March 29

What bird is this? The file id's it as a Javan hawk-eagle but I think it looks more like a black eagle. Said to be photographed in Mount Halimun Salak National Park on Java where 10 Javan hawk-eagles are said to have been found. Rmhermen (talk) 01:51, 29 March 2016 (UTC)[reply]

Yes, it does look more like the black eagle. While the black eagle isn't listed as being native to Java, it does extend down to the Malay Peninsula. I wonder if that would put them close enough to be within flight distance (they could stop off at Sumatra on the way). StuRat (talk) 03:37, 29 March 2016 (UTC)[reply]

It does look more like the black eagle (not a terribly high bar as the pictured bird is pretty clearly not a Javan hawk-eagle), but I'm not so sure that's what it is; the barring on the tail is rather distinctly different. Note: I'm no expert; that could easily be due to normal variation. Matt Deres (talk) 15:20, 30 March 2016 (UTC)[reply]

According to the article on Wikipedia, do exist real uni-polar neuron ans this is in contrast (apparently) to what I saw in the books that I read that there are three is only pseoudo-unipolar neuron. So in fact indeed do exist a real uni-polar neuron? 93.126.95.68 (talk) 02:32, 29 March 2016 (UTC)[reply]

Yes, there are real unipolar neurons. They are very common in insects and other invertebrates but also exist in vertebrates. If you are asking anything more, I can't understand it. Looie496 (talk) 00:58, 30 March 2016 (UTC)[reply]

I'm not even sure why they is asking, as in order to link us that picture, they had to have seen the article explaining that they are real and what they are. --OuroborosCobra (talk) 01:49, 30 March 2016 (UTC)[reply]

Question

Remark

As BenRG (talk) 17:39, 20 March 2016 (UTC) said here https://en.wikipedia.org/w/index.php?title=Wikipedia:Reference_desk/Science&action=edit&oldid=711808143, digital displays cannot lie. But as we know from Special relativity , clocks in moving reference frame are not synchronized. E.g. our clock shows ${\displaystyle t\sec }$ from Big Bang (or from any other moment). Another frame passing us with speed ${\displaystyle \mathbf {u} }$ has 3 clocks: front-end, hind-end and middle clock. When middle clock passing our clock , our clock and middle clock are synchronized manually. Man in moving frame synchronizes front-end and hind-end clocks with middle manually. Then we see from our frame:[reply] our clock shows ${\displaystyle t\sec }$; middle clock shows ${\displaystyle t\sec }$; front-end clock shows ${\displaystyle t-4\sec }$; hind-end clock shows ${\displaystyle t+4\sec }$. When moving reference frame will stop instantly (${\displaystyle a=-\infty }$), what will we see on front-end and hind-end clocks and why?

https://en.wikipedia.org/w/index.php?title=Wikipedia:Reference_desk/Science&action=edit&oldid=711808143 Null result of Michelson–Morley experiment extrapolation https://archive.org/download/PhysicsForTheEnquiringMind/Rogers-PhysicsForTheEnquiringMind.djvu , page 492

37.53.235.112 (talk) 05:04, 29 March 2016 (UTC)[reply]

You've specified that the moving clocks differ by 4 seconds when measured simultaneously wrt the lab frame, and I'll assume that they all stop simultaneously wrt the lab frame. Then after stopping they will still differ by 4 seconds wrt the lab frame, because acceleration doesn't affect ideal clocks; their reading doesn't jump forward or backward at the moment of the acceleration, they just keep ticking. This is an example of ideal clocks losing their synchronization when accelerated, which happens in special relativity but not in Newtonian physics. (Another example is the twin paradox/effect: the twins' initially synchronized ages are desynchronized when they meet again.) -- BenRG (talk) 05:59, 29 March 2016 (UTC)[reply]

When moving ref. frame stops, it merges with ours. Man from moving frame saw clocks sinchronized. An when his frame will stop he will see difference, or what? But digital display can't do such things.37.53.235.112 (talk) 10:30, 29 March 2016 (UTC)[reply]

