Wikipedia:Reference desk/Archives/Science/2010 February 18

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February 18

A greenhouse on Mars

If you put a large air-tight greenhouse on Mars and fuilled it with air, how warm would it get? Would earth-plants grow in the Martian "soil"? Maybe with some human 'compost' mixed in? 78.146.206.38 (talk) 00:00, 18 February 2010 (UTC)

To answer the first part, there is no theoretical barrier to designing a solar greenhouse on Mars that operated at comfortable temperatures of plant life. Dragons flight (talk) 00:14, 18 February 2010 (UTC)

The temperature would depend on the latitude. Climate of Mars#Temperature indicates that temperatures do get up to nice warm temperatures (27 °C max), presumably on the equator. With the (literal) greenhouse effect, the temperature inside your greenhouse would be higher than the surroundings, so anywhere reasonably close to the equator should be fine temperature-wise. The Martian soil might be suitable for the growth of Earth-plants - it seems from that article that more research is required. --Tango (talk) 00:18, 18 February 2010 (UTC)

The temperature would drop quite low, especially at night in the winter, unless there was sufficient thermal mass and a high insulation level. Mars gets on average only 43% the solar intensity or "insolation" received on Earth per [1]. The mean surface temperature (outside the greenhouse) is only -63C (per the NASA site, varying with season and latitude). The air pressure is very low, so a greenhouse dome would have to be quite strong to hold in enough airpressure for earth type plants. See also a NASA project looking at a Mars greenhouse. Apparently 1/4 of earth normal pressure would suffice. Actually sounds doable. The Mars Society also has some suggestions how to build the greenhouse. Edison (talk) 00:21, 18 February 2010 (UTC)

As that Mars Society link says, you can keep it warm with what is essentially an enhanced greenhouse effect - they suggest a silver compound in the plastic that will allow visible light and near-IR (ie. sunlight) through but stop far-IR from getting out. --Tango (talk) 00:45, 18 February 2010 (UTC)

a little help with equilibrium constants and partition coefficients as they relate to solubility

Admittedly, the whole "mole product / mole reactant" thing goes over my head when I imagine scenarios like adding more reactant or taking away more product. So ... like take these titration results.

There's approx 0.013 mmol of dissolved iodine (without iodide) in 50 mL solution.

When I add 5 mL cyclohexane, approximately 0.008 mmol of it escapes into the cyclohexane, resulting in a concentration-in-cyclohexane figure of 0.0016 M. From the new concentrations I calculate a partition coefficient of approximately 16.

When I add 8 mL cyclohexane, despite a 60% increase in the amount of lipophilic solvent there's only a 25% increase in extraction ... Indeed, the concentration of iodine in cyclohexane is now 0.00125 M, a concentration fall. This sort of makes sense since as the iodine gets less-concentrated in the water phase, iodine becomes harder and harder to extract from aqueous solution. The concentration in aqueous phase is now 5.6 * 10^-5 M and the new partition coefficient is 22.

This is clearly experimental error right? Shouldn't the partition coefficients remain roughly the same? (We measured concentrations by titration.) Let's take an ideal case with no experimental error and say I'm adding more water or more cyclohexane. How would I use the partition coefficient to predict new concentrations? What does the partition coefficient really mean, as an equilibrium constant? How likely is it that mistitration or something like that is the source of my error? Basically, I don't know how to think of "mol product over mol reactant" when I say add more of one type of solvent to the mixture. Are the concentrations going to rearrange themselves such that the ratios of concentrations in the different solvents will somehow remain constant.

I get even more lost when I deal with saturated solutions and there's apparently an equilibrium constant between the pure solid phase and a dissolved phase as a solute, because if I add more excess solute to a saturated solution, clearly the ratios cannot adjust themselves to the equilibrium constant since the solution is already saturated. John Riemann Soong (talk) 01:34, 18 February 2010 (UTC)

To address your last question, you need to remember that the concentration of a pure solid substance (or even a pure phase of a liquid) has a constant concentration. TenOfAllTrades(talk) 04:03, 18 February 2010 (UTC)

Coming back to the first part of your question, what's the precision with which you're measuring the amount of iodine (to start with, and in each phase)? If that 0.008 mmol iodine is ± 0.001 mmol (for example), then what is the range of partition coefficients that you could calculate (based on 0.007 or 0.009 mmol in the cyclohexane phase)? TenOfAllTrades(talk) 13:38, 18 February 2010 (UTC)

Angular momentum in collisions

The angular momentum of any object can be divided into the angular momentum of it's center of mass, and it's angular momentum with respect to the center of mass. When analyzing collisions, the total angular momentum is said to be conserved. However, sometimes both the angular momentum of it's center of mass and it's angular momentum with respect to the center of mass are considered, but other times only the angular momentum with respect to the COM is used. Why would there be this discrepancy? And when finding the rotational kinetic energy, are both types of angular momentum used? Thanks. 173.179.59.66 (talk) 04:24, 18 February 2010 (UTC)

I think the OP really means ITS center of mass and ITS angular momentum, without apostrophes in the ITS. Cuddlyable3 (talk) 12:50, 18 February 2010 (UTC)

