1

General Science / Re: Does All Of Mathematics Come Down To Simple Arithmetic?

« on: 12/10/2024 10:39:31 »

I have seen an undesirable side-effect of introducing computers to the classroom. In my kids' primary school I met a group of 10-year-olds who defined a circle as 360 straight lines.

2

Question of the Week / Re: QotW - 24.06.14 - Why do we retain maths ability better than facts?

« on: 10/06/2024 19:00:14 »

to be truly experienced within a certain craft, you must practice it for ten thousand hours.

But there's a saying in aviation:

After 100 hours you know everything
After 1000 hours you know that you don't know everything
After 10000 hours you know that you can't know everything

3

Physics, Astronomy & Cosmology / Re: Are quantum systems dynamic?

« on: 06/01/2024 23:13:08 »

- I am assuming that the calculations involved treating light from the star as a classical light wave passing through a classical medium with time-varying refractive index, so no quantum effects need to be calculated.

At some point in the system, you need to generate an electrical signal representative of the wavefront distortion. This involves a photon-electron converter (usually a CCD), which, like all photon detectors, employs quantum phenomena!

4

Physics, Astronomy & Cosmology / Re: Are quantum systems dynamic?

« on: 30/12/2023 18:52:26 »

Hi.

I essentially agree with what  @evan_au  just said,  with just some minor differences.
So, that does mean that @Origin was also correct - in that if you kept measuring the state of the quantum system (let's say at every minute, on the minute), then you would see evolution in discrete steps.  Indeed if you were to be continuously making a measurment of the quantum system (so that is at all times and not just one minute apart with no measurement made in between) then it may be effectively frozen into one qunatum state and not show any evolution.
Some of this was discussed in another fairly recent forum post but it will be easier just to give a relevant reference here:

    The quantum Zeno effect (also known as the Turing paradox) is a feature of quantum-mechanical systems allowing a particle's time evolution to be slowed down by measuring it frequently enough with respect to some chosen measurement setting.
    Sometimes this effect is interpreted as "a system cannot change while you are watching it".  One can "freeze" the evolution of the system by measuring it frequently enough....
 

[quote taken from Wikipedia, "Qunatum Zeno Effect":      https://en.wikipedia.org/wiki/Quantum_Zeno_effect ]

     I appreciate that it is difficult to understand quantum mechanics, so I will repeat what @evan_au has said but in a slightly different way  -  in the hopes that another way of saying it may be useful.    Note that I am not really attempting to explain "why" quantum mechanics works,  I am just stating "how" it works.

This is what you do, or how quantum mechanics works:

   1.    You have a wave function.   This is something that effectively describes the system at a time t.

   2.   The wave function evolves as determined by the time dependent Schrodinger equation   (TDSE).   That is, it will evolve like this provided you do not measure some observable for the system.   Provided the potential in which the particle exists is sensible then this time evolution will be continuous and even perfectly "smooth".   (I put "smooth" in quotes because it will be as smooth as the potential allows - but we have a tendancy to look at simplified situations like square well potentials where the potential can make abrupt and infinite jumps).
     I would write down the formula to determine  Ψ(x, t+δt)   given   the initial  Ψ(x,t) and the TDSE but this forum doesn't currently support mathematical notation in LaTex script.
     The main thing is that the time evolution of the wave function is continuous (it has no sudden jumps or discontinuities),  this is simply because the wave function at a later time Ψ(x,t+δt) is determined as a definite integral  from  t to  t+δt   of the TDSE.

