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October 31

Does an accelerated body at (instantaneous) rest, emit gravitational waves?

According to LIGO, "every physical object that accelerates produces gravitational waves". But how can the GWs, emitted from the accelerated body, carry away momentum from the body being at (instantaneous) rest? HOTmag (talk) 14:30, 31 October 2024 (UTC)[reply]

Perhaps because "instantaneous rest" is a mathematical abstraction, not a real-world condition that applies to a real-world accelerating body? Others more expert can doubtless address this concept better.

My impression is that you are just making up puzzles using random concepts, as you have previously been doing under this and your previous User name HOOTmag for more than ten years. I'm not intending to play anymore. {The poster formerly known as 87.81.230.195} 94.6.86.81 (talk) 15:39, 31 October 2024 (UTC)[reply]

Sorry, but my question is serious (like all of my questions here). I really don't know how to anwser it correctly (I don't remember I ever made up puzzles using random concepts). HOTmag (talk) 16:11, 31 October 2024 (UTC)[reply]

Zeno of Elea c. 490 – c. 430 BC tried "seriously" to divide time into instants. In his arrow paradox, Zeno states that for motion to occur, an object must change the position which it occupies. He gives an example of an arrow in flight. He states that at any one (durationless) instant of time, the arrow is neither moving to where it is, nor to where it is not. It cannot move to where it is not, because no time elapses for it to move there; it cannot move to where it is, because it is already there. In other words, at every instant of time there is no motion occurring. If everything is motionless at every instant, and time is entirely composed of instants, then motion is impossible. Philvoids (talk) 16:18, 31 October 2024 (UTC)[reply]

Zeno's paradoxes are well known, but the mistake hidden in them has already been discovered, actually after Calculus was discovered. The way to remove Zeno's paradox is achieved by the concept of mathematical limit.

But my question has nothing to do with those old paradoxes, because as opposed to them, I can phrase my question without using any instantaneous velocity, but rather with the rigorous concept of mathermatical limit. I've only used the concept "instantaneous rest" for letting you grasp my question intuitively. If I had used the concept of mathermatical limit, I would have phrased my question otherwise, but the question would have still remained. For example I could ask: What's happening to the GWs emitted by the body, when the body's velocity approaches to zero? Does the GW emission approach to a zero radiation emitted by the body? HOTmag (talk) 16:48, 31 October 2024 (UTC)[reply]

Given that the emission of a GW is continuous process, the question does not really make sense. Note that for a continuously accelerating body, the instant of zero velocity has a length of precisely zero seconds, and hence the amount of momentum transferred during this instant is also zero. But that is true for every (length zero) instant. You need to integrate over a non-zero time period if you want to see a real transfer or momentum (or energy). --Stephan Schulz (talk) 22:31, 7 November 2024 (UTC)[reply]

Some days ago, I already referred to integration: See below, my paragraph beginning with the (green) words: "the accelerated object". HOTmag (talk) 02:17, 8 November 2024 (UTC)[reply]

Yes, but if you integrate over a zero time span, the result is zero. If you integrate over a non-zero time span, the object is not at rest during that time. --Stephan Schulz (talk) 14:17, 8 November 2024 (UTC)[reply]

The statement is simplified and not strictly true. A linearly accelerating body without rotation (or with rotational symmetry around the axis of rotation) does not emit gravitational waves — as discussed in another thread the quadrupole moment of the mass distribution needs to change. This is true for almost any real-life body or mass distribution, and this argument could be used to justify making that statement in a non-technical web page for the lay public. The instantaneous rest thing is fine, by the way and in a frame-dependent way, but not particularly relevant here. --Wrongfilter (talk) 16:50, 31 October 2024 (UTC)[reply]

Thank you for this important clarification. I'm quite amazed. Previous threads mentioned the quadrupole moment of the mass distribution, but none of them mentioned what you're caliming now, that "A linearly accelerating body without rotation...does not emit gravitational waves". On the contrary, some users claimed that an acceleration was sufficient for emitting GWs, and nobody disagreed, so I thought they were correct. Now you're surprising me.

Anyway, according to your clarification, I wonder now why our article Gravitational wave claims "An isolated non-spinning solid object moving at a constant velocity will not radiate". Aren't the words "moving at a constant velocity" redundant? HOTmag (talk) 17:02, 31 October 2024 (UTC)[reply]

"A linearly accelerating body without rotation (or with rotational symmetry around the axis of rotation) does not emit gravitational waves ... This is true for almost any real-life body or mass distribution." (emphasis mine).

This made me curious. There are exceptions to this rule? Would you be willing to give some examples, please? (Asking as a member of said lay public.)