Angular momentum will be separately conserved around any point. Often you only need to consider it around one point to get the information you need. --Tango (talk) 18:02, 18 February 2010 (UTC)

I'm sorry, I don't see what mean... —Precedingunsigned comment added by 173.179.59.66 (talk) 20:38, 18 February 2010 (UTC)

Angular momentum around point P will be conserved and angular momentum around point Q will be conserved and around point R, etc. etc. You often only need to apply one of those conservation laws to get the answer you are looking for. Sometimes it will be easier to use P, sometimes Q. That's why you see different ones used in different problems - you just use whatever is easiest. --Tango (talk) 21:08, 18 February 2010 (UTC)

Hmmm, I guess my question wasn't well worded. Let's say two billiard balls are heading towards each other and collide (with some impact parameter). When doing all the calculations, the angular momentum due to their pure rotation (ie due to w=V/r) is used, while the angular momentum due to their net velocities are ignored. However, in a problem where a superball is bouncing off a wall, both the objects rotation and net velocity are taken into accound when detailing how the angular momentum is conserved. Why would we look at both in one, and only look at rotation in another? If I'm not being clear, let me know. Oh, and he talk about rotational kinetic energy, is it just equal to (1/2)I*w^2, or does the angular momentum of the center of mass need to be considered as well?

Ok, I don't understand that either. The rotation of an individual ball about its centre of mass won't be conserved during such a collision, because it isn't a closed system (the ball interacts with the other ball). --Tango (talk) 01:03, 19 February 2010 (UTC)

But if the rotation of one ball slows down, the other will speed up, keeping angular momentum conserved, right? —Preceding unsigned comment added by 173.179.59.66 (talk) 03:48, 19 February 2010 (UTC)

To be strictly correct, the 'orbital' angular momentum always should be included along with the 'rotational' angular momentum, but under certain circustances one or the other might be considered negligible.Dauto (talk) 15:40, 19 February 2010 (UTC)

Math in physics

Out of curiosity, do most physicists (outside of ultra theoretical fields like string theory) spend time manipulating equations and such by hand, or is basically all the math handled by computers. The reason I'm asking is that I'm sifting through a textbook on quantum mechanics, and it's evident that to make any progress, physicists back then had to be able to work out equations and model situations by pen and paper...is that still true today? —Preceding unsigned comment added by 173.179.59.66 (talk) 04:40, 18 February 2010 (UTC)

Yes, even experimental physicists still need to keep the pen and paper at arms length. But computers play a increasingly indispensable role on more advanced calculations. Dauto (talk) 05:04, 18 February 2010 (UTC)

To turn the (usually continuous) models of classical physics into discrete computer-solvable models which have the same properties as the original continuous model (conserved quantities, stability, ...) is mathematics in itself. In short: you need to use mathematics in order to get the computer to do the mathematics you want it to do. http://wiki.riteme.site/wiki/Numerical_methods —Preceding unsigned comment added by 157.193.173.205 (talk) 10:57, 18 February 2010 (UTC)

There is really three parts to this - there is the business of arithmetic - plugging actual numbers into the equations to get answers in practical, numeric form - and for that, a simple calculator will sometimes suffice - but computers are better for repetitive stuff. Another aspect is taking raw data and performing statistics or fitting equations to those numbers (for which computers are pretty indispensable). But the other part is in the development of the equations themselves. Computers are making inroads into that too - we have 'symbolic math' packages that can manipulate equations symbolically - but it still takes a human mind to spot some of the subtle changes that can change a sheet of paper covered with hieroglyphics into something small, elegant and memorable. SteveBaker (talk) 15:52, 18 February 2010 (UTC)

In my experience, "it depends whether the physicist knows how to tell a computer to solve their problem." That depends on the individual physicist's level of interest and formal training with computers. I hang around mostly with "numerical physicists", which is sort of techno-jargon for "computer programmer". The types of problems we know how to solve with computers reach pretty far and wide. Categorically, we use more computational power than a theoretical physicist. For example, this last week I have been solving the wave equation for residuals in a numerical inversion scheme. Now, when a theoretical physicist "solves the wave equation," they write down some variables on paper and call it "solved." When I "solve the wave equation," that means that I arrange the equation to the simplest form that I can use to represent. Then I formalize an algorithm, and write computer code to represent that scheme. When the computer "solves the wave equation," it reads input data, performs calculations on that data following my instructions (that hopefully represent some physical process), and spits out numerical output data representing a model of experimental observations. So in some sense, both me and the computer are doing math - but I save the redundant calculations for the computer, and the mathematical formalism for myself. We blur the line as far as where the mathematics is actually being "done," and me and my team of powerful computers really solve the wave equation together.

Obviously, the simplest case to consider is basic arithmetic. I'm mathematically inclined, but even I have limits - so when I need to determine a value like (36 + (50*2)/1024)/4, it's a waste of my time to do that in my head or on paper. A computer gives me the correct answer, the first time, as quickly as I can type that in. But in my brain, I have done math to estimate the acceptable range of answers I expect. In that case, the computer has pretty much done most of the math.

