KJW

Talking about linear vs non-linear physics, below is a connection between linear quantum mechanics and non-linear classical mechanics: Klein-Gordon equation: \[\frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2} - \frac{\partial^2\psi}{\partial x^2} - \frac{\partial^2\psi}{\partial y^2} - \frac{\partial^2\psi}{\partial z^2} = -\frac{m^2c^2}{\hbar^2}\psi \] \[\psi = \psi_0 exp(\frac{i}{\hbar} S(t,x,y,z))\] where [math]S(t,x,y,z)[/math] is the action. \[\frac{\partial^2\psi}{\partial t^2} = \frac{\partial}{\partial t}(\frac{\partial}{\partial t} \psi_0 exp(\frac{i}{\hbar} S)) = \frac{\partial}{\partial t}(\frac{i}{\hbar} \psi_0 exp(\frac{i}{\hbar} S) \frac{\partial S}{\partial t}) = \frac{i}{\hbar} \psi_0 exp(\frac{i}{\hbar} S) \frac{\partial^2 S}{\partial t^2} - \frac{1}{\hbar^2} \psi_0 exp(\frac{i}{\hbar} S) (\frac{\partial S}{\partial t})^2 = \frac{i}{\hbar} \psi \frac{\partial^2 S}{\partial t^2} - \frac{1}{\hbar^2} \psi (\frac{\partial S}{\partial t})^2\] Similarly: \[\frac{\partial^2\psi}{\partial x^2} = \frac{i}{\hbar} \psi \frac{\partial^2 S}{\partial x^2} - \frac{1}{\hbar^2} \psi (\frac{\partial S}{\partial x})^2 \;\;;\;\; \frac{\partial^2\psi}{\partial y^2} = \frac{i}{\hbar} \psi \frac{\partial^2 S}{\partial y^2} - \frac{1}{\hbar^2} \psi (\frac{\partial S}{\partial y})^2 \;\;;\;\; \frac{\partial^2\psi}{\partial z^2} = \frac{i}{\hbar} \psi \frac{\partial^2 S}{\partial z^2} - \frac{1}{\hbar^2} \psi (\frac{\partial S}{\partial z})^2\] Thus: \[-i\hbar (\frac{1}{c^2}\frac{\partial^2 S}{\partial t^2} - \frac{\partial^2 S}{\partial x^2} - \frac{\partial^2 S}{\partial y^2} - \frac{\partial^2 S}{\partial z^2}) + (\frac{1}{c^2}(\frac{\partial S}{\partial t})^2 - (\frac{\partial S}{\partial x})^2 - (\frac{\partial S}{\partial y})^2 - (\frac{\partial S}{\partial z})^2) = m^2c^2\] In the classical limit of [math]\hbar = 0[/math], the linear second-order Klein-Gordon equation becomes the non-linear first-order Hamilton-Jacobi equation: \[\frac{1}{c^2}(\frac{\partial S}{\partial t})^2 - (\frac{\partial S}{\partial x})^2 - (\frac{\partial S}{\partial y})^2 - (\frac{\partial S}{\partial z})^2 = m^2c^2\]

