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16th November 2012, 01:53 PM | #641 |
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Actually, Clinger asserted that Farsight's interpretation failed at the eighth grade level (by misapplying the algebraic transposition). Post 627 reaffirms this assertion, and expands on the justification for this assertion first made in post 590.
The correctness of the algebraic transposition itself doesn't really matter much to the discussion, which is actually about how Farsight applied it. |
16th November 2012, 02:15 PM | #642 |
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It's a matter of historical record that Einstein near the end of his life was trying to formulate a unified field theory that would explain both gravity and matter. One of his primary goals? Understanding the origin of mass, so as to account for the electron/proton mass ratio (the other particles were just beginning to be discovered in the 20s and 30s when he started on this).
Did Einstein's own ideas about the origin of mass also violate E=mc^2, Farsight? |
16th November 2012, 02:52 PM | #643 |
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Haven't caught up on the whole thread yet so forgive me if this is covered already. Momentum is conserved in the motion of the cannonball and the earth at all times. At no point is any of it "hidden". At the time it's fired it's momentum is X, at the top of the trajectory it's 0, and when it gets back to the ground it's -X. And Earth's momentum (relative to the cannonball) starts at -X (recoil from the cannon), hits 0 along at the same time as the cannonball, and is X when the ball hits the ground. At all times, they add up to 0, and none of the momentum is "hidden" in the cannonball.
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17th November 2012, 08:04 AM | #644 |
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a personal progress report
Disclaimer: I am not a physicist, and I know nothing about particle physics.
In an earlier post, I quoted the first sentence of the Higgs paper and asked for help in understanding it. Lorentz-covariant field theories, symmetry, Lie groups? I was okay with that. The Goldstone theorem, spontaneous breakdown of symmetry, gauge fields? I hadn't a clue. Perpetual Student and edd offered some helpful suggestions, which I read. Those readings reminded me of a book ben m mentioned in another thread, which has been sitting on my bookshelf for over a year, unread: Francis Halzen and Alan D Martin. Quarks & Leptons: An Introductory Course in Modern Particle Physics. John Wiley and Sons, 1984.The Higgs mechanism is covered in chapter 14, and the Higgs particle in chapter 15. Sections 14.6 (on spontaneous symmetry breaking) and 14.7 (spontaneous breaking of a global gauge symmetry) were most helpful to me. Since we've been speaking of algebra, here's a personal anecdote. Section 14.7 starts with the following Lagrangian: ℒ = (∂μφ)*(∂μφ) - μ2φ*φ - λ(φ*φ)2That's equation (14.48). If you write the complex scalar field as φ = (φ1 + iφ2)/(√2), with φ1 and φ2 real, then the potential energy part of that Lagrangian is minimal on the circle with (φ12 + φ22) = v2 = - μ2/λ.That's equation (14.49). Picking φ1 = v and φ2 = 0 as the values of those real fields at some convenient representative point on that circle, we can examine the Lagrangian in the neighborhood of that point by substituting φ(x) = (1/√2) (v + η(x) + iξ(x))into the Lagrangian, where η(x) and ξ(x) are infinitesimal real fields that model the variation in φ as you move away from the representative point. The result of that substitution is equation (14.51): ℒ' = ˝(∂μξ)2 + ˝(∂μη)2 + μ2η2 + constant + cubic and quartic terms in η, ξThat's what they claim, anyway. Ignoring the use of μ to mean two distinct things in that equation, my eyeball substitution said there should be terms linear in η and quadratic in ξ. When I worked through the algebra, however, those terms cancelled. The authors immediately explain the geometric reason, shown in Figure 14.5, but I hadn't read that far when I did the algebra. Section 14.8 (the Higgs mechanism) is basically the same calculation for a local gauge symmetry, so the partial derivatives of the Lagrangian in section 14.7 are replaced by covariant derivatives, which introduces a vector gauge field. To keep the algebraic manipulations from creating the appearance of an unphysical real field, a different substitution is used. We end up with a massive vector boson and a Higgs particle. That, at least, is what I understood from skimming chapter 14 last night. I'm going to have to read most of the book before I can do justice to chapters 14 and 15. That will take me a while. (The authors' preface suggests chapters 3 through 6 could be the basis for an undergraduate course on QED.) On the other hand, I am no longer stuck on Higgs's first sentence. I can now read the entire paper, noting the details I still don't understand. |
17th November 2012, 08:21 AM | #645 |
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Judging by the rest of your post, the second statement at least is false.
