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Carrock last won the day on December 10 2018

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  1. Your unattributed quote is from a blog written by a Psychiatric Nurse Practitioner and posted by Janet. Not necessarily a reliable source. X-posted : Charony may have found a different reference. Edit: many different possible sources, but not, it seems, your original references.
  2. From https://www.physiology.org/doi/full/10.1152/japplphysiol.01133.2006 That's the only occurrence I could find of “advertisement” in the reference. From https://news.virginia.edu/content/biomedical-engineering-study-demonstrates-healing-value-magnets If that doesn't convince your pseudoscience friends I doubt any actual science will either.
  3. H spins can normally be easily flipped: Isospin monatomic hydrogen is interesting and unique in that two atoms cannot combine into H2 since with isospin electrons that combination is energetically unfavourable due to Pauli exclusion. Let it warm up a little and some spins will be thermally flipped; heterospin H atoms will combine into H2 with enormous energy release... H2 atoms have no net spin so spin can't be flipped. Offtopic, liquid isospin monatomic hydrogen is so chemically unreactive that it is supposed (unlike helium) to be impossible to freeze. Multielectron atoms which would normally be diatomic can't be made stable in this way. e.g. isospin fluorine atoms would have plenty of opposite spin electrons for interatomic interactions. I read (most of) that in Scientific American years ago and haven't been able to find a good reference since, but it all seems very plausible..... An earlier Eagle comic had a more limited discussion, with Dan Dare accidentally blowing up a few spaceships, which inspired my interest.
  4. As all I had to go on with Cuthber was 'you are wrong' I attempted to concentrate on the thing I felt sure was correct, the inability of a Faraday cage to prevent an applied D.C. voltage from appearing on its inner conductive surfaces. The voltage etc of any insulated object in the cage is irrelevant. Against John Cuthber any hint I might have made a mistake elsewhere (as I realised I had after my first post or two - I should have only mentioned voltage, not fields) would be a rhetorical blunder. The Cuthber/Carrock/Swansont part of this thread is a rhetorical debate, not physics. I'm not sure of the etiquette of quoting someone who changes the meaning when he quotes himself so I'll leave this quote as John Cuthber wrote it save with the original struck out. I was not surprised John Cuthber did not specifically say I was wrong on any of the issues you mentioned as I have learned his authority is such that his opinions are not open to challenge and I was not going to mention anything more about those issues. 'A voltage is not a field, so there’s an issue with this' Not the outside field. IOW, it blocks DC as well.' So the OP should have asked about fields and currents, not voltage as there’s an issue with this./s I note you twice mention current being blocked but not voltage. The clear implication is that John Cuthber is right with his deniable claim that e.g. constant voltage is blocked from the (conductive) inside of a Faraday cage. 'The field inside a conducting sphere depends only on the charge inside the sphere. Not the outside field. IOW, it blocks DC as well.' A bit ambiguous. Anything non conductive inside the sphere blocks D.C.. If there's a constant voltage across the sphere I'd expect some direct current on the inner conductive surface of the sphere, dependent on the magnitude of the internal field. I don't see how the internal field would prevent such current flow. Your (correct) selective facts are much more effective than Cuthber's empty rhetoric.... You can ignore my report of Cuthber's last post as it was clearly pointless. My rhetorical skills are clearly inadequate for this forum so I'm taking an indefinite sabbatical. +1 to Cuthber and you for his win.
  5. OK. I looked at a few of your posts and I finally get it. Anyone who begins a sentence with is immune to criticism.
  6. No. Does wiki say this a Faraday cage blocks changing fields, and constant voltage ? No. I'm not familiar with the concept of blocking/not blocking constant voltage in a conductor. I can live without this knowledge. Does wiki say this? Yes. Fun/scary video. You can see the arcing at about 16s as the linesmen connect themselves and the helicopter to the live high voltage line. As the linesmen are wearing Faraday cages and the frequency (60Hz) is too low for conditions inside the cages to differ much from D.C., I presume you claim the voltage inside their cages is pretty near zero. Ask one of them to drop an insulated wire from inside her cage to you on the ground. Good luck persuading her. Linesmen are so conservative and overcautious that not one living linesman has ever tried this experiment. Three posts in this thread and you've provided no useful information. Why bother?
  7. I should have known you'd take me literally. Will you edit Wikipedia's incorrect assertions?
  8. No. In the spirit of this site, do not provide the correct answer.
  9. As a Faraday cage only blocks changing fields, not constant voltage, what is the point? Why not just apply the voltage to a wire or capacitor? That is, er, potentially just as lethal.
  10. In the good old days, someone talking with an object as though it was a person would risk being tried as a witch. Slightly more seriously, anyone who doesn't use Siri and completes the questionnaire is lying, which may devalue your results slightly......
  11. Probably they were thinking of this: https://en.wikipedia.org/wiki/Water_memory IMO this is a classic example of research which cannot or should not be reproduced.
  12. Carrock


