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Where/How do I find "the more complicated exact equation" is "derived without using any approximations" enticingly alluded to in the last paragraph of
http://en.wikipedia.org/wiki/Doppler_effect#Analysis
At LEAST a computer accessible reference NEEDS to be provided !!
What does the "gadget at the end of the URL mean ? I would like to contact the author of this URL to be sure to get the complete derivation. RARE —Preceding unsigned comment added by 83.226.97.246 ( talk) 03:10, 12 October 2010 (UTC)
I think that for purposes of sci-fi, there are very few things so splendidly versatile as nuclear isomers. Even so, this application is likely a bridge too far. Yet...
Now what all this means is that, while it is by no means certain, it is possible that exposing a compound of lead to just the right excitation in a flame, for example, might cause the nucleus to jump to a slightly higher metastable state. Or perhaps lead as we know it is in a metastable state and it can be decreased in energy to some as yet unrecognized "stable" state? And if the effect is to increase the chance of alpha-decay, maybe it wanders over to be mercury... and maybe (perhaps with more excitations) it can emit a positron and then an alpha and end up as gold?
It's all very unlikely sounding, and yet, based on what I've read of these nuclear isomers, it seems like modern science can't preclude the conceivability of chemical alchemy, i.e. the possibility of rare but meaningful effects of chemical ionizations and other electron transitions on the breakdown of the nucleus. Can you prove me wrong? Wnt ( talk) 06:22, 12 October 2010 (UTC)
The main problem here may be one of definitions. You say you want "chemical" transmutation, but when exactly does something stop being "chemistry" and start being "nuclear technology"? I cannot imagine how you would excite a nucleus to the tune of even a single keV (~ about 10 million kelvin!), by methods that anyone would reognize as "chemical". – Henning Makholm ( talk) 19:18, 12 October 2010 (UTC)
Ha, why gold? Gold is easy to procure. I need a process that will give me rhodium or osmium. I'm really glad that no chemical process can cause transmutation though -- if that were so, biological life would be very much under threat. John Riemann Soong ( talk) 19:37, 12 October 2010 (UTC)
so they didn't do it (but that was in 2002). In order to use an optical-to-keV "amplifier", you'd have to have an excited isomer already (as your second SpringerLink article says), in which case you're not using true "base metal"; moreover, the lower energy states should be less likely to then conveniently emit alphas. As for ionization, it's just that the core electrons have much higher ionization energies than the valence electrons, as well as that each successive ionization energy is (typically, at least) higher. But you'll have to use un-alchemical processes like lasers or ion beams to preferentially excite/remove the inner electrons; normal chemistry will always go after the outermost ones. -- Tardis ( talk) 14:25, 13 October 2010 (UTC)`Nuclear light', or the gamma radiation emitted by an atomic nucleus in the optical range, will probably be discovered experimentally in one or two years.
This example problem in my class XI chemistry book proceeds as follows : A swimmer coming out from a pool is covered with a film of water weighing about 18g. how much heat must be supplied to evaporate this water, if it is at 298K? given, enthalpy of vapourisation of water at 373K = 40.66kJ/mol Solution : 18g is equivalent to 1 mole. heat supplied is equal to 40.66kJ/mol * 1mol = 40.66kJ
I feel that this solution to the problem is wrong, as the water on his body is not at 373K, but only at 298K. So, to raise the temperature of the water from 298K to 373K, 18*(373-298)*4.19 = 3771J = 3.771kJ of heat must also be supplied. So the final answer would be 40.66 + 3.771 = 44.431kJ. Am I right?? This is not a homework problem, I'm just asking out of doubt.. Can the enthalpy of vapourisation of a liquid be used at any temperature, or only at its boiling point? How can we assume that the enthalpy of vapourisation of water is same at both 373K and 298K?? Thank You. harish ( talk) 10:57, 12 October 2010 (UTC)
What you need is the heat of evaporation at 298 K. If the only figure you have is for 373 K, you can adjust for the temperature difference using conservation of energy: It should take the same energy to evaporate the water at 298 K and then heat the vapor to 373 K as it should to heat the water to 373 K and evaporate it there. At the precision you're working at here, you can probably get away with assuming that the specific heat of liquid water (resp. water vapor) is not temperature dependent. – Henning Makholm ( talk) 12:32, 12 October 2010 (UTC)
