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Physics, Astronomy & Cosmology / Re: Why are some isotopes fissionable?

« on: 13/11/2023 00:53:14 »

Hi.

I have another question for you: u235 on exposure to a thermal neutron what ratio of u236/ u236* would one expect and are there conditions that affect this ratio?

    I'm not really an expert on Nuclear reactions.  I've just been reading some lecture notes and a textbook on it recently  (they weren't my lecture notes, one of the family just kindly made them available to me).

   U236* should be so short lived, you won't find any significant amount of it.  As fast as you may produce it, it will decay and usually by fission.   So you'll just find the later decay products.   You ( @paul cotter ) mentioned it should have a half life of 10^-12 seconds.   I previously mentioned it may not even be a real particle that exists, it's a useful model and/or may be something like U236 with a deformed nucleus  etc.   There should be very little you can do to increase the proportion of U236*,  the main problem would be keeping it alive (undecayed).

   U236*  is what you shoud be producing  (as the first step) from U235 and a thermal neutron.   You shouldn't produce U236 (the ordinary ground state).   However, there probably are some routes where the excited state  U236* can lose its surplus energy (say by emitting a gamma ray) and become U236 (the ground state).   There are probably also some chains (of neutron absorption and/or decay events) that would also lead to the production of U235.   For example, we can already use neutron bombardment to change one element into another, it's a bit random and the yield isn't great - you produce all sorts of stuff and only a small portion was what you wanted - but it happens.

   Nothing goes absolutely as it should in the textbooks inside a nuclear reactor.   Sometimes a contaminant will be present, that might absorb some neutrons and form something unexpected.   There are various things that could happen,  I'll avoid presenting examples but I'm sure you can imagine the situation.  We know that you end up with traces of all sorts of stuff being left in the reaction vessel and spent fuel rods when reactors are finally decommissioned.

    More significantly, you never really start with just 100% U235.    Natural Uranium is a mixture of isotopes and it's difficult to enrich the proportion of the (desirable for reactors) U235.  Getting to 5% U235 is very good and is considered a high quality fuel.   So, although on paper we can write down what should happen to U235 with a thermal neutron,  in practice you never really had pure U235 to start with, there was quite a lot of other Uranium isotopes in it already.

Best Wishes.

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Physics, Astronomy & Cosmology / Re: Why are some isotopes fissionable?

« on: 12/11/2023 02:07:04 »

Hi.

I have a quick question since you seem to be up to speed on these matters. u235 when it captures a neutron becomes u236 with a subsequent existence of ~10 to the power of minus twelve seconds before scission occurs OR such is what I had previously believed. In my copy of the Merck index an isotope of uranium, namely u236 is listed with a t1/2 of 2.342 x10 to the power of 7 years! This apparent contradiction confuses me.

    I haven't read the original descriptions you saw.   However, it seems that when  "they" say  U235 captures a neutron and becomes U236 with a half-life of 10^-12 s    they were really taking about it becoming the excited state  ²³⁶U*   and how short lived that would be.      It was just slightly sloppy notation on their part,   the ²³⁵U didn't really become ²³⁶U (which is the ground state of U236) by capturing a neutron.   In all fairness, our understanding and choice of models has changed over the past few years and the notation ²³⁶U* may not have been in common use at the time that text was written.

      The usual (ground state) ²³⁶U  is reported as having a half-life of 23.42 million years exactly as you have stated.   The excited state  ²³⁶U*  is a completely different kettle of fish - that will be incredibly short lived.

     Just for your information (and to the best of my knowledge) the existence of a particle like ²³⁶U*  isn't even a certainty or required.   It is just a convenient model, it's helpful to imagine a particle like that existed for a short while after ²³⁵U absorbed a neutron.   The incredibly short half life they are reporting is probably based on theoretical prediction rather than actual measurement and observation.   I mean, you have 10^-12 seconds in which you must   (i)  identify that you really are looking at a particle like ²³⁶U  - which I suppose means confirming that 92 protons and 144 neutrons can be recovered by dismantling the nucleus  and then also  (ii) confirming that it isn't just your regular ground state ²³⁶U you were looking at - which I suppose you would do by confirming that it has mass (rest energy) which is ever so slightly greater than regular ²³⁶U.    This is a lot of precision and not a lot of time to do it in.   
    You don't really need the existence of ²³⁶U*  to make sense of how or why fission may occur following neutron absorption by ²³⁵U.   All you need to know is that you had rest mass  (energy)  equal to that of
     [ the mass of regular ²³⁵U ]   +    [ the mass of a separate neutron ]   
to start with,  so you must have that much energy after the absorption.   This is slightly more than the rest mass of  (ground state)  ²³⁶U  because we know that some binding energy would have been released when the neutron was incorporated into that nucleus.   This surplus energy is what you have available to overcome the activation barrier of the nucleus which would be much like ²³⁶U.    The ineterim stages are less important and may or may not involve there ever having been a ²³⁶U* particle.   

    The latest models that I'm aware of are still based on what was called "the liquid drop model" of a nucleus - which does model the thing as a drop of incompressible fluid.   The process of fission caused by an incoming neutron is thought to be explained by a deformation in the shape of the nucleus as the neutron comes in.   This deformation can either increase or decrease the binding energy of the nucleus at that moment.   There are two competing terms in the binding enery formula for a nucleus,  "the surface term" and the "the Coulomb term".   When you deform a nucleus (e.g. make it a bit prolate instead of spherical) then you increase the average distance between charged particles (the protons), this can increase the binding energy or equivalently lower the mass you would observe when you go and measure the mass of that deformed nucleus.   Meanwhile, "the surface term" represents the reduction in benefit of being surrounded by other nucleons when you are only at the surface of the nucleus instead of in the middle of it.   This "surface term" comes from considering the nuclear force which is effectively only a nearest neighbour force (it's so short range that you only feel the effect from nearest neighbours and nucleons on the surface have less nearest neighbours then those in the middle of the nucleus).   When the spherical nucleus is deformed in any way then the amount of surface it has inceases.   So this increase in the surface reduces the benefit from the attractive nuclear force.  Since that was an attractive force, it was increasing the binding energy.  So reducing that will lead to the mass (energy) of the deformed nucleus being more than it would be for the spherical un-deformed nucleus.   Hopefully you see what I mean about there being competing terms.   Which one wins out can depend on exactly how the nucleus is deformed and what shape it has.
      So to reconcile the model where we consider ²³⁶U* as being something that really exists with the deformation and liquid drop model we would be saying that   ²³⁶U*   is what you have when the nucleus is  "this much deformed" and in "this shape",   it has a rest mass which is slightly more than the rest mass of an undeformed  ²³⁶U nucleus.     However, the main point is that the deformation in the liquid drop model considers the process as being something that is continuous,  as the nucleus deforms the rest mass changes smoothly and continuously,  all values of the rest mass between that for ²³⁵U  and that for  ²³⁶U* are obtained   (and quite possibly some interim shapes that had rest mass below or above that).  Meanwhile the kinetic energy of the incoming neutron was being changed (usually reduced) as it starts impacting on the surface of the nucleus,  so that there was (in general) a smooth and continuous exchange of energy from k.e. in the neutron to the rest energy of the deforming nucleus.  A bit later the neutron is mostly inside the nucleus and it's a bit of grey area how much of its mass is now the mass of the nucleus.  You just don't have a clearly identifiable separate neutron and a separate nuclues on which it is impacting any longer.
     By comparison, the model involving the existance of ²³⁶U*  just breaks it down into a series of discrete steps,  you have    this, then this, then this  :   you start with ²³⁵U  plus a separate neutron,   then you have   ²³⁶U*,   which is like ²³⁶U with enough energy to overcome the activation barrier and so then you will quickly have the two separate daughter nucleii in existance.
      Anyway, I hope you can see what I mean about the existence of a particle like ²³⁶U* being just a model we can use.   Whether the real process is a smooth and continuous one like deformation in the liquid drop model  OR  a series of discrete steps remains uncertain.  Indeed it doesn't even have to be one or the other, it could be both.   Quantum mechanics can model the situation as a wave function where the discrete states or eigenfunctions are the discrete stages suggested by the discrete model.   The wavefunction is just a superposition of those eigenfunctions where the coefficients of the eigenstates evolve over time.   What this means is that at time t = 0 units = initially the wavefunction is 100% that of there being a separate neutron and a separate undeformed nucleus,  as time increases the blend of eigenstates starts to shift so that at time t=1 there is  a 50% chance of finding the thing in the initial state if you observed it and a 50% chance of findining it in the ²³⁶U* state.   As time evolves the proportion of the eigenstate for ²³⁶U* increases and  the amount of the initial state decreases   etc. etc.     In this way you still have only the existance of discrete states when you go and observe the system   BUT the transition from initial state to final state does progress in a smooth and continuous way,  so that the average or expectation value for the rest mass of the nucleus would be a continuous function of time.

