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Messages - Atomic-S

Pages: [1]
1
Physics, Astronomy & Cosmology / Re: Is the brightness of Venus changing?
« on: 08/12/2016 04:00:47 »
Here, we observe it to be very bright, which is not necessarily an error because Venus does get very bright when it is approaching the Earth.  If you are in the southern Hemisphere, you probably have less atmospheric obscuration, which would make it appear especially bright.
The following users thanked this post: Alex Dullius Siqueira

2
Physics, Astronomy & Cosmology / Re: EM wave direction
« on: 11/11/2016 07:17:11 »
The quantity (E x H)  [ H being an experession for the magnetic field strength] is called the Poynting vector, and its divergence  ∇•(E x H) equals the rate of change of electromagnetic energy density at the place evaluated. Thus the Poynting vector can be interpreted as an energy flux. Correspondingly, an electromagnetic wave propagating from east to west will have a Poynting vector that points in the opposite direction of one propagating from west to east. And correspondingly, the associated cross product, being of opposite sign, must have one of its factors also of opposite sign. Accordingly, if E points up, H must point south for an eastbound wave, but north for a westbound wave. So yes, the direction of propagation can be determined by examining the relative directions of E and H.  However, the problem is somewhat more complex than this analysis would suggest. In the simple case described, the energy densities associated with E and H are equal, and the propagation direction is unambiguous.  However in general a wave may be more complex than that, E and H not necessarily having equal energy densities at a specific point, and not necessarily being at right angles. Two plane waves, for example, could be propagating through one another in opposite directions under the principle of superposition, and in that situation, E x H will be dancing around a lot in nodes and antinodes, with energy jumping from odd-numbered positions to even- numbered positions, sometimes being fully electric and other times being fully magnetic. At any given moment at a specific point,  E x H will have a definite value and direction, which could be referred to as the direction of energy  at that time and place, but it will not match the magnitudes and directions elsewhere, so that one could not speak of the whole wave as having a single direction of energy movement.  All that is a classical view. Trying to interpret this quantum-mechanically raises complications that I am unable to answer, such as, what, under these conditions, is the statistical distribution of photon directions and energies?  A big problem with trying to answer that question is that the question itself may be meaningless in view of the fact that such a wave function may well describe a single photon.
The following users thanked this post: nilak

3
Cells, Microbes & Viruses / Re: Does legionella have the chance to grow in combi/mains water systems?
« on: 18/08/2016 04:33:56 »
It seems to me that when the subject of water  purification through temperature has come up, it has been in the course of recommending that it be boiled.  That might be good advice. I am not familiar with your type of water system, but if water is sterilized by heat and then flows into other parts of the system in such a way that it remains sealed against the outside, the drop in temperature at that time would not re-introduce bacteria, so it should still be good. But if heated water uses the same outlet as unheated water, there would seem to be the possibility of contamination. That problem might be solved by flushing out the common section of piping (e.g., faucet) thoroughly, by letting the water from the heated side run vigorously for a short time before actually beginning to use it. 
The following users thanked this post: JojoP

4
Physics, Astronomy & Cosmology / Re: How small can a gamma wave, or gamma knife beam be collimated.?
« on: 09/08/2016 05:26:48 »
To focus on an area of 20 microns width, a good wavelength would be 2 microns or shorter. That would put it in the infrared band. Potentially therefore you can use any radiation shorter than that, including visible, ultraviolet, X, and gamma. If you want to ionize stuff using reasonable power levels, the quantum energy has to be sufficient, and ultraviolet may be sufficient.  It is probably possible to do what you want using presently-available lasers. The choice of wavelength would also be dictated by possible effects on the surrounding environment.  Lenses to handle radiations that are in the longer ultraviolet band and longer are no problem.  However,the nature of the source must also be taken into account. Not even the best lens can focus the light of an overcast sky into an intense pinpoint. Laser light is extremely focusable. But lasers tend to be available mainly in the visible and longer wavelengths. I don't know if ultraviolet lasers are available, but ultraviolet LEDs are readily available, but whether the power would be adequate is another question.  For X-rays, as far as I know we are still limited to vacuum tubes, but these certainly exist and are capable of producing a small emission source compact enough for imaging, which may well be workable with additional optics to give you the tiny impact zone you require.   With shorter wavelengths, things get more difficult, although X-rays have been successfully focused using grazing-incidence mirrors. I don't know any way to do pinpoint application of gamma rays except by direct insertion of a tiny radioactive source at the point of application.  I  think you should look first at visible lasers for brain surgery. However, if the point of application is inaccessibly within tissue, then X-rays may have to be considered. Here, the geometries could get complicated because the farther the point of application is from the lens or mirror structure, the harder it will be to get a narrow spot. This could be a problem if the source is an X-ray tube and the optics consist of grazing-incidence mirrors. (But of course, under such conditions, would you even be able to tell, to within an accuracy of 20 microns, exactly where to position it?) 
The following users thanked this post: Nicholas Lee

