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How about this intuition? Photons are massless and travel at light speed. But that is our perspective. From the photons perspective time is standing still. It i

by ozy 10y ago

How about this intuition? Photons are massless and travel at light speed. But that is our perspective. From the photons perspective time is standing still. It is "touches" all space/time points on its path at once, and passes its quanta to only one point, depending on some chance function. Entanglement is the same, from our perspective it is spooky action at a distance, or over time. But from the two photons (and thus the qm phenomenon that created them) perspective it is the same instance. That is a consequence of time is relative. As mind bending as that might be. Or is this totally off?

raattgift 10y ago
It's wrong, but it's hard to explain precisely why without reference to group theory. Imprecisely why, the issue is that you can't find a global frame in which a massless particle is at rest. In General Relativity that's because there are no global frames at all; in Special Relativity, it's because global frames are Lorentz-invariant; a common tool in General Relativity, the Local Inertial Frame, is also Lorentz-invariant. The spacetime of Special Relativity is Minkowski spacetime (AKA Minkowski space AKA flat space) and at every point in that spacetime the Poincaré group (developed by Minkowski) is the isometry group. The Poincaré group has several subgroups, and here we can focus on the Lorentz group. The Lorentz group has three rotation generators J_[x,y,z] (Cartesian coordinates) or J_[\rho,\theta,\phi] (Spherical coordinates) etc, or in general coordinates, J_[1,2,3] or just J_i. It also has three boost generators along the same axes, K_i. There is a commutation relationship which is satisfied between these generators. (The Poincaré group also has translations. In Cartesian coordinates: P_[x,y,z,t]). For a massive spin-1 particle at rest at a single point in the Lorentz frame, the particle's angular momentum is invariant under rotations on any set of axes, but the direction of the angular momentum is changed. However, the photon is a _massless_ spin-1 particle, by definition under most theories and with excellent experimental support. As noted above, we can't find a Lorentz frame in which such a particle is at rest. However, by careful choice of coordinates, we can pick a single axis along which the whole of the massless spin-1 particle's momentum is found. The four-momentum ends up being invariant under the subgroup generated by [J_1, K_2-J_3, K_3+J_2], which produces a transformation matrix giving the rules for rotations around the particle's momentum [given in detail in Wigner, 1939]. Because this sets up a gauge in which _locally_ the momentum is contained, we can rule out the possibility that a photon is smeared out globally, even when considering a very very low-momentum photon (remembering the Einstein relationship for a massless particle, E = pc = \hbar\omega = hc / \lambda; so we're talking a wavelength comparable to the size of the universe). Intuitively, this is because you can always find (nearby, Lorentz) observers for which the photon's momentum is always contained within a single axis, and those observers are no less privileged than any other. Now, to consider your '"touches" all space/time points on its path at once', we need to do a bit of defining. The worldline is the entire path through the whole block universe spacetime Special Relativity. [http://backreaction.blogspot.co.uk/2008/05/block-universe.html http://backreaction.blogspot.co.uk/2008/05/block-universe.ht...] Every fundamental object has its own worldline. So you're verging on a tautology; the photon is at all its (own) spacetime points along the worldline. However, we can apply some 3+1 formalisms to foliate the block universe of Special Relativity by taking 3d spacelike hypersurfaces along surfaces at a constant _coordinate_ time. In each of these hypersurfaces, there is at each point a probability of interaction between the photon and anything that feels electromagnetism. Because of Lorentz invariance, the exact probability distribution is observer independent. However, as noted above, it is possible to choose observers who will see probabilities extremely close to zero except near a small 3-volume. Conversely, this rules out observers who see identical probabilities everywhere. So in that sense, at any given point, a photon does not "touch" all space at each point in time along its worldline. From this we can address your second last paragraph. A pair of correlated (entangled) photons whose worldlines develop a spacelike separation after entanglement are not in non-Lorentz-invariance-violating contact. A discovery of _any_ (local) Lorentz-invariance violation would be extremely exciting new physics, but there is a mountain of evidence suggesting that doesn't happen in our observable universe. [https://www.wikiwand.com/en/Modern_searches_for_Lorentz_violation https://www.wikiwand.com/en/Modern_searches_for_Lorentz_viol...] That is, if we do a 3+1 formalism and do a foliation on a timelike coordinate across the region(s) where the partial worldlines are spacelike separated, we can find no hypersurface on which the nonzero parts of the probability distributions of the pair of photons overlap.
- ozy 10y ago
  Thank you for the extensive reply! Maybe my intuition is: the world is round. And your reply is: no, there are mountains, rotation flattens the poles, the moon causes bulges (the tides) due the the earth's constant falling to the center of the earth/moon system. An intuition is an extremely zoomed out/abstract version of the real thing. Not saying that this is the case. Perhaps in the above analogy the world is actually square, or donut shaped, and so my intuition is way off. But I am out of my depth to judge. So I am hoping you can. If a high schooler had my kind of intuition, or no intuition at all, with which situation would you be happier? I can say that "Conversely, this rules out observers who see identical probabilities everywhere." is a good point, but also what I meant by "chance function", as apposed to an equal chance for all the points. In the quantum eraser experiment, even if you erase (or not) on a much longer path, the result is the same. But it is no ftl comm, because the other observer needs the eraser data to filter the photons and find the signal. But what causes that randomness, knowing that we can rule out hidden state? In my intuition, because it is all evaluated at once. But what is the birds eye view of that answer if it is better not to have that intuition?

