concrete example, suppose two spacecraft are moving away from each other at some constant speed v, where v is a sizeable fraction of the speed of light. Somewhere in the rough vicinity of the two spacecraft, a flashbulb goes off. Associated with each spacecraft is an inertial frame of reference such that in that frame of reference, the associated spacecraft is stationary, and the flashbulb goes off at the origin of the associated 4-D coordinate system.

The flashbulb going off is an event in spacetime. Associated with the flashbulb going off is a future light cone, which consists of the set of all events in spacetime such that light from the flash reaches that point in space at that point in time. Call the two spacecraft A and B, with associated inertial frames A and B. In inertial frame A, the future light cone associated with the flashbulb going off consists of all events (points in spacetime) such that r=c t, where r is the spatial distance from the flashbulb event as measured in inertial frame A, t is the time since the flashbulb event occurred as measured in inertial frame A, and c is the speed of light. In inertial frame B, the future light cone associated with the flashbulb going off consists of all events such that r’=c t’, where r’ is the spatial distance from the flashbulb event as measured in inertial frame B, t’ is the time since the flashbulb event occurred as measured in inertial frame B, and c again is the speed of light. The speed of light is the same constant value in both frames of reference.

The two observers agree as to which events are on the light cone, but they will disagree as to what the coordinates are of the events on the light cone. E.g., observer A might say that a given event on the light cone occurred 1 second after the flashbulb event, and is 1 light-second away from where the flashbulb event occurred. Meanwhile, observer B might say that the same event on the light cone occurred 2 seconds after the flashbulb event, and is 2 light-seconds away from where the flashbulb event occurred.

More importantly, although observers A and B agree as to which set of events are on the light cone, they do not agree as to which set of events are on the axis of the light cone. Observer A will say that the axis consists of all the events for which r=0, i.e., the world line of an object that was at the flashbulb event, and which is stationary as measured in inertial frame A. In contrast, observer B will say that the axis consists of all the events for which r’=0, i.e., the world line of an object that was at the flashbulb event, and which is stationary as measured in inertial frame B. The only event that observers A and B will agree is on the axis is the origin of the two coordinate systems.

In short, observer A will say that the axis of the light cone is parallel to spacecraft A, i.e, spacecraft A is not moving relative to the axis of the light cone, and spacecraft B is moving relative to the axis of the light cone. In contrast, observer B will say that the axis of the light cone is parallel to spacecraft B, i.e, spacecraft B is not moving relative to the axis of the light cone, and spacecraft A is moving relative to the axis of the light cone. It’s an exactly symmetrical situation, and neither observer is the "correct" one. Red Act (talk) 05:26, 23 June 2009 (UTC)

"the light cone does not have a unique axis" and "you can’t just ask if something is "moving", or talk about "the axis" of a light cone, without specifying which inertial frame of reference "motion" is defined relative to, and "the axis" of a light cone is defined in".

I see. Then the speed of light is the same for all the observers. And no observer moves relative to the light cone, or rather have an axis not parallel to the axis of the cone in the refernce frame of himself.

Thank you.

But as I tried to think about it, I would like to give this thought some space. Is it possible? For example, that an observer in a reference frame can have an axis of motion not parallel to the axis of the light cone in another reference frame, or something?.

Like sushi (talk) 06:20, 23 June 2009 (UTC)

I’m not sure I understand your question. As long as inertial reference frames A and B aren’t stationary relative to each other, then the world line of something that’s stationary according to observer A will for sure not be parallel to a world line that’s the axis of a light cone according to observer B, if that’s what you mean. Red Act (talk) 08:34, 23 June 2009 (UTC)

Well, come to think of it, if the observers have to have an axis of motion parallel to axes of light cones in all reference frames, then the axes of the light cones and (as the axis of a light cone is an axis of motion to at least one observer) of the motion have to be all parallel. That means no motion possible. The one I have come up with is only (a tiny bit) useful when trying to imagine the picture of a light cone and a path of an observer in a different reference frame (because I can recall the image I made for these posts, which hopfully can be used for other purpose).

Like sushi (talk) 08:54, 23 June 2009 (UTC)

Again, you have to be careful to always specify which observer you’re talking about, whenever you talk about any observation. And I’m having a hard time understanding your question, because I’m not sure what you mean by an "axis of motion". Do you mean the world lines traversed by objects which are stationary according to a given observer, or the world lines traversed by any object that’s moving at some fixed speed according to a given observer?