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Suppose you have a symmetry of the laws of physics, like translation invariance. Translation invariance means the laws of physics are the same everywhere in space. Now suppose we consider a state containing some object that is localized at some position in space. Physicists say that such a state "spontaneously breaks" translation invariance, because the state itself is not invariant under translations even though the laws of physics are. But translation invariance still tells us something - it tells us that the energy (and charge and momentum and every other property) of that object cannot depend on where we put it. Consider the a degree of freedom that corresponds to the location of that object, let's call it x. What is the energy of associated with x? Clearly, if the object is in motion, there will be extra energy (the kinetic energy of the object, proportional to the time derivative of x, squared). But there cannot be any energy if the object is at rest, no matter where it is. Therefore, there cannot be a term like x^2 or x^4 in the Lagrangian or Hamiltonian that describes this system. Therefore, x is "massless" - in field theory, x would be a field, the symmetry would be a symmetry of the field space, and the absence of x^2 terms would mean x is a massless field. And that's the basic idea behind Goldstone's theorem - broken symmetries give rise to massless particles, or so-called Goldstone bosons.
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What's less trivial is the presence of a mass term for η, and the lack of one for ξ. The latter is a consequence of Goldstone's theorem. |
17th November 2012, 08:46 AM | #646 |
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17th November 2012, 10:39 AM | #647 |
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I can't stop long guys. So in brief:
No I don't. I refer to energy-momentum, and say energy and momentum are two aspects of energy-momentum. I've made this clear by describing how you can't remove the cannonball's kinetic energy without also removing its momentum. Straw man argument from a guy with a sincerity bypass. Enough. |
17th November 2012, 10:42 AM | #648 |
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Not professionally, but honestly, I talk to professional physicists, and they're forever calling me Professor Duffield. I guess that's because I obviously know so much physics. I always correct them of course, and tell them that I'm a well-read physics amateur with a Computer Science degree who's spent decades being analytical and logical and empirical.
No, go and look at what I said to sol again: Yes no problem. I made a mistake. I was thinking of electromagnetic waves and other transverse waves such as wind waves. A longitudinal wave such as a sound wave exhibits only a back-and-forth motion. In mitigation: we aren't talking about sound waves here. We're talking about light waves here. They convey angular momentum. Stop trying to waste everybody's time with trivia because you can't fault my argument. No thanks. And I won't be checking up on negative mass squared either. |
17th November 2012, 10:47 AM | #649 |
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So, if momentum is the same thing as energy, and particles are made of energy, it stands to reason one could as easily say that particles are made of momentum.
When a particle's momentum is measured to be zero, what is it made of then? |
17th November 2012, 11:13 AM | #650 |
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It doesn't contradict it. I gave the full equation and said how only the momentum term applied for photons, and later talked about cannonball momentum.
It's how much of the book's mass-energy the gravitational field will convert into the book's kinetic energy. A deeper gravitational field will convert more. Which converts a portion of the book's mass-energy that we label potential energy into the book's macrosopic kineitc energy. When you lift a book you do work on it. You give it potential energy. You increase its mass-energy just as surely as you do when you warm it up. Hurl it straight up at 11km/s and it takes that potential energy away with it. That energy doesn't go into the gravitational field. Conservation of energy applies. No. It doesn't change. Your measurement devices run faster when they're higher, that's all. Conservation of energy applies again. I said the book on the shelf has more potential energy than the book on the floor, and that this is hidden kinetic energy inside the book that is converted into the falling book's macroscopic kinetic energy. I illustrated this using a standing wave in a box. I can support this in the case of the book because in atomic orbitals "electrons exist as standing waves". It's akin to a whole collection of standing waves in boxes. Sorry guys, I have to go. |
17th November 2012, 11:30 AM | #651 |
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You're talking to at least three at a minimum professional physicists here who do not hold that opinion of you and very probably a maximum of zero here that do (if its any more than zero they're staying extremely improbably quiet).
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17th November 2012, 12:02 PM | #652 |
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No one else caught this, but it's worth jumping in to say *how thoroughly contra-Einstein this is*.