5
Physics, Astronomy & Cosmology / Re: What's The Real Probability Of Life Existing Elsewhere In The Universe?
« on: 03/08/2016 05:33:51 »
Science does not know enough about the possible ways life might create itself to calculate the probability with any accuracy, but it is evident that the probability per planet is very low.  That is clear from trying to calculate the probability for a one-celled organism to originate on Earth, where we at least have a fairly good acquaintance with the environment and the chemistry. What is clear is that such an organism is quite a complex system whose occurrence by chance is quite unlikely.  We would be assuming that life elsewhere in the universe would necessarily be, if not like that on  Earth, nonetheless chemically complex or we would not be justified in calling it "life", and because it must be such, we can say that the probability of its occurrence elsewhere is likewise very low.  But we really can't be more precise in the calculation in view of our extremely limited knowledge of the composition of environments elsewhere in the universe.
The following users thanked this post: Solium

6
Chemistry / Re: Will removing a neutron from nucleus change electron voltage requirement.?
« on: 26/07/2016 08:19:48 »
Neutrons have no net electric charge, so that to a first approximation, inserting them or removing them from a nucleus has no effect on the attraction of the nucleus to the surrounding electrons, and therefore would result in no change in the spectral characteristics of radiation absorbed by or emitted by the electrons as they jump between orbits. Actually, however, there could be a slight effect because the removal of a neutron would change the nuclear mass, which enters into the state equations governing how electrons orbit it.  This change would be very slight.  Also, the neutron has a magnetic dipole moment and appears to have a possible electrostatic dipole moment also, both of which would exert slight effects on the behavior of electrons. 
The following users thanked this post: Nicholas Lee

7
Technology / Re: How Powerful a Microwave would you need for an Effective Weapon and Radiation-?
« on: 19/07/2016 07:15:08 »
10 KW would no doubt take out any human intruder quickly. Electronic disruption may require quite a bit less. Ionization of missiles may require quite a bit more.
The following users thanked this post: Simple Simon

8
Chemistry / Re: If the hydrogen atoms electron COULD be kept in shell 2 without going back down
« on: 16/07/2016 06:10:06 »
I think the idea is unworkable for several reasons. One is that the human body has elements other than hydrogen to consider.  The human body could not be made transparent without dealing with iron, and in iron, simply making the lowest energy state unavailable would not do much good in the visible wavelengths because in iron that energy state is separated by others with much more energy than that of a visible photon, as are several other states.  Electrons in these states will affect the absorbtion or emission of ultraviolet or X-rays.  Another is that to achieve the necessary electron rearrangement would probably end biochemistry as we know it, killing the patient.  (Pushing electrons out of the lowest orbit and keeping them out would, if the atom is to remain neutral, cause the atom in the case of hydrogen to effectively become the next element up the periodic table, or for all other elements the second-next element up the periodic table.  That would change the whole shape of the chemistry radically). If, however, one wishes to study the spectral effects of making the lowest orbit unavailable, a good place to start is to study lithium, in which two electrons pair in the lowest energy state, leaving the third to do its business in other states only. 
The following users thanked this post: Nicholas Lee

9
Chemistry / Re: How can I test the difference between bone and stone?
« on: 21/06/2016 07:43:42 »
I would start with magnification. If you say that they do not look alike, they should differ. However, to use this as legal proof, you would have to be able to describe accurately the difference. I suspect however that for legal action, some sort of expert testimony would  be required.
The following users thanked this post: NonGeekSeeksHelp

