1

New Theories / Re: Could quantum mechanics be wrong?

« on: 08/09/2024 02:49:37 »

Hi.

Can quantum mechanics explain the generation, propagation, and reception of radio waves?

Short Answer:
   Probably not in an easy way for human beings to understand. 
   It probably is possible to use QM and apply a suitable interpretation of probabilities or the frequency of detection of photons to re-create the usual gross-scale classical radio wave propagation results - but this would be complicated.
   At the fundamental level, in most QM models there aren't any "radio waves".   So the problem starts with the way the question was phrased.

Medium detail / further discussion:
     The name "radio wave" that you are using in the question describes a phenomena that we (human beings) have come to ascribe some meaning to.   This thing,  "a radio wave", is an essentially classical physical phenomena.
    We imagine a radio wave as a ...  (you know) .....  spread out and continuous wave thing.  We expect to be able to measure an Electric and a Magnetic field and a radio wave will be a suitable oscillation in these fields.  There are many versions of Quantum Mechanics but most of them don't have that sort of thing as their subjects or objects of interest.   The simplest QM models consider all objects as if they are particles.   For example, we might consider a particle of light, a photon, as a suitable QM object but we do not consider an e-m wave as a suitable QM object.    QM objects just happen to have a wave function associated to them BUT they are essentially discrete particles and would always be detected as a discrete particle.
      More advanced QM theories, such as the mainstream versions of QFT, will consider the most fundamental objects to be fields.  For example, there will be a photon field that exists everywhere in space.   The thing we call a photon would not be treated as a fundamental QM object, it is just how we may interpret a certain oscillation in that underlying photon field at some place.    So, in QFT, there is a photon field but there is NOT any fundamental Electric field or Magnetic field.   These things, the electric and magnetic fields are just gross-scale fields that can emerge from the more fundamental fields of the particles in the standard model of particle physics.   QFT doesn't aim to deal with oscillations in the Electric and Magnetic fields, it only deals with the fundamental particle fields.   You need to add a layer of interpretation on to the top to translate what happens in these fundamental particle fields into something that can be observed in the gross-scale Electric and Magnetic fields.   That can be done in many versions of QFT but is rarely simple.
    When you are asking about "a radio wave", you have moved attention to something in terms of the ordinary (or gross-scale) Electric and Magnetic fields and not on any fundamental object (a photon in simple QM, or the photon field in QFT).   So the question cannot be answered without translating the results of QM back into something classical involving the Electric and Magnetic fields.   One of the simplest or most naive ways of making this translation is to just analyse the probablities or frequencies with which a QM object would be found at some place and time.   So, we might imagine that an e-m wave is a collection of many photons or that an experiment involving one photon has been done many times and the e-m wave we observe is the combined average of all of these.

Example:    If you could fire just one photon through the usual double-slit experiment apparatus,   then you probably won't get an interference pattern on the screen at the end.   Instead we think you'll get just one spot to glow on the screen.   However, if we repeat the experiment many, many times then we expect to see that the places where the screen glows and how often that bit of the screen glows will correspond to the usual interference pattern for a classical wave being put through that apparatus.    So, one individual photon does NOT behave like an e-m wave - but a thousand or a million of them would seem to reproduce classical radio wave-like patterns and behaviour.

NOTE:   I'm really not convinced anyone has actually done a genuine version of Young's double slit experiment where only a single photon is fired with a clear interval of time before the next is fired.   However, this demonstration gets close and claims to be using such a thin and dim stream of light that only "a few" photons per second will be produced.
https://sciencedemonstrations.fas.harvard.edu/presentations/single-photon-interference


   So, the important point is simply that "a radio wave" isn't the normal subject for any QM model.  It demands some form of interpretation to put the results predicted by any QM model back into terms which would make sense when you consider "a radio wave".   
   If we wished to be controversial we would say "there are NO radio waves... just QM objects that get detected at certain times and places.  On a gross scale, that looks like a radio wave is there".    ... However... I'm not that controversial or convinced that a simple QM model is some ultimate truth....  there are many versions of QM, all have some use and classical e-m theory is also extremely useful.   

Another minor note:    Your (Hamdani) extracts produced by Gemini seem to be blending many different versions of Quantum Mechanics.    For example,  QFT is a specialised theory and there are still about 3 versions or flavours of QFT that are in current use.   I don't think Geminii is reliably identifying where things can and cannot be combined.   For example, the simplest QM models are not like QFT.   They will involve the original Schrodinger equation, which is shown here:



That has a mass term, m, in the Hamiltonian for the QM object under consideration    (it's in the square bracket, h squared over 2 m).   This means it's actually useless for photons which should have 0 rest mass.   None the less, we (physicists) do generalise a lot of results from such a QM theory for massive particles to massless particles like photons.  For example, the Youngs double slit experiment can be modelled for electrons or even bigger particles like carbon bucky-balls.  We can (and have) then done experiments with these things and obtained a good match with the predictions from the simple QM model.   It is natural to assume a similar thing for photons, even though the zero mass term would technically make the simple QM model invalid.   If you read what I've said (above) about firing single photons then you'll hopefully notice that I've got phrases like "we think".   
     I believe that how you reconcile a classical radio wave with a suitable QM object ("a photon") is subject to a lot of interpretation. 

[Disclaimer:     I'M NOT AN EXPERT.  ]

Best Wishes.

2

Just Chat! / Re: Fed up with the lack of reasoned discourse

« on: 29/06/2024 02:41:15 »

Hi.

         A spinning magnet - that can be interesting.   Indeed we had some discussion here (TNS forum) about a year ago.   You ( @paul cotter ) contributed to that old discussion but it was a long time ago.   I'm going to recap some of it.

       In general, a thing that is rotating should just keep rotating at the same rate, that's basic angular momentum conservation.   However, in outer space that doesn't seem to be what happens....    well... at least not if the spinning thing is actually a magnet.    In that situation it does seem that the object emits e-m radiation and MUST eventually stop spinning.

      Let's put conservation of momentum and angular momentum to one side for a moment.    If a spinning magnet emits e-m radiation then it is losing energy.   Energy conservation MUST apply.    If energy is being emitted or lost from the spinning object then it came from somewhere.   Either the spinning object is losing some mass over time, or else it is losing kinetic energy by slowing down its rate of spin.    It doesn't seem that a lot of mass is being lost, so the object must be slowing down its rate of rotation even though there is no obvious source of external torque being applied to the object.

     A long time ago, I started writing a new forum post based on this - but it became too long and had to be split into many parts.    In the end, I decided it was a monologue rather than a discussion so I never actually posted it.

   (Part II) had much of the mathematics in it.   Not a lot of that displays well now that LaTEX markup isn't available.     In words, it showed that the far field produced by a spinning magnetic dipole has the right characteristics to be considered as e-m radiation.        "Does a spinning magnet in space emit e-m radiation" --->  yes, it does.

    (Part III)  was based on an objection that @alancalverd raised in the (very old) original forum thread concerning the notion that the spinning object would ever slow down.   Where is the torque coming from to slow the magnet down if it's in outer space?     How is angular momentum conserved?   etc.
    The answer mainly seems to be that the e-m radiation that is emitted will be circularly polarised.   This allows the e-m radiation itself to have or "carry away" some angular momentum.   Well.... that explains "where" the angular momentum has gone but it still doesn't explain the source or cause of any actual torque acting on the spinning object.
      The torque acting on the spinning magnet seems to arise by consideration of the near field (near to the dipole object) as opposed to the far field approximation.   The mathematics here is just too complicated (for me anyway) and the other references I chased up seem to use only numerical (computational) approximations instead of finding exact analytical solutions.   In spirit, it seems that the near field changes at a rate slightly slower than the dipole is rotating,  so that the B field in the local space differs from the magnetisation, M vector, within the object itself.   As such the B field pulls on the magnetised object slightly and effectively puts a drag or retarding torque on it.   So, let's re-phrase this because it's quite weird when you think about it.......   What exerts a torque on the spinning magnet in free space?  Well... it seems that space does or whatever it is that we recognise as the B field in space.  Space isn't exactly "nothing", it has B fields in it.    The spinning object will spin at the speed it wants to spin, of course, and the Magnetisation vector M of that object should change at that rate - however space a few centimetres away cannot instantly change its B field when the magnetised object has moved.   The maximum propagation speed of a new B field is c,  so the B field just a small distance away is still the old B field.    This is a very simplified and wordy rather than mathematical version of what appears to happen but I hope it makes some sense and provides the gist of what seems to be going on.    It raises a number of other questions... space doesn't seem to be just empty space, it's almost as if the B fields in it are something tangible - but that's probably a whole other thread and not something we need to consider here.