@Petrochemicals , I think your question will be shorter to answer.  I'll be writing that now, after a cup of tea.

Best Wishes.

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Physics, Astronomy & Cosmology / Re: Why are some isotopes fissionable?

« on: 11/11/2023 17:20:23 »

Hi.

    That's a good answer from @paul cotter.   I can add a bit more detail and did start writing last night but it got long and the computer crashed.

   To try and keep it short, I'll just take the comments you made in the OP and reply to each.

Why for example is U235 fissionable yet the more unbalanced u238 not?

   Good question.  There is an answer but it's quite long, see later.

U238 has a greater number of neutrons, so I would assume it would be quicker to decay and be more unstable?

    "Stability"  can be considered in two main ways.

1.   Energetic stability:    When some nucleons (protons and neutrons) come together to form a dense nucleus there is some binding energy released.  A nucleus is more stable when the binding energy is higher.   
   Let's say this another way - some protons and neutrons have a particular mass when they are well separated from each other.   When they come very close together (nuclear separation distances), the strong nuclear force really kicks in and takes over, the potential of each nucleon drops as they approach, so that energy is released.  As such the mass of a nucleus is less than the mass of the individual component nucleons  (by Einsteins famous E = mc^2).
   In general, any two nucleons that approach each other to nuclear separation distances will result in some binding energy being released.   So we expect the toal binding energy of a nucleus to be proportional to the number of nucleons in that nucleus.  As such we usually consider the binding energy PER NUCLEON.   If a large nucleus (say U-235) splits into two smaller nucleii then the binding energy per nucleon increases slightly and as a result the total energy (or mass) that remains in the daughter products is less than it was initially.   
   A nucleus can be considered to be more energetically stable when the binding energy per nucleon is increased.  As you may know, the highest binding energy per nucleon occurrs when a nucleus has around 60 nucleons in total (with about half of those protons and half of them neutrons - so an element like Iron on the periodic table).    U-235 of course has 235 total nucleons, so that splitting in half makes energetic sense.   Indeed it would be reasonable to split in half and then split in half again.

2.   Activation Energy:   A nucleus can't just reach a new final state from an initial state.   U-235 doesn't just spontaneously split because there is some activation energy required, to paraphrase this the nucleus must get over a small hill before it can roll down the big slope on the other side to an energetically stable final state.   This is quite different from another type of radioactive decay (alpha decay). 
   We can consider a simple model, where the two daughter nucleii (well, packages of nucleons that will go on to become the individual daughter nucleii once they separate from each other) pre-form inside a region of space that we are still considering as the nucleus of the original parent.   The strong nuclear force more or less switches off entirely once these two packages of nucleons separate from each other enough that we can consider them to be just touching rather than overlapping.  When this happens the Coulomb force (the electrostatic force between the positively charged protons in each package of nucleons) becomes the important dominating force.   There are textbooks and other sources of information that explain this in more detail.  For now we'll just accept that the Coulomb potential at a critical distance = r₁ + r₂   where  r₁ and r₂ are the radii of these daughter packages (the daughter nucleii that will form), is representive of the height of a potential barrier that would confine these two daughter packages to the parent nucleus region.   While the daughter packages are confined, it all stays together as one parent nucleus instead of splitting into the two daughter nucleii.
    Classical physics suggests that the only way to get over that barrier is to have the daughter nucleon packages with enough energy to get over that barrier.   Quantum mechanics presents another option, the daughter nucleon packages could "tunnel" through the barrier even if they didn't have sufficient energy to legitimately get to the top of that barrier.    Tunnelling is extremely dependant on the width of that barrier, which in turn will be dependant on the height of that barrier at the critical separation distance r₁ + r₂  since outside that parent nucleus region that potential is falling off  like  1/r   (the usual Coulomb potential).   For an alpha particle (which is just 2 protons and 2 neutrons, so  "not much"),  the other package of nucleons  is   Z - 4   nucleons   and for large  Z (mass number or nucleon number,  like we do have in Uranium-235  where  Z= 235)   we have that    Z-4  is approximately Z anyway.    The height of the Coulomb potential  is  q₁.q₂ /  (4πε r)    where r = r₁ + r₂ = the critical separation distance   and  q₁ , q₂  are the charges of the two daughter nucleon packages.  The details aren't too important,  we can simplify and assume r = r₁ + r₂ = the critical separation between the centres of charge is much the same for the situation where we have two daughter packages of about  235/2 ≈ 118 nucleons each,   or when we have  one big daughter package with 231 nucleons and a little alpha particle of 4 nucleons.    So that the height of the potential barrier is then only dependant on the product q₁ . q₂  that appears.   

     For the situation where an alpha particle and a big daughter nucleon package form,  we have  q₁ . q₂  =  2 . P     where 2 is the charge on the alpha particle and P is the charge on the big daughter nucleii left.   (So  P ≈ 90  for Uranium and it's large daughter nucleus left after alpha decay).    For simplicity we have  barrier height ~ P.    Tunneling of the alpha particle out of the parent nucleus does sometimes happen.

      For the situation where the Uranium 235 splits into two nucleii of about the same size,  we have that the height of the potential barrier  =  (P/2) . (P/2)     ~   P².        For large nucleii like Uranimum 235 with P ≈ 90,   P²   is a shocking lot more than  P.   The barrier width is MUCH larger and  tunneling probability falls of exponentially with this barrier width rather than just linearly, so this probaility takes another big hit.    As a result the barrier is just too wide and tunneling almost never happens.

   What this means is that, although U-235 will sometimes spontaneously split into two roughly equal sized daughter nucleii (after millions of years - exactly as quantum tunneling would predict),   the more prevalent decay is just to have an alpha particle emission.

   Anyway, let's get back to the main point....   Quantum tunnelling is not going to allow fission into two roughly equal sized daughter nucleii.   The rule of classical physics does almost hold - the nucleon packages must have enough energy to legitimately get over that barrier,  or at least get sufficiently close to the top that the remaining barrier width is now so small that tunneling is again realistic.   This is the origin of the "activation energy" that is required for fission to proceed.  The daughter nucleon packages must have enough energy to get over the potential barrier that confines them to the parent nucleus.

- - - - - - - - - - -
   OK, so that's the preliminaries all done.

   Now U-235 and U-238 are almost the same things as far as nuclear physics is concerned, they differ by just a few neutrons.   Both can decay by nuclear fission, both can produce two daughter nucleii of about half the original size of the parent.   The total change in binding energy when this fission happens is about 200 MeV for both  U-235  and  U-238.   They are almost the same when you consider the relative energetic stability of the parent nucleus compared to the daughter nucleii.   So, energetic stability isn't the issue here.

   That means that the difference in stability or fission-ability (if fission-ability is a word) must be due to activation energy.  This is the energy required to lift the daughter nucleon packages over the barrier that confines them to the parent nucleus.

   Now it would be nice if we could say that U-238 just has a higher activation energy than U-235.  That would end our discussion because we could argue that the activation energy for U-238 is too big and that's why U-238 is more stable.   However, it's NOT quite that simple....