10
Physics, Astronomy & Cosmology / Re: What will the James Webb telescope reveal?
« on: 21/05/2016 06:35:06 »
Well, if we knew what it was going to see, there would not be much point in sending it up.  One thing it might do is give us a fuller view of the outer Solar System, including hitherto undiscovered dwarf planets.
The following users thanked this post: Bored chemist, PmbPhy

11
Chemistry / Re: How does Barium Hydroxide form?
« on: 21/05/2016 06:29:33 »
An oxygen atom has 6 electrons and 2 vacancies. Connect one of its vacancies to the electron in hydrogen, forming OH.  Do likewise with another OH pair. Each OH pair still has one remaining vacancy. Connect each of these vacancies to the two electrons available in barium.
The following users thanked this post: chris, Eman7

12
Physics, Astronomy & Cosmology / Re: Is there any evidence for aether?
« on: 30/04/2016 05:09:06 »
Here's another thing that needs to be  explained:  If the waves associated with gravity, with the double-slit experiment, and electron diffraction are all waves in the aether, then there needs to be some explanation as to why the three waves in question behave differently. Specifically, the mathematics of the gravitional wave indicate a quadrupolar character, which, if a linearly polarized beam of such waves were to be scattered at right angles to the direction of propagation by an appropriate target, the scattered energy would vary in intensity with respect to angle in the scattered plane according to cos2(2Θ), Θ being the directional angle. If a beam of linearly polarized electromagnetic waves is scattered at right angles by a suitable target, we find that its intensity varies with angle according to cos2(Θ) .  If a polarized beam of electrons is similarly scattered, the scattered intensity varies as cos2(Θ/2) .  So it appears the vibrations are not alike.  Interestingly, the spins of the associated particles are, theoretically for the gravtion if it exists: 2; for the photon: 1; for the electron: 1/2.  So that in general, we have the situation that the wave associated with a paricle of spin n, will scatter as cos2(nΘ) .   A right understanding of aether must be able to account for this.
The following users thanked this post: jeffreyH

13
Chemistry / Re: Can any invisible EM waves that get absorbed by electrons, pass through solid ma
« on: 27/04/2016 08:30:46 »
If waves are absorbed, the part that is absorbed will not pass through. Neither will the part that is reflected back immediately upon encountering the block. The part that passes through will be neither reflected nor absorbed. The tendency of the block to reflect, absorb, or pass a specific wavelength may be affected by temperature, but of course, the hotter the block, the more it will do its own emitting.
The following users thanked this post: Nicholas Lee