You mentioned the following (which is why this example is relevant):
 

The discussion was about possible exemptions to the conservation of momentum because of the finite speed of em propogation.

     The spinning magnet in free space seems to be an example where instead of being a problem that may break the principle of conservation of (angular) momentum,  the finite speed of waves propagating through the E and B fields is actually NECESSARY, indeed it is ABSOLUTELY ESSENTIAL to maintain the principle of conservation of angular momentum.   It is the (tiny) delay between the magnetisation vector of the magnet changing its orientation and the changing of the B fields in nearby space that puts the retarding torque on the spinning magnet.   So, a magnet spinning in free space will slow down and eventually stop spinning and no laws or principles of conservation are violated.

Best Wishes.

3

Question of the Week / Re: QotW - 24.06.28 - Are there stars which aren't part of a galaxy?

« on: 21/06/2024 21:02:53 »

Hi.

Possibly.
     The definition of what is a galaxy, or exactly where one ends is vague.   Hence, under one definition of a galaxy, a given star may be in a given galaxy but under another definition it is oustide any galaxy.

One reasonable definition of a galaxy:
    Galaxies consist of stars, planets, and vast clouds of gas and dust, all bound together by gravity.

[Taken from NASA website.     https://science.nasa.gov/universe/galaxies/ ]

    The key here is that we have some understanding of what it means for gravity to be generally keeping a collecton of things (stars, gas, plantets etc.) together.    Gravity follows an inverse square law with distance.   This means that at some distance from the centre of what we would define as the galaxy, gravity will be too weak and the star(s) this far out are not considered to be part of this galaxy any longer.

The definition of a galaxy is then mired by the existance of dwarf galxies apparently inside or just on the periphery of a larger galaxy.   For example, our Galaxy (the Milky Way), has the (Large and Small) Magellanic Clouds as dwarf galaxies.   They can be though of as small galaxies in their own right but they have become trapped in the gravitational well of the Milky Way and now orbit around the periphery of our galaxy.    So, that they are gravitational bound to our galaxy and could reasonably be considered as just a part of one big galaxy instead of being called galaxies on their own.      We can consider ever larger structures (regions of space) until we identify regions we call Clusters or Super-Clusters.   All the mater in these regions is gravitationally bound together, they do not tend to drift apart, but they appear much as many seperate galaxies orbiting around each other in some complicated way (literally "a cluster of  galaxies").    However, since everything in this cluster is gravitationally bound together, it forces us to question whether we should just define the whole cluster as one really big galaxy.

-------
   So, there probably isn't going to be a definitive answer to the original question.   If there's a star in space somewhere then it is probably inside someones definition of some galaxy.
   None the less we can still try to address some of the spirit of the question:    Stars don't tend to form in complete isolation from other stellar bodies.    Based on our simulations of cosmology and the evolution of the universe,  matter tends to distribute itself into filaments spaning through space.   The overall structure is often referred to as a cosmic web   (you can Google the term Cosmic web to find more info if you wish).    Stars will form only where there is matter for them to be formed from.   So stars tend to form along these filaments rather than in the voids in between them.   Once stars have formed, gravity naturally tends to start attracting other material and vice versa (other material will attract the star), so that stars will usually end up with other stars orbiting around them and/or they are pulled into orbit around some existing cluster of stars.

Exceptions
   Can there be a star that is NOT gravitationally bound to any galaxy?   Yes.    Over enough time, it is likely that every star would eventually be attracted to some some galaxy and orbit in some stable fashion.    However, for a short while (in Astronomical terms that can be a million years), some stars can be "flung out" of a galaxy and fly through the voids for a while.
     Although stars cannot form in the voids between galaxies (since the density of matter is far too low), there are in fact large numbers of intergalactic stars. It has been estimated, for example, that 10 per cent of the mass of the Virgo galaxy cluster is in the form of these stellar interlopers........
     ..... stars can be expelled from their parent galaxy if it collides, merges or passes close to another galaxy.

[Taken from:   BBC Science Focus.    https://www.sciencefocus.com/space/are-there-any-stars-between-galaxies]

Best Wishes.

4

Technology / Re: Does regenerative braking have a hazardous complication?

« on: 27/04/2024 14:33:10 »

Hi.

In Ireland when changing down in a manual the recommended method is to lightly touch the brakes

    As far as I know that's not written in the "highway code" booklet (for the mainland UK) or generally expected - but I have once had such advice from my driving instructor.

    There are also alternatives, like double de-clutching   (double clutching in the U.S.):   The strongest slowing down effect is caused within moments of lifting your foot off the clutch pedal when you change to a lower gear.   This is when the engine, drive shafts on the engine side of the gear box and gear wheels in the gear box which were only spinning at (for example) 2000 rpm has to increase to 3000 rpm since you are initially at the same road speed but will now be in a lower gear.    An option then is to shift gear to neutral, ensure the clutch pedal is then released, briefly touch the accelerator pedal to spin everything (engine and the gears in the gear box) and then clutch and shift again to the new lower gear.   This isn't in the highway code either but it can be done and you will soon get quite good at it.
https://en.wikipedia.org/wiki/Double_clutching_(technique)       (that article talks about reducing wear-and-tear rather than application in racing).

    Another level up again is to just change gear without even using the clutch pedal.  You ease your foot slightly on the accelerator to reduce the pressure or contact force between gear teeth, then move the gear stick to neutral, then increase pressure on the accelator again (but it won't be any seconds, just milli-seconds because without any load the engine will pick up rpm almost instantly), then when the new gear is spinning at about the right rpm you'll push the gear stick again and engage the new gear while everything is still spinning.   With a lot of practice (which involves a lot of terrible noises from the gearbox while you fail) you can perfect this.   To practise this, the usual requirement is to use your dads car and be about 17 - 18 years old.  Your dad will have enough money to get their gearbox repaired and re-built after all the times you failed and have broken gear wheel teeth while trying to enage a gear wheel spinning for 4000 rpm to another gear wheel that was only spinning for 2000 rpm  etc.  (and after-all, new drivers do make mistakes even while making less extreme racing gear changes).   Once you own your own car, you won't try it again because the technique of double de-clutching is actually good enough and considerably cheaper.

    Anyway, all of these driving techniques are fast becoming obselete because most new cars will be fully automatic.   These days, they (automatics) are better and don't result in a loss of performance (e.g. fuel economy). 

     I suppose we should address the original question:
Does regenerative braking have a hazardous complication?
   Possibly but the driver behind shouldn't be following that close.   If they are then according to the highway code you must now adjust your own driving appropriately.  For example, reduce your speed so that they can stop in their braking distance.   It's not exactly what they deserve but it's what is safe and recommended.   Her Majesty's Secret Service takes a dim view of it if we keep deploying the special counter measures on ordinary civillian drivers who were just too close.

Best Wishes.
[ These days     Her = His       but all the films are based on Her Majesty's Secret Service ]

5

Physics, Astronomy & Cosmology / Re: Does time dilation and length contraction apply to light?

