1

Physics, Astronomy & Cosmology / Re: Have you ever seen a FIREBALL?

« on: 16/03/2023 19:40:36 »

Hi.

Welcome to the site, I'm not sure I've spoken to you before.

Have you ever seen a FIREBALL?
   No, sorry.
....But your enthusiasm is apparent from your description and it was a pleasure to read.   Thank you for sharing it. 
(I'm not entirely sure it needed to be in this section rather than "Just Chat" or somewhere else - but that's a separate issue and I'm not a moderator). 

Best Wishes.

2

Just Chat! / Re: Generally speaking, how would Brits respond if...

« on: 14/03/2023 00:29:33 »

Hi.

I really don't know but overall the question could have been stopped after the first half...

How would most Brits react if King Charles summoned a press conference?

     Unless it was Christmas, they may not even switch on the TV set to listen.   Are you from the U.S. by any chance, you do seem to use some US spellings.   It's been a few years since the King really acted like a head of state for the United Kingdom.   Personally I'm reasonably happy with the Royal Family but there's a good number of Brits who are not and are described as "anti-Royalists".  Don't get me wrong, they aren't going to put the Royal Family on a Guillotine but they believe there's no need to support the Royal Family in the style that we currently do when tax-payers money can be spent on other things.

....to announce that he has begun taking GnRH modulators, estrogens, antiandrogens, and progestogens so he can become a woman and become Queen Charles?

    To be honest, the first thing to cross my mind would be that this was a bit of an over-inflated PR exercise.   I mean it would do no harm to have the Royal family seen to be involved in current issues being explored in society.   There would no doubt be books written and films made about their struggle with gender identity but they were never allowed to express themselves because of their duties and role in the Royal Family.   I can see the end scene - a real tearjerker.... they heroically stand up in front of the whole nation and say "I am what I am"   (...so you just know that the later stage versions will use the Gloria Gaynor song of that name but you're going to have wait for that).
   Would it cause the majority of U.K. citizens to recoil in horror?  I wouldn't have thought so.  Get a good ghostwriter for the book and some good actors and screenwriters for the film version and it will probably put Britain back on the maps again, an international "tour de force".

Best Wishes.

3

Physics, Astronomy & Cosmology / Re: Do you change the spectrum of radiation inside an oven if you change its shape?

« on: 10/03/2023 01:13:20 »

...and thanks @Zer0 ,  I must have been still writing when you posted.

4

Geology, Palaeontology & Archaeology / Re: Why do insects fly?

« on: 02/03/2023 01:45:56 »

Hi.

The information given in post #5 by @evan_au  seems to check out.
So that's the first part of your question sorted:    Flying insects did seem to evolve before flowering plants.

    The question about what the advantage of flying would have been,  must then relate to something other than eating nectar.   For example, when you fly you aren't so easily eaten by other animals that don't fly etc.
      Yes, there is typically a high energy requirement for flying but you can get it in ways that doesn't involve nectar eating.   For example, just eat more of the other stuff and metabolise the stuff inside the insect body into something that can be used up quickly to enable flying.  The lack of nectar is not a bar to flying (and it apparently wasn't).

   The entire  "why did that evolve?" question seems to come down to the usual thing:   There was a niche available and random mutations seem to be sufficient explanation for why life generally evolves to fill all available niches given time.

Best Wishes.

[Spoiler:]

5

That CAN'T be true! / Re: How can academia tolerate such error?

« on: 28/02/2023 02:59:16 »

Hi.

He then states that the term grad div E represents the "scalar wave"

   Grad  (of a thing) is always a vector not a scalar.   So on the face of it I would be concerned about it being called the "scalar wave"  - but I haven't looked at exactly what this professor did.

PS I apologise for alternating between the del operator and grad, div as I cannot get consistent use of symbols on this computer/operating system.

   Obviously what you have done is OK anyway.    However, if you ever really wanted some mathematical notation then it can be done on any computer with any keyboard (e.g. a mobile phone)  if you use  LaTeX coding.   I'll put all of this under a pull-down spoiler, since it's not essential for the original post or reply.

Spoiler: show

You have probably noticed already that this forum software places something you want to appear in BOLD  inside  tags.   One very useful set of tags is the start and end     tex    tags because everything between those will be processed as LaTex.