The clocks are just clocks. They aren't aware of each other, they just follow their own local laws. If they end up not Einstein-synchronized, and you still want to call them a reference frame, that's fine as long as you're willing to accept the consequences (more complicated-looking physical laws), but the universe doesn't care one way or another what you use as a reference frame. -- BenRG (talk) 19:58, 29 March 2016 (UTC)[reply]

Could someone clearly and simply state what's being asked here? I can't figure it out...but find the topic interesting so would like to understand what's being discussed...68.48.241.158 (talk) 15:26, 29 March 2016 (UTC)[reply]

I think you need first check links in Remark column. Don't worry, 99% of people don't understand special relativity.37.53.235.112 (talk) 16:47, 29 March 2016 (UTC)[reply]

I think the question is simply whether, if you accelerate a bunch of comoving Einstein-synchronized clocks, they will still be Einstein-synchronized afterwards. The answer is no in general (yes in special cases, but the question wasn't about one of those cases). -- BenRG (talk) 19:58, 29 March 2016 (UTC)[reply]

37.53.235.112 (talk) 16:47, 29 March 2016 (UTC)[reply]

Is the observer actually "seeing"/detecting light from three clocks that (at the moment) displays +4, 0, -4? Or are there three observers that happen to be beside the three moving clocks, at rest relative to one another, who at the same moment (from their perspective) see these readings and later communicate them? In the second case, instantaneous deceleration should not much affect the clock reading as said above. But in the first case, light emitted by the clocks will continue to tick forward after the deceleration, showing them keeping time at a different rate due to time dilation; once the deceleration happens they tick at normal speed. But exactly when the deceleration occurs, there's a bit of a mystery; bear in mind that if the three clocks decelerate at the same time in your frame of reference, that's a different time in theirs, which is why the Lorentz contraction becomes a real crunching together in space (as it must for the Lorentz contracted object all to stop at the same time as you see it). Wnt (talk) 18:14, 29 March 2016 (UTC)[reply]

Is the observer actually "seeing"/detecting light from three clocks that (at the moment) displays +4, 0, -4? Or are there three observers that happen to be beside the three moving clocks -- As in E.Rogers book [page 493, note 19], I think observer at rest allows for time-delay of seeing a clock that is far away. To do that he can use 2 helpers or put 2 additional clocks, yes. But question isn't with observer at rest, but with both observers. Before stop moving observer sees 0, 0, 0. Observer at rest sees +4, 0, -4. After stop observer at rest still sees +4, 0, -4. Stoped observer sees ... what? also +4, 0, -4? 37.53.235.112 (talk) 18:39, 29 March 2016 (UTC)[reply]

The stopped observer sees one clock stop at -4, then he keeps going on for a while and stops later, then the last clock keeps moving at him and then stops at +4. Note that the actual difference in time is generally small and the distance large - you would never see a

timelike slice of someone else's light cone in normal space at any speed - so there is always enough distance for the clocks to tick that far before they change speed. Wnt (talk) 10:18, 30 March 2016 (UTC)[reply

]

I urge you to forget about "observers". The experiment here consists of the train car, the light, and the clocks. There needs to be an experimenter looking at the digital readouts of the clocks, but that person doesn't need to be moving in any particular way, doesn't need to look only at certain clocks and ignore others, etc.

