Rob McEachern
Senior Members
Content count
88 
Joined

Last visited
Content Type
Profiles
Forums
Calendar
Everything posted by Rob McEachern

Is "Schrodinger's Cat" the wrong interpretation for Copenhagen Interpretation?
Rob McEachern replied to Elite Engineer's topic in Quantum Theory
It should be even more obvious, that if you change V, V will no longer be constant, as stipulated in the text book quotes, as the defining property of a free particle. 
Is "Schrodinger's Cat" the wrong interpretation for Copenhagen Interpretation?
Rob McEachern replied to Elite Engineer's topic in Quantum Theory
Yes it is. E= (Kinetic energy + Potential energy) = (K+V). So (E  V) = K+VV = K; the potential exactly cancels out in the special case, where V is a constant, regardless of what the value of the constant is. Note also that, given the expression for p, p2/2m = (EV) = K = mv2/2, just as one would expect, for a free particle with only kinetic energy. The momentum cannot change if the particle is free. 
Is "Schrodinger's Cat" the wrong interpretation for Copenhagen Interpretation?
Rob McEachern replied to Elite Engineer's topic in Quantum Theory
Of course  because the potential has no effect, thus, the particle is free. Here is a quote from Merzbacher, "Quantum Mechanics", second edition, pages 8081: . 
Is "Schrodinger's Cat" the wrong interpretation for Copenhagen Interpretation?
Rob McEachern replied to Elite Engineer's topic in Quantum Theory
It is not possible. The solution that you gave and claimed; "Which I think is more useful as you can plug numbers into it." Is not a solution to Schrödinger's equation, if the energy changes. You will find the same thing in any other text book on the subject. The author of the text cited, was David Bohm  a rather highly regarded figure, in quantum theory. 
Is "Schrodinger's Cat" the wrong interpretation for Copenhagen Interpretation?
Rob McEachern replied to Elite Engineer's topic in Quantum Theory
In the book, there is a twostep process to evaluate a square potential: (1) solve the equation when the potential is constant everywhere (Swansont's question), which is done in the first paragraph, and then (2) evaluate what happens when a second level is introduced. But, if there is no second level (as in the question that was asked), then all the rest of the book, after the first paragraph is irrelevant: There is no second level that is less than the first. Potential wells exist when there is a relative change in potential. The partial derivative of kx with respect to x, depends upon both the derivative of k and the derivative of x, with respect to x. The derivation of your solution assumed the derivative of k with respect to x is identically zero  there is no dependency on position. It also assumed that the derivative with respect to time is also identically zero  there is no dependency on time. Hence, k is being assumed to be a constant, right from the start; it depends on neither of the two variables (position and time) in the equation. I fail to see why you find that to be an interesting result; all it says is that a static sinewave exists. How does that revelation enlighten your knowledge of reality? 
Is "Schrodinger's Cat" the wrong interpretation for Copenhagen Interpretation?
Rob McEachern replied to Elite Engineer's topic in Quantum Theory
There can be no well, if the potential is constant everywhere; wells require more than one level of potential. Where there is a well, there can be no free particle. Nothing can change in the solution you gave, because, in the derivation of that solution, "k" was unwittingly assumed to be independent of both time and position. 
Is "Schrodinger's Cat" the wrong interpretation for Copenhagen Interpretation?