Boiling is always a kinetic phenomenon, because that's when vapor pressure of the liquid equals vapor pressure of the atmosphere, allowing bubbles to form within the water -- below this temperature, nucleation is simply unfavourable because bubbles cause surface energy. Evaporation is a thermodynamic phenomenon. The heat of vaporisation should be nearly constant (it probably changes dramatically when you have really large differences, but within the range of 0-100C I suspect it's rather the same). Notice that you cool down when your sweat evaporates! Your body has to supply heat for it to evaporate -- evaporation and boiling are inherently, endothermic processes. John Riemann Soong ( talk) 15:53, 12 October 2010 (UTC)
Is it a contradiction if you prove, physically/mathematically, that we cannot know ANYTHING about a given "other Universe", and go on to prove, physically/mathematically, something that must be logically true in any "other Universe" -- I'm talking pure logic there, so that if a mathematician in that other Universe were to explore the property of the natural numbers in that Universe... -- , implying that you DO know something about it?? Thank you. 84.153.253.103 ( talk) 13:18, 12 October 2010 (UTC)
Why do electrode potentials change at high temperatures? For example, charcoal can reduce sodium carbonate to sodium at a high temperature. Oxide can reduce protons to hydrogen at high temperatures, decomposing water. Iron(III) chloride releases chlorine when heated. Why do oxidizing agents and reducing agents get so much stronger at high temperatures? -- Chemicalinterest ( talk) 14:10, 12 October 2010 (UTC)
How much did the rescue operation for the 33 Chile miners cost to date? I realize it is ongoing, but is there an estimate for a final cost? Googlemeister ( talk) 15:30, 12 October 2010 (UTC)
Back to the original question, the Santiago Times has estimated the cost of the rescue at US$20 million, mostly in the hire of equipment. Physchim62 (talk) 07:55, 14 October 2010 (UTC)
Much like |±z> and |±x>, imagine we have two orthonormal set of eigenvectors given by |αi> and |βi> with i = 1, ..., n. They could be thought of as eigenvectors of two operators A and B respectively. Using these two set of eigenvectors, assume that I construct an operator U such that it has the following matrix elements: (U)ij = <αi|βi>. Show that U is unitary.
This was my homework question, but I've done a pretty good job of convincing myself that U is not unitary. I figure that (U†U)ij = Σ<βi|αk><αk|βj>, which, if U is unitary, is supposed to equal δij. But I don't see why it should. Did I do something wrong? 74.15.136.172 ( talk) 18:48, 12 October 2010 (UTC)
Emulsion paint goes hard after drying but does not soften when it gets wet. Why? What would make it soften - what solvents? Emulsion paint is used in the UK for painting interior walls and ceilings and may have a different customary name in other countries. Thanks 92.15.31.184 ( talk) 20:59, 12 October 2010 (UTC)
I'm very doubtful that applies to humble emulsion paint, despite having one word in common, as emulsion polymerisation appears to involve a lot of heat. Whereas emulsion paint just dries off at room temperature. 92.15.31.184 ( talk) 21:57, 12 October 2010 (UTC)
why do we get wrinkles around the mouth (like a cartoon beard)? I'm in my late twenties and it is one of the first places I'm getting wrinkles. I can't think of particular facial gestures / expressions I make that stress that part of the skin, unless it is from brushing my teeth... Is there something I can avoid doing so as not to exacerbate that... Thank you. 85.181.51.248 ( talk) 22:14, 12 October 2010 (UTC)
There is no way to avoid wrinkles, but if you smile more often than you frown then you will have pretier wrinkles. 169.139.219.254 ( talk) 22:25, 12 October 2010 (UTC)
How feasible is it to transition from the natural sciences to finance? A lot of the skills seem similar. Because I haven't taken an econometrics class, I am curious when people complain about all-nighters in finance or accounting (this is at the McIntire School of Commerce) whether their complaints are really legit, compared to all-nighters in the study of something also intensively quantitative like ecology or physical chemistry for example.