Thanks! -- Avocado (talk) 12:51, 1 November 2024 (UTC)[reply]

It's a word of caution. If I had written "all real-life bodies" somebody would have blasted me for that. If you wish you can read it as almost every. --Wrongfilter (talk) 13:06, 1 November 2024 (UTC)[reply]

Yes, this is also what I wondered about, but I finally didn't ask you about that, because I guessed you had only wanted to use a word of caution, as you say now. So it seems you don't rule out NadVolum's reservation "a constant linear acceleration doesn't generate gravitational waves", do you? HOTmag (talk) 13:18, 1 November 2024 (UTC)[reply]

Just a little extra on that - a constant linear acceleration doesn't generate gravitational waves. NadVolum (talk) 18:40, 31 October 2024 (UTC)[reply]

Now you add: "constant". But if the acceleration is not constant, then my question in the header comes back... HOTmag (talk) 18:45, 31 October 2024 (UTC)[reply]

A single, accelerating body doesn't even exist: conservation of momentum says that there must be at least second body, accelerating in the opposite direction. And although a single body has no quadrupole moment, the pair of two bodies has. So the issue is avoided. PiusImpavidus (talk) 20:24, 31 October 2024 (UTC)[reply]

This thread hasn't mentioned a "single" body, and I can't see how a universe containing more than one object avoids the issue. My question is actually: how can the GWs, emitted from a given body accelerated by a jerk (i.e. by a non constant acceleration), carry momentum away from the body being at (instantaneous) rest? HOTmag (talk) 08:52, 1 November 2024 (UTC)[reply]

The question is poorly phrased and possibly based on a misunderstanding.

"But how can the GWs, emitted from the accelerated body," The GWs are produced by the body, but the body doesn't do that on its own. The GWs are emitted by the space surrounding the body and its reaction mass. The waves are, as usual, a far-field approximation. The near-field is a bit more complex. "carry away momentum from the body" Who said that? In the discussion a few topics up on spinning rods I mentioned angular momentum. "being at (instantaneous) rest" Here you make the same error as Zeno (good he was mentioned). The accelerated object is at rest for a time interval of zero, so it must emit zero waves during that time interval, as waves are a continuous phenomenon. You have to consider the emission of waves over a time interval equal to the inverse of the wave's frequency. That's rather basic.

BTW, you won't find a constant acceleration in the universe. Also, a speed of zero is physically irrelevant. You can always make the speed zero by coordinate transformation, which cannot change the physics.

Now I'm wondering, you ask questions on general relativity, which I consider academic master's level of physics, yet make such basic errors, third year secondary school. I can't squeeze seven years of physics education in an answer here; that's a pile of physics books. If you aren't making fun of us, then you started reading that pile from the wrong end. I like to assume good faith and love a good physics question, but that's why I don't always respond to your questions. PiusImpavidus (talk) 17:12, 1 November 2024 (UTC)[reply]

I like to assume good faith. Thank you, and please keep assuming good faith. Yes, I graduated secondry school not long ago, so I may make mistakes sometimes. Anyway, when I state any statement, I only rely on articles in Wikipedia. If you think my wording is wrong, don't hesitate and please tell where my mistake is, and I will thank you from the bottom of my heart.

The GWs are produced by the body, but the body doesn't do that on its own. When I wrote "GWs, emitted from the accelerated body", I used a wording used in our article Gravitational wave: "This gives the star a quadrupole moment that changes with time, and it will emit gravitational waves". Anyway, if you're trying to claim that the wording in Wikipedia is wrong and that a single body cannot emit GWs, then please don't hesitate to say that (I'm still not sure if that's the case because you haven't said this yet), and I will thank you from the bottom of my heart for removing this error - not only from Wikipedia - but mainly from me, because I've always thought that also a single body can emit GWs (provided that its quadrupole moment varies).

"carry away momentum from the body" Who said that? Again, I'm only relying on Wikipedia. Please see our article Gravitational wave: "Water waves, sound waves, and electromagnetic waves are able to carry energy, momentum, and angular momentum and by doing so they carry those away from the source. Gravitational waves perform the same function. Thus, for example, a binary system loses angular momentum as the two orbiting objects spiral towards each other – the angular momentum is radiated away by gravitational waves".

The accelerated object is at rest for a time interval of zero, so it must emit zero waves during that time interval, as waves are a continuous phenomenon. You have to consider the emission of waves over a time interval equal to the inverse of the wave's frequency. That's rather basic. AFAIK, you can always use the well known mathematical operation called integration, for "collecting (uncountably) infinitely many infinitesimals" of instants at which the accelerated system was at rest, and then you receive a non zero quantity of energy of GWs produced by the accelerated system at those instants of rest. The question is, where was the energy/momentum of those GWs carried away from? But maybe this integration is actually impossible, not only physically - because (as you say): "waves are a continuous phenomenon", but also mathematically - because the set of those infinitely many instants (at which the system producing the GWs is at rest), is always a countable set only. Am I right?

BTW, you won't find a constant acceleration in the universe. Yes, but AFAIK you can always conduct an experiment which can artificially create a constant acceleration. Additionally, AFAIK a constant acceleration is important in theoretical physics, so you can regard my question as a theoretical one.

