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7th January 2013, 07:37 AM | #801 |
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7th January 2013, 08:20 AM | #802 |
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I looked that up:
"The big difference is that the background value of the Higgs field is Lorentz-invariant — it doesn't define any absolute standard of rest. This is difficult to explain to somebody who doesn't know what a Lorentz transformation is, but it must be possible. Even without knowing mathematics, it is at least plausible that there could be a 'substance' which appears exactly the same to any two observers, regardless of their relative velocity. In fact, this applies to empty space, and it is not unreasonable to say that the value of the Higgs field is just a property of empty space. The problem, of course, is that none of this gives people any idea of what the Higgs field has to do with mass (but in my opinion, neither does the molasses analogy)." -- http://x-sections.blogspot.se/2012/0...mechanism.html Ok, the Higgs field is not an absolute frame of reference. |
7th January 2013, 09:50 AM | #803 |
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It is a bit like some aether theory, particularly since some CERN physicists refer to the Higgs substance. But that doesn't cause an issue for general relativity. Check out Einstein's 1920 Leyden Address where he was talking about the aether of general relativity, and arXiv for papers with aether in the title. Aether isn't the problem. See Einstein's 1905 E=mc˛ paper where he said the inertia of a body is a measure of its energy-content, and that the electron is a body. Saying that the inertia of some bodies is down to something else is the problem.
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7th January 2013, 10:14 AM | #804 |
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Anders, you probably should not listen to anything that Farsight says on these forums. As you can see from other responses, Farsight has a history of presenting inaccurate, false, and insulting statements about science, scientists, and other members of this forum.
His last post provides a great example of his inability to understand the issue at hand. He writes that, "the inertia of a body is a measure of its energy-content," and he imagines that this is somehow a problem for the Higgs theory. However, as you can read from other sources, the Higgs theory is that some of the energy content of some particles is due to this Higgs field. Thus the inertia of some particles is measuring, in part, the energy contributed by the Higgs field. That Farsight continues to miss this point (or that he consciously misrepresents this point) is a mark against trusting him. |
7th January 2013, 10:30 AM | #805 |
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7th January 2013, 10:35 AM | #806 |
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7th January 2013, 12:01 PM | #807 |
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If the Higgs field explained photon momentum and electron mass it would be more reasonable.
Remember that both the photon and the electron have a wave nature because you can diffract them. Now check out Light is Heavy by van der Mark and 't Hooft (not the Nobel prizewinner 't Hooft). Light is "heavy" in that it causes gravity and has an active gravitational mass which is equivalent to inertial mass, but it doesn't have rest mass. However if you trap light as a standing wave in a mirror-box, it adds to the mass to that system, and as a result the box is harder to move. If you were to open the box the photon would come flying out at c, but until you do so, it's effectively at rest so rest mass does apply. The higher the photon frequency the more the mass, and E=hf and m=E/c˛ but the Higgs mechanism is not involved in the slightest. Note that you can make an electron (and a positron) out of a photon in pair production, and in atomic orbitals (see wiki) "electrons do not orbit the nucleus in the sense of a planet orbiting the sun, but instead exist as standing waves". When you say an electron exists as a standing wave even when it's not in an orbital, the situation is just like the photon in a box, where the mass depends on how much energy is there. It's like the electron is a photon in a "box" of its own making. Then when you do electron/positron annihilation it's like opening one box with another. Two 511keV photons fly out as per Einstein's a radiating body loses mass, only afterwards there's no boxes left.
Originally Posted by Kwalish Kid
Originally Posted by Kwalish Kid
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7th January 2013, 12:13 PM | #808 |
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7th January 2013, 12:56 PM | #809 |
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Why can't I start with a 1022 keV photon and produce a 100 keV electron, and a 100 keV positron, flying apart with 411 keV of kinetic energy?
Why can't I start with a 1022 keV photon and produce a zero-mass electron with 511 keV kinetic energy, and a zero-mass positron with 511 keV kinetic energy? Heck, why can't I start with a 100 keV photon, and produce zero-mass electrons/positrons with 50 keV kinetic energy? Because the Higgs field, by interacting with the electron, forces the electron to have a 511 keV rest mass. That's what it does. That's all it does. The rest mass obeys all of the previously-known energy/momentum conservation laws. The rest mass is determined by the Higgs mechanism. There is no conflict between these two statements. |
7th January 2013, 01:15 PM | #810 |
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Doesn't the mirror-box experiment prove that photons indeed have rest mass? And that the Higgs field should interact with photons just like the other particles? If a similar experiment was done where an electron is trapped in a magnetic field inside a box, what's the difference between the electron and the photon except how large the rest mass is?
