Wikipedia:Reference desk/Archives/Science/2010 March 21

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March 21

Intermittent blood supply

Hi. While all organs of the body are intact and connected by the nervous and lymphatic system, are there any organs that would either not survive, suffer permanent damage or organ failure if blood supply to the organ was cut off or significantly reduced for say, one minute? This is not a request for medical advice. Of course, this question does not include the heart, which transmits the blood supply, and "organ" in this case could mean intermittent or interrupted supply to a portion of the organ rather than its whole. Thanks. ~AH1^(TCU) 02:12, 21 March 2010 (UTC)

I took the liberty of fixing your red link. Dauto (talk) 02:22, 21 March 2010 (UTC)

So did I. - Nunh-huh 02:28, 21 March 2010 (UTC)

But history says that Dauto didn't fix any red links...can my question be answered? ~AH1^(TCU) 02:44, 21 March 2010 (UTC)

Of course, we have an article that deals with this. As that section states, highly aerobic organs like heart and brain sustain irreversible damage after 3-4 minutes. Our Cardiac arrest article is also relevant, and highlights the importance of body temperature and other variables. So, I don't think one minute of ischemia would predictably cause irreversible damage to any human organ under normal circumstances. Of course, I'm not recommending that anyone try this at home... -- Scray (talk) 04:06, 21 March 2010 (UTC)

Indeed not! Brain ischemia says that 10 seconds without blood supply to the brain will cause unconsciousness and 20 seconds shuts down all electrical activity. [1] says "Interruption of blood supply to the brain tissue for 2-5 minutes may result in permanent damage." - so in theory, a one minute interruption might not cause lasting damage. SteveBaker (talk) 04:19, 21 March 2010 (UTC)

To what is "Indeed not!" directed? My answer addressed irreversible damage, and nothing in your response refuted what I said. -- Scray (talk) 15:06, 21 March 2010 (UTC)

I presume the disclaimer (don't try this at home) hence why SB first pointed out the potential negatives before agreeing that it might not cause lasting damage Nil Einne (talk) 15:12, 21 March 2010 (UTC)

Thanks for clarifying - makes sense! One also has to worry whether our ability to detect irreversible damage might be too insensitive to detect microscopic foci of damage that, cumulatively over the years, result in changes known as "aging" or "dementia". -- Scray (talk) 21:04, 21 March 2010 (UTC)

Lorentz invariant

I have seen the Lorentz transformations used to derive the Lorentz invariant, but can the Lorentz invariant be used as a postulate to derive the transformations? 173.179.59.66 (talk) 06:36, 21 March 2010 (UTC)

Yes, in fact that's the way it's uasually done. You star with the relativistic interval which is a lorentz invariant. Assuming $dy=dz=0\,$ for simplicity,

$dS^{2}=(cdt)^{2}-dx^{2}$, and

$dS'^{2}=(cdt')^{2}-dx'^{2}$.

Now chose an interval for which $dS=dS'=0\,$ and you get

${\frac {dx}{dt}}={\frac {dx'}{dt'}}=c$

In other words, if an object moves at the speed of light in a coordinate system ${\frac {dx}{dt}}=c$, then it is also moving at the speed of light in the $S'\,$ system ${\frac {dx'}{dt'}}=c$. But $S'\,$ is completely arbitrary so the speed of the object must be the same in all reference systems. which is the usual starting point of the theory of relativity

Dauto (talk) 15:41, 21 March 2010 (UTC)

But it seems that the Lorentz invariant is justified on the grounds of c being the same in all reference frames. If we consider a beam of light emitted at the origin in two reference frames, x^2 + y^2 + z^2 = (ct)^2 and x'^2 + y'^2 + z'^2 = (ct')^2. Then x^2 + y^2 + z^2 - (ct)^2 = x'^2 + y'^2 + z'^2 - (ct')^2 = 0. So for any arbitrary displacement (x, y, z), we should have x^2 + y^2 + z^2 - (ct)^2 = A[x'^2 + y'^2 + z'^2 - (ct')^2] (there's no exponent to preserve linearity). But because y' = y and z' = z (justified under the grounds of symmetry), x^2 + y^2 + z^2 - (ct)^2 = x'^2 + y'^2 + z'^2 - (ct')^2 for any (x, y, z). My question is how we can go from this, to the equations x' = (x - vt)/sqrt(1 - (v/c)^2) and so on. 173.179.59.66 (talk) 00:58, 22 March 2010 (UTC)

I see that I missunderstood your question. Let me have a second stab at it.

Due to spacetime homogeneity we can assume that the transformation is linear so we can write (using Einstein's summation convention).

$x'^{\mu }=\Lambda _{\lambda }^{\mu }x^{\lambda }$

Where the $\Lambda _{\lambda }^{\mu }$ are to be considered as yet unknown.