Also, a speed of zero is physically irrelevant. You can always make the speed zero by coordinate transformation, which cannot change the physics. AFAIK, physics does consider a speed of zero, for many purposes, e.g for deciding whether there is some momentum, and whether there is some kinetic energy, and the like. Additioanlly the speed of zero is important for establishing many relativistic concpets, e.g. proper frame, proper reference frame, proper length - being the longest length a given body can have, proper time - being the shortest life-time a given body can have, rest mass - being the smallest mass a given body can have (for those physicists who make a distinction between a rest mass and a relativistic mass), and likewise. HOTmag (talk) 19:03, 2 November 2024 (UTC)[reply]

The multiverse of science

Superhero fiction loves the concept of a "multiverse", that is, infinite universes that exist somewhere and that completely similar to the main universe, except for some details. It can be something big (as in, all heroes are villains instead, something turned the world into a dystopia or a post-apocalyptic wasteland) or something minor (as in, the radiactive spider does not bite Peter Parker but someone else), but in the grand scheme of things the history of the Solar System, Earth, life on Earth and human history are all basically completely the same. Needless to say, that's just a narrative device, one that has led to some awesome stories (and other so-so ones), but no more than that.

But lately I have noticed in actual science publications people who talk about the multiverse, in the real world. And we do have an article about that, Multiverse. Before breaking my head trying to understand the fine details, just a quick question: would the real multiverse be, at least in principle, similar to the multiverse as seen in fiction, or is it a completely different idea that got distorted? And if there are parallel universes, where would they physically be? I dismiss such a question with fiction because of the willing suspension of disbelief, but in science you can't get away that easily... Cambalachero (talk) 18:40, 31 October 2024 (UTC)[reply]

Try and wrap your mind round things like Elitzur–Vaidman bomb tester and if you really can explain it all well I'll be glad to listen! :-) NadVolum (talk) 19:06, 31 October 2024 (UTC)[reply]

In Multiverse § Types you can see different – sometimes very different – meanings of this term. Authors of pop-science articles who bandy the term accordingly do not all use the term with the same meaning. All of it is purely speculative and in most versions has the problem that the theory is unfalsifiable because it makes no testable predictions that differ from current theory, and is therefore generally deemed to fall outside the scope of science proper. --Lambiam 20:04, 31 October 2024 (UTC)[reply]

It is among the Interpretations of quantum mechanics#Influential_interpretations. Card Zero (talk) 23:33, 31 October 2024 (UTC)[reply]

If there was only one universe, where would it physically be? For multiple universes, we can use the same answer. If you own a car, where do you keep it? It's a deep question, but it's not an obstacle to the concept of a person owning two or three cars. Card Zero (talk) 23:47, 31 October 2024 (UTC)[reply]

So the multiverse would be like a multi-car garage? ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:56, 31 October 2024 (UTC)[reply]

OK, if you have more than one parking spot it raises the question "how does one parking spot relate to the next, in physical space?", and that's a reasonable question if you know the first parking spot's location in physical space. But when the parking spots are universes, the answer might be "they don't", because nobody ever said the first parking spot was located anywhere anyway. Card Zero (talk) 00:08, 1 November 2024 (UTC)[reply]

Yes. It would be in some other dimension beyond mere physicality. ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:23, 1 November 2024 (UTC)[reply]

They are all in your mind. --Lambiam 07:06, 1 November 2024 (UTC)[reply]

I recall one professor contradicting the famous "Cogito ergo sum / I think, therefore I am" as "Maybe he only thinks that he thinks." ←Baseball Bugs ^{What's up, Doc?} carrots→ 14:04, 1 November 2024 (UTC)[reply]

As a few people have said, "The problem with thinking about the universe is that there's nothing to compare it to". That's based on the assumption that universe = "everything that can possibly exist, anywhere". And yet, the human mind can comprehend the concept of an infinite number of different universes, each containing everything that can possibly exist, anywhere. It's no more difficult to work with such an idea than to make great use of the square root of -1. But is it actually true? I'm glad you asked ... -- Jack of Oz ^{[pleasantries]} 17:29, 1 November 2024 (UTC)[reply]

Now imagine a (strongly) inaccessible cardinality of universes. --Lambiam 20:46, 1 November 2024 (UTC)[reply]

I reckon the multiverse concept was inspired by the failure of the universe that we know to contain everything that can possibly exist. This assumes it is bounded and doesn't contain, for instance, versions of itself at all other ages, and versions of itself where the laws of physics are different such that life is impossible. Then the other universes are the other possibilities. They're often synonymous with moments of time, in which case we are constantly moving through universes (or perhaps "featuring in a causally related series of moments" rather than moving - same difference). The parallel universes are moments of time that we don't go to, or in many cases couldn't possibly go to. On the other hand, especially in fiction, the term more often means a causally related series of moments, a timeline or "environment", where subsequent moments are constantly branching and diverging but share a common history. Card Zero (talk) 21:44, 1 November 2024 (UTC)[reply]

Is there a common term indicating that the current value of every derivative (of any order) of the position over time is zero?