Non-arithmetic calculation is a little harder to type into a computer. If I want to know the instantaneous phase of a function or a scale-factor for a fourier transform, I often have to decide whether it's easier to use a symbolic algebra system, a numerical approximation, or a paper-based analytic solution. It depends on the problem. But I'm a physicist, not a "number-cruncher" - so it'd be a mischaracterization to say that I spend all day typing out equations and hitting "enter." So let me diverge a little and explain a bit about what a physicist like me actually does.

People come to physicists with quantitative problems that they would like to answer. In my particular field, they come to us with field-recorded geophysical survey data, and ask us to generate images of the Earth from it. We want to perform this process faster and better, creating clear pictures even if the source data is crappy. So, I look at the physics that represents the field-data collection; I model the physical phenomena, observe the effects of unknowns and interferences, and use those insights to design an algorithm to convert input data into output data, while preserving the physical constraints that we know apply.

I would categorize the design of algorithms as a subcategory of "mathematics." Now, whether we design a particular algorithm with a computer or not depends on its complexity. Having some training in formal software engineering, I find UML diagrams to be a great way to set forth a large numerical physics scheme - especially since I like modular code. So, I can use a CAD tool to help me draw out the mathematical operations I plan to do. But, I also keep a stack of blank paper at my desk to scratch work on - diagrams are easier drawn by hand than by mouse. When I have to do geometry, I do it on paper with a pencil. When I need the value of a tangent, I get that answer from a computer (or desk calculator). When I want to do a coordinate transform for an integral kernel, I often use a computer algebra system, but more often than not, I need to do it by hand anyway. In this way, I am "doing math" - I am applying the structure and formalism of analysis to solve a physics problem. The overwhelming majority of this stage of work is by hand and in my brain.

By the time I have formalized my problem, I inevitably write it out as a series of program statements, in the form of a standalone subroutine, a full-blown application program, or a short script for simple arithmetic - all depending on the scale of the problem. The repetitive calculations are performed by machine - and sometimes, intelligent mathematical decision-making is programmed in as well. I would say that almost everybody (physicist or not) knows how to perform arithmetic by computer; almost all physicists know how to perform algebra and calculus by computer; and many specialized physicists and mathematicians (and others) know how to perform matrix mathematics, optimization (mathematics), and so on.

One of the turning points in formal physics education is being able to distinguish the subtle, qualitative difference between "math" and "physics." In other words, when a physicist sees a new form of the wave equation, they aren't looking at the values of the constants - they're looking at the qualitative interactions and relationships between components. (Ask a real physicist whether Schroedinger's equation, which defines a wave-function, is actually a wave equation! The way they approach that question will astound you). And if you sit in on enough physics seminars, you'll inevitably hear some stodgy old guy grumbling something to the tune of "That's all great, but where's the physics!?" What they mean by this is that despite a load of impressive mathematical maneuvering or experimental observation, the presenter has not identified any qualitative physical principle. In the same way, when you ask whether a physicist manipulates math by hand or computer, they can really do either - whether their physics needs a numerical result will modulate the way that they interact with their computers. Nimur (talk) 22:35, 18 February 2010 (UTC)

Great, thanks for the detailed response. PS Is the Schroedinger's equation a wave equation? —Preceding unsigned comment added by 173.179.59.66 (talk) 00:03, 19 February 2010 (UTC)

Yes. See the article Schroedinger's equation. Please sign your posts. Cuddlyable3 (talk) 17:02, 19 February 2010 (UTC)

Actually it is a diffusion equation ([2], [3]), because the time derivative is first-order and complex (a complex parabolic partial differential equation); while a wave equation is a hyperbolic partial differential equation). Ultimately, you can define a "wave equation" a lot of ways - but most commonly, the defining factor is whether you can construct an invariant of the form (x_i +/- v*t) - in other words, propagation - which cannot be done for Schroedinger's equation! It's the defining state-equation for the wave function - but it is not a wave equation! To some extent, this is semantics and a matter of definition - but the idea is that we care about the physics that the mathematics represents - in other words, the algebraic term that represents wave propagation is decidedly absent in the Schroedinger solution. This means a lot of things - there is no effective velocity to propagate perturbations of the wave-function. This has implication to quantum entanglement, because in the absence of a term to define a velocity, technically there is no speed limit on the propagation of quantum information - hence the paradox of faster-than-light propagation of quantum entanglement information! So, by playing with the maths analytically, we can try to peel away at the actual physics implications. Nimur (talk) 18:14, 19 February 2010 (UTC)

Your explanation for quantum entanglement is fishy. The Schroedinger equation is a non-relativistic approximation. A fully relativistic wave equation such as the klein-gordon equation is a hyperbolic equation with a speed limit associated to it and yet(!) quantum entaglement exists. Dauto (talk) 05:07, 21 February 2010 (UTC)

Of course, there's always room for a more complicated model. Klein-Gordon_equation#Derivation explains the extension/conversion to the relativistic model. Anyway, I'm not really sure how to use the Klein-Gordon model, but as I understand it, it does not correctly solve for the observables in a simple case such as a hydrogen atom. It seems that the Klein-Gordon model only works for certain spinless particles, pions. (I really haven't ever used it). As far as I know, the Dirac Equation is the standard form, relativistic-corrected wavefunction equation for atomic physics - and it too is a diffusion equation. Nimur (talk) 06:13, 21 February 2010 (UTC)