I guess I want to ask is -- what do undergrad students of finance and accounting study? Is it really hardcore? Looking at some appropriate articles, I see things like stochastic models and fractals -- so I wonder if I'm missing out. John Riemann Soong ( talk) 22:30, 12 October 2010 (UTC)
“ | After the war, physicists were often asked to go to Washington and give advice to various sections of the government, especially the military. What happened, I suppose, is that since the scientists had made these bombs that were so important, the military felt we were useful for something. | ” |
— Richard Feynman |
Does the correlation of greater speed at which bodies travel away from us and the greater distance just mean that the greater the distance away from us not only the older the body is but the faster that time is ticking? -- 96.252.213.127 ( talk) 22:53, 12 October 2010 (UTC) In other words does a second near a Quasars or other distant object that is moving away at faster speed than closer objects, ticking faster than a second here? -- 96.252.213.127 ( talk) 04:19, 13 October 2010 (UTC)
You will perceive that any object with a GR redshift (from motion or gravity or the expansion of the universe) has a "slower clock" than yours, by definition. Let's take a very common clock, which is light itself. Red light "vibrates" slower than blue light does. So any light that has been red shifted will vibrate less quickly than its source would otherwise indicate. Hence the events encoded in that red shifted light will take the same number of "light vibrations" as usual, only those vibrations will be spread over a longer period of time from the observer's point of view. Hcobb ( talk) 04:39, 13 October 2010 (UTC)
Okay then that seems to confirm that whatever is going on now where the object is, is occurring at a faster rate than we can observe. In other words explosions that occur at greater distances will appear to take longer to complete. Can we use percent redshift to determine how much slower our observation tiem frame is for these events versus the time they actually take to complete? -- 96.252.213.127 ( talk) 05:37, 13 October 2010 (UTC)
"...just a symmetric difference in our observations." okay that is the terminology and understanding I was looking for. In other words a second is a second here and a second is a second there but an even that takes a second to happen there that is observed from here only appears to take longer than a second when observed from here. -- 96.252.213.127 ( talk) 13:11, 13 October 2010 (UTC)
Science desk | ||
---|---|---|
< October 11 | << Sep | October | Nov >> | October 13 > |
Welcome to the Wikipedia Science Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
Where/How do I find "the more complicated exact equation" is "derived without using any approximations" enticingly alluded to in the last paragraph of
http://en.wikipedia.org/wiki/Doppler_effect#Analysis
At LEAST a computer accessible reference NEEDS to be provided !!
What does the "gadget at the end of the URL mean ? I would like to contact the author of this URL to be sure to get the complete derivation. RARE —Preceding unsigned comment added by 83.226.97.246 ( talk) 03:10, 12 October 2010 (UTC)
I think that for purposes of sci-fi, there are very few things so splendidly versatile as nuclear isomers. Even so, this application is likely a bridge too far. Yet...
Now what all this means is that, while it is by no means certain, it is possible that exposing a compound of lead to just the right excitation in a flame, for example, might cause the nucleus to jump to a slightly higher metastable state. Or perhaps lead as we know it is in a metastable state and it can be decreased in energy to some as yet unrecognized "stable" state? And if the effect is to increase the chance of alpha-decay, maybe it wanders over to be mercury... and maybe (perhaps with more excitations) it can emit a positron and then an alpha and end up as gold?
It's all very unlikely sounding, and yet, based on what I've read of these nuclear isomers, it seems like modern science can't preclude the conceivability of chemical alchemy, i.e. the possibility of rare but meaningful effects of chemical ionizations and other electron transitions on the breakdown of the nucleus. Can you prove me wrong? Wnt ( talk) 06:22, 12 October 2010 (UTC)
The main problem here may be one of definitions. You say you want "chemical" transmutation, but when exactly does something stop being "chemistry" and start being "nuclear technology"? I cannot imagine how you would excite a nucleus to the tune of even a single keV (~ about 10 million kelvin!), by methods that anyone would reognize as "chemical". – Henning Makholm ( talk) 19:18, 12 October 2010 (UTC)
Ha, why gold? Gold is easy to procure. I need a process that will give me rhodium or osmium. I'm really glad that no chemical process can cause transmutation though -- if that were so, biological life would be very much under threat. John Riemann Soong ( talk) 19:37, 12 October 2010 (UTC)
so they didn't do it (but that was in 2002). In order to use an optical-to-keV "amplifier", you'd have to have an excited isomer already (as your second SpringerLink article says), in which case you're not using true "base metal"; moreover, the lower energy states should be less likely to then conveniently emit alphas. As for ionization, it's just that the core electrons have much higher ionization energies than the valence electrons, as well as that each successive ionization energy is (typically, at least) higher. But you'll have to use un-alchemical processes like lasers or ion beams to preferentially excite/remove the inner electrons; normal chemistry will always go after the outermost ones. -- Tardis ( talk) 14:25, 13 October 2010 (UTC)`Nuclear light', or the gamma radiation emitted by an atomic nucleus in the optical range, will probably be discovered experimentally in one or two years.