   3.  The wave function actually determines everything that can be known (or observed) about the system.  So it is a complete description of the system.    However, there are only certain things that can be known or observed.   Something is an observable if it can be formulated as a Hermitian operator acting on the space of wave functions.  We don't need to worry too much about the details - something like position is an observable,  something like whether the electron likes strawberry milkshake is not an observable.    As you may know, there are a few things that may seem like you should be able to know them but actually cannot be formulated as an observable:   For example, in a classical system you would expect to be able to measure the angular momentum vector  (ω_x , ω_y, ω_z ) = the x, y and z components of the angular momentum  of some particle.   However, in a quantum mechanical system you can only measure the x, y or z components seperately and could not measure all three simultaneously.   The total angular momentum vector (with x, y, z components) is evidently a vector and so it would be represented as something like a vector-valued operator, specifically it would map the space of wave functions, H,  to  the cartesian product space  HxHxH.   To say that more clearly, the vector-valued angular momentum operator takes H  --> HxHxH.    As such there is no reason to assume the angular momentum vector would be "an observable", it cannot be described as an operator just on H , i.e.  mapping  H -> H.     It takes a little while to get used to but, in general, we just need to accept that in QM,  something is an observable   if and only if    it can be formulated as a Hermitian opertaor acting on the space of wave functions   (specifically taking H -> H  and being Hermitian with respect to the inner product defined on that wave function space H).
    In the case of angular momentum, we can find a perfectly satisfactory Hermitian operator acting on H for any one of the components   (say the z-axis component),   so that component is an observable.   Now you may think that all we need to do is go and measure the  x, y and z components seperately and then we can put all this information together and have the complete angular momentum vector BUT it's not quite that simple.  As we will discuss in a moment, once an observation has been made the wave function is said to collapse (ie. it changes),  as such any previous value for another component is no longer valid.   So, you can certainly measure the x component and get a definite value but then when you go on to measure the y component you will change the wave function and mess up the previous x component value....  it's no good trying to go back and measure the x component again because that will mess up the y-component...... etc. etc.

4.    As we have already hinted at in the discussion above,  when an observation is made then the evolution of the wave function changes its rules completely.  Under the Copenhagen interpretation of QM, a new set of rules applies:
     (i)  The evolution upto this time, t, which was determined as described in Step (2) above is not completely ignored.  Specifically the evolution as determiend by the TDSE is taken into some consideration.
     (ii)  Only certain values of the observable could be obtained.
     (iii)  Having obtained that measurement, the wave function immediately changes.  It is now entirely just the eigenfunction for the observable corresponding to that measurement.

Now let's explain each of these rules in turn:

   (i)  The wavefunction evolution by the TDSE upto this time t is not completely ignored:    The wave function is actually going to be a combination of eigenfunctions for the observable that is just about to be measured.  This combination is important.  If there was 0 amount of eigenfuction F₁  in the wave function then there is no opportunity for the wave function to collapse to that eigenfunction  F₁.   If the wave function was a lttle bit of eigenunction F₁, then there is a chance it will collapse to become entirely F₁.    The chance or probability of collapse to the function   F_i  increases as the amount of  F_i   increases.
   I would like to say the probability of collapse to F_i  is proportional to the amount of F_i  in the wave function but that isn't quite right or would need some further defintion of what we consider as "the amount" of the eigenfunction.     The wavefunction  Ψ(x,t)  at the given time t is a linear combination of the eigenfunctions:

      Ψ(x,t)  =    a F₁    +   b F₂  + c F₃ + ...... more...
where a,b, c, .....  are called the coefficients of the eigenfnctions and I have called the eigenfunctions  F₁, F₂, ...

The probability of collpase to the eigenfunction F_i isn't proportional to the coefficient for F_i,  it's proportional to the square of that coefficient. ** LATE EDITING:  Actually the square of the modulus of the coefficient.   We are always interested in the size squared and we can allow complex valued coefficients.   If you're an engineer then you are probably familiar with Fourier Analysis as applied to some waveform that is a combination of sinusoidal waves of different wavelengths.   The eigenfunctions are like the individual sinusoidal waves present in the total waveform and the square of the modulus of the coefficients is much like the amplitude of the power spectrum.  What we are saying then is that the combined wavefunction collapses to an elementary waveform F_i that was present in the power spectrum and it will do this with a probablity that is proportional to amplitude of the power spectrum for that elementary waveform.  To put it more casually, it collapses to the elementary waveform F_i with probability proportional to the amount of F_i that was there.**   

   (ii)  Only certain values for the measurement are possible:   The only values possible are the eigenvalues for the eigenfunctions.  The wavefunction is about to collapse to precisely one of the eigenfunctions present in its eigenfunction expansion shown just above.  When this collapse occurs, the eigenvalue corresponding to that eigenfunction is returned as the value for the measurement that was made.
    As outlined above - it is seen that the evolution of the wave function by the TDSE upto the time of measurment was important:  It determines which values for the measurement could be obtained and how likely each such value would be.   For example, if there was no amount of an eigenfunction F_i then there will be no chance to measure the value corresponding to that eigenstate.  (I'm going to assume the simple situation where the eigenfunction expansion is non-degenerate,  i.e. you will not have an identical eigenvalue for two different eigenfunctions).