« on: 08/04/2024 01:56:22 »

Hi.

   I basically agreed with @Halc and therefore didn't need to say anything.   However, it's always pleasant to hear from a moderator (Kryptid) and since you give up a lot of time to answer others, we ought to give up some of our time to reply to your post.

I had been under the assumption that time is frozen for light because it travels at the speed of light.

    This is usually a result people obtain or propose just by extrapolating the Lorentz Transformation with v→c.   So it can be useful to have as many ways as possible to guard against it or see where it might fail.   It's especially helpful to go along with their line of reasoning and identify some potential flaws in it, which is what I'll try to do first (below).  It's also possible to go straight against the line of reasoning but then people will wander around trying to figure out why their line of reasoing wasn't right and whether you might just be wrong or trying to brush a potential flaw in physics under the carpet.

Situation 1:    People often personify light and assume that this result (extrapolating LT with v→c) tells you how things would look or seem to the light.
    Problem 1:  We shouldn't personify light but let's be tolerant.
    Problem 2:    Whenever people make the claim 'this is how it seems to "Fred" ',  they are always telling you how Fred reports things with co-ordinates of an inertial frame where Fred is at rest.    For example, no one would dream of telling you how Fred would report things or identify events in a co-ordinate system where Fred himself is still moving at a speed of  0.133 c,  or  c/2, or  -0.3c   why would any of those frames seem natural to Fred?   It is clear that we're interested in how Fred reports things with co-ordinates derived from a frame of reference where he is at rest and we would probably be prepared to accept that a frame where Fred is at rest is the one that seems most natural to him.   Provided Fred does use the co-ordinates of his own rest frame, then his wrist watch is measuring the co-ordinate time that he will use, that seems natural.   If he used any other frame then his wrist watch is not counting the co-ordinate time he would be using (and that isn't a natural choice for Fred, if he's anything like me and you).
     So the problem with light is that we have no guarantee that light is at rest in any inertial frame.    People have usually just tried to extrapolate  the LT  with v→c but if you start trying to do this, there is no evidence that light is ever slowing down.   Go along with a simple approach of just following a pattern....

      With boost parameter  v =0.9c   ,   light would move at speed c in the primed co-oridnates.
      with v = 0.99c,  light would move at c in the primed co-ordinates.
      with v =0.999c,  light moves at c in the primed co-ords.
      with v =0.9999c,  light moves at c in the primed...
      .....
      with v =0.999999999999c,  light still moves at c in the primed co-ordinates.

   The important point is:  There's nothing here suggesting that light is slowing down in the primed co-ordinates as you increase the parameter v towards c.
    Now you could even use some algebra and pull out the relativistic velocity addition formula.   You can easily show that light is always travelling at c in the primed co-ordinates, regardless of how close that LT boost parameter v  gets to c.

    This is very diferent to what happens with Fred.   If you tried the same game with Fred,  then as you increase the v parameter in the Lorentz transformation (LT),  his speed in the primed co-ordinates is slowing down.   Moreover you can even over-shoot the mark and have the parameter v set so high that Fred is travelling backwards.   Specifically, we know that we can always find an inertial frame where Fred would be at rest.    However, we have no such guarantee with light,  it never seems to slow down in the primed co-ordinates regardless of how high we set the v parameter in the LT.

     Now when people (who have just extrapolated the LT with v→c) suggest that   "this set of things"  (like time being frozen and length contacting to 0) is how it would seem to the light,   you at least have something you can come back with as a reply or counter-argument.    You'd say  "but this isn't how it seems to Fred, even though he could have chosen to use the same frame.  I'll assume that's because Fred would naturally choose to use a frame where he is at rest.    However, we have no guarantee that light would be at rest in this frame you are proposing for light and so you cannot claim that this frame would be the natural one for light to use or that this is in some way how things would seem to the personified light."

- - - - - - - - - - - -

Situation 2:     You just try to remind them of the limits of SR.
    SR can be derived in various ways,  however the most common method is just to assume light has a speed c in all inertial reference frames.
    So it is clear that any frame where light is at rest cannot be an inertial frame and therefore isn't a situation that SR ever aimed to or claimed to say anything about.
     They probably will have arguments about extrapolating the LT with v→c and suggest or insist that it ought to mean something or be valid.   You could try to remind them that it's valid while v<c but abruptly stops being valid when v=c.   They won't like it but you can remind them that this happens often in physics.   
   There's a whole set of favourite examples you might have,  here's just a few you can use:
  (i)  A position on planet earth can be uniquely defined by spherical polar co-ordinates everywhere and perfectly continuously right up until you hit the North or South poles.   At these poles, the situation abruptly changes,  the azimuthal angle can be anything you like and yet you're still describing the same position and not a different position.   You can't do this anywhere else, a different azimuthal angle would certainly be describing a different position but at the poles, it's fine.   This isn't a continuous change, it's a discontinuous or abrupt one. 
   (ii) A graph of  y=1/x  behaves perfectly sensibly with x gettting as small as you like.   However at x=0 it's undefined,  worse than this it's very badly behaved.    It might seem that it's becoming positive infinity but you only have to look at small negative values of x and you realise it's a strange and slippery thing.   It somehow doesn't come back from positive infinity, it seems to come back from  negative infinity.   Any pattern you may try to follow as x → 0 has got to break at x=0.
   (iii)  The helicity of a Dirac particle will be a fixed unvarying quantity (+/-  1 for an electron)  in all frames of reference right up until you overtake the particle, in those frames the helicity changes sign (would be - or resp. + 1).  It's an abrupt change, it doesn't slowly slide down to +1 from -1.  Extrapolating the behaviour for frames of reference with boost parameter v < particle speed (in the lab frame) would not have predicted this behaviour.  That'll do as an example and you could leave it there.   If you know what helicity is then it's apparent that it must change sign when you overtake the particle.  However, this is why I picked this example because we can push this one a bit further:  Light is an interesting particle in that it really does always have a constant helicity in any inertial frame of reference.   This is a result you can obtain from Quantum Mechanics.  We can only understand that result classically by recognising that light cannot be overtaken (or even brought to rest) in any inertial reference frame.

- - - - - - - - - - - - -

Situation 3:    Again you go straight against any attempt to extrapolate the LT with v→c and their claim that this is an indication of what would happen in a "frame" where v=c and light might be at rest.   However, this time you target the tacit or implicit assumption that you would have anything you can call a valid frame.
    A co-ordinate system can be almost anything but it MUST serve to uniquely identify different events, this is part of the formal definition of what a co-ordinate system must be and is non-negotiable.   Just take all of their claims at face value.    So they claim that light starting it's journey over here and the event where the light ended it's journey over there (which could clearly be identified as different events in the lab frame) are given the same  x,y,z,t co-ordinates in the rest "frame" of the light.   That automatically and necessarily dictates that this "co-ordinate system" obtained from the "rest frame" of the light is actually NOT a valid co-ordinate system.
     We have no experience of using a degenerate pseudo-co-ordinate system and it is not a natural choice for Fred to try and use a degenerate pseudo-co-ordinate system.   There are plenty of degenerate systems Fred could have chosen to use but he didn't,  he chose a co-ordinate system where time was the thing on his watch and distance was something which would also help to keep different events separated, only that seemed like a natural choice of co-ordinate systems to Fred.   So it's not sensible to suggest that anything reported in the degenerate pseudo-co-ordinate system you propose for the "rest frame" of light indicates how things would seem for the light.
     We then have two possibilities:
     (i) There may be some co-ordinate system for light where its notion of time is the thing that passes on its hypothetical wrist watch and its notion of space stuff also helps to seperate different events in some way.   However, the Lorentz Transformation does not seem to be the thing connecting these co-ordinates for the rest frame of light to the co-ordinates of space and time that we use.  (We can see that precisely because it would suggest that light is left with a set of numbers or pseudo-co-ordinates that fail to separate different events).
    (ii) There just may not be any such co-ordinate system or a suitable rest frame for light.