Now we can use the metric tensor $g_{\mu \sigma }$

$g={\begin{bmatrix}1&0\\0&-1\end{bmatrix}}\$

And build invariants (note that I will be working with 2-dimentional spacetime for simplicity)

$dS^{2}=g_{\lambda \rho }dx^{\lambda }dx^{\rho }$

and

$dS'^{2}=g_{\mu \sigma }dx'^{\mu }dx'^{\sigma }$

$dS\,$ is an invariant so we have

$dS'^{2}=dS^{2}$

$g_{\mu \sigma }dx'^{\mu }dx'^{\sigma }=g_{\lambda \rho }dx^{\lambda }dx^{\rho }$

$g_{\mu \sigma }\Lambda _{\lambda }^{\mu }dx^{\lambda }\Lambda _{\rho }^{\sigma }dx^{\rho }=g_{\lambda \rho }dx^{\lambda }dx^{\rho }$

$g_{\mu \sigma }\Lambda _{\lambda }^{\mu }\Lambda _{\rho }^{\sigma }=g_{\lambda \rho }$

which gives us four equations, but only three are independent, namely

$\Lambda _{0}^{0}\Lambda _{0}^{0}-\Lambda _{0}^{1}\Lambda _{0}^{1}=1$

$\Lambda _{0}^{0}\Lambda _{1}^{0}-\Lambda _{0}^{1}\Lambda _{1}^{1}=0$

$\Lambda _{1}^{0}\Lambda _{1}^{0}-\Lambda _{1}^{1}\Lambda _{1}^{1}=-1$

Now we square the middle equation and make a couple of substitutions from the other two

$\left[\Lambda _{0}^{0}\Lambda _{1}^{0}\right]^{2}=\left[\Lambda _{0}^{1}\Lambda _{1}^{1}\right]^{2}$

$\left[1+\left(\Lambda _{0}^{1}\right)^{2}\right]\left(\Lambda _{1}^{0}\right)^{2}=\left[1+\left(\Lambda _{1}^{0}\right)^{2}\right]\left(\Lambda _{0}^{1}\right)^{2}$

$\left(\Lambda _{1}^{0}\right)^{2}=\left(\Lambda _{0}^{1}\right)^{2}$

$\Lambda _{1}^{0}=\Lambda _{0}^{1}$

We take the positive square root here because the negative one leads to either time reversals or space reflexion which we are not interestead in.

Plugging that back into the equations above (after the word "namely"), we get

$\left(\Lambda _{0}^{0}\right)^{2}=\left(\Lambda _{1}^{1}\right)^{2}=1+\left(\Lambda _{1}^{0}\right)^{2}$

Which allows us the parametrizations

$\Lambda _{1}^{0}=\Lambda _{0}^{1}=sinh\phi$

$\Lambda _{0}^{0}=\Lambda _{1}^{1}=cosh\phi$

$\phi \,$ is called the rapidity.

Another (more familiar) possible parametrization is

$\Lambda _{1}^{0}=\Lambda _{0}^{1}=\beta \gamma$

$\Lambda _{0}^{0}=\Lambda _{1}^{1}=\gamma$

in which case $\beta ={\frac {\Lambda _{1}^{0}}{\Lambda _{0}^{0}}}=tanh\phi$

Also

$\gamma ^{2}=cosh^{2}\phi ={\frac {1}{1-tanh^{2}\phi }}$

Which finally leads to

$\gamma ={\frac {1}{\sqrt {1-\beta ^{2}}}}$

Short-pulse NMR

I'm trying to understand the principle/concept behind short pulse-NMR (not the original continuous-wave (CW) method). I read the NMR article's relevant section but there are several things in the article that don't make sense. Please clarify:

"A short square pulse of a given "carrier" frequency "contains" a range of frequencies centered about the carrier frequency, with the range of excitation (bandwidth) being inversely proportional to the pulse duration" >>>Is this because period is just inverse of frequency? Is the pulse duration the same as period of the pulse? Also, how can a group of frequencies give a single, unified period? I thought frequency and period of a wave were in a 1-to-1 relationship.

"Applying such a pulse to a set of nuclear spins simultaneously excites all the single-quantum NMR transitions." >>>After resonance happens, how do we actually know which frequency among many others contained in the pulse was the resonance frequency? Say we treated a sample with a pulse containing waves of 3 frequencies: 1nm, 2nm, or 3nm. If 2nm were the resonance frequency, how do we know if it's absorbed and then emitted?

"In terms of the net magnetization vector, this corresponds to tilting the magnetization vector away from its equilibrium position (aligned along the external magnetic field)." >>>What is this magnetization vector? I have no intuition about it and I've never seen it mentioned when I learned the CW method NMR

"The out-of-equilibrium magnetization vector precesses about the external magnetic field vector at the NMR frequency of the spins. This oscillating magnetization vector induces a current in a nearby pickup coil, creating an electrical signal oscillating at the NMR frequency. This signal is known as the free induction decay (FID) and contains the vector-sum of the NMR responses from all the excited spins. In order to obtain the frequency-domain NMR spectrum (NMR absorption intensity vs. NMR frequency) this time-domain signal (intensity vs. time) is FTed." >>>My understanding is that the pulse that has the resonance frequency among its many other non-resonance frequencies will cause the magnetization vector to move out of the equilibrium and the out-of-equilibrium vector will precess in a different space, sending signal to a coil. The coil measures the signal and plots its intensity VS time. Then we FT this to turn this into the typical NMR spectrum we know. The intensity VS time graph only has one frequency so if we FT this, then the intensity VS resonance frequency (chemical shift) graph will have only one peak, at the resonance frequency (x-axis). Please correct me if anything I just said is wrong. —Preceding unsigned comment added by 70.68.120.162 (talk) 07:15, 21 March 2010 (UTC)

This excellent blog entry answers many of your questions, with (sound) examples : http://scienceblogs.com/builtonfacts/2010/03/hearing_the_uncertainty_princi.php 83.134.177.217 (talk) 08:36, 21 March 2010 (UTC)

You might find impulse response a very useful concept also. Note that a short pulse has a large bandwidth, so it captures a lot of information in the frequency domain. Nimur (talk) 18:46, 21 March 2010 (UTC)

This image shows how a square wave consists of the sum of many frequencies, as can be demonstrated by taking its Fourier transform. In pulsed NMR a single-frequency carrier is gated by a lower-frequency square wave, effectively multiplying the carrier frequency by all the component frequencies of the square wave. Cuddlyable3 (talk) 23:02, 21 March 2010 (UTC)

speaker and receivers

what is the main difference in structure of receivers and speakers. can speakers be changed into receivers by just altering some connections?

