User talk:Sbyrnes321

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Your transmission line animation

It's great.

I'd like to suggest an animation of a pulse reflecting off an open termination would be very interesting.Constant314 (talk) 22:01, 4 August 2012 (UTC)

Thanks! The pulse is a good idea ... I put it on my list of things to do someday. --Steve (talk) 23:18, 4 August 2012 (UTC)

--- I loved it too - Nice work ! — Preceding unsigned comment added by Fentlehan (talkcontribs) 19:21, 8 April 2013 (UTC)

And I love it, too :) --Raphman (talk) 01:41, 6 August 2013 (UTC)

I disagree with you revert at Fourier transform infrared spectroscopy in Conceptual introduction

I really understand what you mean but the problem is that viewing a Michelson interferometer as a wavelength filter is not right at all. In fact, that is how a spectrometer works. More over the usual mistake about the Michelson interferometer is to compare it to a spectrometer which it is not. You must know that all wavelengths are passing through a Michelson interferometer and that what you see on one of the output ports is the sum of the interference pattern of each wavelength for a certain optical path difference. I know that the way a Michelson interferometer works is not easy to understand but it may be better to avoid false description (I am sorry for my poor english). This is a description taken from Michelson interferometer : "The Michelson interferometer's detector in effect monitors all wavelengths simultaneously throughout the entire measurement, increasing the integration time and the total number of photons monitored"

Oh, I already wrote a message at Talk:Fourier_transform_infrared_spectroscopy#Conceptual_introduction ... I will copy this message and we can keep talking there :-)

Your Angular frequency animation

It too is great! Out of Phase User (talk) 19:32, 26 April 2013 (UTC)

Thanks!! :-) --Steve (talk) 20:39, 26 April 2013 (UTC)

On electron holes

I saw your stackexchange writeup linked from Talk:Electron hole -- nice explanation! It's surprising how common the misconception is (e.g. [1] "Electrons in the valence band also move in the same direction."). I looked for a similar textbook explanation to yours and actually didn't have to look far: Kittel's Introduction to Solid State Physics has something very close, just more technical. So, I think you can go ahead and make the edits without being guilty of WP:OR. :) --Nanite (talk) 23:03, 24 May 2013 (UTC)

Thanks, I just did. --Steve (talk) 16:13, 7 June 2013 (UTC)

Maxwell's Equations : Limitations

If you read "Quantum Mechanics , 3rd edition"; Eugen Merzbacher; 1998; Chapter 23: Photons and the Electromagnetic Field it is demonstrated that a Hermitian operator (A) may be constructed from the photon creation and annihilation operators. This operator satisfies the Coulomb gauge condition and the wave equation. Further, the electric potential (E) and magnetic potential (B) may be constructed from (A) as it usually is for the electromagnetic field. With (E) & (B) defined this way a simple matrix product factorization of the wave equation establishes Maxwell's equations, where (E) & (B) are differentiable. The dynamics of interacting fields must be analyzed via these defined operators due to the non-differentiability. I feel expressing the situation this way is a more correct expression, than to simply call Maxwell's equations approximations. Cloudmichael (talk) 11:26, 4 July 2013 (UTC)

I agree, you can find QED equations that are written like ∇⋅E=ρ, ∇⋅B=0, etc. The question is, are they still "Maxwell's equations"? When people describe Maxwell's equations in classrooms, 99% of the time they're talking about equations like "∇⋅E=ρ where E is a vector at each point and ρ is a real number at each point". Now you're writing down an equation that looks the same, ∇⋅E=ρ, but this time E is a triplet of operators at each point and ρ is an operator at each point. If you say "This is also Maxwell's equations", well, you're entitled to say that and I admit that people do often say that. But for most people -- who don't know the big-picture relationship between classical and quantum mechanics -- they could be easily misled by the statement "This is also Maxwell's equations". The symbols are the same, but they're not the same equations because the symbols mean something different.
I think it's worth emphasizing that "∇⋅E=ρ where E is a vector at each point and ρ is a real number at each point" is an approximation, not an exact description of reality, because my experience suggests that a lot of people are really surprised to learn this and it's a very important thing to know.
I think "Maxwell's equations are an approximation" is a pretty good way to succinctly say "It's an approximation to say that ∇⋅E=ρ where E is a vector at each point and ρ is a real number at each point", although I'm sure there are other wordings that are also good.
I think the equations like "∇⋅E=ρ where E is a triplet of operators at each point and ρ is an operator at each point" are important equations (whether or not we call them "Maxwell's equations") and I think it's an excellent idea for you to add some text to the Maxwell's equations article to tell people that these equations exist. Again, I think it would confuse people to say "These are also Maxwell's equations", but a slightly different wording would undoubtedly be OK. Maybe "Quantum field theory equations that superficially resemble Maxwell's equations" :-P Anyway, with the appropriate wording, I think it would be a valuable addition to the article to say something like you wrote above, and I hope you do so :-D --Steve (talk) 12:48, 4 July 2013 (UTC)