- -  - - - - - - -
   That's too long, even I'm fed up with listening to myself.   I hope some of it is of some use.

Best Wishes.

6

Question of the Week / Re: QotW - 24.04.12 - What law governs the distance of an electron from the nucleus?

« on: 06/04/2024 16:41:20 »

Hi.

     Why are we arguing over fine details when the basics haven't been esatblished?

'What law governs the distance of an electron from the nucleus of an atom?'

     The basic answer is that it depends on the model(s) you are using.

    In a very simple model, the electron is attracted by electrostatic forces to the nucleus but the electron is assumed to be in some motion.   So the electron orbits the nucleus like a planet orbiting a star.    The laws governing everything are similar to Kepler's laws for planetary motion but where Newtonian gravity is replaced by Couloumb's law of electrostatic attraction.

     There is now a whole set of more advanced models we can work through to suggest slightly different answers to the question....
     Eventually, you reach a level of sophistication in the models where the basic question needs to be completely re-examined.   For example, an electron may not be an ordinary particle as we might imagine, it may be a wave in an underlying field and thus "spread out" all over the place.   The question becomes impossible to answer because there are no definite positions for any of the things mentioned in the question and hence no way to determine a distance as the difference between these two positions.

     I propose that the best answer for the question is to take things only as far as you can so that words like "electron", "nucleus" and  the distance between them   continue to have the same meaning as that which is suggested in the question.
    In practice, a fairly classical model of an atom works quite well and adequately describes the macroscopic or bulk properties of most materials.   We can determine the typical distance between atoms by performing experiments involving X-ray crystallography or surface electron scattering (often called  LEED = Low Energy Electron Diffraction).   We can reasonably assume that in the solid state of this material, the atoms pack together as closely as they can so that they are effectively in contact with each other.   Indeed we find that they are only a few crystal lattice structures for solids and simple cubic crystals   (such as  b.c.c. [body centred cubic] ,  f.c.c. and just  s.c.) are prevalent.   
    It is found that most atoms have an effective radius in the order of 1 Angstrom  =  10^-10 metres.   For example,  Copper has a crystal lattice constant ≈  3.6 A   (Editing:   The website won't support an A with a circle over it,  A means Angstrom  and any ? character is probably an Angstrom that I didn't edit)    =>   atomic radius ≈ 1.8 A  (note that there are two atomic radii between the centre of two atoms that are in contact).     
   Glance through this website (   https://periodictable.com/Properties/A/LatticeConstants.html   -   I'm not guaranteeing the reliability, it's just the first site that came to hand and readily displays the info we want).   Although there is a little variability the lattice constants are all between 2 - 7 Angstrom.
     Back this up with our favourite model of an atom,  the Schrodinger Hydrogen atom solution,  where the tpyical distance of the electron from the nucleus is the Bohr radius ≈  0.5 A  and note that this is one of the smallest atoms around.
     Overall, it seems that as regards the bulk properties of materials,  all atoms tend to have a radius of about 1 Angstrom.    At first consideration, this is a bit surprising,  e.g. Iodine has 53 electrons so you would have thought they occupy electron shells that are much further out than the  1s orbital of a Hydrogen atom.   However, we need to balance this against the idea that the total charges involved are 53 times greater than that in Hydrogen and all the orbitals are effectively pulled in and are of much smaller radius.    Indeed we get a fair explanation of the result by just considering the last electron of any atom.   In a simple (often called a Hydrogenic model), the last electron is essentially outside all the other electrons and the nucleus,  so it "sees" a total charge (in units of e) interior to its orbit that we can express in terms of Z, the atomic number:
  (Total charge seen, in units of e)  =   (charge from the nucleus)   +    (charge from the other electrons)
                                                     =   Z   +   -(Z-1)     =   1.
   We can approximate the behaviour of the outermost electron by modeling it as though it sees a total charge of just +1e at the centre of the atom.    To say that more simply,  the outermost electron would behave like an electron around a Hydrogen nucleus and so it's not at all that surprising it always has a typical distance from the centre of the atom equivalent to the distance displayed in the simple Hydrogen atom.

    That's where I'd leave the answer.   Summary follows:

1.    The standard rule of thumb is that atoms have a radius ≈ 1 Angstrom   irrespective of the atomic number of that atom, i.e. what that atom actually is.   If you want a "simple law", you can't get much simpler than this.   The distance between an outer electron and the nucleus is about 1 A.
2.    This is what we see when we perform practical experiments on bulk materials  (e.g. LEED experiments on crystals).  We have the real-world evidence.
3.    A simple explanation is that the outermost electron of any atom will behave and therefore take an orbit as if it is orbiting a Hydrogen nucleus.
4.    It's possible to estimate the radius of orbit of an electron around a Hydrogen nucleus.   This depends on the model you might use but most of them do suggest a radius of about 1 Angstrom.   So the theory agrees reasonably well with the evidence.   You could use models (of varying sophistication) to estimate the distance of any other (more interior) electrons from the nucleus, if you wish.   However, this has negligible relevance for any properties of the bulk material.   We could leave all interior electrons as being "some cloud" rather than assuming they must have a definite radius of orbit.

Best Wishes.

7

General Science / Re: Is Absolute Zero Possible?

« on: 04/04/2024 20:34:16 »

Hi.

    The simple answer is the one your High School chemistry teacher gave you:   No.   That would be sufficient for most exams at most levels.

   It looks like you've already had a few comments or read some replies in other places (you mention other forum boards). 

    A lot of arguments will consider temperature with definitions based on kinetic theory (hot things have fast moving particles in them,  cool the thing and the particles slow down,  at zero Kelvin all motion of the particles should have stopped).   This is not the only way you could define "temperature", for example we can take a thermodynamic approach like the one suggested by @alancalverd involving Carnot engines.  We can also take an approach based on statistical mechanics:  Entropy can be directly defined as proportional to the number of microstates a system can occupy at a given energy   and temperature can then be defined as a derivative of entropy with respect to a change in energy.

The details are not important, it's just important to note that "temperature" is NOT a simple thing to define and there are several approaches that are all useful and equally valid.

    In typical situations, all of these definitions of "temperature" will be equivalent, they will all be measuring and describing the same thing.   However, they each describe what it would mean to be at zero Kelvin slightly differently.
The Kinetic Theory definition of temperature is often the easiest one to work with and has become the approach incorporated into the standard set of units for describing quantities  - the S.I. units.   S.I.  = "Syst?me International d'Unit?s"   [French.  Late editing: this website doesn't want to reproduce accents over the e].    So this is the definition of "temperature" that we would normally assume but it is not without problems.    For example, the basic models associated with this kinetic theory are all based on classical mechanics and not quantum mechanics.    So,  when you obtain a numerical value, say a temperature of 100 K in the SI units, for a substance it's telling you about some properties related to a classical model of the substance only.

     I expect you've already seen lots of discussion and arguments to state that a tempertaure of 0 kelvin could never be reached.  So I won't repeat that here.   It's true enough when you recognise that temperature was always set up to tell us only about a classical property of a substance under a conceptualised (and fully classical) kinetic theory model.   Instead we'll try to go around the problem and consider a situation where quantum mechanics is going to win the day.