To answer your third question first, ${\displaystyle t'}$ is a coordinate. You're probably familiar with Cartesian coordinates from Euclidean geometry, and you know that the transformation between different coordinate systems (with a common origin) looks like

${\displaystyle x'=x\cos \theta -y\sin \theta }$

${\displaystyle y'=y\sin \theta +x\cos \theta }$

where ${\displaystyle \theta }$ is the angle between (say) the x axes. There's no reason we have to use the angle; we could instead use the slope of one x axis with respect to the other coordinate system. Call the slope ${\displaystyle v}$. Then ${\displaystyle \theta =tan^{-1}v}$, and plugging that in and doing some trig substitutions gives you

${\displaystyle x'=\gamma (x-vy)}$

${\displaystyle y'=\gamma (y+vx/c^{2})}$

where ${\displaystyle \gamma =1/{\sqrt {1+v^{2}/c^{2}}}}$ and ${\displaystyle c=1}$. The reason this looks almost the same as the Lorentz transform is that they are almost the same thing. The similarity is hidden by the completely arbitrary way in which special relativity has always been taught. By convention, we use the slope ${\displaystyle v}$ between the coordinate systems instead of the angle ${\displaystyle \alpha }$. By convention, we use different units for distance and time intervals, which necessitates adding in some factors of ${\displaystyle c}$. None of this accomplishes anything; the universe doesn't care. Time is actually just another dimension, and the Lorentz transformation is just a coordinate transformation.

When you transform Cartesian coordinates, you plug in the coordinates of a point (x,y), and get new coordinates of the same point, (x',y'). Either pair of coordinates, by itself, identifies the point. You don't need both pairs. You don't need to pick the coordinate system that the point is "in". All points are "in" all coordinate systems. You should pick the coordinate system that's most convenient for the problem you're trying to solve. Any coordinate system will work, because they're equivalent descriptions of the same geometry.

Coordinate pairs like (x,y) or (x',y') must always be kept together because the components are meaningless by themselves. An equation like x=3 can define a line, but if you want to transform that equation into a different coordinate system, you can't just plug the x coordinate into the transformation equations and get an x'. The line doesn't have an equation of the form x'=anything. You can, for example, write the line parametrically as x(q) = 3, y(q) = q, or better as (x,y)(q) = (3,q), and plug the pairs (3,q) into the coordinate transformation, and get a correct parametric equation for the line in terms of the other coordinates. Likewise, pairs like (x,t) and (x',t') must be kept together. You can't just transform time. You must always use both (or all four) of the Lorentz transformation equations together, otherwise you will get nonsensical results (unless you really understand what you're doing).

In your problem, say the light is emitted at (0,0) (in either coordinate system). It reaches the walls at ${\displaystyle (x',t')=(\pm L/2,L/2c)}$, or equivalently ${\displaystyle (x,t)=({\tfrac {\pm Lc}{2(c\pm u)}},{\tfrac {L}{2(c\pm u)}})}$. Your first clock has the parametric equation (x,t)(τ) = (0,τ), which you can plug into the Lorentz transformation to find (x',t')(τ). To find the digital reading of that clock at a particular (x,t) or (x',t'), you just solve for τ. Your other three clocks have the parametric equations (x',t')(θ) = (0, θ) and (x',t')(θ^[]) = (±L/2, θ^[]), and you can plug those into the Lorentz transformation to find (x,t)(θ^whatever). Any question you have about what someone sees at a given (place,time) can be answered in this way. They are simple questions about

coordinate geometry. -- BenRG (talk) 19:58, 29 March 2016 (UTC)[reply

]

Last Sunday, March 27, 2016 we had to put our beloved cat to sleep. She was 21 and sick for a year. She weighed only 4 pounds with her premorbid weight of 12 pounds, skin and bones. She could not eat at all for the last week or so. She lived with us for 18 years. Two weeks before she died our long-functioning wall clock stopped and it seems we must throw it out. On Monday I found out to my surprise that it is a common belief that if your clock stopped, somebody will die in your family and this cat was a true family member. It is a true story. Read this Daily Mail article. Is it a Special Relativity or what? --AboutFace 22 (talk) 22:11, 29 March 2016 (UTC)[reply]