1 MORE THING

i have a receiver with 2 wires coming out, and a speaker with 2 wires coming out, both will work with 2 pencil cells (i suppose so), how should i connect them to make a mic {receive my voice and give it out too}.... thanx--Myownid420 (talk) 10:16, 21 March 2010 (UTC)

I suspect there may be a language problem here. A receiver normally means a device that receives radio wave transmissions and passes them on to the amplifier and speakers. However, I think you are using "receiver" to mean microphone. Is that right ? StuRat (talk) 14:26, 21 March 2010 (UTC)

As for whether a single device can work as both a microphone and speaker, I think it's possible, but the sound quality is quite low, so it isn't normally done that way. For your 2nd Q, the wires are inputs in the case of the speaker and at least one is an output for the microphone, so it would require switching where they plug into the recording device. StuRat (talk) 14:33, 21 March 2010 (UTC)

oh! yes i mean microphone no receivers.. sorry--Myownid420 (talk) 16:45, 21 March 2010 (UTC)

Now, as to why you get worse sound quality when using a combination microphone/speaker, I don't know the details, but you do need larger speakers to get good bass, while this isn't as much a factor for microphones. The speaker and microphone on a telephone are of similar size, but they pretty much just cover the treble frequencies. So, if you don't care about bass, it's more doable. StuRat (talk) 17:51, 21 March 2010 (UTC)

Loudspeakers were used for decades with great success as both microphone and reproducer in intercoms. An electrodynamic microphone is basically a loudspeaker. Alexander Graham Bell's primitive telephone used them interchangeably (without any amplification) before the carbon microphone was invented by others. Edison (talk) 03:21, 22 March 2010 (UTC)

thanx for answers but how could i change a speaker in to a microphone —Preceding unsigned comment added by Myownid420 (talk • contribs) 08:17, 22 March 2010 (UTC)

It will work (perhaps poorly, depending on the precise speaker) if you plug it into the 'in' rather than 'out' connection. Try taking some headphones and plugging them into your computer's microphone jack, for example. 94.168.184.16 (talk) 21:09, 22 March 2010 (UTC)

I've used a piezoelectric earbud as a microphone simply by plugging it into a computer's microphone jack. It works, but not very well -- you need a great deal of amplification, and even then, the sound is distorted. --Carnildo (talk) 00:36, 23 March 2010 (UTC)

guns in video games

I remember those days when i would play a bird shooting video game. a simple gun having one lens and a chip and a led and a tree type thing made of glass in it {i figured it out when i broke it}. it was connected to the video game keyboard by a fifteen pin socket, what you have to do is to just aim the flying bird {in tv screen} and shoot, and the bird is gone. that wasa a simple tv and even if precessor of video game come to know that the player had shot the shot, how would it come to know the aim was correct?, tv dosent had any sensors.

also note that it would not work if i touch the barrel of the gun and shoot

please explain how it worked..............thanx--Myownid420 (talk) 10:36, 21 March 2010 (UTC)

There are a few different ways to make this work- see Light gun. Staecker (talk) 11:06, 21 March 2010 (UTC)

You probably were using a NES Zapper. Basically whenever you press the trigger the screen would turn black except the targets (ducks) which would turn white. The zapper had a little chip in it that could detect if it was aimed at something white (of the right intensity of light). That way the gun would know if it was pointed in the right direction or not. Then the screen would go back to being in color again. Not the most accurate or foolproof method but it worked pretty well for children's games. You could notice the quick change if you looked at it out of the corner of your eye, I recall. --Mr.98 (talk) 15:05, 21 March 2010 (UTC)

oh! ya i remember it flashed whenever i shoot. but why it dosent work from snall range(very near) why?--Myownid420 (talk) 16:48, 21 March 2010 (UTC)

This is rather old tecnology. Our Sears Telegames Pong system had a gun for shooting a white dot on a black screen. This would have been in about 1978 or so. 75.41.110.200 (talk) 18:22, 21 March 2010 (UTC)

There are two ways these things could work - the simplest (as described above) just had a photocell in the barrel of the 'gun' which determined if you were pointing at something white when you pressed the trigger. That's really crude though because you only know that the gun was pointed at something bright - you don't know WHICH something. Also, you have to have your targets be white against a dark background - you can't draw clouds and trees and such. A more sophisticated version displays a light color on the screen all the time (grey, probably) and measures the amount of time between the vertical refresh signal going to the TV and light entering the barrel of the gun. Knowing that time lets you know where the gun is pointing because the progress of the raster scan is known. That approach allows the computer to show misses graphically - and if you hit a bird, then you know which one. There can also be other things on the screen, allowing you to draw anything you like as a 'background'. SteveBaker (talk) 19:17, 21 March 2010 (UTC)

Note that you could have clouds and trees with games like Duck Hunt, because of the aforementioned "flash" to the black/white mode. Which could have multiple targets, too? I wonder if they had two separate black/white flash modes. That would probably put an upper limit on the number of targets, though. --Mr.98 (talk) 19:20, 21 March 2010 (UTC)

The NES Zapper article suggests that different targets flashed separately, so the computer only has to match the gun-detector to the frame display. The idea of watching the raster-scan to know directly "where is the thing pointing" on an arbitrary screen is the basis for light pen devices. DMacks (talk) 19:25, 21 March 2010 (UTC)

Copperhead lookalike?