     There are some things and situations where classical models just aren't useful or relevant any more.   In particular there are some substances (like a simple gas) that can form a Bose-Einsten condensate.   This is a situation where all the particles are occupying the lowest possible energy state.   This can happen at what we describe as a low "temperature" using the SI units for quantifying temperature   (e.g. close to but NOT exactly at 0 Kelvin).   When the gas has reached this state,  trying to cool it further is irrelevant, everything is in the lowest energy state, there is no lower energy condition it can have.     The particles do not have a "kinetic energy" that varies in proportion to the "temperature" in the region of space that some other instruments or thermometers you may have are reporting this "temperature" to be.   Whether it was 1 K  or  0.1 K  doesn't make the slightest bit of difference to the condensate, nothing about it can change, all particles are in the lowest energy state.
     So we are in a situation where the conventional definition of "temperature" just doesn't apply to the condensate.   It doesn't have this temperature as any sort of intrinsic property.   For example, lowering this "temperature" doesn't make the particles move more slowly.  Similarly, increasing the temperature (provided you keep it below the critical condesation temperature) doesn't make the particles move any faster.
     The "temperature" which your other equipment in the region is reporting and we nominally assign as the temperature of the condensate is just an artefact of the way we define "temperature" in SI units.  It's saying  "  if the condensate was represented by a classical system in kinetic theory, then the particles would have this much energy"  -  but the key point is that the condensate is NOT in any way behaving like a classical system any longer.   It is absurd to assign a value for the temperature of the condensate as if it was a classical system, because it just isn't one.    There is no physical difference in the condensate at 1 K,  compared to the condensate at 0.1 K.   Whatever this nominal temperature is describing it isn't any physical property that the condensate has.   I have probably laboured the point but I just wanted to try and make it clear.   When we consider quantum mechanical behaviour, very low temperatures can lose their meaning and just be artefacts of something that would happen if the system was a classical one.

    Anyway, I hope that's a positive note to end on.   We have cooled some things down enough to form a Bose-Einstein condensate and so they have achieved their absolute zero temperature  - the point at which no lower energy or physically meaningful internal temperature could be assigned to the substance.

Best Wishes.

8

Just Chat! / Re: Can an interest in space exploration lead to a greater interest in science?

« on: 02/04/2024 00:41:02 »

Hi  and welcome.

    A lot of us are cautious over first posts because there are a lot of instances where things just become advertising.
However, your post looks interesting and looks like you've put some time into it rather than just copying some text from somewhere.

Can the space program help inspire people to become supportive of other science or has the evolution of my beliefs been due to other things besides the revival of the space program and a TV show?

    Obviously, I don't know what caused your interest in science better than you would know yourself.
A lot of people are interested in space and the science fiction genre for books and TV shows usually attracts a big audience.   I'm reasonably certain this stirs some interest in science among many people quite often.

Looking back I am embarrassed.....

   It's less important what you did some years ago and perfectly fine to acquire or re-acquire an interest in science now.

Best Wishes.

9

New Theories / Re: How does light propagate? By what means does light get from Point A To Point B?

« on: 01/04/2024 17:24:42 »

Hi.

Some minor questions from @Zer0  seem to have been missed.

Please do Allow a few randomly absurd thoughts...

   I'll assume you want short replies.

Can " Space " bend over itself?

    I'm not sure what you were asking.    Here are 2 possibilities:
1.     Can space act to bend space?   Space isn't as empty as you might think,  there is always vaccum energy in every cubic metre of space.    All energy contributes to gravitation (the curvature of spacetime).   So space will bend spacetime and some of that curvature will be purely in the spatial part of spacetime,   so we could reasonably say "yes, space can bend space".  Indeed "vaccum energy" is an important component in modern cosmological models.
2.      Can space roll up like a carpet or fold over itself somehow?    It's best that you don't think of space as some 2-dimensional rubber sheet that curves in 3-dimensional space,   so that whatever image you have of space bending over itself, it probably isn't like that.   Black Holes are an example of a region where curvature becomes extremely degenerate as you approach the singularity.   You might declare this as a region where space has folded over on itself but you'd need to be working in higher dimensional flat space to "see" that.
     String theory sometimes considers the existence of additional dimensions that are described as being tightly rolled up and generally go un-noticed.  However, this isn't really a consequence of space doing anything to roll these other dimensions up,  they just are tightly curled up for some reason.

Can ' Relativistic mass ' not produce a similar, but not the same, effect such as Gravity?

     Yes and it essentially does produce a gravitational effect broadly equivalent to having more mass.
    1.   We don't really use "relativistic mass", it's an old term and discouraged - but we don't need to, we capture the important information about the movement in other ways.
    2.   In GR various things are sources of gravitation.   Sources of gravitation are all the forms of stress-energy that exist and all this information appears in the stress-energy tensor.   So, mass is certainly a source of gravitation but so is energy, momentum and pressure.    So a moving particle contributes a lot more to gravitation (the curvature of spacetime) then a stationary particle.    A moving particle has mass, kinetic energy and momentum.   It can also elevate the pressure that exists in a region of space  (e.g. a gas has pressure due to the movement of particles in it).    Overall, the curvature of spacetime will be more significant when a moving particle is in that region than when a stationary particle is in that region.   
     If you have Newton's ideas of gravity in the back of your mind then it is broadly equivalent to a moving particle acting as a source of gravity according to its relativistic mass rather than its rest mass.   (Why only "broadly equivalent"?  At relativistic speeds, where relativistic mass would be significantly different to rest mass, we'd be better off abandoning Newtonian mechanics and Newtonian gravity entirely and using General Relativity.   Patching up Newtonian gravity with relativistic mass is only going to be a rough or crude adjustment you could make).

Could the ' Higgs field ' twist, tangle, entangle in on itself?

   I don't know.   The Higgs field isn't an ordinary physical thing and I don't suppose it's very useful to think of it as a thing that can twist and tangle in any ordinary sense.   However, such terminology may exist in some very specialised models and areas of Physics that I haven't seen.

Best Wishes.

10

Just Chat! / Re: My History.

« on: 30/03/2024 15:01:26 »

Hi.

You also have my sympathy and I'm sure the same applies from a great many people.

"Your history" isn't just a collection of medical things that may have happened to you.  You won't be judged (or shouldn't be judged) on your medical history.   You've been involved in several discussions here.   I would have thought you're welcome to continue asking questions or join discussions in this website.

Best Wishes.

11

Physics, Astronomy & Cosmology / Re: Why do the Earth's satellites orbit longer than the Moon's?

« on: 21/03/2024 23:29:10 »

Hi.

planetary satellites do not have satellites

    There are many moons around Saturn  (146 according to the NASA website today - but new bodies get identified often).   Of those some were thought to be moons that were in a fairly stable orbit around other another moon - but carefull examinations seem to have shown they weren't.     (Reference:  https://en.wikipedia.org/wiki/Subsatellite    under discussion of "possible natural instances" of sub-satellites).
     None-the-less  there are so many things up there around Saturn that some moon will have some other body that will orbit it,  even if that sub-satellite isn't much bigger than a dust particle.

    This article  ("Can a moon have a moon?",   BBC sky at night magazine),   https://www.skyatnightmagazine.com/space-science/can-moon-have-moon       also discusses the issue.    Their main conclusion was:.......

    ....the moons probably don?t have a large enough gravitational field to control the motions of a smaller object and compel it into orbit.   The planet would likely still be the controlling influence and create an unstable orbit. So the answer is probably no: planetary moons cannot have natural satellites.

    In general, it is known that the mechanics of a gravitational 2-body system will usually result in the formation of a stable orbit of the two bodies around a common barycentre.   However, the 3-body problem is much more complicated.   For example, one of the bodies is often ejected or thrown out of the system over enough time.  It is presumably rare for a moon to retain a sub-satellite in a stable orbit around itself.

Why do asteroids have satellites,

    Many of them stay far enough away from large sources of gravitation (like a planet) that they are effectively only two-body problems,  the asteroid and its satellite.

Best Wishes.

12

Physics, Astronomy & Cosmology / Re: How many spots are produced from Stern-Gerlach apparatus that rotates?