That is, the current velocity iz zero, and so is the currect acceleartion, and so is the current jerk, and so forth...

I thought about "static" as a sufficient condition, but I'm not sure, so I'm also asking: Should every "static" body be considered to satisfy the property mentioned in the header? HOTmag (talk) 21:56, 31 October 2024 (UTC)[reply]

Something that doesn't move at all is described as a fixed point or fixed object. This usually means that the object won't move even if a force is applied to it. --Amble (talk) 23:34, 31 October 2024 (UTC)[reply]

Having all time derivatives zero at a particular instant is not the same as having constant position. The standard example is

e^{-{\frac {1}{t^{2}}}}

, but there are lots of other possibilities.

This is important to know when formulating differential topology, as it enables finding bump functions and partitions of unity in the smooth (

C^{\infty }

) category. --Trovatore (talk) 05:31, 1 November 2024 (UTC)[reply]

A term used by mathematicians for the function giving such a position as a function of time is "flat function". --Lambiam 07:03, 1 November 2024 (UTC)[reply]

So my original question can be phrased as follows: Is there a common adjective, describing an object, and indicating that the object's location with respect to time is a flat function? Additionally, can the body's adjective "static" be a sufficient condition, for the above location to be a flat function? HOTmag (talk) 08:28, 1 November 2024 (UTC)[reply]

Extended content

1. immovable 2. immoveable 3. fixed 4. immotile 5. unmovable 6. fast 7. nonmotile 8. stiff 9. firm 10. stabile 11. amovable 12. unmoveable 13. static 14. immoble 15. moveless 16. irremovable 17. rooted 18. stationary 19. nonmobile 20. standing 21. dead 22. nonmoving 23. rigid 24. motionless 25. unbudgeable 26. inamovable 27. unimmobilized 28. unshiftable 29. nonimmobilized 30. unmoving 31. staid 32. immoved 33. nonmutable 34. nonchangeable 35. undeposable 36. untransmutable 37. unrelocatable 38. nonremovable 39. inflexible 40. untranslocatable 41. unfluid 42. nonchanging 43. nonrotatable 44. non-mobile 45. unmigratable 46. unmobilized 47. unstationary 48. non-stationary 49. nonflexible 50. incommutable 51. nonmodifiable 52. nonrelocatable 53. unresizable 54. unfixed 55. unmodifiable 56. unchangeable 57. untransformable 58. untransportable 59. unadjustable 60. intransmutable 61. unflexible 62. nontransportable 63. sedentary 64. invariable 65. nonmigratable 66. nonvariable 67. non-animate 68. noncommutable 69. nondisplacable 70. nondisplaceable 71. immalleable 72. unvariable 73. unmechanizable 74. inanimate 75. unlocomotive 76. confined 77. unmanipulatable 78. nonadjustable 79. undisplaceable 80. uninclinable 81. nondetachable 82. unalterable 83. undislodgeable 84. intransformable 85. unmutable 86. inanimated 87. restagnant 88. nonstationary 89. torpid 90. semistationary 91. unfluidizable 92. unfixable 93. unmaneuverable 94. nonportable 95. unbending 96. nonrotative

Philvoids (talk) 17:16, 1 November 2024 (UTC)[reply]

You need a preferential reference frame to get zero velocity, so the property is not intrinsic but observer-dependent. Have objects with this property been the subject of studies in theoretical physics? If not (and I can't think of a reason why they should be of interest to physicists), it is very unlikely that there is a term of art for the property. --Lambiam 20:00, 1 November 2024 (UTC)[reply]

Time crystal? 176.2.70.177 (talk) 18:03, 2 November 2024 (UTC)[reply]

November 1

Midnight sun in Norway

Directives for military officers and military commanders in the event of an armed attack on Norway has a completely uncited section discussing an alleged event from 1968 on the Russo-Norwegian border:

On the evening of 7 June, the garrison heard the noise of powerful engines coming from the manoeuvres...Actual observations were not possible over the border in the dark...At daybreak the impressive numbers of the Soviet forces staged along the entire border became visible.