(I put this on the herp project page this morning, but then noted that the project has low activity)

I'm having a few worries about a species of snake that keeps popping up near our springhouse/smokehouse (the smokehouse has been re-purposed as a playhouse for my daughter). It certainly resembles a copperhead, but the markings are slightly more muted and they don's seem to get quite as large. I shot one of them last year, but another has moved in, and I don't want to kill it if it's not venomous (we have black northern watersnakes and garter snakes living in the same pond, and the garters actually "snuggle" with this snake on warm days). However, if it is venomous, it can't stay!

I didn't see any list of lookalikes on the article, but I've been told that there's a snake called a rat snake or king snake that is similar. Anyone know what they're talking about? I'm in eastern Pennsylvania. --SB_Johnny | ^talk 09:46, 21 March 2010 (UTC)

You might find this site [2] interesting, it actually discusses copperhead lookalikes. Richard Avery (talk) 13:36, 21 March 2010 (UTC)

Awesome, thanks! The one by the house is not a copperhead, thankfully. (We do have them in the woods, but that's not as alarming). --SB_Johnny | ^talk 14:12, 21 March 2010 (UTC)

hairdryers

Hypothetically were I to tape over most of the bit of a hairdryer where the air goes in, leaving only a little hole, would the air be sucked through there much faster or would it blow out the other end slower, or would the whole thing just explode? And would that make it louder or quieter? 148.197.114.158 (talk) 15:35, 21 March 2010 (UTC)

The fans are not fixed displacement types, so it would...

• Blow out slower
• Blow out much hotter air
• (as one did at work last week, when the fan fell off the shaft) maybe catch fire!

 Ronhjones  ^(Talk) 15:45, 21 March 2010 (UTC)

I'd say it would catch fire. The lubricating oil for the fan, in particular, as well as dust inside from previous usage. StuRat (talk) 17:37, 21 March 2010 (UTC)

Should have a high temp cut out, really. --BozMo talk 20:24, 21 March 2010 (UTC)

Eventually you would burn out or wear out the motor. It would be quite noisy and not blow out much air until then. Googlemeister (talk) 13:24, 22 March 2010 (UTC)

Why would it catch on fire? Just from that lack of air flowing through it? also, is there anything with a different sort of fan then, that might be affected differently? 148.197.114.158 (talk) 15:27, 22 March 2010 (UTC)

The heating elements must be cooled by blowing air or they get hot enough to ignite a fire. As BozMo pointed out, many countries require a temperature sensor to shut it down if it gets too hot, to prevent this. Some hair dryers also have a "fan only" setting, where no heat is supplied. Then the only heat would be that generated by friction in the fan motor. StuRat (talk) 15:34, 22 March 2010 (UTC)

Would a rediculously cheep hairdryer have only the fan and no extra heat?148.197.114.158 (talk) 20:30, 22 March 2010 (UTC)

I don't think they make them that cheap, because any potential customers would just put their head in front of a fan, instead. StuRat (talk) 00:41, 23 March 2010 (UTC)

Natural Gas

What is the chemical compund for Natural gas which is supplied by utility companies to businesses and residences? —Preceding unsigned comment added by 71.84.229.190 (talk) 17:28, 21 March 2010 (UTC)

It's mostly methane with some ethane, propane, and butane. Dauto (talk) 17:31, 21 March 2010 (UTC)

Odorants like t-butyl mercaptan are usually added too so that you can notice leaks. 75.41.110.200 (talk) 17:57, 21 March 2010 (UTC)

Fixed redlink 94.168.184.16 (talk) 18:21, 21 March 2010 (UTC)

And a possibly confusing typo. --Anon, 20:30 UTC, March 21, 2010.

For further information, we have a whole article with the obvious title of natural gas about the contents, odorants, etc. DMacks (talk) 22:54, 21 March 2010 (UTC)

Hmm, I wonder why the odorant in natural gas smells similar to a durian? ~AH1^(TCU) 23:34, 21 March 2010 (UTC)

Durian#Flavour and odour says there are sulfur compounds in it. I've never smelled one (and from that article, doesn't sound like I want to) so no idea myself. The talk-page mentions some disagreement about whether H2S in particular is involved, and the article relies heavily on a wide-ranging secondary/review article that doesn't exist (!) so can't find the underlying actual supporting refs. DMacks (talk) 09:30, 22 March 2010 (UTC)

The smell is mildly reminiscent of it, but not to the point where you'd confuse one with the other or anything. The durian's scent has a fruity, sweet, kind of undercoat to it. It's not so much that it smells inedible, so much as it smells like something very edible that's already been partly digested... Difficult to describe, really, but you can almost taste a sweet sliminess in the air. If you're curious, many Thai and Vietnamese restaurants will have, if not the fruit itself, at least a frozen milkshake version which will give you a feel for it. Matt Deres (talk) 16:42, 22 March 2010 (UTC)

You may also be able to find frozen durians in Asian grocery stores if you're really interested although don't expect them to be cheap. (You definitely see them here in Auckland so I would be surprised if you can't find them in most other countries with a resonable number of East Asian and South East Asian immigrats like the US, UK, Canada, Australia etc.) Of course the flavour and smell probably won't be as strong as fresh and this depends on the cultivar anyway Malaysia durians for example generally considered more pungent then Thai AFAIK. Nil Einne (talk) 14:15, 23 March 2010 (UTC)