   (iii) Having obtained that measurement, the wave function immediately changes.   This is the single most important thing that happens when a measurement is made (under the Copenhagen interpretation).   The wave function just before measurement was   
Ψ(x,t)  =    a F₁    +   b F₂  + c F₃ + ...... more...
as stated under  subsection (i) just above.    After measurment we have:
   Ψ(x, t+)  =     F_i   =   1 . F_i
where I have written  t+  to denote an arbitrary value of time that is (even just a tiny bit) greater than t (the instant when the measurment was made).   It is no longer a combination of several different eigenfunctions for the observable, it is entirely just one part the eigenfunction corresponding to the value obtained from that measurment.

(iv)   OK, I know there wasn't a sub-section (iv) but you'll want to know what happens next....   We just go back to subsection (i),   the new  wavefunction   [which was just one eigenfuction at time  t(+a little bit) ]  now continues to evolve as directed by the TDSE  until a new time is reached when another measurement is made.

- - - - - - - - - - - -
  That's it,  that's how QM works  (under the Copenhagen interpretation).   I'm not saying it seems reasonable or suggesting we know "why" it should work like this.   Indeed "wave function collapse" and the "measurement problem" is something that is extremely weird.   Why is it that the wave function evolves one way (by the TDSE) until a measurement is made and then a new set of rules comes into effect?   However, let's just be in the camp of people who just do the calculation rather than worrying about why it works like this.
 

Can quantum system be said to evolve (I think so)?

    Yes.   Without measurement the wavefunction evolves by the TDSE.   Since the wavefunction determines everything that can be known or measured for the system,  we can reasonably say that the system evolves (changes with time) if the wavefunction changes with time.
    Note that there could be some situations where the wavefunction would actually not change with time.   This is comparable to a classical system like a ball that has no initial velocity and is placed at the bottom of a valley -  the ball just sits there and is the same for all time.  We could say the ball system was showing no evolution with time if you wanted to be annoying.   I would prefer to say that we can determine the evolution with time, it just so happens that there is no change.
    If a measurment is made on a QM system then there is also some evolution of the system.  It is much more immediate, abrupt or discontinuous compared to evolution by the TDSE  (usually).   
    Note that we can prepare a QM system so that it was initially in an eigenstate for a particular measurment and have a potential such that this eigenstate is undisturbed by evolution under the TDSE.  Then provided we only measure that observable for which the system was in an eigentstate then we will change nothing (the wavefunction collpases to percisely what it was anyway).    A concrete example would be "a particle in a box" system,  let's say an electron in an infinite square well.  We can prepare the system so that initially it had precisely the energy required to be in an energy eigenstate and then it will evolve under the TDSE to remain in that state.  We can also go and measure the energy again at some later time, the wavefunction is only that energy eigenstate and nothing else so it can only collapse to that same energy eigenstate.   So, leaving it to evolve un-measured or just measuring the energy does not change the wavefunction in any way.   However, if you measured something else, let's say the position of the electron then you would immediately change the wavefunction and hence disturb the evolution in a discontinuous way.
    Overall summary:   Yes, we can determine how a QM system evolves with time.  It has different rules depending on whether you have measured something or not.  When un-measured, the evolution of the system changes smoothly and continuously with time  (for sensible potentials).  When measured the change can be sudden or discontinuous.

Do they ,then evolve in discrete ways and not in a continuous fashion except when viewed as part of a classic system?