Rob McEachern replied to Elite Engineer's topic in Quantum Theory
It does not change in your solution. That solution was derived by assuming the potential is zero, and E is a constant. You can find a simple derivation of that solution, in the last answer given on this page: https://physics.stackexchange.com/questions/347996/whensolvingtheschrodingerequationhowdoweknowwhatfunctionstouseexp Just set the potential V0 = 0 and solve for "k". But note that now, you have left out the "x" in the solution. It should also be pointed out, that the very first step in this "derivation", assumed that the derivative of "k" with respect to x is zero. Think about the consequences of that assumption. Everything. Because it demonstrates that Schrödinger's equation is just the expression for the Fourier transform of a wave function, transformed into a differential equation. Consequently, all the properties associated with wave functions (such as the uncertainty principle, superposition, entanglement, vacuum fluctuations, the Born rule etc.), turn out to be purely mathematical properties of Fourier transforms and have nothing to do with physics per se. In other words, these properties are properties of the mathematical language being used to describe physical phenomenon  not properties of the phenomenon being described. Consequently, attempting to interpret any of these properties as a physical phenomenon, rather than as mere sideeffects of a specific mathematical technique (Fourier transforms) being employed, is a pitfall, that has confused the physics community for 90 years. See: https://books.google.com/books?id=hEHCAgAAQBAJ&pg=PA232&lpg=PA232&dq="In+any+region+where+V+is+constant,+the+solution+of+the+wave+equation+is"&source=bl&ots=n0Kf89ulEE&sig=b_mFvKHojT3NTPKOew0CugXg&hl=en&sa=X&ved=0ahUKEwiG2cTrvdnaAhWjrVkKHTgiCDoQ6AEIKjAA#v=onepage&q="In any region where V is constant%2C the solution of the wave equation is"&f=false 
Is "Schrodinger's Cat" the wrong interpretation for Copenhagen Interpretation?
Rob McEachern replied to Elite Engineer's topic in Quantum Theory
You stated you were only concerned with solutions "for a single particle translating freely in space": a potential, that has any effect whatsoever on a particle, implies that the particle is not free. is not even a function of time, so it does not represent anything "translating" anywhere. 
Is "Schrodinger's Cat" the wrong interpretation for Copenhagen Interpretation?
Rob McEachern replied to Elite Engineer's topic in Quantum Theory
Equation (2) is the general solution! 
Is "Schrodinger's Cat" the wrong interpretation for Copenhagen Interpretation?
Rob McEachern replied to Elite Engineer's topic in Quantum Theory
You have reversed both causeandeffect and history. Schrödinger derived his equation, for a free particle, from the a priori known Fourier Transform of the wavepacket that he (following de Broglie) "associated" with a particle, as a sort of tracking mechanism. The derivation of the Schrödinger equation for a free particle, consists of little more than computing the first partial derivative of the Fourier transform with respect to time, then the second partial derivative with respect to position, and then equating the two. 
Is "Schrodinger's Cat" the wrong interpretation for Copenhagen Interpretation?
Rob McEachern replied to Elite Engineer's topic in Quantum Theory
Exactly my point. Superpositions only exist as a mathematical abstraction, and abstractions do not collapse. So was the entire physics community. Schrodinger set out to develop a physical model of the trajectory of a particle, by using Fourier transform based superpositions, to describe a wavepacket, that "accompanies" a particle as it moves. But he eventually realized that while his equation provided a good computational model, for producing results, it resulted in an absurd physical model. Thus, he introduced his cat, in an attempt to convince people that one should never attempt to "interpret" a wavefunction as anything other than a useful computational device, which has no known relevance to a physical model, such as an actual propagating wave. In the case of the coin, as in the case for spin and polarization, the state being observed, is not a property of the entity being observed. It is a property of how the observer decides to observe it  the observer will obtain different results, if the entity is observed from different "directions". In other words, "calling" an entity "spin up" or "heads" is not a statement about the state of the entity. It is merely a statement about what the observer happens to "measure" when observing the entity from a particular direction. 
Is "Schrodinger's Cat" the wrong interpretation for Copenhagen Interpretation?
Rob McEachern replied to Elite Engineer's topic in Quantum Theory
Yes. A coin is in a superposition of being both "heads" and "tails". Always. Even after you "call it" one or the other. 
The Logical solution to the Twin Paradox Explained comprehensively
Rob McEachern replied to TakenItSeriously's topic in Relativity
I never said it was. I said "imply the existence of a force". Do you really wish to maintain the a gravitational potential does not imply the existence of a gravitational force, and that that in turn implies that objects at different heights from the center of gravity would experience different accelerations, if not preventing from falling by some other force? 
Is there any reason this Quantum Telegraph couldn’t work?