Klein-Gordon equation: \[\frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2} - \frac{\partial^2\psi}{\partial x^2} - \frac{\partial^2\psi}{\partial y^2} - \frac{\partial^2\psi}{\partial z^2} = -\frac{m^2c^2}{\hbar^2}\psi \] \[\psi = \psi_0 exp(\frac{i}{\hbar} S(t,x,y,z))\] where [math]S(t,x,y,z)[/math] is the action. \[\frac{\partial^2\psi}{\partial t^2} = \frac{\partial}{\partial t}(\frac{\partial}{\partial t} \psi_0 exp(\frac{i}{\hbar} S)) = \frac{\partial}{\partial t}(\frac{i}{\hbar} \psi_0 exp(\frac{i}{\hbar} S) \frac{\partial S}{\partial t}) = \frac{i}{\hbar} \psi_0 exp(\frac{i}{\hbar} S) \frac{\partial^2 S}{\partial t^2} - \frac{1}{\hbar^2} \psi_0 exp(\frac{i}{\hbar} S) (\frac{\partial S}{\partial t})^2 = \frac{i}{\hbar} \psi \frac{\partial^2 S}{\partial t^2} - \frac{1}{\hbar^2} \psi (\frac{\partial S}{\partial t})^2\] Similarly: \[\frac{\partial^2\psi}{\partial x^2} = \frac{i}{\hbar} \psi \frac{\partial^2 S}{\partial x^2} - \frac{1}{\hbar^2} \psi (\frac{\partial S}{\partial x})^2 \;\;;\;\; \frac{\partial^2\psi}{\partial y^2} = \frac{i}{\hbar} \psi \frac{\partial^2 S}{\partial y^2} - \frac{1}{\hbar^2} \psi (\frac{\partial S}{\partial y})^2 \;\;;\;\; \frac{\partial^2\psi}{\partial z^2} = \frac{i}{\hbar} \psi \frac{\partial^2 S}{\partial z^2} - \frac{1}{\hbar^2} \psi (\frac{\partial S}{\partial z})^2\] Thus: \[-i\hbar (\frac{1}{c^2}\frac{\partial^2 S}{\partial t^2} - \frac{\partial^2 S}{\partial x^2} - \frac{\partial^2 S}{\partial y^2} - \frac{\partial^2 S}{\partial z^2}) + (\frac{1}{c^2}(\frac{\partial S}{\partial t})^2 - (\frac{\partial S}{\partial x})^2 - (\frac{\partial S}{\partial y})^2 - (\frac{\partial S}{\partial z})^2) = m^2c^2\] In the classical limit of [math]\hbar = 0[/math], the linear second-order Klein-Gordon equation becomes the non-linear first-order Hamilton-Jacobi equation: \[\frac{1}{c^2}(\frac{\partial S}{\partial t})^2 - (\frac{\partial S}{\partial x})^2 - (\frac{\partial S}{\partial y})^2 - (\frac{\partial S}{\partial z})^2 = m^2c^2\]

Klein-Gordon equation: \[\frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2} - \frac{\partial^2\psi}{\partial x^2} - \frac{\partial^2\psi}{\partial y^2} - \frac{\partial^2\psi}{\partial z^2} = -\frac{m^2c^2}{\hbar^2}\psi \] \[\psi = \psi_0 exp(\frac{i}{\hbar} S(t,x,y,z))\] where S(t,x,y,z) is the action \[\frac{\partial^2\psi}{\partial t^2} = \frac{\partial}{\partial t}(\frac{\partial}{\partial t} \psi_0 exp(\frac{i}{\hbar} S)) = \frac{\partial}{\partial t}(\frac{i}{\hbar} \psi_0 exp(\frac{i}{\hbar} S) \frac{\partial S}{\partial t}) = \frac{i}{\hbar} \psi_0 exp(\frac{i}{\hbar} S) \frac{\partial^2 S}{\partial t^2} - \frac{1}{\hbar^2} \psi_0 exp(\frac{i}{\hbar} S) (\frac{\partial S}{\partial t})^2 = \frac{i}{\hbar} \psi \frac{\partial^2 S}{\partial t^2} - \frac{1}{\hbar^2} \psi (\frac{\partial S}{\partial t})^2\] \[-i\hbar (\frac{1}{c^2}\frac{\partial^2 S}{\partial t^2} - \frac{\partial^2 S}{\partial x^2} - \frac{\partial^2 S}{\partial y^2} - \frac{\partial^2 S}{\partial z^2}) + (\frac{1}{c^2}(\frac{\partial S}{\partial t})^2 - (\frac{\partial S}{\partial x})^2 - (\frac{\partial S}{\partial y})^2 - (\frac{\partial S}{\partial z})^2) = m^2c^2\] In the classical limit of [math]\hbar = 0[/math], the linear second-order Klein-Gordon equation becomes the non-linear first-order Hamilton-Jacobi equation: \[(\frac{1}{c^2}(\frac{\partial S}{\partial t})^2 - (\frac{\partial S}{\partial x})^2 - (\frac{\partial S}{\partial y})^2 - (\frac{\partial S}{\partial z})^2) = m^2c^2\]

1c2∂2ψ∂t2−∂2ψ∂x2−∂2ψ∂y2−∂2ψ∂z2=−m2c2ℏ2ψ Let ψ=ψ0exp(iS(t,x,y,z)ℏ) ∂2ψ∂t2=∂∂t(∂∂texp(iS(t,x,y,z)ℏ)) \[\frac{\partial^2\psi}{\partial t^2} = \psi_0 \frac{\partial}{\partial t}(\frac{\partial}{\partial t} exp(i\frac{S(t,x,y,z)}{\hbar})) = \psi_0 \frac{\partial}{\partial t}(\frac{i}{hbar} exp(i\frac{S(t,x,y,z)}{\hbar}) \frac{\partial S(t,x,y,z)}{\partial t})\]

Ok. Now I suppose you're either from Australia or a neighbourhood close to a zoo with very poor security. I meant "observer effect" is a misnomer. Probably a mild one. It's the apparatus that does it. Not consciousness. Oh, I know what you were saying. I just thought it amusing that you used the example of a koala to someone who not only lives in the country of koalas, but actually had koalas living in their backyard, and therefore has experience of koalas that even other Australians might not have. It is sad that the koalas are on longer around, but I am glad that I was able to experience them for a period of time.