   This is where it gets a bit interesting.   The wavefunction can evolve smoothly BUT you cannot go out and see or observe the wavefunction.    The tea cup infront of me has a wavefunction but I cannot see it, hear it or smell it.  I have no equipment with which I could measure it and indeed no Physics laboratory has the equipment to measure and record the wavefunction of my tea cup.  It is generally thought that the wavefunction is not any sort of physical thing.
     As previously mentioned the only things I can see or measure are the values of observables.    Observables are Hermitian operators acting on the space of wave functions.    When I observe something, a value is reported to me.  An observable can only report or be measured as having a value that is a single real number  (it can only be an eigenavlue for that operator and it's Real valued because the operator was Hermitian).  So we know instantly that there is no valid observable that would report the entire function  Ψ(x,t=now)  because that is a function of  x and would need to be given as an infinite set of values.   (For example,  I need to know what  value  Ψ(at x=0, t=now) is  and also what value  Ψ(at x= 1, now) is and also what value  Ψ(x=√2, now) is supposed to be  etc.  etc...   I need to know an infinite set of values for Ψ because there is an infinte set of positions, x, that I need to know about).   The best I can hope for is that there are observables that may report the real and imaginary parts of Ψ  at just one given x position and time,  t = now).    Just as with the previous discussion about angular momentum and its components,  we have no reason to beleive that we could measure Ψ(at a given x, t=now) and then go on to measure  Ψ(at another x position, t=slightly later)  without messing up the previous value for Ψ at the other x position.
    So what does this mean?   You cannot see the evolution of the wavefunction happening as it unfolds according the TDSE.  In theory it is happening, you just cannot see it.   What you can see are only the valid QM observables.  These may be things like the position of my tea cup.   Depending on the situation (e.g. the potential in which my tea cup exists), the position of my tea cup may very well be qunatised.  So it may only ever be found at x=1 , 2, 3, 4, ... positions  and never at   x = √2 position.    This means that if the position of my cup does change with time, then it could only take discrete hops from one place to another.   For typical systems, the discretisation is so fine grain that you would not notice it is discrete rather than being continuous.
    That's OK and we could leave it there but I'm sure that someone (that'll be @Halc with his "counterfactual" comments about Qunatum Mechanics) will raise an argument.    It is very human to assume my QM tea cup which I used as an example must always have some definite position but that is not actually required or suggested by Quantum Mechanics.  Just to be clear, there is no need to assume the cup had to be occupying one given position when it is not observed.  When un-obsserved the wavefunction can be in a superposition of states,  so that if the position was measured then it could be found to be at x=1 or x =2 with weighted probability such that the average or expected position did turn out to be x = √2.   This does not mean that the cup was actually at x=√2 at this time,  as stated, we do not know where it is until the position is measured so it just does not have a definite position.   However, for classical systems where many (millions) of particles may be involved, the ensemble will behave according to the average or expected positions.   So that a classical system (let's say a model for the structural integrity of my table which is supporting the tea cup) will show stress in various parts of the table as if it was supporting a mass equal to my tea cup at position x=√2 at that time.     To say that another way:   Although the tea cup could never be found at x=√2  when the position is observed,  it is possible to have a classical system that responds as if the tea cup does slide smoothly and continuously along the table rather than making discrete "hops" to different positions.

     I know that was a lot of words and may not be simple.   Here's an alternative summary:    We all want to say "yes, the wavefunction evolves smoothly and continuously when un-observed"  and then quickly suggest that's all there is to it, the system can evolve smoothly and continuously when un-observed.   However, the wavefunction just isn't a valid observable,  the only things you can observe may actually be discrete and therefore could only change (evolve in time) to take on other values in that discrete set when you do measure them.   When you don't observe them, those observables don't even need to have a definite value and we can therefore say very little about how they would "appear" to change because you just aren't observing them, they do not "appear" at all.    However, a classical system coupled to it can behave as if the observable value exists at all times (taking a value ~ the expected value of that observable) and that would show a continuous or smooth evolution.

   I hope that makes some sense and explains some of what you were asking about.

Best Wishes.

5

Physics, Astronomy & Cosmology / Re: Why isn't my salt water evaporating?

« on: 15/09/2023 20:13:10 »

Thanks, Evan_au. As I explained above I answered similarly in a more verbose but most likely less clear answer that evaporated.

6

Physics, Astronomy & Cosmology / Re: Does helium make all sounds higher pitched?

« on: 05/09/2023 23:04:38 »

length of the air column and the timbre of the harmonics are determined  by the wood (or metal in most cases nowadays) .

So, a brass flute would sound more  like a tin whistle than like a wooden flute.
And a wooden flute would, I guess, sound like a recorder, rather a metal flute.

Or... not.

I suspect that the reason why plastic musical instruments exist is that, to a first approximation, the timbre depends on the shape of the instrument.

It would certainly be interesting to  fill a flute with helium (or hydrogen, if you are concerned that He isn't a renewable resource) and close it with cling-film so you don't blow any air through it.