   As previously mentioned, U-238 and U-235 are almost the same things as far as nuclear physics is concerned.   They both have similar activation energies.   Typically (say in a nuclear reactor) we provide that activation energy by supplying a neutron to the nucleus.   We have some high energy neutrons produced by earlier reactions and they crash into another nuclues of U-235 or U-238.   Now this actually makes  U-236 (or respectively U-239) for a brief while so it is really the activation energy of U-236  (or  U-239) that we are considering.   It's the U-236 (or U-239) that we want to undergo fission.
   First of all, the energy is not really supplied by the kinetic energy (or speed) of the neutron when it hits the parent nucleus.  It's natural to think that to smash apart a nucleus you must use a neutron with some speed but it's not really like that.  Most of the energy is supplied to the parent nucleus just because you have supplied it with a neutron.   We have already mentioned binding energy much earlier in this post.   Initially we had a neutron that was not in a nucleus, so it had mass (energy)  ~   the usual mass of an individual neutron.   Once that neutron gets into a nucleus then we have some binding energy that can be released,  the mass of the new nuclues is less than the sum of the initial nucleus plus the individual neutron.   This binding energy that can be released is usually much more important than the kinetic energy of the original inbound neutron.     This is important so we'll say it again another way:  Most of the energy you supply to a Uranium nucleus is just beause you have supplied it another neutron, the amount of kinetic energy that the neutron also carries in is just a bit of "fine tuning".    Indeed, for U-235 we have a better cross-section (a probability of causing a fission event) when we present it with a low speed neutron.  Much of a nuclear reactor is concerned with slowing down (moderating) the high speed neutrons that are produced and presenting a nice slow neutron to the U-235 nucleii.   Ideally, if you could present U-235 with a neutron that had 0 kinetic energy then that would be just great, that's almost exactly what it wants - but the neutron obviously won't move into the nucleus unless it does have some velocity relative to the U-235 nucleus.   Give the neutron too much speed and it tends to go past or "through" the U-235 nucleus and not result in a fission event at all.

   So, the activation energy for U-235 to undergo fission after acquiring a neutron (so that's really the activation energy of U-236) is 6.2 MeV  (that one I know).   The activation for U-238 for fission after acquiring a neutron is also about 6 MeV.   the exact figures for U-238 you can find in .... books and stuff,  they may differ by about 1 MeV.     To keep it simple we'll just say the activation energy is approximately 6 MeV for either the U-238 and U-235 that you use as the fuel.   That is to say, they are almost exactly the same.  As previously mentioned U-235 and U-238 just are much the same from the point of view of nuclear physics, they differ by just a few neutrons.   So it's NOT that the activation energy for U-238 is much higher than U-235,   that isn't the case and it does NOT explain why one is readily fissile after being supplied with a neutron and the other is not.

   The difference in fissionability is mainly explained by the availability of that 6 MeV just from absorbing a slow neutron.  The simple model for fission following neutron absorption is that for a brief while we can imagine that ²³⁵U will form a particle we label as  ²³⁶U* ,  the asterisk * that appears after the symbol is important.  ²³⁶U*  is  ²³⁶U  in an excited state,    while   ²³⁶U  is just the ground state (normal state) of U-236.   You can find information and data about  ²³⁶U  from...... books and stuff.   They got their information by doing experiments with U-236.   For example, you can measure the mass of one mole of U-236 and thus obtain the mass of one atom of U-236.  A little subtraction for the electrons in that atom and you have the mass of the (ground state) U-236 nucleus.   Meanwhile the mass of ²³⁶U* is not something you can measure in practice,  even if a particle like ²³⁶U* does exist,  it doesn't exist for long.  It will undergo fission or revert back to ²³⁵U plus a neutron quickly.   The particle  ²³⁶U*  is really just a model or idealisation that is used.   The idea then is that ²³⁶U* has a mass which is equal to the mass of ²³⁵U  plus the mass of one individual neutron,   i.e.   it has a mass which has not been adjusted by the release of binding energy that would result when the ²³⁵U  picks up a neutron and adds it to its nucleus.

   Now the activation energy available when ²³⁶U is formed from the addition of a (slow) neutron to the ²³⁵U nucleus is equal to the difference in energy (mass)  of  ²³⁶U*  and ²³⁶U    (the difference in energy between the excited state that is briefly formed and the rest state that would be possible after the binding energy from incorporating the neutron into the nucleus is released).    A similar thing applies for  U-238   (where the activation energy is the difference between  ²³⁹U*  and ²³⁹U ).     It turns out that this difference in energy between the excited states and the ground state is just over 6 MeV for the U-235 starting nucleus   and just under  6 MeV for the U-238 starting nucleus.   That is what matters, when U-235  absorbs a slow neutron this is just enough to lift it over the 6 MeV activation energy barrier,  while for the U-238 absorbing a slow neutron it isn't.
We actually have a good explantion for this small difference in excitation energy and it is precisely due to some of the stuff that @paul cotter has alluded to.   We have studied the binding energy of a nucleus and we have a fair model of how it depends on the number of protons and neutrons.   There is a "pairing term" that is important in this binding energy formula.  It turns out that the binding energy gets a boost when neutrons come in pairs and also when protons come in pairs,   so that there is an overall boost when both the neutron number and proton number is even,  there is no net adjustment when only one is even and the other is odd,  there is negative adjustement when both are odd numbers.   So the key thing to note is that with our U-238 and U-235, they are both Uranium and so both have proton number = 92 which is even.   However one has an odd number of neutrons while the other has an even number, this makes a small adjustment to the binding energy.  Indeed U-233  is just like U-235,  it is readily fissile with thermal neutrons.  Meanwhile, U-236 is just like U-238, it is not readily fissile with thermal neutrons.   It's just whether the number of neutrons they have is odd or even which makes a small, but as it turns out very important difference.

   Now, I mentioned that the kinetic energy of the neutron is some fine tuning.  U-238 will undergo fission after absorbing a neutron but that neutron must be carrying in the extra needed in the form of kinetic energy.  It must be fast.   However, for reasons that we won't explain (and I'm not sure if we have a good explanation), fast neutrons tend to have a lower cross-section for interaction with a nucleus, so fast neutrons just do not get absorbed by a nucleus as often as you would like.   In the case of U-238 and U-235 the cross-section for interaction is important for nuclear power generation and it has been studied in detail.  The cross-section for interaction with U-238 is greatest when the neutron has a kinetic energy of about 1 MeV,  below this and there just isn't enough energy to get over that activation barrier and fission to occur,   above that and we have the usual effect where fast neutrons just tend to go through or past the nucleus and not interact.  1 MeV is the sweet point where there is enough energy but the reduction in chance of interaction due to speed hasn't become too important.    Meanwhile,  U-235 has a preference for the slowest neutrons it can get because just having been supplied with the neutron is enough to get it over the activation energy.     Over-all  U-235 has a much higher cross-section for interaction at all neutron energies ranging between  0  (stationary) and 10⁷  eV   =  10 MeV.    I have attached a diagram.


fission of Uranium.jpg (50.9 kB . 809x507 - viewed 123 times)


   I hope that helps a bit.   The fissionability of  U-235  and U-238 is so different just because one will get just over the activation energy for fission by absorbing a neutron,  while the other doesn't quite get there.

Best Wishes.   

4

Physics, Astronomy & Cosmology / Re: Which bit of the Shell theorem is not working?

« on: 19/10/2023 21:46:06 »

Hi.

   Yes I can see what you're saying @Halc .
I'm deeply sympathetic.   I don't suppose that will help.

Let's try and be a bit more useful.    You seem to be trying to do the impossible, so don't.
You discuss time dilation and gravitational potential.   Most of the results we have for that would tend to come from using the Schwarzschild metric in GR.    Around a massive spherical body, the Schwarzschild metric is a fair approximation of what you have and how spacetime behaves.    For that solution of the EFE (Einstein Field equations), the co-ordinate time is exactly as you have stated.   It is the rate at which clocks progress when you are infinitely far from the mass.
    After that you are generalising the result and imagining that time would always run more slowly in a region of lower potential than some other region you are comparing against.   It might, it does seem reasonable but we do not have the Schwarzschild solution applying across all of space.   We know the universe is not a vacuum around one spherical mass.