This code produces  some usefull stuff:

Code: [Select]

"Del" or "nabla"  symbols are called with a "backslash" \  and the text "nabla" written straight after it   like this   [tex]  \nabla    [/tex]
Most special LaTex sybols are called with the backslash \ in front but note that a forward slash / MUST be used in the end tex tag.

Vectors with the arrow over the top   [tex] \vec{A}   [/tex].      Vectors as underlined characters    [tex] \underline{A}  [/tex]

The times symbol [tex] \times [/tex]  looks neater than using a letter  [tex] X [/tex]  for multiplication or vector cross products.

Write the curl of a vector like this  [tex] \nabla \times \vec{A}  [/tex]    but I prefer underlined vectors [tex] \nabla \times \underline{A} [/tex]

Exponentials can be done with the ^ character. So the Laplacian operator is written as   [tex] \nabla^2  [/tex]   
Just type that directly into the forum reply box and you will get this produced when you hit  "preview" or "post":

                - - - - - - - - - -
"Del" or "nabla"  symbols are called with a "backslash" \  and the text "nabla" written straight after it   like this   
Most special LaTex sybols are called with the backslash \ in front but note that a forward slash / MUST be used in the end tex tag.

Vectors with the arrow over the top   .      Vectors as underlined characters   

The times symbol   looks neater than using a letter    for multiplication or vector cross products.

Write the curl of a vector like this      but I prefer underlined vectors

Exponentials can be done with the ^ character. So the Laplacian operator is written as   
 - - - - - - -  - -  -

   Back to your original post:

   The identity         will hold for any sensible  vector  E.
   See half-way down the page here:   https://en.wikipedia.org/wiki/Vector_calculus_identities#Second_derivative_identities

By sensible vector, I mean it's smooth enough - all the derivatives you would want actually do exist.   You could have deliberately chosen a vector that had some discontinuity in some first derivatives if you wanted to invalidate that identity.
Note also that the operator ∇² on the right hand side is understood to be the vector Laplacian  (it acts on a vector to produce another vector).   Assuming E is the electric field and satisfies Maxwell's equations then you automatically have most of what you'll need to be confident the vector function is smooth enough for that identity to hold.   I haven't spent too long looking at it but it seems you would have to be deliberately trying hard to find a most unusual E field if you wanted to defeat that vector identity - so we'll just go along with the assumption the vector identity will hold.

   So nothing much is wrong with what that professor said to this point.   The only bit, as I said previously, that concerns me is the terminology "scalar wave".   That's not terminology I'm familiar with and the thing isn't a scalar, it would be a vector.

   Just from Maxwell's equations,  we have Grad(E) = 0 in free space without any charges exactly as you stated,  so yes  all of what he/she is calling the "scalar wave" would become a 0 vector there.    I'm going to assume the professor knows that and their  "scalar wave" idea is only interesting in regions where there are charges.   Overall impression  --->  I wouldn't be rushing off to find out what the professor was trying to say (or sell) based on what you have written.   

    As regards the fuel thing.   "An ice" can have a meaning other than the usual solid water ice.   Here's an example:
     In astrophysics and planetary science the term "ices" refers to volatile chemical compounds with freezing points above about 100 K, such as water, ammonia, or methane, with freezing points of 273 K (0°C), 195 K (−78°C), and 91 K (−182°C), respectively (see Volatiles).           [Extract from Wikipedia:     https://en.wikipedia.org/wiki/Ice_giant ]
     Without reading the full details,  maybe their ice thing could work.

Best Wishes.

6

Physics, Astronomy & Cosmology / Re: How Do We Know The Universe is 13.8B yrs Old If We Can Only See The Observable ?

« on: 25/02/2023 22:59:35 »

Hi.

   You've got to like @Halc 's answer.   It keeps things simple and gets the major point across.