« on: 16/03/2024 14:54:29 »

Hi.

how are the silver atoms entrained in a beam?

    Put silver in a furnace chamber,  it will eventually start to vaporise.   Put a small hole in the side of the chamber and a beam of essentially atomised silver streams out of it.
    Probably not something that passes health and safety standards but it was all done in a vaccum chamber that was well sealed.   I don't know the full history but I'm sure there are some full explanations and accounts of what was done.   There are a few critera they wanted to meet  (e.g. I think they wanted a thin flat beam so they used slits inside instead of round holes,  probably also ensured the beam was reasonable collimated by passing it through two slits etc.)

on Earth- which is rotating.

   It is,  although not in the way we (possiby just I) want to rotate it.   The centre of rotation and axis are in the wrong places.

Best Wishes.

13

Physics, Astronomy & Cosmology / Re: Gas discharge tube hydrogen

« on: 16/03/2024 14:10:50 »

Hi.

The initial excitation is by collision with cations (in this case protons) streaming from the anode.

    Some if it is but there are also always going to be some free electrons to be found.   
For one reason or another, there will be some atoms in the gas tube that are ionised so that we have a free electron and a positive ion (a "cation").   This might be a consequence of cosmic rays striking some atoms or something else.   One way or another you have a tiny number of cations and free electrons in the gas, even though it might be a very small number.

Most gas discharge tubes work by causing an "avalanche effect".   See  https://en.wikipedia.org/wiki/Townsend_discharge   for a description of the "Townsend Avalanche".



   Basically if you have a small number of free electrons to start with and then accelerate them in an electric field, they have collisions with other atoms and quickly start to liberate more electrons and leave more positive charged ions.   Under the right conditions the numbers of electrons increase geometrically (typically doubling on each collision).   So fairly quickly you have a large number of free electrons and positive ions in the gas tube.   There is a more detailed discussion in the Wikipedia page that was linked to.

Now let's go back to your ( @theThinker ) comments:

But QM requires that the electrons could only be excited by the exact frequency light

     That's half right.    If electrons were going to be moved from one orbit to another by absorbing a photon then those photons should have exactly the right amount of energy corresponding to the energy gap between the orbits.    However, QM never said that absorbing a photon was the ONLY way to move electrons from one orbit to some other  (or in this case to liberate them from the atom completely).
      It is quite possible to liberate electrons through impact between the atom and some other particle.   In our case, we have free electrons (and also some cations which can also be accelerated and get involved) these collide with the (un-ionised) atoms and liberate electrons from them.
    Actually, if you just want to completely free the electron then you CAN also do this with photons and it's a lot less restrictive then trying to move them from one orbit precisely to another.   See "the photo electric effect",   https://en.wikipedia.org/wiki/Photoelectric_effect ,   provided the photons have energy greater than the ionisation energy then the electron can be freed and the surplus energy will usually just become kinetic energy that the electron leaves the atom with.   However, that isn't what is done in gas discharge tubes - we just have electrons liberated as a consequence of collisions and impact energy.

In the gas discharge tube experiment for H2,  we have four emission lines in the visible spectrum.

    Once you have quite a few free electrons these will start to fall back into the ionised atoms.   They don't usually go from being free electrons straight to being bound in the lowest energy ground state of the atom.   Indeed they will usually start at some higher orbit energy and steadily fall further down to lower energy orbits.   Exactly as you have stated,  some of these transitions will be to the final level n=2  from higher energy orbits (this is the Balmer series of emission lines).   These are the ones which produce visible light.   More discussion and diagrams available here:  https://www.chemguide.co.uk/atoms/properties/hspectrum.html

 

I hope that helps.
Best Wishes.

14

Physics, Astronomy & Cosmology / Re: What is the relationship between an object and its field?

« on: 13/03/2024 20:59:28 »

Hi.

   Original title question:
What is the relationship between an object and its field?

A reasonable reply is much as you ( @geordief ) have already stated.    For QFT,  a particle is just an excitation in the underlying field corresponding to that particle.

In QFT the former (an electron) is described as an excitation of the latter (the electric field),I believe.

   Not quite.   The electron is an oscillation in the electron field, not an oscillation in the electric field.  The electron field is NOT the same as the electric field,   the start of the words looks the same but the ending is very different.    In QFT there isn't an electric field, or at least it's not anything that you need to assume exists as a fundamental object.    The Electric field will just be something that appears on a macroscopic scale.   However, the electron field is fundamental - you must have an underlying field for electrons if you wish to have electrons as fundamental particles in your model.

As an example an electron and its electric field?

    That's possibly where the confusion may start.   In QFT,  an electric field is not something that belongs just to an electron.   On a macroscopic scale, an electrostatic interaction actually involves a coupling between two fundamental particle fields  (the electron field and the photon field).   The overall interaction does look like or emerge as something we can describe as if there was a classical field, the Electric field, in existance.

Are there other ways of looking at it?

    I'm sure there are.    If it's just a change of pictures or vocabulary,  then it's just window dressing.   These sorts of changes are only useful or important if the new vocabulary and pictures are better for human understanding - even then the changes are not always made.
   Examples:  Methanoic acid is a 1 Carbon organic acid you can make in a lab.   Formic acid is a thing you can extract from ants (Formic comes from the Latin name for an ant, Formica).  It turned out that they were the same thing but both names still persist, we never insisted that we all start calling it the same thing. 
     In Physics, we thought that current was the movement of positive charges.  It turned out it's usually electrons that are negatively charged and moving the other way.   We didn't change the vocabulary or diagrams.

    There may also be some other ways of looking at things that would really involve changing the mathematics or mechanics of how things work instead of just changing some words or pictures.

Can we imagine the field as an extension of the object(the two being aspects of the greater whole)?

    Yes ?   I guess that's the opposite way around to how we usually imagine or phrase things with QFT.  The thing is, does it help in some way and will it change the mathematics or mechanics at all?   Is it just a change of vocabulary?

Could there be (is there?) a mathematical equation  where the object is defined in terms of the field? (along the lines of energy and mass in relativity)

     Does it have to be one equation?   QFT is a collection of results, mainly equations, that quite sucessfully treats objects that look like particles as oscillations in fields.   
     If you mean is there some simple and short equation that does the job - well, I don't know but I don't think so.   If we try hard we could fit most of QFT on a T-shirt but it may be a bit longer than E=mc².   Looks like some people have already tried.

15

Physics, Astronomy & Cosmology / Re: What is the Hopf gauge?

« on: 08/03/2024 17:28:57 »

Hi.

    Thank you @Origin and @paul cotter .    You're probably right about the magnetic vector potential being the best candidate to explain how your toroidal transformer works.     
    It's also noteworthy that although Maxwells equations do work when you consider relativistic changes in the co-ordinate frame,  the E fields and B fields will be different and it's a little tricky to write equations that would show what the E fields and B fields would be in a different frame. For example, there is no 4-vector where the usual 3 component E field vector would be its spatial components.
      It's far easier to construct a 4-vector potential where the usual magnetic vector potential A will be the 3 space co-ordinates and, almost as if it couldn't just be coincidence, the scalar potential Φ for the Electric field turns out to be exactly what we needed as the time component to build this 4-vector.  We can transform this 4-vector potential easily and, if we wish, extract information about the E fields and B fields in each frame from that.   The potentials for the fields and not the fields themselves seem to be the thing that will combine to make a proper spacetime 4-vector.
     Without spending too much time going over the details (which are in several textbooks anyway), this suggests that the electromagentic potential is the natural physical or physically meaningful quantity when you consider Electromagnetism in a frame independent way.  The 4-vector electromagnetic potential will change only precisely as you would expect a vector to change when you change co-ordinates or reference frames.  The actual E and B fields are not so well behaved, they will change in strange ways when you change co-ordinates.  If we paraphrase this further:  The electromagnetic 4-vector potential is a quantity that really does seem to describe a physical thing, a thing that we can describe and talk about in a frame independent way.    By comparison, the actual E and B fields are completely frame dependent and may have no physical significance outside of the frame where they were measured.   So (in my opinion) it probably shouldn't be all that surprising that what may be happening in the potential field is going to matter and possibly matter far more than what appears to be happening with the E or B field.