I suspect the tone would not be improved by a bit of thin plastic, but it would illustrate a point.

: if you filled your trombone or bagpipe with helium, would that alter its pitch?

One way to find out; check on YT and see if it's been done.

Apparently, if you put helium into a bagpipe it turns into laughing gas.

7

Chemistry / Re: Is nitrogen based liquid fuel a viable energy storage solution?

« on: 29/05/2023 11:24:19 »

toxicity is a problem at high concentrations

Not if you are comparing it to lead-acid or lithium batteries.

8

Physics, Astronomy & Cosmology / Re: BLC 1 Signal from Proxima Centauri?

« on: 18/05/2023 23:09:32 »

- Half of the time, the Sun is above the horizon for a radiotelescope on the far side of the Moon

True, but for on the "near" side of the moon, the earth is in line of site 100% of the time and on earth the earth is in line of site 100% of the time.

So having the sun in your eyes only 50% of the time looks like a really good option.

9

Plant Sciences, Zoology & Evolution / Re: Where would I find King Kong?

« on: 15/05/2023 18:32:56 »

Where would I find King Kong?
Alphabetically, between Harald V of Norway and  Letsie 3 of Lesotho.
https://en.wikipedia.org/wiki/List_of_current_monarchs_of_sovereign_states

10

Just Chat! / Re: Can a being with a strength like Superman exist?

« on: 01/04/2023 12:21:08 »

Reversing time by flying around the Earth is not compatible with our understanding of General Relativity,

But if you fly eastwards from Samoa you can arrive yesterday, and 20,000 miles to the west,  even in a light aircraft.

11

COVID-19 / Re: can i have a full run down on spike protiens please?

« on: 18/03/2023 22:10:46 »

I think Ebola has spike protein
https://www.reuters.com/article/health-ebola-dc-idUKN0937961120080709

, are they god/bad/neutral

From the viruss perspective they are very good.
From the patient's point of view, they aren't.

3, how are they made

The same way a virus makes any other protein; they hijack the biochemical pathways of the cells they infect.

4, do they occur naturally

Yes; they are part of many viruses.

5, if they are bad, how do we get rid of them

Our immune system destroys them (if we are lucky)
To completely eliminate them I think we need to wait until the sun goes red giant and bakes them all.

12

New Theories / Re: What is the real readshift in the Cosmic Microwave Background (CMB)

« on: 08/02/2023 13:10:02 »

My simplistic understanding of what they said is:
- Roughly 15 (or 14) billion light years is the distance from "where the source was when the light was emitted" to "where we are now"

It's actually about 1/10000th of that. In cosmic coordinates (the only coordinate system I know that describes the universe), the oldest light we see (that of the CMB) was emitted at a proper distance of about 1.5 million LY away. The reason it took 13.8 billion years to get here is due to the very high expansion rate of the universe back at the time of the recombination event, perhaps 3M km/sec/mpc compared to 70 km/sec/mpc today.

Roughly 100 billion light years is the distance from "where the source is now" to "where we are now"

That would put it beyond the size of the visible universe which means we could not see it. So around 45 billion light years is the proper distance from "where the source is now" to "where we are now".

- There is an even smaller number which represents the distance from "where the source was when the light was emitted" to "where we were back then"

Well since we have not moved significantly in that time, that distance is also that 1.5 MLY figure. Cosmic coordinates has the Earth at the center, unmoving. You have to assign the origin somewhere.

As relativity illustrated, all times and distances are relative to which frame of reference you are talking about

Yes, which is why I carefully specified the cosmic frame and not say some inertial frame, which isn't valid at all at large distances since spacetime isn't Minkowskian.

And anything outside your light cone is irrelevant to you (eg if some object is now 100 billion light-years away, light from that object will probably never reach us, due to the expansion of the universe).

Correct. Any recombination light emitted from what is currently over about 58 BLY away will never reach us. Any light emitted today from over 16 BLY away will also never reach us. That latter figure is the current distance to the event horizon.

13

Technology / Re: How were images transferred before the digital age?

« on: 08/01/2023 18:31:53 »

Ahh video, hours and hours of people recording their holidays on camcorders,

Before Facebook, you had to photograph your lunch, rush to the chemist, get the film developed, make 25 postcard prints,then go to the post office and send them to your friends, by which time the meal was cold.
But whilst Evan can't see his recent photos on  his new computer, I still have black and white prints of equipment I built nearly 60 years ago, and if you want a good picture of an extinct bison there are plenty of 40,000 year old cave paintings that show you how to hunt them with spears.