Rob McEachern replied to TakenItSeriously's topic in Speculations
However, interpretations do have a significant impact on where, or even if, one looks for answers. For example, last summer, I posted a link to a detailed demonstration, on this website, falsifying the supposed "loopholefree" verification experiments of Bell's inequality theorem. That was motivated by a different interpretation. I have noticed that the two of you, in particular (and many hundreds of others) have been noticeably silent about all the errors you have found in that demonstration (and before you ask, yes, the results have been replicated by others). Physicists like to disparage the "god of the gaps", all the while talking about wavefunctions and superpositions, as if the "wavefunction of the gaps" is more real than the "god of the gaps". But wavefunctions and superpositions are no more necessary to explain observations than god is. And the continued belief in all such things has hampered scientific progress. That is the difference interpretations make; bad ones, based on unobserved gods, superpositions and wavefunctions, induce dogma, which in turn stifles progress. So let me put this to you. Above I just told you why quantum theory is entirely probabilistic  not because reality is, but because quantum theory amounts to nothing more than a very complicated procedure for constructing histograms, in an attempt to describe that reality. That is not what was intended by its developers. But it is what they inadvertently developed, for better or worse. So why not take that newfound insight and try to develop a theory, that is not based on histograminducing, Fourier superpositions. Communications engineers (the inventors of information theory) did this a long time ago and it resulted in a revolution in communications technology. Maybe it is time the physics community gives it a try too, rather than "shutting up and calculating" the same old histograms over and over again and wondering why so little progress is being made, in fundamental physics. 
The Logical solution to the Twin Paradox Explained comprehensively
Rob McEachern replied to TakenItSeriously's topic in Relativity
No. You made an incorrect interpretation of my comment. You seem to have interpreted my use of the word "acceleration" as implying only to a change in velocity (or speed?), rather the the more inclusive use of the term to imply the existence of a force (such as a gravitational potential). You seem to have assumed that I believed that the effect is solely due to relative motion. But I said no such thing. 
The Logical solution to the Twin Paradox Explained comprehensively
Rob McEachern replied to TakenItSeriously's topic in Relativity
Obviously. Please read what I said. My original comment stated that the spacetraveler experiences 1g, total, not 1g from gravity plus 1g from thrust or anything else. The whole point is to arrange for the traveler to experience the EXACT same experience as the nontraveler. That cannot be done unless the thrust is continually adjusted to compensate for the change in the gravitational potential, due to increasing distance from the earth. 
The Logical solution to the Twin Paradox Explained comprehensively
Rob McEachern replied to TakenItSeriously's topic in Relativity
See my response to swansont. 
The Logical solution to the Twin Paradox Explained comprehensively
Rob McEachern replied to TakenItSeriously's topic in Relativity
A difference in gravitational potential, is a difference in acceleration (the acceleration due to gravity)  that was my point and that is the equivalence principle. But I'm sure you already know this. What causes the difference in acceleration, gravity or enginethrust, is irrelevant to the effect. 
Truth, Right, and Wrong: Are They Related?
Rob McEachern replied to Gees's topic in General Philosophy
As when billions of people, for centuries, all believed that the sun and planets orbited the earth? Ever hear of the Liar paradox? 
The Logical solution to the Twin Paradox Explained comprehensively
Rob McEachern replied to TakenItSeriously's topic in Relativity
So is the equivalence principle and the acceleration due to gravity. If a spacecraft accelerates outbound at 1g and back inbound at 1 g, etc, the space traveler experiences the same acceleration (except from a brief reversal of thrust) as someone remaining on earth. However, if you leave one atomic clock on earth and have another circle the earth in an aircraft, the two clocks experience different accelerations and thus display differing elapsed times when they are finally brought back together and compared. 
The Logical solution to the Twin Paradox Explained comprehensively
Rob McEachern replied to TakenItSeriously's topic in Relativity
That is a rather dubious assumption. The outgoing twin can never return unless he or she reverses direction, which requires acceleration, to slow down and reverse course. If this acceleration is too great, it will kill the twin. If it is too small, the twin will die before returning, because it will take too long to slow down and reverse course. A one g acceleration is about 10m/s2. The speed of light is about 3x108 m/s. So at 1g acceleration (earth gravity), how many seconds do you have to accelerate in order to reach 80% of the speed of light? How many do you have to accelerate in the opposite direction, to return at 80% of c? How many to slow down and stop, upon arrival? It's going to be a long trip, even if you accelerate the entire time. 
