User talk:Vze2wgsm1

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Hello, Vze2wgsm1, and welcome to Wikipedia! Thank you for your contributions. I hope you like the place and decide to stay. Here are some pages that you might find helpful:

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Your edits

Your edits to capacitance, LC circuit and Meissner effect make little technical sense to me. Please discuss them first, or ask questions at the ref desk if you are not sure. SpinningSpark 16:45, 10 June 2009 (UTC)

Today (20-July-09) I accidently deleted everything from this discussion page, without reading it. Please resubmit anything you may have sent.

Electric potential

I seem to have given you the impression that I require you to give me personally a preview of your proposed edits. My apologies, that is not what I meant, I have no more authority than any other editor. What I meant was to post the proposed changes on the article talk page so that any interested editor could comment. If the material is lengthy, it is better to construct it in your own userspace and just provide a link on the talk page to avoid cluttering it up. Anyway, as you have asked me, here are my comments;

• Most of the material is going off-topic for the electric potential article, only a small part of it belongs there. The rest belongs in capacitance and LC circuit as far as I can tell. Also, you should not wikilink the article the material is to be inserted in.
• The whole piece is totally devoid of any references. This is essential for anything on Wikipedia.
• Within a charged capacitor, a minus charge (q − ) has the same potential energy as a plus charge (q + ). Therefore, Potential energy ($\phi _{\mathbf {E} }$) is the same within both capacitor plates. This is very confused: first of all, the energy resides in the electric field, not in the plates. Secondly the implication of that statement is that the total PE should be taken as 2qV which is four times the correct answer of ½qV.

1. You do realize that the negative half of a capacitor contains half the total charge and half the total voltage. Therefore:

$PE={\frac {1}{2}}q_{\mathbf {t} }V_{\mathbf {t} }$.

Where is the implication that PE = 2qV?

2. I suggest that you call $\phi _{\mathbf {E} }$ electric potential ($\phi _{\mathbf {E} }={\frac {U_{E}}{q}}$) instead of potential energy ($U_{E}$ )

3. You claim “all the energy resides in the electric field, not in the capacitor plates”. Ok, I agree, there is no electric field within the capacitor plates.

In accordance with http://en.wikipedia.org/wiki/Electric_field, a capacitor’s electric field is limited to a Coulomb field, because a capacitor does not contain a time-varying magnetic field. In other words, elementary charges cause a capacitor’s electric field. Should I also agree that the capacitor plates do not contain concentrations of charges?

• Capacitor energy versus separation of charge. The phenomenon you seem to be describing there is called self-capacitance and belongs at that location. The capacitance formula you give there is only valid for a parallel-plate capacitor with no fringing. This is only approximately true if the plate dimensions are large compared to d. As the plates are moved apart fringing starts to dominate, the field eventually becoming a closer approximation to a dipole than parallel plate and that formula becomes totally invalid.

The existing equations for a two-plate capacitor are Ok for a capacitor with small plate area and a large distance between plates. The energy required to charge a capacitor with a low plate area to distance between plate ration corresponds to capacitance, not to dipole energy.

For example: Both the grid-cathode and plate-cathode of a vacuum tube triode function as a capacitors. Albeit overlapping plate-cathode area is much larger than grid-cathode area, the nearby grid greatly influences electron flux from thermionic emission. If the grid has a fringe field, the fringe field enhances grid-cathode capacitance, instead turning the grid-cathode into a dipole.

• Relative polarity of magnetism in the coil and voltage in the capacitor (not amount of capacitor voltage) controls direction of current flow. I can't make any sense at all of this entire section, but just picking up on that point, the current in a capacitor is entirely determined by the rate of change of voltage across it regardless of anything else happening in the external circuit.
• I really cannot make out what your point about superconductivity is, I don't think it is relevant. You certainly need to cite a reference for any statement there.
• Therefore, moving a coil through a magnetic field can cause current in the coil w/o increasing electron kinetic energy. If electrons are moving they must have kinetic energy by definition. You need a reference for this statement and also need to explain how this relates to electric potential.

The following equation describes the above relation between voltage and current in an LC circuit.

$i_{C}(t)=C{\frac {dV_{C}}{dt}}.\,$

Voltage stops changing into magnetism twice during a 360-degree oscillator cycle. Logically, at these points, voltage and magnetism should be at equilibrium and magnetism should not convert back to voltage. In reality, magnetism begins converting back to voltage.