"The invariant mass, rest mass, intrinsic mass, proper mass, or (in the case of bound systems or objects observed in their center of momentum frame) simply mass, is a characteristic of the total energy and momentum of an object or a system of objects that is the same in all frames of reference related by Lorentz transformations." -- http://en.wikipedia.org/wiki/Invariant_mass |
7th January 2013, 01:49 PM | #811 |
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Even I can tell that the article is wrong. This is what it says:
"In the case of light, the rest mass is zero, but the gravitational mass equals the inertial mass, which is identical to the relativistic mass." As an example, two photons are sent parallel to each other at the same time in air, so the relative velocity of the photons is slightly less than c. If, as the article claims, the photons have gravitational mass, then that would pull the photons closer together! And relative to the first photon, the second photon is at rest! Hence, the gravitational pull would prove that the photons have rest mass. |
7th January 2013, 02:05 PM | #812 |
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Hi Anders Lindman, Farsight has retained a fantasy that the Higgs mechanism violates SR for some time. That means that Farsight will never accept that the Higgs field is Lorenz invariant.
From 19th November 2012: Farsight: What does Higgs mean by Lorentz-covariant and relativistic? as a follow-up to: Farsight: Is the Higgs mechanism a relativistic quantum field theory? 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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7th January 2013, 02:14 PM | #813 |
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Not to mention it's speculations, boldly presented as fact, that "rest mass never applies to a system at complete rest, because such systems do not exist; there will always be internal dynamics" (implying that electrons somehow have stuff whizzing around inside them) and the claim that elementary particles all have non-zero spin (implying that the Higgs observed by CERN cannot be spin-0 like the standard model Higgs).
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7th January 2013, 02:31 PM | #814 |
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But here is a curious quote that also claims that photons have gravitational mass:
"Since light has energy, it is also a source of gravitational effects on other objects, although not a very strong one under ordinary circumstances." -- http://van.physics.illinois.edu/qa/listing.php?id=19800 If that is true, then photons must also have rest mass, as I showed with the example: http://www.internationalskeptics.com...&postcount=811 |
7th January 2013, 03:05 PM | #815 |
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Firstly, photons are indeed deflected by, and are a source of, gravitational fields. However, like anything that goes at c they have zero rest mass (indeed, they cannot even be brought to rest). Going back to your thought experiment above, and keeping it reasonably simple, bear in mind that the objects involved are not just photons, but photons interacting with gas molecules. If you move to what you're calling the "rest frame" of a photon moving through air, what you'll see is not a bare photon at rest but a photon jiggling around in an extremely high speed wind. As such you can't draw the conclusion that you've observed the photon's rest mass.
(As an aside, here's something that might amuse you: in a vacuum, light beams going in parallel directions do not interact gravitationally, while those going anti-parallel do.) Going back to the first article, briefly, they say:
Quote:
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7th January 2013, 03:11 PM | #816 |
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7th January 2013, 04:25 PM | #817 |
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I did a poor job of explaining myself. The point is that you are not observing the behaviour of free photons, you are observing the mean behaviour of a photon+gas system over the course of many, many interactions. So although you might be able to observe an attractive force (perhaps the photons' paths converge in the long term) it would not be due to the gravitation caused by free photons alone. In fact, free photons moving in parallel do not interact gravitationally (as I mentioned earlier).
Let me try a different angle, comparing electrons with photons without the complicating factors of moving through air. Go into the vacuum of deep space, and you find that you can move to the rest frame of any free electron. You will also find that an electron's energy E at any momentum p follows a simple law, E = √(m2c4 + p2c2). where m is a parameter called the rest mass, or often simply the mass. If you try the same trick with a free photon, you find that you cannot move to its rest frame. Free photons in a vacuum always whiz around at c, as judged by any observers in uniform motion. What's more, the law relating energy to momentum is different: E = pc. Note that this can be obtained from the previous one by setting m to zero, and so photons have zero rest mass (and are often just called "massless"). ............ On a technical note: I suppose we don't really know for certain that photons have zero rest mass, though there are very compelling theoretical and empirical reasons to believe they do. According to the standard model, they are indeed exactly massless, while the best experimental upper limit I know of is m < 3 × 10−27 eV/c2, some 2 × 1032 times lighter than the electron. |
7th January 2013, 04:34 PM | #818 |
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Then your claim is different than the article I quoted:
"So light is definitely affected by gravity. Since light has energy, it is also a source of gravitational effects on other objects, although not a very strong one under ordinary circumstances." -- http://van.physics.illinois.edu/qa/listing.php?id=19800 |
7th January 2013, 04:54 PM | #819 |
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7th January 2013, 04:55 PM | #820 |
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No. They don't have rest mass in the normal sense because you can't make a photon go faster or slower. Trapping a photon in a box is just a trick way to change its speed from c to an effective speed of zero, even though inside the box it's still going back and forth at c.