Nuh-uh. Dirac's equation is not a parabolic equation. Don't let the first order time derivative fool you. Parabolic equations are second order differential equations while the Dirac's equation is a first order differential equation. It can be easily shown that every solution of the Dirac's equation also is a solution of a second order differential equation but it turns out to be a hyperbolic equation. That's not surprising at all because it is the limited speed characteristic of hyperbolic equations that make them consistent with relativistic speed limit. Schroedinger's equation is not a relativistic equation so it is free to violate that principle.

Starting with Dirac's equation

$\left(\beta mc^{2}-i\hbar \sum _{k=1}^{3}\alpha _{k}\nabla _{k}\,c\right)\psi (\mathbf {x} ,t)=i\hbar {\frac {\partial \psi (\mathbf {x} ,t)}{\partial t}}$

And making use of $\beta ^{2}=1\,$, $\{\alpha _{i},\alpha _{j}\}=2\delta _{ij}$, and $\{\alpha _{i},\beta \}=0$ we can find an expression for $(i\hbar )^{2}{\frac {\partial ^{2}\psi (\mathbf {x} ,t)}{\partial t^{2}}}$

$(i\hbar )^{2}{\frac {\partial ^{2}\psi (\mathbf {x} ,t)}{\partial t^{2}}}=(i\hbar ){\frac {\partial }{\partial t}}(i\hbar ){\frac {\partial \psi (\mathbf {x} ,t)}{\partial t}}=(i\hbar ){\frac {\partial }{\partial t}}\left(\beta mc^{2}-i\hbar \sum _{k=1}^{3}\alpha _{k}\nabla _{k}\,c\right)\psi (\mathbf {x} ,t)$

$=\left(\beta mc^{2}-i\hbar \sum _{k=1}^{3}\alpha _{k}\nabla _{k}\,c\right)(i\hbar ){\frac {\partial }{\partial t}}\psi (\mathbf {x} ,t)$

$=\left(\beta mc^{2}-i\hbar \sum _{k=1}^{3}\alpha _{k}\nabla _{k}\,c\right)\left(\beta mc^{2}-i\hbar \sum _{l=1}^{3}\alpha _{l}\nabla _{l}\,c\right)\psi (\mathbf {x} ,t)$

$=\left(\beta ^{2}m^{2}c^{4}-i\hbar \sum _{k=1}^{3}\{\alpha _{k},\beta \}\nabla _{k}m^{2}c^{3}+(-i\hbar )^{2}{\frac {1}{2}}\sum _{k=1}^{3}\sum _{l=1}^{3}\{\alpha _{k},\alpha _{l}\}\nabla _{k}\nabla _{l}c^{2}\right)\psi (\mathbf {x} ,t)$

$=\left(m^{2}c^{4}+(-i\hbar )^{2}\sum _{k=1}^{3}\sum _{l=1}^{3}\delta _{kl}\nabla _{k}\nabla _{l}c^{2}\right)\psi (\mathbf {x} ,t)$

$=\left(m^{2}c^{4}+(-i\hbar )^{2}c^{2}\nabla ^{2}\right)\psi (\mathbf {x} ,t)$

so

${\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}\psi -\nabla ^{2}\psi +{\frac {m^{2}c^{2}}{\hbar ^{2}}}\psi =0.$

Which is the Klein-Gordon equation!!

Dauto (talk) 22:21, 21 February 2010 (UTC)

Hydrogen bond position transition

I was trying to understand the File:Glycine-zwitterion-2D-skeletal.png variant of Glycine but then I realized that File:Glycerin Skelett.svg might be subject to the same hydrogen bond position transition.

What reasons are there to believe that glycerin hydrogen bonds are stationary? 99.60.3.241 (talk) 05:02, 18 February 2010 (UTC)

In glycine (the neutral form), there is an "acid" part (a hydrogen can be released easily) and a "base" part (has a high affinity for free hydrogen). Acids and bases react with each other pretty well, and the result is the zwitterionic form. Glycerin does not have any part that is particularly acidic or basic, so it does not change to an alternate hydrogen attachment pattern. DMacks (talk) 18:25, 18 February 2010 (UTC)

Black hole

Hello i have read article on black hole but i do not understand how in some documentary space time is shown to be warped so much that it is such one layer is underneath or over lap with another layer such that a large enough distortion of gravity though to another place in space time that is space can be warped by mass but how can it tunnel be created or what causes space to warp in such that the infinite steep sides of a black hole gravity impression comes out on space time instead of just going forever (Dr hursday (talk) 06:29, 18 February 2010 (UTC))

It sounds like an Einstein-Rosen bridge. Check out that article and see if it answers your questions.  :) Mac Davis (talk) 06:54, 18 February 2010 (UTC)