The complete exhaustion of voltage or magnetism before polarity reversal raises questions about the role of electrons in voltage and magnetism. If voltage represents pressure, then a capacitor is equivalent to 2 gas containers, one with high pressure, the other with low pressure. Connecting the two containers together with a tube will result in equal pressure in both containers. Unlike voltage, high pressure does not oscillate between the two containers.

SpinningSpark 10:12, 26 July 2009 (UTC)

Plane/Plane mirrored lasers

Hi Vze2wgsm1,

Thanks for your input on Srleffler's talk page. My own experiments with lasers seems to indicate that curved mirrors actually cause an increase in the gain. I assume this is because more photons become "trapped" within the resonance cavity and the different ray paths cause less spatial hole burning, allowing greater stimulated emission. (With my mechanical way of thinking, I often find it easier for these discussions to think of light as photons moving in ray paths instead of waves, although I know it's rather simplistic.)

What I've noticed about a laser with plane mirrors is an increase in coherence, a decrease in divergence, and an extremely narrow bandwidth at a cost in the overall gain. I believe the low divergence is due to the fact that only the photons transversing at exactly a 0 degree incidence angle stimulate the emission. All other photons simply "walk" off of the beam path. The narrow bandwidth and good coherence is probably due to the fact that, with the standing wave generated, the only photons that can resonate have a wavelength exactly some 1/2 multiple of the cavity length. As I understand it, the wavelength can actually be tuned within a very narrow margin by adjusting the cavity length, say, like with a piezo, (which is not possible with curved mirrors, which tend to produce that entire narrow margin). A high degree of spatial hole burning and diffraction losses probably cause the loss in gain

But I'm no physicist, and beyond flashtubes my knowledge begins to drop off, so I found your input to be very insightful. Thanks. Zaereth (talk) 22:14, 27 October 2009 (UTC)

Desaneis stimulated emission rules

1. Standing waves can increase the efficiency of stimulated emission between the parallel mirrors of a CO₂ laser.

The laser’s output wavelength can be tuned by adjusting the cavity length

A distance change equal to one wavelength between a laser’s mirrors does not normally change the number times non-resonant light waves traverse between mirrors.

2. Stimulated emission within standing waves requires molecular gain media.

Molecules contain the energy that can convert to photons. Photons within a vacuum do not add or subtract energy from each other.

Any gap in media path is a wave boundary. Gap creation is equivalent to the creation of an open circuit within a superconductive magnet (resulting in magnetic quench). Think of a gap in this way: An antenna is the gain media of RF. An open circuit within the antenna would cause the antenna to lose resonance.

3. Optical amplification within a monochromatic laser results when atoms release electromagnetic energy without absorbing electromagnetic energy with the same wavelength.

An excited atom that absorbs a photon cannot emit two photons, each with the same energy. This violates the Pauli Exclusion Principle. Receipt and emission of a single photon is not optical amplification.

Photon emission from an atom is a decay product and the time between receipt and emission of a photon would follow exponential decay rules. During this time, the atom could lose all memory of the direction of the impinging photon.

— Preceding unsigned comment added by Vze2wgsm1 (talk • contribs)

Discussion

Rule #1 doesn't take into account optical amplifiers, which don't have any mirrors. An optical amplifier is a laser gain medium with no mirrors. It doesn't lase, but if you pass a laser beam through it the beam is coherently amplified by stimulated emission.

Rule 1 is not even true in lasers. A laser will only lase at wavelengths where a standing wave can occur, but stimulated emission occurs at all wavelengths where there is gain and photons. This is important in the theory of how a laser starts up. If you turn on a laser, the gain medium begins amplifying spontaneously generated photons in the cavity; these generally have completely random direction and wavelength. Stimulated emission amplifies whatever photons are there; the wavelength selection is a function of the cavity, which acts as a filter selecting photons with certain wavelengths and directions.--Srleffler (talk) 06:51, 4 November 2009 (UTC)

I poorly worded the first rule, therefore I revised the rule to exclude unintended phenomenon. I was trying to say that a standing wave (caused by parallel mirrors) receives more amplification than a similar traveling wave. Vze2wgsm1 (talk) 19:29, 5 November 2009 (UTC)

Regarding rule 3: I agree with the rule, but note that generally atoms can absorb a photon and emit two photons with energy totalling the energy of the one that was absorbed. When we pick gain media for lasers, we choose atoms that won't do this for the wavelength at which the laser will operate. Or were you just trying to say that an atom can't absorb a photon and emit two that each have the same energy as the original? Obviously that is prevented by conservation of energy. The Pauli Exclusion Principle does not apply to photons.--Srleffler (talk) 06:55, 4 November 2009 (UTC)

By analogy, creating a closed circuit instantly causes conversion of chemical energy within a battery into electrical energy (w/o absorption of energy).