Google Maps says that the southernmost point of the border is about 69°N, and based on Midnight sun, it looks like anywhere north of 67°13'N experiences midnight sun by the end of May. Consequently, this means that all points on the Russo-Norwegian border experience midnight sun on 7-8 June, so the whole scenario is impossible. Am I understanding rightly, or have I missed something? This isn't some recent vandalism; it's present in the first version of the page history, apparently translated from the corresponding article in the Norwegian Bokmal Wikipedia. Nyttend (talk) 18:59, 1 November 2024 (UTC)[reply]

The article on the Norwegian bokmål Wikipedia ascribes the difficulty in observing the cause of the hubbub to dårlig vær, bad weather. In the original version on the Norwegian bokmål Wikipedia, the difficulty is said to have been, specifically fog. BTW, in this original bokmål version the alleged incident took place on 7 June 1967. Half a year later, "1967" was changed to "1968" by a user whose only contribution was this change. --Lambiam 20:33, 1 November 2024 (UTC)[reply]

User:Lambiam, there's also a no:Sovjets demonstrasjon av militær styrke ved den norsk-russiske grensen i 1968, with several sources. Do the sources confirm the year, or is 1968 an error? Maybe Theohein was just fixing a typo. Nyttend (talk) 21:25, 1 November 2024 (UTC)[reply]

The sources confirm June 1968 but appear to name 6 June 1968 as the date when Norwegian soldiers stationed along the border with the Soviet Union became alarmed by a sudden advance of Soviet tanks and heavily armed soldiers, stopping only within metres of the border. There is no mention of any difficulties in observing this. Apparently, the information has been kept classified for 40 years. --Lambiam 07:00, 2 November 2024 (UTC)[reply]

November 2

Is there a name for 0.001 miles?

I need to work on a software that works in units of one one-thousandth of a mile internally. Is there a name for such a unit? --193.83.24.42 (talk) 00:35, 2 November 2024 (UTC)[reply]

American silliness? HiLo48 (talk) 00:45, 2 November 2024 (UTC)[reply]

As opposed to French silliness, such as the met-ray. ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:09, 2 November 2024 (UTC)[reply]

The Roman mile was by definition a thousand paces (milia passuum). catslash (talk) 01:24, 2 November 2024 (UTC)[reply]

Actually, that's 1000 double steps, a bit under 1500m. --Stephan Schulz (talk) 22:40, 7 November 2024 (UTC)[reply]

The obvious millimile gets a little use. (The author there can't redefine span (unit) so easily though.) I also found it in a more modern book about chemistry, where it seems to be part of a quiz designed to test the reader's understanding of units: Which length is longer, a millimile or a decameter? but archive.org has stopped showing snippet views of in-copyright books. Millimole tends to pollute search results, which is perhaps why a chemist would be inspired to invoke millimiles. Card Zero (talk) 01:46, 2 November 2024 (UTC)[reply]

See pace (unit). When I worked as a surveyor, we often used this informal unit and with practice it became 99% accurate, good enough for most purposes. Shantavira|^{feed me} 09:21, 2 November 2024 (UTC)[reply]

There is an existing more accurate unit, used by the railways, known as the "link". Your unit is 8 links. The link is divided decimally and contains 7.92 inches. Your unit is therefore (7.92 x 8) = 63.36 inches. Before metrication, Ordnance Survey maps were scaled at 1 inch to a mile. The scale was therefore 1/63 360. 2A00:23D0:FFC:3901:90DF:CC65:72B0:11FF (talk) 14:51, 2 November 2024 (UTC)[reply]

Can I determine distance from size of a photographed object

If a photograph contains an object of known real size, can I use that to determine how far it was from the camera or other device that took the picture? Also, would this question fit better here or on the math reference desk? Primal Groudon (talk) 16:18, 2 November 2024 (UTC)[reply]

For one thing, you would have to know the type of lens. A wide-angle or fisheye lens makes things look farther away than they actually are. ←Baseball Bugs ^{What's up, Doc?} carrots→ 17:55, 2 November 2024 (UTC)[reply]

Even with a pinhole camera you couldn't, unless you know the distance inside the camera between the pinhole and the photographic plate or film. Define four variables:

$h$ _ext is the real size of the object;
$d$ _ext is the distance between the pinhole and the object;
$h$ _int is the size of the object's image;
$d$ _int is the distance between the pinhole and the photographic plate or film.

Then

h

_ext :

d

_ext =

h

_int :

d

_int.

If you have the values of three of these variables, you can determine that of the fourth. With simple fixed lenses, this also gives a reasonable estimate.

To determine the value of

d

_int, you don't have to look inside the camera. Just take a picture of an object of known size at a known distance and measure

h

_int. You can now calculate

d

_int. --Lambiam 20:16, 2 November 2024 (UTC)[reply]

Empirical calibration is surely the best approach, but if your camera has a zoom, then you will need to ensure the same level of zoom is used for the measurement and the calibration. You can do this by checking the 35 mm-equivalent focal length in the exif meta-data in the jpeg files. catslash (talk) 23:49, 2 November 2024 (UTC)[reply]

Lambiam, what if the object is a building of known dimensions? If you can see 3 corners couldn't you solve for the relative position of the camera? Or do you need 4 points? fiveby(zero) 04:32, 9 November 2024 (UTC)[reply]

In general, we have a projective transformation that projects a 3D point

(x,y,z)

in real space to a 2D point

(\xi ,\eta )

on the film by a transformation of the form

\xi ={\frac {d+ex+fy+gz}{1+ax+by+cz}},~~\eta ={\frac {h+ix+jy+kz}{1+ax+by+cz}}.