Originally Posted by Anders Lindman
Originally Posted by Anders Lindman
Originally Posted by Anders Lindman
Originally Posted by Ander Lindman
Originally Posted by Anders Lindman
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7th January 2013, 04:55 PM | #821 |
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7th January 2013, 05:23 PM | #822 |
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They aren't called spinors for nothing, and the electron magnetic dipole moment isn't magic. People tend to insist that no motion is involved in intrinsic spin, and say there's no classical equivalent. But they forget about tornados. A tornado has intrinsic spin. Without it, it isn't a tornado any more. They forget about moebius strips too.
Originally Posted by ctamblyn
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7th January 2013, 05:45 PM | #823 |
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I'm not sure whether you caught my link earlier:
http://en.wikipedia.org/wiki/Bonnor_beam Basically, and perhaps intuitively if you imagine graviton exchange as a model for gravity (just a thought, not a statement of fact), two photons both heading off in the +x-direction (say) will not interact gravitationally. On the other hand, if one photon is going in the +x direction while the other is going in the -x direction, then they will interact. |
7th January 2013, 05:45 PM | #824 |
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7th January 2013, 05:47 PM | #825 |
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So the effect of gravity depends on the photon's relative velocity? I doubt that. The gravity is dependent on the photon's frequency, not its velocity. So two photons traveling side by side parallel to each other will attract each other gravitationally dependent on their energy, i.e. their frequency.
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7th January 2013, 05:50 PM | #826 |
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7th January 2013, 05:53 PM | #827 |
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7th January 2013, 05:56 PM | #828 |
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Electron "spin" is a different beast to the rotation of a tornado. There is no evidence that the electron is literally a rotating ball of charge, for example.
That's not how it works. The spin of whatever decayed would leave its imprint on the distribution of decay products, regardless of how short-lived it was. It will be hard to detect, but not impossible. ETA: Someone correct me if I'm wrong, but I believe there are far shorter-lived particles than the Higgs which have been detected (e.g. the W and Z), and have had their non-zero spins determined. So short life does not force spinlessness. |
7th January 2013, 06:02 PM | #829 |
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To be specific: I'm claiming that, according to an exact solution of GR's field equations, the gravitational interaction of two photons depends on their relative directions.
I linked to the Wikipedia article twice already, but here's Bonnor's original paper: http://link.springer.com/content/pdf...7%2FBF01645484 It deals with beams rather than individual particles, but the same result holds. |
7th January 2013, 06:09 PM | #830 |
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That's a bit over my head, lol, but it talks about gravity waves. If gravity is caused by gravitons (which I doubt, but for the sake of argument), then if the photons are traveling in OPPOSITE directions for example, then how will the gravitons be able catch up with the photons? Faster-than-light gravitons? LMAO.
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7th January 2013, 06:26 PM | #831 |
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The article talks about extended beams, like two infinite parallel rods if you like. In that case the interactions go as I described.
Similarly, if you start off with two point-like photons far apart which then approach each other, you will get an interaction. In your scenario though, where I picture two individual point particles being emitted at distinct points and moving directly apart, each at the speed of light, they could not interact by any means, gravity or otherwise (excluding space being wrapped up like a cylinder, wormholes etc.). So, that's an interesting exception to what I stated earlier. For some reason I didn't think of that. Now, In case I got that wrong, perhaps one of the resident physics-gods could chime in. |
7th January 2013, 07:37 PM | #832 |
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7th January 2013, 08:23 PM | #833 |
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What are "they"? The context seems to be electrons but they are never called spinors.
Spinors are mathematical objects that are used to describe quantum spin which has nothing to do with classical spin, e.g. electrons do not actually spin around an axis like your silly example of a tornado.
Quote:
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7th January 2013, 11:51 PM | #834 |
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8th January 2013, 12:31 AM | #835 |
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Ok, if the Higgs field is Lorenz invariant, then it can ONLY give rest mass to particles. Otherwise the mass would be dependent on the particles' velocity relative to the Higgs field as an absolute frame of reference.
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8th January 2013, 05:32 AM | #836 |
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8th January 2013, 07:56 AM | #837 |
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8th January 2013, 08:28 AM | #838 |
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8th January 2013, 08:47 AM | #839 |
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The gravitational attraction between any two objects or particles is dependent on their relative velocity - or more precisely, the dot-product of their 4-momenta. That's a Lorentz invariant quantity.
Gravity is more than just Lorentz invariant. Lorentz invariance is the set of symmetries of flat spacetime. Gravity determines the spacetime, which (away from singularities) is locally flat... so gravity is locally Lorentz invariant, but globally can have other symmetries. That's correct. |
8th January 2013, 10:54 AM | #840 |
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Sol already answered, so I'll just add that I came across this page on John Baez's site that you might find interesting (I think it answers your question, in a roundabout way):
http://math.ucr.edu/home/baez/physic...k_gravity.html (It's worth remembering, though, that no-one has observed a graviton; they are hypothetical particles, for the time being at least. GR itself already describes in classical terms how light is affected by gravity and why it can't escape from inside the event horizon of a black hole.) |
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