You just said that, if you fire a cannonball upwards, its rest mass will vary along with its distance from Earth. Imagine an observer in a sealed capsule who comes along and finds that a cannonball has punctured their hull. "Either we just flew very fast past a stationary cannonball, or we're at rest and someone fired a cannonball at us," he says. "Although, since the capsule has no rockets, the only reason it would be moving fast would be if we're deep in a gravity well." The fundamental principle of GR is that they can't tell the difference. All of the laws of physics are invariant in all free-falling reference frames. That's why it's a problem when Farsigh beams aboard, saying, "No, I can tell you quite a lot about your reference frame. Using this specially-designed spring scale for moving objects, let's measure the mass of the cannonball as it flies by. If the mass is large, we're deep in a gravity well. If the mass is small, we must be far from the well." Thus Farsight contradicts Einstein on the indistinguishability of free-falling reference frames. |
17th November 2012, 12:55 PM | #653 |
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Very droll, sol. Nice try. But it won't work because the mass of of body is a measure of its energy-content will never square with the mass of a body is a measure of its interaction with the Higgs field. It's one or the other. You can't have your cake and eat it, not when for a linear wave, momentum is a measure of resistance to change of motion, and for a standing wave, mass is a measure of resistance to change of motion. Not when you can create an electron (and a positron) from a photon in pair production, and when you can diffract an electron, and when in atomic orbitals "electrons exist as standing waves". What do you think a free electron consists of? You know, that thing that has magnetic moment and spin angular momentum? A point particle? No, it's a wave, and it isn't going past you at c. It's just sitting there right there in front of you. So it's a standing wave.
Oh how so very convenient when mass is a measure of energy content. Here, try disproving this: mass is a measure of the interaction with fairies and it's always proportional to energy content. You can't disprove it when mass is a measure of energy content. And you know it. So that little bit of sophistry is a busted flush, n'est pas? Yep. Take a look at Susskind's lecture. Two minutes fifty seconds in. He refers to Planck's constant of action, the h in E=hf. Action has the same dimensionality as angular momentum. And Susskind said angular momentum is quantized. Come on sol, this is kid's stuff. Mass is just a measure of energy content. It's never negative, and it certainly isn't imaginary. I know all about imaginary numbers. In physics, they're asscociated with rotation. Funnily enough, so is mass. When the energy-momentum is locked into rotational motion instead of linear motion at c, the result is standing-wave mass instead of linear-wave momentum. I experience it all the time sol. One day, you will too. |
17th November 2012, 01:08 PM | #654 |
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Thank you both for the above posts. As someone who has only recently been making a serious attempt to get a handle on quantum field theory, the above discussion is challenging to the extreme for me. My pursuit of QFT is quite genuine, so I hope to have a better understanding of this discussion in the coming months. In contrast, it's quite ludicrous that Farside, with demonstrably no understanding of QFT -- and making no attempt to correct that ignorance -- has the unmitigated gall to continue to babble about his pretend physics. Considering the significance of the Higgs mechanism, these discussions (excluding Farside) are very much appreciated.
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17th November 2012, 01:12 PM | #655 |
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You are right smartcooky. Stick to your guns, and don't think this is all miles over your head. You can comment, and you can understand it. However there are people here who are trying to persuade you that you can never understand it, and that you should just roll over and believe what they say, even though they can't explain anything at all. Don't fall for it. Be skeptical instead.
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17th November 2012, 02:09 PM | #656 |
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I wonder if Farsight realizes that the mechanism by which the Higgs field gives mass to particles in the Standard Model is essentially the same as the mechanism by which photons acquire a non-zero rest mass in a superconductor? Or that the tachyonic field issue he dismissed out of hand also applies there?
I wonder if he also thinks that photons acquiring mass in a superconductor also violates E=mc^2? Or if he accepts the empirically verifiable fact that photons do acquire rest mass in a superconductor, does he insist that the Standard Model's explanation for how it does so must also be wrong? After all, a photon with non-zero rest mass is a "body", and a body's mass is due to its energy content, so it can't be due to interaction with a Cooper pair condensate. |
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17th November 2012, 03:35 PM | #657 | ||
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17th November 2012, 03:47 PM | #658 |
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That is simply not true. You said:
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17th November 2012, 03:54 PM | #659 | ||||
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When I talk to professional physicists, they usually call me "Will".