What mathematics are you referring to? @joigus said: which is in complete agreement with what I said.

I know what a koala is. They used to be in my backyard until they disappeared as a result of urban development. I know what quantum measurement is. And I know that they are about interactions and not observation. But there is an aspect that is about observation that goes beyond the objective nature of measurement. When you asked me why we need observers, I said we don't need observers, we are observers, I don't think you understood why I said that. Being an observer does have its consequences. You can't assume that the world would run the same way without observers. Not every manifestation of significance is objectively real. For example, what in the objectively real world is the present? There are no "this is now" markers in spacetime. But as observers, we can't help but see that the present is different to the past and that the future doesn't yet exist. Where is the objective reality in that? I am saying that the collapse of the wavefunction to a single eigenvector is not objectively real and requires a conscious observer to explain. But note that there is another significant issue involved that also needs to be explained, and that is the notion of intrinsic randomness. The many-worlds interpretation explains it, the Copenhagen interpretation does not. That's not necessarily true. I do my own analyses beyond what I read. I don't necessarily accept the exact version of the many-worlds interpretation. My view is that something like the many-worlds interpretation is necessary, even if it is a modified version of the original.

Γijk = ½ gku (∂gui/∂xj + ∂guj/∂xi – ∂gij/∂xu)

If one has a conformally flat metric, the corresponding Friedmann-Lemaître-Robertson-Walker (FLRW) metric can be obtained from it by a coordinate transformation: (ds)² = A(t)² ((c dt)² – (dx)² – (dy)² – (dz)²) t = ƒ(t') ; x = x' ; y = y' ; z = z' dt = ∂t/∂t' dt' + ∂t/∂x' dx' + ∂t/∂y' dy' + ∂t/∂z' dz' = dƒ(t')/dt' dt' dx = ∂x/∂t' dt' + ∂x/∂x' dx' + ∂x/∂y' dy' + ∂x/∂z' dz' = dx' dy = ∂y/∂t' dt' + ∂y/∂x' dx' + ∂y/∂y' dy' + ∂y/∂z' dz' = dy' dz = ∂z/∂t' dt' + ∂z/∂x' dx' + ∂z/∂y' dy' + ∂z/∂z' dz' = dz' Note that the primed coordinates (t', x', y', z') and unprimed coordinates (t, x, y, z) have reversed roles compared to the earlier posts in this thread. dƒ(t')/dt' = 1/A(ƒ(t')) A(ƒ(t')) dƒ(t')/dt' = 1 Let F(t) be such that: dF–1(t)/dt = A(t) Then: dF–1(ƒ(t'))/dt' = A(ƒ(t')) dƒ(t')/dt' = 1 F–1(ƒ(t')) = t' – t'0 where t'0 is an arbitrarily chosen value of t' ƒ(t') = F(t' – t'0) Let a(t' – t'0) = A(ƒ(t')) Then: dƒ(t')/dt' = 1/A(ƒ(t')) = dF(t' – t'0)/dt' = 1/a(t' – t'0) And therefore: (ds)² = (c dt')² – a(t' – t'0)² ((dx')² + (dy')² + (dz')²) .............................. For example: (ds)² = exp(k t)² ((c dt)² – (dx)² – (dy)² – (dz)²) A(t) = exp(k t) 1/A(ƒ(t')) = dƒ(t')/dt' = exp(–k ƒ(t')) exp(k ƒ(t')) dƒ(t')/dt' = 1 (1/k) exp(k ƒ(t')) = t' – t'0 where t'0 is an arbitrarily chosen value of t' ƒ(t') = (1/k) ln(k (t' – t'0)) 1/a(t' – t'0) = dƒ(t')/dt' = 1/(k (t' – t'0)) a(t' – t'0) = k (t' – t'0) Therefore: (ds)² = (c dt')² – (k (t' – t'0))² ((dx')² + (dy')² + (dz')²) as would be expected from the earlier post.