14

Physics, Astronomy & Cosmology / Re: Has the speed of light been tested in a vacuum?

« on: 08/01/2023 10:26:07 »

Hey...  Does anyone have any comments on my back-of-the-envelope calculation of air being 99.85% vacuum?

Especially the part where I try to derive the refractive index of air from the refractive index of liquid nitrogen
- The method I used for that calculation used a lot of hand-waving
- What would be a better method of calculating it?

While your calculation isn't precise, it's a whole lot better than lots of the ideas put forward in this thread.
Ignoring the oxygen is not a serious problem because the refractive indices of O2 and N2 are pretty similar.


It's possible to measure the speed of light in a gas at varying pressures and then plot a graph and extrapolate to zero pressure, which will give you the speed of light in a vacuum.

15

Physics, Astronomy & Cosmology / Re: why do a lot of people confuse between interference and diffraction?

« on: 28/12/2022 09:50:26 »

(400 kHz homing beacon)...interference and diffraction are distinct phenomena, but both are effects of superposition

The behaviour of radio waves from a homing beacon can be described quite well by Maxwell's equations. This is "classical" physics.
- Superposition derives from quantum theory, and cannot be described by classical physics.
- You don't need superposition to describe radio waves

Superposition principle has been widely used long before quantum theory.
https://en.wikipedia.org/wiki/Superposition_principle

The superposition principle,[1] also known as superposition property, states that, for all linear systems, the net response caused by two or more stimuli is the sum of the responses that would have been caused by each stimulus individually. So that if input A produces response X and input B produces response Y then input (A + B) produces response (X + Y).

A function F(x) that satisfies the superposition principle is called a linear function. Superposition can be defined by two simpler properties: additivity


and homogeneity

for scalar a.
This principle has many applications in physics and engineering because many physical systems can be modeled as linear systems. For example, a beam can be modeled as a linear system where the input stimulus is the load on the beam and the output response is the deflection of the beam. The importance of linear systems is that they are easier to analyze mathematically; there is a large body of mathematical techniques, frequency domain linear transform methods such as Fourier and Laplace transforms, and linear operator theory, that are applicable. Because physical systems are generally only approximately linear, the superposition principle is only an approximation of the true physical behavior.

The superposition principle applies to any linear system, including algebraic equations, linear differential equations, and systems of equations of those forms. The stimuli and responses could be numbers, functions, vectors, vector fields, time-varying signals, or any other object that satisfies certain axioms. Note that when vectors or vector fields are involved, a superposition is interpreted as a vector sum. If the superposition holds, then it automatically also holds for all linear operations applied on these functions (due to definition), such as gradients, differentials or integrals (if they exist).

16

New Theories / Re: Not-Quite-So Elementary, My Dear Fermion

« on: 28/12/2022 04:30:44 »

However, I hope you will not be offended. If I said, we could use a second opinion of a physicist with regards to this particular point. because I believe it's important, I don't remember you addressing it anyway

Sure, anyone who wants should feel free to chime in.

Literally existing!? partly existing!? There are no such things, and they have no meaning in this universe, it's either you exist, or you don't.
I think the name quasi-particles has nothing to do with them being real or imaginary or partly existing. They have mass and they interact, they are a form of matter.

I'm not trying to say that quasi-particles aren't real, only that they aren't the same as true elementary particles, and thus aren't "real" particles (in the same sense that "horny toads" aren't "real" toads): https://en.wikipedia.org/wiki/Quasiparticle

Although am not sure if they can split the positron in a different experiment into anti-quasi-particles.

In principle, it should be possible to do so, but you would need bulk antimatter in order to accomplish that. Put a positron in normal matter and it annihilates.

what mattered is the question: would there be any matter left of the Electron? which you did not address as well.

Assuming you could somehow get all three quasiparticles to interact with a positron at the same time, then I think it would annihilate and thus become radiation. In that case, nothing would be left of the electron.

Maybe, they don't need a medium like light, or maybe when they are bound, they are each other's medium, maybe if there is a force that governs their bound state (external factor) maybe, it would be their medium, can you say for sure it is impossible to be?