A more complex answer includes the following:

     Use another method and check for the same answers:    We can estimate the age of stars.    This is done by using spectroscopy to get information about what the star contains along with other information like the mass and luminosity of the star.  All that information effectively puts the star on only one place in our stellar models, so we can determine how long it has been a star.   The universe should be older than the oldest star.   Conversely, our models of stellar evolution suggest that, if all things were random (i.e. there's nothing special about where we are in space), then we should find stars older than 15 bn years within the range of our telescopes BUT WE DON'T.    While it is possible we've just been unlucky and not looked in the right places, it's more likely that the universe as a whole is just not that old.
    We usually add on a little bit more time for the period of the universe before Hydrogen clumped together and formed the first stars but overall the age of stars is in pretty good agreement with the age of the universe.
   This is another example where we have just used a model to estimate the age BUT this model is reasonably independent of the sort of model we would use with the Hubble constant as discussed by Halc.

     This article:    https://www.space.com/how-can-a-star-be-older-than-the-universe.html     discusses what I believe is still the oldest star we know of,  Methuselah   and how it was once thought to be 16 bn years.    Note that modern estimates and refinements have seen that it is possible to get the age of that star under 13.8 bn years (you still need to apply the full limit of the error bars).

Best Wishes.

7

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

« on: 28/12/2022 15:59:52 »

Hi.
   Now just some basic tests to see if everyone really has thought this through:

When monochromatic light passes through a single slit and produces a pattern on a screen.  What do you call that pattern on the screen?
   (a)   A diffraction pattern.    (Note that Feynmann suggested diffraction applies only for multiple sources interacting.  However, light does spread out when passing through a slit, I mean this single slit thing is the text-book example of a diffraction process isn't it?).
   (b)   An interference pattern.   (Can you have interference with one source of light?  Yes you can and you do @alancalverd  whenever that light passes through an aperture).

When monochromatic light passes through two slits and produces a pattern.... What do you call that?
   (a)  A diffraction pattern.
   (b)  An interference pattern.

When ...light  ... passes.... diffraction grating... screen.  What do you call that?
    (a)  A diffraction pattern.   (I mean, it had the name "diffraction" in "diffraction grating" didn't it?)
    (b)  An interference pattern.    (Feynmann suggestion is that you really should be calling this diffraction by now).

   I think common usage of the terms would have the answers   a (for 1 slit),  b (for 2)  then back to a (for diffraction grating).  @hamdani yusuf  asked about school pupils and what they are taught in schools.  I've just spent 20 minutes going through the AQA syllabus for physics at "A" level.  They do describe the patterns with these terms.

   Meanwhile the Feynmann suggestion should give   b, b, a.

    If I've got the gist of what @hamdani yusuf  wants the definitions of diffraction and interference to be.   (One is about bending light, the other includes producing bright and dark bands in a pattern)  then the answers   are  b, b, b.

    My view is... what the figgy pudding does it matter?

Best Wishes.

8

General Science / Re: How would you explain or define calculus in a sentence or two?

« on: 27/12/2022 22:56:24 »

Hi.

What is the real purpose for starting this thread?   Is it a social experiment to see how much time people will spend writing something?

   Anyway, I suppose we just get on and oblige.  You asked for a sentence or two.

Calculus is something quite simple and yet it has an inexplicable and unreasonable effectiveness.  It can be taught to school children and they can use it to solve a very large set of problems.

Best Wishes.

9

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

« on: 27/12/2022 18:24:54 »

Hi.

    Yes, OK, it's possible to try and keep diffraction and interference separate.
However,   is it necessary to try and do so    and    is it accidentally concealing the possibility that they are both due to the same underlying phenomena?