This example problem in my class XI chemistry book proceeds as follows : A swimmer coming out from a pool is covered with a film of water weighing about 18g. how much heat must be supplied to evaporate this water, if it is at 298K? given, enthalpy of vapourisation of water at 373K = 40.66kJ/mol Solution : 18g is equivalent to 1 mole. heat supplied is equal to 40.66kJ/mol * 1mol = 40.66kJ
I feel that this solution to the problem is wrong, as the water on his body is not at 373K, but only at 298K. So, to raise the temperature of the water from 298K to 373K, 18*(373-298)*4.19 = 3771J = 3.771kJ of heat must also be supplied. So the final answer would be 40.66 + 3.771 = 44.431kJ. Am I right?? This is not a homework problem, I'm just asking out of doubt.. Can the enthalpy of vapourisation of a liquid be used at any temperature, or only at its boiling point? How can we assume that the enthalpy of vapourisation of water is same at both 373K and 298K?? Thank You. harish ( talk) 10:57, 12 October 2010 (UTC)
What you need is the heat of evaporation at 298 K. If the only figure you have is for 373 K, you can adjust for the temperature difference using conservation of energy: It should take the same energy to evaporate the water at 298 K and then heat the vapor to 373 K as it should to heat the water to 373 K and evaporate it there. At the precision you're working at here, you can probably get away with assuming that the specific heat of liquid water (resp. water vapor) is not temperature dependent. – Henning Makholm ( talk) 12:32, 12 October 2010 (UTC)
Boiling is always a kinetic phenomenon, because that's when vapor pressure of the liquid equals vapor pressure of the atmosphere, allowing bubbles to form within the water -- below this temperature, nucleation is simply unfavourable because bubbles cause surface energy. Evaporation is a thermodynamic phenomenon. The heat of vaporisation should be nearly constant (it probably changes dramatically when you have really large differences, but within the range of 0-100C I suspect it's rather the same). Notice that you cool down when your sweat evaporates! Your body has to supply heat for it to evaporate -- evaporation and boiling are inherently, endothermic processes. John Riemann Soong ( talk) 15:53, 12 October 2010 (UTC)
Is it a contradiction if you prove, physically/mathematically, that we cannot know ANYTHING about a given "other Universe", and go on to prove, physically/mathematically, something that must be logically true in any "other Universe" -- I'm talking pure logic there, so that if a mathematician in that other Universe were to explore the property of the natural numbers in that Universe... -- , implying that you DO know something about it?? Thank you. 84.153.253.103 ( talk) 13:18, 12 October 2010 (UTC)
Why do electrode potentials change at high temperatures? For example, charcoal can reduce sodium carbonate to sodium at a high temperature. Oxide can reduce protons to hydrogen at high temperatures, decomposing water. Iron(III) chloride releases chlorine when heated. Why do oxidizing agents and reducing agents get so much stronger at high temperatures? -- Chemicalinterest ( talk) 14:10, 12 October 2010 (UTC)
How much did the rescue operation for the 33 Chile miners cost to date? I realize it is ongoing, but is there an estimate for a final cost? Googlemeister ( talk) 15:30, 12 October 2010 (UTC)
Back to the original question, the Santiago Times has estimated the cost of the rescue at US$20 million, mostly in the hire of equipment. Physchim62 (talk) 07:55, 14 October 2010 (UTC)
Much like |±z> and |±x>, imagine we have two orthonormal set of eigenvectors given by |αi> and |βi> with i = 1, ..., n. They could be thought of as eigenvectors of two operators A and B respectively. Using these two set of eigenvectors, assume that I construct an operator U such that it has the following matrix elements: (U)ij = <αi|βi>. Show that U is unitary.