Being in the right medium is what gives rise to them. That's what makes them quasiparticles. An analogy is phonons, which are another type of quasiparticle. Phonons are quantized vibrational modes in a lattice of atoms. If I had a solid that was filled with phonons and then pulled it apart atom-by-atom, then all of the phonons would be gone because there is no longer any way for lattice vibrations to exist in individual atoms. That's what I mean by quasiparticles being an emergent phenomenon. Take away what's needed for emergence, and you can no longer have them. If you still think that quasiparticles are elementary particles, then I think you still have misconceptions of what a quasiparticle is. Here is a video that explains them:

The takeaway here is that spinons, orbitons and holons are not more fundamental than electrons. Rather, they are more complex than electrons.

I don't see that; they can't cease to exist in any sense other than "cease to exist"

If you pluck an electron out of a solid, then it's just an electron. If it was an orbiton, holon and spinon before you did that, then yes, you can choose to say those three quasiparticles ceased to exist when you did that.

17

Just Chat! / Merry Christmas

« on: 25/12/2022 00:56:53 »

Merry Christmas everyone.

   
  {Image from Wikimedia - should be copyright free, more or less}

Stay cheerful and we'll see you in the new year.

18

Physics, Astronomy & Cosmology / Re: How dense is neutronium and how many stars are in the Milky Way?

« on: 21/12/2022 05:28:08 »

Hi again.
    Looks like everyone was writing a post at the same time.
Somehow everyone forgot to mention the important point about degeneracy pressures.

From one of the very early posts:

There is space between the particles, held at distance by the nuclear strong force which is strong enough to resist the pressure due to the gravity.

   Which is not entirely true and also uses the term "pressure" in a very informal way (I think "compression due the gravity" might have been better.  You can see that calling it pressure instead of compression has started a bad trend because @evan_au has gone right down the same road).

(i)  To keep it simple, gravity produces a net inward force, pressure does not.  Pressure is a force in all directions and the gradient in pressure is what can give rise to an outward force that can help to fight against gravity and prevent further collapse.
(ii)   It isn't just the strong nuclear force that stops the collapse.   The degeneracy pressure is a significant contributor.   Just to be clear then for anyone else reading, the collapse of the star is not caused by pressure, quite the opposite, pressure is on our side fighting against the collapse.

   What is degeneracy pressure and why would it form a gradient (With P decreasing from the centre to the outer edge of the star)?
[LATE EDITING - surplus discussion removed]  ...Well no-one asked and it's not entirely relevant to the OP, so I won't bore everyone with that.

If the electrons could get to the protons, they'd likely be turned into neutrons as had occurred to most of the proton/electron matter, but the picture below has electrons quite deep where the protons are, so go figure.

     The Chandrasekhar limit describes the point where the density of electrons would exceed that permitted by the Pauli exclusion principle.  If the star has mass above the Chandrasekhar mass, then gravity is too strong and the pressure due to electron degeneracy is insufficient to stop it.  At that point, either the temperature has to increase, so that the degeneracy of the electron gas is lifted and higher energy states are available to the electrons  (which doesn't happen*)  OR ELSE  the usual thing happens.
   *I'm going to pause here and just mention that there really was a good way out of the problem - electrons could have just occupied ever-higher energy states (which is just saying they could have higher momenta).   However, there was already high enough temperatures and pressures so that electrons and protons can fuse and that is the escape route from the Pauli exclusion principle that will be taken.   Pop Sci articles often skip that and just imply that violating the Pauli exclusion principle was inevitable at this point.
   As the star continues to collapse (and since promoting electrons to even higher energy states isn't done) the Pauli exclusion principle is about to be violated and that just cannot happen,  that number density of electrons has to be brought down somehow.   What seems to happen is that protons in the region pick up some electrons and form neutrons.  This is where or why we get a Neutron star,  the electrons are being depleted and only Neutrons tend to persist.   However, there is no need to deplete ALL the electrons,  Pauli is quite happy with their being some density of electrons,  it just can't exceed a certain density.
   In a simple model we have:    At any place in the universe,
n_e(p) dp  ≤     
  where n_e(p) dp =  number density of electrons (number per unit volume) with momentum between p and p+dp.
   There's nothing really special about a Neutron star in this respect, it abides by the same limits imposed by the Pauli exclusion principle.   It's just that the density can be so high that the limit of n_e(p) dp  was reached.
   To say this another way,  an ordinary lump of iron on your desk at home can have a free electron density of about 10²⁸  electrons per cubic metre.   So a Neutron star can also have an electron density of 10²⁸ electrons per cubic metre anywhere in it, even right at the centre.   
    Actually the momentum distribution of the free electrons in the lump of iron on your desk followed a Maxwell-Boltzmann distribution (more or less) which meant that the temperature had to be quite high to support that density (room temperature say, otherwise at some value of momentum p, the quantity  n_e(p) dp  would have exceeded the limit).  Meanwhile the electron gas in a Neutron star is almost fully degenerate (which means all of the lower energy states are filled only) and so that has a very different probability distribution for momenta.  Being fully degenerate is the optimum way to accomodate all of the electrons while having the lowest total energy.  This means, a degenerate gas of electrons would support a total electron density of 10²⁸ at a much lower temperature  (an equivalent way of saying this is that an electron gas with that total number density of electrons would not be fully degenerate unless the temperature is close to 0 kelvin).  I'm mentioning this because a simple model would treat the electrons in a Neutron star as an electron gas which is assumed to be fully degenerate - and yet the temperature is many thosuands of kelvin.  So that the total number density of electrons is necessarily way above 10²⁸ electrons/m³ everywhere.

