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Schrödinger’s Cat

Why has Schrödinger’s Cat, the experiment, not actually been performed?
Basically because we have no way of putting a system as complex as a cat into a pure quantum state.
Schrödinger’s cat thought experiment was intended to challenge the idea that a system can have undetermined values of some observables in a way that does not just correspond to a lack of knowledge on our part. If the killing of the cat were the result of a purely classical source of uncertainty such as a coin toss we could just say that the cat is either truly alive or truly dead and we just don’t know which until we look. But quantum mechanics includes uncertainties of a kind that cannot be interpreted as just the result of incomplete information.
An example on a scale much smaller than that of a cat is if a single electron has a known value of its vertical spin component (up or down), and if we subsequently divert it by a device that sends it in different directions depending on its horizontal spin component, then there is a 50% chance of seeing it go in either of the two directions and it appears (from an analysis of the observed probabilities in various other directions) that there was no way of predicting the horizontal component (say from some other initial observations) before we actually measured it.
Schrödinger’s point was that there’s something weird about this business of not having a property until we measure it, and he used the cat as an extreme example. But in practice, if we are good enough detectives, there will always be evidence in the box telling us exactly when the cat died; and so, although there may have been some time before we knew the outcome, we never see anything that looks significantly different from what might have happened if the killing of the cat had been triggered by some classical probabilistic event (such as say the first double six in a series of dice throws).
In order to experience the weirdness of having a cat that is neither alive nor dead (sometimes referred to as being both at once) we would need to keep the system of cat and triggering nucleus (and everything else that they can interact with) in what is called a ‘pure’ quantum state, which requires having complete information about the quantum states of all its constituent elementary particles.
This is obviously impossible for a cat, but experiments have been done in which larger systems such as complex molecules are put into superposition states where something like the shape of the molecule (which perhaps seems more substantial to us than the spin of a single electron) does not have a value until we observe it. These will undoubtedly get larger and more impressively weird seeming as technology improves, but I am pretty sure that they will never reach the scale of an actual cat.

Source: (1001) Alan Cooper’s answer to Why has Schrödinger’s Cat, the experiment, not actually been performed? – Quora

Is Decoherence Reversible?

It depends on what you mean by the word “decoherence”.

The conventional use of “decoherence” to describe part of what happens in a measurement process refers to the interaction of a pure state of an experimental system (which may be a superposition of eigenstates of some observable) with a statistical mixed state of a complex environment (which includes some kind of measurement apparatus for that observable) such that, after the interaction, the relative state of the system is a statistical mixture of eigenstates of the observable, each of which is linked to some indicator state of the environmental apparatus. This is typically NOT reversible for thermodynamic reasons (basically due to the fact that the final state is not known in sufficient detail needed to determine the actions needed in order to reverse the process).

BUT, as another answer notes, if the word “decoherence” is being used to describe interaction of the system with an ancilliary system that is in a pure state, then after the interaction the combined system is still in a pure state and the unitary evolution of pure states is reversible.

Source: (1001) Alan Cooper’s answer to Is it possible to reverse quantum entanglement decoherence? – Quora

Watching Fall Into Black Hole

If we on Earth are now observing an object that we see as near a black hole and in free fall on a path that intersects the event horizon of that black hole, then what we are seeing will be extremely red shifted version of the object that is therefore both very dim and ageing very slowly. So the apparent (to us) progress of everything in the object’s frame (including its rate of fall) is very slow. As time (for us) progresses, we will see the object’s clocks and apparent rate of fall to get progressively slower as it also dims towards invisibility.

Source: (1001) Alan Cooper’s answer to An event horizon is a boundary beyond which events cannot affect an observer. What would we be witnessing if a spacecraft was to cross the event horizon of a black hole? The spacecraft disappearing, or eternally approaching the black hole? – Quora

Bergson vs Einstein

After reading this article twice, and yet again the paragraph where the author purports to show that “it’s wrong to think that Bergson’s idea of duration can be assimilated into the idea of psychological time”,

I am still unable to find any explanation of the difference between our internally experienced psychological time (which, by the way can not necessarily always be “aligned with external clock time”) and “the first-person experience of (Bergson’s unmeasurable) duration” (which they appear to identify as the “lived time” in terms of which “An hour in the dentist’s chair is very different from an hour over a glass of wine with friends”).