14
Physics, Astronomy & Cosmology / Re: What is a tensor?
« on: 26/04/2016 08:31:58 »
Another way to look a tensors is as follows:  Consider an object having a specific location and simple numerical value. That is a scalar. Now imagine that a second like object is placed in its near vicinity but with the opposite value. The difference between them multilied by the displacement is a dipole moment, and has not only a magnitude but also a direction, and in 3 dimensional space would be expressed by 3 components. Assuming linear mathematics, this would be a vector.  Vectors can of course be combined by the usual procedures to give other vectors, that likewise each have (in this case) 3 components and are equivalent to a magnitude times a displacement.  Now imagine you have a vector and then displaced from it another like vector except of opposite sign. This configuration is now associated with the underlying original value, the magnitude and direction of the first displacement, and the magnitude and direction of the second displacement. This object is vector-like but is more complex than a vector, and sums of such objects cannot necessarily be expressed as a simple replica of the exact style of object that one such object would be based only on the two-displacement model.  (Although the similarity of this to a quadrupole moment is obvious, it is not correct to state that the structure is a quadrupole because one needs to distinguish between the case of starting out with a vector in the x direction followed by displacement in the y direction for the opposite vector, as distinguishted between starting out with the vector in the y direction and then displacing for the opposite vector in the x direction. These two procedures are not in general equivalent, and to have a true quadrupole, as in an electromagnetic sense, it is necessary that both processes be present simultaneously, something that is not true of all processes).   In general, in 3 dimensional space, an object of this type require 9 components for its expression and admits of additional complexities, specifically, that tensors (of the second rank, which is what we are talking about here) can be expressed as combinations of (using ordinary geometrical language such as one would use in things such as mechanical engineering) symmetric, shear, and antisymmetric forms, or some combination of these. An alternative expression is in terms of dyads, none of which fit these descriptions but all of which are aligned along the coordinate axes of interest (here assumed rectangular) and can be combined to form the aforesaid forms.  An example of a symmetric tensor is the surface of a lens, the gradient of whose surface constitutes an (effectively linear) vector field, so that the value of that gradient (which is a vector) as a function of the distance from the center (which is also a vector)  is simply a numerical constant times the distance vector. This is a special kind of symmetric tensor, known as a unitary tensor, and is represented by a matrix having 0 as all off-diagonal values and a constant for the diagonal values. The multiplication whereby this tensor would actuate upon the position vector to give the gradient vector is simply to do the matrix multiplication of the diagonal-only matrix by the vector, which, of course, in this case simply multiplies each component of the position vector by the corresponding diagonal element in the matrix, resulting in simply a constant multiplication everywhere. If the lens is astigmatic, having different curvatures on perpendicular axes, the vector field varies from place to place in both magnitude and direction, so that in general the gradient is neither simply proportional to the distance from the center nor is it necessarily parallel to that displacement. It would be expressed by a matrix whose non-diagonal elements are still 0 but whose diagonal elements are unequal. That is of course if the coordinate system used is aligned with the principle axes of the lens. If they are not, then we will end up with something having nonzero  off-diagonal components but they will be symmetrically balanced across the diagonal (hence the term symmetric tensor).  In general, symmetric tensors can be associated with ellipsoidal or hyperboloidal configurations , whether in 2, 3, or more dimensions, and as such will be characterized by principle axes, and when these axes are aligned with the coordinate system being used, will result in a matrix expression having only diagonal values, the off-diagonals being zero.  If the values are all of one sign, then the configuration will be ellipsoidal; otherwise hyperboloidal.  (If some of the diagonal elements are zero, you get a configuration with cylindrical characteristics). Then we have the antisymmetric tensor. That is characterized by, when aligned with the coordinate axes, of a matrix expression that has zero for all diagonal positions, and in the off-diagonal positions, values on opposite sides of the diagonal are of opposite sign. An example of this is the velocity distribution of a spinning object.  The dyadic tensor, mentioned earlier, is one that is aligned strictly on one pair of coordinate axes, so that it is like a vector along the x direction, and another like but opposite vector displace from it along the y direction, but antiparallel to the original. For some purposes these are the easiest and simplest to work with, and whenever you have a matrix expression for the tensor, each value in the matrix actually corresponds to one such dyad.  A physical example of a pure dyadic tensor is the stress present in a beam when one face of it is being pulled in a direction parallel to its length while the other face is being pulled in the opposite direction.
The following users thanked this post: Arthur Geddes

15
Chemistry / Re: Can any INVISIBLE EM waves get electrons in solid matter, to the same energy lev
« on: 15/04/2016 06:28:49 »
Speaking generally, something like what you describe is probably possible. However, the availability of energy states for electrons is not dependent only upon what radiation may be present, but also upon the crystallography of the material. Black carbon, which is graphite, is of a different configuration than diamond. For this reason, the two materials do not have identical energy levels available. It would be highly doubtful that one could convert graphite into diamond by simply irradiating it.
The following users thanked this post: Timemachine1

16
Chemistry / Re: Why does a rough surface encourage water to form droplets?
« on: 25/01/2016 05:42:59 »
The following may be involved:  The more synclastically curved a surface is, the lesser area of it is required to accumulate a given force per unit length of periphery because the lay of the surface deviates from flatness over a shorter distance. Therefore, if a sphere of liquid encounters a hydrophobic surface that is flat, that deforms the surface the same as does placing a basketball on the floor, resulting in tighter curvature in certain areas that repels it from the floor enough to stabilize it there. However, if you take the same basketball and place it on a surface having irregularities that are substantial but smaller than the diameter of the ball, and try to force the ball into the same amount of contact that it had with the flat surface, it will resist more vigorously because the curvature required to make it  conform to the surface is greater.  That suggests that hydrophobic liquids will tend to have less contact with a rough surface than a smooth one at any given pressure, because the surface tension resists deforming so as to match the roughness.