Questions about File:ResistanceHydraulicAnalogy2.svg

I've been looking at this illustration, and am wondering if either it needs adjustment or if my understanding of the nature of resistors needs adjustment (the latter is more likely, but let me explain). If a resistor, big or small, acts like a hair clog in a drain, then additional pressure on the clog from higher-pressure water (more voltage) wouldn't result in any change (necessarily) to the hair clog itself: if the clog holds or slows down water flow at high pressure, it also does this at low pressure; if it slows it at low pressure but not high pressure, water (voltage) passes through the pipe at a rate higher than the resistor can "resist." But this doesn't seem to take into account what happens to the water which gets blocked by a clog. It is my understanding that one of the things that resistors do is they heat up in response to voltage passing through them, and that the voltage "passes" through the system regardless of the presence or absence of a resistor but some of it can be "drained" or "siphoned off" by the resistor and turned into heat. What I am getting at is that the analogy as you have shown it suggests that the "water" that is being blocked by the clog ends up just being conserved, like in a tank, possibly for some alternative use, but that in an electrical circuit, a resistor doesn't actually "save" the excess current, it takes it and converts it into heat (which is technically wasteful though often necessary). Do you think the illustration could be modified to reflect this conserved-vs.-wasted concept or (drum roll) am I not getting something about the nature of electrical circuits? I will be looking forward to your response. Thanks! KDS4444Talk 13:04, 15 December 2013 (UTC)