    Let's take an example and consider Huygens principle of secondary wavelets.   It's not the best ever explanation of how or why em radiation travels and exhibits certain phenomena,  for example you can always ask why the secondary wavelets only travel forward,  however it is one reasonable model that can be used.
    Let's just assume that something like Huygens principle is happening.   According to many references (e.g. see any reference provided below), the continued propagation of a plane wave or spherical wave can be explained just by using Huygens principle.   Specifically, once a wavefront has travelled away from it's original source (any tiny distance) then it is possible to forget about that source entirely.   The next position of the wavefront at a later time, t+δt, can be determined just by considering every point on the wavefront at time t to act as a new spherical source.   This is an important point so it's worth saying it again in a slightly different way...
    Nature does not need to know or be concerned about what the original source of the wave might have been.   It is not trying to keep the wave progressing or propagating away from that original source.  As far as Nature is concerned there is now a set of sources (one new source at every point on the wavefront) and the new position of the wavefront will be determined just by propagating each of those spherical waves away from those sources.
   A simple version of Huygens principle is provided by reference [2],  where it is stated that the new wavefront would be on the surface which is tangential to all the spherical wavelets.   Other references (e.g. [3] ) make the situation more easily expressed by referring to "the envelope" of the spherical wavelets but this is the same surface.  I think - but I'm not certain - that the original statement by Christiaan Huygens used the concept of an envelope but it probably wasn't in English and the translation is subject to some interpretation.
   Now, most versions of Huygens principle suggest that the secondary sources do behave like genuine sources of a new wave in most respects.   In particular there will be some superposition and interference.   It is very possible that the placement of the new wavefront on the envelope of the set of spherical wavelets is just a simplification of the underlying process that is actually happening.  Examples:
a)    From reference [1]:   ...every point on a wavefront is itself the source of spherical wavelets, and the secondary wavelets emanating from different points mutually interfere. The sum of these spherical wavelets forms a new wavefront.
b)    From reference [4]:   Wave propagation is linear so superposition holds: it should be possible to decompose an impulsive propagating wave front into its constituent points, then consider the impulsive wavelets radiating from each of those points at a future time, and combine those wavelets in a simple direct geometric manner to obtain the progressing wave front at that future time.

    I have included reference [4] just because it's one of the few articles I've seen where a very direct attempt is made to remove the notion of placing the new wave front "on the envelope" and there is an attempt to show that the new wavefront could just be a result of something like conventional superposition and interference.   There is a different type of wave generated by Huygens secondary sources (a Dirac delta function) but with that assumption you no longer need to exclude a backward propagation, there is just an automatic cancellation when a sum of all the waves is computed.  In a similar way, there is no effect often described as a "wake" left behind.   What remains from the superposition is precisely the same as putting the new wavefront right on the envelope (i.e. a lot of the problems and apparently arbitary rules for Huygens principle have been resolved).
   Anyway the key issue is that what is seen as a conventional straight line or spherical propagation of a wave could actually be the result of interference from many secondary wave sources on the wavefront.

Let's take this point mentioned in one of the much earlier posts:

I don't think that diffraction is a subset of interference. We can produce diffraction without interference pattern.

   
   It is quite possible that you are seeing the results of interference.  The propagation of those waves just doesn't look like a typical interference pattern but wherever there is straight line or spherical propagation of a wave then this has been the result of interference from Huygens secondary spherical wavelets.   More-over the "diffraction" effect from the knife edge on the mountain is entirely due to the phenomena of Huygens secondary wavelets.   The radio waves are entering the geometric shadow only because there were no wavelets below them that were able to provide the interference necessary to keep the radio wave progressing in a straight line from the original transmission aerial.

Summary:
   This is NOT an attempt to state that the propagation of an e-m wave has to be this way.  I have provided reference [4] because it's an article in a respected publication ("Nature", see * LATE EDITING) and it simply suggests that Huygens principle and ideas along the lines of conventional superposition and interference could play an important role in all wave propagation phenomena.
   As such what we call "diffraction" could very well be just an "interference" effect.  When light seems to bend around an obstacle, that could just be due to the lack of the interference required to keep it going in a straight line from the original source.  The opening sentence of this post can be re-stated:   Is it necessary to keep the terms "diffraction" and "interference" entirely separate    and    is it accidentally concealing the possibility that they are both due to the same underlying phenomena?


Best Wishes.

References:
    1.    Wikipedia:    https://en.wikipedia.org/wiki/Huygens%E2%80%93Fresnel_principle
    2.   Byjus:    https://byjus.com/physics/the-huygens-principle-and-the-principle-of-a-wave-front/
    3.   University of New Mexico, physics course:   https://physics.unm.edu/Courses/Fields/Phys2310/Lectures/lecture16.pdf
    4.   Nature*:   https://www.nature.com/articles/s41598-021-99049-7

* LATE EDITING:   On closer inspection, reference [4] is properly an article from "Scientific Reports" which is a sub-journal affiliated to "Nature".   It still has a peer review process but the readership and general academic reference value is not the same as the major journal "Nature".