    Test of Mathpix Snipping Tool... Select a partial screenshot, paste from clipboard.... From my last post here S=N k\left(\ln \left(\frac{V_{x}}{N}\left(\frac{4 \pi m U}{3 N h^{2}}\right)^{\frac{3}{2}}\right)+\frac{5}{2}\right) ie. [math]S=N k\left(\ln \left(\frac{V_{x}}{N}\left(\frac{4 \pi m U}{3 N h^{2}}\right)^{\frac{3}{2}}\right)+\frac{5}{2}\right)[/math] From this image at https://www.scienceforums.net/topic/117535-probability-interpretation/ [math]\begin{array}{l}{P_{1}\left(E_{1}\right)=\left|\left\langle\phi_{1} | \psi\right\rangle\right|^{2}=\frac{3}{5}} \\ {P_{3}\left(E_{3}\right)=\left|\left\langle\phi_{1} | \psi\right\rangle\right|^{2}=\frac{3}{10}}\end{array}[/math] and [math]\begin{array}{l}{\text { Why not }} \\ {P_{1}\left(E_{1}\right)=\left|\left\langle\phi_{1} | \phi_{1}\right\rangle\right|^{2}=?}\end{array}[/math] Pretty good for free software...
  13. I regard this as a definition issue. I had assumed that, as in many other fields, thermodynamics could have two or more subsystems comprising one system, but all I could find was systems. From https://en.wikipedia.org/wiki/Thermodynamic_system I'll refer to your diagram and labelling to avoid confusion. The system A (i.e. the contents of chamber A) has a volume of x cubic units. The system B (i.e. the contents of chamber B) has a volume of y-x cubic units. The system C, which is the contents of chamber A plus the contents of chamber B, has a volume of y cubic units. System A contains ideal gas in equilibrium at a temperature [math]T= \frac {2U}{ 3Nk}[/math]. System B is a vacuum. Just before the start of my scenario the barrier between system A and system B is almost instantaneously removed and taken outside both systems with no significant effect (at that instant) on any of the three systems. More plausible scenarios for this action can be devised. Classically, such things can be done with an arbitrarily small effect on the system. Without the barrier, at the start of my scenario, system A, still instantaneously in equilibrium, has entropy [math]\displaystyle S = Nk \Bigg(\ln\bigg(\frac{V_x}{N}\Big(\frac{4 \pi mU}{3Nh^2}\Big)^\frac{3}{2}\bigg)+\frac{5}{2}\Bigg)[/math] System B has entropy [math]0\frac{J}{K}[/math]. As entropy is an extensive property, the entropy of System C is the sum of the entropies of system A and system B. Do you agree? Ignoring intermediate steps for now, system C eventually reaches thermal equilibrium. Its entropy is [math]\displaystyle S = Nk \Bigg(\ln\bigg(\frac{V_y}{N}\Big(\frac{4 \pi mU}{3Nh^2}\Big)^\frac{3}{2}\bigg)+\frac{5}{2}\Bigg)[/math] The change in entropy is [math]\Delta S = nK \ln\big(\frac {y}{x}\big) [/math] Its temperature is the same as system A's original temperature i.e. [math]T= \frac {2U}{ 3Nk}[/math], since neither U nor N has changed. No net work has been done. Intermediate steps: and You seem to be saying that if e.g. the left wall of the chamber was rigid, adiabatic and movable, entropy increase would happen since compressing the gas from volume y back down to volume x would be possible i.e. reversing the process; if the left wall is not movable the expansion process is IMO unchanged but entropy cannot increase since the process is irreversible.... From your source http://www.splung.com/content/sid/6/page/secondlaw i.e. since my example is neither perfectly efficient at producing work nor a closed thermodynamic cycle entropy increases. IMO some of the above quotes are inconsistent with this: I'm not sure if all these are your views or partly representation of Timo's views. I'm avoiding quoting Timo since we seem to agree with him but draw different conclusions. The above quote is arguably inconsistent with I took that a bit casually; of course the gas cooling was not work. All work done ends up becoming heat as the system approaches equilibrium. In short, the system evolution is not in practice describable with any accuracy but the entropy increases monotonically from the initial state to the final equilibrium state. More I could say but not now...
  14. Carrock


    inline [math]\mathrm{[math][/math]}[/math] test newline [math]\displaystyle S = Nk \Bigg(\ln\bigg(\frac{V_x}{N}\Big(\frac{4 \pi mU}{3Nh^2}\Big)^\frac{3}{2}\bigg)+\frac{5}{2}\Bigg)[/math] newline [math]\Delta S = Nk \ln (\frac{y}{x}) [/math] [math]\displaystyle \Delta S = Nk \ln\big(\frac {y}{x}\big)[/math]
  15. I'll respond to this and the previous post eventually (tomorrow at earliest). I have had limited time, combined with getting back into latex (inspired by comments from you a few months ago); there's some good looking perfectly safe free software but sorting out dependencies etc is a pain and I've been putting it off.
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