Medical Advice

Is there somewhere on the internet where I can ask medical advice questions? I think I have a medical problem, but I cannot just go to a doctor easily and it'd be such a wase of time to travel weeks to see one if my problem isn't serious. Thank you —Preceding unsigned comment added by 94.4.255.173 (talk) 17:41, 21 March 2010 (UTC)

You should be able to phone a doctor, even if it isn't practical to visit them. There are lots of places on the internet where you can ask for medical advice, but I know of none where you can have any confidence in the answers. You need to talk to a doctor. --Tango (talk) 17:47, 21 March 2010 (UTC)

The answer to this is highly dependent on the person's location and resources. The IP address is London, UK, unless he/she is hiding his location. That does not match the "travel weeks to see one" phrase. Almost no doctors will give free advice by telephone without an institutional or already established relationship. There are for-profit internet medical consultation sites (just google "online medical consulting" and you will get dozens of choices), so if the questioner is willing to pay for the consultation, he can get answers from a doctor without a face-to-face visit. Obviously this would work better for some types of problems than others. alteripse (talk) 18:29, 21 March 2010 (UTC)

In the UK, almost all doctors will give free advice over the phone (to patients registered with them, anyway, and I'm assuming the OP has a specific doctor weeks away that they could call). An alternative is phoning NHS Direct (if the OP really is in the UK, this would be my recommendation - NHS Direct is really good. You'll get to talk to a nurse and they'll get a doctor to call you back if necessary). --Tango (talk) 18:34, 21 March 2010 (UTC)

I had no idea medical practice in the UK is that different from North America. Do British doctors have telephone hours, as I assume the volume could be horrific? Will they talk to patients not assigned to them by the NHS? alteripse (talk) 18:51, 21 March 2010 (UTC)

Exact procedures vary from GPs surgery to GPs surgery. Talking to a doctor on the phone would normally be used in emergencies, rather than for routine stuff. You would either phone the surgery and talk to the receptionist, who would choose to give you an emergency appointment or get a doctor to call you back, or you can go through NHS direct as I mentioned above, and a nurse will either advise you themselves or get a doctor to call you back. Patients aren't really assigned by the NHS. You register with a GP of your choice (assuming they have space on their books, anyway). For routine stuff, you have to go through your GP (who may refer you to a specialist), but for emergency stuff you can just call the nearest surgery (eg. if you fall ill while on away from home). --Tango (talk) 19:02, 21 March 2010 (UTC)

Thanks. Perhaps the reality is not as different as it seemed. In the US, if you are an established patient of a practice, you can usually call and talk to a nurse who will decide if you need an appt or might get an answer from the doctor or in some cases might have the doctor call you back. No US doctors will simply provide telephone consultations to new patients who may not be coming in. It seems unlikely to me that even a UK doctor would provide a telephone consultation to a patient who has not been seen and will not be seen in his practice. What else would they have time to do? alteripse (talk) 19:23, 21 March 2010 (UTC)

Tango represents things correctly but in my experience if you are away on holiday in a different part of the UK a local GP will generally return a call and try to be helpful, without requiring a temporary registration (which they will organise if they actually have to see you). A typical rental cottage with list the local GPs phone number. I think most GPs in the UK would return a call but I guess they would be circumspect about giving some types of advice. Most of such advice is probably about children anyway. Lots of things though can be answered on the phone. --BozMo talk 20:20, 21 March 2010 (UTC)

There are many docs in the U.S. who provide a call-in number to patients, along with instructions regarding when to use it. For some of us, that means under certain circumstances, for others that means at certain times. This is entirely up to the individual physician, but enhanced access can be very reassuring to the patient, and most patients respect the privilege. This is not just "boutique" medicine. -- Scray (talk) 21:00, 21 March 2010 (UTC)

Not for people who have not been their patients because there is no mechanism to charge, unless you are talking about an online or telephone only practice for profit. You ignored a key part of my assertion. alteripse (talk) 11:32, 22 March 2010 (UTC)

Sorry, I misunderstood the last part of your statement; I did not ignore it. When you said they were not assigned by the NHS, I assumed that this meant that they were known but not primarily assigned (as can happen in the U.S.). The words you used were not obvious to me. No disrespect meant! BTW, your comparison does not entirely hold because most docs I know in the U.S. don't bill for phone time - it's too hard to get paid enough to justify the paperwork, even if they could get paid at all. Just part of being a doctor (in a system that has poorly-aligned incentives). -- Scray (talk) 02:18, 23 March 2010 (UTC)

No offense taken, and you are right that all the incentives prevent telephone care. The point i was making is that exactly because doctors in the US cannot charge for telphone calls they will not make them except to established patients. Every telephone call costs time, requires paperwork, and incurs legal responsibility and risk. I cannot imagine anywhere in the world where doctors will treat a patient by telephone who is not included in their practice population (and hence the doctor is paid at least indirectly to provide care for them). alteripse (talk) 03:34, 23 March 2010 (UTC)

Are you trying to treat or diagnose? If you are trying to diagnose then I'd say definitely go to a doctor. No one over the internet will be able to diagnose you as well as a doctor in person and it is too easy to misdiagnose yourself just from what you think your symptoms are, confirmation bias and all that. If you are trying to treat then we have articles for a lot of illnesses which include typical treatment regiments and there are support groups and forums for a whole range of illnesses online. Vespine (talk) 23:44, 21 March 2010 (UTC)