10

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.

11

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.

12

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

« on: 20/12/2022 21:05:21 »

Hi.

Just thought I'd mention a couple of minor points, relating to the original questions in the first post.

1.   You can Google to get estimates of the number of stars in the Milky Way.  Estimates are between 1~4  x10¹¹ stars     (checked several references for consistency:   https://en.wikipedia.org/wiki/Milky_Way   ;
    https://earthhow.com/milky-way-galaxy/          and     https://asd.gsfc.nasa.gov/blueshift/index.php/2015/07/22/how-many-stars-in-the-milky-way/  ).
   The NASA reference describes in more detail precisely where some of the problems are and why this is JUST AN ESTIMATE and provides another 3 references with more discussion of the problem.
    Note that we do sometimes "un-discover" a star.  For example, a black hole can distort space and create a lensing effect.  What can happen is that you see two points of light on each side of the black hole which you may have thought were two stars in different places but they were actually light from just the one star that has been bent around both sides of the black hole.    Meanwhile, we also frequently discover two stars where we thought there was only one.  Two close stars forming a binary system can look like just one point of light until you get a better resolution image of the region of space.
   The key point then is that this is a very rough estimate, don't be surprised if it's an order of ten bigger or smaller.   If we do revise our estimate then your book can start to look quite dated.

2.  "Neutronium" seems to be a term used mainly in Science Fiction and some Pop Sci articles.  It's not really used in more formal scientific literature.  It can mean the material a Neutron star is made of,  it could mean something else.  Historically it was a hypothetical element on the periodic table with 0 protons in the nucleus.

...I take it from what you (Halc) are saying that the strong force is repulsive at very short ranges?

   It can be desirable to try and keep two terms slightly separate.  The strong interaction or "strong force" is a fundamentally quantum mechanical interaction between quarks.   The nuclear force or "strong nuclear force" is a residual effect from that,  where a classical Newtonian force is identified as something that exists between any two nucleons  (which, as I'm sure you know, is a collection of quarks).   All too frequently the terms are used differently and often interchangeably, which is just a shame.  Just go about reading an article with the idea that they are different, one acts only on quarks, the other acts on whole nucleons.   If you were trying to consider all things in a very Newtonian way (which isn't always a great idea) then the Strong force always acts so as to keep some quarks bound together.   However, the residual strong nuclear force is different.   There can be an attractive force between nucleons at large distance and a repulsive force at small distances.
   
   See this description and diagram in Wikipedia for a discussion of the nuclear force between nucleons.   (Especially the sub-section titled "The nuclear force as a residual of the strong force")

   image taken from  https://en.wikipedia.org/wiki/Nuclear_force

  That "force" is, of course, just treating the nucleons as particles and the interaction between them in a very mechanical or classical Newtonian way.  A fully quantum theory just has some other mathematics describing the more fundamental strong interaction and there is no need to assign a mechanical force or assume the nucleons were ordinary solid particles on which a classical Newtonian force can act.

   I hope that helps a bit.

Best Wishes.

(Another post has just been added by @Halc before I completed this.   I'll read that next and hope this isn't repeating anything).

13

Physics, Astronomy & Cosmology / Re: Is there a substance where entropy decreases at higher temperature?

« on: 16/12/2022 05:51:33 »

Hi.

    I've had a bit of time to think.   As you ( @chiralSPO ) implied:  It's very much a case of needing to keep our definitions of temperature straight (and it does become a little dull then).

    The situation with a simple two-energy level system does illustrate a situation where temperature can increase (in this case it is increasing by becoming less negative) while the entropy decreases.   As you ( @chiralSPO ) implied it's clear that energy is always being added to the system as more particles are driven to the higher energy state,  however that wouldn't guarantee that the temperature was increasing.   
    For a typical substance like an ideal gas held at constant volume, it's apparent that increasing the internal energy will increase the temperature.   For a more arbitrary substance and where statistical temperature is used, you really do have to directly check the gradient  ∂S/∂E.   A diagram will save 1000 words here:

Suppose the entropy, S, of a system of N particles (each particle having only two energy levels) varied with E, the total energy of the system, as follows:


Untitled.jpg (49.3 kB . 1152x648 - viewed 1228 times)
   An increase in the the total energy of the system would not produce an increase in temperature once you've entered the region with most of the particles in the higher energy state.  The temperature is negative in that region but perfectly constant.