There is another limit: If the total mass of the neutron star exceeds about 2.5 times the mass of the Sun (in a ball only 10km across!), it is thought that even the Strong Nuclear Force will not be able to withstand the pressure, and it will collapse into a black hole.

   That comment from evan_au was just another example where it reads as if "pressure" is what was causing the inward collapse and not the thing opposing it.  Additionally, degeneracy pressure has to be mentioned again.  It's not just the strong force that is keeping the Neutrons apart.  Neutrons are also fermions - they behave much like electrons and must comply with the Pauli exclusion principle.  The degeneracy pressure of the Neutron soup is the last bastion that gravity has to overwhelm.   This one is a much more genuine last great barrier where degeneracy pressure and the Pauli exclusion principle really is everything because there is no alternative this time.  The electrons could combine with protons and be removed.  The neutrons don't seem to fuse with anything and be removed.

Best Wishes.

LATE EDITING:  Changed notation   n(e) dp   to    n_e(p) dp     which is more sensible to describe the number density of electrons at a given momentum p.

19

Physics, Astronomy & Cosmology / Re: Orbiting or descending into the black hole at the centre of the Milky Way?

« on: 28/11/2022 16:30:16 »

Mainly the MW is in orbit around itself.

And the mass of the Milky way is dominated by the (so far) invisible Dark Matter halo.

Though a great bit of that halo extends beyond the Galactic disk, and thus, due to the shell theorem, has no effect on the stars' orbits around the Sun.  Only that DM at a lesser distance from the center than the star is question would.
 So for the Sun, DM has an effect on its orbit, but doesn't dominate, as the DM closer to the center than we are, while a significant fraction of the total mass affecting our orbit, it isn't the majority of it.
If you look at the predicted and observed rotation curves here:
https://www.astronomy.ohio-state.edu/thompson.1847/1101/RotCurve2.gif
Note that close in to the center, you see little to no difference, as the mass of visible matter heavily dominates there. You don't see the deviation until you get out of the central bulge region and into the disk.

20

General Science / Re: Would a pipe made of synthetic diamond erode and be corroded with seawater?

« on: 24/11/2022 23:24:36 »

if the tetrahedral diamond carbon structure could coated with something like a graphene carbon surface,

The graphene would act as kindling.
https://en.wikipedia.org/wiki/Graphene#Chemical

graphene carbon surface, which has no loose bonds

Graphene is entirely covered with loose bonds and is rather reactive.

But at the surface are "loose" carbon bonds, which are much more reactive. Dissolved oxygen will react with these bonds, exposing yet more unpaired carbon bonds, and slowly eroding the diamond.

If that was significant, the same process would occur in the absence of water  and indeed, it does. Diamonds will burn in air.
But the reaction is so slow that you need to work on it.

Adding water will slow down the oxidation.

Diamonds in seawater are pretty much "forever".
The sun will boil the oceans before the diamonds dissolve (unless you make them impractically thin).