A light wave is a media that FACILITATES conversion of excess energy within the atom into a photon(s). An excited atom cannot absorb a photon and emit two that each have the same energy as the original. Vze2wgsm1 (talk) 19:29, 5 November 2009 (UTC)

Thanks for the vote of confidence on Srleffler's talk page, as I find this conversation very interesting. I should state here that lasers to me are just a hobby, and other than the two experimental results which I stated above, most of that is mere conjecture on my part, trying to reason why.

By checking out a few sources, it appears that Srleffler is correct. According to the book Laser Fundamentals, a laser with curved mirrors exhibits a much, much greater gain due to fewer diffraction losses. The book Principles of Lasers describes this further in great detail, using both natures of light, ray and wave. To look at it from a ray pont of view, with a flat mirrored cavity, the beam itself is comprised of only those photons which strike the mirror at a normal incidence angle. But photons are traveling in all directions, stimulating emission as they go. This leads to a very high degree of "scatter", especially from those photons than can still transverse the cavity multiple times, say by reflecting off of the mirrors in a zig-zag path. Flat mirrors do not contain these slightly off normal reflections, so a great deal of the stimulated emission is lost in the scatter. What remains of the beam is not much, but the beam is highly collimated and exhibits very low divergence.

Curved mirrors tend to "trap" most of that scatter to be used to stimulate even more energy. The drawback is that these often exhibit a high degree of divergence and need to be lensed or collimated to achieve a straight beam. (Collimating is usually better, which consists of expanding the beam with a lens, focusing it with another lens, and straightening it with a third.) There is also a great difference between a spherical cavity, a focal cavity, a concentric cavity and a confocal cavity. (According to Principles of Lasers, only a flat mirrored cavity is referred to as Fabry-Perot.)

There are some flaws in your rules. Stimulated emission does not always require molecular gain media, but often atomic. A laser can operate without standing waves, which ring lasers do, which eliminates spatial hole burning and the wavelength dependancy on length. Stimulated emission is not an atom absorbing one photon and emitting two. It's more a case of an atom getting ready to emit just as a photon passes very close by, causing the emitted photon to pick up the same wavelength and direction as the passing one. Zaereth (talk) 18:57, 4 November 2009 (UTC)

I should have thought of ring lasers. They are a perfect counterexample to rule 1—no standing wave at all, just a travelling wave. And yet they lase, and at discrete frequencies too.--Srleffler (talk) 23:37, 4 November 2009 (UTC)

Wave propagation

The Wave propagation article seems to indicate that peaks and valleys of electromagnetic energy move in the direction of propagation. The following indicates the possibility of a different conclusion.

A consequence of movement of peaks and valleys of RF intensity during propagation is that once emission begins at a point along its travel direction, emission must continue at that point. Release of RF energy that has not yet reached that point is premature. Note: emission from a single point on an antenna removes dependence on antenna length.

At each (external) destination point that receives the above traveling wave, RF will have a (sinusoidal) variation of intensity that varies with time. Therefore, average intensity at that point and at all points along the distance along the direction of propagation would be constant, albeit instantaneous intensity per distance would be sinusoidal.

Similarly, if a light wave’s peaks and valleys move toward propagation direction, the human eye would not be able to detect an interference pattern. The eye would integrate the intensity levels at each point along the length of the interference pattern. The interference pattern from two travelling waves would only be detectable if the frequency of detection snapshots is greater than the RF frequency.

The human eye’s ability to detect coherent light’s interference patterns indicates that light waves have a stationary sequence of intensities along the direction of propagation. In other words, each sine wave is stationary (does not travel in the direction of propagation). Coherent RF is equivalent to coherent light. Peaks and valleys remain stationary along propagating distance.