Let

y

denote the vertical coordinate in the 3D space and

\xi

the horizontal coordinate on the film. Then, assuming the camera is not tilted, the value of

\xi

is invariant under changes of

y

while

x

and

z

remain constant, which implies that

f=b=0.

Let furthermore

x

denote the horizontal coordinate parallel to the film in the camera. Then the value of

\eta

is invariant under changes of

x

while

y

and

z

remain constant, which implies that

i=a=0.

Every point in the straight line of sight from the camera, as well as in the vertical plane through that line, has the same value for

x

, and its image on the film is on a vertical line with constant

\xi .

We can set both equal to

0,

which implies that

d=g=0.

The third 3D coordinate

z

denotes the horizontal coordinate in the direction of sight of the camera. We can set

\eta =0

for the height of the horizon on the film. This means that

\eta \to 0

as

z\to \infty ,

which implies

k=0.

Finally, at a constant distance

z,

real-world squares remain squares on the image, so we know that

e=j.

A judicious choice of coordinates has led to the simpler projective transformation

\xi ={\frac {ex}{1+cz}},~~\eta ={\frac {h+ey}{1+cz}}.

Three unknown coefficients still remain:

c,e,h.

Relating a real-world point with known coordinates

(x,y,z)

to a point on the film with known coordinates

(\xi ,\eta )

gives you two equations, so if my reasoning is correct, in general doing this for two points should suffice. It becomes more complicated if you don't know the coordinates of real-world points but only the distances between them. --Lambiam 07:34, 9 November 2024 (UTC)[reply]

Furthermore, putting

\lambda ={\frac {e}{c}},~~z_{0}=-{\frac {1}{c}},y_{0}=-{\frac {h}{e}},

we can rewrite the transformation as

\xi =\lambda {\frac {x}{z-z_{0}}},~~\eta =\lambda {\frac {y-y_{0}}{z-z_{0}}}.

Fixing the value of

y

for the ground level as being zero,

y_{0}

should be the eye height of the camera, which is presumably easily measured, eliminating one more unknown. This might suggest just one real-world point

(x,y,z)

needs to be related to an image point

(\xi ,\eta ),

but the equations you get are not independent, since, if the values of

\xi ,\eta

and

y

are known, the value of

x

can be computed without measuring it, using

x={\frac {\xi (y-y_{0})}{\eta }}.

. --Lambiam 12:18, 9 November 2024 (UTC)[reply]

Here is a scenario, if you have an uncropped picture of the White House and you know the Camera Model and the Fixed Len used. You can buy the Camera and Len and take the same Picture at vairous distance of the White House until the your picture matches up to the target Picture. Then you can estimate the distance. 2001:8003:429D:4100:5CCA:727C:C447:3877 (talk) 22:07, 3 November 2024 (UTC)[reply]

There are two factors that change the size of an object in a photo when you fix the lens and zoom. One is the distance to the object. The other is the scaling of the photo itself. If I print a photo as 4x6 it will be much smaller than the same photo printed on 8x10. To measure distance, you really need another photo with known distance from the same camera with same lens and same zoom so you can measure the width of an object of known size on the size of the photo you printed. Then, you know the scaling factor to work out the distance to an object in another photo. Another option is to have two (or more) objects of known size at different distances where you know the distances between them. When in the same photo, you can use the sizes of the objects and the known distance between them to estimate the distance to the camera. 12.116.29.106 (talk) 18:26, 5 November 2024 (UTC)[reply]

Projective Geometry can help if you know the length of something in a picture and the shape of something, e.g. a square on the ground. NadVolum (talk) 18:47, 5 November 2024 (UTC)[reply]

Magenta vs. purple

How come mixing blue and red with equal gets magenta in modern color theory but a bluer purple (more violet-like) in RYB color theory?? Georgia guy (talk) 16:36, 2 November 2024 (UTC)[reply]

No colour theory is perfect, but, more importantly, it is ill-defined what it means to "mix" colours. Different physical procedures will also give different outcomes. --Lambiam 20:20, 2 November 2024 (UTC)[reply]

it depends! Red in RYB is something "not (subtracting) blue" and blue is somewhat "not (subtracting) red". So the red takes the role of the blue in the other system and vice versa. If you constructed a special "blue pigment" with some green in it and a special red pigment with some orange in it, you would mix either the pigments and have a tone between magenta and purple or you would mix the reflected light from both pigments, and have a more bright version of exactly the same tone. I don't know if you can make the brightness the same too. So it doesn't have to be a different colour, but that only works with purple because of the complementarity. Generally the colours in both systems are very differently mixed. 176.2.70.177 (talk) 20:48, 2 November 2024 (UTC)[reply]