Question

Quote

So ${\displaystyle (x',t')(\tau )}$ are space-time coordinates of 1st clock in rest frame seen from moving frame, right? ${\displaystyle (x',t')(\tau )=({\tfrac {x-ut}{\sqrt {1-u^{2}/c^{2}}}};{\tfrac {t-ux/c^{2}}{\sqrt {1-u^{2}/c^{2}}}})}$ ${\displaystyle (x',t')(\tau )=({\tfrac {0-u\tau }{\sqrt {1-u^{2}/c^{2}}}};{\tfrac {\tau -0}{\sqrt {1-u^{2}/c^{2}}}})}$ At time ${\displaystyle \tau _{1}}$ 1st clock will show ${\displaystyle (x',t')(\tau _{1})=({\tfrac {-u\tau _{1}}{\sqrt {1-u^{2}/c^{2}}}};{\tfrac {\tau _{1}}{\sqrt {1-u^{2}/c^{2}}}})}$, seen from moving ε' frame. Is it correct? Thank you. But I will study Special relativity as rotatings next month. Now I must understand all without coordinates, angles, etc. I follow Feynman Lectures and chapters 15-1...15-6 don't use rotatings.

In your problem, say the light is emitted at (0,0) (in either coordinate system). It reaches the walls at ${\displaystyle (x',t')=(\pm L/2,L/2c)}$, or equivalently ${\displaystyle (x,t)=({\tfrac {\pm Lc}{2(c\pm u)}},{\tfrac {L}{2(c\pm u)}})}$. Your first clock has the parametric equation (x,t)(τ) = (0,τ), which you can plug into the Lorentz transformation to find (x',t')(τ). To find the digital reading of that clock at a particular (x,t) or (x',t'), you just solve for τ.

37.53.235.112 (talk) 09:05, 30 March 2016 (UTC)[reply]

So ${\displaystyle (x',t')(\tau )}$ are space-time coordinates of 1st clock in rest frame seen from moving frame, right? – ${\displaystyle (x,t)(\tau )}$ is where the clock actually is, and ${\displaystyle (x',t')(\tau )}$ is also where the clock actually is, just expressed in a different way. To figure out what people see, you need to trace light from the clock to someone's eye (and you can use either coordinate system to do that, and get the same result, since they describe the same world).

At time ${\displaystyle \tau _{1}}$ 1st clock will show ${\displaystyle (x',t')(\tau _{1})=({\tfrac {-u\tau _{1}}{\sqrt {1-u^{2}/c^{2}}}};{\tfrac {\tau _{1}}{\sqrt {1-u^{2}/c^{2}}}})}$, seen from moving ε' frame. Is it correct? – I think the algebra is right, but the description is sort of backwards. The clock shows ${\displaystyle \tau _{1}}$ on its face when it's at ${\displaystyle (x',t')=({\tfrac {-u\tau _{1}}{\sqrt {1-u^{2}/c^{2}}}},{\tfrac {\tau _{1}}{\sqrt {1-u^{2}/c^{2}}}})}$. -- BenRG (talk) 17:20, 31 March 2016 (UTC)[reply]

What is the expression of the kinetic energy of moving charged particles?--5.2.200.163 (talk) 15:38, 29 March 2016 (UTC)[reply]

You'll need to provide more context. In many cases, the classical formula - the one you probably know as ${\displaystyle KE=1/2mv^{2}}$ - is good enough for the job. In Griffith's book, he advances the definition to account for special relativity, using the formuka ${\displaystyle E_{kin}=mc^{2}({\frac {1}{\sqrt {1-{\frac {u^{2}}{c^{2}}}}}}-1)}$ for relativistic velocity u. In Bittencourt's book, which makes extensive use of energy analysis for charged particles, the author uses a more statistical approach, defining an energy density just for kinetic energy: ${\displaystyle W_{\alpha }=1/2\rho _{m\alpha }u_{\alpha }^{2}+1/2\rho _{m\alpha }<c_{\alpha }^{2}>}$ for the mean kinetic energy density of particles of species α, mass density ρ, with velocity u and peculiar velocity c (representing sort of "Brownian motion" separately from bulk particle transport velocity). The author proceeds to analyze other statistical measures of the kinetic energy distribution - there's more information than just the average value! In a sense, this is a classical treatment: so if you wanted to deal with something awful, like a multi-species, hot, dense, relativistic plasma, with each charged specie interacting with every other charged specie - say, modeling the charged particles inside the sun or inside a nuclear bomb - you'd have to synthesize a corrected total energy and total kinetic energy equation that combines a lot more terms and forms!