Judging by the OP's wording I'm guessing they do not live in the US/UK since it would take weeks to get to a doctor? I have no idea in which countries this would be the case. Regards, --—Cyclonenim | Chat  09:21, 22 March 2010 (UTC)

They do not have to be in a country. Couldn't they be say on a slow car carrier ship in the middle of the Pacific or something? Googlemeister (talk) 13:31, 23 March 2010 (UTC)

If you're on a slow carrier ship, I'm not sure how you'd travel for weeks to go and see a doctor. If there's an emergency, the would I presume be ways for dealing with that e.g. travelling to the nearest port (probably wouldn't have to be one the ship can dock with), meeting up with another ship, calling in some sort of air rescue all of which would be problematic and may take several days but wouldn't seem to entail travelling several weeks to see a doctor. In a non emergency, it might take several weeks before you're in a port and can see a doctor, but you wouldn't be travelling several weeks to see a doctor. In any case, wouldn't most ships have some sort of protocol for dealing with these sort of things already? So yeah like Cyclonemin I'm having difficulty imagining many places where you'd have to travel for weeks to see a doctor (days yes) particularly places with decent internet access and would be interested in hearing where the OP is. Nil Einne (talk) 14:38, 23 March 2010 (UTC)

All commercial and military ships (even slowpoke car-carrier ferries) have at least one medic as part of the crew. If the OP is on a ship, he/she could just go to the ship's dispensary and talk to the medic about his/her problem. 24.23.197.43 (talk) 23:43, 23 March 2010 (UTC)

Meaning of retrolateral and prolateral

I see the terms retrolateral and prolateral used all the time in entomology papers, but I haven't been able find out what these terms actually mean. I've already checked anatomical terms of location, but they aren't mentioned there. Anyone know the answer? Kaldari (talk) 19:08, 21 March 2010 (UTC)

Since retro means behind and pro means forward and lateral means side, I would assume that retrolateral means along the side toward the rear of the body and prolateral means along the side toward the front of the body. However, IANAE. alteripse (talk) 19:30, 21 March 2010 (UTC)

Hmm, I think that's close, but not quite right. The terms are usually used in reference to drawings of pedipalps, which are basically the equivalent of a "hand" or "arm". I'm thinking that one refers to "side view from the side closest to the body" and the other refers to "side view from the side away from the body". Perhaps retrolateral is the side towards the body (behind), and prolateral is the side away from the body (forward). Kaldari (talk) 21:00, 21 March 2010 (UTC)

In this article about scorpion'a teeth (behind a subscription wall) the author explains "I use the terms prolateral and retrolateral respectively". Cuddlyable3 (talk) 22:19, 21 March 2010 (UTC)

After some more reading, I think alteripse might be correct as the term is also applied to arachnid legs, not just pedipalps. Kaldari (talk) 22:52, 21 March 2010 (UTC)

Reaction quotient, dependent on standard state?

Resolved

I am trying to understand some quantitative chemistry, but continue to be stumped by quantities that are ostensibly dimensionless but whose values nevertheless depend on essentially arbitrary choices of "standard states". One of these is "activity"; another is "reaction quotient". But the latter seems to be used in contexts where it cannot be allowed to depend on our choice of standard states.

For example, imagine that we fill a vessel with an equal (by volume) mixture of diatomic ozygen and ozone at standard temperature and pressure, and consider the equilibrium $3\,O_{2}\leftrightarrow 2\,O_{3}$. In gases, the activity (chemistry) of a species is roughly equal to its partial pressure divided by the standard pressure, so we have $\{O_{2}\}=\{O_{3}\}=0.5$, and the reaction quotient $Q_{r}$ becomes $(0.5)^{2}/(0.5)^{3}=2$.

Now, if "standard pressure" had been defined as 25 kPa instead of 100 kPa, the numbers would be different. Namely, then for the same situation as above, $\{O_{2}\}=\{O_{3}\}=2$ and $Q_{r}=0.5$.

The article Gibbs free energy tells us $\Delta _{r}G=\Delta _{r}G^{\circ }+RT\ln Q_{r}$, so the sign of $\ln Q_{r}$ (i.e. whether $Q_{r}$ is less than or greater than unity) determines whether $\Delta _{r}G$ will increase or decrease if we change the temperature a small bit. But how $\Delta _{r}G$ varies with temperature is something that should be discernible by an appropriate experiment and cannot be allowed to depend on which standard state we use for our computations!

Clearly I am missing something here – but what? Does one of the articles have it wrong? Is my math wrong? Is $\Delta G$ not actually a physical quantity (even though the dimension $[\Delta G]=\mathrm {J} /\mathrm {mol}$ looks respectable enough)? Or does $\Delta _{r}G^{\circ }$ secretly vary with the temperature (even though the "$X^{\circ }$" notation is supposed to mean, among other things, "at standard temperature")? –Henning Makholm (talk) 21:55, 21 March 2010 (UTC)

$\Delta _{r}G^{\circ }$ changes depending on what the standard states are defined to be: It's the standard state Gibbs energy change, and does of course depend on what you consider to be standard. Typically, we choose standard states to be 1 atm (or 1 bar, I guess; they're nearly the same]] for gasses, 1 M for solutes in solution, and 298 K (?) for temperature, but those are pretty arbitrary. $\Delta _{r}G$, on the other hand, does represent a constant physical value that can be measured, and is the same under given conditions no matter what you define your standard states at. Buddy431 (talk) 14:51, 22 March 2010 (UTC)

OK, so $\Delta _{r}G^{\circ }$ is really a constant once I have chosen standard states. That elminiates one possibility, but I still haven't found my misunderstanding. If I can measure $\Delta _{r}G$ for my sample gas mixture at 298 K and 299 K, the difference between those measurements is independent of $\Delta _{r}G^{\circ }$. Of my two calculations, one says that $\Delta _{r}G$ must be higher at 299 K, and the other says that it must be higher at 298 K. But those can't both be the case.