   Indeed there is a completely different metric that is often used when discussing the Universe,  the Robertson-Walker metric.    That is obtained as a solution to the EFE under a very different set of assumptions.   In particular the universe is not a vacuum, it's actually too much unlike any vacuum.  Nowhere is less dense than anywhere else,  the universe is filled homogeneously with whatever cosmological fluids you have chosen to have (matter, radiation, dark energy  etc.)    With the Robertson-Walker metric, time dilation does not happen when you move from one gravitational potential to another because   (i)  gravitational potential is ill defined anyway,   (ii) nowhere is any different to anywhere else, there is no place of different potential.    That is not to say that time-dilation isn't an effect you can observe at all, you certainly can.   Special relativity is automatic in the machinery of GR, so movement will cause time dilation.
    The time co-ordinate in the Robertson-Walker metric is not and does not need to be related to how time may flow when you are infinitely far away from all mass.   Indeed it turns out that there are lots of places in a FRW universe where time flows at the same rate as the time co-ordinate.   Provided a clock is co-moving with the expansion of the universe then it will progress at a rate identical to the time co-ordinate.
    The passage of co-ordinate time is not abstract, it is something that a real clock can measure.  There is no requirement to be in some special place, for example to be in a place with the lowest gravitational potential, or infinitely far from all mass.   For the Robertson-Walker metric, the co-ordinate time is not that.   

     There is a natural tendancy to blend the two metrics together and assume that our real universe will have properties of both,  with more of one in some places (close to a body the Schwarzschild metric prevails) than other places.
None-the-less it is an error to assume that the co-ordinate time used in the Schwarzschild metric is precisely the same as the co-ordinate time in the FRW metric,   or that either of these is precisely the same as the underlying co-ordinate time that our real universe with its real metric would use.   
   

The absurdity of trying to express this is one of my pieces of evidence against models with flowing time.

    Maybe a clock infintely far from all mass would progress infinitely faster than all real clocks in the universe.   That doesn't need to matter and it doesn't demonstrate the absurdity of time as something that flows.   The underlying co-ordinate time of the real universe and its real metric is probably not the thing that such a hypothetical clock in this hypothetical place is presumed to show.

Best Wishes.

5

Physics, Astronomy & Cosmology / Re: Which bit of the Shell theorem is not working?

« on: 16/10/2023 07:58:30 »

Hi.

Has my post been deleted, and Paul's?

   I don't know.

If an object within a shell has no gravitational  effect on any object within it, surely that is from the point of view of the outside Observer? One would assume that the shell would indeed have gravitational attraction, if the she'll where 100, 000km thick and 10 million in diameter?

    There may be some minor errors in the English in that question.   I think I know what you're asking.
Wikipedia gives a short definition or explanation of the shell theorem:

1.         A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its center.

2.   If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.

 - - - - - - - - - -

   So the shell is made of some material and that material does gravitate - it attracts other stuff to it.    When you're outside that shell it behaves exactly as you would expect .   You would have a force of attraction identical to the force of attraction you would have if you imagine all of the shell was just squahed down into one really small ball and placed at the centre position of that shell.
    It's only when you're inside the hollow shell where it may seem you get an unexpected result.   You are still attracted to each part of that shell.   If you happen to be at the centre of the shell then you are equally attracted in all directions and there is no net force, as you may expect.    However, even when you move away from that centre and toward one piece of the shell,  you still feel no net force of attraction in any direction.   Provided you stay inside the shell, you find no net force acts on you.
    There are many articles and YT videos that provide explanation in varying levels of detail and sophistication.   The simplest explanation is merely suggestive or hand-wavy:     When you move toward one piece of the shell,   the distance between you and that piece decreases,   so by Newton's law the force of attraction to that piece (the bit infront of you) should increase.   That does happen.   However,   since you have moved off-centre, there is now a bit more of the shell behind you rather than infront of you.   That tends to increase the attraction to the direction behind you.   It just so happens that both things happen in the right proportion and the net (or total) force stays at 0.
      The acceleration can be determined in various ways,  you could use any inertial reference frame you want   OR   you can just identify that acceleration more directly by giving the person insde the shell an accellerometer.    To get the right result with an accelerometer in the real world you'd need a really, really big shell and a person who is actually floating in outer space so that they weren't experincing any forces other than the gravitational attraction of the shell.

Best Wishes.

6

Physics, Astronomy & Cosmology / Re: How Can We Know The Cosmos Age If We can Only See The Observable ?

« on: 03/10/2023 19:37:26 »

Hi.

    I started writing this after post #11 but it got too long and too late so I left it and went to bed.   Since then @paul cotter  has shown some interest and @Halc has written more.   This is hasty re-editing of the reply I was going make but it now seems even more neccessary to say it because I think you ( @Halc ) are at risk of confusing some readers.   

                     - - - - - - - - - - - - - - - - - - - - - - - -
   It's always a pleasure to hear from you, @Halc.  I'm also fairly sure you won't be too offended if I oppose some of your comments.  It's a discussion and I wouldn't bother to spend the time making this argument if I didn't think you would be able to appreciate it.

Well, if we're going to go strictly by what relativity says, it posits the entirety of 4D spacetime as 'the universe', and therefore it has no meaningful age at all.

   Yes, that's OK.   However, let's not get too hung up on whether we adopt the view of presentism or of a block universe (or something else).   The phrasing I used was just more like natural English language and that does tend to utilise a tacit assumption of presentism.

   We (Astronomers / Scientists) do impose some common understanding on what we mean by the age of universe.   We understand that the 13.8 bn years refers to the use of co-moving co-ordinates.   Specifically,  the 13.8 billion years is the proper time interval that would be recorded on a planet that remains (or past tense: remained, maybe but I'm trying to avoid implying presentism) at constant (co-moving) spatial co-ordinates while the sclae factor evolves (the universe expands).   To paraphrase that, the planet stays still in as much as it possible for anything to stay still in an expanding universe. 

   One advantage of this understanding is that no planet (or anything) in the universe would find that more than 13.8 bn years of their proper time had elapsed since the big bang.    The Robertson-Walker metric  (which is what is used in main-stream Cosmology)  has the form

        dT²   =   dt² - a(t) ds² 

with dT being a proper time interval (a differential)   while  dt and ds are co-ordinate time and space  differentials respectively.  a(t) is the scale factor (which is some function of co-ordinate time).   

     The important point is that a co-ordinate space interval is SUBTRACTED from the co-ordinate time interval,  much as in flat Minkowski space.
   So, if an object follows a path where the spatial co-ordinates vary with t,  then something is subtracted from the change in time co-ordinate.   To say this another way, it's just like the travelling twins paradox of SR.    The longest proper time path between  Event A (say the big bang)   and Event B  (say the planet being at a given position, s₀ and at co-ordinate time, t₀),  will be obtained when that planet had just stayed as still as possible - it must have 0 peculiar velocity relative to the given co-ordinates s,t that appear in the metric.    Those co-ordinates are often called co-moving co-ordinates and it's common enough to say that the the planet must be "co-moving with the expansion of space" to obtain this maximal amount of elapsed proper time since the big bang.   For an expanding universe, being co-moving is the equivalent of being the "stationary" twin in the twin paradox.   A planet that had a non-zero peculiar velocity relative to the co-moving co-ordinates and arrived at the same event (the given position s₀ and time t₀) would have experienced LESS proper time since the big bang.   Such a moving planet is like the twin who did the travelling in the twin paradox, their path through spacetime was different and it had a shorter proper time length.