Hello yes this is what I am talking about but i do not understand how the "U" curve at the left of this picture http://wiki.riteme.site/wiki/File:Worm3.jpg occurs. what causes this? (Dr hursday (talk) 07:01, 18 February 2010 (UTC))

The U is only there because of the way the image is drawn. The same thing is happening in this image (imagine the plane "above" the wormhole extending on forever instead of only in two spots). The only difference is the way it was drawn. The U is there because two places in spacetime are connected by a jump through a higher dimension. If you were on any part of the "U" you would not notice any bends and it would appear perfectly "flat." Mac Davis (talk) 07:43, 18 February 2010 (UTC)

The "higher dimension" has no physical relevance, it may be useful to "visualize" the situation but it need not "exist" in any physical sense. The last paragraph in http://home.fnal.gov/~skent/cosmo/cosmo3.pdf makes this point too. —Preceding unsigned comment added by 157.193.173.205 (talk) 08:28, 18 February 2010 (UTC)

Strange bug

Omg what is this. Can someone tell me? --✶♏‭ݣ 06:31, 18 February 2010 (UTC)

Belostomatidae. -- kainaw™ 06:35, 18 February 2010 (UTC)

Ew ew ew. But thank you. Ew. *shudders* --✶♏‭ݣ 06:37, 18 February 2010 (UTC)

Sometimes fear of the unknown is the scariest thing of all!!! What the Jesus God Hell ever happened to curiosity. 86.4.186.107 (talk) 06:58, 18 February 2010 (UTC)

Given that the sting of the Belostomatidae is considered "one of the most painful that can be inflicted by any insect" and thus is beyond the highest on the Schmidt Sting Pain Index (which is limited to Hymenoptera) "4.0+ Bullet ant: Pure, intense, brilliant pain. Like fire-walking over flaming charcoal with a 3-inch rusty nail in your heel" perhaps Belostomatidae should be rated "5.0+ Jesus God Hell that hurts!" 58.147.58.28 (talk) 08:33, 18 February 2010 (UTC)

Why the profanity? Did your question get a better or quicker answer because of it? Just curious. Kingsfold (talk) 14:51, 18 February 2010 (UTC)

Did the profanity ofend you?Dauto (talk) 15:27, 18 February 2010 (UTC)

No. The answer was quick because all it took was a simple web search. Instead of trying to make a highly juvenile joke by seeing how profane I could be, I went to http://tineye.com and pasted in the URL of the photo. It showed multiple results. The second one was to a Russian site that had a link below the photo right back to the Wikipedia page for the bug. Then, all I had to do was put a link to the article here. I didn't complain because many people seem to prefer to ask questions instead of searching for themselves. -- kainaw™ 14:58, 18 February 2010 (UTC)

Thanks for that link. I didn't know that site. Dauto (talk) 15:27, 18 February 2010 (UTC)

There's a Firefox extension that lets you just right-click on an image and search for it in Tineye. Unfortunately, Tineye seems to have a very tiny index. I wish Google would just buy them and do it right. --Sean 16:10, 18 February 2010 (UTC)

I think it's obvious that the OP was humorously incorporating the sort of reaction one might make upon seeing this beast into his/her question. "Jesus", "God", and "Hell" seem pretty mild profanities for such a sight. --Sean 16:13, 18 February 2010 (UTC)

If you think that's bad, you should meet my Italian girlfriend. Imagine Reason (talk) 16:59, 18 February 2010 (UTC)

I hope you're talking about the use of colorful language and not about the Belostomatidae! (I also hope she doesn't read this page!) SteveBaker (talk) 20:17, 18 February 2010 (UTC)

I'll point out, just because nobody else yet has, that the stuff on the back is a mass of eggs. Looie496 (talk) 00:42, 19 February 2010 (UTC)

Although Ferrofluid smeared on the back of a magnetic beetle would appear similar. I have changed the question title for clearer reference. Cuddlyable3 (talk) 16:52, 19 February 2010 (UTC)

I agree with that but for clarification for any future readers, the profanity discussed above was largely in the title [4] Nil Einne (talk) 22:44, 19 February 2010 (UTC)

What would happen to photon

Hello if I shine a flash light out into space in such a direction that it never encounters anything what happens to the photon over time? (Dr hursday (talk) 06:55, 18 February 2010 (UTC))

It keeps going and will get redder due to the metric expansion of space according to Hubble's law. Mac Davis (talk) 07:20, 18 February 2010 (UTC)

If the photon gets redder where does the energy go? Ariel. (talk) 07:44, 18 February 2010 (UTC)

Nowhere. Hubble flow implies that the farther away one looks, the faster the local matter is moving away from you. By extension, if one travels to those distant places, then you have to subtract the effect of the average local velocity when considering your motion. Hence the farther the photon travels the more it will appear doppler shifted with respect to the local standard of rest. Dragons flight (talk) 08:15, 18 February 2010 (UTC)

You may be concerned that energy does not seem to be conserved in this scenario. That is correct. In general relativity, energy is not conserved. There's a new blog post at Cosmic Variance today explaining this. -- Coneslayer (talk) 17:21, 22 February 2010 (UTC)