Is there any reason this Quantum Telegraph couldn’t work?
Rob McEachern replied to TakenItSeriously's topic in Speculations
I too am talking about science. Can you devise an experiment to distinguish between two interpretations, one requiring ghostly superpositions to exist and one that does not? No. Occam's razor states that the simplest explanation is to be preferred, over otherwise indistinguishable hypotheses. As Laplace once said, "God is an unnecessary hypothesis". Likewise, superpositions are an unnecessary hypothesis. They explain nothing that needs to be explained. Everything that needs to be explained (the probabilities of observations) can be explained  perfectly, by a simple, but unappreciated, histogramming process  a mathematical identity. 
Is there any reason this Quantum Telegraph couldn’t work?
Rob McEachern replied to TakenItSeriously's topic in Speculations
It is not an analogy. I'm talking about establishing the meaning of the Fourier transforms (AKA wavefunctions) at the heart of quantum theory. Have you ever actually measured a wavefunction? Of course not. So what makes you think they exist? What I am trying to communicate to you, is that fast Fourier transforms "work", by rearranging most of the equation completely "outofexistence". In other words, the vast majority of the terms in the equation can be rearranged such that they will always be identically equal to zero. So why bother to ever even implement them? The fast algorithm is fast, precisely because it never bothers to compute anything that fails to make any difference in the end result. Now here is the important part, concerning the standard interpretations of quantum theory: Just as it is possible to rearrange most of the computations outofexistence, it is also possible to rearrange ALL of the superpositions outofexistence. There are no superpositions left, whatsoever, in the rearranged equation. All that is left, is a mathematically identical, description of a histogram process  which is why the whole procedure yields only probability estimates  sans anything that can possibly be interpreted as a wavefunction. So why worry about Copenhagen and Multiverse interpretations, or mysterious, undetected, wavefunctions, wafting through spacetime, when a mathematical identity enables you to simply say that the whole theory boils down to nothing more than the description of a histogram, counting the arrivals of "things" at the times and places described by the histogram? The histogramming does not care what path the "things" took to the detector, or what the things are (particles or waves or waveparticle dualities)  it simply counts them, whatever they are and wherever and whenever they arrive at the detector. And that is all the "rearranged" mathematically identical equations of quantum theory, actually describe  a histogramming process. 
Is there any reason this Quantum Telegraph couldn’t work?
Rob McEachern replied to TakenItSeriously's topic in Speculations
A physical circuit that has two physical multipliers (ab+ac) has a different physical mass and physical energy consumption than a circuit with only one a(b+c). But you cannot distinguish which is inside a blackbox, from only observing the output computation. It can make a huge difference in cost and size (as well as physical interpretation). For example, discreet Fast Fourier transforms, implemented in hardware, may require many orders of magnitude less size and power (think about putting it on a spacecraft) than a direct computation of the Fourier transform equation, as it is usually written. This does not involve making any approximations. It is an exact mathematical identity, with very, very real consequences in the real world. Thus, if you were to study the physical implementation of a verylarge, direct Fourier transform computation, you might be prone to the conclusion that it would be absolutely, physically impossible for such a transform to be implemented, with existing technology, in something as constrained as a spacecraft  but you would be very wrong. Details (which side of a math identify is being manifested) matter. Similarly, the structure of a Fourier transform has a very different physical interpretation (based on an identity) than the only standard, superposition interpretation that most physicists are aware of, and have consequently embedded into all the standard interpretations of quantum theory. 
Is there any reason this Quantum Telegraph couldn’t work?
Rob McEachern replied to TakenItSeriously's topic in Speculations
Precisely my point. a(b+c) = ab+ac is a math identity, but not a physical identity; One side of the equation has twice as many physical multipliers as the other. If all you could ever observe is the end result of either computation, there is no way to ever deduce which physical manifestation produced the result. In other words, it is generally not possible to ever deduce the correct physical manifestation, from an equation, because the existence of mathematical identities, enables one to construct another equation, that will make identical predictions, but not be the same physically. Thus, very different physical interpretations (AKA manifestations) of any mathematical equation, will almost always exist.