This was my homework question, but I've done a pretty good job of convincing myself that U is not unitary. I figure that (U†U)ij = Σ<βi|αk><αk|βj>, which, if U is unitary, is supposed to equal δij. But I don't see why it should. Did I do something wrong? 74.15.136.172 ( talk) 18:48, 12 October 2010 (UTC)
Emulsion paint goes hard after drying but does not soften when it gets wet. Why? What would make it soften - what solvents? Emulsion paint is used in the UK for painting interior walls and ceilings and may have a different customary name in other countries. Thanks 92.15.31.184 ( talk) 20:59, 12 October 2010 (UTC)
I'm very doubtful that applies to humble emulsion paint, despite having one word in common, as emulsion polymerisation appears to involve a lot of heat. Whereas emulsion paint just dries off at room temperature. 92.15.31.184 ( talk) 21:57, 12 October 2010 (UTC)
why do we get wrinkles around the mouth (like a cartoon beard)? I'm in my late twenties and it is one of the first places I'm getting wrinkles. I can't think of particular facial gestures / expressions I make that stress that part of the skin, unless it is from brushing my teeth... Is there something I can avoid doing so as not to exacerbate that... Thank you. 85.181.51.248 ( talk) 22:14, 12 October 2010 (UTC)
There is no way to avoid wrinkles, but if you smile more often than you frown then you will have pretier wrinkles. 169.139.219.254 ( talk) 22:25, 12 October 2010 (UTC)
How feasible is it to transition from the natural sciences to finance? A lot of the skills seem similar. Because I haven't taken an econometrics class, I am curious when people complain about all-nighters in finance or accounting (this is at the McIntire School of Commerce) whether their complaints are really legit, compared to all-nighters in the study of something also intensively quantitative like ecology or physical chemistry for example.
I guess I want to ask is -- what do undergrad students of finance and accounting study? Is it really hardcore? Looking at some appropriate articles, I see things like stochastic models and fractals -- so I wonder if I'm missing out. John Riemann Soong ( talk) 22:30, 12 October 2010 (UTC)
“ | After the war, physicists were often asked to go to Washington and give advice to various sections of the government, especially the military. What happened, I suppose, is that since the scientists had made these bombs that were so important, the military felt we were useful for something. | ” |
— Richard Feynman |
Does the correlation of greater speed at which bodies travel away from us and the greater distance just mean that the greater the distance away from us not only the older the body is but the faster that time is ticking? -- 96.252.213.127 ( talk) 22:53, 12 October 2010 (UTC) In other words does a second near a Quasars or other distant object that is moving away at faster speed than closer objects, ticking faster than a second here? -- 96.252.213.127 ( talk) 04:19, 13 October 2010 (UTC)
You will perceive that any object with a GR redshift (from motion or gravity or the expansion of the universe) has a "slower clock" than yours, by definition. Let's take a very common clock, which is light itself. Red light "vibrates" slower than blue light does. So any light that has been red shifted will vibrate less quickly than its source would otherwise indicate. Hence the events encoded in that red shifted light will take the same number of "light vibrations" as usual, only those vibrations will be spread over a longer period of time from the observer's point of view. Hcobb ( talk) 04:39, 13 October 2010 (UTC)
Okay then that seems to confirm that whatever is going on now where the object is, is occurring at a faster rate than we can observe. In other words explosions that occur at greater distances will appear to take longer to complete. Can we use percent redshift to determine how much slower our observation tiem frame is for these events versus the time they actually take to complete? -- 96.252.213.127 ( talk) 05:37, 13 October 2010 (UTC)
"...just a symmetric difference in our observations." okay that is the terminology and understanding I was looking for. In other words a second is a second here and a second is a second there but an even that takes a second to happen there that is observed from here only appears to take longer than a second when observed from here. -- 96.252.213.127 ( talk) 13:11, 13 October 2010 (UTC)