   We don't need to know exactly where we are right now in the co-moving co-ordinates.  We are at some event that we can write as (s₀,  t₀)  in those co-ordinates.   However we do know something about when we are in those co-moving co-ordinates.   The cosmological model implies that t₀ = 13.8 bn years.   It has to be this value (with a conventional understanding of a "year" as a measure of time) for the current expansion rate or Hubble constant to be what we do observe as we look out into space.   (Given the cosmological model we do have that is based on GR and especially the Robertson-Walker metric and the Friedmann equations etc.)

   If you take the presentism view of things then you would paraphrase that and say that we know that 13.8 bn years of the co-moving co-ordinate time have elapsed since the big bang.   (But that's optional, you can assume a block universe or something else if you wish,  all we need to agree on is that we are at an event we will write as (s₀, t₀) in the co-moving co-ordinates).     We know that the proper elapsed time along the path taken from big bang to (s₀, t₀) cannot be more than 13.8 bn years.  If we have some peculiar velocity through the universe then our elapsed proper time since the big bang could be less than 13.8 bn years BUT, for certain, it cannot be more.   The way you ( @Halc ) have phrased things it looks as if the age of the universe is quite arbitrary and frame dependant but it is not like that, we can make a much more objective statement.

   

That's not the age of the universe, it's just the time coordinate of humanity in that frame.

    We can make an objective, frame independent statement about the maximum proper time that could have elapsed between the big bang and the event where I find myself while I am writing this forum post.    That maximum elapsed proper time is 13.8 bn years.   It does not matter if you are a human being living on planet earth or a space alien whizzing through space at great speed in a starship who just passed by planet earth while I was writing this forum post.   Any being that finds themselves at the event (s₀,  t₀)  - which I'll say is at planet earth and at a time the presentism-ists would call "now"  will agree that the maximum elapsed proper time from the big bang event to this event (s₀,  t₀) for everyone is 13.8 bn years regardless of the motion or path they may have taken through spacetime to get there.   
    If we lapse into more informal English Language which does tend to be more aligned with a presentism view of things, then we can just say that every being who finds themselves at (s₀,  t₀) agrees on the maximum age of the universe.

- - - - - - - - - - - -
   That's already getting too long, let's just move on to look at one example you mentioned about frame dependance...

As for the 'Earth's frame' bit, one can make the 'current' time as large as you want by picking a different frame. That lets relativity of simultaneity work for you.

   No, not quite.    For a start there are no global (or universe wide) inertial frames of reference in an expanding universe - nothing that satisfies the conventional criteria of being an inertial frame for Special Relativity.   In GR with the Roberston-Walker metric there are only going to be Local Inertial Frames,  they are local in spatial terms but also local in the range of time over which they remain valid.   None the less we do often approximate the situation and I'll assume that is what you have done.   What I have concerns with is that 13.8 bn years is not a small amount of time, it's the entire lifetime of the universe, it is certainly not keeping anything to a local region in the time co-ordinate.   As such using SR as an approximation is probably little better than wild speculation.
   
   More generally, I feel sure that the readers could easily be confused by some of your comments.   They aren't necessarily wrong, just easily mis-understood - but time is short and this is too long, so let's just look at one statement that is more easily identified as being wrong than right.

So say you want to compute the current age of the universe relative to the (inertialish) frame of some galaxy relative to which our galaxy is moving away at say .97c. In that frame, our clocks are dilated by a factor of 4, so the event on that far galaxy simultaneous with humanity here is somewhere around 55 billion years since the big bang.

    I'm sorry but you have trimmed too many corners here and ended up with a statement that is just wrong even in SR.  Time dialtion on its own is not the master of disagreements in what would be said to be happening now or simultaneously across the universe. 
    Consider a frame S and another frame S' which is a standard Lorentz boost from frame S.   Specifically assume the origin   (s, t) = (0,0)    in frame S and the origin  (s',t') =(0, 0) coincide  at time t=0  (or at t'=0 if you prefer) but frame S' just has some velocity v relative to S.     Now it does not matter what that velocity v is and hence what the time dilation factor of frame S relative to S' might be,   the origins of the two spacetime frames   (s, t) = (0, 0)   and (s', t') = (0,0) just do coincide.   The appropriate Lorentz transformation  takes  (s,t) = (0,0) in S  ---->  (s', t') = (0,0) in S'.    So that event will lie along a line of simultaneity for "my time = 0" that a person would draw across a spacetime diagram regardless of whether they are at rest w.r.t. frame S or frame S'.  To say this another way, both observers agree that the event (0,0) is happening at "my time =0" or what they might call "now".   You can change the value of v, the offset velocity and have a time dilation factor as large as you like and repeat the calculation, it isn't going to matter because there was no physical spatial distance between the origins at time t=0 (or time t'=0).
    This is the essence of what is often called "the Andromeda paradox" in SR.   It is not the time dilation factor on its own that determines the discrepency in which events in the universe are considered to be happening "now", it is a combination of the time dilation AND ALSO the distance between the two observers that matters.   So, in the case of the Andromeda paradox, a human being on earth can make the invasion force from Andromeda have departed 1 week ago or not having been sent yet just by walking slowly in one direction or the other on planet earth.   They are walking slowly, the time dilation factor is not important,  however the distance between Earth and Andromeda is large.   Now, as it happens, Andromeda is moving towards the Milky Way.   If it turned out that the Andromedons were only going to invade Earth when Andromeda was in the same place as planet Earth, then the Andromedon paradox stops working.   A human being can walk (or run as fast as possible) in one direction or the other on planet earth but it will not change the invasion launch from being in what they consider as the future or the past.

Summary

    "The age of the universe"  is not arbitrary and changed just by boosting to another frame.   We have a very specific understanding of what is meant by the age of the universe and we can make a powerful and objective statement:

     We are at co-moving co-ordinate event  (s₀, t₀ ) and we know that t₀ = 13.8 bn years.   You do not need to assume that this co-moving co-ordinate system is especially important or a way of describing the universe that is more truthful than some other co-ordinate system.    However, we can assert that the maximum proper time that could have elapsed (on the path we have taken) since the big bang event is 13.8 bn years.   If we have had some peculiar velocity through space then the proper time elapsed since the big bang could be less than 13.8 bn years but it cannot be more.

Best Wishes.

7

General Science / Re: Is there a simple way to make fresh milk drinkable?

« on: 07/09/2023 01:46:28 »

Hi.

Yes, OK, you could drink fresh milk and expect to be OK most of the time.   However, this is a science forum and someone needs to issue the government health warning and toe the party line on this one.

bTB  (Bovine Tuberculosis) is one example of a disease that has been studied in some detail.  The UK has experimented with things like culling badgers beacuse they could be one vector for spreading bTB.  As it happens, the evidence that badger culling makes any diference is far from conclusive.   In some cases, badger culling seems to have actually increased the spread of bTB.     One possible explanation for this is that badgers will travel further to find a mate when the population of badgers is small,  so that culling badgers can actually make an infected badger spread their disease over a larger area than they otherwise would.

bTB is caused by a bacterium  (Mycobacterium Bovis) which is most active in cattle but it can infect humans and cause symptoms similar or identical to TB  (human TB). 

As long as the animal that is milked is healthy.......

   Yes, but a cow can have bTB and appear healthy for a long time, quite possibly reaching old age and dying from unrelated conditions.

bTB is known to be found in mucus, puss and some other body fluids of cattle.   It is NOT usually directly produced in the milk but it only takes a minor injury (say an ulcer or infection in the mammary tissue of the cow) and some other body fluids can get into the milk.   

    One of the triumphs of the compulsory pasteurisation of milk in the UK is that (to the best of my knowledge when I studied this about 6 years ago) we have not seen ANY cases of TB in humans that seems to have been contracted by consuming milk.

   In conclusion, although it is very much a belt-and-braces approach and you probably could drink unpasteurised milk, it is very likely that the pasteurisation of milk has benefited public health and saved lives since it was made compulsory.

Best Wishes.

8

Physics, Astronomy & Cosmology / Re: Weird physics with a guitar string

« on: 07/09/2023 00:30:32 »

Hi.

