Schneibster
Registered Member
Well, then don't mind me while I satisfy myself you're right.
Misleading Phrase: Collapse of wave function
- Thread starter Dinosaur
- Start date
hilarious-- not at this moment--
but again, i suspect it was at the site that is now shutdown and you are very well aware of the truth of my accusations. i am sure there are some here whom remembers and can show it.Since I think this last statement is incorrect, let's agree that....
Spacetime is homogeneous if the metric is the same everywhere - in the jargon the metric tensor field is constant. OK? (This is the essence of Farsight's Einstein quote.)
And since, at least in GR, curvature is given by the Ricci field, which is a second order derivative of the metric field with respect to local coordinates (albeit indirectly via the connection), then all that is required for the curvature tensor to vanish everywhere (i,e for spacetime to be flat) is for the metric field to be constant, by the definition of second order derivatives.
NOTE that it need not be everywhere zero, only the same everywhere. This is the GR equivalent of the famous Laplace equation.
Schneibster
Registered Member
That would be a tensor field that has the same value everywhere, not just an ordinary real tensor field like you find in our universe. But OK. I don't know if I agree whether that's what Farsight's Einstein quote means; and since it's immaterial I won't bother taking a position on that.
I'll point out that it's not just the Ricci term in the curvature; the scalar curvature and the metric tensor come in as well, and so does the cosmological term. Perhaps I am not familiar with what you mean when you talk about the "Ricci field." Yes, I agree with the part about derivatives and second-order derivatives, but I'm pretty sure that the total curvature is given by either side of the whole equation, not just by one term of it.
We'll see if any of that affects what you're trying to say when you get to it. Please do that soon.
Yeah, got that.
That doesn't mean anything to me; I'm not familiar with the Laplace equation. It's aside from the point though.
Can you show this in his science or just in popular writing where he had to be imprecise?
Yes, and you are quite adept at ignoring the specifics of what he wrote in his science and cherry-picking specific quotes from his imprecise popular works.
Einstein used specific mathematical abstractions to talk about space. He also loved quantum mechanics, as one can read in his letters.
No, QuarkHead's definition of field is exactly how Einstein defined a field. You cannot find a contrary definition in any of the scientific literature that Einstein wrote.
In one popular work talking in a very specific context. In practice, Einstein discussed many homogeneous and isotropic gravitational fields; e.g., the various cosmological models he explored.
The sad thing is that all of these errors have been pointed out to Farsight, repeatedly, by dozens of people, for about a decade. This is sad because it means that there is some deficiency in Farsight that keeps him from simply reading Einstein.
Your analogy is flawed. It's the quantum state that collapses that they're talking about when they speak about collapse. I.e. suppose the wave function is
|Psi> = |S1> + |S2> + |S3> + |S4>
If a measurement is made and the state is found to be in the state |S2> then the state goes from
|S1> + |S2> + |S3> + |S4>
to
|S2>
The its said that the wave function collapsed from |Psi> to |S2>. There's nothing wrong with that statement. See more of this at: https://en.wikipedia.org/wiki/Wave_function_collapse
exchemist
Valued Senior Member
Aren't you a sock of pmb, who made a grand trampling exit in 2013, changed his mind and then got banned anyway? And who set up a rival forum, newenglandscience.org which has since conked out?
Or am I mistaken?
I cannot parse this - you seem to be saying that it is not worthwhile to talk about the properties of something in the limiting case, in this case the metric tensor field, if the limiting doesn't occur in Nature.
It's a point of view, but leads quickly to logical difficulties
I have already "shown" you that the Ricci curvature tensor at a point depends on the metric at the same point. And surely you shouldn't have to be told that the trace of the matrix representation depends on the matrix itself - in other words the Ricci scalar depends on the Ricci curvature tensor, which in turn depends on the metric.
We can ignore the cosmological constant as long as we are not doing cosmology
The "=" sign in the field equations tell you that everything on the RHS is the same as everything on the LHS - it's true that these equations are non-linear, but not so extravagantly as to make the LHS depend on the RHS and vice versa
It's not aside from the point. The Laplace equation says that, where a (scalar) field exists, in the absence of a source the divergence of its gradient is zero. Or $$\nabla^2\phi =0$$
Yes, because the wavefunction is the particle. A photon is a wave in space, not some probabilistic point-particle. When you measure the wavefunction you're measuring it with more of the same. We make electrons (and positrons) out of photons in pair production, and we can diffract electrons. An electron is a wave in space too, not some probabilistic point-particle.
Because these waves in space propagate at c.
Are you thinking of somebody else? I haven't aggressively stated that I don't want to do any maths, and I didn't think I was distinguishing space and spacetime "vehemently".Schneibster said:
Right it is not a matter of not wanting to do the math, it is a matter of not being able to do the math.
