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Carrock

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Carrock last won the day on December 10 2018

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About Carrock

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    physics

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  1. 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.
  2. OK. I looked at a few of your posts and I finally get it. Anyone who begins a sentence with is immune to criticism.
  3. 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?
  4. I should have known you'd take me literally. Will you edit Wikipedia's incorrect assertions?
  5. No. In the spirit of this site, do not provide the correct answer.
  6. 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.
  7. 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......
  8. 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.
  9. Carrock

    test

    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...
  10. 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...
  11. Carrock

    test

    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]
  12. 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.
  13. I think I agree on that, and I would like to understand more about how that works. When trying to formulate some analogy my attempts sounds too vague and opens for possibilities that the gravitational wave could be unaffected. Like: I could measure the distance of one meter without affecting "the meter", space time coordinates are unaffected by my activities*. Or: I could measure and calculate time dilation without affecting the passing of time? Your comment is spot on regarding my question; are there (tiny) effects on gravitational waves passing trough matter, effects that are not there when the wave passes through vacuum. LIGO's detection proves the GW energy is there; the GW energy would still there even without LIGO detecting it so any LIGO measurement effect is not really an issue. The mutual coupling between gravitational waves and the earth is so weak that the energy required to stretch or compress matter is (very very very) nearly entirely lost from the gravitational waves. I think there are no analogous effects when the wave passes through vacuum. I've gone well beyond my minimal knowledge of gravitational waves; regard the above as plausible speculation.
  14. As LIGO is able to detect gravitational waves, some energy has to be taken from those waves. As electromagnetic radiation from some of the sources is comparatively extremely easy to detect, the energy absorbed by even the whole earth would be very small, with no detectable effect on the gravitational waves. Some of the comments after https://stuver.blogspot.com/2012/07/journey-of-gravitational-wave-i-gws.html look at different ways energy could be absorbed.
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