    We seem to be side-tracking the original thread, so I'll stop.

Best Wishes.

16

Physics, Astronomy & Cosmology / Re: What is the Hopf gauge?

« on: 06/03/2024 20:41:21 »

Hi.

You made some statements that I didn't understand so I asked for some clarity.

   I didn't originate the posts but this is what I think was meant:

What is a "matter field"?

     In QFT (Quantum Field Theory) it is any field for which oscillations in that field or energy in that field could be considered as a particle of matter.   Usually we're thinking of fermions - typical matter particles like an electron, a quark or a composite particle like a proton.   

    I'm not entirely sure that @varsigma really needed to consider QFT for what was being discussed.    The Aharonov-Bohm effect is explainable or can be described with just a non-relativistic, plain vanilla version of quantum mechanics.

   

What is the phase change you are referring to?

     In the standard experiment, it would be a shift in what is called the complex phase of the wave function.   Basic idea:   Wave functions have a value that is a complex number,  they have real and imaginary parts.    So  a wave function might be something that looks like this:



     If you look at the place where the wave function is being measured by the... disc shaped speedometer looking thing with the big arrow spinning around....  you'll see the magnitude of this wave function never changes.  However, the portion of it that is in the imaginary axis  (say running in the up-down direction of the page) compared to the portion of it that is the real axis (say running in the direction that looks like it's coming out of the page or going into the page) does change.  Sometimes all of the magnitude of the value is in the imaginary axis, sometimes all of it is alog the real axis.  Overall the complex valued wave function seems to spin or rotate when you view the bare complex value of the wave function.   The modulus, or overall size of the wave function is what is usually important,   the modulus squared is proportional to the probability of finding the particle at this place,  so in this diagram that never changes and the particle is as likely to be found at place X now as it will be likely to be found there in 20 seconds or tomorrow.    The fact that the complex value of the wave function is rotating in the complex plane doesn't matter in the slightest.   Indeed for almost all purposes, the phase or exact way in which the magnitude of the wave function is spread between the real and imaginary axis doesn't affect any physical observation you could make.   All we usually care about is the overall magnitude of the value and the complex phase of the wave function was, historically, thought to be essentially irrelevant and something that was only there as an artifiact of the mathematics.

     However, the individual oscillations in the real and imaginary axis WILL matter if you brought two quantum mechanical wave functions together,  i.e. made one wave packet approach another wave packet and merge.   On merging, you just add the wave functions much as you would for any other sort of wave mechanics.   However, it is not the modulus or magnitude of the two wave packets that you would add, you add the direct or bare complex value of the two wave functions.   If the two wave packets had their rotations in the complex plane synchronised or "in phase",  then the real part adds to the real part,  imaginary part will add to the imaginary part,   all of it gets made bigger and you end up with one really big valued wave function that still spins or rotates around the Complex plane as before (it's just got a bigger magnitude or longer spinning arrow on the speedometer). 
     On the other hand if the two wave packets that merge had their rotations in the complex plane precisely 180 degrees,  or one half a cycle, out of phase,  then when you add them you end up with the oscillation tending to cancel out:  Whenever one was going big in the positive direction along the real axis, the other is going big in the negative direction of the real axis  etc.   When you add the two wave packets together, you get only destructive interference and 0 oscillation all together.   Hence 0 probability of finding the particle(s) in this place where the wave packets merged but were so badly out of phase.

   

What electron beam?  Where did that come from?

    What the Aharanov-Bohm experiment did was to take a beam of electrons and split it into two.  Pass one of those groups of electrons around the left side of a barrier and the other electrons around the right side of the barrier.  While travelling, the complex phase of any beam doesn't matter at all.   Now re-combine the two beams after they get around the barrier and let the beam hit a screen that will glow on contact with an electron.
     If the two split beams remained in phase,  then you get a certain pattern on the screen.   If the two beams went out of phase then you'll get a different pattern on the screen due to all the destructive interference as discussed just above.

    Here's a diagram (the magnetic field will be explained next):


Aharanov Bohm.jpg (64.35 kB . 978x663 - viewed 443 times)

How do you shield a magnetic potential?

    Well you don't.     What you can shield is an ordinary magnetic field (the thing that should exert force on a moving charge).    The potential, it seems, you can do nothing about.    What was done in the Aharabov-Bohm experiment was to use a special sort of barrier that they sent electrons around.   The simplest one to use is just a very long coil.   As you may know, inside a coil of wire carrying an electric current, there is a very strong magnetic field.   Outside of the coil the magnetic field is really small and effectively negligble at most distances.    The magnetic field wasn't really "shielded", although that could have been done, but it was just easier to use a coil and assume the magnetic field exists only inside of it and not around the outside of it.
     The thinking of the time was that since there wasn't any magnetic field outside the coil and in the space where the electrons were being passed,  the behaviour of the split beams should be much the same regardless of which side of the coil they were passed around.    In particular, the complex phase of their wave functions were in phase before they were split and they ought to just stay in phase.  So, on being recombined at the other side of the coil,  you'll get only the usual pattern on the screen.
    As you may have guessed... you don't.... the complex phase of the beams does seem to be different depending on which side of the coil they were passed around and you end up with a pattern on the screen that suggests destructive interference has occurred when the two beams were recombined.   

      For quite a while this was quite a serious issue:   There was no magnetic field where the electrons were travelling, what on earth was affecting one beam differently to the other?   Turn off the coil and everything is back as it should be... good old pattern 1 on the screen when you recombine the electrons.   Somehow the magnetic field that existed only inside the coil seemed to be influencing the electrons that were only ever outside of the coil.    Sure, if some charged particle was moving and it went through a magentic field you'd expect to see some effect but you need to realise that the magnetic field is precisely 0 valued outside of the coil, there is no magnetic force of any sort to be felt where the electrons were travelling.

    Let's use gravity as an example because that's usually easier for us to appreciate.   Let's assume I can put two equal sized planets next to planet earth,  one on the left and one on the right,  equal distance from the earth,  everything nice and symmetric.   You would have a new gravitational force pulling to the left but also a new one pulling right, so overall no net force.   You would carry on with your life as usual and expect to notice no change in anything whether these pair of planets were there or not.    What the Aharanov-Bohm experiment suggests is that... actually there may be a difference.   Although there is no net gravitational force acting on you, you have certainly changed your gravitational potential.   There are loads of things for which we still think "potential" is fairly arbitrary and all we need is some arbitrary reference point we ca call 0 potential.    In electronics, the earth line is reference 0 volts and everything works provided the +5 Volt line is at a potential 5 Volts higher.   We don't try to guess what the actual potential of the earth line was and we assume it doesn't matter.   If we took the box of electronics to another planet,  drove an earth pylon into the ground there and used their "earth" as our electronic earth, we would still expect the box of electronics to work.   

     The Aharanov-Bohm experiment suggests that for magnetic potential (and possibly other potentials) things are not as we once thought.   It made a difference when the coil was turned on because the magnetic potential of the space the electrons were in had changed, even if there was actually 0 magnetic field (no magnetic force if you prefer) in the space.   It's maybe not that interesting these days where we flippantly add another field whenever required - but back in the day it was important to imagine that magnetism (and possibly every other force) may demand that two fields existed throughout all of space.   There is a magnetic field, which can exert a force on moving particles but there is also a magnetic potential field.   More importantly, the potential field may actually have physical effects rather than just being a mathematical tool.   This alone is important, it means that maybe knowing that the +5V line in your box of electronics is 5 volts higher than the earth line is NOT enough,  it may actually matter what the absolute potential at the earth line really is.