  Fortunately the entropy of the simple 2-energy level system doesn't look like that,  it has this sort of shape:


entropy2.png (24.19 kB . 882x680 - viewed 1270 times)
   Where the gradient =  1/T   is  monotonically decreasing with E.

  With that established:   The simple 2-energy level system does have the usual relationship where increasing its energy remains synonymous with increasing its temperature.   The note of caution in your ( @chiralSPO ) first sentence is then substantially laid to rest:

I think you need to be careful about specifying whether you mean temperature or thermal energy.

   More generally, the simple 2-energy level system ticks all the boxes that I was asking for in the original post.  So, well done and thank you.   You can have the best answer award.
-------
     I don't think I need to ask anything else.   If you restrict attention to just non-negative temperatures, then I'm now fairly sure that the answer rests entirely upon finding a substance with specific heat capacity, C_V < 0.
     While I don't know of any substance with that property, in any situation (e.g. in any range of temperature, pressure or other conditions),  if someone else does -  Please let me know.

Best Wishes.

14

Physics, Astronomy & Cosmology / Re: Is there a substance where entropy decreases at higher temperature?

« on: 15/12/2022 02:39:37 »

Hi and thanks.

   Lasers and population inversion was exactly what I had in the back of my mind.

This sentence

One reasonable definition of energy is: T = ∂U/∂S (constant volume).

   might be a typing error.   That's a perfectly good thermodynamic definition of temperature.

Obviously I'm going to have to take a bit longer to consider everything.  Thank you very much for your time and it's exactly the sort of thing I was looking for.

Best Wishes.

15

Physics, Astronomy & Cosmology / Re: Does time have more than one direction?

« on: 12/12/2022 04:51:25 »

Hi.

Some of the formating of quotes has gone wrong in your last post. So I'll just use something sensible here:

ES said:   How long did you take to come up with that?
Halc replied:  Quite some time ago, when first asked how to point in the 4th direction.
   The original comment was about the time axis and was sufficiently inspired it's worth repeating here:

The time dimension is like that. You pick two events (points in spacetime). You say 'this' event and clack two rocks together to define the event.  Then you wander off some arbitrary place and clack the same two rocks a 2nd time to define a second event. The one unique line through spacetime connecting those two events is now defined. You've chosen a totally arbitrary orientation for your time dimension since the two events you chose are completely arbitrary.

   
   Now, I did mention in a post some time that it was "more or less" correct.   I'm just going to discuss the "less" part here because the "more" part speaks for itself, it's an amazing way of describing how you could choose a time axis.  So, there's genuinely no offence intended.  You've obviously spent some serious time replying to @Dimensional and might just want a little feedback or something to think about.

   Let's start by considering a person, we'll call her Roxanne ("Roxy") because she's the one who will be doing the rock clacking.   Now Roxy starts by clacking some rocks at  x=0, t=0 in whatever co-ordinate system she was originally using.   Supposedly, Roxy can clack more rocks at any arbitrarily chosen event in spacetime.  Let's say she clacks the rocks at the event   x=10, t=0   in her original co-ordinate system.   So that's just two events that were at the same co-ordinate time and just lay along the x-axis of her original co-ordinate system.....   do you see where I'm going with this?
    The spacetime interval between those two events =  +100   (in a suitable convention with  ΔS² =  Δx² - c²Δt²),  in particular they are most definitely spacelike separated.  Moreover, the spacetime interval is a conserved quantity under any Poincare transformation we apply.  So whatever new co-ordinates t'  and x' we try to generate,  we will still have    ΔS² = +100 = (Δx')² - c²(Δt')² .   It's the being positive thing which is the real sting in the tail.  The two events remain spacelike separated.   In particular there is no way those two events can lie along the new t'  time axis.
   More generally, the two events with the rock clacking can't be completely arbitrarily chosen,  the first one is arbitrary.  The second one must be timelike separated.
   **Also,  Roxy is only metaphorically allowed to "wander off" to the other rock clacking event.  Since that event must be timelike separated from the other one, she has to run faster than anything else in her world.  That's not too bad though, there doesn't need to be a Roxy and even if there was she was only metaphorically wandering off to the other event.**

Best Wishes.   