Put another way, the photon appears redder because whoever is observing it happens to be moving away from us. Someone moving towards us would see a bluer photon. I think this effect is indistinguishable from that of the space itself between the objects having expanded.. EverGreg (talk) 09:22, 18 February 2010 (UTC)

It's not only indistinguishable. It is the same thing. Dauto (talk) 13:32, 18 February 2010 (UTC)

No. The redshift due to the expansion of space depends on how much space has expanded during the photon's free flight. The usual picture of this is that it's the light-wave that's stretched along with the space it travels in. These are distinctions with real consequences. Consider for instance that distant objects are more red-shifted than nearby ones, so their apparent speed is greater, but an object does not feel acceleration as it recedes from us in this way. Redshift due to expansion also allows an object to speed away from us faster than the speed of light, which is impossible with doppler shift. (Yeah, I read up on the redshift article :-P) EverGreg (talk) 19:32, 18 February 2010 (UTC)

This difference you are pointing out is just an illusion. All redshift derives from the same principle no matter whether it is a gravitational redshift, doppler redshift or cosmological expansion redshift. In fact those distinctions are not as relevant as they sound since what is a doppler redshift for an specific choice of coordinates will be a gravitational redshift for a different choice of coordinates and vice-versa. The cosmological expansion redshift is an artifact from our choice of coordinates. specifically our choice of using comoving coordinates. Dauto (talk) 20:49, 18 February 2010 (UTC)

You cannot do a Lorentz transformation such that an object is accelerating in one choice of coordinates but not in another. EverGreg (talk) 09:13, 19 February 2010 (UTC)

Yes, true but

• A) Lorentz transformations explain the bulk of the effect. See that[5] paper for an explanation on how to build an expanding universe without matter or cosmological constant.
• B) There is no reason to restrict yourself to Lorentz transformations - they do not hold any special status within General Relativity.

Dauto (talk) 15:21, 19 February 2010 (UTC)

All of this fancy color shift stuff only happens from the point of view of the observer at the source of the light. From the point of view of the photon - nothing whatever happens - it just keeps on going. If space is infinite (we're not 100% sure of that) then it'll keep going forever completely unchanged. If space is finite then perhaps it 'wraps around' and comes back towards you from the opposite direction - but it's still completely unchanged. Photons can't "degrade" over time because they are travelling at the speed of light - and for them, the whole of eternity passes by in zero time. If they don't hit something (which seems highly implausible), then nothing can change because for them, time isn't advancing at all. SteveBaker (talk) 15:43, 18 February 2010 (UTC)

If by "never encounters anything" you mean your beam of light misses large objects like stars, galaxies and space rocks, its fate will likely be extinction in the interstellar medium. --Sean 16:19, 18 February 2010 (UTC)

Perhaps its information content will be absorbed into the Omega point. Graeme Bartlett (talk) 03:24, 19 February 2010 (UTC)

OK, I know I'm an ignoranimus about physics, but the OP stated "it never encounters anything". If you observe it, hasn't it encountered something, namely your eyeball? ←Baseball Bugs ^{What's up, Doc?} carrots→ 05:40, 19 February 2010 (UTC)

Not sure, but I suspect all photons are only produced with a matching absorbtion elsewhere.114.75.18.3 (talk) 06:48, 19 February 2010 (UTC)

Not sure what that means. In any case, the earlier discussion seemed to be confusing photons with galaxies. Obviously, galaxies can produce gazillions of photons, and those photons can appear red or blue depending on whether the galaxy is approaching or receding. But talking about photons that way implies that photons are ejecting other photons that would be visible without seeing the original photon. Maybe I'm wrong, but I don't think it works that way. ←Baseball Bugs ^{What's up, Doc?} carrots→ 06:54, 19 February 2010 (UTC)

How much sodium in 1g of Baking powder?

How much sodium is there in 1g of baking powder, after it has been used to make a cake for example? Thanks 89.243.151.96 (talk) 14:50, 18 February 2010 (UTC)

this suggests that there are 520mg per teaspoon. (In that brand, anyway.) I don't know how many teaspoons are in a gram of baking powder, though. Sorry. APL (talk) 15:29, 18 February 2010 (UTC)

It says on your picture that 1/8 teaspoon = 0.6 grams, so about 1/5 teaspoon per gram. --Sean 16:24, 18 February 2010 (UTC)

Most baking powdersoda is "sodium hydrogen carbonate" which is NaHCO₃ - so there is one atom of sodium, one hydrogen, one carbon and three oxygen. You have to figure out the atomic weight of each atom - add them up and then you know the ratio of sodium to the rest of the elements. Then you can figure out the fraction of a gram that is sodium. According to List of elements: Roughly, Hydrogen is 1, Carbon is 12, Oxygen is 16 and Sodium is 23. So one molecule of this compound has a total atomic weight of 23+1+12+3x16 = 84. So for every 84 grams of baking soda, 23 grams is sodium which means that about 27.4% of this stuff is sodium. There is therefore 0.274g of sodium in every gram of sodium hydrogen carbonate - if your brand of baking powder mixes that with some other 'stuff' (which is possible, for example to stop it clumping) then the answer will be different - check on the ingredients list on the packet. SteveBaker (talk) 15:35, 18 February 2010 (UTC)