Reading Subtractive color and Additive color might help. Klbrain (talk) 01:08, 11 November 2024 (UTC)[reply]

November 6

Pornography addiction in animals

Has any scientific research ever been conducted showing that non-human animals are capable of addiction to pornographic imagery, in the same or similar manner as mice have been shown to be capable of addiction to cocaine, in that they will prefer cocaine over food and refuse food? 104.171.53.110 (talk) 23:44, 6 November 2024 (UTC)[reply]

I'm not sure there is unambiguous evidence that mere imagery can elicit sexual arousal in any non-human animal species. --Lambiam 08:01, 7 November 2024 (UTC)[reply]

"Unambiguous evidence" sounds like the challenge. We do have an article about panda pornography if we're willing to weaken the standard. And rhesus like to look at photos of certain body-parts of the opposite sex of their species (see Animal sexual behaviour#Others for ref). I assume the male researchers involved in that study made plenty of "look at this picture of macaque!" jokes. DMacks (talk) 08:44, 7 November 2024 (UTC)[reply]

Somewhere in one of our journals is a study on turkeys. Male turkeys get excited and attempt to mate with anything that makes them think it is female. The researchers began with a wooden female turkey and eventually ended up with a wooden female turkey head on a stick and the males still attempted to mate with it. If it is of interest, I can see if it is still in our collection and get a better reference. It isn't a photo of a turkey, but it is still an artificial substitute for a live turkey. 68.187.174.155 (talk) 18:26, 7 November 2024 (UTC)[reply]

They just gobble, gobble, gobble it up. Clarityfiend (talk) 01:06, 8 November 2024 (UTC)[reply]

Anyone wanna help me develop a RealHen? DMacks (talk) 04:46, 8 November 2024 (UTC)[reply]

November 8

Street lights, rain drops and windows

A few nights ago it was pouring down with rain. I looked out a window to take a look, and I noticed that really beautiful patterns of light appeared as I put my eyes right in front of rain drops that were in front of a street light. The rain drops had interesting 'arms' surrounding them, but the most important part I noticed was that there were so many black lines covering the entirety of the drops.

I was able to take a picture of them with my phone, but unfortunately most of the the lines do not appear in the photos. You can see some on the sides but most of them are missing.

What caused these lines to appear?

―Panamitsu (talk) 05:08, 8 November 2024 (UTC)[reply]

See Caustic (optics). --Lambiam 09:37, 8 November 2024 (UTC)[reply]

The repeating black lines are an example of Newton's rings. They occur due to internal reflections in a thin wedge of fluid and are most apparent when the source light is monochromatic e.g. yellow sodium light. The article shows a more reliable way to view the rings using a thin convex lens than relying on chance raindrop spreading. Philvoids (talk) 10:49, 8 November 2024 (UTC)[reply]

Yes, can confirm that this is what the lines looked like, although they were not as round as in the article's images. Thanks. ―Panamitsu (talk) 22:11, 8 November 2024 (UTC)[reply]

November 9

Black Body emissive power in medium

The black body emissive power in a medium $E_{m}$ is equal to the product of the square of its refractive index $n$ and the emissive power in vacuum $E_{v}$ with the formula:
$E_{m}=n^{2}E_{v}$ .
What does this mean in terms of the energy emitted, respecting the principle of conservation of energy and in the case where the energy is emitted in a vacuum, then enters a medium with refractive index $n$ ? Malypaet (talk) 23:16, 9 November 2024 (UTC)[reply]

Power is energy emitted over time. So energy is conserved as it is emitted more slowly. Heat energy turns into electromagnetic energy. Graeme Bartlett (talk) 10:15, 11 November 2024 (UTC)[reply]

Yes, but here, if you use the SI units in

Joules\cdot s^{-1}\cdot sr^{-1}\cdot m^{-2}

for emissive power as radiance, you have

E_{m}>E_{v}

. So, to conserve energy, you cannot use only the velocity for power, as you suggested. Malypaet (talk) 23:31, 12 November 2024 (UTC)[reply]

November 10

What is a pipehead dam?

I've seen various things described as a "pipehead dam" (common noun), as well as some specific instances of dams named "... Pipehead Dam", eg Serpentine Pipehead Dam, which is separate to Serpentine Dam. I gather from the text of Serpentine Pipehead Dam that a pipehead dam is a smaller dam fed from a larger dam, with the smaller (pipehead) dam then feeding water into the pipe into the water supply system - but I cannot find anything (including with a Google search) that specifically says that. Mitch Ames (talk) 01:03, 10 November 2024 (UTC)[reply]

Wow! This was hard to hunt down. Deep in the results for probably the same set of searches you did, I finally found on page 77 of [https://sitecore9-cm-prod.watercorporation.com.au/-/media/WaterCorp/Documents/Our-Water/Regional-Water-Supplies/water-forever-south-west-final-report.pdf]: "Pipe-head dam — a diversion dam that takes streamflow

from the catchment to another dam for storage." --jpgordon^{𝄢𝄆𝄐𝄇} 01:22, 10 November 2024 (UTC)[reply]