You don't tell them about your A-level maths and tutoring experience?
Speaking of dishonesty and creepy disregard for evidence, let's look at this example: In the following quotation, Farsight appears to suggest that energy and momentum aren't different things: When someone goes out of his way to argue that two things aren't different, classical logic tells us he's suggesting they're the same. Combining that inference with what Farsight wrote above, we might conclude that Farsight thinks of himself as a "straw-man-monger". That chain of reasoning fails because the final step implicitly assumes a fact not in evidence: that Farsight is capable of applying logic or other objective criteria to his own words and arguments.
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17th November 2012, 04:24 PM | #660 |
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Quote:
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17th November 2012, 06:06 PM | #661 |
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Gibberish.
The energy of electrons at rest depends on the Higgs field. Therefore, so does their mass.
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Tell us, Farsight - what's the angular momentum of a beam linearly polarized light? What's the angular momentum of a beam of unpolarized light? What's the angular momentum of two photons with equal momentum and equal and opposite helicity? (The correct answer to all of the above is "zero".) |
18th November 2012, 02:11 AM | #662 |
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No. His ideas about the origin of mass gave us E=mc˛. As far as I know he never worked out the proton/electron mass ratio, which is c^˝ / 3π with a small binding-energy adjustment:
c^˝ = 17314.5158177 3π = 9.424778 c^˝ / 3π = 17314.5158177 / 9.424778 r = 1837.12717877 Actual = 1836.15267245 |
18th November 2012, 02:43 AM | #663 |
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At which point the cannonball isn't moving and its momentum is zero along with its kinetic energy. Conservation of energy tells you the energy-momentum you gave it hasn't just vanished.
A 1kg cannonball fired upwards at 1000m/s has momentum of 1000 kg m/s. But at the top of its trajectory its momentum is zero along with its kinetic energy. So where has all that energy-momentum gone? Focussing on total vector momentum does not allow you to claim that the energy-momentum of the cannonball was always zero. You gave that cannonball energy-momentum, and regardless of whether it's blasting upwards at 1000m/s, or is at a great height momentarily motionless before falling back to Earth, it's got it. |
18th November 2012, 03:23 AM | #664 |
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Momentum isn't the same thing as energy, it's a different aspect of energy-momentum, and we tend to say a particle is "made of energy" rather than is "made of energy-momentum". We don't say a particle is "made of momentum" even though the particle has spin angular momentum. If we consider a fast-moving electron, we say it has kinetic energy x and momentum y in addition to its rest-mass energy-momentum of 511keV. To slow down the electron and bring it to rest we have to exert a force on it. Its kinetic energy x is our force times distance measure of its energy-momentum on top of the 511keV. Its momentum y is our force times time measure of its energy-momentum on top of the 511keV.
Note that rather than exerting a force ourselves we can slow down the electron using a series of Inverse Compton scatters. The electron's kinetic energy-momentum is converted into the energy-momentum of photons. Photon energy is E=hf whilst momentum is p=hf/c, the former being a scalar and the latter a vector associated with direction. Once we've got the electron at rest, we can annihilate it with a 511keV positron, which usually results in two photons. The electron and positron rest mass energy-momentum is converted into the energy-momentum of photons. It isn't very different to what happened to the electron kinetic energy-momentum. The electron at rest is made out of the same thing that makes an electron move fast: energy-momentum. You can make an electron move fast using Compton scattering, whereupon the electron absorbs a portion of the photon energy-momentum. When you make an electron and a positron via photon-photon pair production, the electron and positron each absorbs all of the photon energy-momentum as they are created. |
18th November 2012, 03:45 AM | #665 |
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If people like ben can't put up a counterargument and instead offer only ad-hominem abuse, then I have to say that their opinion of me isn't that important. It's never important anyway. What's important is the scientific evidence.
No. But they are all associated with angular velocity. And as I said to sol, take a look at Susskind's lecture. Two minutes fifty seconds in he refers to Planck's constant of action, the h in E=hf that applies to all photons. Action has the same dimensionality as angular momentum. And Susskind said angular momentum is quantized. Whether all transverse waves convey net angular momentum is trivia. You're trying to use it to distract attention from the point of discussion because you have no adequate counter to Einstein's the mass of a body is a measure of its energy-content and his radiation conveys inertia between the emitting and absorbing bodies. Sorry edd, but it just won't work. |
18th November 2012, 03:49 AM | #666 |
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18th November 2012, 03:53 AM | #667 |
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18th November 2012, 04:33 AM | #668 |
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Mindful of the above, the next post from ben appears to offer a counter-argument, so I'll address it and shoot it down in flames.