Steve, that's baking soda. Baking powder is different. -- Flyguy649 ^talk 15:50, 18 February 2010 (UTC)

Baking powder says "Most commercially-available baking powders are made up of an alkaline component (typically baking soda), one or more acid salts, and an inert starch". So I guess we need to account for the other stuff too. SteveBaker (talk) 17:22, 18 February 2010 (UTC)

(EC)It would depend on the brand. In addition to the sodium bicarbonate (baking soda) and acid, like cream of tartar, baking powders contain various amounts of cornstarch. The acids used seem mostly not to contain sodium. So the answer is in how much baking soda is in baking powder. While this article gives some suggestions on how to make your own baking powder, it uses teaspoon measurements (not weight) for the ingredients. That article gives ratios ranging from 1 1/4:1 to 2:1 cream of tartar:baking soda. You should be able to figure out a rough answer by using the densities of each reagent. -- Flyguy649 ^talk 15:42, 18 February 2010 (UTC)

(edit conflict) Baking powder will certainly be mixed with other stuff; that's what makes it different from baking soda. Most baking powder is sodium hydrogen carbonate as well as an acid, like cream of tartar or monocalcium phosphate. They also typically contain an inert filler, like cornstarch. Look on the ingredient label, for both the serving size (hopefully they give a value in grams) and the amount of sodium per serving. Divide the second by the first to find grams sodium (or probably milligrams sodium) per gram powder. Buddy431 (talk) 15:49, 18 February 2010 (UTC)

This particular brand, for example, appears to contain 65 mg of sodium per 0.6 g serving, which is 108 mg sodium per gram baking powder. Buddy431 (talk) 15:53, 18 February 2010 (UTC)

The brand in my kitchen (Magic baking powder)has 45 mg of sodium per 0.6g (1/8 tsp) serving or 75 mg per gram of baking powder. This plus Buddy431's contribution above shows that there is a huge variation by brand. -- Flyguy649 ^talk 15:59, 18 February 2010 (UTC)

Yes and given all this, I think it's clear that unless the OP provides the precise brand and name of baking powder, we can't answer the question. It would surely be easier for the OP to just look on the label which is probably on his/her product anyway Nil Einne (talk) 16:38, 18 February 2010 (UTC)

None of the brands of baking powder I have in my possession or have seen in supermarkets have mentioned how much sodium they have in them. I presume they are trying to hide that they have a lot. 89.243.197.22 (talk) 15:19, 21 February 2010 (UTC)

File:refraction.jpg

what should be the minimum distance of the sun from the horizon so as to enable the observer to see it's image? —Preceding unsigned comment added by 117.201.65.74 (talk) 16:37, 18 February 2010 (UTC)

I converted your header into a proper subject header with the image as a link Nil Einne (talk) 16:38, 18 February 2010 (UTC)

Even only a part of the Sun need be above the sea horizon for its image to be visible. Cuddlyable3 (talk) 16:47, 19 February 2010 (UTC)

Monarch Butterflies

Where do Monarch butterflies go in the winter? —Preceding unsigned comment added by Dredfern (talk • contribs) 17:10, 18 February 2010 (UTC)

They migrate. "The most famous Lepidopteran migration is that of the Monarch butterfly which migrates from southern Canada to wintering sites in central Mexico. In late winter/early spring, the adult monarchs leave the Transvolcanic mountain range in Mexico for a more northern climate. Mating occurs and the females begin seeking out milkweed to lay their eggs, usually first in northern Mexico and southern Texas. The caterpillars hatch and develop into adults that move north, where more offspring can go as far as Central Canada until next migratory cycle." --Mr.98 (talk) 17:19, 18 February 2010 (UTC)

Addiction

Hi,

I'm very worried. In fact I think I'm addicted. It's a highly dangerous drug; an overdose can lead to death and the substance constitutes 98% of all cancer cells. So, should I be going to Waterholics Anonymous? —Preceding unsigned comment added by 86.150.210.228 (talk) 18:04, 18 February 2010 (UTC)

It's not as funny if you call it water. Try Dihydrogen monoxide - that sounds really scary. --Tango (talk) 18:06, 18 February 2010 (UTC)

Oh my God, you're not inhaling it, are you? AlexHOUSE (talk) 18:54, 18 February 2010 (UTC)

Think this is bad? I'm far more concerned about oxygen being a mutagen. Regards, --—Cyclonenim | Chat  21:39, 18 February 2010 (UTC)

Ah, but the antidote is red wine - have enough of that, and you won't be concerned at all! --Tango (talk) 23:05, 18 February 2010 (UTC)

I hate wine. I'm f**ked, but I guess I could just start on those vitamins... Regards, --—Cyclonenim | Chat  23:52, 18 February 2010 (UTC)

Bear in mind the kind you take intravenously is usually cut with salt. AlmostReadytoFly (talk) 09:57, 19 February 2010 (UTC)

waters not a drug —Preceding unsigned comment added by 67.246.254.35 (talk) 05:56, 19 February 2010 (UTC)

