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Why don't photons experience time? |
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| Jan28-13, 03:00 PM | #69 |
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Why don't photons experience time?Um, that's why I said analyzed in the limit as v=c. Perhaps better wording would have been as v goes to c. |
| Jan28-13, 03:12 PM | #70 |
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But say that massless particles do not sense the passage of time, and you get interminable threads about how this means photons don't move in time at all, only in space, how a photon can see the entire Universe all at once, etc., etc., leading to all sorts of further inferences that are just false. Then you have to patiently go back and explain how, when you said massless particles do not sense the passage of time, you didn't really mean that, but something else. |
| Jan28-13, 03:30 PM | #71 |
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Post #57 by DaleSpam in this thread and post #59 by me might be helpful at this point.
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| Jan28-13, 03:45 PM | #72 |
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Combine that with the fact that neutrinos would not able to undergo neutrino oscillations if they had zero mass and what does that suggest. It all suggests that "massless particles do not sense the passage of time" In short, I think that saying the phrase in quotes is dead wrong would be as misleading as saying it is technically exact. |
| Jan28-13, 03:55 PM | #73 |
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It seems like the term "passage of time" is being thrown around so loosely I can't even ascertain how it is being defined in this context. If you want to ascribe a quantity / notion of time that is frame independent then you could talk about [itex]\int_{\gamma } d\tau [/itex] (where [itex]\gamma [/itex] is the time - like curve the massive particle is traveling on). What would "passage of time" even mean for light when you can't use proper time as an affine parameter along a null - like path? Are you wanting to use coordinate time? Coordinate time isn't frame independent so what kind of physical significance of "passage of time" can you even define for that?
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| Jan28-13, 04:24 PM | #74 |
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Neutrinos come in three "flavors", electron, muon, and tau, corresponding to the three kinds of "electron-like" leptons. Neutrino oscillation means that a neutrino that starts out as one flavor can change to a different flavor--more precisely, the quantum mechanical mixture of flavors of neutrinos changes over time: the amplitudes for the different flavor eigenstates oscillate. Oscillating amplitudes in themselves don't require timelike objects: photon amplitudes can oscillate and photons are massless. The point is that the flavor eigenstates of neutrinos--the states in which only one flavor amplitude is nonzero--are *different* than the mass eigenstates--the states in which a neutrino has a definite invariant mass. But for this to lead to neutrino oscillations as defined above, there must be more than one mass eigenstate, so that the amplitudes for different mass eigenstates can oscillate with different frequencies, which in turn means that the amplitudes for each flavor eigenstate (which are just different linear combinations of the mass eigenstates) also oscillate. That means at least one neutrino mass eigenstate must have a nonzero mass. It does *not* require that *all* of the neutrino mass eigenstates have nonzero mass; there could still be one such state with zero mass. AFAIK the current belief is that all of the mass eigenstates have nonzero mass, but that's based on experimental data, not theoretical requirements. So I would say that the statement "neutrino oscillation requires neutrinos to have non-zero invariant mass" is, while technically correct, a little misleading since it invites the false implication that *any* kind of "oscillation" requires a non-zero invariant mass. |
| Jan28-13, 05:33 PM | #75 |
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There is one exception - virtual particles. I really wish they invented a different way to talk about those guys! Even graduate level QFT physics texts could do a better job here. |
| Jan28-13, 05:41 PM | #76 |
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| Jan29-13, 07:40 AM | #77 |
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| Jan29-13, 11:18 AM | #78 |
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This is like saying that biology has no empirical evidence about whether humans can reproduce by fission. Biology says that bacteria can reproduce by fission and humans can't. We observe that bacteria reproduce by fission, and we never observe a human to do so. What experiment would satisfy you, even in principle, that massive particles *can't* move at c? If the only experiment you'll accept is one in which we accelerate a massive particle to c and see what happens, then there is no experiment, even in principle, that would convince you that motion at v=c doesn't exist. This would be like saying that you want to see a human to reproduce by fission so that you can test whether humans can reproduce by fission. |
| Jan29-13, 01:04 PM | #79 |
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bcrowell
I seem to be making things worse rather than better....[That's what my wife always claims!!] yeah, we seem to have good evidence massive particles can't get to v =c..... I have never considered that an issue. This below seems to be one example which I had not seen before....I just stumbled across it....but it conveys the concept I am attempting to describe already: I'll start a new thread...that may enable you guys to help me understand "what happens when a null like path [a photon] intersects a null like surface [an event horizon]. [just a first thought as a problem statement] let's do that separately after [if] I collect my feeble thoughts!! |
| Jan29-13, 01:07 PM | #80 |
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If two photon travel parallel in empty universe, what will happen? Gosh, don't give me warning because of this. |
| Jan29-13, 01:48 PM | #81 |
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| Jan29-13, 01:49 PM | #82 |
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| Jan29-13, 02:22 PM | #83 |
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Now I understand time dilation is 0 for v=c under SR. Will it be also 0 under GR? |
| Jan29-13, 02:38 PM | #84 |
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| Jan29-13, 02:57 PM | #85 |
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