On the other hand Steven Savitt’s “solution” does not address the subjective nature of duration and appears to just identify it with the non-subjective proper time associated with a possible observer’s world line – which seems to be just giving up on the idea of any special “philosophical” time as this has always been the only kind of time that is ever discussed in relativistic physics.

Source: Who really won when Bergson and Einstein debated time? | Aeon Essays

Is Heat Radiation Just Infrared? 

It’s really no different than with any other kind of energy. Heat is basically just energy that is disorganized ie distributed in a way that is too complicated for us to keep track of – like the random motion of molecules in a gas as opposed to bulk motion or pressure waves in that same gas.

With regard to electromagnetic radiation, every hot body emits a broad spectrum of disorganized electromagnetic radiation at all frequencies – not just infrared (though that is the dominant range at the temperatures we normally experience); but it is also possible to produce organized electromagnetic radiation (eg from electronic circuits and lasers) and this is not “heat” even when it is infrared (though it can be used to create heat when absorbed by a body – just as organized mechanical energy can be converted to heat by friction).

Source: (1002) Alan Cooper’s answer to Is all electromagnetic radiation considered heat? Or just infrared radiation? – Quora

Why is KE quadratic?

Conservation of Momentum (or equivalently Newton’s Law of action and reaction) tells us that when two particles interact just with one another, the acceleration of each multiplied by its mass (which we refer to as “Forces”) are equal and opposite. For any motion, the accumulated product of acceleration during a time interval times distance travelled over that interval is equal is equal to half of the change in the square of the speed in the direction of acceleration.

(This is just the calculus identity [math]\int{x’’ dx}=\int{x’’ x’ dt}=\int{x’ x’’ dt}=\int{x’ dx’}=\Delta(x’^2/2)[/math]; but if that’s not familiar to you, then a more elementary version is the fact that for constant acceleration with [math]v=at[/math] at time [math]t[/math], the average speed from time [math]0[/math] to time [math]t[/math] is [math]\frac{1}{2}at[/math] , so the distance travelled is given by [math]x=(\frac{1}{2}at)t=at^2/2[/math] , and acceleration times distance is [math]ax=a(at^2/2)=(at)^2/2=v^2/2[/math].)

So the quantity that is increased by applying force through a distance (to do “work” on a particle) is quadratic in its speed.

But why is this important enough to give it a special name? That is because it allows us to define a quantity that is conserved throughout the evolution of any physical system in which the Forces between particles depend on their relative displacement only(*), and have no other dependence on time. This ensures that the speed lost when moving against forces (such as when a projectile moving away from a planet slows down) can be recovered if the motion is later reversed. So in the motion of any such system of particles, the sum of [math]mv^2/2[/math] for all the particles plus the net work done against forces is a constant. We call this the “Energy” of the system and identify the part involving the speeds as the “kinetic” part of that energy – and the work done against forces (which includes an arbitrary constant depending on what we take as the starting point) is a called “potential” energy since it could in principle be returned to the system in future interactions.

(*)- If we allow velocity-dependent forces such as friction then the process might not be reversible and we might have to include also other kinds of energy such as heat in order to still have a conserved quantity.

Source: (1002) Alan Cooper’s answer to Why does kinetic energy increase quadratically, not linearly, with speed? – Quora

Bertlmann’s Gloves (yet again on Quora)

Can quantum entanglement can be looked at as a example of glove manufacturing? They are produced in pairs, and if without looking we send them to different parts of universe, once opened, one will always be left and other right? If not why not?

NO. The idea that every electron has a spin direction of its own, which does not become apparent until we view it but is always there (like the handedness of a glove), IS compatible with what we observe if we only ever measure spin components along one axis, but it is NOT compatible with what we observe when we measure spins in directions that are not either parallel or perpendicular.

