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Carrock last won the day on December 10 2018

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  1. 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 [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...
  2. 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 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 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...
  3. 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]
  4. 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.
  5. Carrock


  6. 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.
  7. 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 look at different ways energy could be absorbed.
  8. I don't agree with part or all of this but I suppose I can see its relevance for the following post. I tried to indicate I was only concerned with initial and final conditions in my example of how an isolated system can increase its entropy from one well defined value to another well defined value. The total work done in getting from the low entropy state to the final high entropy state, which is an "equilibrium state" where the entire volume y (or A plus B) is in equilibrium, is zero. While transitioning between states work is done. Also the temperature of the gas is lowered and it acquires kinetic energy. As the volume y comes to equilibrium all this work is transformed to heat. At equilibrium the temperature of the gas in the volume y is the same as it was in x.
  9. I thank you also Timo. I reread this thread in light of your post and saw a comment I missed and should have addressed. I disagree with 'must.' It is IMO a very good calculation method. I don't think its impossibility re my isolated system would prevent its use with the working assumption that my system is temporarily not isolated. But it's not the only method. Save mouse wheel.... I used this example for simple maths. The crucial issues. The gas in x and the empty rest of y are each in equilibrium states until the first atom leaves the x volume. I believe the entropy of the entire system can be calculated when each part is in self equilibrium. The following seems to confirm that view. After the side is removed but before any atom has passed the position of the missing side calculate the entropy of the gas still in equilibrium in the volume x. See e.g. (latexphobia.). The entropy of the empty part of y is 0. The subsequent expansion is not isothermal but the final equilibrium temperature of the gas is the same as before it expanded. The new entropy of the gas in volume y can be calculated.(latexphobia.) Entropy change: delta S = nK ln(y/x) The important difference between my and timo's examples is that in mine there is equilibrium only in the initial and final states and there are always equilibrium states in his. IMO this isn't a problem... Definitely agree.
  10. When you change a system how does that become a new system rather than a modified system? You seem not to be distinguishing between a system and its state. This is unconventional and confusing. Would you disagree with "So the modified state of the system (being a list of values of all state variables) differs from the original state of the system?" I specified an isolated system with entropy increasing so that these and some other issues would be irrelevant.
  11. Carrock

    Some Thoughts on Air Conditioning

    Yes. From a brief look the reference you gave is clear and accurate. In it 'temperature' is frequently used but not derived. e.g. So this version, probably edited/simplified for students, of the equipartition theorem can't be used in a definition of temperature. I hope this response is relevant. I don't doubt that you're right and I'm wrong, not least from your work on clocks. Understanding how you're right is still an issue for me. Thanks for your help and patience; I've now reached a point where I have to do some actual reading. I'm now rather dubious about reader friendly Wikipedia, so I'm going to have to look for relevant peer reviewed papers, or at least preprints.
  12. Really? The second sentence is true whether or not the process is irreversible. Entropy has been increased so of course. Did you mean: the original state of the isolated system cannot be restored after the irreversible process has changed the state of the system? Entropy has been increased so of course. Clarify please. I don't see any sense in the second or third sentence.
  13. Carrock

    Some Thoughts on Air Conditioning

    I agree with your post with caveats - see below. I've been looking through Wikipedia and it doesn't exactly help. My problem is how do you tell if e.g. Z1 is in equilibrium with itself and with Z2 without assuming heat flows from hot to cold? An answer seems to be that if Z1 and Z2 are in contact for a long time without changing they are in equilibrium. From But if heat doesn't flow from hot to cold (or from cold to hot) they presumably wouldn't change from whatever weird states they were in i.e. mutual equilibrium would be a meaningless concept. So heat flows from hot to cold seems to be a necessary postulate for the zeroth law.... And from Heat flows from hot to cold assumed..... More Wikipedia : Heat flows from hot to cold is three laws down from the second law?! From Wikipedia is usually reliable but all I can really say is that my knowledge of what I don't know about thermodynamics has significantly increased.
  14. Maybe a misunderstanding? If the original system is in equilibrium, then it can never gain entropy if isolated. In my example the side of the small box was removed 'instantaneously,' taken out of the larger box and the larger box sealed 'instantaneously,' before any ideal gas atoms could escape. That was the original isolated system, which then spontaneously changed itself and increased its entropy. No and yes. Rather than pretend I didn't refresh my understanding, I'll just quote a bit of Wikipedia e.g.: The value of this concept can often be seen during problematic, non cyclic engine startup, before the startup transients die down. Enough for today!
  15. Carrock

    Frequency multipliers

    Most (digital) frequency meters can also be configured to measure time. A divide by 60,000 or more device would at least be simpler than a multiplier. Noise and jitter in the MSF signal would be a problem; a very large division ratio would reduce those but maybe not enough. Getting high accuracy is the big issue whatever you do.