As for hydrophilic surfaces: Because they tend to attract the liquid, the liquid will tend to cover the surface. It is energetically favorable for the liquid to contact the entire surface rather than just the high points of it, which it will try to do. But there is also the issue of the surface tension on the exposed side of the liquid. The liquid will try to keep that surface flat.   The volume of the available region is determined by the thickness of the region of roughness. If the amount of liquid is less than this volume, it will not be able to assume a fully flat surface on the outside, and will develop a negative pressure (relative to the environment) because of the conflict, and if additional liquid  becomes available, it will be pulled into the system as the system tries to flatten its outside surface. Thus, the rough surface will tend to adsorb liquid until its roughness is filled.  On a flat (smooth) hydrophilic surface, of course there is no roughness and therefore no roughness volume, so there is no energy advantage for a thin layer of liquid more than several molecules thick to become thicker.
The following users thanked this post: Roju

17
Physics, Astronomy & Cosmology / Re: Is the Copenhagen Interpretation correct interpretation of quantum mechanics?
« on: 09/11/2015 04:37:38 »
Quote
The particle travels through a single slit and the associated wave passes through both.
I believe this to be in error, and that a quantum object is not a wave paired with an associated particle, these two things being thought of as associated but distinct objects; but rather a quantum object is one indivisible reality that can exhibit wavelike or particlelike properties depending upon circumstances.  The widely accepted view that a quantum object is a "wave associated with a particle" seems to explain a number of observations but runs into significant conceptual problems if taken literally. Among these is the fact that if there is a distinct particle which moves in a manner defined by the separate wave so that the probabilities turn out the way they should, then that motion must be in principle be position as a function of time, and as such must have certain mathematical properties. For example, the X component of the function will of necessity have some frequency spectrum , as will the Y and Z. Such a spectrum must have a mean.  But no one has ever to my knowledge has any clue as to how to calculate it. Another problem is that if the quantum object is emitted and absorbed only with respect to its particulate nature, then there must be some mechanism showing how the particle interacts at these times, and a theory to this effect is altogether lacking.

We can discard this entire picture if we adopt the following ideas:  First, discard the idea that when a blip is recorded in an instrument (e.g., Geiger counter), that a little hard ball, the classical sense, has collided with something.  We need to recognize the crudity of observational processes.  They simply do not have enough precision and finesse to actually show us in detail what has happened. Second, we must understand that this crudity is not simply a matter of limited technology, but is actually inherent in the order of things -- we must observe quantum objects by using instruments that are themselves composed of quantum objects, and it appears that there exists as a fundamental property of the order of things that an instrument composed of quantum objects is incapable of establishing its state with sufficient precision that the exact state of another quantum object can be fully examined. Thirdly:  we must reunderstand the concept of a wave as it applies to this situation.  We toss out the classical concept of a wave (a field that varies according to a differential equation, such that when boundary conditions are imposed as by spatial confinement, only certain modes are possible but amplitudes remain arbitrary), and replace it with one in which amplitude itself  becomes one of the coordinates in which the function exists, and, as such, when limited by boundaries, limits the function to discreet modes as pertains to amplitude.  This is what is known as quantization of the radiation field, and has the interesting effect that it automatically creates "particles" which consist in the differences in energy between different possible amplitude modes.  Thus, with this revised understanding of wave properties, "particles" follow automatically and do not have to be added as an addendum.  The quantum object we thus end up with is neither a classical wave nor a classical particle, but a different animal than both.

With this understanding, the whole way many people look at quantum experiments can be revised.  One consequence is that in the double slit experiment, both the "wave" and the "particle" travel through both slits, because the "wave" and the "particle" are in fact the same thing.  When it hits the screen, only one "particle" is detected because it corresponds to the transition of the state from one energy level to the one below it, which, because the function is quantized not only spatially but also amplitudinally, is necessarily a discrete transition. Remaining to be understood is why, then, the "particle" shows up at only one location.  I believe the answer to that is that the nature of the screen is such that only localized reception is possible (mainly because of the way it is composed of localized molecules), and that if the screen were not so constituted ,  the arriving "particle" could indeed be detected impacting in several places at once or over a wide area; but that such an occurrence would not constitute detecting several particles because the nature of such a screen would have to be such that such a possibility is excluded. It is still only one "particle".  To make such a screen, one needs to overcome the conventional realities of material structure, and that may be possible by using a Bose-Einstein condensate.
The following users thanked this post: jeffreyH

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