Many times, especially in quantum mechanics, kinetic energy is defined as an operator, and is usually expressed in relation to the momentum (not the mass and velocity). This is particularly helpful when laboratory measurement of mass or velocity is muddied by technical difficulties: for example, it is often easier to measure the energy of a charged particle than to measure its mass, charge, or velocity: we have lots of machines that can directly measure electron energy, like an electron spectrometer or an electron energy loss spectrum analyzer.

The point is, lots of advanced physics just builds on top of the basic physics you already know, and adds complications one step at a time.

Nimur (talk) 16:01, 29 March 2016 (UTC)[reply]

More context would be connected to what the article momentum#Electromagnetism#Particle in field says and the relation between kinetic energy and momentum within electromagnetism.--5.2.200.163 (talk) 16:39, 29 March 2016 (UTC)[reply]

If you look closely at the subsection you linked, you can see that the first row of the first column contains a classical kinetic energy term, in vector coordinates. The next column applies a

Lorentz transform

to that equation. The trick to extract the kinetic energy term is to carefully observe and recognize its general form. If you make simplifying assumptions, the equation is exactly the same as the one you learn in basic physics. You can go very far with Newtonian-style equations - and unless you like making extra work for yourself, in many cases you should use the simplest form of the equation.

But, the authors of the section you linked are being very cautious in order to be generally correct. For example, they write ${\displaystyle {\frac {m}{2}}\mathbf {\dot {r}} \cdot \mathbf {\dot {r}} }$ instead of ${\displaystyle {\frac {m}{2}}v^{2}}$. Why have they written it in this way?

Can you think of any condition in which the time derivative of the position vector is not the strict velocity vector? Or, more likely: where would the inner product of the time derivative of the position vector, multipled by itself, not be equal to the square of the scalar velocity?

I can think of a case: a non-

van Allen belt

)! We don't even have to dive deep into relativistic physics or quantum physics - just classical cases - where the general form of the equation is more correct. Throw in the effects of relativity, or the wacky nonsense called "quantum mechanics," and the equations just get more and more complicated. Yet, this allows us to solve equations of motion - and conserve energy, and so forth - even in non-inertial reference frames, or in cases of non-conservative forces, and so forth.

Nimur (talk) 17:47, 29 March 2016 (UTC)[reply]

Hey @Nimur: are you saying that for a pendulum whose pivot undergoes simple harmonic oscillation, the equation ${\displaystyle {\frac {m}{2}}v^{2}}$ does not yield the right answer for kinetic energy, but the r-dot form does (assuming both the v and r are with respect to the same non-moving frame in which the pivot oscillates)? Also, is "non-scleronomous" just a synonym for "rheonomous" in this case (or in general)? Both those articles could use some improvement... SemanticMantis (talk) 18:34, 29 March 2016 (UTC)[reply]

The point is that a double pendulum provides an easy-to-visualize case of a complicated non-inertial reference frame. If you are not very careful about the definition of the position (and therefore, the velocity), you end up computing very strange (incorrect!) quantities of energy. For example: you can fully treat the double pendulum as a two-dimensional system; you may, therefore, represent the position of the end-effector as either Cartesian (x,y) pair, or you may use the angular position of each hinge (θ₁, θ₂). Both representations are of equal rank, so they fully describe all the degrees of freedom; but if you use the x, y, position of the second segment of the arm, you are probably forgetting to account for the mass, momentum, and energy carried by the first segment of the arm - or you're setting yourself up for some terrible algebraic transforms! One representation provides a much more natural way to represent and compute the kinetic energy.