Do note that it's space that's homogeneous or inhomogeneous. And that the metric field is an abstract thing derived from your measurements. For example you could place optical clocks at various locations throughout an equatorial slice through the Earth and the surrounding space. When you plot the clock rates, your plot ends up looking like the Riemann curvature depiction because lower clocks tick slower and higher clocks tick faster. Also note that the clocks don't tick at different rates because spacetime is curved. Instead they tick at different rates because a concentration of energy in the guise of a massive planet "conditions" the surrounding space altering its metrical properties, this effect diminishing with distance in a non-linear fashion.
Note that the second derivative relates to the tidal force, whilst the first derivative relates to the force of gravity. In theory you could have flat-but-tilted spacetime where there's no curvature but the force of gravity is not zero.
Space is a sub-manifold of spacetime. As such they MUST have the same topological properties
This is ass-backwards, as always - the metric field ALLOWS measurements, not the other way around.
So gravity is a "force"?
Ok. If I understand you right, you're saying that a measurement on one photon can't affect the wavefunction of another, spacelike-separated photon because all the waves involved only propagate at c. Does the same argument apply to entangled photons? If it does, please explain Bell inequality violations. If it does not, please re-answer the original question about superluminal communication for the case where entangled photons are used.
(As much as I realize this is somewhat talking past the GR discussion that's dominating the rest of the thread, I feel like it's a lot more relevant to the OP and the original topic, so I'll keep exploring the issue.)
Space has its metrical properties. You typically measure things using the motion of light. There's no motion in spacetime.
There is no metric field per se. There's space, and light moving through it. When space is homogeneous light moves uniformly and there's no gravitational field. The massive planet alters the surrounding space.
Yes. See the Einstein digital papers. There's plenty of instances where Einstein referred to gravitational force.
Am I saying that? Note this from Wikipedia:
"When a space-like interval separates two events, not enough time passes between their occurrences for there to exist a causal relationship crossing the spatial distance between the two events at the speed of light or slower."
But a photon isn't some pinprick event that happens at some specific time and location. Think of a photon as something like a seismic wave. When a seismic wave moves from A to B across a gedanken plain, it isn't only the houses on the A-B line that shake. The wave takes "many paths". As you walk away from the A-B line the shaking diminishes, but there's no actual edge to it. It isn't some localized event that exists at one place in space and time. It's an extended entity.
Yes, because you can't send information instantly. There is no superluminal communication.
I can't explain it simply. And now I have to go I'm afraid.
Listen to yourself, Farsight! You are claiming that space has a metric but spacetime does not.
This is clearly false
But their IS - this is the definition of a (semi-)Riemann manifold. This is precisely the model that your hero employs. You should not cherry-pick like this
And so? You will forgive my sacrilegious view , but if he does he is wrong. According to his own General Theory
Schneibster
Registered Member
No, I'm trying to relate what you're saying to your claim that spacetime with a bunch of gravity fields in it is homogenous and isotropic when it's got all these bumps in it. I'm not finding that what you've claimed (spacetime + field is homogenous and isotropic) is supported. Flat spacetime is homogenous and isotropic; flat spacetime +fields is not.
Your claim basically boils down to a claim that spacetime without fields is equal to spacetime with fields, and this is obviously ridiculous.
This looks like you're obfuscating and the rest of your post confirms it. I don't bother much with people who obfuscate. Good luck.
On edit: I can see why Farsight is so frustrated.
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I have concluded that it cannot be physical because I view time symmetry as being too valuable to discard.
Agreed on the first point, disagreed with the rest. The Transactional Interpretation is an example of one which would not require wave function physicality, and there are others. The Bell hidden variables are simply "hiding" in the future measurements.Schnieibster said:
Schneibster
Registered Member
Actually we know that QM is not time symmetric; it is CPT-symmetric. You can reverse the Feynman diagram, but do do so you have to reverse the handedness and charges of all the particles. Time symmetry alone is only a feature of classical mechanics, not QM. But that is aside from the point.
Time symmetry does not mean that no interactions can occur; and once an interaction has occurred things have changed. This is the root of the irreversibility of thermodynamics. Time symmetry only means you can run the interaction backwards and it will still obey all the physical laws (I'm ignoring CPT symmetry here for didactic purposes). And run backwards, there's still an interaction, and things are still different after the interaction than they were before. For time symmetry as you seem to mean it to apply there could be no interactions.
This in turn implies, due to my next point in that post, that superposed states must be physical, and that the "wavefunction collapse" represents the point at which an interaction occurs and the formerly superposed state changes to a definite state, and its Heisenberg complement changes from a definite state to a superposed state. It is impossible to measure a particle without interacting with it. It is only the (classically and intuitively derived) misconception that a measurement can occur without an interaction that drives the debate about whether superposition and "wavefunction collapse" represent real states (and real changes of state) of the particle. If one accepts quite simply that a superposed state is just as real as a definite state, then what was unclear becomes clear.
To be very clear, I think you are saying that you agree that superposed states are physical, but denying that particles change state when they interact. I find this very confusing. To me, the rest is logically required if you accept the first point. I'm also not at all clear on how particles cannot change state when they interact with other particles.
Bohm's interpretation is one of the others and the hidden variables hide in the "pilot wave."