Well I don't quite understand everything you're asking. See if this helps:
  • In the analogy, water flows through the pipe. Every water molecule above the pipe winds up below the pipe, none of it is "siphoned off". This is a good analogy: The flow of water is analogous to the flow of current. In real electric resistors, the current flowing into one side of the resistor equals the current flowing out the other side of the resistor. No current is turned into heat.
  • In the analogy, there is a high water pressure on the top side of the pipe, and low (actually zero) water pressure on the bottom side of the pipe. The pressure difference pushes water downward through the pipe. This is a good analogy: Voltage difference pushes electrons through a resistor.
  • In the analogy, a pipe will always slightly heat up when water is pushed through a clog. The amount of heating is really really small, too small to measure, and very few people are aware of this obscure trivia about water flow. This is an OK analogy: In real electric resistors, when current flows through them, they heat up, but that fact usually doesn't matter. I mean, 99% of the time, a resistor in a circuit heats up a negligible amount and you don't think about the fact that it is heating. The other 1% of the time, the heating might be very important -- maybe the resistor is a fuse, or an electric stove, or a thermistor, or creates so much heat that it can start a fire. Anyway, I would say that the analogy is OK but not perfect in the category of Joule heating.
  • In the analogy, there is a big pile of water that accumulated above the clog, and which is generating the pressure (through gravity, i.e. the water's own weight pushing down). That's not a very good analogy: In electric circuits, voltage drop is not normally generated by a big pile of electrons, and certainly not through the weight of electrons being pulled down by gravity! (Capacitors sort of have a big pile of electrons, but it's different.) A closer analogy would be to show a water pump pushing water through the clogged pipe, in a closed loop; this would be analogous to a battery + resistor circuit. Unfortunately, most people have never seen a water pump and don't understand what it does, in my experience. So the "big pile of water generating pressure through gravity" is probably the clearest way to show the concept of "water pressure". (Ask a random person "What is water pressure", and you'll probably find that they have only a vague idea, even though they have heard of it.) Even though this part of the analogy is not very good, I think it's OK ... I think an analogy can be helpful without being faithful in all of the details.
If I haven't answered your question yet, can you ask it again? Thanks! :-D --Steve (talk) 16:44, 15 December 2013 (UTC)
Steve, I follow your talk page because interesting discussions occur here. I hope you don’t mind me adding a comment. I think KDS4444 has confused current with energy.Constant314 (talk) 18:37, 15 December 2013 (UTC)
Oh dear. I may have done just that. I have no formal training in electronics (trying to teach myself what a MOSFET does is leaving me dazed and confused so far) but I thought I had a pretty good understanding of resistors. Your explanations are very clear, thank you, and I even already had the concept of the infinitesimal amount of heat generated when water pressure enters a pipe (2 point for me!). So let me go to a real-world example and see what you think:
I have a string of tiny LED Christmas tree lights in part of a holiday display at my house. They came battery-operated, but the batteries never lasted more than a day or two and I wanted these mini lights to run for more than two months. So I hooked them up to a wall socket, and put a DC transformer in between. I did not know what the collective voltage drop of all of the LEDs was, because they didn't exactly come with a data sheet, so I had to improvise on resistors, making sure I never allowed too much current to run through them and ruin them all in a flash. My power supply was/ is 12V.
Of course, they were running on 2 AA batteries before, only 3 volts, but I did not know how many amps. The configuration that I ended up using involves a series of I believe it's five 27Ω 5-watt resistors in series, and all of those resistors have to be hooked up to an aluminum heat sink to make sure they don't overheat! And they are already 5 watt resistors! But this combination allows me to run the tiny LED lights without making them burn out.
Now, I realize at this point that what I should have done is gotten a power supply with a lower voltage, and then not needed to use so many high-watt resistors. But they illustrate the confusion I am having: my understanding has been that it is because the resistors are so large that they are able to handle "resisting" large voltage drops, and that they do this by converting the undesirable energy from the power supply into heat which, if substantial, must then be heat sunk away or the resistor could blow. But you tell me that resistors do not get hot as a result of resisting... Then how come mine are all so warm? I do not intend for them to be, but I am glad that my lights work and I don't plan on messing with my lighting system again this year (maybe next year). I started wondering if the analogy to the water tank and the clog might not be taking into account this "extra" heat being shed by my numerous resistors, and if the analogy should either be expanded or my understanding corrected. All analogies are flawed if you push them far enough, I know, but I like the water analogy for many reasons and wanted to see if I could fit my own understanding/ experience into it somehow. I understand much of what you are saying, and had a pretty good grasp on it before I contacted you (just don't ask me anything about MOSFETs)— how do my "hot, high-watt, low-amp" resistors fit in? Is this an "energy" problem and not a "current" one? Does the analogy fail already? Am hoping not. KDS4444Talk 21:25, 18 December 2013 (UTC)
Certainly resistors can get hot if you run current through them (P = I2R). I'm not denying that!
Incidentally, if your description is accurate, each resistor should be dissipating at most 0.21W, only 4% of its rated power. I find it surprising that they were getting dangerously hot ... But anyway it never hurts to use a heat sink!
I think it's problematic to imagine that the voltage supply is the source of a certain amount of energy or power, and then you analyze the circuit to decide how that energy is partitioned. The amount of energy is not fixed! For example, if a 9V battery is sitting in a closet, not connected to anything, then the energy is zero. Likewise, if you took your christmas lights setup and drastically increased the resistance of the resistors, it's not correct to say "the resistor would turn even more of the undesirable energy into heat". There would actually be less energy in the first place ... The resistors would be colder, not hotter.
When people analyze circuits, either intuitively or mathematically, they figure out the voltages and currents, not directly the energy flows. (E.g., There's no Kirchhoff's law for energy.) You can figure out the energy flows later on (if you're curious), by multiplying a voltage drop by a current.
The better analysis is to say that the voltage supply is the source of a certain amount of voltage, and think through how the voltage is partitioned. If you choose the resistance just right, 9V out of the 12V will drop across the resistors, leaving the desired 3V across the christmas lights. Yes, the resistors will heat up when you do this, but that's just a side-effect -- it's the 3V consequence that is most important. Your circuit turned 75% of the energy into heat, but a more complicated circuit might be able to put 3V across the christmas lights while turning only 5% of the energy into heat (i.e., if you make a DC voltage transformer). So once again, the heating of the resistors is a side-effect, not an essential goal.
Hope that helps ... As before, I'm sorry if I'm misunderstanding you :-D --Steve (talk) 01:19, 19 December 2013 (UTC)
No, no, I think you understand me just fine. I think it is me that is having the difficulty. I have read over your comments above, and on first read I still am certain that I need to, er, read it again (because for one thing I don't know what Kirchoff's law is... yet... though thank goodness for Wikipedia, eh?). Then I will need to rethink how these fat resistors of mine are really operating. And then, if I still have questions, I will get back to you. Thank you for taking the time to try to explain all of this to me! Much appreciated (and to quote Homer Simpson: "There's so much I don't know about astrophysics.") KDS4444Talk 10:54, 19 December 2013 (UTC)







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