For a given observer, the axes of all light cones according to that observer, and the world lines traversed by all objects that are stationary according to that observer, are all parallel. Those lines all point in the direction that’s "forward in time" according to the given observer. All those lines are not, however, parallel to the world lines traversed by objects that are moving at some nonzero fixed speed according to the given observer, or parallel to the axes of light cones according to a different observer, or parallel to the world lines traversed by objects that are stationary according to a different observer. Red Act (talk) 10:41, 23 June 2009 (UTC)

I meant by an "axis of motion", the world lines traversed by objects which are stationary according to a given observer.

Like sushi (talk) 12:28, 23 June 2009 (UTC)

No. I think I have found another problem. That is when the observer is out of the light cone and eventually touches its surface. The first question now makes sense? or is it still faulty? I am confused. Like sushi (talk) 10:18, 23 June 2009 (UTC)

An "observer" in special relativity actually refers to a whole inertial reference frame, and shouldn't be pictured as being one person sitting at one point, receiving light from all around. That's a common misperception among beginning students of special relativity.

The world line traversed by an object moving at some fixed speed (including 0) can certainly intersect a light cone that emanates from some event. For example, the time axis of an inertial frame can certainly intersect a light cone. And the time axes of various inertial frames are not in general parallel, if that's the question. Even if two inertial frames share the same origin, the time axes of the two inertial frames will point in different directions, as long as the two inertial frames are moving relative to each other.

Where's Steve Baker? He's good at explaining special relativity in a way that's understandable to beginners. Red Act (talk) 10:41, 23 June 2009 (UTC)

I have managed to get out of the confusion. All light cones assume axes parallel to the axis of motion of the observer in the reference frame of him. Like sushi (talk) 11:10, 23 June 2009 (UTC) Or do they not? Like sushi (talk) 11:15, 23 June 2009 (UTC)

It sounds like you might be thinking about the problem correctly. According to an arbitrary observer A, the axes of all light cones extend in the "forward in time" direction in spacetime, and hence are parallel to the world lines traversed by all objects which are stationary (according to observer A), which also extend in the "forward in time" direction. Those axes and world lines identified by observer A will also appear as being parallel to observer B, although they will not extend purely in the "forward in time" direction according to observer B. Furthermore, observer B will say that the world lines identified by observer A are for objects which are not stationary, and will say that what observer A has identified as being the axes of the light cones are not the actual axes of the light cones. Red Act (talk) 13:06, 23 June 2009 (UTC)

Sorry,I didn't notice Mr.or Ms. Red Act has written. The axes of all light cones according to that observer are parallel. So the contents of a light cone is always the same for any observer, i.e. the light cone occupies the same spacetime, and only difference is the coordinate system which gives the event the time and location? In fact, I have posted to the math desk to confirm that two points at which observers moving parallel to the source of the cone, are the same distance away, and in a plane with the same y (the spacial axis perpendicular to the direction of motion of the observer) and the light cone meet always make a line of the same tilt, if the velocity of observers are the same (in mathematical form). But I could not be answered somehow. (About for what reference frame, I think I didn't specify. I was thinking only in one reference frame.) If all planes sharing simultaneity is inclined at the same rate in any one reference frame, no matter from what plane you start, the resulting planes are the same if any numbers of observers travel for the same time in their common reference frame, and as such, with the same velocity. Is it that it does not matter if it's the section with the light cone or not, but "from and onto the same plane" and "along parallel lines", and if not we can not simply compare them? Like sushi (talk) 12:28, 23 June 2009 (UTC)

I can answer the first question in this paragraph easily: Two observers, A and B, will agree about what set of events constitute the events within a light cone, although they will label those events with coordinates differently. However, the two observers will not agree about what set of events constitute the axis of the light cone. Beyond the first question in your paragraph, I’ll have to think a bit to see if I can understand what you’re asking. Red Act (talk) 13:18, 23 June 2009 (UTC)

I'm afraid I have gone a little too wild. It does matter if it's the section with the light cone or not, and although "from and onto the same plane" and "along parallel lines" motion has always the same properties other than the starting point and end point, the problem is not that, but whether or not the shape made in the distorted coodinate system by the light cone is the same to the light cone in right coordinate system if the distorted coordinate system is transformed to the right one.