**LATE EDITING:  I've already spotted that the last paragraph was backwards,  Roxy can't get to spacelike separated events but timelike is no problem.   I've left the error in because we're all human.  **

16

Physics, Astronomy & Cosmology / Re: Is there an experiment that shows the oscillation in the E field of light?

« on: 07/12/2022 19:43:49 »

Hi, thanks and well spotted @paul cotter

   It was a typing error,   
       Where we have used the relationship c = 1/(μ₀ε₀).
 Should have been
       Where we have used the relationship c² = 1/(μ₀ε₀).

Fortunately that was what was actually done.  The original post has now been edited.

Best Wishes.

17

Physics, Astronomy & Cosmology / Re: Is there an experiment that shows the oscillation in the E field of light?

« on: 06/12/2022 06:04:48 »

Hi.

Really, I don't see that beat frequencies in a linear medium can excite events at the beat frequency...

     Rather than worrying about the medium that the frequencies are travelling through or were generated in behaving in a non-linear way, the object which the beat is incident upon can't always respond linearly, in particluar it can't always "keep up" with the speed of oscillations.  An example is an object in the region that is behaving as a damped oscillator while some pair of frequencies that resulted in a beat (an amplitude modulated signal) are incident upon it.

I expect you've seen something like this already but I'll put some rough mathematics down here anyway (I don't know who might be reading this and I need a distraction from housework).

     The object  (the damped oscillator) has this equation of motion:

[Eqn 1]

with b, k constants (b for the damping and k for the spring constant), x= displacement from equilibrium position    and F(t) = driving force which varies with time.   There's an m (a mass term) multiplying the d²x/ dt² usually but I've just divided by m and so my constants k,b and Force F are "per unit mass" if you want to be fussy.

  The oscillator is driven by the vibrations in ...whatever...  the vibrations in the air for your guitar string examples.   So we will take 
      F(t)  =  A(t) .  Cos Ωt 
     With   A(t) =   Cos (ωt+Φ)     (with Φ = some constant)
as usual.     
A(t) describes the oscillating amplitude envelope for the beat incident on our damped oscillator, while the Cos Ωt describes the carrier waves within it.    For a beat,  Ω = π (f₁ + f₂)    and   ω = π(f₁ - f₂),   with f₁ and f₂ being the original two frequencies that generated the beat (where we assume they had equal amplitude, or at least equal where they became incident on our object).   All that matters is that Ω >> ω.  (Minor note: You can set Φ which appeared in the Amplitude wave form to 0 if you like, it doesn't make any difference to what follows).

When Ω >> ω we can solve the equation of motion over a small interval of time  t ∈ (t₀, t₁) such that we can assume  A(t)  ≈  A₀ = constant  while the Cos (Ωt) term shows much more variation.      Over that small interval the equation of motion then becomes:
      A₀. Cos Ωt

Which is the equation of motion for a damped oscillator with a sinusoidal driving force at an angular frequency Ω (which is assumed to be far above the natural frequency of this oscillator).  We have a standard solution for this.  The damped oscillator hardly responds to that driving term and we have that x(t) is almost constant throughout that interval,  x(t) ≈ x(t₀).    For convenience we can take  t₁ - t₀   =  2π/Ω  (then Ωt moves through 2π over our interval but with ω << Ω  we will have that ωt  hardly changes giving us A ≈ A₀ throughout the interval as we wanted).