by ozy 10y ago

raattgift 10y ago

It's wrong, but it's hard to explain precisely why without reference to group theory. Imprecisely why, the issue is that you can't find a global frame in which a massless particle is at rest. In General Relativity that's because there are no global frames at all; in Special Relativity, it's because global frames are Lorentz-invariant; a common tool in General Relativity, the Local Inertial Frame, is also Lorentz-invariant. The spacetime of Special Relativity is Minkowski spacetime (AKA Minkowski space AKA flat space) and at every point in that spacetime the Poincaré group (developed by Minkowski) is the isometry group. The Poincaré group has several subgroups, and here we can focus on the Lorentz group. The Lorentz group has three rotation generators J_[x,y,z] (Cartesian coordinates) or J_[\rho,\theta,\phi] (Spherical coordinates) etc, or in general coordinates, J_[1,2,3] or just J_i. It also has three boost generators along the same axes, K_i. There is a commutation relationship which is satisfied between these generators. (The Poincaré group also has translations. In Cartesian coordinates: P_[x,y,z,t]). For a massive spin-1 particle at rest at a single point in the Lorentz frame, the particle's angular momentum is invariant under rotations on any set of axes, but the direction of the angular momentum is changed. However, the photon is a _massless_ spin-1 particle, by definition under most theories and with excellent experimental support. As noted above, we can't find a Lorentz frame in which such a particle is at rest. However, by careful choice of coordinates, we can pick a single axis along which the whole of the massless spin-1 particle's momentum is found. The four-momentum ends up being invariant under the subgroup generated by [J_1, K_2-J_3, K_3+J_2], which produces a transformation matrix giving the rules for rotations around the particle's momentum [given in detail in Wigner, 1939]. Because this sets up a gauge in which _locally_ the momentum is contained, we can rule out the possibility that a photon is smeared out globally, even when considering a very very low-momentum photon (remembering the Einstein relationship for a massless particle, E = pc = \hbar\omega = hc / \lambda; so we're talking a wavelength comparable to the size of the universe). Intuitively, this is because you can always find (nearby, Lorentz) observers for which the photon's momentum is always contained within a single axis, and those observers are no less privileged than any other. Now, to consider your '"touches" all space/time points on its path at once', we need to do a bit of defining. The worldline is the entire path through the whole block universe spacetime Special Relativity. [http://backreaction.blogspot.co.uk/2008/05/block-universe.html http://backreaction.blogspot.co.uk/2008/05/block-universe.ht...] Every fundamental object has its own worldline. So you're verging on a tautology; the photon is at all its (own) spacetime points along the worldline. However, we can apply some 3+1 formalisms to foliate the block universe of Special Relativity by taking 3d spacelike hypersurfaces along surfaces at a constant _coordinate_ time. In each of these hypersurfaces, there is at each point a probability of interaction between the photon and anything that feels electromagnetism. Because of Lorentz invariance, the exact probability distribution is observer independent. However, as noted above, it is possible to choose observers who will see probabilities extremely close to zero except near a small 3-volume. Conversely, this rules out observers who see identical probabilities everywhere. So in that sense, at any given point, a photon does not "touch" all space at each point in time along its worldline. From this we can address your second last paragraph. A pair of correlated (entangled) photons whose worldlines develop a spacelike separation after entanglement are not in non-Lorentz-invariance-violating contact. A discovery of _any_ (local) Lorentz-invariance violation would be extremely exciting new physics, but there is a mountain of evidence suggesting that doesn't happen in our observable universe. [https://www.wikiwand.com/en/Modern_searches_for_Lorentz_violation https://www.wikiwand.com/en/Modern_searches_for_Lorentz_viol...] That is, if we do a 3+1 formalism and do a foliation on a timelike coordinate across the region(s) where the partial worldlines are spacelike separated, we can find no hypersurface on which the nonzero parts of the probability distributions of the pair of photons overlap.

ozy 10y ago
Thank you for the extensive reply! Maybe my intuition is: the world is round. And your reply is: no, there are mountains, rotation flattens the poles, the moon causes bulges (the tides) due the the earth's constant falling to the center of the earth/moon system. An intuition is an extremely zoomed out/abstract version of the real thing. Not saying that this is the case. Perhaps in the above analogy the world is actually square, or donut shaped, and so my intuition is way off. But I am out of my depth to judge. So I am hoping you can. If a high schooler had my kind of intuition, or no intuition at all, with which situation would you be happier? I can say that "Conversely, this rules out observers who see identical probabilities everywhere." is a good point, but also what I meant by "chance function", as apposed to an equal chance for all the points. In the quantum eraser experiment, even if you erase (or not) on a much longer path, the result is the same. But it is no ftl comm, because the other observer needs the eraser data to filter the photons and find the signal. But what causes that randomness, knowing that we can rule out hidden state? In my intuition, because it is all evaluated at once. But what is the birds eye view of that answer if it is better not to have that intuition?