Like sushi (talk) 14:50, 23 June 2009 (UTC)

I think, in one reference frame, a tilted world line (I am not used to this term) of another observer moving relative to you which shows another axis of the light cone goes through the center of the ellipse which is the section of the light cone and the plane of simultaneity for that observer in the first reference frame. If that distorted cone (it is not distorted in outer cone though) can be transformed in to the right cone, making the right cone the distorted cone, then the reciprocity is shown. Like sushi (talk) 13:27, 23 June 2009 (UTC)

I’m falling asleep. I’ll have to respond to this this evening, if someone else doesn't step in and respond before then. Red Act (talk) 14:50, 23 June 2009 (UTC)

Did nobody link to Lorentz transformation? Look at it this way, Lorentz transforation is just rotation, but we don't live in euclidean geometry. In our geometry, no matter how you rotate a light cone it stays exactly the same. If you were to rotate a line, it would not. As such, you can't say it has an axis any more than you can say an infinitely long line has a midpoint. Look at the third picture from the top (the moving one) to see how Lorentz transforation works, and how it doesn't affect the light-cone. That animation is also moving through time, and the light cone is moving too. If I could find one without that, I'd show that. Just try to ignore the fact that all the events are moving backward. — DanielLC 15:42, 23 June 2009 (UTC)

I have noticed that the diagonal lines (the edges of the light cone in the plane y=z=0),are not affected by the transformation. And the outer cone is the same with and without the transformation. As I think (if it is arbitrarily set) the axis of the cone or the world line of the observer moving in respect to the right coordinate system goes through the center of the ellipse which is the section of the light cone and the plane of simultaneity for the distorted coordinate system, it just need to adjust the x-axis to make the distorted cone (of which the surface is not distorted) right.

But does the transformation to make the distorted cone the right one makes the right cone the distorted one with the opposite tilt? And does the y- or z- width of the ellipse section of the transformed light cone appears to be small than the radius of the circle section of the right cone with the centers at the same time in the reference frame of the right coordinate system? (I think that means if the observer regarding himself stationary measures the travel distance of light for the other observer moving relatively to him, in y or z in his right coordinate system, at the same time in his reference frame, it is less than his.)

(I think I have managed to understand , with the help of all of you, that the light cone is always the same if the coordinate system is adjusted to the right one, though the distance to the origin of the light cone in spacetime is different. The questions I am asking now is on the side of the original one, which was about if the shape of light cones, and thus the speed of light are always the same in all (right) coordinate systems, or for all observers who regard themselves at rest.)

Thank you very much for all of you! And I hope you will answer the side questions. Like sushi (talk) 23:39, 23 June 2009 (UTC)

Alhough it is difficult to imagine, caliculating Lorentz transformation,

$t'=\gamma (t-{\frac {vx}{c^{2}}})$

$x'=\gamma (x-vt)$

for t and x gives

$t=\gamma (t'+{\frac {vx'}{c^{2}}})$

$x=\gamma (x'+vt')$.

So coordinate system A from perspective of B is coordinate system B from perspective of A just with the opposite v. It seems a light cone for A from perspective of B has just the opposite tilt of the light cone for B from perspective of A. (I changed the way of describing coordinate systems and light cones, accoring to (I hope) Mr. or Ms. Red Act's comment below. In the former way of writing, "the transformation to make the distorted cone the right one makes the right cone the distorted one with the opposite tilt.") Like sushi (talk) 06:42, 25 June 2009 (UTC)

I’m having a bit of a problem with some of the terminology you’re using. You talk about a "right" coordinate system vs. a "distorted" coordinate system, and a "right" light cone vs. a "distorted" light cone. There is no one "right" coordinate system, in either the sense of "correct" or "orthogonal". In general, whether two lines or two planes in spacetime are at "right angles" to one another depends on which coordinate system you are using. And there is no preferred or "correct" coordinate system, except that we are here limiting our discussion to inertial frames of reference. All inertial frames of reference are equally valid, in that the laws of physics are the same regardless of what choice of coordinate system they are expressed in. If you have two observers A and B moving relative to each other, the coordinate system used by A, which has axes that are all at "right angles" according to observer A, will not in general have axes that are at right angles according to observer B. And the coordinate system used by B, which has axes that are at right angles according to observer B, will not in general have axes that are at right angles according to observer A. When working with only two observers, it’s convenient to define the coordinates used by observer A such that the motion of observer B relative to observer A is parallel to one of observer A’s spatial coordinate axes (usually observer A’s x axis), and define the coordinates used by observer B such that the motion of observer A relative to observer B is parallel to one of observer B’s spatial coordinate axes (usually observer B’s x axis). That way, at least the two observers will agree that each other’s spatial coordinate axes are orthogonal. The two observers won’t, however, agree as to whether each other’s spatial coordinate axes are orthonormal, and won’t agree as to whether the "time" axis of an observer is at "right angles" to that observer’s spatial axes. Instead of "right" vs. "distorted", a better description would be coordinate system "A" vs. coordinate system "B", or quite commonly, the "primed" coordinate system vs. the "unprimed" coordinate system. However, which coordinate system is chosen to be labeled as being "primed" and which is chosen to be labeled as being "unprimed" is really an arbitrary choice.