I don't know.   Limitations of human vision need to be considered.   Are you absolutely certain these sticky-tape flags were spinning in opposing directions?

Compare with "the wagon wheel effect", where any spinning wheel can appear to spin backwards - this is an effect called "aliasing".   I'm not suggesting your eye is a camera with limited fps (frames per second), only that your brain is trying its best to interpret the images as a smooth and consistsent motion which may be difficult for something spinning that fast.

Best wishes.

9

Physics, Astronomy & Cosmology / Re: Talking about Physics

« on: 08/07/2023 00:31:10 »

Hi.

I'd like to canvas an opinion or two about the likeness between two well-understood things, namely energy and information.

   You could start a new thread and put a Poll on it.  It appears right at the top of the thread and is automatically updated.  People can easly select a response and a bar chart of results is automatically generated (or hidden until some deadline is reached).  Experiment with the options and you'll see for yourself.

See  https://www.thenakedscientists.com/forum/index.php?topic=85028.msg681214#msg681214   for an example of a poll and some screenshots showing you exactly which buttons to push to create one.

Best Wishes.

10

Physics, Astronomy & Cosmology / Re: How do we know that particles are entangled?

« on: 06/07/2023 04:11:29 »

Hi.

@Zer0   --->   Yes you could make a quantum coin toss in some way and use it for the purposes you described much earlier   (having one person sacrifice themselves to the Borg Armada to buy time for the other,  while the other tries to get away).    However, you could do this without any quantum entagled particles ever being needed.
     Just use bits of paper and a hat.   Cut two equal sized squares of paper.   Write on one   "You will sacrifice yourself",   write on the other   "You will try to get away".    Put them in a hat and shake the hat.   Have one person pull a piece of paper out (as usual without looking).   The other person takes the last remaining piece of paper.   Just to keep it in the style you wanted, let's say no-one unfolds their square of paper and reads it,  they just put it away in their top pocket.   Then they separate and go off to do whatever,  with the understanding that if and when a Borg Armada does arrive they must read their bits of paper and do what is written.
     Now, provided no-one looks at their paper until (or unless) a Borg Armada arrives, neither person did know what they were supposed to do and when the Aramda does come into sight they do both know what to do without needing to relay messages between themselves.   Although it's a big Armada if the two people really were 10 billion light years away from each other and they did both see it - but that's a different issue.

Best Wishes.

11

Just Chat! / Re: Is "new theories" getting worse?

« on: 01/07/2023 19:57:17 »

Hi.

   I'm wondering if the Forum is due for complete re-design or re-development.    Try to sort out what the forum is supposed to do and how it is supposed to be used.   For example,  is it a place for people to try and present or publish an article at all?   
   
   I have increasingly thought that one of the defining features of a forum post should be that it seeks discussion or input from other people.   If anyone found themselves just publishing something and merely hoping it would be read and used or incorporated into someone else's knowledge -  then the entire article is just not something that should be in the forum at all.    For example,  asking a question is obviously fine - you're seeking input (answers)   - but keeping and updating a chronological account of the activities of the Hubble telescope may not be.   You see what I mean about how tricky it is to sort out what is reasonable from what isn't.     Most of us would agree that the activities of the Hubble telescope should be regarded as perfectly good science.   A regularly updated post may be interesting to amateur astronomers and all sorts of people that TNS might consider as their target audience -  but if the article is not inviting or promoting discussion then it's not really something that needs to be on the forum.   There are already websites providing all the latest news about Hubble (e.g.   https://www.nasa.gov/content/news-latest-hubble-news ).

   There are evidently some issues concerning the forum at the moment and its long term future.   Until or unless those are sorted out I don't suppose it's worth creating surveys to see what people want in a forum etc.   A more pressing issue might be sending out requests for additional financial support.      i.d.k.

     Another thing that (I find) has to be thought about is that we do actually want some questions or something to do.   There are days when I log onto the forum and hope to find a discussion or question I can spend a moment on and get involved in.   So it's not as if the forum will work without people asking questions or putting something in new theories.  We can't all be answering or replying, someone has to ask a new question, start a new thread, or update a long running thread with some new material etc.   So maybe there is a case to argue in favour of threads like a chronological account of the Hubble telescopes activities,  a just for fun quiz,  a jokes thread   etc.    Anyway, that's an option, try to look at the glass as being half-full:    When there's a new theory presented that you may think is bad, at least there's a new theory to look at and something you could do today which may be useful.     I'd recommend politely engaging with the author rather than anything else.

Best Wishes.

12

General Science / Re: Would a mobile with messages from the 5 Titan victims survive for recovery ?

« on: 01/07/2023 04:50:41 »

Hi.

Anyone know Anything bout the whereabouts of B.C.?

   Sorry, no.    It's been less than a week,  it could be holiday.

Best Wishes.

13

General Science / Re: Infinity...

« on: 29/06/2023 18:51:26 »

Hi.

I was wondering. Can anything physically literally go onto into infinity?

    Maybe, although it's harder to see or imagine if you're thinking of something having infinite length or dimensions.   The universe might be like that, we don't know.

   However, you can more easily imagine situations where an infinite number of things can happen within a finite amount of space  (or time).
   Example:    Have a race between a tortoise and a hare.     The tortoise moves at half the speed of the hare but let the tortoise have a head start,  let it start half way up to the finish line.
    You should be able to do the calculation:      Time taken to reach the finish line   =   Distance travelled / Speed.
The tortoise had half the speed but also they only half the distance to cover.  So both the tortoise and the hare will cross the finish line at the same time.   More importantly that will be some finite amount of time.
   Now just consider how many times the distance between the tortoise and the hare has been halved.   At the start of the race   the distance between them  was   L/2    (with  L being the length from the hare's start line to the finish line),    allow a bit of time to pass and both of them move a bit but the hare is faster so it will close the distance between them.    At some time the distance between them would only be L/4,  at a later time it would only be L/8,  a later time the distance would be L/16    ..... etc..etc....        You can keep on identifying moments of time where the the distance between tortoise and hare has been halved.
    Once the tortoise and hare actually do reach the finish line (which they will do together and in finite time), the distance between them is precisely 0,    so the distance between them does seem to have been halved an infinite number of times.   The situation hopefully seems fairly concrete or realistic to you, i.e. it really does seem that an infinite number of events (the halving of distance between them) can happen.   More importantly this does seem to happen within a finite total amount of time and space (you do not need an infinite length race course or an infinite time to pass  etc.)

    There are several real world situations where it seems that an infinite number of things can happen within a finite amount of time or a finite amount of space.

But there is a bigger multiverse, isn't there, of many universes.

    Maybe.   There are theories involving "many worlds" or many universes.   How much time or discussion did you want?    There's no agreement on whether there is a multi-verse.

Best Wishes.

14

Physics, Astronomy & Cosmology / Re: Talking about Physics

« on: 26/06/2023 17:50:41 »

Hi.

   For a superconductor,  you're (@alancaverd) right.   I'm going to take for granted that for a more ordinary wire, you'd agree that it does have some resistivity.

Say you have a superconducting wire delivering power to a load what would the Poynting vector be?

   
   That's actually a very interesting question.  You just won't be able to establish a potential gradient across a length of superconductor.   It has precisely 0 resistance and will not support an infinite current in steady state  (So  V=IR always gives V=0 since R=0  etc.).   So there is no need for surface charges to accumulate on the surface of the superconducting wire (similar to the diagram a few posts ago).  There is zero E field inside the conductor (in steady state), any current that was there will just always keep flowing, it does not need any E field to overcome any resistivity.  In practice, if you tried connecting a battery then you'll have a problem, the current will escalate rapidly and the material will get warm fast,  so in practice the superconductivity will be lost or something else in the circuit (like the wires to the battery) will break  etc.