Doing the math in more detail:

${\frac {\partial \Delta _{r}G}{\partial T}}={\frac {\partial }{\partial T}}(\Delta _{r}G^{\circ }+RT\ln Q_{r})=R\ln Q_{r}+{\frac {RT}{Q_{r}}}\,{\frac {\partial Q_{r}}{\partial T}}$

ought to be a measurable physically quantity (and so not depend on our arbitrary choice of standard state). The final term is the same under all choices of standard states, because another choice just shows up as a linear factor in the activity of each species, which applies to the same power in the $Q_{r}$ denominator and the $\partial Q_{r}$ numerator. Therefore, by choosing sufficiently extreme standard states, we can make $R\ln Q_{r}$ so large (of whatever sign) that it dwarfs the final term, and thus both sign and magnitude of $\partial \Delta _{r}G/\partial T$ seem to depend entirely on which standard state I choose for my calculations. Which is absurd.

Again, what am I doing wrong here? –Henning Makholm (talk) 17:33, 22 March 2010 (UTC)

Hmm. Upon reading the Gibbs free energy article, it appears that $\Delta _{r}G^{\circ }$ does change depending on temperature; specifically, $\Delta _{r}G^{\circ }=-RTln(K)$, and both T and K (the equilibrium constant) will depend on temperature. This is borne out by the fact that my CRC Handbook of Chemistry and Physics lists standard Gibbs energies of formation at different temperatures for some compounds. So you were right, that $\Delta _{r}G^{\circ }$ does secretly scale with temperature. And at this point, I must defer to someone who's more knowledgable in thermodynamics. Buddy431 (talk) 22:51, 22 March 2010 (UTC)

Thanks – that actually explains everything. Then $\Delta _{r}G=RT\ln(Q_{r}/K)$, and because $K$ is just a $Q_{r}$ taken at equilibrium, the dependencies on choice of standard state cancel each other out.

I see now that the temperature dependence of $\Delta _{r}G^{\circ }$ is actually in the article immediately above the line I got hung up on. Oops. But I have made a clarifying edit to the part of Nernst equation that originally sent me out chasing geese. –Henning Makholm (talk) 08:04, 23 March 2010 (UTC)

Teleportation

This video with physicist Michio Kaku, http://www.youtube.com/watch?v=-FqLCLooayM&feature=PlayList&p=1B829DF36754F91C&playnext=1&playnext_from=PL&index=2, talks about teleportation. He mentions that we were able to teleport photons of light some distances. He also states that we are working on teleporting particles of matter some distances as well in the future. How is this possible? Our article on teleportation doesn't explain anything at all. Wouldn't teleportation constitute as a FTL (a circumvention of it)? I guess my confusion is, my understanding is that we have never solved the way to circumvent the light speed barrier. All of our ideas were merely theoretical. If Michio Kaku is telling the truth, didn't we already solve it? ScienceApe (talk) 22:09, 21 March 2010 (UTC)

Extraordinary claims need extraordinary evidence. Saying "We have teleported photons 600 meters across the Danube" doesn't do it. Wikipedia has an article about Michio Kaku. Cuddlyable3 (talk) 22:30, 21 March 2010 (UTC)

He is almost certainly talking about quantum teleportation (see that article for an explanation). As the first paragraph of that article says, it doesn't allow FTL travel or information transfer. --Tango (talk) 22:41, 21 March 2010 (UTC)

(EC) The obvious question is did you search, the answer I presume is no since two simple searches [3]/[4] [5]/[6] would have lead you to [7] and [8] or similar articles and eventually Quantum teleportation which should get you started Nil Einne (talk) 22:45, 21 March 2010 (UTC)

Oh no, you're wrong. I did search the answer. ScienceApe (talk) 23:05, 21 March 2010 (UTC)

What terms? In this specific case, I can find the answer very easily, e.g. even 'teleport photons of light some distances' finds one of the earlier links as does 'teleportation light & 'teleportation particles'; meanwhile and 'teleporting particles of matter some distances' finds [9], 'Michu Kaku teleportation' finds [10], 'teleportation ftl' leads to [11]. In fact even 'teleportation light speed barrier' in Google Bing show as the first and fourth result respectively the wikipedia article Faster-than-light both of which highlight this quote "Quantum teleportation transmits quantum information at whatever speed is used". So yeah, I'm having great difficulty finding any likely search terms that doesn't fairly quickly lead you to the answer, i.e. it'll be more helpful for you to learn where you went wrong then anything else Nil Einne (talk) 23:15, 21 March 2010 (UTC)

This is relevant. —Akrabbim^talk 23:15, 21 March 2010 (UTC)

The real question you should be asking is not that, but why did I ask the question, when I already knew the answer. :) ScienceApe (talk) 23:22, 21 March 2010 (UTC)

Don't need to. If you have a working teleporter, you can use it as a time-machine, so I just naturally presumed that you came here and read the answer before asking the question. SteveBaker (talk) 03:35, 23 March 2010 (UTC)

Fourier Transform

When we FT a time-domain function made up of only one frequency, why do we see two peaks (at + and -)? (Eg. FT of a sine function has a peak at freq=1 and -1.)