Two parts of what you seem to be interested in pertain to the "axis" of a light cone, and the intersection of a light cone with a hyperplane of simultaneity. I think it’d be helpful to examine those two things mathematically from the point of view of two observers. Choose the origins of the two coordinate systems involved to be the event at the vertex of whatever light cone you’re talking about. Then choose the orientation of the two coordinate systems to be based on the motion between the two observers, such that the coordinate systems are in what the Lorentz transformation article calls the "standard configuration". I’ll adopt that article’s notation that there’s an observer O that uses the coordinates $(t,x,y,z)$, and an observer Q that uses the coordinates $(t',x',y',z')$. I’ll follow that article’s notation closely, but I’ll also take the additional step of choosing units of distance and time such that the speed of light is one, i.e., $c=1$.

Observer Q would describe the light cone as being the set of all events such that ${t'}^{2}={x'}^{2}+{y'}^{2}+{z'}^{2}$. Observer Q would consider the "axis" of the light cone to be the set of all events such that $x'=y'=z'=0$. What observer Q considers to be a hyperplane of simultaneity is the set of all events such that $t'={t'}_{0}$ for some constant value of ${t'}_{0}$. The intersection of that hyperplane of simultaneity with the light cone is the set of all events such that $t'={t'}_{0}$ and ${{t'}_{0}}^{2}={x'}^{2}+{y'}^{2}+{z'}^{2}$.

Now we’ll express all of the above geometrical objects in terms of Observer O’s (unprimed) coordinates, by substituting the Lorentz transformation article’s expressions for the primed coordinates in terms of the unprimed coordinates into all of the above equations, and simplifying:

Observer Q’s description of the light cone, ${t'}^{2}={x'}^{2}+{y'}^{2}+{z'}^{2}$, gets expressed in the unprimed coordinates as $[\gamma (t-vx)]^{2}=[\gamma (x-vt)]^{2}+y^{2}+z^{2}$, which simplifies down to $t^{2}=x^{2}+y^{2}+z^{2}$. Thus, what observer O considers to be the light cone for a given event is the exact same thing as what observer Q considers to be the light cone for the given event, and is expressed in the exact same way.

Observer Q’s description of what observer Q considers to be the axis of the light cone, $x'=y'=z'=0$, gets expressed in the unprimed coordinates as $\gamma (x-vt)=0$, $y=z=0$, which simplifies to $x=vt$, $y=z=0$. Thus, what observer Q considers to be the axis of the light cone is different from what observer O considers to be the axis of the light cone, which is the set of all events such that $x=y=z=0$.

Observer Q’s description of the hyperplane of simultaneity, $t'={t'}_{0}$, gets expressed in the unprimed coordinates as $\gamma (t-vx)={t'}_{0}$, which can alternatively be expressed as $t=vx+{t'}_{0}/\gamma$. Since the right-hand side of this equation is not a constant, the hyperplane that observer Q considers to be a hyperplane of simultaneity is not a hyperplane of simultaneity according to observer O.

Observer Q’s description of the intersection of what observer Q considers to be a hyperplane of simultaneity with the light cone, $t'={t'}_{0}$ and ${{t'}_{0}}^{2}={x'}^{2}+{y'}^{2}+{z'}^{2}$, gets expressed in the unprimed coordinates as $\gamma (t-vx)={t'}_{0}$ and ${{t'}_{0}}^{2}=[\gamma (x-vt)]^{2}+y^{2}+z^{2}$. The first equation can be expressed as $t=vx+{t'}_{0}/\gamma$, which when used to substitute for $t$ in the second equation leads after simplification to ${{t'}_{0}}^{2}={(x-t_{0}v\gamma )^{2}}/{\gamma ^{2}}+y^{2}+z^{2}$. This is an ellipsoid, whose center is at $t=\gamma t_{0}$, $x=t_{0}v\gamma$, $y=z=0$. This center point of the ellipsoid, as you seem to have guessed, is the same event that’s at the intersection of the hyperplane of simultaneity (according to observer Q), and what observer Q considers to be the axis of the light cone. Red Act (talk) 06:06, 24 June 2009 (UTC)

Thank you. Although I do not know much about the formulation of the things we are considering now, I could somehow follow the mathematical examination.

So, if I want to express what I called "right" coordinate system, "coordinate system A in coordinate system A", and for "distorted" coordinate system, "coordinate system A in coordinate system B" are prefered?

Like sushi (talk) 08:07, 24 June 2009 (UTC)