    Unlike the electric or gravitational fields,   the magnetic potential field cannot be described just as a scalar field,  we require a vector valued field for magnetic potential (i.e. it has both a magnitude and also a direction).   This was another reason why the notion of a magnetic potential was contested or considered to be just a mathematical artifact.   An explanation for the Aharanhov-Bohm experiment is not attainable if the magnetic potential is just a scalar field.   There is a perfect symmetry around the coil in the experiment ,  an electron passing on an equivalent path on either side of the coil should have been put into all the same potentials for all the same amounts of time,  so even if it did affect the complex phase of their wave functions, they should be affected equally.   On re-combining the beams at the other side of the coil, you'd still expect to have both beams remain synchronised or in phase - and get a perfectly standard pattern for constructive superposition being shown on the screen.
   If we allow the magnetic potential to be vector valued instead of just a scalar, then we can break this symmetry.   The potential field has direction in addition to numerical value and there can be a net circulation of the vector field around the coil:   We imagine that electrons passing around one side can be travelling against the direction of the magnetic vector field,  the ones travelling around the other side would be travelling in the same direction as the magentic potential vector.

    That's probably all I know about the Aharanov-Bohm effect,  I hope it's of some use to you.

 Best Wishes.

17

Physics, Astronomy & Cosmology / Re: What is included in a configuration space?

« on: 01/03/2024 19:44:43 »

Hi.

   I've probably had enough time to answer my own question.

    The formal definition of "configuration" for a system where only classical mechanics will be used is just the position of every particle in the system.    It doesn't seem to matter in the slightest if the system is fully described just by positions or not.
    So the "configuration space" is just the space of all possible positions of the particles.

    The forum was still useful.   Just trying to write out the question or problem precisely actually helps to formulate what the resolutions may be.   So thank you to everyone.

Definition 2.2.1. The set of all possible (in principle) instantaneous configurations for a
given physical system is known as configuration space. We will denote it by C.
It is important to note that this includes positions, but it does not include velocities.
One informal way of thinking about configuration space is as the space of all distinct
photographs one can take of the system, at least in principle.
     [Taken from some lecture notes for Durham University:  https://www.maths.dur.ac.uk/users/inaki.garcia-etxebarria/MPII/ch/2.2a.pdf  ]
     The article does go on to emphasise that configuration and configuration space is not sufficent to describe the dynamics of most systems and therefore is woefully inadequate to fully describe what I would have considered "the state" of the system.   Thus, making the definition of "generalised co-ordinates" given in Wikipedia essentially misleading.

From WIki (and given in the earlier post):
In analytical mechanics, generalized coordinates are a set of parameters used to represent the state of a system in a configuration space.
   If you took away the last few words in red, the definition is just plain wrong.  Knowing the generalised co-ordinates for a system is simply inadequate to know the state of the system.   If you leave the words in red, then it's unhelpful because you have only a circular definition since the term "configuration space" is only defined by reference to generalised co-ordinates in Wiki.

Best Wishes.

18

Just Chat! / Re: Is Category Theory the well of mathematical philosophy

« on: 29/02/2024 02:07:55 »

Hi.

Anyways, does anyone here have any interest in this stuff

     Some interest but not a lot of expertise, sorry.

or some understanding of how far it all goes, this attempt to codify all mathematical objects and relations between them?

      Mathematicians will always be keen to identify what their most fundamental mathematical objects must be.   Physicist's are keen to know what their most fundamental physical object or "atoms" would be, mathematicians are much the same.

or how doing this appears to simplify things considerably, in that mathematics is actually mind bogglingly simple?

   It is simple and Mathematicians are probably only working with and manipulating a very small class or category of structures, all built from what are probably just a small set of atomic objects.
    On the other hand, the English language is just 26 different characters with some rules on how to combine them - how much more simple can you get?   Yet, you can still generate propaganda that will start wars between nations, poetry that can help to end them and works of philosophy that examines the most profound issues of existence.    Computers require only two atomic elements, a 0 (or "false", "off", or whatever you wish to call it) and a 1.  With that they can capture and display photographs or the audio from an orchestra,  or seemingly understand and respond to questions as demonstrated by ChatGPT.
     How many different categories or types of structures do you think we would need?

Best Wishes.

19

New Theories / Re: What caused inflation?

« on: 29/02/2024 00:28:11 »

Hi.

    We may need to take some care when developing a model with virtual particles.  You've taken a notion that is far from certain and was originally intended only to serve as a picture to simplify explanations and then you've run as far as you can with that illustration.
    When PopSci articles describe things they do tend to use phrases like "virtual particles" because it's easier for most people to imagine something in terms of particles doing something.    "Virtual particles" may be nothing more than a metaphor, simplified model or just an illustration of an idea.    In much the same way that gravity isn't a rubber sheet that gets stretched when a planet rests on top of it even though this representation is shown in many diagrams, virtual particles are just something happening in quantum fields.    It can be helpful to imagine interactions as an exchange of virtual particles but most experts stop short of declaring that virtual particles would actually be exactly like some sort of real particle.

The first thing you need to know is that virtual particles ...  ...aren't really particles at all.  The language makes people misunderstand 
   2nd paragraph,  New Scientist,  Why virtual particles don't exist but do explain reality,   https://www.newscientist.com/article/mg25834383-000-why-virtual-particles-dont-exist-but-do-explain-reality-for-now/

The term (virtual particle) is somewhat loose and vaguely defined....   ....The accuracy and use of virtual particles in calculations is firmly established, but as they cannot be detected in experiments, deciding how to precisely describe them is a topic of debate.
     Wikipedia.    https://en.wikipedia.org/wiki/Virtual_particle

Best Wishes.

20

Physics, Astronomy & Cosmology / Re: If Red Light cures everything, why aren't hospital lightbulbs red?

« on: 27/02/2024 01:04:00 »

Hi.

    Some of the phrases in that article you linked to include:

more research is necessary to fully determine its potential benefits. ;

it's important to keep in mind that many of the published studies using red light as therapy are often small (less than 30 people), animal or lab studies. As a result, most medical professionals agree that more quality research with more human participants is needed.;

They're not dramatic treatments, ;

    You should also add your own experience and assessment of Magazine or website articles.   No one wants to write an article that says "there's not much evidence here".... you can sell more if you are able to say  "there may be some effect here".
    However, some of article was interesting.    The problems of wound healing in zero-G environments and the possible use of red light was interesting and not something I was previously aware of.   Apparently this has been identified many years ago and there are quite a few articles about this already.  Here's a short one:
https://www.newscientist.com/article/mg16322053-100-space-healing/

why are hospital lights still white?

I don't know and I'm not an expert.   These reasons seem sensible:

    (i) There isn't enough evidence for the benficial effects yet.
    (ii)  You need to do more with the light than just heal surface skin injuries.  Patients want to read a book.   Older and partially sighted people want to be able to walk to the bathroom without crashing into obstacles.   Nurses want to be able to see skin rashes in a natural white light not some low intensity red light   etc.
    (iii)   Red is an inherently low-energy and hence low-intensity light.   It's not bright or easy to see with.   It's not cheerful or motivating to get up and do something - like a jigsaw puzzle in the day room etc.    Keeping patients optimistic and motivated is at least as important for getting them well as healing their minor skin complaints.
      (iv) As mentioned in your Forbes article, if you had very powerfull red light emitters all along the hospital wards, then you would also need to issue eye protection to every patient, nurse, doctor, visitor etc.

Let's also do something similar to that magazine article and close with a nugget of information:

  Did you Know?         There already is some red light in white light.   

Best Wishes.