The problem with this reply is that it depends on precisely what an "unplugged" or "on" detector does to the light, the term "detector" not being sufficiently specific. It may be useful to consider the following: Suppose one places a horizontally oriented polariser in one slit, and a vertically oriented polariser in the other slit. Then there is no interference pattern even though no which-slit detection has been performed. Each photon has which-slit information by virtue of its polarisation, which is sufficient to prevent the interference pattern even when the polarisation is not measured. The quantum states from both slits are orthogonal, and there is no interference between orthogonal states. Perhaps you should also explore quantum eraser experiments. I'm quite confident about what I said... or are you telling me you can mathematically predict quantum outcomes?

It wasn't on "The Science Forum" but on an Australian forum that hasn't been running for years. The thread was started and had run its course before I had even joined the forum. I had looked at a few posts in the thread, but I don't think I contributed to it. It was a running joke on the forum to mention this thread because of the size of it. As you can imagine, it covered every conceivable aspect of the question. I doubt that it came to any definite conclusion, though.

On another now defunct science forum, there was a thread that was quite notable among forum members for being the longest thread in the history of the forum. It was titled: "What colour is an orange in the dark?"

This thread is titled "The Observer Effect", so in order to meaningfully discuss the observer effect, one needs to distinguish between observers playing a passive role and observers playing an active role. An active role means that the conscious observer somehow affects the measurement of the quantum system so as to collapse the wavefunction. A passive role means that the quantum system affects the measuring device in a way that produces a macroscopic state that through perception affects the mind of the conscious observer. As conscious observers, we have a somewhat solipsistic subjective point of view that needs to be taken into consideration when considering an objective reality. I'm not applying any "old misguided" concept of observing. I'm certainly not suggesting that it is some kind of physical influence. Indeed, I said precisely the opposite. That is why I consider it important to distinguish between the active and passive role. I hadn't really considered whether or not the measuring apparatus is doing anything to the quantum system. Wavefunction collapse is ultimately about what the quantum system is doing to the measuring apparatus. In the many-worlds interpretation, there is no wavefunction collapse and that a superposition of quantum states leads to a superposition of measuring apparatus states which leads to a superposition of observer states. It is the superposition of observer states, entangled with the superposition of measuring apparatus states and quantum states that is the essence of the many-worlds interpretation. The role of the observer is then to answer the question: why do we not observe the superposition of observer states, measuring apparatus states, and quantum states? You've indicated that the many-worlds interpretation is not falsifiable. The reason it is not falsifiable is because the many worlds are not observable (if they were observable, there would be no question about existence). Therefore, it is about observation. At the heart of the matter is where we draw the line between objective reality and subjective observation. There is nothing in the mathematics that says that a measurement selects one eigenstate from the set of possible eigenstates. The mathematics only identifies the set of eigenstates. The selection of a particular eigenstate from the set of eigenstates is artificially imposed on the basis of observation. The mathematics ultimately implies the many-worlds interpretation, or as it was originally called, the "relative state formulation".

As one attempts to separate the quarks in a proton, the force between them increases. Thus, there comes a point where there is so much energy that one or more quark-antiquark pairs are created. This results in one or more mesons being released, leaving behind some form of baryon (not necessarily a proton).

Yes, the observer does have an important role, though it is a passive role. We don't need observers, we are observers. Without observers, the measuring devices would still do their thing, only there would no longer be any need for the Copenhagen interpretation. The Copenhagen interpretation suggests that it is physics that determine the particular eigenstate that we observe, whereas the many-worlds interpretation implies that it is not physics but the observer that determines the particular eigenstate that we observe, simply through the act of observation. Because the microscopic quantum state, the macroscopic state of the measurement device, all the observers observing the measurement device, and the world that contains all of these are in an entangled state. Considering Schrödinger's cat, the cat becomes entangled with the radioactive source, and when I open the box, I become entangled with the cat and the radioactive source. By becoming entangled with the cat, I no longer observe the cat state and the radioactive source state as a superposition of states. Any other observers who observe the cat becomes entangled with the same cat state as me, as well as becoming entangled with me, and therefore observes the same cat state and radioactive source state as me.

It's not really an interpretation of an interpretation, but more like saying that the notion of the splitting of realities is a strawman that should not be used to reject the many-worlds interpretation of quantum mechanics. In truth, all I am saying is that in the Born rule, all the eigenstates exist, and that it is merely a first-person perspective that gives rise to the observation of only a single eigenstate. Although I am saying that the many worlds is the ontological reality, and that the Copenhagen interpretation is the subjective observation of the many worlds, I am not claiming to fully understand the precise details of the many worlds.