The usual Fourier transform does not use sines and cosines as base functions, but $e^{ikx}$, and you need two of those to make up a real sine. See Euler's formula#Relationship to trigonometry. –Henning Makholm (talk) 23:18, 21 March 2010 (UTC)

If we FT a sine function, why do we see a peak at frequency=1 instead of at 2pie which is the period of a sine function?

Because the k axis of the transformed function is actually angular velocity, not frequency. –Henning Makholm (talk) 23:18, 21 March 2010 (UTC)

If we FT a sine function that is shifted up, why does its freq-domain graph have a peak at frequency=0 in addition to at -1 and 1?

Because the way to get a constant term is $c\cdot e^{i\cdot 0\cdot x}$. –Henning Makholm (talk) 23:18, 21 March 2010 (UTC)

Also, where can I get some practice questions on guessing the result of doing an FT on a function (whether time-domain or freq-domain)? —Preceding unsigned comment added by 70.68.120.162 (talk) 22:50, 21 March 2010 (UTC)

• 1) Think about the identity $cos(\theta )={\frac {e^{i\theta }+e^{-i\theta }}{2}}$. The negative frequency is there
• 2) $\omega =2\pi f\,$ there is your $2\pi \,$
• 3) Shifting up is adding a constant which has zedro frequency. Think $e^{0}=1=const.$

Dauto (talk) 23:23, 21 March 2010 (UTC)

For practice working fourier transform problems, I recommend Signals and Systems, by Simon Haykin and Barry Van Veen, (available at Amazon, $52). There are hundreds of Fourier transform problems, with answers, and many worked examples, ranging from theoretical to applied to numerical/computational problems. There is also a great MATLAB tutorial for FFTs. This is a great textbook if you have never seen Fourier theory before, but want a complete mathematical and conceptual introduction to it. Nimur (talk) 00:21, 22 March 2010 (UTC)

are thermal baths bad for the skin

are thermal baths and mineral baths like salt baths bad for the skin, and is the sauna —Preceding unsigned comment added by 82.113.121.95 (talk) 23:26, 21 March 2010 (UTC)

Take a look at sauna#Health Risks & Benefits and Epsom salt#Applications. ~AH1^(TCU) 23:37, 21 March 2010 (UTC)

CO2 in Antarctic Ice Sheet

Hi. How many tons of CO₂ is stored in the Antarctic Ice Sheet, including gas frozen in the ice within bubbles of air, carbonic acid in precipitation that piled up into ice, and the carbon dioxide frozen onto the surface of the ice over time due to the cold temperatures and katabatic winds? Also, if all of this carbon dioxide was to be released into the atmosphere, how much of an increase in atmospheric CO₂ concentrations in ppm would result? In addition, could the cold temperatures over East Antarctica be freezing the anthropogenic additions of CO₂ out of the immediate troposphere, reducing the greenhouse effect and resulting in the observed slight cooling in the region? Thanks. ~AH1^(TCU) 23:42, 21 March 2010 (UTC)

Also, could there be deposits of carbon dioxide or methane clathrate hydrates frozen into the ice, and could there be any carbon-based soot in the ice that could either enter the ocean or speed up melting on the continent as the ice begins to melt? Thanks. ~AH1^(TCU) 23:43, 21 March 2010 (UTC)

Yes there could be (I don't see any reason why not); the real question is, are there? 24.23.197.43 (talk) 05:18, 22 March 2010 (UTC)

According to File:Vostok-ice-core-petit.png, the average CO₂ Antarctic ice core content is about 240 ppmv. (I don't have any immediate data on ice core porosity, which would be needed to calculate bulk CO₂ content.) Bear in mind that the ice has trapped bubbles or atmospheric air (and not just carbon dioxide), so the release of gasses due to the melting of ice sheets may not change the overall percentage of CO₂ in the atmosphere. caknuck _° needs to be running more often 07:41, 22 March 2010 (UTC)

According to [12] the volume of the antarctic ice cap is around 30 million cubic kilometers. At 240 parts per million by volume of CO2 - that could contain 7200 cubic kilometers of CO2. Presuming the bubbles are at normal air pressure (because that's the pressure they were at when they were formed) - then that's something like 10¹⁰ tonnes. That sounds like a lot - but it's about the same as the total CO2 output by US coal fired power plants in a couple of months. But as Caknuck points out - it won't make much difference to the percentage of CO2 in the atmosphere - and that's what matters from a global warming perspective. The amount of CO2 in the air has varied over history - but the long term average probably isn't much different from today.

As for the other things - I seriously doubt that much CO2 is freezing into the antarctic ice in the long term - that requires temperatures down at -78C which only happens for very short periods of time in the center of the continent during the depths of winter. As soon as the temperature warms up again, it would sublime back into a gas before it could get too deeply buried and form long-term deposits. Katabatic winds and the kinds of deep snowfalls needed to bury frozen CO2 only happen near the coast of antarctica - and it doesn't get cold enough there for CO2 to be a solid.

Black, sooty deposits in the ice are likely to be one reason for the melting we're seeing - but that stuff is coming from forest fires and things like diesel engines that are swept inland in modern times. It's possible that some major earthquake or forest fires in the distant past laid down layers of dark material that could eventually become exposed - but I've not seen any mention of that from the ice-core studies. SteveBaker (talk) 10:54, 22 March 2010 (UTC)

I think you lost a factor of ten. Glacial ice is about 10% air by volume. Dragons flight (talk) 07:18, 23 March 2010 (UTC)