Author: alan

Source: (1002) Alan Cooper’s answer to Why does a mirror reverse things horizontally but not vertically? – Quora

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

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

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

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

Quantum Mechanics “explains” the lack of radiation from electrons in the ground state of an atom by telling us that our idea of electrons (or anything else) as discrete particles with well-defined positions and momenta is wrong – and that the bound electron is not in fact accelerating on a curved path around the nucleus, but rather has a range of possible values for our attempts to measure its positions and momenta at different times (with the property that the probability distribution of these values is concentrated around the nucleus but invariant with respect to time).

Source: (1002) Alan Cooper’s answer to I still don’t understand why an accelerating electron doesn’t emit electromagnetic radiation as it’s quantized motion around the nucleus. How does quantum mechanics explain the prevention of this radiation? – Quora

The difference between a density matrix and a state vector is that the latter corresponds to having more complete information about the system. Let’s see how this plays out in a simple example.

A density matrix that is diagonal with entries of {1}/{2} means that we are talking of just a two dimensional state space (such as that of a single electron whose position we are ignoring), and that we have chosen a particular observable, such as spin component in the z direction, to determine the basis with respect to which you are representing the state.

In this context there are actually many different pure states for which the probability of having spin up or down in the z direction are equal. It could for example be certain to have a particular value in (say) the x direction, or the y, or any other direction perpendicular to the z axis. Each of these corresponds to a “ket” whose components (relative to the z eigenstates) are of the form {1}/{\sqrt{2}} multiplied by complex phase factors (and the different relative phases correspond to different directions of spin in the xy plane). On the other hand, representing the state by that density matrix means that we have prepared the electron in such a way that its spin has equal probability of being measured up and down in the z direction, but also that all ways of getting that result are equally likely so there is also equal probability of getting up or down in any other direction.

Source: (1002) Alan Cooper’s answer to Why can’t the density matrix with 1/2 in their diagonals be equal to a ket wave function with 1/2 in each entry. Do they not both describe how each 0 or 1 state has a 1/2 probability? – Quora