    If we now consider [Eqn 1] at several  discrete values of time  t₀,  t₁, t₂, .., t_n, ....
where the gap between t_n   and t_n+1   is always  2π/Ω    then we obtain   F(t_n) =  Cos (ωt_n + Φ) . Cos (Ωt₀ + n.2π)   =   c .  Cos(ωt_n + Φ)    with c = constant = Cos (Ωt₀).
We can then imagine a new damped oscillator with displacement y(t) and this equation of motion:
        c .Cos (ω t + Φ)   

[Eqn 2]

Confine your attention to just numerical solutions of this O.D.E.   Since the RHS of [eqn 2] will be precisely the RHS of [eqn 1] at every value t= t_n, we know that numerical solutions of  [Eqn 2], y(t_n) =  f(t_n)  for some function f, will be precisely the same as numerical solutions of [Eqn 1],  x(t_n) = f(t_n)  whenever we take uniform intervals Δt = t_n - t_n-1= 2π/Ω and start the numerical technique at the same place t=t₀.   However, [Eqn 2] is just the equation for a damped oscillator being driven at the angular frequency ω << Ω.   We have an analytical solution for this.   The numerical solution we generate for [Eqn 1] is then well approximated by the anlytical solution y=y(t) for [Eqn 2] evaluated at the points t= t_n.   Finally we make use the earlier paragraph (about x(t) being reasonably constant over the small intervals because the damped oscillator hardly responds to a driving frequency of Ω) and conclude:  The exact solutions of [Eqn 1]  take the form    x(t) =  y(t) + δ(t)    where  y(t) is the exact solution of [Eqn 2]  and  δ(t) is a function that is some small correction,  i.e.  |δ(t)| ≈ 0.

   Anyway, the general idea or the "bottom line" of the argument, is that the damped oscillator will respond as if it is seeing a driving force of the form Cos (ωt)  (as seen in [Eqn 2] ) where ω is the angular frequency of the amplitude modulation.

Best Wishes.

18

Physics, Astronomy & Cosmology / Re: Is there an experiment that shows the oscillation in the E field of light?

« on: 05/12/2022 17:53:05 »

Hi.

...I don't want to derail ES's question....

    That's fine, discuss anything that's even vaguely relevant,  it won't trouble me.   I'm still busy doing housework of one type or another, so I'll apologise in advance for not replying but I'll try and keep up where and when I can. 

Best Wishes.

19

New Theories / Re: How Many Numbers Exist?

« on: 02/12/2022 04:44:58 »

Hi.

You can't jump to use root operator for the first iteration of number classification because it requires exponentiation by rational number.

  Not necessarily.
   For example,  the Taylor series for √(1+x)  is just a series involving only exponentiation by natural number powers (which is just repeated multiplication if you didn't even want to use that exponentiation).   Admittedly, I would like some use of division.   Anyway set x = 1 and then you have a series  which can be the motivation for them wanting to build a bigger algebraic structure that includes the limit of that series.    (Just to be clear, they don't need to know anything about Taylor series - they can just observe that there is a series which seems to be leading somewhere and be motivated to build a bigger algebraic structure that would include the limit of such a series).
   
Best Wishes.

20

Physiology & Medicine / Re: Is fever supposed to hinder the pathogenes or cause our cells to go on strike?

« on: 24/11/2022 23:11:49 »

Hi.

    Presumably, you know already that the reasons for raising body temperature are complicated and the consequences are numerous.   We're still not entirely sure about all of it.   Not all of it has an explanation in the sense of being helpful - some of it seems to be just a side-effect of fighting infection and doesn't offer any direct advantage.

    The podcast was a very simplified explanation of just one thing that raising the body temperature can do.   Your comments about reducing the replication rates that can be achieved by viruses is another.   (Just for clarity,  viruses and bacteria are not the same thing, they attack by slightly different methods and hence slightly different things are helpful against each of those attacks.   I seem to recall the podcast mainly considered a bacterial infection).

    I might leave it for some others who are more like experts in Biology to say some more.   You can find many explanations from Google (stick to the more credible sources of information).   You can get a slightly different set of consequences and effects from each website and they are ALL at least partially right.   As stated in the beginning of this post, there are numerous consequences of raising the body temperature and many mechanisms by which we think that will help to fight infection.

     If you were trying to simplify the situation and pull as many of the mechanisms together as possible then it's considered reasonable to say that raising the body temperature tilts the playing field in your bodies favour.   You aren't offering a like-for-like penalty between yourself and the bacteria or virus.   Yes, some of your physiological functions are impaired as much as any bacteria or virus* but the key activities you need like your immune system, that usually gets stimulated and works at least as well at higher temperatures.

* physiological machinery that a virus has hijacked, it has little of it's own.

Best Wishes.