It isn't contra-Einstein at all. See this section of the wikipedia Mass in general relativity page: In special relativity, the invariant mass of a single particle is always Lorentz invariant. Can the same thing be said for the mass of a system of particles in general relativity? Surprisingly, the answer is no. A system must either be isolated, or have zero volume, in order for its mass to be Lorentz invariant. While the density of energy momentum, the stress-energy tensor is always Lorentz covariant, the same cannot be said for the total energy-momentum. (Nakamura, 2005). Non-covariance of the energy-momentum four-vector implies non-invariance of its length, the invariant mass. Yes it will. See above. Its invariant mass varies by virtue of conservation of energy. When you fire a cannonball straight up at 1000m/s, the kinetic energy you gave to the cannonball is converted into potential energy in the cannonball. At the top of its trajectory the cannonball is momentarily motionless, at which point all of its kinetic energy has been converted into potential energy. In the cannonball. The cannonball at rest five miles up comprises more energy than the cannonball at rest on the ground. All they know initially is that the cannonball had relative motion compared to them. A spring scale doesn't work, because the mass/energy of the spring also varies with gravitational potential. No I don't. And what you've forgotten is that Einstein used infinitesimal reference frames. See this Einstein Online article on the equivalence principle: "Realizing that what matters are the size of the region, and the duration of our observations, we are led to a formulation in which the equivalence principle is not just a useful approximation, but exactly true: Within an infinitely small ("infinitesimal") spacetime region, one can always find a reference frame - an infinitely small elevator cabin, observed over an infinitely brief period of time - in which the laws of physics are the same as in special relativity. By choosing a suitably small elevator and a suitably brief period of observation, one can keep the difference between the laws of physics in that cabin and those of special relativity arbitrarily small." The principle of equivalence is only exactly true in a region of zero extent where measurements take zero time. And it's only a principle, not a golden rule. It doesn't actually say that if you're in a box you can never hope to find out whether you're in free space or in a gravitational field. If you can measure say tidal force or the fine structure constant with adequate precision you can tell the difference. Doing so doesn't mean general relativity is wrong, it just reminds you that the principle of equivalence is only exactly true in a region of zero extent where measurements take zero time. Gotta go. |
18th November 2012, 05:30 AM | #669 |
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I note in the very next post you're having to respond to one of his counterarguments. edit to add: I note you noted that! Never mind.
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18th November 2012, 05:34 AM | #670 |
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Random not-really-science question: will the Higgs boson continue to be called the Higgs boson? I was just thinking that the other fundamental particles aren't named after people - neither their discoverers nor the people that predicted their existence. Should the Higgs be any different in that respect?
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18th November 2012, 05:46 AM | #671 |
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Fermions and bosons are both names derived from people, and many other classes of particles have names coming from people so I don't see why not.
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18th November 2012, 06:17 AM | #672 |
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So you're denying history now as well as physics? Would it matter if I posted quotes from his papers on that?
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18th November 2012, 07:52 AM | #673 |
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18th November 2012, 08:40 AM | #674 |
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The momentum is transferrred to earth, via gravity. It never disappears and is never converted to anything else. Every bit of momentum the cannonball loses is gained by the earth. When the cannonball loses it's initial momentum X and reaches 0 speed, that's because the earth has gained exactly X momentum.
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You seem to think that momentum and energy are the same thiing, but they are not. Momentum is related to kinetic energy, but not potential energy. Potential energy is not hidden momentum, because that momentum is never hidden. It's only hidden in your mind because you only consider the cannonball and not the earth and the force of gravity between them. It comes down to what you denied in the other thread, that gravitational potential energy is stored in an object. It is not, it is a property of a system of objects. GPE is relative. One object can have different amounts of GPE, depending on your frame of reference, just like it can have different velocities. This is because GPE is not a property of the object, it's a property of the relationship between multiple objects. |
18th November 2012, 12:05 PM | #675 |
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Farsight: Is the Higgs mechanism a relativistic quantum field theory
Once again you did not!