That's the problem with the requirement of falsifiability: Reality is not obligated to reveal itself in its entirety. It may be that the many worlds actually do exist, but are unobservable because they are mathematically orthogonal to each other. By rejecting the many worlds on the grounds that they are not falsifiable, one then has a reality that is inexplicable. I personally don't interpret the many-worlds interpretation as a splitting of realities each time a quantum decision is made. Instead, I regard all the spacetimes as existing from the very beginning, at least conceptually (I don't think reality is as simple as multiple spacetimes).

When a pure quantum state is measured, it is decomposed into a complex-number weighted sum of eigenstates. However, only a single eigenstate is observed as the result of the measurement, and this is chosen randomly with a probability based on the complex-number weighting of the eigenstate. If a large number of identical pure quantum states are identically measured, then the various eigenstates will be observed with their respective probabilities. For example, in the double-slit experiment, each spot corresponds to a random position eigenstate, but the accumulation of spots is an interference pattern resulting from each quantum state passing through both slits. In the Copenhagen interpretation, the pure quantum state collapses randomly to a single eigenstate, and the other possible eigenstates simply vanish. But in the many-worlds interpretation, the different versions of the observer in the different worlds observe all of the possible eigenstates. And because all the eigenstates are mutually orthogonal to each other, there is no interference between them, and thus the observer can observe only a single eigenstate and not the other versions of the observer.

\[ y = \int f(x) dx \]

Also, in the animal world, brightly coloured means dangerous. This is called "aposematism".

Continuing from my previous post: This can be further developed by manipulating the arbitrary constant to obtain: (ds)² = exp(k (t' – t'0))² ((c dt')² – (dx')² – (dy')² – (dz')²) where t'0 is an arbitrarily chosen value of t' at which the metric is locally Minkowskian For the general case: dƒ(t')/dt' = a(ƒ(t')) 1/a(ƒ(t')) dƒ(t')/dt' = 1 Let F(t) be such that: dF–1(t)/dt = 1/a(t) Then: F–1(ƒ(t')) = t' – t'0 where t'0 is an arbitrarily chosen value of t' ƒ(t') = F(t' – t'0) dƒ(t')/dt' = dF(t' – t'0)/dt' = A(t' – t'0) (ds)² = A(t' – t'0)² ((c dt')² – (dx')² – (dy')² – (dz')²)

[math]{R_{pqr}}^s[/math]

Fractional distillation under a vacuum is all I can offer. Yes, fatty acids found in nature do tend to have an even number of carbon atoms because they are biosynthesised from two-carbon units (acetyl-CoA).

No, it doesn't. Only the time coordinates are involved in the coordinate transformation. If the original metric describes a flat three-dimensional space, then the transformed metric will be a scalar function multiple of the Minkowskian metric: guv = ƒ(t) ηuv Consider: (ds)² = (c dt)² – a(t)² ((dx)² + (dy)² + (dz)²) t = ƒ(t') ; x = x' ; y = y' ; z = z' dt = ∂t/∂t' dt' + ∂t/∂x' dx' + ∂t/∂y' dy' + ∂t/∂z' dz' = dƒ(t')/dt' dt' dx = ∂x/∂t' dt' + ∂x/∂x' dx' + ∂x/∂y' dy' + ∂x/∂z' dz' = dx' dy = ∂y/∂t' dt' + ∂y/∂x' dx' + ∂y/∂y' dy' + ∂y/∂z' dz' = dy' dz = ∂z/∂t' dt' + ∂z/∂x' dx' + ∂z/∂y' dy' + ∂z/∂z' dz' = dz' dƒ(t')/dt' = a(ƒ(t')) Solve for ƒ(t'), then let A(t') = dƒ(t')/dt' = a(ƒ(t')) Then: (ds)² = A(t')² ((c dt')² – (dx')² – (dy')² – (dz')²) .............................. For example, let a(t) = k t. Then: (ds)² = (c dt)² – (k t)² ((dx)² + (dy)² + (dz)²) dƒ(t')/dt' = k ƒ(t') ƒ(t') = exp(k t' + C) where C is an arbitrary constant. A(t') = k exp(k t' + C) (ds)² = (k exp(k t' + C))² ((c dt')² – (dx')² – (dy')² – (dz')²)