Fish reproduce in it (as per W.C. Fields). ←Baseball Bugs ^{What's up, Doc?} carrots→ 06:34, 19 February 2010 (UTC)

On the contrary, you could very broadly define water as a drug. It can alter bodily functions. If you're dehydrated, it's a drug which can relieve symptoms. Regards, --—Cyclonenim | Chat  11:28, 19 February 2010 (UTC)

Water intoxication is real. DMacks (talk) 11:38, 19 February 2010 (UTC)

Twin Cities

What is the most twin city to New Bedford, MA in Europe, geographically and climate-wise? Note I am not referring to sister cities. --Reticuli88 (talk) 18:49, 18 February 2010 (UTC)

The article about New Bedford does not give any info about its climate. I'm not sure if places having the same Hardiness zone would feel as if they had the same climate from a human point of view, since the seasonal daylight, summer temperatures, and rainfall may differ. What hardiness zone is it please? 92.24.96.55 (talk) 22:21, 18 February 2010 (UTC)

I think it is 6a. --Reticuli88 (talk) 22:51, 18 February 2010 (UTC)

Assuming New Bedford has a Hardiness Zone of 6 and a Heat Zone of 4, then the places listed in Europe with the same figures are Bratislava Slovakia, and Vienna Austria. They are both inland. Kaliningrad in Russia is more coastal, but it is only heat zone 2 and I think its latitude is higher so the seasonal daylight will be different. Do not know about the rainfall. 89.240.61.50 (talk) 23:51, 18 February 2010 (UTC)

Of course there is nothing close to a perfect match, but in terms of setting, population, and climate my choice would be La Rochelle, France. Looie496 (talk) 00:33, 19 February 2010 (UTC)

The climate would be much warmer in the winter than it would be in New Bedford, and probably cooler in summer. I assumed that the "geographical" similarity specified by the OP did not include population. 78.146.181.195 (talk) 00:49, 19 February 2010 (UTC)

Thanks everyone. Just wanted to know if I traveled to Europe today from New Bedford, MA, what city would seem like I never left MA - meaning the landscape (hills n such) and weather-wise. --Reticuli88 (talk) 13:20, 19 February 2010 (UTC)

WAG here, but maybe Sheffield? --TammyMoet (talk) 15:14, 19 February 2010 (UTC)

No, that would be one of the least similar places in all of Europe. 78.147.225.78 (talk) 20:36, 19 February 2010 (UTC)

If you are willing to accept somewhere with a warmer winter and a cooler summer then you have much more choice. With this in mind, if you want to limit yourself to Britain, then Nairn in Scotland would be a possibility, but with a smaller population. Inverness has a similar population. But even in Scotland, the urbanisation of the surrounding area is probably going to be much more than I expect it is around New Bedford. 78.147.225.78 (talk) 20:43, 19 February 2010 (UTC)

It's really unlikely that you'll find a place with similar weather in Europe. New Bedford has that vast continental landmass off to the west and only ocean to the east. Nowhere in Europe has that much continental mass behind it and such as there is tends to be to the east - not to the west. All of that results in very different climates in Europe and the USA. Also, the prevailing ocean currents are quite different. New Bedford gets the gulf stream bringing warm water up from the south. There isn't an equivalent thing in Europe. SteveBaker (talk) 01:01, 20 February 2010 (UTC)

I'm shocked - wernt you brought up in Britain? I thought every schoolboy knew about the Gulf Stream coming across the Atlantic and keeping Britain, plus Scandinavia and western Europe, much warmer than it would otherwise be for its latitude. When I was at school it was called the Gulf Stream, now people call it the North Atlantic Drift. Where the OP is does I believe have what meteorologists call a Continental climate, while Britain for example is a maritime climate (Oceanic climate) with less extremes of temperature. Given that climate difference, plus the different latitudes, and the far greater population density, makes it as people have said impossible to match. 78.149.241.220 (talk) 16:54, 20 February 2010 (UTC)

Not quite all true. As the Gulf Stream article to which you linked says, the Gulf Stream proper does indeed warm all of the USA's eastern seaboard, and the North Atlantic Drift is not a recent synonym for it, but an offshoot from it. 87.81.230.195 (talk) 00:40, 21 February 2010 (UTC)

You've got the wrong end of the stick. I was referring to the North Atlantic Drift which was, in a classroom in the UK many years ago, referred to as the Gulf Stream. 78.146.167.216 (talk) 01:44, 21 February 2010 (UTC)

Freezing Point Depression and Boiling Point Elevation

After reading one of the above posts I came across this picture. I understand what the unbroken segments of the lines are. But how can one understand the dashed segments of the lines? For example on the right hand side of where the black "solid" line crosses the dark blue "Liquid (pure solution)" line? •• Fly by Night (talk) 22:55, 18 February 2010 (UTC)

It appears that the dashed lines are merely the continuation of the trends for the chemical potential versus temperature. That is, if the solid didn't melt at its melting point, its chemical potential vs. temperature curve would follow the dashed black line. I guess that the dashed lines are just in there to make the intersections more clear. Buddy431 (talk) 04:49, 19 February 2010 (UTC)