Stationary waves do exist within an RF antenna and within coax during propagation of RF. An apparatus that can show voltage levels along (the equivalent of) a length of coax is located at Bird Technologies in Cleveland Oh. A voltage probe can slide along the center wire through a slot cut into the outer shield. When the probe is stationary, voltage is constant. As the probe slides along the cable, voltage per distance forms a sinusoidal pattern that corresponds to (RF) wavelength.

Note: stationary sine waves in antennas and coax differ from RF voltage applied to a coax cable’s center wire and shield. This voltage must convert to an RF wave before stationary sine waves appear along the length of a cable’s center wire.Vze2wgsm1 (talk) 23:02, 20 November 2009 (UTC)

Sand

My sandbox

Your submission at Articles for creation: sandbox (October 18)

Your recent article submission to Articles for Creation has been reviewed! Unfortunately, it has not been accepted at this time. Please check the submission for any additional comments left by the reviewer. You are encouraged to edit the submission to address the issues raised and resubmit when they have been resolved.

• If you would like to continue working on the submission, go to User:Vze2wgsm1/sandbox and click on the "Edit" tab at the top of the window.
• If you need any assistance, you can ask for help at the Articles for creation help desk or on the reviewer's talk page.
• You can also get real-time chat help from experienced editors.

Origamite^ⓣⓒ 14:49, 18 October 2014 (UTC)

Hello! Vze2wgsm1, I noticed your article was declined at Articles for Creation, and that can be disappointing. If you are wondering or curious about why your article submission was declined please post a question at the Articles for creation help desk. If you have any other questions about your editing experience, we'd love to help you at the Teahouse, a friendly space on Wikipedia where experienced editors lend a hand to help new editors like yourself! See you there! Origamiteⓣⓒ 14:49, 18 October 2014 (UTC)

AfC notification: User:Vze2wgsm1/sandbox has a new comment

I've left a comment on your Articles for Creation submission, which can be viewed at User:Vze2wgsm1/sandbox. Thanks! Origamite^ⓣⓒ 02:27, 4 December 2014 (UTC)

Your draft article, User:Vze2wgsm1/sandbox

Hello, Vze2wgsm1. It has been over six months since you last edited your WP:AFC draft article submission, entitled "sandbox".

The page will shortly be deleted. If you plan on editing the page to address the issues raised when it was declined and resubmit it, simply and remove the {{db-afc}} or {{db-g13}} code. Please note that Articles for Creation is not for indefinite hosting of material deemed unsuitable for the encyclopedia mainspace.

If your submission has already been deleted by the time you get there, and you want to retrieve it, you can request its undeletion by one of two methods (don't do both): 1) follow the instructions at WP:REFUND/G13, or 2) copy this code: {{subst:Refund/G13|User:Vze2wgsm1/sandbox}}, paste it in the edit box at this link, and click "Save page". An administrator will in most cases undelete the submission.

Thanks for your submission to Wikipedia, and happy editing. JMHamo (talk) 11:25, 26 July 2015 (UTC)

May 2017

Welcome to Wikipedia. Everyone is welcome to contribute constructively to the encyclopedia. However, talk pages are meant to be a record of a discussion; deleting or editing legitimate comments, as you did at Wikipedia:Reference desk/Science, is considered bad practice, even if you meant well. Even making spelling and grammatical corrections in others' comments is generally frowned upon, as it tends to irritate the users whose comments you are correcting. Take a look at the welcome page to learn more about contributing to this encyclopedia. Thank you. Jim1138 (talk) 03:46, 15 May 2017 (UTC)

Your submission at Articles for creation: sandbox (August 10)

Your recent article submission to Articles for Creation has been reviewed. Unfortunately, it has not been accepted at this time. The reason left by Theroadislong was: Please check the submission for any additional comments left by the reviewer. You are encouraged to edit the submission to address the issues raised and resubmit after they have been resolved.

• If you would like to continue working on the submission, go to User:Vze2wgsm1/sandbox and click on the "Edit" tab at the top of the window.
• If you do not edit your draft in the next 6 months, it will be considered abandoned and may be deleted.
• If you need any assistance, or have experienced any untoward behavior associated with this submission, you can ask for help at the Articles for creation help desk, on the reviewer's talk page or use Wikipedia's real-time chat help from experienced editors.

Theroadislong (talk) 11:04, 10 August 2023 (UTC)