   For one thing, you'll need to be very specific about what the load is and how it was attached without itself being a superconductor:  For example, if it's a resistor (or lamp filament or something) which is also supercooled and superconducting then you just can't establish a potential across the resistor (or filament) either,  so no work can be done when charge passes through that resistor.   Specifically, it's not sufficient to say  "you have a superconductor delivering power to..(whatever)....", in some cases you just would not deliver power or have power consumed by what you thought was the load.     If you dunk a superconducting circuit complete with a lamp filament that can also become superconducting into a tub of liquid Nitrogen then you may find that when you connect a battery, the lamp filament just would not light up.

   To connect a battery and have it all work (hopefully),  there must be some regions where you don't have a superconductor.  For example the battery and some wires to it are NOT in liquid nitrogen, they are at room temperature,  similarly the load  (lamp filament or whatever) is not superconducting and it has some wires to/from it which are also not in the liquid nitrogen.    Now you're in a position where there are some regions of the circuit where E and B fields can be set up fairly conventionally.   So there are some E fields outside the superconducting piece of wire just from the bits of circuit that were not superconducting.   
    That's my best estimate anyway:
1.   There will still be a magnetic field outside the superconducting piece of wire because there were moving charges in it.
2.    there will still be E fields outside the superconducting piece of wire but they were sourced from the non-superconducting bits of the circuit rather than directly from surface charges on the superconducting piece.
3.    If you prevent (2) and don't have E fields produced in space from anything or anywhere in the circuit  (e.g. dunk the whole circuit into liquid nitrogen) then the bulb does not light and NO power is delivered anywhere or consumed by any part of the circuit.

Best Wishes.

[Spoiler:]

15

Physics, Astronomy & Cosmology / Re: What is the exact cause of the time dilation of the twin?

« on: 24/06/2023 18:16:55 »

Hi.

I'm a bit surprised this thread is still going - but it's an interesting topic and can be examined from many angles.
Some of the older posts have lost their special symbols and especially any equations but there's a diagram that survives.
   Post #45 had this diagram which is a great diagram:



   I would say a diagram like that explains most of what people want to know.   I'd just like to colour all the events that happen at Earth between the highest blue line and the lowest red line .   Here is that diagram:

Twins-green.jpg (41.62 kB . 511x421 - viewed 767 times)

    Looking at the blue lines of simultaneity we can label those lines  (from bottom to top)   t'=1 unit,     t'=2   and t'=3 (where t' is the time elapsed for the travelling twin).   We see that the events in the green area are still in the future for the travelling twin on their outbound journey.    For example, assume the twin had NOT turned around at t'=3  (at spatial location = Alpha Centauri ) but instead had just continued on in uniform motion.   Here's the diagram for that:

Twins2.jpg (25.32 kB . 511x421 - viewed 563 times)
   We see that events in the green area (which are spatial located at Earth) fall along a blue line of simultaneity with  t' > 3   (that is later than t'=3,  when the twin reached Alpha Centauri and should have turned around).

    However the travelling twin does turn around at t'=3.    Which immediately changes the slope of the lines of simultaneity,  we have the red lines of simultaneity.   We can start labelling those from bottom to top  as   t'=3,   t'=4 and t'=5.    Note the lowest red line is t'=3   exactly as for the latest blue line of simultaneity when the travelling twin reached Alpha Centauri -  we assume the turn-around took negligible time for the twin. 
    So on the return portion of the travelling twins journey all the green events are effectively in what they would consider as the past  -   e.g.  the green events would lie along the red lines of simultaneity for t' < 3.
Here's the diagram for that:

Twins4.jpg (25.48 kB . 511x421 - viewed 539 times)

    Anyway, that hopefully explains, with diagrams, what was discussed in earlier posts.   When the travelling twin changes their direction of motion at Alpha Centauri,  they change rest frames.   This has the consequence of making all the green events that were located at Earth, no longer be events that the travelling twin would consider as being in their future  (t' > 3).  When the travelling twin changed direction, the green events immediately change into events that the travelling twin would consider as being in their past  (t' <3).

Some events (located on earth) were in the travelling twins future (in the old rest frame) but they abruptly changed to being events that were in their past (in the new rest frame), they were never in their present or "now" , they were skipped over entirely.

   or, as Janus wrote:

An analogy would something along these lines:
Two men are back to back and then separate. According to each of them, the other is a given distance "behind" him. Man 1 then turns 180 degrees.  By his perspective, Man 2 goes from being behind him to being in front of him.

   
- - - - - - - - - - -

    Anyway, I quite like the diagram approach and just seeing where the "green events" will map to on the outward journey compared to the return journey.   It reduces the Twins paradox to a situation that is almost identical to "the Andromeda paradox"  (where an invasion fleet has or has not yet been launched from Andromeda depending on which direction you are moving in).    It is all seen to be a consequence of the idea that spacetime is like a loaf of bread and there is no preferred or correct way to cut it into time slices.  We can foliate it (slice it) at different angles to obtain very different "now" slices:  Boost to another frame of reference and we change the angle at which we slice through that bread,  so that events that were at some spatial distance from us can sometimes be moved into what we would consider as the past, the future or "now" just by changing your motion.
   So you (Hamdani and/or Dimensional who started the thread) can have that as another explanation of the "exact cause" of different aging of the twins if you want it:   It's a consequence of our ability to foliate spacetime in different ways just by changing the motion of one twin together with there being some physical distance between them when that change of motion occurred.  So you can recover all the criteria that has previously discussed:  For example, the distance between earth and Alpha Centauri is important because you're going to pivot around Alpha Centauri and hence sweep out a greater amount of what has been drawn as green events on Earth.  Similarly the speed of travel for the travelling twin is important because that determines the angle at which you will be foliating spacetime (if the travelling twin has high speed relative to the Earth twin the blue and red lines of simultaneity are severely angled from the horizontal on the diagram).   The acceleration of the travelling twin at Alpha Centauri is important but not numerically important:   The traveling twin HAD to change the direction of their motion at Alpha Centauri but exactly how that was done didn't matter much,  for convenience it was assumed to be almost instantaneous.   If the acceleration had been done over  0.1 s  instead of  1 s then the size of the acceleration would have been ten times larger but it doesn't matter and doesn't affect the age of either twin much.  All that matters is the total angle swept through when you foliate spacetime for the outbound journey compared to the return journey (the angle between the blue line of simultaneity vs. the red line at t'=3 when you pivot around spatial location= Alpha Centauri).
     
    Already too long.... I'll hide the next bit.   It may only add confusion anyway.   If you're happy with events being moved from the future to the past then that's fine, leave it there.
 

Spoiler: show

  Now, when you spend even more time considering Special Relativity you may start to realise that it is actually silly to assume an event is "definitely in my past" if it's spatially separated from me and has not yet entered my past light cone.  If an event has not yet entered my past light cone then I have almost complete control over whether it will get into my past.  I can accelerate away from the event and delay it getting into my past light cone and if I keep accelerating rapidly enough then it will never get into my past light cone.   Conversely, if I do what the travelling twin did,  I can turn to move towards the event instead of accelerating away from it.   Then those events that were spatially remote from my original starting position will rapidly start entering my past light cone:   The travelling twin can have much more than 1 earth second's of the green events entering their past light cone for every 1 second of their own elapsed time.   Of course they can change direction again and start accelerating away... only those events which have entered their past light cone are definitely "locked in":  They are most certainly in their past and cannot be prevented from having some influence on what they may experience locally.
   Anyway,  that's almost a completely different discussion about Special relativity.....  the causal structure of the universe is based on light cones and not on a time co-ordinate.  So some would say that we aren't even allowed to ask if the green events shown on the diagram above were in the travelling twins past or future when they were at Alpha Centauri -  they were outside the twins light cone and therefore not uniquely fixed or required to be in that twins past or future at that point in their journey.   The travelling twin had almost complete control over how quickly those green events would enter their past light cone,  they just change their motion to achieve that.

Best Wishes.

16

Technology / Re: Will thousands of undeleted e-messages break my Gmail account on cellular phone?

« on: 24/06/2023 14:45:14 »

Hi.