There is nothing “spooky” about the fact that measuring the spin of one electron from a pair immediately tells us (but not a distant observer of the other electron) what the distant observer will see IF they measure the spin in the same direction. But what IS “spooky” is that it allows us to predict the result of a measurement of the remote electron at say 45 degrees to the one we measured with greater confidence than would be possible with ANY pre-assigned set of spin values in all directions.

See this famous paper by John Bell for a colourful illustration of these ideas.

Source: (1002) Alan Cooper’s answer to Can quantum entanglement can be looked at as a example of glove manufacturing? They are produced in pairs, and if without looking we send them to different parts of universe, once opened, one will always be left and other right? If not why not? – Quora

Gas with Temperature Gradient

Source: thermodynamics – Is there a pressure gradient in a stationary gas with a temperature gradient? – Physics Stack Exchange

The answer to the posed question is indeed a simple “no”, but to establish this does not require any analysis of internal gas dynamics.

If there is ever an overall pressure gradient, then in the absence of external forces the container will accelerate. (By conservation of momentum the gas within will accelerate in the opposite direction until the pressure gradient is eliminated and eventually reversed, and in the absence of friction this would result in an oscillation, but in any case it won’t be an equilibrium until the forces on the walls are in balance.)

In the equilibrium situation the constant pressure means that the density will (in the ideal gas approximation) be inversely proportional to temperature, and I think that any student familiar with the gas laws would accept that each layer of gas can thus remain in balance with its neighbours (one being cooler and denser and the other hotter and more rarified).

The analysis of how this is derived from kinetic theory (with molecules not being confined to layers etc) is more interesting, but does not appear necessary for an answer to the original question.

With regard to what was probably the really intended question, namely how to reconcile the $n \sqrt{T}$ flux out of each layer with the constancy of $nT$, it might be sufficient to tell a student who is not ready for the full analysis just that the flux of those incoming molecules which interact with the layer does not come just from the neighbouring layers but is a mix from various distances which turns out to give zero net flux when $nT$ is constant.

Source: thermodynamics – Is there a pressure gradient in a stationary gas with a temperature gradient? – Physics Stack Exchange

 Light Mill applet — Greg Egan

 Crookes radiometer – Wikipedia

 Thermal transpiration – Wikipedia

 How does a light-mill work?

 Light Mills | The n-Category Café

 The Chapman-Enskog closure

 XVIII. On certain dimensional properties of matter in the gaseous state. – Part I. Experimental researches on thermal transpiration of gases through porous plates and on the laws of transpiration and impulsion, including an experimental proof that gas is not a continuous plenum. – Part II. On an extension of the dynamical theory of gas, which includes the stresses, tangential and normal, caused by a varying condition of gas, and affords an explanation of the phenomena of transpiration and impulsion

 On Stresses in Rarified Gases Arising from Inequalities of Temperature

 1512.02590.pdf

 0402011.pdf

 Thermal conductivity of an ideal gas.pdf

Why is CO2 a “Greenhouse Gas” when N2 is not?

Bonds that are larger and looser tend to vibrate at lower frequencies than those that are small and tight. This rule of thumb applies both to classical mechanical systems and to quantum transitions, and so it “explains” why CO2 tends to respond to longer wavelengths than O2 and N2 (and much longer than those required to change the energy levels of electrons within those molecules).

The measured locations and widths of the corresponding absorption bands fit very closely with calculations based on quantum and statistical mechanics, so it can reasonably be said that we understand very well why they are where they are.

It is, however, just a fluke that the vibrational excitation modes of CO2 (and H2O and CH4) happen to fall near the peak intensity of thermal radiation at the Earth’s temperature of around 300K. (And they might be less effective than other choices as “greenhouse gases” on a planet that was either white hot or not illuminated by the Sun.)

Source: (1002) Alan Cooper’s answer to Why is CO2 transparent to incoming shorter infrared wavelengths of light, but absorbs outgoing longer infrared wavelengths from Earth’s surface? Are they certain these IR wavelengths that are more affected than others, and if so, why? – Quora