Yes, rheonomous and scleronomous coordinate systems are opposites. These terms show up in robotics, physics, and so on. These terms refer to whether your generalized coordinates are time-invariant.

You might also find holonomic and non-holonomic coordinates, especially in the contexts of robotics. These terms refer to whether your generalized coordinates are orthogonal and whether they inherently depend on the time-history.

Nimur (talk) 20:21, 29 March 2016 (UTC)[reply]

[1] and [2] makes the following claims:

(Cracked is not a reliable source for anything but bad jokes, but they are right surprisingly often.)

• The Antarctic krill is the most abundant animal species, with a population of 500 trillion. I would have guessed nematodes or one of the bacteria.
• Cows are the animal species with the highest total weight -- 170 million tons. I would have guessed one of the ant species.

Has anyone made an educated guess as to the most abundant species by number of individuals or by total weight? --Guy Macon (talk) 21:26, 29 March 2016 (UTC)[reply]

if google 'total weight all ants' a bunch of articles will come up...apparently there's no scientific consensus as to how many ants there are on earth (like not even a good guess)...only that there are a lot....obviously too bacteria and such are not considered animals..68.48.241.158 (talk) 21:36, 29 March 2016 (UTC)[reply]

What you are looking for is the animal with the largest biomass, the article has a list. Vespine (talk) 00:05, 30 March 2016 (UTC)[reply]

Actually, the list is at Biomass (ecology) - the Biomass article is about plant crops used as a source of fuel. 217.44.50.87 (talk) 09:32, 30 March 2016 (UTC)[reply]

Ants are commonly estimated to be roughly equal to human biomass.

Formicidae are a family. In contrast, we could probably claim that all the breeds of cattle still constitute a single species, and thus the cattle would out mass any single species of ant. Back to the krill, keep in mind they live in a much thicker layer than any other candidates on the list. Sure there are arboreal ants, and the nematodes go down a few feet maybe, but that's nothing compared the zone Antarctic krill inhabit, which is up to 3 km deep. Anyway, if you want to restrict to single species, a specific ant is probably not going to hold any records. If you want to compete all ranks, clades, and unranked groups then all the records are going to be set by one of the three domains. SemanticMantis (talk) 13:37, 30 March 2016 (UTC)[reply

]

It's also worth noting that it is often claimed that 1/4th of all living things are

beetles. (like here) However, that claim is usually based on the fact that about 1/4th of all identified species are beetle species, which is yet another way to count living things. About 1/4th of all species are species we classify as beetle species. Some will confuse this with the above definitions of "most", so while "ants" as a classification contain more individuals than "beetles" would, there are more different kinds of beetles than there are kinds of ants. --Jayron32 14:38, 30 March 2016 (UTC)[reply

]

And just to emphasize it again, that's only described species. Amateur and semi-professional collecting of beetles goes back more than a century and the differences between beetles are rather easier to discern with the naked eye. In other words, more attention was historically paid to describing and cataloging beetles than ants, which skews the numbers. Matt Deres (talk) 15:25, 30 March 2016 (UTC)[reply]

There is also the question of whether it is fair to lump multiple species of ants or beetles together when determining "most individuals" or "largest biomass". Even the "individual" bit is somewhat ambiguous. Is the "largest organism" listed at Clonal colony#Record colonies one individual or thousands? Is a Volvox one individual or many? See Slime mold#Behavior for a colony that is "no more than a bag of amoebae encased in a thin slime sheath, yet they manage to have various behaviours that are equal to those of animals who possess muscles and nerves with ganglia – that is, simple brains." Also see: Largest organisms. --Guy Macon (talk) 15:46, 30 March 2016 (UTC)[reply]

Even, to add to the excellent points you are making, the

species problem. --Jayron32 18:41, 30 March 2016 (UTC)[reply

]

Note that bacteria aren't animals, so even if they are more numerous, they wouldn't count for the original stat. Iapetus (talk) 16:12, 31 March 2016 (UTC)[reply]