That definition appears to be the reverse of what the Serpentine articles say. The articles say water goes from main dam to pipehead dam, but the Water Corp definition suggest the water goes from pipehead to another (main?) dam. Mitch Ames (talk) 10:01, 10 November 2024 (UTC)[reply]

The point that might not necessarily come from the easy picking of the water authority or google online materials, is that in the history of the dams, the water can be moved either from the main dam to the pipehead, or vice versa - and in turn can also be distributed to other parts of the system, there is no one way only part of the system, maybe not easily found online but nevertheless the current water corp web space is very poor on the intracies of the dynamics of the water supply system. There could well be a range of security issues attached to the lack of information . JarrahTree 10:55, 10 November 2024 (UTC)[reply]

...for the over two centuries that pipe head dams have existed? --jpgordon^{𝄢𝄆𝄐𝄇} 16:37, 10 November 2024 (UTC)[reply]

The higher the pitch of the instrument the longer the bow: why?

As everyone has probably noticed, the violin has a longer bow than the viola, which has a longer bow than the cello, which has a longer bow than the double-bass. Why? I'm guessing a given length of bow (irrespective of the instrument) takes the string through a given number of vibrations. Therefore to make the string vibrate for a given amount of time at a higher frequency requires more bow length. But is this correct? Another consequence would be that no matter what the instrument the bows make the string vibrate for roughly the same amount of time and that the violin requires a higher bow speed than the viola which requires a higher bow speed than the cello which requires a higher bow speed than the double bass. Again, is this correct? 178.51.16.158 (talk) 08:15, 10 November 2024 (UTC)[reply]

To make the string vibrate with a nice sound, there has to be sufficient (but not too much) friction between the bow and the string, which requires the bow to move at the same speed or just slightly faster than the top speed of the vibrating string, 2π times the product of amplitude and frequency. So higher frequencies at a given level of dynamics require a higher bow speed. --Lambiam 09:17, 10 November 2024 (UTC)[reply]

Ok. But leaving aside variations of the amplitude of the vibration, of the tension of the string, of the tightness or looseness of the bow (which the player can adjust), of the mass of the string and of the bow, of the thickness of the string and of the material it is made of, of the thickness of the bow, of the length of the string, of the force exercised by the hand, of how carefully the player has rubbed his bow with rosin, of the quality of the rosin, etc. etc. is it nevertheless the case that (things being roughly equal) to sustain a string's vibration at a higher frequency for a given unit of time requires more bow length? Clearly in practice there wouldn't be a linear relation between increase in frequency and increase in length. 178.51.16.158 (talk) 17:12, 11 November 2024 (UTC)[reply]

(Only) slightly pertinent to this query, you might be amused by Kingsley Amis's 1971 novel Girl, 20, in which a would-be avant-garde classical composer and violinist performs a controversial concert with rock musicians (an actual thing at the time, see for example Concerto for Group and Orchestra). Someone has secretly greased both his violin bows, but he impresses with his technical skills (though not with his actual music) by borrowing and using a double-bass bow. {The poster formerly known as 87.81.230.195} 94.7.95.48 (talk) 17:54, 10 November 2024 (UTC)[reply]

The articles about Bow (music) and the archetien who makes them say little about bow length. My survey below does not support the OP's observation. Lengthwise the bows for viola, violin and cello seem nearly interchangeable. The wide variation in longer bows for the double bass is due to the sitting players' preferences and arm lengths.


                |  Viola |   Violin |  Cello  |  Double bass
                |        |          |         |
bow    strings  |        |          |         |
cm   LOW    TOP |        |          |         |
----------------+--------+----------+---------+-------------
80   196    659 |   GE   |          |         |   x
79     .      . |        |          |         |   x
78     .      . |        |          |         |   x
77     .      . |        |    CA    |         |   x
76     .      . |        |   x      |         |   x
75     .      . |  x     |   x      |         |   x
74     .      . |  x     |          |   CA    |
73     .      . |        |          |  x      |
72    41     98 |        |          |  x      |    EG
 cm    Hz     Hz

Philvoids (talk) 12:15, 11 November 2024 (UTC)[reply]

Your chart would make the relationships clearer if Violin were in the first column, reflecting the order of relative sizes (hence string lengths and usual ranges) of the instruments. I can see a clear correlation between increasing size and decreasing bow length for the first three instruments. The double-bass may be anomalous because, unlike the other three, it is usually played standing.

I am also puzzled by your quoted figures, as my full-sized violin bow is only 65cm (ribbon length), and I am sure I have seen double-basses played with bows less than 50cm. {The poster formerly kown as 87.81.230.195} 94.7.95.48 (talk) 17:53, 11 November 2024 (UTC)[reply]

November 13