It was a repeat of you ignoring the Higgs mechanism and restating your fantasy (so far) that it is inconsistant with E=mc^2. You then complain about putting relativistic in front of QFT ! There is nothing in that post about whether that Higgs mechanism is relativistic or not. The answer is either
i.e. is it is based on special relativity and is thus consistent with E=mc^2. First pointed out 1 November 2012 |
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18th November 2012, 01:34 PM | #676 |
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18th November 2012, 02:04 PM | #677 |
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I'm not denying history or physics. It's me quoting from Einstein's papers here. I'd be only too pleased if you quoted some yourself.
He did explain where mass comes from - from the kinetic energy of the body. When the body is at rest and emits kinetic energy in the form of radiation, its mass reduces and the radiation conveys inertia to the absorbing body, which gains mass. Don't forget that the "Higgs boson" has a gamma-gamma decay channel. Just think of it as a body. He didn't say where energy originally comes from, that we don't know. Energy is the one thing we can neither create nor destroy. Sure. The c^˝ / 3π expression sits on top of another expression λ = 4π / n c^1˝ metres where n is a dimensionality conversion factor n with a value of 1. It's all to do with harmonics and ratios and spin ˝, and everything is based on the motion of light. If you change your definition of c everything else changes too, but the sense of E=mc˛ and E=p/c still holds. It's the same for these expressions. The thing we call c isn't so much a speed as a conversion factor between our units of distance and time. They're both defined using the motion of light. Everything relates back to the motion of light. Check out the watt balance section of the wikipedia Kilogram article and note the bit that says this: "The Planck constant defines the kilogram in terms of the second and the meter. By fixing the Planck constant, the definition of the kilogram would depend only on the definitions of the second and the meter." The article goes on to say "the definition of the second depends on a single defined physical constant: the ground state hyperfine splitting frequency of the caesium 133 atom". However there's a little flaw in that in that you can't define the second using a frequency, which is cycles per second. Anyway, SI is the kilogram-metre-second system, and will end up being more of a metre-second system where everything relates back to the motion of light. Interesting stuff I think. A bit off topic mind, but I think we've almost exhausted it anyway. |
18th November 2012, 02:37 PM | #678 |
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I missed this one:
No it isn't. This is the heart of it. The mass of of body is a measure of its energy-content will never square with the mass of a body is a measure of its interaction with the Higgs field. Not when the Higgs mechanism is responsible for only 1% of the mass of matter and doesn't apply to a standing wave in a box. Not when the electron exists as a standing wave and the kinetic energy of the LHC protons was used to create the Higgs boson. Not when there's a gamma-gamma decay channel and it's a body too, just like the electron. Saying that the Higgs interaction is exactly proportional to energy-content is just a cop-out. It doesn't. It depends on h. I haven't got it wrong. Anybody can look this up: "The photon also carries spin angular momentum that does not depend on its frequency.[17] The magnitude of its spin is √2ħ and the component measured along its direction of motion, its helicity, must be ±ħ. These two possible helicities, called right-handed and left-handed, correspond to the two possible circular polarization states of the photon.[18]" Also see this where you can read: "We can therefore think of the spin angular momentum of the photon being quantized as well as the energy. This has indeed been experimentally verified.[2] Photons have only been observed to have spin angular momenta of ±ħ." Saying the angular momentum of two photons with opposite helicity is zero is like saying the momentum of the Earth and cannonball is zero. Ergo the cannonball doing 1000m/s has no momentum. |
18th November 2012, 02:51 PM | #679 |
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Join Date: Jul 2006
Posts: 6,387
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Why is this so hard, Farsight? The presence of the Higgs field alters the allowable energies of Standard Model particles. Because it alters the energies, it alters the masses. There is no conflict between the Higgs mechanism and E^2=m^2c^4+p^2c^2. The Higgs mechanism just helps determine m.
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18th November 2012, 02:54 PM | #680 |
Philosopher
Join Date: Oct 2007
Posts: 8,613
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That's not what Einstein thought. He devoted years of his life to explaining the origin of mass. I'll look for excerpts of those papers when I have time.
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You're wrong. And on top of that, after pages and pages no one has any idea why you think that (except ben, but even his take on your psychology doesn't really explain it).
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