I don't know for sure.   You'd have to read all of Gmail's terms and conditions carefully.   Details can vary depending on whether you have a free account or have paid for more space etc.   At a quick glance there seems to be a storage size limit rather than a limit on the NUMBER of emails stored.    So you could store many small emails but less emails of huge size.

   A glance at information on the internet suggests the storage size limit is about 15 GB - that's the free account specification but you can pay more to get more.    Let's assume you use your Gmail account only for emails and a typical email might be 50KB   (it's a lot bigger if it has attachments like photos etc.).    So you should fit  ~   300,000  emails  of 50KB  before the storage space is used up.   That's how much you could store online or "in the cloud",  some email software will also download some emails directly onto your device.   Note that if you use other Google services, like their online file storage for Google documents etc. this may share some of that 15 GB storage limit and therefore seriously reduce the number of emails you could store.

    In the old days  if you tried to send a new message to someones account that was already full, then the message would be returned to the sender with an added error message (e.g.  "this could not be delivered...... with some error codes / reasons").

    Overall, the good news is that having too many old emails stored online won't "break" your phone or device in any dramatic way - it just may stop you receiving emails from people who were trying to send them to you.

   FINAL NOTE and WARNINGS:    Some email providers don't enforce their storage size limits too rigorously.   If they have it to spare (and they usually do) they may let you go over the 15 GB size limit.   However, when you read the t&c  you'll see that they reserve the right to delete old messages or old files without warning if they ever needed to recover that surplus storage space they have allowed you.   What that means is that if you have gone over the 15 GB then some old messages are not safe, you could lose them at any time.      More generally,  you must read the "terms and conditions" for yourself -  the advice given above is personal opinion only and I can take no responsibility for the loss of any data you may experience  etc.

Best Wishes.

17

Physics, Astronomy & Cosmology / Re: Distribution of Earth's mass?

« on: 23/06/2023 20:55:22 »

Hi.

Follow-up...

Does Earth's Gravity modulate or fluctuate with Time?

    I think this question may have got overlooked or missed.

Anyway,  the rough or practical answer is "no":   It's fairly constant and consistent.

More complicated answers exist.   For example, the earth rotates on it's own axis every 24 hours (while the moon stays fairly still by comparison).   The most obvious effect is the ocean tides (bulges in the ocean) which move around the earth.    Beneath the outer crust of the earth there will be a lot of molten rock which can move (e.g. due to convection currents).   We know the magnetic field lines of our planet do move around instead of staying constant   (see for example:   https://en.wikipedia.org/wiki/North_magnetic_pole    which discusses the movement of the Magnetic North pole thought to be due to changes in the movement of molten ferro-magnetic material in the outer core).    So there will be some movement of various material  (surface water and molten rock inside the earth) which can affect the gravitational field local to a fixed place on the surface of the earth.

Best Wishes.

18

Geek Speak / Re: Why is Windows 7 still updating 3 years after updates were discontinued?

« on: 17/06/2023 23:30:21 »

Hi.

I switch the router off before I put the computer on....

    It could be an old one your computer downloaded sometime ago.    It may be a failed update - something went wrong.     Rather then just break or stop everything completely, the system is designed to roll-back to the state before it attempted to install the update and then it will re-attempt installation at a later time.   Microsoft often end up releasing patches for their patches and/or replacing a patch with one that actually works.   Therefore, your computer might try and install the (presumably old and faulty) update several more times (maybe once a week for ever more).   Sadly, it may never find a later update that actually works because Microsoft are under no obligation to make one any longer.
    If it's annoying you can probably roll back your system to an even earlier state, where it was stable and did seem to have all the latest updates (at that time), then switch off the automatic updates completely.   BEFORE you start experimenting make a System restore point now,  (see for example:  https://www.dell.com/support/kbdoc/en-uk/000132661/system-restore-in-windows-7     an old help page that told you how to create a Restore point in Win 7) so you can get back to what you have now.  With some hours of work you can probably identify exactly which update is the one causing a problem BUT it's probably not worth the time - it will be quicker to just keep rolling back a bit further and trying it until the problem is gone.   You have an outdated system anyway, it hardly matters if you roll it back another month or two.

Best Wishes.

19

Geek Speak / Re: Why is Windows 7 still updating 3 years after updates were discontinued?

« on: 17/06/2023 14:17:27 »

Hi.

1.   From what you're describing, that was finishing up an update that was downloaded from the last time you switched the computer on  (a year ago etc.)

2.    Microsoft didn't actually stop all updates.   They merely didn't guarantee to provide any more.   However, your computer is still going to look for updates unless you specifically switch that feature off.   If it's in Microsoft's interests then they can continue to provide updates.     For example,  it is in their interests to collect data they can use.  So running their anti-virus tool on your computer and collecting data is something they are likely to do.

3.     The Windows Operating System is a collection of programs rather than one just program.   For example, you can probably run the .NET framework which is still used by Windows today.    If there are updates for any bits of Windows that you can use, it's likely that your computer will still get those updates.

- - - - - - - -
    As you probably know, the main problem is just that Microsoft don't guarantee to provide updates, which means they don't guarantee to fix or patch exploits that might have found since support for the Operating System was discontinued.   As such it's not particularly safe to connect your old computer to the internet.   You may also want to switch off any wireless connections to the outside world  (e.g. Bluetooth and Wi-Fi if it has that).    However, if you do keep the computer isolated ("air-gaped") from any external networks then the computer can go on being used for ever more.   Indeed you can connect it to the internet if you wish - just be aware that it isn't safe, you could be handing information out to anyone  and/or  receiving information that has been maliciously corrupted.

Best Wishes.

20

New Theories / Re: Where does quantization of energy of electromagnetic radiation come from?

« on: 13/06/2023 17:33:05 »

Hi.

About the videos:
   I've only glanced at some bits of the videos you (Hamdani) presented.   They seem to support the idea that Planck focused on the oscillators in the walls.  See, for example,  around 6:10 ~ 6:30 in the first video where the narrator states that light came from those quantised oscillators. 

   I found at least one statement that was made which isn't necessarily true.   In the second video at about 7:00 to 7:30  the narrator discusses the use of a cuboidal shaped cavity for the mathematical model (which is the typical choice).   There is a fairly casual statement made that any shaped cavity could have been used, the result would be the same but the maths would just be harder.   That is not at all obvious and I don't think it should be stated as established fact.  There are several adjustments to the method that would normally be made.  In particular, there are some extra assumptions that seem essential and must be added to the mathematical model to correctly predict that very small and peculiarly shaped cavities will still ultimately produce the usual BB spectrum.
   There are some pieces of equipment which work purely because there is a cavity (including a suitable material for the walls of the cavity) with the right shape and dimensions to seriously influence the spectrum or frequency of radiation that is supported.   (See, for example, the discussion of lasers, resonance and cavities  in this post  https://www.thenakedscientists.com/forum/index.php?topic=86164.msg700593#msg700593   which you (Hamdani) seemed to approve of at the time).   
 
   The videos do seem to take a fairly conventional approach and walk you through how you must have stationary e-m waves that fit inside the cavity,   only these will be considered as supported "modes" of radiation.    Overall, the mathematical model discussed (which is a typical derivation for the BB spectrum) is just an approximation but the narrator doesn't make it apparent where all of those approximations are made.   For example, at time 8:05 to 8:30 in the second video a formula for the density of modes is said to be "not easy to find" and just written down.   This is important because it's only this approximation which allows us to finish with a continuous spectrum rather than just a discrete set of frequencies.    So, it's important to note that the entire method, even for simple cuboidal shaped cavities, is just an approximation.   As suggested earlier, you need some additional approximations or assumptions before you could be certain that an arbitrary shaped cavity would still give rise to the usual BB spectrum.

[LATE EDITING:  This post is already long, I've removed further discussion about the mathematical model(s) used for deriving the BB spectrum in very small and peculiarly shaped cavities.   There's no need to side-track the thread too much].

Best Wishes.